Given the triangle $ ABC$ with side $ a$ opposite $ \angle{A}$, side $ b$ opposite $ \angle{B}$, and side $ c$ opposite $ \angle{C}$, find $ \sin{A}\plus{}\sin{2B}\plus{}\sin{3C}$ if $ a\equal{}3,b\equal{}4,$ and $ c\equal{}5$.

I don't get how to find $ \sin{3C}$.

Thanks in advance!

[quote="The Zuton Force"]What kind of triangle is ABC? What's the measure of angle C, in degrees?[/quote]

$ ABC$ is a right triangle.  I'm trying to do the problem without calculating the actual angle measures.  [hide="The solution says"]$ \sin{A}=a/c=3/5 \ \checkmark$ - I get this.

$ \sin{2B}=2\sin{B}\cos{B}=2(4/5)(3/5)=24/25 \ \checkmark$ - I get this.

${ \sin{3C}=\sin{3\pi/2}=-1}$ - I don't get this.[/hide]

Well, you know C is a right angle, so its measure is 90 degrees. You want the sine of 270 degrees.

[url=http://en.wikipedia.org/wiki/Unit_circle#Trigonometric_functions_on_the_unit_circle]Refer to this link[/url] to see how to extend the trigonometric functions to contexts beyond right triangles.


These questions are all the same please stop posting unneccessary topics, you could have put these in 1 post as they concern the same topic.

I agree with riddler- the topics you have just posted are very similar, so you should combine them.

symmetry, I believe many people (including me) have given you some advice on posting multiple topics, learning LaTeX, etc. (I don't want to mention all of them). Since this place is not only for you but for many people, we all should follow some rules. :)

No offence, but as far as I can see, you seem to be posting multiple topics with different levels of difficulty.  A good way to practice problem solving/math is to organize what you are studying; from least difficult to most difficult.

Masoud Zargar

[quote="symmetry"]Convert questions 1 and 2 from degrees to radians.  Express your answer in decimal form, rounded to TWO decimal places.

1) 73 degrees =

2) 350 degrees =[/quote]

this seems suspiciously like homework

look on mathworld

when you have, i will unlock the 3 topics you posted and do please combine them all in one topic

[size=200][color=red]LOCKED!!!![/color][/size]

Symmetry,

As some users have pointed out, your posting looks a bit like homework.

Finding AoPS could be very valuable to you if you take a slightly different approach.  If you're looking for a bit of homework help, it's probably best to post that up front ("Could somebody help be understand degree/radian conversions?").  If you focus on the typically generous explanations of AoPSers, perhaps you will learn how to solve the problems on your own.  If not, this isn't the forum for you.

To other students:

While it's good to recognize and point out when a student seems not to be using the forums for its intended purpose (learning to understand mathematics and problem solving), try to do your best to simply convince that student.  If all else fails, you can contact an admin and we can disable the ability of that user to post.

Remember that most students have never really been encouraged to think deeply in order to learn the mechanics of mathematics.  When you see a student posting what appears to be their homework, try to act as a guide encouraging them to see the thinking side of mathematics.  Perhaps one such student will catch on and then bring good ideas to their school.  Encouraging a positive culture of thinkers is a great deal of our role as a community.

P.S. I have unlocked this post for the sake of discussion, but if the problem continues, I will allow the moderators to lock threads.

[quote="MCrawford"]Symmetry,

As some users have pointed out, your posting looks a bit like homework.

Finding AoPS could be very valuable to you if you take a slightly different approach.  If you're looking for a bit of homework help, it's probably best to post that up front ("Could somebody help be understand degree/radian conversions?").  If you focus on the typically generous explanations of AoPSers, perhaps you will learn how to solve the problems on your own.  If not, this isn't the forum for you.[/quote]

Yes, that's true.  However, it seems that he says he's not in high school and that he's practicing for a math contest.  Anyway, I believe, you, symmetry, should practice problem solving/mathematics in a more organized fashion as oppose to trying to solve problems at very different levels of difficulty.

Masoud Zargar

OK symmetry. So this stuff is just really basic, you need that to solve trig problems right?

$ 1degree=\frac{\pi}{180}radian $

$ 1radian=\frac{180}{\pi}degree $

Hope it helps.

Yoosam,

yes, this is what I was looking for.  I can take it from here.


conics
parabola
geometry
trapezoid
geometry proposed

Let $ F$ and $ V$ denote the focus and vertex, respectively, of the parabola $ x^2 \equal{} 4py$. If $ PQ$ is a focal chord of the parabola, show that:

$ PF \cdot FQ \equal{} VF \cdot PQ$

Let $T$ be the intersection of the focal axis $FV$ with the directrix $\tau$ and let $M,N$ be the projections of $P,Q$ onto $\tau.$ In the right trapezoid $PQNM,$ we have:

$FT = \frac {PM \cdot FQ + QN \cdot FP}{PQ}.$

But since $FT = 2 \cdot VF$ and $PM=FP, \ QN=FQ,$ it follows that

$2 \cdot VF = \frac{PF \cdot FQ + PF \cdot FQ}{PQ}  \Longrightarrow \ PF \cdot FQ = VF \cdot PQ.$

Since the parabola is of the form $ x^2\equal{}4py$, the focus F is at $ (0,p)$, vertex V $ (0,0)$, directrix eqn is $ y\equal{}\minus{}p$. Then P, Q should be $ (\minus{}2p,p),(2p,p)$ since $ PQ$ is the focal chord (parallel to the directrix and passing through the focus), and $ PF\cdot FQ\equal{}2p\cdot2p\equal{}4p^2$ while $ VF\cdot PQ\equal{}p\cdot4p\equal{}4p^2$...

Oh I was informed that my proof is wrong, what's the definition of focal chord then? (I researched that before and it returned the previous understanding, but maybe I misread).


superior algebra
superior algebra unsolved

is there any elementary solution to equation $ x^5\equal{}x$ in an arbitary ring?

Besides [b]x=0[/b]?

Also, $ x \equal{} \pm 1$ does the trick if you happen to take all of your rings as rings with 1.

I dont think so, at least unless you consider $ x\equal{}0$. Since $ x\equal{}1$ does work but not every ring (as it is arbitrary) has a multiplicative identity :\

I don't understand what the word "elementary" means here.


probability
number theory
relatively prime

Alfred and Bonnie play a game in which they take turns tossing a fair coin.  The winner of a game is the first person to obtain a head.  Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game.  Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n$, where $m$ and $n$ are relatively prime positive integers.  What are the last three digits of $m + n$?

[quote="shyong"]If both person tossed tail in first attempt , who is the loser ? And who goes first in second round ?[/quote]
I think in each game, one goes first and they take turns and keep going until one gets head, then they go to the next game

For each round, the probability that the starter will win is 2/3, the probability that 2nd person will win is 1/3. You go from there to figure out the probabilities for all the cases. It's long, and i am lazy, so i won't figure it out.

i worked out this problem, and it is easy to show that after each game the probability is a fraction with denominator as a power of three and the fraction is as close as possible (integrally) to one half. it fluctuates between games from just over one half to just under.

[hide]
The chance of the first person winning a game that he starts is equal to $\frac{1}{2} + (\frac{1}{2})^3 + (\frac{1}{2})^5+ ... = \frac{2}{3}$. The chance of the second person winning is then $\frac{1}{3}$

Probability of Alfred winning on...

Game 1
$\frac{2}{3}$

Game 2
Either WW, or LW
$\frac{2}{3}\times \frac{1}{3} + \frac{1}{3} \times 2{3} =\frac{4}{9}$

Game 3
WWW
LWW
WLW
LLW

Probability: $\frac{14}{27}$

Game 4
WWWW
LWWW
WLWW
WWLW
LLWW
LWLW
WLLW
LLLW

Probability is then equal to $\frac{41}{81}$

Looking at the probabilities for the first four games $\displaystyle \left(\frac{2}{3}, \frac{4}{9}, \frac{14}{27}, \frac{40}{81} \right)$

The $nth$ term is formed by subtracting $(-1)^{n+1}\frac{2}{3^n}$ from the previous term, where $n$ is the current game.

So using this pattern, the probability of Alfred winning the 6th game is equal to $\frac{364}{729}$
[/hide]

[hide]
let the probability of winning the nth game be $a_n$.

