AMC
AMC 12
AIME
modular arithmetic
complex numbers

How difficult do you think they are compare to this year's AMC12s?

The mock tests were harder...that's why the AIME qualifying scores were lowered.

[quote="Rep123max"]Well, I havent look at this years AMC 12s (I'd like to look at them if someone texed it), but I'll tell you about my test.

The first ones were about the level of a regular AMC-12, but then the last 6 were easier than a regular AMC-12. There really was no "killer." #25 could take forever, unless you realize that the denominators are 3-r, 3-s, and 3-t, respectively.[/quote]

I dissent: there was [u]a[/u] killer (in terms of AMC-12 level).  #23 (the 2002nd row question) was definitely harder than any problem I've seen on an AMC-12; the recursion solution was well hidden.  I will concede that the other problems did follow from reasonably "standard" methods.

Yay! I actually got 23. I've only seen a couple AMC 12s, but I think the mock tests were definitely harder than those.

That is exactly my point, when nothing is obvious look for stuff like that.  (I actually did solve it in a rigorous complex numbers way.)

[quote="takuanitromars36"]The mock AMC B was by far harder than the mock AMC A. My score was higher on A than on B, and I qualified for the mock AIME in A, but not B. The mock AMC A was like this year's AMC 10A, and the mock AMC B was like this year's AIME I.[/quote]

It is interesting you thought so.  I thought second one was supposed to be way easier (at least that was what I thought before taking the test)  I bombed both,  :(  and I am not happy (I am gonna cry  :?

[quote="beta"][quote="Mildorf"][quote="beta"]#17 of Mock AMC A was a killer, I am not sure how someone can come up the idea of complex numbers fast. However, my scores on both mock AMCs are higher than my scores on 2004 AMC12s, so I am not sure if they are easier or I just improved.[/quote]

I wouldn't call that particular problem a killer. In any answer-based contest, it is important to consider cheap tactics. If you are a resourceful test taker, you probably didn't need to actually "solve" the problem.  It is clear that #17 was a special case of

[tex]\displaymode{
\sum_{k=0}^{n} (-1)^{k} {2n+1 \choose 2k}
}[/tex]

If we consider $n = 0, 1, 2, 3, 4, 5, \dots$ we find the values $2^0, -2^1, -2^2, 2^3, 2^4, -2^5 \dots$, and it is pretty clear that the expression is $-2^n$ for $n \equiv 1, 2 \pmod {4}$ and $2^n$ for $n \equiv 0, 3 \pmod {4}$. As $49 \equiv 1 \pmod {4}$, we choose the answer: $-2^{49}$.

When I reviewed the Mock AMC B, I did not notice such a cheap trick for #23 - nothing really distingiushed the answers in the 20's.  Did anyone have non-rigorous inspiration to pick the correct answer?[/quote]

no offense, but that is kind of cheap ;), at least you should prove it with induction.[/quote]

Actually I think his answer is very practical.  I got the question right because I had this "feel" on the answer.  I am the cheap one :)


MATHCOUNTS
AMC
AMC 10
AMC 12
AIME
USA(J)MO
USAMO

I'm still in middle school, but I think Mathcounts should create a high school competition, since as a middle school competition it's one of the best ever organized... although it may be really difficult to pull of with the AMCs being the real only important contest in Hich school (although there are others). What do you all think?

I think mathcounts focuses a lot on speed, and 3 years is just right for this type of competition to benefit you.  I believe I improved a lot in math from mathcounts between grades 6 and 8, but I don't believe taking it now would help me much, since it's only speed at this point.  .`. I will join the naysayers ;).

...and can you submit your reasons for your votes?  ... I just thought it would be interesting...

well, there are math teams and also academic league(which is everything including math) but it would be really nice to have mathcounts for high school!

It is impractical for the math-inclined high school students to train similarly as they have done in middle schools for MathCounts. There are already many established, and meaningful, math contests at the high school level such as AMC10, AMC12, AIME, USAMO, USAMTS, Mandelbrot, etc., to occupy a bit of their available time.

For those who still crave for the good feelings of MathCounts, they should look seriously into joining (or putting together) an ARML team. Many ARML teams do have regular training sessions (but not likely on a weekly basis). There are also many local math events that resemble MathCounts, though they are not available everywhere in USA. I think there is already a thread on AoPS that attempts to identify these local events.

Mathcounts in high school would be okay, except with harder problems of course. There are already so many competitions for high school math lovers. they should have mathcounts be everyday in my school, instead for once a week

[quote="MathFiend"]I think it would be cool, but highly unlikely. There is much more range of "smartness" in high school than in middle school. In high school classes range from Alg. I to CalcAP, whereas not that many middle schoolers participate in the higher level math courses. I know there are those awesome middle schoolers who are in calculus, but it would be much more difficult to write a competition for high schoolers due to the wide range that would have to be accomadated.[/quote]
Of course, they can split high school into upper and lowerclassmen, which lowers the division a bit.

Of course dude, if I don't win this time, there's always next year!

I did my first high school math comp. yesterday. They did split it up like you said, and had some really easy problems compared to MATHCounts. They also had ciphering, which is like countdown. It [i]was[/i] a multiple choice for the first part, but you had to do numerical answers for the ciphering. No calculators at all were allowed for the competition, but I did get a really nice TI-84+Silver for being the top 8th-grader! :lol: :)  This competition also had proofs, and I moved up a place among the 8-10th graders because of them. :lol: :lol: :):) :lol:

Billy

i think they should have them for people who want to try again in high school, but i wouldn't die of disappointment if they don't.

I voted (an EMPHATIC) no. (this is coming from a former national champion, so don't think I'm biased against MathCounts or something)

There is absolutely no reason for MathCounts to be extended beyond junior high. There are already math leagues and ARML and other junk for starters, and the entire style of MathCounts isn't suited toward high school. If you changed the style of MathCounts, you might as well rename it AMC/AIME/USAMO (oh wait, that already exists).

As much as I'm sure you love MathCounts (which you absolutely should; I loved it too and will always hold MathCounts in the highest regard as the program that got me interested in math in the first place), the style is simply ... too simple.

 Too much time dedicated toward MathCounts style stuff in high school (including ARML, Math Leagues, etc) is time taken away from USAMO. ARML, Mandelbrot, Math Leagues, etc. are all great building blocks toward the furthering of your ability to do well in the USAMO. But they are not comparable competitions to the USAMO. Unfortunately, too many peoples' goals are to make USAMO, and not to do well at USAMO. It's the ability to do a decent portion of USAMO problems that should be everyone's ultimate goal, because that's where the true beauty in mathematics begins to appear.

As great and fun as single answer contests can be, often times the limiting factor is time, which is too bad.

Say ARML individual was extended to 20 minutes per set, or 30 minutes. I think getting an 8 would be pretty easy then.

Say AIME was extended to 6 hours. I would put my chance of getting a 15 at about 80% then.

Say USAMO was extended to 24 hours ... I think I would have collapsed by then and there's no way I would have gotten all the problems.

That's the difference, and the beauty.


How many ways are there to arrange $5$ identical red, $5$ identical blue and $5$ identical green marbles in a straight line such that every marble is adjacent to at least one marble of the same colour as itself ?

How did I get that answer, or how do you solve 3(5 C 2).

Well, since you have 5 marbles of each color, you have to choose two of them that can line adjacent to each other. So you have 5 C 2, but you have to multiply by 3 since there are 3 different kinds of marbles. So you get 3(5 C 2). Did I do it right?

I might be wrong but
you can split it each color as 2-3, one group of 2 and one of 3 or keep it 5
[u]With 2-3[/u]
you have 6 total groups
so $6!$ ways

[u]With 5[/u]
You don't have to do 5 because if the 2 and 3 group of one color are next to each other thats 5

so $6!$?

Keep track of what you're counting though.
For example, how many times does your method count RRRRRGGGGGBBBBB?

heh i realized that before you posted...


does anyone know a formula for finding the sum of consecutive integers starting with 1 and going to any number?
ex.
1+2+3+4...

i used to know it but i forgot, i'm just wondering if anyone could remind me. :oops:

[hide](a+z)*n/2

a = first number
z = last number
n = number of numbers[/hide]

Actually, starting with 1,

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

I can derive this from the formula that LuCy4EvA wrote.

The formula can be seen as an animation under the toolbar on this site.  :?

say you want to find the sum of the numbers 1-10.
Use the formula x(x+1)/2
plug in the numbers...10(11)/2
ANSWER: 55


I still desperately need a personal recommendation.  Does anyone have any ideas as to who might be suitable?  Can it be a teacher?  Thanks in advance.  It is very late to ask for a recommendation now but I'm willing to give anything a try...

Well, I don't think MathCamp will be that restrict on recommendations. I'm homeschooled so it was rather difficult for me to obtain an academic and a personal recommendation. Instead, both of my recommendations are about half personal and half academic I think. So...it all works out. Or at least I hope it does. *gulp* It would be bad if I wasn't accepted b/c my recommenders didn't put the correct content in those letters.

[quote="probability1.01"]I still desperately need a personal recommendation.  Does anyone have any ideas as to who might be suitable?  Can it be a teacher?[/quote]

At this point you should probably go ahead and ask another teacher (but give them instructions for the personal letter, if possible).

[quote="litoxe2718281828"]My teacher is telling me that the online recommendation submission is not working.  If I don't see it received at 11:45 pm, I'm going to go ahead and submit my application... will this be all right? I can later ask her to e-mail the recommendtion via e-mail. If it comes to this, I will send an e-mail explaining the problem.[/quote]

You should definitely go ahead and submit your application.  If the rec letter comes after your app is submitted, it is still added to your file.  The order doesn't matter.  This seems to be a common misunderstanding -- we must not explain it well.

Dave


analytic geometry
graphing lines
slope
MATHCOUNTS
geometry

[quote="236factorial"]A line contains the points (0,4) and (2,0). Show that the point (-25,81) does or doesn't lie on this line.[/quote]

[hide]A line that goes through the points (0,4) and (2,0) has a slope of $-2$. It's y-intercept is $4$.

So the equation of the line is $y=-2x+4$

Plugging in the point (-25,81):

$81=-2(-25)+4$

$77\not=50$

So that point isn't on the line.[/hide]

say we have two distinct points $A(0,4)$ and $B(2,0)$, the colinearity rule sais thet if C lies on the line defined by A and B than the fallowing determinent (i think that this the word in english) must be zero:
| $x_{A}$ $y_{A}$ 1 |
| $x_{B}$ $y_{B}$ 1 | =0
| $x_{C}$ $y_{C}$ 1 |

so we have
|    0   4 1 |
|    2   0 1 | = 54 which is not equal to $0$
| -25 81 1 |

This is the proof that A, B, C or not collinear.

No matrices please (this is easily solved with basic geometry).

The only time you can use matrices in the Mathcounts forum is in solving complex area problems.


inequalities
trigonometry
three variable inequality

Oh yes,this one is really stronger than mine :) .However,I have a solution for it now,but it is not nice.
So I really want to see a beautiful one first :D

It's Schur's type inequality, so we can use Schur's trick:
WLOG suppose that $a \le b \le c$ and put $ b = a + x, c = a + x + y$. Put in our ineq, expand (!!!) and we get

$LHS - RHS = 4a^4x^2+4a^3x^3+4a^4y^2+12a^3y^3+4a^4xy+y^6+6a^3x^2y+26a^3xy^2+2a^2x^3y+45a^2x^2y^2+44a^2xy^3+32ax^3y^2+48ax^2y^3+28axy^4+a^2x^4+7x^4y^2+14x^3y^3+13a^2y^4+6ay^5+13x^2y^4+6xy^5 \ge 0$

Not so nice, but effective.

Namdung

Yes sir,it is really a very effective method.However,it is different from mine.How about you,Harazi? :?
It is too interesting when we can use that method for a inequality,but hard to use for a weaker one. :D

My solution is completely different and I really think it is nice. Of course, expand (this is the ugly part, but not monstruous) and use the identity $ x^3+y^3+z^3+3xyz-xy(x+y)-yz(y+z)-zx(z+x)=x(x-y)(x-z)+... $. Now, the inequality becomes $ a^2(a^2-b^2)(a^2-c^2)+...+2abc[ a(a-b)(a-c)+...]\geq a^2b^2(a-b)^2+... $. And a second identity makes the job:
   $ (a-b)^2(b-c)^2(c-a)^2=(ab(a-b)+bc(b-c)+ca(c-a))^2= a^2b^2(a-b)^2+...-2abc[ a(a-b)(a-c)+...] $. And this is how we reduce it to $ a^2(a^2-b^2)(a^2-c^2)+...\geq (a-b)^2(b-c)^2(c-a)^2 $ . Now, take $ a\geq b\geq c $ and it suffices to prove that $ a^2(a^2-b^2)(a^2-c^2)+b^2(b^2-a^2)(b^2-c^2)\geq (a-b)^2(b-c)^2(c-a)^2 $ which is the same as $ (a^2+b^2-c^2)(a^2-b^2)^2\geq (a-b)^2(b-c)^2(c-a)^2 $ and from now on it's trivial.

[quote="Ji Chen"][quote="harazi"]Prove that for all nonnegative reals $ a,b,c$ we have:

$ \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2\geq 2(bc \plus{} ca \plus{} ab)\left[bc\left(b^2 \plus{} c^2\right) \plus{} ca\left(c^2 \plus{} a^2 \right) \plus{} ab\left(a^2 \plus{} b^2\right)\right]$.[/quote][quote="Namdung"]It's Schur's type inequality, so we can use Schur's trick:[/quote]$ \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2 \minus{} 2(bc \plus{} ca \plus{} ab)\left[a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right]$

$ \equal{} \sum (a \minus{} b)(a \minus{} c)\left[a^4 \plus{} a^3(b \plus{} c) \minus{} a^2\left(b^2 \minus{} 3bc \plus{} c^2\right) \plus{} (b \minus{} c)^2 \left(b^2 \plus{} bc \plus{} c^2\right)\right]$

$ \geq 0,$

which is true for

$ a^4 \plus{} a^3(b \plus{} c) \minus{} a^2\left(b^2 \minus{} 3bc \plus{} c^2\right) \plus{} (b \minus{} c)^2\left(b^2 \plus{} bc \plus{} c^2\right)$

$ \equal{} \frac {\left[4a(2a \plus{} b \plus{} c) \minus{} 5(b \minus{} c)^2\right]^2 \plus{} (b \minus{} c)^2\left[40a(b \plus{} c) \plus{} 3\left(13b^2 \plus{} 38bc \plus{} 13c^2\right)\right]}{64}$

$ \geq 0,$

$ \frac {\left[b^4 \plus{} b^3(c \plus{} a) \minus{} b^2(c^2 \minus{} 3ca \plus{} a^2) \plus{} (c \minus{} a)^2(c^2 \plus{} ca \plus{} a^2)\right] \minus{} \left[c^4 \plus{} c^3(a \plus{} b) \minus{} c^2(a^2 \minus{} 3ab \plus{} b^2) \plus{} (a \minus{} b)^2(a^2 \plus{} ab \plus{} b^2)\right]}{b \minus{} c}$

$ \equal{} a^3 \minus{} a^2(b \plus{} c) \plus{} a(b \plus{} 2c)(2b \plus{} c) \plus{} bc(b \plus{} c)\geq 0,$

and the [url=http://en.wikipedia.org/wiki/Schur%27s_inequality]Vornicu - Schur's inequality[/url].