As discussed earlier, the first person to go has a 2/3 chance of winning in any given game, and the second person has a 1/3.  So, the probability that someone will win the n+1th game is the probability they won the last game times 1/3 (since they're going 2nd in the next game) plus the probability they lost times 2/3.

$a_n = \frac{1}{3}a_{n-1} + \frac{2}{3}(1-a_{n-1})$

we know that since Alfred goes first, $a_0 = 0$ and we need to find $a_6$.  It's easy to compute that this is 364/729.  Since $364 \equiv 1 \mod 3$, this will not reduce, and our answer is the last 3 digits of 364 + 729, which is 093.

EDIT: by the way, $a_2 = 4/9, a_3 = 14/27, a_4 = 40/81, a_5=122/243, a_6 = 364/729$.  It looks like the denominators are powers of 3, and the numerator is the floor or ceiling of the denominator divided by 2.

EDIT 2: Fixed typo on the subscripts
[/hide]

Technically, to answer the question,
[hide]364+729=1093, and the last three digits=093[/hide]


Is there a special property for point in the figure:

angles APB, APC, and BPC are congruent.

Yes, this is called the [url=http://en.wikipedia.org/wiki/Fermat_point]Fermat Point[/url] of the triangle.

[i](read the second entry under the "Properties" section in the linked article)[/i]

[quote="lingomaniac88"]Here's another question:

Find a way to construct it.[/quote]
[hide="Construction"]
Construct $ M$, $ N$, and $ Q$ such that $ \triangle ABM$, $ \triangle BCN$, and $ \triangle ACQ$ are equilateral.  The Fermat Point, $ F$, is the intersection between the circumcircles of these equilateral triangles.  [/hide]


algebra
polynomial
quadratics
algebra proposed

The sum of the three discriminants is $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq 0$, so one at least is $\geq 0$ and one at least of the three equations has a real root.

--
Patrick

Let the polynomials be $f_1(x),f_2(x),f_3(x)$ respectively.Now see that $f_1(0)+f_2(0)+f_3(0)=0$ since being a quadratic $f_i(x)\to +\infty$ as $x\to \infty$. And we see that at least one of $f_i(0)\le 0$. If equality holds then we have a root at $0$ otherwise a root in $(0,\infty)$. I know this is not much of a unique solution but it gives that we can ensure the existence of a non-negative root. I hope this one is correct.

Say none of the equation has a real root then , $\sum (a-b)^2<4\sum (b-c)<0$ contradiction.

[quote=N.T.TUAN]Given reals numbers $a$, $b$, $c$. Prove that at least one of three equations $x^{2}+(a-b)x+(b-c)=0$, $x^{2}+(b-c)x+(c-a)=0$, $x^{2}+(c-a)x+(a-b)=0$  has a real root.
[i]O. Podlipsky[/i][/quote]

Let $f_{i}(x)$ be the discriminant of i-th equation.
$f_{0}(x) = (a - b) ^ 2 - 4(b - c)$.
$f_{1}(x) = (b - c) ^ 2 - 4(c - a)$.
$f_{2}(x) = (c - a) ^ 2 - 4(a - b)$.
Suppose none of the equation has a root $\implies$ $f_{i}(x) < 0$ $\implies$ $\sum_{i=0}^{2} f_{i}(x) < 0$ and we get $\sum (a - b) ^ 2 < 0$,contradiction.

[quote=godfjock]Wlog take $a\leq b \leq c$ And let $A,B,C$ be the three equations
If $c\ge 0$ Then $A$ has a real solution if $c< 0$ then $B$ has a real solution[/quote]

A minor doubt,  Can we assume $a\leq b \leq c$ here , as the equations are cyclic, not symmetric.

Sorry If I am being dumb here.

[quote=N.T.TUAN]Given reals numbers $a$, $b$, $c$. Prove that at least one of three equations $x^{2}+(a-b)x+(b-c)=0$, $x^{2}+(b-c)x+(c-a)=0$, $x^{2}+(c-a)x+(a-b)=0$  has a real root.
[i]O. Podlipsky[/i][/quote]
Suppose all the equations have non-real roots,then then we must have $$\begin{aligned}(a-b)^2 -4(b-c) < 0 \\ (b-c)^2 -4(c-a) <0 \\ (c-a)^2 -4(a-b) <0 \end{aligned}$$ Adding these we get  $(a-b)^2 +(b-c)^2 +(c-a)^2 <0$ which is a contradiction as $a,b,c \in \mathbb{R}$

If all the equations have non real roots,

$$\begin{aligned}(a-b)^2 < 4(b-c) \Rightarrow b > c \\ (b-c)^2 < 4(c-a) \Rightarrow c > a \\ (c-a)^2 < 4(a-b) \Rightarrow a > b \end{aligned}$$
 Which is a contradiction.


MATHCOUNTS
geometry
perimeter
Pythagorean Theorem

I need to know how to do this problem, not the answer.

[b][i]An isosceles right triangle is inscribed in a circle of radius 4 inches.  What is the perimeter of the triangle, expressed in inches as a decimal to the nearest tenth.
[/i][/b]

Is there a formula or something that relates the triangle to the radius/diameter/perimeter of the circle?

PLEASE HELP!!!

[hide]Well, since the triangle is isoceles and right, the hypotenuse is the diameter of the circle, which is 8. Use pythagorean theorem c^2=2a^2 (only works for isoceles right triangles) and get the side lengths. Add together and get the perim.[/hide]


inequalities
trigonometry
calculus
inequalities proposed

[quote="silouan"]I think it can be solved by
http://www.mathlinks.ro/Forum/viewtopic.php?t=118306&sid=669bf46efeacb34ef9125f278389e2a6[/quote]

How?

I try to solve this problem with your advice, but I can't solve this problem.

Show me how to solve this problem with your advice.  :)

[quote="silouan"]I think it can be solved by
http://www.mathlinks.ro/Forum/viewtopic.php?t=118306&sid=669bf46efeacb34ef9125f278389e2a6[/quote]
I don't think so. arqady's ineq is very different from this ineq

[quote="zaizai-hoang"][url]http://www.mathlinks.ro/Forum/viewtopic.php?p=672068#672068[/url]
My problem have left expression the same with your problem  :)[/quote]
No it's not the same! :P
Unfortunately your links do not provide a soultion to the problem so try more :wink:

yes, my problem isn't true  :D But I found a new inequality reference with above inequality. I sure nonce it is true  :wink:

any solution :D Albanian Eagle, Can you talk about your inequality?  :D

I can wait  :lol:

A similar problem i created sometime ago (not homogenous but maybe easier ) :
find the best constant $k$ so that $(a^{5}+\frac{1}{a})(b^{5}+\frac{1}{b})(c^{5}+\frac{1}{c})\geq k\sqrt{(a+b)(b+c)(c+a)}$ for all real a,b,c

[quote="me@home"]Please tell me, is there any nicer way than derviatives?[/quote]
What do you mean ? :)

[quote="toanhocmuonmau"][quote="Albanian Eagle"]Let $a,b,c$ be positive reals . Show that :
$3\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}\geq 4\sqrt[4]{ab+bc+ca}$[/quote]
Setting $a=\tan{A},b=\tan{B},c=\tan{C}(0<A,B,C<\frac{\pi}{2})$. Then, by some computations you will obtain the result. :)[/quote]
interesting but I don't think it makes it easier...
can you show how do you proceed ? my solution doesn't use trigonometric substitution.

Since there is no solution till now i decided to write a (major) hint then I will wait for comments if any.
$\frac{(x^{2}+1)(y^{2}+1)}{x+y}\geq \frac{8\sqrt{3}}{9}$

[quote="Albanian Eagle"]
$\frac{(x^{2}+1)(y^{2}+1)}{x+y}\geq \frac{8\sqrt{3}}{9}$[/quote]
And using the Carlson's inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27766
Beautiful inequality, Albanian Eagle!

The following particular case is very nice:

If $a,b,c$ are positive numbers such that $ab+bc+ca=1$, then
$(a^{2}+1)(b^{2}+1)(c^{2}+1) \geq \frac 4{3}$.

Find someone a particular solution to this case?

For the special case we can dierctly apply carlson because $a^{2}+1 = (a+b)(a+c)$

You are right, Soarer.
Moreover, the following more general statement is true.

[i]If $a,b,c$ are positive numbers such that $ab+bc+ca=3p$, then for any [/i]$0<p\leq 1$,

$(1+a^{2})(1+b^{2})(1+c^{2}) \geq (1+p)^{3}$.

[quote="silouan"][quote="Albanian Eagle"]
$\frac{(x^{2}+1)(y^{2}+1)}{x+y}\geq \frac{8\sqrt{3}}{9}$[/quote]

Albanian Eagle what is your solution for this ? I solved it in an ungly way with calculus .Or anyone else has a nice solution to this ?[/quote]
$(p^{2}+3)(q^{2}+3)=(p^{2}+1+1+1)(1+q^{2}+1+1)\geq(p+q+1+1)^{2}\geq8(p+q)$
and let $p=\sqrt3x,$ $q=\sqrt3y.$ :wink:

[quote="Vasc"]
[i]If $a,b,c$ are positive numbers such that $ab+bc+ca=3p$, then for any [/i]$0<p\leq 1$,

$(1+a^{2})(1+b^{2})(1+c^{2}) \geq (1+p)^{3}$.[/quote]
I think, it's true for all $0<p\leq3.$
In the case $p=3$ it's a very interesting inequality. :wink:

[quote="arqady"][quote="Vasc"]
[i]If $a,b,c$ are positive numbers such that $ab+bc+ca=3p$, then for any [/i]$0<p\leq 3$,

$(1+a^{2})(1+b^{2})(1+c^{2}) \geq (1+p)^{3}$.[/quote]
I think, it's true for all $0<p\leq3.$
In the case $p=3$ it's a very interesting inequality. :wink:[/quote]
We have
\[(1+a^{2})(1+b^{2})(1+c^{2}) =(a+b+c-abc)^{2}+(ab+bc+ca-1)^{2}=(a+b+c-abc)^{2}+(3p-1)^{2}\]
Otherwise, we have
\[a+b+c-abc \ge 3\sqrt{p}-p\sqrt{p}\]
We are done. :)


i realize that my ineptitude at writing proofs is my largest weakness, now that my propensity for careless errors has lessened considerably.  i certainly wish to make the usamo next year, and perhaps do marginally well on it.

so the question: where can i find resources regarding proper proof-writing techniques or examples of well-written proofs on the net?

Right here [url]http://artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php[/url] is one place

I would also recommend the USAMTS and any similar contests you can find for proof-writing practice. Having your work graded is a good way to develop the skill.


algebra
polynomial
calculus
calculus computations

You have numerical coefficients; that means you want a numerical method. There is no general formula for solving a sixth-degree equation, which is another reason to go numerical.

Those coefficients would be more readable in scientific notation.

Newton's method is a very effective way to approximate roots of polynomials.


Find all solution if any to the following:

$ q_1,q_2,p$ be primes

$ q_1q_2\equiv 1 \mod p$
$ q_1 \plus{} q_2 \equiv 2 \mod p$

$ (q_1 \minus{} q_2)^2 \equal{} (q_1 \plus{} q_2)^2 \minus{} 4q_1q_2 \equal{} 4\minus{}4 \equal{} 0 \bmod{p}$
$ \implies q_1 \equal{} q_2 \equal{} 1 \bmod{p}$

So all solutions are characterised by pairs of primes $ q_1, q_2$ which are both of the form $ np \plus{} 1$
By Dirichlet's theorem there are infinitely many such primes for any $ p$


calculus integration geometry inequalities function logarithms trigonometry

In the $x$-$y$ plane, for the origin $ O$, given an isosceles triangle $ OAB$ with $ AO \equal{} AB$ such that $ A$ is on the first quadrant and $ B$ is on the $ x$ axis.
Denote the area by $ s$. Find the area of the common part of the traingle and the region expressed by the inequality $ xy\leq 1$ to give the area as the function of $ s$.

[quote="kunny"]In $ x \minus{} y$ palne, for the origin $ O$, given an isosceles triangle $ OAB$ with $ AO \equal{} AB$ such that $ A$ is on the first quadrant and $ B$ is on the $ x$ axis.
Denote the area by $ s$. Find the area of the common part of the traingle and the region expressed by the inequality $ xy\leq 1$ to give the area as the function of $ s$.[/quote]

not sure, but this is what I got so far

[hide]if at most one point touches the curve, then the answer is $ s$.

otherwise, let $ A(x,y)$ $ x,y>0$. then the area is $ 1 \plus{} \sinh^{\minus{}1}x \minus{} \frac{1}{2}\log\frac{x}{y}$.[/hide]

What do you mean by $ \sin h^{\minus{}1} x$?

[quote="kunny"]What do you mean by $ \sin h^{ \minus{} 1} x$?[/quote]

$ \log (x \plus{}\sqrt{x^{2}\minus{}1})$. :D

[quote="kunny"]In $ x - y$ palne, for the origin $ O$, given an isosceles triangle $ OAB$ with $ AO = AB$ such that $ A$ is on the first quadrant and $ B$ is on the $ x$ axis.
Denote the area by $ s$. Find the area of the common part of the traingle and the region expressed by the inequality $ xy\leq 1$ to give the area as the function of $ s$.[/quote]

[color=blue]
Let $ \ B$ be $ \ (2a,\ 0)$ and $ \ A$ be $ \ (a,\ b)$ where $ \ a,\ b > 0$

$ \implies$ Area of the triangle $ \ s = ab > 0$

Note: $ \ xy = 1$, represents rectangular hyperbola $ \implies xy\le1$ is the region outside the hyperbola. To be more clear I am following a convention that in case of hyperbola the region which contains centre of the hyperbola is called outside region $ ($ as centre is $ \ (0,\ 0)\implies 0\times0 < 1)$

Equation to line $ \ AB$ is $ \ bx + ay - 2ab = 0$ and equation to line $ \ OA$ is $ \ bx - ay = 0$

Consider the following cases:

[b]case 1:[/b] Line $ \ AB$ never meets the Hyperbola or just tangent to hyperbola $ \implies bx^2 - 2abx + a = 0$ will have nonreal roots or equal roots

$ \implies D\le0\implies 4ab(ab - 1)\le 0\implies s(s - 1)\le 0\implies s\in (0,\ 1]$

$ \implies$ In this case $ \ \triangle ABC$ entirely out side the Hyperbola$ \ \implies$ The area of the common part of the triangle and the region expressed by the inequality $ xy\leq 1 = s,\ s\in (0,\ 1]$

[b]case 2:[/b] Line segment $ \ AB$ intersect the Hyperbola at $ \ Q \implies bx^2 - 2abx + a = 0$ will have real and distinct roots

$ \implies D > 0\implies s(s - 1) > 0\implies s\in (1,\ \infty)\implies$ Point $ \ A$ is inside the hyperbola.

Now $ \ x$ coordinate of $ \ Q\ > x$ coordinate of $ \ A\implies x$ coordinate of $ \ Q\ = \frac {s + \sqrt {s^2 - s}}{b}$

Let $ \ P$ be the point of intersection of the hyperbola and line segment $ \ OA\implies P\equiv \frac {\sqrt s}{b}$

Now Required area $ = s - (\int_{\frac {\sqrt s}{b}}^a(\frac {bx}{a} - \frac {1}{x}) dx + \int_{a}^{\frac {s + \sqrt {s^2 - s}}{b}}(2b - \frac {bx}{a} - \frac {1}{x}) dx)$

$ = s - \sqrt {s^2 - s} - \ln(\sqrt s - \sqrt {s - 1})$

$ \boxed{\boxed{\text{Hence Required area} = \begin{cases}{s} & {0 < s\le 1} \\
{s - \sqrt {s^2 - s} - \ln(\sqrt s - \sqrt {s - 1})} & {s > 1}\end{cases}}}$
[/color]

Perfect!
Sorry for my delay reply, makar.

kunny


Show that if $\sum_{k}^{\infty}|a_{k}|$ converges, then $\sum_{k}^{\infty}a_{k}$ converges.

Compare terms: $a_{k}\leq |a_{k}|$ for all $k$. So suppose $\sum_{k=1}^\infty |a_{k}|=A$ then $\sum_{k=1}^{n}a_{k}\leq\sum_{k=1}^{n}|a_{k}|\leq \sum_{k=1}^\infty |a_{k}|=A$, so $A$ is an upper bound for the partial sums of $a_{k}$. Therefore $\sum_{k=1}^\infty a_{k}$ converges.