By the way, harazi's inequality holds for [b]real[/b] $ a,b,c$ with $ bc \plus{} ca \plus{} ab\geq 0.$[/quote]

But I think you should prove that $ x\geq y$ or $ z \geq y$  to have $ \sum x(a\minus{}b)(a\minus{}c) \geq 0$ while you have proven that $ y \geq z$ and $ x \geq 0$.

Am I right?

[quote="harazi"]Prove that for all nonnegative reals a,b,c we have:
$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2\geq 2(ab \plus{} bc \plus{} ca)( ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2))$.[/quote]
The inequality is equivalent to
\[ \left(\sum a^3 \plus{} 3abc\right)^2 \minus{} \left(\sum ab(a \plus{} b)\right)^2 \ge 2\left(\sum ab\right)\left(\sum ab(a^2 \plus{} b^2)\right) \minus{} \left(\sum ab(a \plus{} b)\right)^2
\]

\[ \Leftrightarrow \left(\sum a(a \minus{} b)(a \minus{} c)\right)\left(\sum a(a \plus{} b)(a \plus{} c)\right) \ge \left(\sum ab\right)\left(\sum ab(a \minus{} b)^2\right) \plus{} \left(\sum ab\right)\left(\sum ab(a \plus{} b)^2\right) \minus{} \left(\sum ab(a \plus{} b)\right)^2
\]

\[ \Leftrightarrow \left(\sum a(a \minus{} b)(a \minus{} c)\right)\left(\sum a(a \plus{} b)(a \plus{} c)\right) \ge \left(\sum ab\right)\left(\sum ab(a \minus{} b)^2\right) \plus{} abc\sum a(b \minus{} c)^2
\]

\[ \Leftrightarrow \left(\sum a(a \minus{} b)(a \minus{} c)\right)\left(\sum a(a \plus{} b)(a \plus{} c)\right) \ge \left(\sum ab\right)\left(\sum a(b \plus{} c)(a \minus{} b)(a \minus{} c)\right) \plus{} abc\sum (b \plus{} c)(a \minus{} b)(a \minus{} c)
\]

\[ \Leftrightarrow \sum a(a \minus{} b)(a \minus{} c)\left(\sum a(a \plus{} b)(a \plus{} c) \minus{} (b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} bc(b \plus{} c)\right) \ge 0
\]
We have
\[ \sum a(a \plus{} b)(a \plus{} c) \minus{} (b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} bc(b \plus{} c) \equal{} a(a \plus{} b)(a \plus{} c) \plus{} (b \plus{} c)(b \minus{} c)^2
\]
Therefore, the last inequality is equivalent with
\[ \sum a(a \minus{} b)(a \minus{} c)\left(a(a \plus{} b)(a \plus{} c) \plus{} (b \plus{} c)(b \minus{} c)^2\right) \ge 0
\]

\[ \Leftrightarrow \sum a^2(a^2 \minus{} b^2)(a^2 \minus{} c^2) \plus{} (a \minus{} b)(a \minus{} c)(b \minus{} c) \sum a(b^2 \minus{} c^2) \ge 0
\]

\[ \Leftrightarrow \sum a^2(a^2 \minus{} b^2)(a^2 \minus{} c^2) \ge (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2
\]
Now, assume that $ a \ge b \ge c$, then we have
\[ a^2(a^2 \minus{} b^2)(a^2 \minus{} c^2) \plus{} b^2(b^2 \minus{} c^2)(b^2 \minus{} a^2) \equal{} (a^2 \minus{} b^2)^2(a^2 \plus{} b^2 \minus{} c^2) \ge a^2(a^2 \minus{} b^2)^2 \ge 4a^2b^2(a \minus{} b)^2 \ge 4(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2
\]
Therefore, our inequality is proved. :)

[quote="Vasc"]I think these inequalities hold, too:
Prove that for all nonnegative reals a,b,c we have:

$ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2\geq 2(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)$;

$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)^2 \plus{} (ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2)^2\geq 4(ab \plus{} bc \plus{} ca)^2(a^2 \plus{} b^2 \plus{} c^2)$.[/quote]
These inequalities are not hard with $ pqr$ :), for example with the first inequality. Assume that $ a\plus{}b\plus{}c\equal{}1$, put $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$, the our inequality becomes
\[ 3(1\minus{}3q\plus{}6r)^2 \ge 2q(1\minus{}q)(1\minus{}2q)\]
Case 1. If $ 4q \le 1$, then we have
\[ LHS\minus{}RHS \ge 3(1\minus{}3q)^2\minus{}2q(1\minus{}q)(1\minus{}2q)\equal{}(1\minus{}4q)(3\minus{}8q\plus{}q^2) \ge 0\]
Case 2. If $ 4q \ge 1$, then by Schur's Inequality for fourth degree, we have
\[ 6r \ge (4q\minus{}1)(1\minus{}q)\]
Therefore
\[ 1\minus{}3q\plus{}6r \ge 1\minus{}3q\plus{}(4q\minus{}1)(1\minus{}q)\equal{}2q(1\minus{}2q)\]
Thus
\[ LHS\minus{}RHS \ge 2q(1\minus{}2q)[6q(1\minus{}2q)\plus{}q\minus{}1] \equal{}2q(1\minus{}2q)(1\minus{}3q)(4q\minus{}1) \ge 0\]
:)

[quote="Vasc"]Prove that for all nonnegative reals $ a,b,c$ we have:

$ 3\left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2\geq 2(a^2 \plus{} b^2 \plus{} c^2)(bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab).$[/quote]$ 3\left(3abc \plus{} \sum{a^3}\right)^2 \minus{} 2\left(\sum{a^2}\right)\left(\sum{bc}\right)\left(\sum{a^2} \plus{} \sum{ bc}\right)$

$ \equal{} \sum{(b \minus{} c)^2\frac {(b \plus{} c \minus{} a)^2\left[2a(5a \plus{} 2b \plus{} 2c) \plus{} 3(b \minus{} c)^2\right] \plus{} 4bc(b \minus{} c)^2}{2}}\geq0.$

By the way, the first inequality of Vasc holds for [b]real[/b] $ a,b,c$  with $ a \plus{} b \plus{} c\geq 0\leq abc.$
[quote="Vasc"]Prove that for all nonnegative reals $ a,b,c$ we have:

$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)^2 \plus{} ( bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2)^2\geq 4(bc \plus{} ca \plus{} ab)^2\left(a^2 \plus{} b^2 \plus{} c^2\right).$[/quote]$ \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\right)^2 \plus{} ( bc \plus{} ca \plus{} ab)\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \minus{} 4(bc \plus{} ca \plus{} ab)^2(a^2 \plus{} b^2 \plus{} c^2)$

$ \equal{} \sum (a \minus{} b)(a \minus{} c)\left[a^4 \plus{} 2a^3(b \plus{} c) \minus{} 2a^2\left(b^2 \minus{} 4bc \plus{} c^2\right) \plus{} (b \minus{} c)^2\left(b^2 \plus{} bc \plus{} c^2\right)\right]$

$ \geq 0,$

which is true for

$ a^4 \plus{} 2a^3(b \plus{} c) \minus{} 2a^2\left(b^2 \minus{} 4bc \plus{} c^2\right) \plus{} (b \minus{} c)^2\left(b^2 \plus{} bc \plus{} c^2\right)$

$ \equal{} (b \minus{} c \plus{} a)^2(b \minus{} c \minus{} a)^2 \plus{} 2a^2(2bc \plus{} ca \plus{} ab) \plus{} 3bc(b \minus{} c)^2\geq 0,$

$ (b \minus{} c)\Big\{\left[b^4 \plus{} 2b^3(c \plus{} a) \minus{} 2b^2\left(c^2 \minus{} 4ca \plus{} a^2\right) \plus{} (c \minus{} a)^2\left(c^2 \plus{} ca \plus{} a^2\right)\right]$

$ \minus{} \left[c^4 \plus{} 2c^3(a \plus{} b) \minus{} 2\left(a^2 \minus{} 4ab \plus{} b^2\right)c^2 \plus{} (a \minus{} b)^2\left(a^2 \plus{} ab \plus{} b^2\right)\right]\Big\}$

$ \equal{} (b \minus{} c)^2\left[a(b \plus{} c \minus{} a)^2 \plus{} a\left(2b^2 \plus{} 9bc \plus{} 2c^2\right) \plus{} 2bc(b \plus{} c)\right]\geq 0.$

By the way, the second inequality of Vasc holds for [b]real[/b] $ a,b,c$ with $ bc \plus{} ca \plus{} ab\geq 0;$

with equality if $ a \equal{} \minus{} 1,b\equal{} 1 \plus{} 2\cos{\frac {\pi}{9}} \equal{} 2.879\cdots,c\equal{} 2\cos{\frac {2\pi}{9}} \equal{} 1.532\cdots.$
[quote="manlio"]But I think you should prove that $ x\geq y$ or $ z \geq y$  to have $ \sum x(a \minus{} b)(a \minus{} c) \geq 0$ while you have proven that $ y \geq z$ and $ x \geq 0$.

Am I right?[/quote]For [b]symmetric[/b] $ \sum S_{a}(a \minus{} b)(a \minus{} c) \geq 0$ :

$ S_{a}\geq 0\models S_{b}\geq 0,S_{c}\geq 0$;

$ (b \minus{} c)(S_{b} \minus{} S_{c})\geq 0\models$ the triples $ (a,b,c)$ and $ (S_a,S_b,S_c)$ are similarly sorted.

And then,  the [url=http://en.wikipedia.org/wiki/Schur%27s_inequality]Vornicu - Schur's inequality[/url] works here.

[quote="Ji Chen"][quote="harazi"]Prove that for all nonnegative reals $ a,b,c$ we have:

$ \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2\geq 2(bc \plus{} ca \plus{} ab)\left[bc\left(b^2 \plus{} c^2\right) \plus{} ca\left(c^2 \plus{} a^2 \right) \plus{} ab\left(a^2 \plus{} b^2\right)\right]$.[/quote][quote="Namdung"]It's Schur's type inequality, so we can use Schur's trick:[/quote]$ \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2 \minus{} 2(bc \plus{} ca \plus{} ab)\left[a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right]$

$ \equal{} \sum (a \minus{} b)(a \minus{} c)\left[a^4 \plus{} a^3(b \plus{} c) \minus{} a^2\left(b^2 \minus{} 3bc \plus{} c^2\right) \plus{} (b \minus{} c)^2 \left(b^2 \plus{} bc \plus{} c^2\right)\right]$

$ \geq 0,$
By the way, harazi's inequality holds for [b]real[/b] $ a,b,c$ with $ bc \plus{} ca \plus{} ab\geq 0.$[/quote]
let$ a\plus{}b\plus{}c\equal{}p$,$ ab\plus{}bc\plus{}ca\equal{}\frac{p^2\minus{}q^2}{3}$,$ abc\equal{}r$
it's equal to $ (pq^2\plus{}6r)^2 \geq \frac{2(p^2\minus{}q^2)}{3}(\frac{p^2q^2\plus{}p^4\minus{}2q^4\minus{}9pr}{9})$
use $ r \geq \frac{(p\plus{}q)^2(p\minus{}2q)}{27}$
we get $ f(p)\equal{}(pq^2\plus{}\frac{2(p\plus{}q)^2(p\minus{}2q)}{9})^2$$ \minus{}\frac{1}{3}(2p^2\minus{}2q^2)(\frac{3p^2q^2\plus{}3p^4\minus{}6q^4\minus{}p(p\plus{}q)^2(p\minus{}2q)}{27})$
$ f'(p)\equal{}\frac{22}{27}pq^4\plus{}\frac{16}{81}p^3q^2\minus{}\frac{20}{81}q^5\minus{}\frac{20}{27}p^2q^3\equal{}\frac{2}{81}q^2(2p\minus{}q)(4p\minus{}5q)(p\minus{}2q)$
if$ p \leq \frac{5q}{4}$ or $ p \geq 2q$
so $ f'(p) \geq 0$
$ f(p) \geq f(q) \geq 0$
if $ 2q \geq p \geq \frac{5q}{4}$
so $ f'(p) \leq 0$
$ f(p) \geq f(2q)\equal{}0$
the equality hold with $ a\equal{}b\equal{}c$ or$ a\equal{}0,b\equal{}c$

[quote="Ji Chen"]
Prove that for all nonnegative reals $ a,b,c$ we have:

[quote="vase"]$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)^2 \plus{} ( bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2)^2\geq 4(bc \plus{} ca \plus{} ab)^2\left(a^2 \plus{} b^2 \plus{} c^2\right).$[/quote]$ \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\right)^2 \plus{} ( bc \plus{} ca \plus{} ab)\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \minus{} 4(bc \plus{} ca \plus{} ab)^2(a^2 \plus{} b^2 \plus{} c^2)$

$ \equal{} \sum (a \minus{} b)(a \minus{} c)\left[a^4 \plus{} 2a^3(b \plus{} c) \minus{} 2a^2\left(b^2 \minus{} 4bc \plus{} c^2\right) \plus{} (b \minus{} c)^2\left(b^2 \plus{} bc \plus{} c^2\right)\right]$

$ \geq 0,$

By the way, the second inequality of Vasc holds for [b]real[/b] $ a,b,c$ with $ bc \plus{} ca \plus{} ab\geq 0;$

with equality if $ a \equal{} \minus{} 1,b \equal{} 1 \plus{} 2\cos{\frac {\pi}{9}} \equal{} 2.879\cdots,c \equal{} 2\cos{\frac {2\pi}{9}} \equal{} 1.532\cdots.$
[/quote]
let$ a \plus{} b \plus{} c \equal{} p$,$ ab \plus{} bc \plus{} ca \equal{} \frac {p^2 \minus{} q^2}{3}$,$ abc \equal{} r$
it's equal to $ (pq^2 \plus{} 9r)^2 \plus{} \frac {1}{3}(p^2 \minus{} q^2)(p^2 \minus{} \frac {2(p^2 \minus{} q^2)}{3})^2 \minus{} 4(\frac {p^2 \minus{} q^2}{3})^2(\frac {p^2 \plus{} 2q^2}{3}) \geq 0$
use $ r \geq \frac {(p \plus{} q)^2(p \minus{} 2q)}{27}$
we get $ f(p) \equal{} \frac {4p^2q^4 \plus{} p^4q^2 \minus{} 4p^3q^3}{9}$
$ f'(p) \equal{} \frac {4pq^2(p \minus{} q)(p \minus{} 2q)}{9}$
$ f(p) \geq f(2q) \equal{} 0$or $ f(p) \geq f( \minus{} q) \equal{} q^6 \geq 0$
the equality hold with $ a \equal{} b \equal{} c$ or$ a \equal{} 0,b \equal{} c$ or
\[ a \equal{} \minus{} 1,b \equal{} 1 \plus{} 2\cos{\frac {\pi}{9}} \equal{} 2.879\cdots,c \equal{} 2\cos{\frac {2\pi}{9}} \equal{} 1.532\cdots.\]

[quote="Ji Chen"][quote="Vasc"]Prove that for all nonnegative reals $ a,b,c$ we have:

$ 3\left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2\geq 2(a^2 \plus{} b^2 \plus{} c^2)(bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab).$[/quote]$ 3\left(3abc \plus{} \sum{a^3}\right)^2 \minus{} 2\left(\sum{a^2}\right)\left(\sum{bc}\right)\left(\sum{a^2} \plus{} \sum{ bc}\right)$

$ \equal{} \sum{(b \minus{} c)^2\frac {(b \plus{} c \minus{} a)^2\left[2a(5a \plus{} 2b \plus{} 2c) \plus{} 3(b \minus{} c)^2\right] \plus{} 4bc(b \minus{} c)^2}{2}}\geq0.$

By the way, the first inequality of Vasc holds for [b]real[/b] $ a,b,c$  with $ a \plus{} b \plus{} c\geq 0\leq abc.$
.[/quote]
let$ a \plus{} b \plus{} c \equal{} p$,$ ab \plus{} bc \plus{} ca \equal{} \frac {p^2 \minus{} q^2}{3}$,$ abc \equal{} r$
it's equal to $ (pq^2 \plus{} 6r)^2 \plus{} 2(p^2 \minus{} 2\frac {p^2 \minus{} q^2}{3})(\frac {p^2 \minus{} q^2}{3})(p^2 \minus{} \frac {p^2 \minus{} q^2}{3}) \geq 0$
use $ r \geq \frac {(p \plus{} q)^2(p \minus{} 2q)}{27}$
we get $ f(p) \equal{} \frac {15p^2q^4 \plus{} 6p^4q^2 \minus{} 24pq^5 \minus{} 16p^3q^3 \plus{} 20q^6}{27}$
$ f'(p) \equal{} \frac {10pq^4 \plus{} 8p^3q^2 \minus{} 8q^5 \minus{} 16p^2q^3}{9} \geq 0$
because if $ f'(p) \equal{} \frac {10pq^4 \plus{} 8p^3q^2 \minus{} 8q^5 \minus{} 16p^2q^3}{9} \leq 0$ we have $ p < 2q$ that is we can have $ r < 0$
so $ f(p) \geq f(2q) \equal{} 0$
the equality hold with $ a \equal{} b \equal{} c$ or$ a \equal{} 0,b \equal{} c$

[quote="Vasc"]$ 3\left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\right)^2\geq 2(a^2 \plus{} b^2 \plus{} c^2)(bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab).$[/quote][quote="Ji Chen"]By the way, the first inequality of Vasc holds for [b]real[/b] $ a,b,c$  with $ a \plus{} b \plus{} c\geq 0\leq abc.$[/quote]$ 3\left(3abc \plus{} \sum{a^3}\right)^2 \minus{} 2\left(\sum{a^2}\right)\left(\sum{bc}\right)\left(\sum{a^2} \plus{} \sum{ bc}\right)$

$ \equal{} abc(a \plus{} b \plus{} c)\sum{(b \minus{} c)^2} \plus{} \prod{(b \minus{} c)^2} \plus{} \frac {5}{3}\left[\sum{a(a \minus{} b)(a \minus{} c)}\right]^2$

$ \plus{} \frac {2}{3}(a \plus{} b \plus{} c)^2\sum{(b \minus{} c)^2(b \plus{} c \minus{} a)^2}\geq0,$

which is true for $ abc(a \plus{} b \plus{} c)\geq 0.$:juggle:
[quote="Vasc"]$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)^2 \plus{} ( bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2)^2\geq 4(bc \plus{} ca \plus{} ab)^2\left(a^2 \plus{} b^2 \plus{} c^2\right).$[/quote][quote="Ji Chen"]By the way, the second inequality of Vasc holds for [b]real[/b] $ a,b,c$ with $ bc \plus{} ca \plus{} ab\geq 0;$

with equality if $ a \equal{} \minus{} 1,b \equal{} 1 \plus{} 2\cos{\frac {\pi}{9}} \equal{} 2.879\cdots,c \equal{} 2\cos{\frac {2\pi}{9}} \equal{} 1.532\cdots.$[/quote]$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)^2 \plus{} ( bc \plus{} ca \plus{} ab)(a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 4(bc \plus{} ca \plus{} ab)^2\left(a^2 \plus{} b^2 \plus{} c^2\right)$

$ \equal{} \left[\sum{a(a \minus{} b)(a \minus{} c)}\right]^2 \plus{} \frac {3}{2}(bc \plus{} ca \plus{} ab)\sum{(b \minus{} c)^2(b \plus{} c \minus{} a)^2}\geq0,$

which is true for $ bc \plus{} ca \plus{} ab\geq 0.$:play_ball:

[quote=harazi]Prove that for all nonnegative reals a,b,c we have:
$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2\geq 2(ab \plus{} bc \plus{} ca)( ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2))$.[/quote]

My bad solution is:
Assume $c=\text{min}\{a,b,c\}$
$\text{LHS-RHS} = M(a-b)^2 +N(a-c)(b-c) \geqq 0$
where,
$M:=4\,{c}^{4}+1/2\, \left( 24\,a-48\,c+24\,b \right) {c}^{3}+1/2\,
 \left( 26\, \left( a-c \right) ^{2}+35\, \left( a-c \right)  \left( -
c+b \right) +26\, \left( -c+b \right) ^{2} \right) {c}^{2}+1/2\,
 \left( 12\, \left( a-c \right) ^{3}+20\, \left( a-c \right) ^{2}
 \left( -c+b \right) +20\, \left( a-c \right)  \left( -c+b \right) ^{2
}+12\, \left( -c+b \right) ^{3} \right) c+ \left( a-c \right) ^{4}+2\,
 \left( a-c \right) ^{3} \left( -c+b \right) + \left( a-c \right) ^{2}
 \left( -c+b \right) ^{2}+2\, \left( a-c \right)  \left( -c+b \right)
^{3}+ \left( -c+b \right) ^{4} \geqq 0$
$N:=1/2\,{c}^{2} \left( {a}^{2}+2\,ca+{b}^{2}+2\,bc+2\,{c}^{2} \right) \geqq 0$

[quote=harazi]Prove that for all nonnegative reals a,b,c we have:
$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2\geq 2(ab \plus{} bc \plus{} ca)( ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2))$.[/quote]
$a^3+b^3+c^3+3abc$
$=a(ab+bc+ca)+\frac{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}{2a}$
$+\frac{(c-a)(a-b)[b^2+c^2-a^2+a(b+c-a)]}{2a}$

$\ge \sqrt{2(ab+bc+ca)[ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)]}$,
where $c \ge a \ge b$.


algebra
polynomial
function
complex numbers
binomial coefficients
functional equation

1)P is a point inside triangle ABC.AP.BP.CP cut
opposite sides at D,E,F.AP,BP,CP cut EF,FD,DE at K,L,M. Prove that
AP/PK + BP/PL + CP/PM >=12.When does equality hold.
2)Find all natural numbers n>=3 such that nC0
+nC1+nC2 +nC3 divides 2^(2001).

now some from the selection tests
3))For any positive integer n,show that there exists
a polynomial P(x) of degree n with integer coeffecients such that
P(0),P(1),..P(n) are all distinct powers of 2.

4)Let p be a prime and Q(x) be a cubic polynomial
with integer coeffecients.Suppose p divides Q(x_j),1<=j<=4,for 4
distinct integers x_1,..,x_4 in the set {0,1,..,p-1}.Prove that p
divides each of the coeffecients of Q(x).

5)Find all functions f from the set of positive reals
to itself such that f(f(x)-x)= 2x for all positive reals x.

6)Let p,q,r be 3 complex numbers such that
p^k+q^k+r^k is an integer for every integer k>=1.Prove that the
polynomial (x-p)(x-q)(x-r) has integer coeffecients.

7)Let <f(n)> be a strictly increasing sequence of
positive integers which are such that gcd(f(m),f(n))=f(gcd(m,n)) for
all natural numbers m,n.Suppose k is the least positive integer for
which there exist positive integers r<k<s such that
f(k)^2=f(r)f(s). Prove that r divides k and k divides s.

think these are enuf .have a good time in solving them

regards

ok, so i dont know if you posted this stuff for answers, but here goes :

1)  of course AP/PK=AP/PD*PD/PK  and we find AP/PD=AF/FB+AE//EC and then if we say AF/FB=gamma and BD/DC=alpha and so on we finnaly get AM-GM
2) what does nC2 mean ?

3)didnt think
4)didnt think

5) if we note g=f(x)-x we get g(g(x))+g(x)=2x. isnt this known? (cause if it is it would clear my mind, i got stuck... :( )

6) we have to prove p+q+r, pq+pr+qr and pqr are integers.
p+q+r ---------- right  :)
pq+pr+rq ( \sum p)^2- \sum p^2
pqr from that known formula that gives  \sum p^3-3pqr=bla bla bla

7) i think u wanted to say  f^2(k)=f(r)f(s) cause otherwise f(s)>f(k)  :(
    i'll get back

ya just posted these problems to get as many solutions as i cud.i tried thses quite sum time bak and got 2 or 3 (dont remember xactly).so i just wanted to see some solutions other than the clumsy ones i find.

now
nCr refers to the binomial coefficients in the expansion of (1+x)^n.(i guess the conventions here and in romania differ)
i'll look for the correction(if any)  in the last question.as for the functional equation i dont know how to solve it
cheers

Ok I think I can shed some light into the functional equation:

f(f(x)-x) =2x , so because f is only defined on positive numbers f(x)>=x should hold , otherwise this doesn't make any sense . Now , if you have g(x) = f(x)-x , as K already noticed , you obtain :

g(g(x)) + g(x) = 2x  , with g(x) >= 0 for all x (this is very important)

Now , consider a fixed x0 and the sequence : a0 = x0 , a_{n+1} = g(a_n) . This is the so-called 'orbit' of g around a0 . Doesn't really matter ..

Anyway , a_{n+2} + a_{n+1} = 2 a_n , so the characteristic equation has the roots 1 and -2 . So , a_n is of the form a_n = x * (-2)^n + y . Solve for x = (x0-f(x0))/3 from the initial conditions , y doesn't matter . Now , if x is non-zero , we can find a large enough n such that a_n < 0 . But this is a contradiction since a_n = g(g(g( .. (x0) ... ) (n times) >= 0 . So x is zero , indeed , and f(x0)=x0 , for all x0 . We're done !

4) Reduce the cubic over the field Z_p[X] . It's a field cause p is prime :) Now there are four distinct values in that field for which the polynomial is zero . But it has degree <= 3 so it must be the zero polynomial . But that means that p divides all its coefficients ..

3) I think that using the Lagrange Interpolation Polynomial for finding a poly such that , say , f(0)=2^0 , f(1)=2^1 , .. f(n)=2^n . It will obviously have degree n and it is easily proved that if a poly of degree <= k-1 takes integer values at k different integers then it takes integer values for all integers (again with the Lagrange thing).

2) I know this one already it's from a Chinese olympiad , can't remember the year but it doesn't have any bright ideas in it . Just lots and lots of calculations .. if you really want to see the sol I can post it in here

7) I didn't think about this one yet

Bye , it's getting late and I have to get back to my own problems :)

thanx for that soln  to the functional equation(duz ur method generalize to similar type of functional equations).as for that chinese problem i have found the soln.
thanx agen
cheers

3) has been solved here :
http://www.mathlinks.ro/phpBB/viewtopic.php?t=1366&highlight=

Pierre.


function
combinatorics unsolved
combinatorics

there  are $ 25$ people sitting  around a table , and each person  has two cards , One of the numbers : $ 1,2,3....25$ is written  on each card , and each number occurs on exactly  two cards , At a signal each person  passes one of his  cards ( the one with the smaller number) to his right hand neighbor ,Prove that sooner or later  one of the players will have two cards with the same number

Thank you  :)

Let $ x_i$ denotes the bigger number in the hand of the $ i$-th person.
then one can easily check that the function $ f: \sum_{i \equal{} 1}^{25}x_i$ is strictly increasing in each $ 25$ consecutive steps.
so at a specific time if no persons have two equal number in the steps,$ f$ will attain its maximum which is easy to check that it is $ 25 \plus{} 25 \plus{} 24 \plus{} 24 \plus{} ... \plus{} 14 \plus{} 14 \plus{} 13$.but when this maximum is reached...there exists still a number $ 13$...when this card arrives in the hand of the person that had $ 13$ in his hand before,then the process will end!
DONE!


topology
real analysis
real analysis unsolved

We denote the set by alphabet $ X$.Consider a arbitrary bunch of open sets $ \{V\}$,where $ V\subset X$ ,and for any$ E
\subset X$,

if for any point $ x \in E$,we can find a set in $ \{V\}$,denote by $ V_{\alpha}$ , such that $ x \in V_{\alpha}$ , then we say

$ E$ has been cover by a bunch of open sets $ \{V\}$. If we can take finite sets from $ \{V\}$, and we denote the finite

subset of $ \{V\}$ by $ \{V^{'}\}$. If for any point $ x\in E$, we can find a set in $ \{V^{'}\}$,denote by $ V^{'}_{\alpha}$,

such that $ x \in V^{'}_{\alpha}$ , then we say $ X$ is a compactness set.

Usually we call the above "Heine-Borel Finite Covering Theorem"

(my english is poor ,so i don't know whether the expression is right :maybe: )

[color=red]My Question is : Is Heine-Borel Finite Covering Theorem only has the connection with open set ? In other words: Is the

above theorem has non-connection with the distance of the set $ X$ ?[/color]

Yes, compactness depends on the topology (the open sets) and not explicitly on the distance (metric).  So if you have two different metrics on a set X, and it happens that the open sets are the same for the two metrics, then X is compact for one of them if and only if it is compact for the other one.