Erm... Not every series whose partial sums are bounded from above converges! ;)

You can use Cauchy Criteria for convergence:

$\sum_{n}|a_{n}|$ converges, then

For $\epsilon >0$ there is a natural number $N(\epsilon)$, suct that for all $n>N(\epsilon)$ and all $p\geq 1$ we have:

$|a_{n+1}|+\cdots+|a_{n+p}|=\parallel a_{n+1}|+|a_{n+2}|+\cdots+|a_{n+p}\parallel <\epsilon$
Now observe that

$|a_{n+1}+\cdots+a_{n+p}|<|a_{n+1}|+\cdots+|a_{n+p}|$ from which it follows that $\sum a_{n}$ also converges.


the following mechanism was proposed for the addition of hydrogen chloride to propene
1)two hydrogen chloride molecules combine to form (HCl)2
2)a molecule of HCl and propene combine to give a complex
3)then the di-hydogenchloride reacts with the complex to give 2-chloro-propane and two molecules of HCl

Find the rate law in terms of reating species

I'm guessing step one is the RDS which makes it prop to [HCl]^2.

sorry i missed out some key steps
alll reactions here are to be assumed elementary
the thirs step is the slowest step :)

Cha.... I have seriously no idea how the 3rd step is RDS, oh well in that case.

$ Rate \equal{} k_1 [(HCl)_2][Complex]$
$ Rate \equal{} k_1k_2 [HCl]^2 k_3[HCl][Propene]$
$ Rate \equal{} k[HCl]^3[Propene]$

where $ k \equal{} k_1k_2k_3$

4th order reaction.

answer $ \frac {k_{1} \cdot k_{3}}{k_{\minus{}1}} \cdot (\ce{HCl})^3 \cdot$ [propene]

[quote="mythili"]the answer is product of rate constants times the conc of hcl raised to third power into concentration of propene...but how does one get that?[/quote]

that's what i said  :huh:
neway its bcos u have to rate the rate eqn for the slow step and then express the intermediates in terms of the given initial reactants  :)

[quote="mythili"]i don't get u ...do assume that the reactiopns have different rate constants[/quote]

Yeah yeah, they are k1, k2 and k3... I am just keeping the constant at each stage as k just for simplicity purpose. Madness has given it clearly.

[quote="mythili"]i didn't get ur symbols...anyway can u explain that in detail please[/quote]

the assumption is that the first eqn is in eqbm with $ k_{1}$ and $ k_{\minus{}1}$ being the rate constants on the forward n backward rns  :)

but the reaction 2 is also an equilibrium reaction and the answer says that the rate doesn't depend on the backward rates....
first i think the answer is only for the induction period....steady state approx. doesn't seem to hold...so i don't think the reverse reaction shud matter u

[quote="mythili"]why is the rate term in each of ur equations the same..... :?:[/quote]
They are not the same. Its just that keeping k makes it simpler for you to understand. I've edited the post.

[quote="mythili"]i meant the rate and note the rate constant[/quote]

I swear I'm unable to comprehend whats in your mind. We know that the rate is proportional to (HCl)2 and the complex which inturn depend on HCl and Propene......

I've edited the proof, so read it again.

Well if u are doing by steady state apprximations then you have to equate the rate of change of the intermediates to 0  :)  and then substitute the conc. of the intermediates in the slowest step i.e. the rate law  :)

Well i prefer doing the sum by finding the conc of the intermediates without steady state apprx but anyway the'll both give the same answer.

Wish carcul was here! He'd do this all right n i hate giving long explanations  :D

and i think it's pretty clear that the steady state approximation doesn't give the required answer...
that's why i suppose the answer is for the initail induction period when the steady state approximtion doesn't hold good

otherwise i don't find a method how to get the required answer . :rotfl:

i dont get what the fuss is all abt. and why cant u use steady state approx here?

u can but then it doesn't match the given answer...
also i think the given answer is supposedly for the initial induction period and hence u can't use steady state approximation which is only valid after the induction period

what is the answer given? and ssa as the name suggests is only an approximation. i dont know how well its use over here can be justified, but if u have to calc the rate during the induction period then it bcoms very complicated. and the induction period is very short, almost negligible.

no actually in this case that short period turns out to be of the order 100 hours....
well the answer is $ r \equal{} k_{1}k_{2}k_{3}[\ce{HCl}]^{3}[propene]$

HCl^2. i got the answer as HCl^3. and how do you know abt the induction period? :huh:
oh sorry did u edit that or are my eyes playing tricks on me?

yes i changed that already ....actually this was an experiment carried out by D.R.Dalton to predict the mechanism what basically was done was the fractional produce of 2-chloro propane with respect to both reactants was found out keeping the other constant. in the initial periods....and so on.
that's why the equilibrium is even not considered
how is ur solution

[quote="mythili"]ok let' s put it straight the first rate gives us
$ \frac {d[(\ce{HCl})_{2}]}{dt} \equal{} k_{1}[\ce{HCl}]^{2} \minus{} k_{3}[(\ce{HCl})_{2}][complex]$
the second
$ \frac {d[complex]}{dt} \equal{} k_{2}[\ce{HCl}][Propene] \minus{} k_{3}[(\ce{HCl})_{2}][complex]$
the third
$ \frac {d[2 \minus{} chloro propane]}{dt} \equal{} k_{3}[(\ce{HCl})_{2}][complex]$

so how does one proceed now??[/quote]
actually i used ssa to arrive at the solution. just equate the first and second rates to 0. and arrive at an exp for the conc of HCl2 and the complex in terms of the others. and put in the last eq. and one more thing ssa doesnt req the intermediate to be in eq. as long as the  conc of the intermediate is low.

using steady state aprrox u get the answer simply with the first equation and that too wrong...
also it really doesn't matter if that's in eqb or smalll concentration in any case d[I]$ \approx 0$
also we can't say anything about their concentrations see the intermediate steps are indeed the fastest steps

i find no problem with ssa  :huh:  It comes fine atleast i think. Please post the exact answer of your book, the answer got on ssa (as u think) and the problem u face with that method  :D

using steady state approximation the first equation of
[quote="mythili"]
ok let' s put it straight the first rate gives us
$ \frac {d[(\ce{HCl})_{2}]}{dt} \equal{} k_{1}[\ce{HCl}]^{2} \minus{} k_{3}[(\ce{HCl})_{2}][complex]$
the second
$ \frac {d[complex]}{dt} \equal{} k_{2}[\ce{HCl}][Propene] \minus{} k_{3}[(\ce{HCl})_{2}][complex]$
the third
$ \frac {d[2 \minus{} chloro propane]}{dt} \equal{} k_{3}[(\ce{HCl})_{2}][complex]$
[/quote]
from the first equiation we get
$ k_{1}[\ce{HCl}]^{2} \equal{} k_{3}[(\ce{HCl})_{2}][complex]$
putting this in the last equation we get the
$ \frac {d[2 \minus{} chloro propane]}{dt}\equal{}k_{1}[\ce{HCl}]^{2}$ which is wrong
the answer being $ \frac {d[2 \minus{} chloro propane]}{dt}\equal{}r \equal{} k_{1}k_{2}k_{3}[\ce{HCl}]^{3}[propene]$

hey, what mythili is saying is rite. i dint see that the rate determinng step was given. :wallbash:  if that is given how can u ever apply ssa. if ssa shud be valid over here, k3>>k1 and k2 or atleast the ratio k3/k1>10. or else the rate of production of the intermediates is much greater than the consumption, so the conc of intermediates. is never low which is the basic criterion for using ssa. also the max conc. of intermediates is given when t = ln(k1/k3)/(k1-k3). here since k1>> k3, hterefore induction period is very high, and use of ssa is not justified.

are there any other conditions provided like any of the steps being in equilibrium?

yeah that's right n i finally got what was trying to be said  :D

yes the first two reactions are equilibrium  reactions

You shud have mentioned that earlier as we have not included those terms while using ssa. Here’s my answer :
K1[HCl]^2  –  K-1[HCl2] – k3[Hcl2][complex]=0
K2[hcl][prpene] - k2-[complex] – k3[hcl2][complex]=0
[hcl2]=k1[hcl]^2/k-1+k3[complex]
[complex]=k2[hcl][propene]/ k2- + k3[hcl2]
And as k2->>k3[hcl2],
And as k1->>k3[complex].
The rate law bcoms K1K2k3[Hcl]^3[propene]
well, this is the closest i can get to the answer. :D
n sorry for not latexing


topology
search
real analysis
real analysis unsolved

Are the rationals in (0,1) and the rationals in [0,1] homeomorphic.My guess is no but I cant prove it.