This has nothing to do with Heine-Borel, though; Heine-Borel is a theorem that asserts that the definition of compactness you've given (by open covers) is equivalent to some other definition when you're working in a Euclidean metric space (i.e., a sub-metric space of $ {\bf R}^k$).  The $ {\bf R}^k$-specific definition (closed and bounded) does require us to know what the metric is, since "boundedness" is a metric (not topological) property.


inequalities
email
inequalities proposed

to quykhtn-qa1

let 0<a,b,c_<1 . X=a+b+c
 compare :A=(x/2-a0(x/2-b)(x/2-c)
 and          B=abc

dao nay chan lam dai ca a!
kheo khong duoc thi hsg mat thoi , chan qua!

do you mean this? let $ 0<a,b,c\leq1$ and $ X\equal{}a\plus{}b\plus{}c$, compare $ A\equal{}(x/2\minus{}a)(x/2\minus{}b)(x/2\minus{}c)$ and $ B\equal{}abc$

[quote="trannguyenquyND_VN"]to quykhtn-qa1

let 0<a,b,c . X=a+b+c
 compare :A=(x/2-a)(x/2-b)(x/2-c)
 and          B=abc
easy:B $ \geq\ 8A$

vao phan email xem,ta gui cho roi ra chat voi ta(kaizenvn)


This was from a USNCO Local Section Exam.

Which substance produces an acidic solution when it is bubbled into water?

(A) $CO_{2}$
(B) $Ar$
(C) $NH_{3}$
(D) $CH_{4}$

I guess the logical answer would be (A), but what is the theory behind this answer?

Interesting that you say (A) is the "logical" answer.  What do you mean by this?

In terms of the theory--first ask yourself, what forms when each of the following reacts with water?  (B), a noble gas, certainly does not react with water.  (C), a base, will form some $\text{NH}_{4}^{+}$ ions and $\text{OH}^{-}$ ions--a basic solution.  (D), an organic nonpolar substance, has no affinity for water, and no real reaction occurs.

That leaves only (A).  Note the reaction $\text{CO}_{2}+\text{H}_{2}\text{O}\leftrightharpoons\text{H}_{2}\text{CO}_{3},$ which you should recognize as carbonic acid.  This reaction actually has a very low equilibrium constant, but the carbonic acid on the right side then dissociates to some extent, forming an acidic solution.

The complete equilibrium being

$CO_{2}+H_{2}O \rightleftharpoons H_{2}CO_{3}$
$H_{2}CO_{3}+H_{2}O \rightleftharpoons HCO_{3}^{-}+H_{3}O^{+}$.


[quote="doicu"]hi,
prove that the equations
\[ x^{n} \plus{} y^{n} \plus{} z^{n} \plus{} u^{n} \equal{} v^{n \minus{} 1},
\]

\[ x^{n} \plus{} y^{n} \plus{} z^{n} \plus{} u^{n} \equal{} v^{n \plus{} 1},
\]
have an infinite sistinct solutions for every interger n > 1.[/quote]
$ (x,y,z,u,v) \equal{} (0,0,0,k^{n\minus{}1},k^n)$
and
$ (x,y,z,u,v) \equal{} (0,0,0,k^{n\plus{}1},k^n)$

This works for every $ k$...

hi,
ohh sorry i didn't express myself clearly  :blush: .
in the first post there is "sistinct" . it is nonsense ... it has to be "distinct".


Αν $ a,b,c>0$ με $ a\plus{}b\plus{}c\equal{}1$ να αποδείξετε ότι

$ \frac{a}{\sqrt{b\plus{}c}}\plus{}\frac{b}{\sqrt{c\plus{}a}}\plus{}\frac{c}{\sqrt{a\plus{}b}}\geq\sqrt\frac{3}{2}$

P.S Έχω βρει μια λύση την οποία θα βάλλω αργότερα αλλά θα ήθελα να δώ και άλλες :)

[quote="silouan"]Άμεσα από Jensen στην συνάρτηση $ f(x) \equal{} \frac {x}{\sqrt {1 \minus{} x}}$

Μπορούμε να την κάνουμε και με Cauchy Swartz  :)[/quote]

Πότε πρόλαβες κιόλας ρε Σιλουαν  :)  Δεν πιάνεσαι με τίποτα. Πάντως και εγώ με Jensen έλυσα :) .

[quote="galois01"] Πάντως και εγώ με Jensen έλυσα :) .[/quote]
I think, the following is better than to differentiate two times:
By Holder we obtain:
$ \left(\sum_{cyc}\frac {a}{\sqrt {b \plus{} c}}\right)^2\cdot\sum_{cyc}a(b \plus{} c)\geq(a \plus{} b \plus{} c)^3 \equal{} 1.$
Thus, it remains to prove that $ 3(ab \plus{} ac \plus{} bc)\leq1,$ which obvious. :)

[quote="arqady"] I think, the following is better than to differentiate two times:
[/quote]

That is true arqady  :)  .So the following solution is dedicated to you .

We have $ \sum\frac{a}{\sqrt{b\plus{}c}}\equal{}\sum\frac{a^2}{a\sqrt{b\plus{}c}}\geq\frac{(a\plus{}b\plus{}c)^2}{\sum a\sqrt{b\plus{}c}}$

But now $ \sum a\sqrt{b\plus{}c}\equal{}\sum\sqrt{a}\sqrt{ab\plus{}bc}\leq\sqrt{(a\plus{}b\plus{}c)(2(ab\plus{}bc\plus{}ca))}\leq\sqrt{\frac{2}{3}}$

and we and done

This thread is dead, right? But the problems are still awesome :D


trigonometry
trig identities
Law of Cosines
geometry proposed
geometry

Let $ ABC$ be a triangle, and $ D$ the midpoint of the side $ BC$. On the sides $ AB$ and $ AC$ we consider the points $ M$ and $ N$, respectively, both different from the midpoints of the sides, such that \[ AM^2\plus{}AN^2 \equal{}BM^2 \plus{} CN^2 \textrm{ and } \angle MDN \equal{} \angle BAC.\] Prove that $ \angle BAC \equal{} 90^\circ$.

[hide="An idea."]
Define  the points $ \{\begin{array}{ccc}
M'\in (AB) & , & AM'=BM\\\\
N'\in (AC) & , & AN'=CN\end{array}$ . From the [b]Carnot's lemma[/b] obtain that

exist the points $ P$ , $ Q$ for which $ \{\begin{array}{c}
PD\perp BC\ ,\ PM'\perp AB\ ,\ PN\perp AC\\\\
QD\perp BC\ ,\ QM\perp AB\ ,\ QN'\perp AC\end{array}$ a.s.o.[/hide]

Let $ A'$ the point such that $ BACA'$ is a paralellogram. Let $ M' \in A'C$ such that $ D$ is the midpoint of $ MM'$, let $ N' \in A'B$ such that $ D$ is the midpoint of $ NN'$. Define $ b \equal{} BM \equal{} M'C$, $ c \equal{} CN \equal{} BN'$, $ m \equal{} AM \equal{} A'M'$, $ n \equal{} AN \equal{} A'N'$, $ p\equal{}MD\equal{}DM'$, $ q\equal{}ND\equal{}DN'$, $ \angle \alpha \equal{} \angle MAN \equal{} \angle MDN$.
By the Law of Cosines:
 in $ MAND$ :  $ m^2 \plus{} n^2 \minus{} 2mn\cdot cos\alpha \equal{} p^2 \plus{} q^2 \minus{} 2pq\cdot cos\alpha$
 in $ BMDN'$:  $ b^2 \plus{} c^2 \plus{} 2bc\cdot cos\alpha \equal{} p^2 \plus{} q^2 \plus{} 2pq\cdot cos\alpha$
Suposse that $ \alpha \neq 90^\circ$, then $ cos\alpha\neq 0$  since $ m^2 \plus{} n^2 \equal{} b^2 \plus{} c^2$ we get $ bc \plus{} mn \equal{} 2pq$... (1)
Since $ [BACA'] \equal{} 2[BMN'] \plus{} 2[MAN] \plus{} [MNM'N']$ we get $ (b \plus{} m)(c \plus{} n)\cdot sin\alpha \equal{} (bc \plus{} mn \plus{} 2pq)\cdot sin\alpha$  we get $ bn \plus{} cm \equal{} 2pq$... (2).
By (1) and (2) we have $ (b \minus{} m)(c \minus{} n) \equal{} 0$, contradiction!

[quote="Valentin Virgil Nicula"][color=darkred][b]Let $ ABC$ be a triangle and $ D$ be the midpoint of the side $ [BC]$ .   Exist $ M\in (AB)$ , $ N\in (AC)$ so that

$ AM^2 + AN^2 = BM^2 + CN^2\ \ \wedge\ \ \widehat {MDN}\equiv\widehat {BAC}$ . Prove that $ AB\perp AC\ \ \vee\ \ AB=AC$ .[/b][/color][/quote]

[quote="Virgil Nicula"][color=darkred][u][b]Lemma (trigonometric form of the Ptolemeus' relation in a cyclic quadrilateral).[/b][/u]

Let $ ABC$ be a triangle. Then for any $ \phi\in R$ exists the identity $ a\cdot\sin\phi + b\cdot\sin (C - \phi ) = c\sin (B + \phi )$ .

[b]Particular cases :[/b][/color] $ \{\begin{array}{ccccc} \blacktriangleright & \phi : = 0 & \implies & b\cdot\sin C = c\cdot\sin B & \mathrm {(Sinus'\ theorem)} \\
 \\
\blacktriangleright & \phi : = \frac {\pi}{2} - B & \implies & a\cdot\cos B + b\cdot\cos A = c & \mathrm {(Cosinus'\ theorem)}\end{array}$ [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \{\begin{array}{c} BM = m\ ,\ CN = n \\
 \\
m(\widehat {BDM}) = x\ ,\ m(\widehat {CDN}) = y\end{array}$ . Thus, $ \{\begin{array}{c} AM = c - m\ ,\ AN = b - y \\
 \\
x + y + A = 180^{\circ}\end{array}$ and

$ AM^2 + AN^2 = BM^2 + CN^2$ $ \Longleftrightarrow$ $ 2(mc + nb) = b^2 + c^2$ $ \Longleftrightarrow$ $ \boxed {\ c(2m - c) + b(2n - b) = 0\ }\ \ (1)$ .

Apply the [b]Sinus' theorem[/b] in the triangles $ BDM$ and $ CDN$ :

$ \{\begin{array}{ccccc} \frac {BM}{\sin\widehat {BDM}} = \frac {DB}{\sin\widehat {DMB}} & \implies & \frac {m}{\sin x} = \frac {a}{2\sin (B + x)} & \implies & 2m = \frac {a\sin x}{\sin (B + x)} \\
 \\
\frac {CN}{\sin\widehat {CDN}} = \frac {DC}{\sin\widehat {DNC}} & \implies & \frac {n}{\sin y} = \frac {a}{2\sin (C + y)} & \implies & 2n = \frac {a\sin y}{\sin (C + y)}\end{array}$ . From the relation $ (1)$

obtain : $ \frac {c}{\sin (B + x)}\cdot[a\sin x - c\sin (B + x)] + \frac {b}{\sin (C + y)}\cdot[a\sin y - b\sin (C + y)] = 0$ .

Using the upper lemma obtain $ \frac {c}{\sin (B + x)}\cdot b\sin (x - C) + \frac {b}{\sin (C + y)}\cdot c\sin (y - B) = 0$ $ \Longleftrightarrow$

$ \sin (x - C)\sin (C + y) + \sin (B + x)\sin (y - B) = 0$ $ \Longleftrightarrow$

$ \cos(2C + y - x) - \cos(x + y) + \cos (2B + x - y) - \cos (x + y) = 0$ $ \Longleftrightarrow$

$ 2\cos (x + y) = \cos (2C + y - x) + \cos (2B + x - y)$ $ \Longleftrightarrow$ $ \cos (x + y) = \cos (B + C)\cos (x - y)$ $ \Longleftrightarrow$

$ \cos A = \cos A\cos (x - y)$ $ \Longleftrightarrow$ $ A = 90^{\circ}\ \ \vee\ \ x = y = 90^{\circ} - \frac A2$ . Prove easily that $ x=y\ \Longleftrightarrow\ AB=AC$

and in this case $ [MN]$ is the $ A$ - middleline of $ \triangle ABC$ , i.e. $ MA=MB$ , $ NA=NC$ . [/color]

My Solution:

Let $ E, F$ be midpoints of $ AC, AB$ respectively.

From the condition of the problem we have $ AM^2 - BM^2 = CN^2 - AN^2$ which is equivalent to $ AB(AM-BM) = AC(CN-AN)$. Since $ M, N$ are different from $ E, F$ we have
$ \dfrac{AB}{AC} = \dfrac{DE}{DF} = \dfrac{CN-AN}{AM-BM} = \dfrac{2NE}{2MF} = \dfrac{NE}{MF}$

From this we derive that $ \triangle NED \sim \triangle MFD$ and this leads to $ \angle{ABC}=\angle{DEC}=\angle{MFD}=\angle{NED}$ But $ \angle{DEC}+\angle{NED} = 180^{\circ}$ We therefore have $ \angle{BAC}=90^{\circ}$.

[quote="mathangel"]My Solution:

Let $ E, F$ be midpoints of $ AC, AB$ respectively.

From the condition of the problem we have $ AM^2 - BM^2 = CN^2 - AN^2$ which is equivalent to $ AB(AM - BM) = AC(CN - AN)$. Since $ M, N$ are different from $ E, F$ we have
$ \dfrac{AB}{AC} = \dfrac{DE}{DF} = \dfrac{CN - AN}{AM - BM} = \dfrac{2NE}{2MF} = \dfrac{NE}{MF}$

From this we derive that $ \triangle NED \sim \triangle MFD$ and this leads to $ \angle{ABC} = \angle{DEC} = \angle{MFD} = \angle{NED}$ But $ \angle{DEC} + \angle{NED} = 180^{\circ}$ We therefore have $ \angle{BAC} = 90^{\circ}$.[/quote]
Why $ \triangle NED \sim \triangle MFD$ ?  :blush:
Thanks

I really liked the problem, despite the problem somewhat hinting towards the use of midpoints. Let $L, K$ be the midpoints of the sides $AB, AC$, respectively. WLOG, let $AM > AL$, $AN < AK$. We'll prove that,
$$ AB/AC = NK/LM $$
which will give that $NK/LM = KD/ LD$ or $NK/KD = LM/ LD$, which we can then use law of sines on $\triangle NKD, \triangle MLD$.