All countable metric spaces with no isolated points are homeomorphic. [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=474774578&t=48347]Here[/url]'s the relevant thread.

However, they are not uniformly homeomorphic (i.e. via a map that is uniformly continuous both ways). Indeed, a uniform homeomorphism would extend to a homeomorphism of $ [0,1]$ onto itself, which necessarily fixes the set $ \{0,1\}$.


geometry
integration
perimeter
analytic geometry
inequalities
calculus
geometry proposed

Let $P$ be a polygon (not necessarily convex) with $n$ vertices, such that all its sides and diagonals are less or equal with 1 in length. Prove that the area of the polygon is less than $\dfrac {\sqrt 3} 2$.

Indeed, but this estimate allows for different solutions then the one you are thinking of Mikhail! :)

I don't understand the aim of this remark.

Anyway, the initial problem is quite easy.

Construct two strips of width 1, which contain the polygon and form angle $\pi/3$ (strips $AB$-$CD$ and $AD$-$BC$). Their intersection is a rombus $ABCD$. We assume that $AC$ is a vertical line. Consider a horizontal strip of width 1, which contain the polygon. Guess: for what position of that strip the area of the thick hexagon is maximal ;) :D

I realized that I don't know any elementary proof for this fact, though I saw the estimate for several times, but everytime without a solution.

Anyway, my proof is the following. Consider a line with polar angle $\varphi$ and let $r(\varphi)$ be a length of projection of the figure to this line. Let's calculate $I=\int_0^\pi r(\varphi)d\varphi$. Since $r(\varphi)\leq 1$ we get $I\leq \pi$. On the other hand $I=P$, where $P$ is the perimetr of the figure :!: . It means that $P\leq \pi$. And we know that the maximal area for all figures of perimetr at most $\pi$ has a circle of radius $\frac{1}{2}$.
I.e. I reduced this problem to a really well known not-easy one (isoperimetrical problem).

That's all.

[quote="Valentin Vornicu"][quote="Fedor Petrov"]An estimate $\pi/4$ is very well known. I guess, a lot of participants were familiar with it.[/quote]I think you are giving the Romanian team too much credit (and you did give the Russian team last year too little :P :) ). It seems that only one student knew this estimate, and only a couple more solved the problem.[/quote] I guess it is a very funny joke. :?
Is my solution correct? If so, then I don't understand...

[quote="Myth"]I don't understand low results on this problem.[/quote]It's simple: the Romanian team is not that good as you think it is :)

I guess that Mikhail does not understand, why such easy problem was so poorely solved by Romanian students. I do not understand this, too, but, to be true, I often do not understand such things.

By the way, an estimate $\pi/4$ may be proved without reduction to isoperimeter problem: let our (convex) set lie in an upper half-plane and contain the origin. Its boundary has equation $r=r(\phi)$ in polar coordinates, where $\phi$ moves from 0 to $\pi$. An area equals $\frac{1}{2}\int_0^{\pi}r^2(\phi)d\phi$. But we have $r^2(\phi)+r^2(\phi+\pi/2)\le 1 \, (0\le \phi\le \pi/2)$ by Pythagor theorem. Now just integrate the last inequality on $[0,\pi/2]$.

Oh, my God!
I have just remembered! I even saw a picture for this solution, but I completely forgot it :(

I guess most of them can solve it if the problem states the exact estimate $\pi/4$, the bound $\sqrt{3}/2$ is too weak that most people don't know how to arrive ;)
Anyway, this problem can be easily done by isoperimetric theorem.

is there an elementary solution to this one?

Which one do you mean?
Isoperimetric problem has a simple solution if you know that such figure exists.

i meant to the initial problem.
is there any solution without integrals?
i'm also interested in a simple solution of the isoperimetric problem

I gave an elementary solution for the initial problem in the very beginning.

hey,

sorry 4 mi ignorance, but wat is the isoperimetric problem?

i didn't understand the elementary solution, which seems to be only for a hexagon

No, it works for every polygon. We construct a hexagon of area $\sqrt{3}/2$, which contains our polygon inside.

[quote="Myth"]From all figures of given perimetr find the the figure having maximal area.[/quote]
Which is circle and i have just find a nice prove for it ha-ha :heli:  :heli: .see if it is right?

Lemma one: If a shape has mor than two lines that the shape is symmetric due to these lines then, these lines are concurrent.

Lets name the figure $\mathcal L$.Consider two points like $A$ and $B$ on its premeter such that these points devide the premeter of the figure  into two equal parts.
Lets name these parts of premeter $\mathcal E$ and $\mathcal L$.
So the [b]lengh[/b]$ \mathcal E$=[b]lengh[/b]$ \mathcal Q $=$\frac12 $ premeter of $\mathcal G$.

Consider these areas:
1.Area between the line $AB$ and $\mathcal E$=$S[\mathcal E AB]$
2.Area between the line $AB$ and $\mathcal Q$=$S[\mathcal QAB]$

Now If $S[\mathcal EAB] \ge S[\mathcal QAB]$ then [b]delete[/b] the $\mathcal Q$ from the figure and replace it by  the symmetric of $\mathcal E$ WRT $AB$ ,And we will reach a new figure with same premeter of our first shap but with a greater area :
Equal premeters:[b]lenght[/b]$ \mathcal E$+[b]lenght[/b]$\mathcal Q=2$[b]lengh[/b]$\mathcal Q$
Bigger Area: $2S[\mathcal E AB] \ge S[\mathcal E AB]+S[\mathcal QAB]$.

by these argumets ,in the shape that satisfies the condition of the isopremeter problem ,  for every two points like $(A,B)$ that besicts the premeter,$AB$ bisects the Area and $AB$ is a symmetric line for the shape .
By using lemma one we can see that all of this lines are congurrent .
So we have changed the problem to this :
Find all of the shapes that are symmetric wrt $\infty$ congurrent lines.

lomos_lupin, I don't want to disappoint you-but your proof to isoperimetric theorem is completely wrong.
Nevermind, I had also the wrong guess. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29793]http://www.mathlinks.ro/Forum/viewtopic.php?t=29793[/url] for more details.

Rest of post 23

Let the symmetric lines be $l_1,...,l_n$ and they be concurrent at $O$ ,its clear that every point of shape has a same distance from the $O$ so the shape is a circle with center of $O$. :D  :D
Sweet home

Could you elaborate your idea and give a formal proof?

That area is less than $\pi /2-\sqrt3/2 < \pi /4$.The proof is not very complicate,if you want it just let me know.

I don't understand. If the polygon has all sides and diagonals length <1, isn't it contained in a circle of diameter 1, which has area pi/4? What am I assuming that is false?

Everything. Is an equilateral triangle of side-length $1$ contained in a circle of diameter $1$? The best result is Jung's theorem, stating that a plane figure where the distances between any two of its points are at most $1$ is contained in a circle of radius $\dfrac {1} {\sqrt{3}}$ (which in fact is best precisely for the triangle).
But that has too large an area for the problem; $\dfrac{\pi}{3} > \dfrac {\sqrt{3}}{2}$, unfortunately :)

Of course, I was not thinking properly. Now my next question is: can we improve the bound in this problem by using more stripths of worth 1, as in myth's solution?

What do you mean by "improve the bound"? the [b]best[/b] bound was [b]shown[/b] to be $\pi/4$, right?


How did u all do guys?

It was very easy and a perfect paper for 2.5 hours I thought. :D I might get 90+ easily. :)

expect 95 to be honest but i think its pathetic  :(   :o

Yes not even 1 conversion in the paper how pathetic? :o  :o  :o

Lol.. No conversions, no mechanism.. Nothing worthwhile in short, but stilll

The paper was straight picked up from the text. :D
In the 2nd set we have 1 question repeated(diazonium + KI). :P

[quote="chemrock"]
In the 2nd set we have 1 question repeated(diazonium + KI). :P[/quote]

Way too theoritical. Chaaa!!!

LOL! I wonder why they rep the question.