By our condition, $AM^2 - BM^2 = AB(AM - BM) = AC(CN - AN) = CN^2 - AN^2$, or since $AM - BM = 2LM$ and $CN - AN = 2NK$, we get our claim.

Let $\angle NDK = \angle MDL = a$, $\angle BLD = \angle DKC = b$. By Law of Sines on $\triangle NKD$ and $\triangle MLD$, we get

$$NK/KD = \sin a/ \sin {(b-a)} = \sin a/ \sin (a + b) = LM/LD$$

or $\sin (b-a) = \sin (a+b)$. Two cases:

$1)$ $b - a = a + b$
Then $a = 0$, contradiction.

$2)$ $b - a = 180^{\circ} - a - b$
Then $b = 90^{\circ}$, which is what we needed to prove.

[hide= Remarks]
 There are also some other shorter solutions, such as the official solution, which uses a somewhat genius trick of just constructing additional right triangles to prove our problem. Had the same idea as some other problems which use these types of sum of squares, USAMO 2019 P2 and Russian problems, to which I've forgotten.
[/hide]


My keyboard won't type anything into TeXnicCenter! I tried Notepad (and this forum post) and it works perfectly.  I clicked "Save As" to account for <Read Only> file stuff, but my keyboard still won't work on JUST THAT APPLICATION.  Could I please have a suggestion on what I should do?

[quote="AIME15"]Wait, so you can't type something here and copy/paste it in either?

Sometimes I have the same problem with my Mac, except it won't type anything into Google Toolbar or the chat on FTW.[/quote]

[quote="stevem"]Try opening a new document File, New. Does that work?[/quote]

Actually, it's working now! Maybe it's just a temporary thing.


calculus
integration
number theory unsolved
number theory

show that the equartion nCk=$m^{l}$ has no integral solution with $l$=>2  and 4=<$l$=<n-4


Given prime number $p$. $a_1,a_2 \cdots a_k$ ($k \geq 3$) are integers not divible by $p$ and have different residuals when divided by $p$. Let
\[ S_n= \{ n \mid 1 \leq n \leq p-1, (na_1)_p < \cdots < (na_k)_p \} \]
Here $(b)_p$ denotes the residual when integer $b$ is divided by $p$. Prove that $|S|< \frac{2p}{k+1}$.

Can anyone help?
I've been stuck on this problem :)

In general, when making some estimates/bounds based on fractional parts, it is always easy to convert the problem into "a sum of fractional parts equal to an integer". [hide=This idea is used in] CTST 23/3 and ISL 19 N6 [/hide].

Solution: Note that $\{ \frac{na_j}{p} \} < \{ \frac{na_{j+1}}{p}\} \iff \{ \frac{na_j}{p} \} + \{ \frac{n(a_{j+1}-a_j)}{p} \} = \{ \frac{na_{j+1}}{p}\}$

Therefore, if $x_1=a_1, x_j=a_j-a_{j-1}$ for $j=2,\cdots,k$ and $x_{k+1}=p-a_k$ then

$$\sum_{j=1}^{k+1} \{\frac{na_j}{p}\} = 1$$

Consider the sum $\sum_{s\in S} \sum_{j=1}^{k+1} \{\frac{sa_j}{p}\}$. On one hand it is equal to $|S|$. On the other hand, it is $\sum_{j=1}^{k+1} \sum_{s\in S} \{\frac{sa_j}{p}\} \ge (k+1) \frac{|S| (|S|+1)}{2p}$. Hence $|S|+1 \le \frac{2p}{k+1}$, as desired.


Find $ x, y \in R$ of this system: $ \begin {cases} xy \minus{} \sqrt {xy} \equal{} 3 \\
\sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 1} \equal{} 4 \end {cases}$

Thank stephencheng for your post.

Root of this system is: ${{ \begin {cases} x=8+\sqrt{\frac{39+\sqrt{13}}{2}}}+4\sqrt{5+\sqrt{\frac{39+\sqrt{13}}{2}}} \\ y=8+\sqrt{\frac{39+\sqrt{13}}{2}}}-4\sqrt{5+\sqrt{\frac{39+\sqrt{13}}{2}}}\end {cases}$


Show that $ \sin(3x) \equal{} (1\plus{}2\cos(2x))\sin x$.

[hide="Solution"]$ \sin 3x \equal{} \sin (x \plus{} 2x) \equal{} \sin x \cos 2x \plus{} \cos x \sin 2x \equal{} \sin x \cos 2x \plus{} 2 \cos^2 x \sin x \equal{} \sin x (\cos 2x \plus{} 2\cos^2 x) \equal{} \sin x (\cos 2x \plus{} (\cos 2x \plus{} 1)) \equal{} \sin x (1 \plus{} 2 \cos 2x)$ [/hide]


function
integration
floor function
calculus
calculus computations

For $ 1<s< \infty$ define $ \zeta(s) \equal{} \sum_{n\equal{}1}^{\infty} \frac{1}{n^s}$. Prove that $ \zeta(s) \equal{} s \int_{1}^{\infty} \frac{[x]}{x^{s\plus{}1}} \ dx$. So consider an interval $ [1,N]$. Let $ g(s) \equal{} s \int_{1}^{N} \frac{[x]}{x^{s\plus{}1}} \ dx \minus{} \sum_{n\equal{}1}^{N} \frac{1}{n^s}$. Now we want to show that $ g(s) \equal{} 0$. Then we take $ N \to \infty$ and conclude that $ g(s) \equal{} 0$ on $ [1, \infty)$. The second series is convergent.

[quote="Tyl"]Why would $ g$ be zero for each $ N$? It seems quite false for $ N \equal{} 1$, $ s \equal{} 1$.[/quote]

Nope, lordmahone123 says that $ s>1$

The key here is the partial summation formula:
\[ \sum_{n \equal{} 1}^Na_nb_n \equal{} a_{N \plus{} 1}B_N \minus{} \sum_{n \equal{} 1}^NB_N(a_{n \plus{} 1} \minus{} a_n) \text{, where }B_n \equal{} \sum_{k \equal{} 1}^nb_k.
\]
Setting $ a_n \equal{} n^{ \minus{} s}$ and $ b_n \equal{} 1$ gives
\[ \sum_{n \equal{} 1}^Nn^{ \minus{} s} \equal{} \frac {N}{(N \plus{} 1)^s} \minus{} \sum_{n \equal{} 1}^Nn((n \plus{} 1)^{ \minus{} s} \minus{} n^{ \minus{} s}).
\]
Now, $ (n \plus{} 1)^{ \minus{} s} \minus{} n^{ \minus{} s} \equal{} \int_n^{n \plus{} 1} \minus{} sx^{ \minus{} s \minus{} 1}dx$, so
\[ n((n \plus{} 1)^{ \minus{} s} \minus{} n^{ \minus{} s}) \equal{} \int_n^{n \plus{} 1} \minus{} s\lfloor x\rfloor x^{ \minus{} s \minus{} 1}dx.
\]
This gives
\[ \sum_{n \equal{} 1}^Nn^{ \minus{} s} \equal{} \frac {N}{(N \plus{} 1)^s} \plus{} s\int_1^{N \plus{} 1}\lfloor x\rfloor x^{ \minus{} s \minus{} 1}dx.
\]
When $ Re(s) > 1$, the first term on the RHS goes to zero as $ N\to\infty$. The result follows.

[quote="BlAcK_CaT"]
Nope, lordmahone123 says that $ s > 1$[/quote]

But later on he claims to want to "conclude that $ g(s) \equal{} 0$ on $ [1,\infty)$".

Also, CaptainCanada's approach seems to incorporate an error term $ \frac {N}{(N \plus{} 1)^s}$.

edit: but his integral goes to $ N\plus{}1$, not $ N$. Hmm.

Hi every body :)

$ if s \in ]1; \plus{} \infty[$

it's very easy to prove that:

$ \int_1^n \frac {[x]}{x^{s \plus{} 1}}dx \equal{} \frac {1}{s}\left( \frac { \minus{} 1}{n^{s \minus{} 1}} \plus{} \sum_{k \equal{} 0}^{n \minus{} 1}\frac {1}{(k \plus{} 1)^s} \right)$

than

$ s \int_1^n \frac {[x]}{x^{s \plus{} 1}}dx \minus{} \sum_{k \equal{} 1}^{n } \frac {1}{k^s} \equal{} \frac { \minus{} 1}{n^{s \minus{} 1}}$

than if $ n \to \plus{} \infty$ than $ g(s) \equal{} 0 \forall s > 1$

PS: if is not clair i will post all prove by the first :)

[quote="Tyl"][quote="BlAcK_CaT"]
Also, CaptainCanada's approach seems to incorporate an error term $ \frac {N}{(N \plus{} 1)^s}$.

edit: but his integral goes to $ N \plus{} 1$, not $ N$. Hmm.[/quote][/quote]

The whether the upper bound of the integral is $ N$ or $ N\plus{}1$ doesn't matter, because the difference goes to zero. All it changes is the form of the error term. Using the integral to $ N\plus{}1$ just makes the approach using partial summation a little easier to work through, which is nice, because this method leads to many other useful formulae.

Ah but I did not comment on the solution you presented. I fully understand it doesn't matter. However, I added that edit to show that the error term is not what I said before, as you also pointed out yourself in your last post. I also see why you chose to let it be $ N\plus{}1$.

My comment was about the OP, which seems to be a bit sloppy about things.

Yeah, I misinterpreted the intent of your post. I understand what you meant now. :blush:


[b]I'm looking for friends~~cuz I'm going to USC next year!!!
Please reply if you're going as well!!!
It's gonna be fun!!! :blush: [/b]

Hi Connie! I'm not sure if I'm going to USC, but I was there for Explore weekend, and it was pretty cool.

If that's a strong criterion, then...

UCSB: 4.

Simon Fraser University: 7

(yes, I know I don't go to a university in southern california)

[quote="Kent Merryfield"]Number of students placing in the top 500 students on the 2005 Putnam:

CSU Long Beach: 1
USC: 0

Just thought I'd mention that.  :)[/quote]
wow... notice that I'm a BUSINESS student...not interested in math at all...
even I was invited to last year's Waterloo camp in Canada--the only good thing is I met my boyfriend there

I think USC business is decent...

My apologies for the continued distortion of the purpose of this thread :-)

[quote="mna851"]Where are the putnam results posted?[/quote]

Kiran Kedlaya's Putnam Archive is available at http://www.unl.edu/amc/a-activities/a7-problems/putnam/


Let $a,\ b,\ c$ be postive numbers such that $a+b+c=3$ and $abc=1$.
Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\leqq 3$.

again this is pointles...
your condition gives us $a=b=c=1$.

I don't think the condition only gives $a = b= c =1$ :)
Why should it be like that? :maybe:

I found something, but without using abc = 1 ...

By Cauchy we have

$(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}\leq(a+b+c)(1+1+1)=9$

Done.

By the Arithmetic Mean-Geometric Mean Inequality, we have $\frac{a+b+c}{3}\geq \sqrt[3]{abc}$.  The conditions tell us that $\frac{a+b+c}{3}= \sqrt[3]{abc}$.  Equality holds if and only if $a=b=c$.  Using the fact that $a+b+c=3$, we have $a=b=c=1$.  As a result, $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$.  Therefore, this is a useless inequality, especially for the Olympiad level.


a)simplify $ (\minus{}2i)^{3}$

b)Hence find all complex numbers $ z$ such that $ z^{3}\equal{} 8i$. Express your answers in the form $ x\plus{}iy$

[hide=book's soln.]It has 3 ways to do it. But im confused with this one.
Expressing 8i in mod-arg form $ \equal{} \equal{} \equal{}\Rightarrow$ $ 2^{3}[\cos\frac{\pi}{2}\plus{}i\sin\frac{\pi}{2}]$

so, $ z^{3}\equal{} 2^{3}[\cos\frac{\pi}{2}\plus{}i\sin\frac{\pi}{2}]$

can this be written as $ z^{3}\equal{} 2^{3}[\cos\frac{\pi}{2}\plus{}i\sin\frac{\pi}{2}]$

$ z \equal{} 2[cis(\frac{2n\pi}{3}\plus{}\frac{\pi}{6})]$

How did they obtain the step above ^^ ?

I thoguht DeMoivre's therom just...$ \equal{} 2[cis(\frac{\pi}{6})]$

-------------------------------
But then another qn....

express $ \sqrt 3\minus{}i$ in mod-arg form.
Hence express $ w^{5}$  in the form x+iy where x and y are real numbers

$ \sqrt 3\minus{}i$ =$ 2(cos\frac{\pi}{6}\minus{}isin\frac{\pi}{6})$

then $ w^{5}$ = ..... and simplify

But how come you don't make use of general soln. in this question? but the question above they have general soln.?

DeMoivre's is with the $ \frac{2k\pi}{n}\ ,\ \{ n | z^{\frac{1}{n}}\}\ ,\ k\in\mathbb{Z}\ \text{and}\ \{ 0,1,2,3,...,n\minus{}1\}$ part, finding all the roots since there are three of them.

don't they both mak euse of De Moivre's theorem?

[quote="bos1234"]a)simplify $ (\minus{}2i)^{3}$[/quote]
This one is easy $ (\minus{}2i)^{3}\equal{} (\minus{}2)^{3}i^{3}$ now simplify.

[quote]
b)Hence find all complex numbers $ z$ such that $ z^{3}\equal{} 8i$. Express your answers in the form $ x\plus{}iy$
[/quote]
The most important part here is to write $ 8i$ in polar form. Notice that $ |8i| \equal{} 8$ and $ \arg(8i) \equal{}\frac{\pi}{2}$.
So,
$ z^{3}\equal{} 8\left[\cos\left(\frac{\pi}{2}\plus{}2\pi k\right)\plus{}i\sin\left(\frac{\pi}{2}\plus{}2\pi k\right)\right]$ for $ k\equal{}0,1,2$.

That means,
$ z_{1}\equal{} 8^{1/3}\left[\cos\left(\frac{\pi}{6}\right)\plus{}i\sin\left(\frac{\pi}{6}\right)\right]$

$ z_{2}\equal{} 8^{1/3}\left[\cos\left(\frac{\pi}{2}\plus{}\frac{2\pi}{3}\right)\plus{}i\sin\left(\frac{\pi}{2}\plus{}\frac{2\pi}{3}\right)\right]$

$ z_{3}\equal{} 8^{1/3}\left[\cos\left(\frac{\pi}{2}\plus{}\frac{4\pi}{3}\right)\plus{}i\sin\left(\frac{\pi}{2}\plus{}\frac{4\pi}{3}\right)\right]$

Couldn't you just divide z^3-8i by z+2i???

[b]@ 1=2:[/b] To answer your triple-question, yes, he could have done that and applied the quadratic formula. bos is learning about De'Moivre's theorem and polar form, so that would've been a useless response.

[b]@ bos:[/b] Remember that while what you did applies neatly for positive integer exponents, it doesn't quite work here because there are three solutions. Here's a parallel: find all numbers such that $ x^{2}\equal{} 4$. $ 4^{\frac{1}{2}}\equal{} 2$ is a solution, but not the only one: there is also $ \minus{}2$.

Keep in mind that the angle measure is periodic - it wraps around every $ 2\pi$. So when you want to take a cuberoot, you must not only consider it's simplest angle by $ 3$, but also add $ 2\pi$ to the angle and divide that result by three, add $ 4\pi$ to the angle and divide that result by three, etc.

What was done here is that $ \minus{}8i \equal{} [\text{cis}0]^{3}(\minus{}8i) \equal{}\left[\text{cis}\frac{2\pi}{3}\right]^{3}(\minus{}8i) \equal{}\left[\text{cis}\frac{4\pi}{3}\right]^{3}(\minus{}8i) \equal{}\ldots$

Why? Because for all $ n\in\mathbb{N}$, note that $ \left[\text{cis}\frac{2 n\pi}{3}\right]^{3}\equal{}\text{cis}2 n\pi \equal{} 1$
So, we need to consider all these possibilities. Each yields a cuberoot. Now by periodicity, there are only three [b]unique[/b] values that this can produce.


Please Help.

  Post it here or mail me at sumitmenon93@hotmail.com.

  Hope for ur reply.

[quote="mstoenescu"]Hi !

See for instance [url]http://en.wikipedia.org/wiki/Geometer[/url] :)[/quote]

Thebault
Casey
Morley
Poncelet
Steiner
Poincare
Hilbert (a big problem of axioms)
Pascal
Archimedes
Gergonne
Euclides

let me add $DAVOOD$$VAKILI$, to the above. :wink:


inequalities
geometry
incenter
inradius
geometry proposed
U sviti matematyky

Let $ I$ be the incenter of a triangle $ ABC.$ Points $ P$ and $ R$, $ T$ and $ K,$ $ F$ and $ Q$ are chosen on sides $ AB,$ $ BC,$ and $ AC$ respectively such that $ TQ\|AB,$ $ RF\|BC,$ $ PK\|AC$ and the lines $ TQ,$ $ RF,$ and $ PK$ are concurrent at the point $ I.$ Prove that $ TK\plus{}QF\plus{}PR\ge KF\plus{}PQ\plus{}RT.$

Let $r$ denote the inradius of $\triangle ABC.$ Substituting $TK=r \cdot (\cot B+ \cot C),$ $PQ= r \cdot \sec \frac{A}{2}$ and the cyclic expressions, the object inequality becomes:

$2 \cdot ( \cot A+ \cot B+ \cot C) \ge \sec \frac{A}{2}+ \sec \frac{B}{2}+\sec \frac{C}{2}.$


hello, we get
$ \cos(\arcsin(1\minus{}\sin(x)))\equal{}\sqrt{1\minus{}(1\minus{}\sin(x))^2}$
Sonnhard.


The complex number $ z$ satisfies $ z \plus{} |z| \equal{} 2 \plus{} 8i$.  What is $ |z|^2$?  Note: of $ z \equal{} a \plus{} bi$, then $ |z| \equal{} \sqrt{a^2 \plus{} b^2}$.

$ \textbf{(A)}\ 68 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 160 \qquad \textbf{(D)}\ 208 \qquad \textbf{(E)}\ 289$

My method is not too slow, but I'm sure there's a faster way.

[hide="My Solution"]
Substituting $ z \equal{} a \plus{} bi$ into the first equation, we have

$ a \plus{} bi \plus{} \sqrt{a^2 \plus{} b^2} \equal{} 2 \plus{} 8i$

Because $ b$ is the coefficient of the only term involving $ i$, $ b \equal{} 8$. Plugging that in and equating to the real part, we have

$ a \plus{} \sqrt{a^2 \plus{} 8^2} \equal{} 2\\
a^2 \plus{} 64 \equal{} a^2 \minus{} 4a \plus{} 4\\
4a \equal{} \minus{}60\\
a \equal{} \minus{}15$

Thus, $ |z|^2 \equal{} (\sqrt{a^2 \plus{} b^2})^2 \equal{} 225 \plus{} 64 \equal{} \boxed{(E) 289}$.
[/hide]

[hide="Hint"] $ |z|^2 \equal{} z \cdot \bar{z}$.  What is $ \bar{z} \plus{} |z|$? [/hide]


abstract algebra
group theory
inequalities
number theory
relatively prime
superior algebra
superior algebra unsolved

[b]Moderator edit: Please discuss problems 1a, 1b at http://www.mathlinks.ro/Forum/viewtopic.php?t=27046 .[/b]1a
Let G be a finite group and suppose the automorphism T sends more than three-quarters of the elemnts of G onto their inverses. Prove that $ T(x) \equal{} x^{ \minus{} 1}$ for all $ x \in G$
1b
Can you find an example of a finite group which is non-abelian and which has an automorphism which maps exactly three-quarters of the elements onto their inverses?

2
If G is a finite group and its p-Sylow subgroup P lies in the center of G, prove that there exists a normal subgroup N of G with P intersect N = identity and $ PN \equal{} G$

3
[b]Moderator edit: Please discuss problem 3 at http://www.mathlinks.ro/Forum/viewtopic.php?t=196356 .[/b]
Let G be a finite group and suppose that T is an automorphism of G such that $ T^3$ is the identity automorphism. Suppose further that $ T(x) \equal{} x$ implies $ x \equal{} e$. Prove that for any prime $ p$ which divides $ |G|$, the p-sylow subgroup is normal in G.

1a and 1b: http://www.mathlinks.ro/Forum/viewtopic.php?t=27046

Thanks for your beautiful solution. What about the other two problems?

[quote="Gockel"]Hi.

I have found a solution for 2, but it's not that elementary... Since P is in the centre of G it is normal. Since P is a p-sylowgroup, |P| and |G/P| are relatively prime, so there is a complementary subgroup N with the Schur-Zassenhaus-theorem. The last thing to prove is that N is normal, but this is easy, because P is in the centre and so gN=Ng for all g in P.[/quote]

Sorry, can you tell me what the theorem is, and if possible, tell me a proof? I have never heard of it before. Thank you.

It states: If $N$ is a normal subgroup of a finite group $G$ s.t.  $gcd(|G/N|,|N|)=1$ and either $N$ or $G/N$ is solvable, then $G$ splits over $N$, i.e. there's a subgroup $U$ of $G$ s.t. $G$ is the semidirect product of $N$ and $U$.

I don't think that this theorem should be applied here. It would be too trivial :D

I don't even know what semidirect product is...  :?  :(

You can look it up ...

http://planetmath.org/encyclopedia/SemidirectProductOfGroups.html

Actually, the statement I've seen doesn't require solvability. What does require the solvability of one of the two groups is a strengthening of the theorem, staing that all such subgroups $U$ of $G$ belong to the same conjugacy class.

Hi.

In fact, the solvabilty of N or G/N (or both) is included in the requirement $gcd(|N|,|G/N|)=1$ because of the theorem of Feit-Thompson, but if one only wants to show, that N has a complementary subgroup, the solvability isn't needed for the proof.

a bump to see if anyone can help. seems like nobody has attempted question 3.


geometry
3D geometry
pyramid
sphere
geometry proposed

A pyramid $ ABCDS$ is given (the base is convex quadrilateral). A sphere is inscribed in this pyramid and it is tangent to side $ ABCD$ at point $ P$.
Prove that $ \angle APB\plus{}\angle CPD\equal{}180^{o}$

can anybody help me with this problem, please? I cannot do it anyway....

Nice problem, but I think it is easy. What are the [b]conditions to inscribe a sphere to such a pyramid??[/b] Maybe the base (this convex quadraliteral is inscribed in a circle??

So, can anybody tell me what are the [b]conditions to inscribe a sphere to such a pyramid?[/b]


inequalities
logarithms
inequalities proposed

For $ a,b,x,y>0$, prove that: $ a^{x\plus{}y}\plus{}b^{x\plus{}y}\geq a^xb^y\plus{}a^yb^x.$

[quote="marin.bancos"]For $ a,b,x,y > 0$, prove that: $ a^{x \plus{} y} \plus{} b^{x \plus{} y}\geq a^xb^y \plus{} a^yb^x.$[/quote]
$ \iff (a^x\minus{}b^x)(a^y\minus{}b^y) \ge 0$

I know! So what?
For $ x,y\in N, x,y\geq 1,$ it's obvious, because: $ (a^x\minus{}b^x)\cdot(a^y\minus{}b^y)$ is equal to
$ \left.(a\minus{}b)^2(a^{x\minus{}1}\cdot b^0\plus{}a^{x\minus{}2}\cdot b^1\plus{}a^{x\minus{}3}\cdot b^2\cdots\plus{}a^0\cdot b^{x\minus{}1})\cdot (a^{y\minus{}1}\cdot b^0\plus{}a^{y\minus{}2}\cdot b^1\plus{}a^{y\minus{}3}\cdot b^2\cdots\plus{}a^0\cdot b^{y\minus{}1})\right.$.
But: $ x,y\in R, x,y>0$.

using subs of logarithms must help exactly, but I can't figure it out.

[quote="marin.bancos"]I know! So what?
For $ x,y\in N, x,y\geq 1,$ it's obvious, because: $ (a^x \minus{} b^x)\cdot(a^y \minus{} b^y)$ is equal to
$ \left.(a \minus{} b)^2(a^{x \minus{} 1}\cdot b^0 \plus{} a^{x \minus{} 2}\cdot b^1 \plus{} a^{x \minus{} 3}\cdot b^2\cdots \plus{} a^0\cdot b^{x \minus{} 1})\cdot (a^{y \minus{} 1}\cdot b^0 \plus{} a^{y \minus{} 2}\cdot b^1 \plus{} a^{y \minus{} 3}\cdot b^2\cdots \plus{} a^0\cdot b^{y \minus{} 1})\right.$.
But: $ x,y\in R, x,y > 0$.[/quote]
If $ a > b$ then $ a^x > b^x$ and $ a^y > b^y$. Since $ t \to t^x$ and $ t \to t^y$ are increasing. So if $ a > b$ we get $ a^x\minus{}b^x,a^y\minus{}b^y > 0$ and if $ a < b$ we get $ a^x\minus{}b^x,a^y\minus{}b^y<0$.

[quote="xzlbq"]Can be extended to n-variables？
BQ[/quote]

Certainly

 :wink:

if $ x_k>0, a_k \ge 0(k\equal{}1,2,...,n)$, we have

\[ \left[ \begin{array}{c} x_1^{a_1 } ,x_2^{a_1 } , \cdots ,x_n^{a_1 } \\
x_1^{a_2 } ,x_2^{a_2 } , \cdots ,x_n^{a_2 } \\
\vdots \\
x_1^{a_n } ,x_2^{a_n } , \cdots ,x_n^{a_n } \\
\end{array} \right] \ge \left[ \begin{array}{c} x_1^{a_1 } ,x_2^{a_1 } , \cdots ,x_n^{a_1 } \\
x_2^{a_2 } ,x_3^{a_2 } , \cdots ,x_1^{a_2 } \\
\vdots \\
x_n^{a_n } ,x_1^{a_n } , \cdots ,x_{n \minus{} 1}^{a_n } \\
\end{array} \right]\]

holds by " Wei-Wei dual inequalities " .

[quote="xzlbq"]$ a^{x \plus{} y \plus{} z} \plus{} b^{x \plus{} y \plus{} z} \plus{} c^{x \plus{} y \plus{} z} \geq (\frac {1}{2}b \plus{} \frac {1}{2}c)^xb^yc^z \plus{} a^y(\frac {1}{2}c \plus{} \frac {1}{2}a)^zc^x \plus{} a^zb^x(\frac {1}{2}a \plus{} \frac {1}{2}b)^y \geq(a^xb^yc^z \plus{} a^yb^zc^x \plus{} a^zb^xc^y)$

BQ[/quote]

$ a^{x \plus{} y \plus{} z} \plus{} b^{x \plus{} y \plus{} z} \plus{} c^{x \plus{} y \plus{} z} \geq (\frac {1}{2}b \plus{} \frac {1}{2}c)^xb^yc^z \plus{} a^y(\frac {1}{2}c \plus{} \frac {1}{2}a)^zc^x \plus{} a^zb^x(\frac {1}{2}a \plus{} \frac {1}{2}b)^y$

this ineq dose not hold.

yon can try $ (a,b,c) \equal{} (\frac {7}{8}, 1, 1)$ and $ x \equal{}2 , y \equal{} z \equal{} 1$.

and this

$ (\frac {1}{2}b \plus{} \frac {1}{2}c)^xb^yc^z \plus{} a^y(\frac {1}{2}c \plus{} \frac {1}{2}a)^zc^x \plus{} a^zb^x(\frac {1}{2}a \plus{} \frac {1}{2}b)^y \geq(a^xb^yc^z \plus{} a^yb^zc^x \plus{} a^zb^xc^y)$

dose not hold too.

you can try $ (a,b,c) \equal{} (9, \frac {3}{4}, 1)$ and $ x \equal{} 2 , y \equal{} z \equal{} 1$.

[quote="xzlbq"]You reply too fast,I have modify it.
at #9.
BQ[/quote]

I have modify too.

8# and 10#.


Find all $(x,y)\in\mathbb{Z}^{2}$ such that $x^{3}-y^{3}=2xy+8$.

Does that mean that X and Y are elements of the perfect squares? Is that what $\mathbb{Z}^{2}$ signifies?

$\mathbb{Z}^{2}: = \mathbb{Z}\times\mathbb{Z}: = \{(x,y)|x,y\in\mathbb{Z}\}$

Generally

$A^{n}: = \underbrace{A\times A\times\dots\times A}_{n}: = \{(x_{1},x_{2},\dots,x_{n})|x_{1},x_{2},\dots,x_{n}\in A\}$


If $ a,b,c,d$ are positive numbers such that
$ a+b+c+d=\frac 1{a}+\frac 1{b}+\frac 1{c}+\frac 1{d}$,
then
$ (1-abcd)(a^{k}+b^{k}+c^{k}+d^{k}-\frac 1{a^{k}}-\frac 1{b^{k}}-\frac 1{c^{k}}-\frac 1{d^{k}}) \ge 0$
for any positive integer $ k$.