And that too no mechanisms and conversions,etc. and in our set 1 problem for 5 in electrochemistry too. :D

Trying to help students pass perhaps?  :alien:  :moose:  :stretcher: Rock, you set 2 da?

Maybe I am late here....(of ourse, I am)

I am expecting sth like 95-97....

hey!
any high schooler here :ninja:
how did you find your social sciences paper?

[quote="mathswizard1992"]hey!
any high schooler here
how did you find your social sciences paper?[/quote]

You could have asked tips from shreyas who got 100 in social in boards  :P   :rotfl:  :D


This question is slightly rhetorical but here goes.

A man is taken into the place of eternal damnation. However, the Devil makes him a deal: At any time the man may flip a special coin; if it lands heads he goes free, and if tails he spends the rest of eternity in eternal damnation. Also, every day that the man waits, the coin becomes more weighted towards the heads side. Specifically the probability that the coin lands heads is $1-(0.5)^{n}$ where $n$ is the number of days that he waits.

Q1: Will a rational person ever flip the coin?
Q2: What would you do?

Um...if you wait for 10 days, it will be quite impossible to get a tail.

But think about it, Hell is like hell. Each day would be more pain, torture, and suffering. So maybe the best possible time could be calculated by making a curve that's like an upside down U, and from there a good date can be determined.

Doesn't $(0.5)^{n}$ become smaller when $n$ becomes bigger?  :maybe:

Yes.  But the probability of heads is $1-0.5^{n}$, meaning as you wait longer, it is more likely to come up heads.

a/1-r might make people go "oh" but not help

This might spur some ideas...

$\frac12+\frac24+\frac38+\frac4{16}+\frac5{32}+\ldots$

1. Well, since you would have a higher chance of going free each day, a rational person might not ever flip it. 9He's really rational)

2. Well of course I would flip it! (After a week or so)

2.  I wouldn't even wait for a day to go by.

Then n = 0.

1 - (0.5)^n becomes

1 - (0.5)^0 =

1 - 1 = 0 means

0% chance of a getting a head

Say, "Devil, I'm not going to be tortured by being

made to wait in here on a future flip, so you

can just go to H_e_l_l.  Wait, you already ARE there."

(I have my doubts about my answer.)
Note:  this system wouldn't just let me spell out the Devil's home
without using spaces; it considers it a "bad" word.

[quote="miyomiyo"]Am I doing something wrong or does this coin become easier and easier to get free with as the days go on?[/quote]

It does, but you are in a bad place while you're waiting.

I'd settle on some reasonable probability to get tails...say....2^-64 (approx 10^-19) and wait until I get that probability.

[quote="randomdragoon"]the coin becomes more weighted towards the heads side. Specifically the probability that the coin lands heads is $1-(0.5)^{n}$ where $n$ is the number of days that he waits.[/quote]
If it's weighted towards the heads side, that makes tails more likely.

Umm... not quite.
[quote="randomdragoon"]Specifically the probability that the coin lands heads is $1-(0.5)^{n}$ where $n$ is the number of days that he waits.[/quote]
It becomes more likely that it comes up heads.

Looking at the expected value, assuming the value of spending an eternity in damnation is negative infinity, the expected value for flipping the coin is always negative so a rational person would never flip it ... which means he spends an eternity in damnation.

This was part of the Devil's plan all along.

[quote="lingomaniac88"][quote="tjhance"]i think "rational" needs to be defined better. A rational person could also argue:

If you never flip the coin, there is 0% chance of escaping.
If you do flip the coin sometime, then there is a chance you will escape.[/quote]
Exactly.  A rational person would probably wait a while, and then flip the coin.  The specific amount of time depends on how tough the person is.  Would he be willing to sacrifice more days of torture in order to improve the chances of his freedom, or would he break down so fast that he feels a need to flip the coin after a day or so?[/quote]
That's not acting rational.

Rational person (same thought process every day):

I have a choice: Wait one more day or flip. Expected value if I wait one more day is -1. Expected value if I flip is -infinity. So I should wait.

It's a good thing most people aren't rational when they're burning in heck

BTW, I would wait 33 days, which would put my probability of losing at lower than 1/(number of people in the world).

No, if spending the rest of your life there is -infinity, then the expected value for flipping or not flipping the coin is both -infinity. -infinity , although in a small probability like 10^-19, weigh out any other things.

Imagine for each day you wait, the probability for tail is halved. However, the Devil says,

Stop waiting, silly guy! For each day you wait, i will double the the duration of "eternity"!


limit
linear algebra
matrix
real analysis
real analysis unsolved

[quote="kunny"][hide="Answer"]$\frac{a+b+c}{3}$[/hide][/quote]
That's right! But what is your solution? Can you post it ?  :wink:

This problem is nice, I think.I have posted problem before as follows.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=exam+entrance+university+gakusyuin&t=89960[/url]

At least I do know that $\lim_{n\to\infty}(a_{n}-b_{n})=0\Longrightarrow \lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}$ is not always hold.
Let's see other's solution.

a general approach for this kind of problem is to write it as $U_{n+1}=A U_{n}$ where $A$ is a matrix and see what is the limit of $A^{n}$. It is often very easy when $A$ is diagonalisable (which is the case here since $A$ is symetric)


2. ii) Find k such that the equations are consistent (that is, they can be solved):

x+2y+z=k
2x+y+4z=6
x-4y+5z=9

hope that i didnt make some idiot mistakes
from $x-4y+5z=9$ we obtain $x=9+4y-5z$ ,
since $2x+y+4z =2(9+4y-5z)+y+4z=6$ , we get $z=\frac{4+3y}{2}$
thus $x+2y+z=9+4y-5z+2y+z=9+6y-2(4+3y)=1$
so we obtain $k=1$


inequalities
geometry
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ be a cyclic quadrilateral,inscirbed in a circle centered at $ I$.Prove the following inequality:
$ (AI \plus{} DI)^2 \plus{} (BI \plus{} CI)^2 \leq (AB \plus{} CD)^2$

Posted at least once here (now I'm 100% sure this is a MY problem, when I've seen the interest for its solution).

[url]http://www.mathlinks.ro/viewtopic.php?t=208043[/url]


calculus
integration
parameterization
function
trigonometry
analytic geometry
real analysis

Calculate
\[ \oint\limits_C {x^2 ds}
\]
where
\[ C: \equal{} \left\{ {\left( {x,y,z} \right): x^2  \plus{} y^2  \plus{} z^2  \equal{} 1} \right\} \cap \left\{ {\left( {x,y,z} \right): x^2  \minus{} 2x \plus{} y^2  \plus{} z^2  \equal{} 0} \right\}.
\]

I found this one getting old...  Thought I'd give it a try.

The 2 surfaces intersect at $ x \equal{} 1/2$.
At $ x \equal{} 1/2$ the intersection of surfaces is a circle, $ z^2 \plus{} y^2 \equal{} r^2 \equal{} \frac {3}{4}.$
Parametric representation of the circle is given by

$ z \equal{} t, z^2 \equal{} t^2, y^2 \equal{} \frac {3}{4} \minus{} t^2$ and $ y \equal{} \pm \sqrt {\frac {3}{4} \minus{} t^2}.$

$ D_tS \equal{} \left|D_tR(t)\right|$

Write C as $ R(t) \equal{} \pm \sqrt {\frac {3}{4} \minus{} t^2}\; \textbf{j} \plus{} t\;\textbf{k}$
$ D_t R(t) \equal{} \displaystyle \frac {t}{3/4 \plus{} t^2}\; \textbf{j} \plus{} 1\;\textbf{k}$
$ \left|D_t R(t)\right| \equal{} \sqrt {\frac {t^2}{\frac {3}{4} \minus{} t^2} \plus{} 1} \equal{} \sqrt {\frac {3}{3 \minus{} 4 t^2}}$

$ \therefore D_tS \equal{} \sqrt {\frac {3}{3 \minus{} 4 t^2}}$ and $ s \equal{} \int^t_0 \sqrt {\frac {3}{3 \minus{} 4 u^2}}du$

$ ds \equal{} \sqrt {\frac {3}{3 \minus{} 4 t^2}}dt$ and $ x^2 \equal{} (1/2)^2 \equal{} 1/4$

$ \frac {1} {2} \oint x^2 ds \equal{} \frac {1}{4} \oint \sqrt {\frac {3}{3 \minus{} 4 t^2}}dt$

$ \oint x^2 ds \equal{} \frac {\sqrt {3}}{2} \oint \frac {1}{\sqrt {3 \minus{} (2 t)^2}}dt$

$ \oint x^2 ds \equal{} \frac {\sqrt {3}}{2} \sinh^\minus{}1{(\frac {2 }{3}}\sqrt {3} z)$

Without having examined anything you did in great detail, your answer is absurd -- what does it mean for a $ z$ to be there?  Your parametrization of the curve of intersection of those surfaces probably has something to do with it, since that's not a parametrization of a smooth curve ($ y$ isn't given as a function of the parameter).  The "usual" way to parametrize a circle is to use trigonometric functions, e.g. $ y \equal{} \frac{\sqrt{3}}{2}\cos t$, $ z \equal{} \frac{\sqrt{3}}{2}\sin t$ for $ 0 \leq t < 2\pi$.