For $ k\equal{}2$ there is a nice and no hard solution.

[quote="Vasc"]For $ k \equal{} 2$ there is a nice and no hard solution.[/quote]
I want to see

[quote="Vasc"]If $ a,b,c,d$ are positive numbers such that
$ a \plus{} b \plus{} c \plus{} d \equal{} \frac 1{a} \plus{} \frac 1{b} \plus{} \frac 1{c} \plus{} \frac 1{d}$,
then
$ (1 \minus{} abcd)(a^{k} \plus{} b^{k} \plus{} c^{k} \plus{} d^{k} \minus{} \frac 1{a^{k}} \minus{} \frac 1{b^{k}} \minus{} \frac 1{c^{k}} \minus{} \frac 1{d^{k}}) \ge 0$
for any positive integer $ k$.[/quote]
by cauchy-swcharzt :
$ \equal{}> a\plus{}b\plus{}c\plus{}d \ge 4$

So how to do continue ?


(256y^4)^.25 i am not sure if i got the right but my answer is

64y^4

not quite: the "^.25" means raising the expression to the 1/4 power, or taking the 4th root of it.


prove that u can never find a natural number d such that 2d-1,5d-1,13d-1 are all be perfect squares.

[hide]
MY solution:
It is obvious that d is of the form 4n+1
Then 2d-1= 8n+1
Then 5d-1 = 20n+4 = 4(5n+1)
Then 13d-1 = 52n+12 = 4(13n+3)
Implies 8n+1,5n+1,13n+3 are perfect squares
Let 8n+1 = $ x^2$
5n+1 = $ y^2$
13n+3 = $ z^2$
Then $ z^2$ - $ y^2$ -1 = $ x^2$
Then $ z^2$ - $ y^2$ = $ x^2$ + 1
But $ x^2$ + 1 is congruent to 2 mod 4
Let $ x^2$ + 1 = 2l where l is odd
$ z^2$ - $ y^2$ = 2l
Then one of z+y and z-y is even whereas other is odd
Which implies z and y cannot be natural numbers
Contradiction.
So no d exists.

[/hide]

Is my solution correct.
Others can post their solutions too :D

befuddler i think the soln orl posted there and ur soln are kind of similar except tht he went by mod 16 directly while u analysed using mod 4. Nice one :rotfl:  :D !1


Find all the numbers $ x,y,k\in \mathbb{N}^*\ ,\ x>y\ ,\ k\ge 2$ , that satisfies : $ x^k\minus{}y^k\equal{}242$ .

$ 242\equal{}2 \times 11^2$
$ x,y$ both odd or both even or $ L.H.S.$ is odd , impossible. So $ 2|x\minus{}y$ and $ x\minus{}y \ge 2$
$ x,y$ can't be both even or $ 4|L.H.S.$, impossible again.

Case 1: $ y\equal{}1$, $ x^k\equal{}243$ give solution $ (x,y,k)\equal{}(3,1,5)$

Case 2: $ y \ge 3$, then mod 7 both sides we have if $ k\equal{}3$ , $ x^k\equal{}\minus{}1$, $ 0$ or $ 1$ for all $ x$ , impossible.

Now $ 242\equal{}x^k\minus{}y^k \ge (y\plus{}2)^k\minus{}y^k \ge 2ky^{k\minus{}1} \ge 2k(3)^{k\minus{}1}$ gives $ k \le 4$

For $ k\equal{}4$, $ (x^2\plus{}y^2)(x\plus{}y)(x\minus{}y)\equal{}242$, impossible as $ 2|x\plus{}y$ also.
For $ k\equal{}2$, $ (x\plus{}y)(x\minus{}y)\equal{}242$, impossible again by the same reason.

So $ (x,y,k)\equal{}(3,1,5)$ is the only solution.

as 4 doesn't divide RHS so both $ x$ and $ y$ are odd.
well if $ k$ is even, let it be $ 2m$

so
\[ (x^m \plus{} y^m)(x^m \minus{} y^m) \equal{} 2.11.11\]
checking all cases leads impossiblity.
hence $ k$ is odd.
\[ (x \minus{} y)\sum_{i \equal{} 0}^{k \minus{} 1}x^iy^{k \minus{} 1 \minus{} i} \equal{} 2.11.11\]
only possible case $ x \minus{} y \equal{} 2$

so
\[ 121\equiv 2^{k \minus{} 1}mody\]
which gives
\[ k \equal{} 3,5,7\]
as both $ x$ and $ y$ are odd, it is easy to observe from
\[ \sum_{i \equal{} 0}^{k \minus{} 1}x^iy^{k \minus{} 1 \minus{} i} \equal{} 121\]
that only $ x \equal{} 3$, $ y \equal{} 1$, $ k \equal{} 3$
satisfy as all rest give values greater than $ 121$.

[quote="alex2008"]Find all the numbers $ x,y,k\in \mathbb{N}^*\ ,\ x > y\ ,\ k\ge 2$ , that satisfies : $ x^k \minus{} y^k \equal{} 242$ .[/quote]
Remember: If $ p \in \mathbb{P}, a,b \in \mathbb{Z}, n \in \mathbb{N}$ and $ (ab,p) \equal{} 1, a \equiv b \bmod p$ then $ e_p(a^n \minus{} b^n) \equal{} e_p(a \minus{} b) \plus{} e_p(n)$.
Obviously $ x \equiv y \bmod 2$. So $ e_2(x^n \minus{} y^n) \equal{} e_2(x \minus{} y) \plus{} e_2(n) \equal{} 1 \plus{} e_2(n)$. But $ e_2(x^n \minus{} y^n) \equal{} e_2(242) \equal{} 1$, so $ e_2(n) \equal{} 0 \iff 2 \nmid n$. $ x > y \ge 1$. Hence $ x^k \minus{} y^k \ge (y \plus{} 1)^k \minus{} y^k \ge 2^k \minus{} 1^k$. Hence $ 2^k \minus{} 1 \le 242 \iff 2^k \le 243$. But then $ k \le 7$. So $ k \in \{3,5,7\}$. Note that $ 242 \equal{} 2 \cdot 11^2$. $ 11 \mid x$ or $ 11 \mid y$ means $ 11^3 \mid LHS$, so $ (11,xy) \equal{} 1$. And in the same way $ (2,xy) \equal{} 1 \Rightarrow (22,xy) \equal{} 1$
$ k \equal{} 3$ then $ x^3 \equiv y^3 \bmod 11 \Rightarrow x^{21} \equiv y^{21} \bmod 11 \iff 11 \mid x \minus{} y$.
$ k \equal{} 7$ then $ x^7 \equiv y^7 \bmod 11 \Rightarrow x^{21} \equiv y^{21} \bmod 11 \iff 11 \mid x \minus{} y$.
Hence if $ k \equal{} 3$ or $ k \equal{} 7$: $ x^k \minus{} y^k \ge 12^k \minus{} 1^k \ge 12^3 \minus{} 1 > 242$. Contradiction.
Hence $ k \equal{} 5$. If $ x \equal{} 2$ then $ LHS < 242$. If $ x \equal{} 3$ we get a solution: $ 3^5 \minus{} y^5 \equal{} 242 \iff y^5 \equal{} 1 \iff y \equal{} 1$. If $ x \ge 4$ we get $ x^k \minus{} y^k \equal{} x^5 \minus{} y^5 \ge 4^5 \minus{} 3^5 > 242$.

So the only solution is $ (x,y,k) \equal{} (3,1,5)$.


symmetry
geometry
geometric transformation
reflection
counting
distinguishability
combinatorics unsolved

Given a positive integer n,  n >= 3, with n noncollinear points in the plane, work
out a formula in terms of n that gives the total number of combinations of line segments that can be drawn between the points from all those cases with one line
segment added to those with two line segments on up.  (You keep drawing in line segments in the latter stages until you run out of points to do so.)

Example)  n = 3

3 where there are individual line segments
3 where there are 2 line segments in each one
1 where a triangle is formed
------------------------------------------------------------
7 total

Example)  There are a total of 63 for n = 4.  [1 line segment placements on up to the
                 maximum you can fit in.]

Note:  Your picked (or random) points need not have any symmetry, but let's look at points of a rectangle's vertices for instance.  It has certain symmetries.  However, translations, reflections  of shapes made by points/line segments are to be counted as different if the collection of points is different by at least one point from all other prior sets looked at.

I just wanna inform you that there is (since it looks like combinatorics) a Combinatorics forum (along with others like Geometry or Number Theory). Don't post verything into Algebra...

To the problem: asks it for the connected graps on $n$ distinguishable vertices where some of them can be neglected (thus a single vertice doesn't disturb connectivity here)¿
I have problems understanding it...

The problem indicates to look at all possible arrangements of vertices taken two at a time, then three at a time, and so on until all the vertices have been used up and to come up with an algebraic formula for a grand total for all of those cases.  You will be "ignoring"
vertices only as they are not included in a count of a combination.

At first, lets count the number of possible line segments between these $n$ points:
One can connect any pair of points, thus there are es many as there are pairs of given points, namely $\frac{n(n-1)}{2}=\binom{n}{2}$.

Now for every segment, we can put it in the constellation or not, giving $2$ choices for each segment, thus $2^{\frac{n(n-1)}{2}}$ in total.
Subtracting the empty one without any segment, the total number is $2^{\frac{n(n-1)}{2}}-1$.


How do they award partial credit on USAMO? Will they award a point or 2 for an interesting observation that can possibly lead to a solution?

For example, here is 2003 #4:

ABC is a triangle. A circle through A and B meets the sides AC, BC at D, E respectively. The lines AB and DE meet at F. The lines BD and CF meet at M. Show that M is the midpoint of CF iff MB·MD = MC^2.

Suppose you just write:

"Since D is the intersection of 3 concurrent cevians inside BFC, by Ceva's Theorem,
(BE/EC)(MC/MF)(AF/BA) = 1. If MC=MF, then (BE/EC) = (MF/MC). Therefore AE is parallel to MC.  "

Will you receive a point or 2 for this(generally)?

Maybe I misread the problem or something, but it doesn't seem that what you said is true.  But just assuming that it is, the answer to your question as far as I know is that it depends.  If this fact happens to help in one of the most straightforward proofs that the AMC folks know about, then you will probably get 1 point.  It is possible that you would get 2 points.  However, the AMC folks are generally very stingy about giving points! (meant jokingly ;) )  Then again, I wouldn't know for sure.  But this is what I've heard and to a limited extent seen.

Often in an iff statement, one of the directions is pretty trivial.  Definitely knock it off if you can.  You should at least get a point, which for 9th graders may be enough to qualify for MOP.  Similarly, on problems in which you are supposed to prove something for all integers, do a couple cases - at the very least the base case for an induction.  You might pick up a point there, too.  Better yet, you might come up with an idea that solves the problem.

[quote="probability1.01"]  If this fact happens to help in one of the most straightforward proofs that the AMC folks know about, then you will probably get 1 point.[/quote]
BTW I meant BFC, not ABC in the previous post(I hope I don't do something like that on USAMO).

That fact does lead to a solution.

Continue: angle AEB =BCF = BDA, this implies triangle MCD similar to MBC. MD/MC = MC/MB, or MD*MB=MC^2. The other way is trivial(the same thing going backward).

[quote="paladin8"]Too bad some USAMO questions are too hard to find out even where to BEGIN... at least for me.[/quote]

Many times, the place to begin is not necessarily the beginning. I've found that solutions can take shape when you start in the middle and then work out from both sides. And of course, working from the back forward is a very common strategy when you're stuck.

[quote="filletwho"]What do you mean by middle?

That observation seems pretty simple, if it did get a point, wouldn't there be a lot less zeros?

If all else fails, write an angry letter to the grader complaining about bad plumbing at school. Or prove the grader is good.[/quote]

There aren't a lot of 0's in the first place, last year there were only 22 people who got a 0(out of 264 qualifiers). I bet a couple of these for some reason did get to take the USAMO, and some just gave up after 1 hour.

[quote="white_horse_king88"]I guess even a low positive score won't guarantee 9th grade MOSP then.[/quote]

Well only 22 ppl may have gotten 0, but the 0-19% range was 0-1 score and 20-46% was 2-3. If you get a low positive score you should easily qualify for 9th grade MOP.

[quote="paladin8"]Yes, I'm familiar with working backwards; it is indeed very useful at times. But from the middle? Interesting idea. I'll remember that.[/quote]

Yea...if you think of something you think may be one part of the proof...or a step somewhere in the middle...run with it. You don't know where you might get. And even if you don't solve it, 1 to 2 points is almost a given if you make some progress.

This is certainly not new advice, but it is entirely true and I feel compelled to share since we are discussing geometry: it helps to draw a large, precise diagram. I cannot tell you how often this immediately leads to divine inspiration, good conjectures, and yes, even complete solutions. If two lines look parallel, it's quite likely that they are. Ditto for cyclic quadrilaterals, concurrent lines, etc.

[quote="xxreddevilzxx"]but how do i go about from there is abig question. I am weak in Geometry.[/quote]

That's where practice and divine inspiration play a huge role  ;)

[quote="zabelman"]That's where practice and divine inspiration play a huge role  ;)[/quote]

Easy for someone with potential of making Winner to say. :P

reddevil, sometimes you just gotta get lucky.


trigonometry
inequalities unsolved
inequalities

I can help
Find min of:

P=$ Sin^2 A\minus{}Sin^2 B\plus{} Sin^2 C$

a,b,c: the sides of a triangle ABC

[/quote][/hide]

[quote="nhat101"]I can help
Find min of:

P=$ Sin^2 A \minus{} Sin^2 B \plus{} Sin^2 C$

a,b,c: the sides of a triangle ABC

[/quote]

$ P\equal{}\sin^2 A \minus{} \sin^2 B \plus{}\sin^2 C$
$ \equal{}\frac{1}{4R^2}(a^2\minus{}b^2\plus{}c^2)$ by sine formula
$ \equal{}\frac{1}{4R^2}(2ac \cos B)$ by cosine formula
$ \equal{}2\sin A \sin C \cos B$

As $ \sin A, \sin C>0$, to attain min. value of $ P$ , $ cos B$ should be made negative.

$ 2\sin A \sin C \cos B\equal{}[\cos(A\minus{}C)\minus{}\cos(A\plus{}C)] cos B$, so for fixed $ B$, i.e. fixed $ (A\plus{}C)$,

min. value of $ 2\sin A \sin C \cos B$ attains when $ A\equal{}C$

Now it remains to find min. value of $ (1\minus{}\cos 2A)(\minus{}\cos 2A)\equal{}\cos^2 2A\minus{}\cos 2A\equal{}(\cos 2A\minus{}\frac{1}{2})^2\minus{}\frac{1}{4}$

So min. value of $ P$ is $ \minus{}\frac{1}{4}$ which occurs when $ (A,B,C)\equal{}(\frac{\pi}{6},\frac{2 \pi}{3},\frac{\pi}{6})$


function
LaTeX
algebra
functional equation
algebra unsolved

[quote="hitech"]find f: R-> R:
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/993290998c538a37c4418be9e0b8608a.gif[/img][/quote]

i believe the first term is seperate from the 2nd on RHS not under same bracket as your latex shows.

No trick seems to be working!?!?!?

I'm sorry!
this is true problem!!
please help me!

find the function f:
f: R -> R :
f(x^5) - f(y^5) = (x^2 + y^2) [ f(x^3) - f(y^3) ] + x^2*y^2*[ f(x) - f(y)] with x,y is real

I tried all the satndard tricks and nothing seems to work Any other ideas .

By the way,
The correct eqn is...

$ f(x^5) - f(y^5) = (x^2 + y^2) [ f(x^3) - f(y^3) ] + x^2 y^2 [ f(x) - f(y)]  $

can you stop spamming Rushil? your two posts are unnecessary. first you ask someone to $\LaTeX$ then you do the latexing yourself? that doesn't make any sense! i see SPAMS, i checked on your profile and looked at your other posts and im assured that a pattern is forming. a little control please.

[quote="Rushil"]I am NOT spamming. I wanted to see the authors latex and found it to be wrong. that's whi I have given my version

Stop being after me all the time

And don't consider THIS to be SPAM[/quote]
there is no reason to change from $(\,)$ into this $[\,]$. if everyone in this forum posts their own version of the problem, we need to read pages after pages of different versions before seeing a solution? just get the idea and solve the problem. if you can't solve it, wait for a solution. if there is no solution for a long time, like a month or so, ask again.

i don't think you understand the issue here. i can stop chasing after you and that's even worse because your case will be passed to the admins for consideration if you don't improve. use common sense when you post, and that's all i ask or else... i don't want to see the  :furious: face of Valentin or another admin. and this concludes our discussion of SPAMS, PM me if you got a problem with it so i can contact admins to arbitrate about this. DO NOT REPLY to this message, just PM me. only post here if you solved problem or asking for a solution. THANK YOU

What's going on here? :? Guys behave :police: Both of you.

Rushil: he is right up to some point. You did post some useless posts around here, so if you could reduce their numbers that would be great, because I don't particularly enjoy doing :cleaning:

Palytoxin: next time you see spamming, either PM the user that is doing it, a moderator and finally me or another admin if nothing is solved. Don't try to solve the problem like this because most of the times you will fail  ;)

Now everybody, let's get back to the function. Please no more offtopic in this topic!

1) I'm still a bit confused about this for the author himself ave the 2 versions.....

2) Is an asuumption for continuity etc. necsaary. I'm not saying I have such a solution.. Just asking!!!

well since this topic is still alive, I think I defined g(x) = f(x) - f(0), then h(x) = g(x)/x (I'm trying to prove h is constant, of course); and I reduced the functional equation to the relation h(x) = h(x^3) = h(x^5), but that's all I could get out of the functional equation (and that relation doesn't seem enough to conclude h is constant).

Great manipulation.. Let's see if this gives something!!!

[quote="enescu"]I have two questions:
1. Is the last paranthesis $(f(x)-f(y))$ or $(f(y)-f(x))$?  Both variants appeared  in this topic and in one case linear functions do the job. Also, there was another post (deleted, I assume) with the same problem, with $x$ and $y$ reversed in the end. (By the way, both posts were from Vietnam and both had "help me!" in the subject line ;))
2. Are you sure there is no aditional condition imposed to $f$? (such as monotony or continuity?)[/quote]

I can answer your questions that:
 1/ the function true is  f(y) - f(x)
 2/Yes I sure that !

The problem is easy but so difficul!
I wish another people sovled it ! thank you


An equilateral triangle with side $n$ is built with $n^{2}$ [i]plates[/i] - equilateral triangles with side $1$. Each plate has one side black, and the other side white. We name [i]the move[/i] the following operation: we choose a plate $P$, which has common sides with at least two plates, whose visible side is the same color as the visible side of $P$. Then, we turn over plate $P$.

For any $n\geq 2$ decide whether there exists an innitial configuration of plates permitting for an infinite sequence of moves.

I don't understand thich you post now
Please post clearly than this
Sorry formy bad English

For the triangle ,we consider 1 if it is black and -1 if it is white
$S_{k}=\sum a_{i}a_{j}$ with $a_{i},a_{j}$ is the number in two triangle which have common side
We have $S_{k+1}>S_{k}$ and $S_{k}<+\infty$

We name the plate which the problem says it's turnable $ good $.
We define $ degree $ of the plate $ P $: the number of its neighbors whose visible color is as the same as the visible color $ P $ ($ D_P $) . And let $ S $ be the sum of all plates' degrees.
It's obvious that $ S\geq0. $ Let $ P $ be a good plate that is not on the sides of the main triangle. There's 2 possible conditions:
$ i) D_P=2 $ or
$ ii)D_P=3 $
In the first condition, when we turn $ P $, its degree wanes $1$ unit and the degree of each of its homochromatic neighbors wanes $1$ unit and the degree of its non-homochromatic neighbor increases $1$ unit. So, $ S $ wanes $2$ units.
In the second condition, when we turn $ P $, its degree wanes $3$ units and the degree of each of its neighbors wanes $1$ unit. So, $ S $ wanes $6$ units.
Now, consider the plates on the sides of the main triangle. The head triangles won't move; because they have only $1$ neighbor.
Assume $ Q $ a good plate on a side of the main triangle. So, it's $2$ neighbors are homochromatic with $ Q $. When we turn $ Q $, its degree wanes $2$ units and the degree of each of its neighbors wanes $1$ unit. So, $ S $ wanes $4$ units.
[b][size=150]So[/size][/b], after each movement, $ S $ wanes. So, in order to $ S\geq0 $ and $ S $ is finite, we get that the number of possible movements is finite.
So, there doesn't exist any initial configuration of plates permitting for an infinite sequence of moves.


trigonometry
function
calculus
calculus computations

Find the incident angle for an air-glass interface such that the transmitted angle is $ \frac {1}{2}$ of the incident angle.

I have trouble manipulating trig functions, is there an identity that will help me here?

$ n_{air} \equal{} 1$ and $ n_{glass} \equal{} 1.7$

so:

$ \sin(\theta) \equal{} 1.7\sin(\frac {1}{2}\theta)$

Can someone point me in the right direction?

Thanks

You have:
 $ \sin \theta \equal{} \sin \left(2\cdot \frac{\theta}{2}\right) \equal{} 2\sin \frac{\theta}{2} \cos \frac{\theta}{2}$.
Hence,
$ \sin(\theta) \equal{} 1.7\sin(\frac{1}{2}\theta)  \implies 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} \equal{} 1.7\sin(\frac{1}{2}\theta) \implies \theta \equal{} 2 \cos^{\minus{}1}(\frac{1.7}{2})$


Does anyone know how to attach an entire folder?  I have several documents stored in a single folder and I don't want to attach each individual document, rather the entire folder.  Is this possible?

You can like zip it and attach teh zip file


calculus
geometry
trigonometry
analytic geometry

I am done with high school (i don't officially graduate until tuesday but there are no more school days for seniors) and have been accepted into California Polytechnic University of Pomona as an Aerospace Engineering major, and i couldn't be more excited. All incoming freshman looking to take MAT 114 (Calculus with analytic geometry 1) are required to take the Pre-Calculus version of the Mathematics Diagnostic Placement Test (MDPT). There is also a "Math analysis" version of this test but that is for a different, non-standard math track, and from what i know, is a significantly easier test. Anyways, most people take the MDPT in July before fall term, but all science and engineering majors had to take it last month, which i did. Now, i passed the test, but not by much. I got 46/60 (6/10 on trigonometry questions and 40/50 on algebra questions). Now the cutoff scores for getting into MAT 114 without having to take prerequisite courses are either 1 of the following 2 combinations:

At least 37/50 on algebra with at least 5/10 on trig

OR

At least 30/50 on algebra with at least 8/10 on trig

So as you can see, i qualify under the first set of conditions, but just by a hair. This worries me because i passed my high school precalculus course (entitled "Trigonometry/Math analysis") with a 99.2% and 1st semester i got a 98 point something percent. I owned that class. Does that mean my high school class underprepared me or does it mean that the test is purposely hard. I noticed that the scores that are considered passing are unusually low. I mean it's possible to get 38/60 (as long as you meet the second set of conditions) and still be considered to be ready to take MAT 114 right away. Well anyways, does this mean i might experience some difficulties in calculus (even though i completely owned both Algebra 2 and Precalculus) or i am just overreacting? Thanks for your input.

A lot of high school math courses underprepare students for college math. That's one reason this forum includes a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]FAQ on College Math Study[/url]. Colleges are used to getting underprepared high school graduates up to speed, but the new college student who has already gotten himself up to speed by self-study usually enjoys college better and has more academic success.

The ALEKS online courses

http://www.aleks.com

include materials for a thorough review of precalculus mathematics, including trigonometry. You can try the forty-eight-hour free trials on that site an unlimited number of times, and you may find it worth your time and money to sign up for the precalculus with trigonometry course there as a review. Reading up about math in books you can get from your local library or other nearby libraries would also be helpful. Another site with some good lecture notes on various mathematical topics you can download for free is

http://math.berkeley.edu/~wu/

Good luck. Enjoy your studies.

We're starting to use the MDPT here at CSULB (which is in the same system as CPP). I have a little bit of data about how our students did on our Calculus I after taking this exam last year. The correlation with eventual grades isn't particularly strong, and it was quite possible to be "unprepared" on the placement test and still do well in the course.

CPP is using a slightly different version of the test (with an exanded number of trig questions) but I think I can do a translation. If you were at my institution, I wouln't hesitate to recommend that you take the "regular" Calculus I course. I'm also going to guess that the number of students who qualify under the second criterion (high trig score, lower algebra score) but not under the first criterion is exceedingly small - perhaps an empty set. I also wouldn't call your qualification "marginal," since you have a fair margin on the algebra side.

So: consider yourself prepared to take Calculus I. But that preparation is not everthing. Be prepared for the fact that college calculus courses move much faster than similar high school courses. And even if we somehow devised the perfect placement test and assured ourselves that everyone on the class was fully prepared, there would still be students who fail or drop the course. It takes a lot of work, especially a lot of work outside of class.

Thanks for your replies. They are greatly appreciated. I am not too worried about the outside effort i will have to put it because first off, i have exposed myself to most of the material on the AP Calculus AB syllabus (which is equivalent to MAT 114). I obviously haven't fully grasped everything but i think i have a good head start. Secondly, this past year i took AP Physics and AP Statistics. These were my first and only AP courses throughout high school. I had to work alot harder in these classes than in trig, or any of my other classes. So i have a feel for what college will be like. I have a very strong work ethic when it comes to mathematics because i enjoy the subject and find it very interesting. If i don't get a problem on the homework, i'll keep trying it until i get it right instead of giving up on it and asking about it the following day. I have never asked a question in math since geometry (sophmore year). I am also a nerd in a way because when i am bored and have nothing else to do, i'll try to teach myself material that hasn't been covered in class yet, just to get the main idea down, even if i don't fully grasp it yet. I have found by doing this, i instantly master the concept when it is formally covered in class. This strategy has helped me alot. I struggled to get B's in Algebra 1 and Geometry, but then i suddenly gained interest in math out of nowhere and adopted this method of learning math, and i got A's in Algebra and Trig without any troubles. I got B's in AP Statistics, but i got lazy in that class :p. Thankfully i will never have to take another statistics class again lol.

Well, i have a good work ethic in math. It's all the other classes i'll have to stop being lazy in XD

[quote="ROLEX4life"]I have never asked a question in math since geometry (sophmore year).[/quote]

This is a habit you are almost certainly going to want to break.

Yeah well that isn't really a habit. I mean i ask questions in my other classes whenever i need to. It's just that i haven't ever needed to ask a question in math for the last two years. And anyways i asked questions in physics and statistics, so it's not really a "habit" i have gotten into.


trigonometry
vector
real analysis
real analysis unsolved

[color=blue]Assume $ x,y,z \geq 0$, and $ x\plus{}y\plus{}z\equal{}\pi$

Find the maximum and minimum of  $ 2\cos x\plus{}3\cos y\plus{}4\cos z$[/color]

Let ABC triangle with $ \angle A \equal{} \alpha,\angle B \equal{} \beta, \angle C \equal{} \gamma$. And let vectors $ e_1,e_2,e_3$  colinear to CA,AB,BC respt.
$ |e_1| \equal{} \sqrt {\frac83},|e_2| \equal{} \sqrt {\frac32},|e_3| \equal{} \sqrt {6}$. Then
$ (e_1 \plus{} e_2 \plus{} e_3)^2\ge 0$ , $ \frac83 \plus{} \frac32 \plus{} 6 \plus{} 2(2\cos \alpha \plus{} 3\cos\beta \plus{} 4\cos\gamma) \ge 0$
Thus $ 2\cos \alpha \plus{} 3\cos\beta \plus{} 4\cos\gamma\ge \minus{} \frac {61}{12}$.

Now, we can construct triangele with sides $ \sqrt {\frac83},\sqrt {\frac32},\sqrt {6}$. For this triangle we obtain $ e_1 \plus{} e_2 \plus{} e_3 \equal{} 0$, so for angles of this triangle we have
$ 2\cos \alpha \plus{} 3\cos\beta \plus{} 4\cos\gamma \equal{} \minus{} \frac {61}{12}$.


p>0 fixed

$ s(n)\equal{}\sum_{k\equal{}0}^{n}( \frac{n!}{(k!)(n\minus{}k)!})^{p}$

find an equivalent of s(n) when n tends +oo

sure this is from Saint petersburg  :)

generalization:
lets' consider:
$ q>0$ and $ p\geq 0$
$ S_{n} \equal{} \sum _{1\leq k < \frac {n}{2}}(C_{n}^{k})^{q}|2k \minus{} n|^{p}$
 show that
$ S_{n}$~$ 2^{nq}n^{\frac {p \minus{} q \plus{} 1}{2}}\Gamma (\frac {p \plus{} 1}{2})\sqrt {\frac {2^{p \plus{} q \minus{} 3}}{\pi ^{q}q^{p \plus{} 1}}}$
if nobody post solution in this week then I will post full solution.
to Moubinool :
show that your problem is particular case of this generalization.