[quote="JBL"]Without having examined anything you did in great detail, your answer is absurd -- what does it mean for a $ z$ to be there?  Your parametrization of the curve of intersection of those surfaces probably has something to do with it, since that's not a parametrization of a smooth curve ($ y$ isn't given as a function of the parameter).  The "usual" way to parametrize a circle is to use trigonometric functions, e.g. $ y \equal{} \frac {\sqrt {3}}{2}\cos t$, $ z \equal{} \frac {\sqrt {3}}{2}\sin t$ for $ 0 \leq t < 2\pi$.[/quote]

I'll try it your way...  I don't understand why my parametrization is no good?

What point on the curve is determined by your parametrization at $ t \equal{} \frac{1}{2}$?

[quote="JBL"]What point on the curve is determined by your parametrization at $ t \equal{} \frac {1}{2}$?[/quote]

$ ds \equal{} \frac{1}{2} \sqrt{3} dt$ ?

[quote="fedja"]This is a deviation from the discussion currently going on, but once you figured out that $ x^2$ is constant on the curve and that the curve is a circle, why do you need any parametrization at all? ;)[/quote]

I assumed $ ds$ was related to distance.  The only problem examples I have to get to $ ds$ is through using a parameter t.  Which I didn't do correctly I might add...

[quote="fedja"]Yes, of course, $ ds$ is related to the arc length. Now, do you need to parametrize the circle to find its length?[/quote]

Circumference x $ x^2$... ?

[quote="fedja"]Only not by $ x^2$ but by the constant value $ \frac 14$ that $ x^2$ take on the circle.[/quote]

I understand now... Thanks both of you...

I gather the original question has finally been finished off, but there's still some unfinished business:

[quote="vector2"][quote="JBL"]What point on the curve is determined by your parametrization at $ t \equal{} \frac {1}{2}$?[/quote]

$ ds \equal{} \frac {1}{2} \sqrt {3} dt$ ?[/quote]
 :huh:  I think you've completely misunderstood my question.  A parametrization is a [i]function[/i]; we usually ask for it to be a smooth function from some interval to our curve.  Your "parametrization" isn't a function: at the value $ t \equal{} \frac{1}{2}$ (and most other values), it gives two points, $ (\frac{1}{2}, \pm \frac{\sqrt{2}}{2}, \frac{1}{2})$.  If this question didn't happen to supply us with such a simple calculation, the alternate parametrization I suggested has the advantages of actually being a parametrization, being relatively easy to work with, and providing a constant-speed parametrization of the curve (so $ dt$ is closely related to $ ds$).

[quote="JBL"]I gather the original question has finally been finished off, but there's still some unfinished business:
What point on the curve is determined by your parametrization at $ t \equal{} \frac {1}{2}$?
:huh:  I think you've completely misunderstood my question.  [/quote]

I appreciated your question as it uncovered my error. When I put $ t\equal{} 1/2$ into what I originally thought was a good parametrization, I found that the coordinates didn't work for the circle.  I understood a circle could be represented by $ y \equal{} \pm \sqrt{r^2 \minus{} t^2}$ and then multiplying by 2 (as I incorrectly used) or by $ x\equal{} a \cos t ; y \equal{}a \sin t$ (which I thought about using but elected the other).

Is it correct to say that when I paramatize a function, the parametrization function needs to be a 1:1 function? Also, is that the point of your question?

Yes, it should be a 1-1 function though you are free to split the curve into several pieces, parametrize each piece separately and then add up several integrals you obtain. Thus, your original parametrization could be used too if you did the "upper" and the "lower" semicircle separately.


Let the vertex of an angle ABC be located outside a circle  and let the sides of the angle intersect equal chords AD and CE with the circle.Prove that $ \angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Thank you.

[hide]
Let $ A\in(BD), \ C\in(BE)$

$ \widehat{DCE}\equal{}\widehat{ABC}\plus{}\widehat{BDC}\Rightarrow\widehat{ABC}\equal{}\widehat{DCE}\minus{}\widehat{BDC}\equal{}$

$ \equal{}\frac{arc(DE)}{2}\minus{}\frac{arc(AC)}{2}\equal{}\frac{1}{2}\left(\widehat{DOE}\minus{}\widehat{AOC}\right)$
[/hide]

[quote="red_dog"][hide]
Let $ A\in(BD), \ C\in(BE)$

[/hide][/quote]

But I do not understand the solution.The question says that the vertex of ABC lies outside the circle.

Can you please explain it more clearly?Thank you.

Let O be the center of the circle. Let D be the point on the circle as well as on line BO with D NOT inbetween points B, O. Let the radius OE intersect BA at E. Let $ \angle OBE \equal{} x,\angle BOE \equal{} a,\angle EOA \equal{} b,\angle AOD \equal{} c,\angle OEA \equal{} \angle OAE \equal{} \theta$

$ a \plus{} b \plus{} c \equal{} 180$
$ \theta \equal{} x \plus{} a$
$ 2\theta \plus{} b \equal{} 2x \plus{} 2a \plus{} b \equal{} 180 \equal{} a \plus{} b \plus{} c$
$ \implies2x \equal{} c \minus{} a\implies x \equal{} (c \minus{} a)/2$

Same argument for A replaced by C on the other side.

[quote="muks"]

But I do not understand the solution.The question says that the vertex of ABC lies outside the circle.

Can you please explain it more clearly?Thank you.[/quote]

If the cords are AD and CE then A and C lies on the circle and the vertex is B which is outside the circle.


For all Korean ppl who live in the US, did anyone take the NMC? Not necessarily even 2007, just ever? I tried this year, but I failed, like I do all competitions... :(

I did last year. My younger brother also did, this year.
The real Korean Math Olympiad is far harder...

Yeah, probably :) . Shouldn't the KMO be somewhere along the difficulty level of the USAMO?

Second round is, but the first round is AIME-style (20 questions, 4 hours, answers 000 to 999) but I think it's harder actually.


suppose a number x satisfies x^2-4x-1=0. How much greater is x than its reciprocals!

lolt, i put an exclamation point instead of an "?". wooooot

*sigh* sorry to reprimand you, but I don't think "wooooot" is appropriate, but it is funny :) oh well.

and I know, I know. I should go worry about somat else, right? :)

no.

hint if you want it [hide]note that the amount asked for in the problem is x-1/x. Hmmmm that looks familiar...I wonder what would happen if we multiplied it by something...[/hide]

Syntax Error wrote:suppose a number x satisfies x^2-4x-1=0. How much greater is x than its reciprocals!

Spoiler:[hide]

so divide x^2-4x-1=0 by x to get x-4-1/x=0, or x-1/x=4[/hide]

..So the answer is true for any x, not just for the roots, right?

correct, but to be absolutely certain [hide]x cannot be 0, because it does not satisfy the equation, which is why we can divide through by x[/hide]

fierytycoon, no, it's just for the roots.

I guess. but it's still kind of feels weird...and I feel weird saying this because I should know this stuff as the comoderator, but don't. :)

its okay, to clear up the 0 thing further, 1/0 is undefined, because if 1/0=0, multiplying both sides by 0, 1=0. and thats not true. i think mystic may have said that also, but i just skimmed it

Actually, I said that, but oh well. I'll go worry about other things now :) (Like the fact that I didn't solve the question :( )

It gets really annoying when you post solutions without the thinking process, [b]especially[/b] when someone else [b]already answered the question.[/b]

Am I sounding too harsh? :twisted:

No, but with Neal's solution, you could do that with any x, not just for the roots.

lolt

fiery...in neal's solution, [hide]he used the equation x:^2:-4x-1=0. In doing so, he restricted himself to just the roots of that equation. (because if you took any old x and plugged it in there, chances are that it won't be 0). do you see what i mean?[/hide]

Oh, the what's the T?
and btw, this discussion was moved to the Etc. for onvious reason :) so let's continue our conversation there, shall we?

divide through by x to get x-1/x -4 =0. x-1/x = 4!! cool


Does anyone have a clue as to why when I make a document while using the amssymb package it gives me $``$ for \angle but when I take out amssymb it gives me $\angle$?

Hm, I don't know what's happening; \angle works for me. What other packages are you using? Compile this document (after inserting the dollar signs), and tell me what happens:
[code]\documentclass{article}
\usepackage{amssymb}
\begin{document}
(dollar sign)\angle(dollar sign)
\end{document}[/code]

I'm working on my USAMTS solutions right now.  So I'm using all the packages in their template.

I compiled what you have and it worked out fine.

I actually figured out the problem... it's the \documentclass[11pt]{article}.  If I take out the [11pt] and just have \documentclass{article} it works fine.  Does that happen to you as well?


simplify the sum \[\tan1^\circ \tan2^\circ+\tan2^\circ \tan3^\circ+\dots+\tan2006^\circ \tan2007^\circ\]

[quote="igiul"]:maybe: ....No.[/quote]

what in the world do you mean by that? :huh:

is this too hard for HSB? :maybe:

I mean :
[hide]
$\sum_{n=1^\circ}^{2006^\circ}{\tan (n)tan (n+1^\circ)}$
That what you wanted?
[/hide]

EDIT 1 : Corrected to $2006^\circ$
EDIT 2 : If you actually want it simplified to a small, general equation that does not requires a sigma, I suppose I might find out HOW, but yes, it would be fairly difficult for me, as inexperienced in number theory as I am.
EDIT 3 : Grammar (GOD, I need to reread [i]before[/i] I submit!)

[quote="igiul"]...$\sum_{n=1^\circ}^{2006^\circ}{\tan (n)tan (n+1^\circ)}$
That what you wanted?[/quote]

no, just find a simpler form of the given sum.

hint: tan 1 (1-tan(n)tan(n+1)) = tan (n+1)- tan (n)

Just a question:

Would one find this problem more in Algebra II or Pre Calc?

[hide="hint:"]$\tan{1}+\tan{2}+\cdots+\tan{360}=0$[/hide]
:wink:

EDIT: nvm.. didn't see the multiplication (assumed it was all addition) XD

this is a bad problem because of that reason...i think your friend took this from the woot class...tried to change the numbers around and did not realize that the sum is not defined

actually my friend gave me the problem with $1995$, i changed it to $2007$. i think he found it in a contest or something (because of the number 1995), its not my fault that the problem is wrong.

well obviously the problem with 1995 did not work either...this is an interesting problem though...one of the few trig sums i have not been able to solve lately...

[quote="t0rajir0u"][quote="Shaister"]Question: what are you supposed to do when the series gets to

... (tan 89)(tan 90) + (tan 90)(tan 91) + ....

Isn't tan 90 undefined?[/quote]

In a sort of roundabout way we can hand-wave this away by saying that $\tan 89 =-\tan 91$ and so the sum of the two terms can be redefined to be $0$.[/quote]

So, does my solution still work, then?

yes i think it does and i'm really happy that the problem is AOK after all! :lol:  my solution is also the same as yours, vishalarul! :wink:

[quote="t0rajir0u"][quote="Shaister"]Question: what are you supposed to do when the series gets to

... (tan 89)(tan 90) + (tan 90)(tan 91) + ....

Isn't tan 90 undefined?[/quote]

In a sort of roundabout way we can hand-wave this away by saying that $\tan 89 =-\tan 91$ and so the sum of the two terms can be redefined to be $0$.[/quote]

or we can go back to like first grade and realize that we have to follow the of operations  :lol:

Or we could just change "degrees" to "radians" and retain all of the interest and none of the problems.  (The problem as stated is [i]not[/i] okay after all -- undefined terms don't "cancel.")


On monday morning, the number of people entering Bugis Junction MRT station per minute has a Poisson Distribution with mean = 3. and the number of people leaving the station per minute has an independent Poisson distribution with mean =2.

Find the probability that during a particular minute in the monday morning exactly one will enter the station given that the number of people entering and leaving the station during the particular minute is exactly 4..

CAN ANYONE HELP ME WITH THIS QUESTION... I AM STRUCK WITH THIS... PLS SHOW ME UR CLEAR WORKING IF POSSIBLE..

First of all, do your own homework.  Second of all, don't YELL IN ALL CAPS!!!  IT"S VERY ANNOYING!!  Third of all, read the descriptions of the forum that you are posting in.

You could at least give the poor guy a few hints, Joel. There's really nowhere to put statistics/probability problems like this one.
Poisson distribution: $P(x)=\frac{\lambda^xe^{-\lambda}}{x!}$
Mean=$\lambda$
It's also useful to know that the difference/sum of two independent variables that both follow Poisson distributions also follows a Poisson distribution.
See where you can get from there.

Edit: Caught some errors. I shouldn't be posting so late at night.

[quote="Bictor717"]You could at least give the poor guy a few hints, Joel.[/quote]

No I can't: I don't know any statistics.  (Well, an exceedingly tiny bit of self-taught stuff, but I haven't the faintest idea what a Poisson distribution is.)  Also, this is the forum I sent this to -- it was in the round table (which was why I was so annoyed; that, and the all caps.)

Bictor, you dominate these statistics questions! Did you learn most of your statistics in high school, in college, or on your own? And what book, or books, did you mostly use?

These aren't much more difficult than a probability course at the college level, but they're not covered in AP Statistics.
I'm currently a high school senior but I took a statistics course at a local college this term. The professor's name is Rothenberg, so we use his book, as well as Strait.
It becomes easy to self-teach after you cover the concepts behind the formulae.


I was wondering if somebody knows some usefull links about polymers and some aplications in informatiscs....
I have no idea what i should write about, so i could use some help....Thanks!!!

I'm not sure how polymers might be used in informatics, except in cases like DNA (certainly a polymer).  The basic case might be polymers of, say, CH2=CHX, where X is some halogen.  By changing X, different information might be encoded.  In reality,  however, it would be impossible to string together such a polymer in a predetermined order.  DNA is a very sophisticated molecule that actually does have the ability to reliably and repeatedly encode information.

As for internet links to polymer resources, I can't offer more than wikipedia.

Thanks,ut cold you poin some resources where i could find some more detalieted information......


abstract algebra
vector
superior algebra
superior algebra unsolved

I was reading something and it says the "category of finite-dimensional $ k[x]$-modules". I interpreted this as $ k[x]$-modules that are finite-dimensional as $ k$-vector spaces. Is this a standard interpretation? It seems that that would be equivalent to finitely generated $ k[x]$-modules, but they use this term as if it were different.

k[x] is finitely generated as a k[x] module, but not finite dimensional as a k-module (assuming x is transcendental).

I usually see "the category of finite dimensional R-modules" when R is a finite dimensional k-algebra, and in that case "finite dimensional" and "finitely generated" coincide.

Probably you will not often see the restriction to only finite dimensional modules over an infinite dimensional ring, but for k[x], the category of finite dimensional k[x] modules is very important (finite dimensional vector spaces along with an endomorphism; the place that Jordan form and rational canonical form are studied these days).  In some sense it is also easy to deal with, since the only indecomposable finitely generated module being left out is k[x] itself.

Another collection of modules that is fairly nice are the modules of finite length.  In the case of k[x] this coincides with finite dimension.


Let $ AB$ be a diameter of a circle $ \Omega$ and $ P$ be any point not on the line $ AB$. The lines $ PA$ and $ PB$ cut $ \Omega$ again at $ U$ and $ V$, respectively. (In the case of tangency $ U$ or $ V$ may coincide with $ A$ or $ B$; also, if $ P \in \Omega$ then $ P\equal{}U\equal{}V$.) Suppose $ s,t \ge 0$ are such that $ PU\equal{}s \cdot PA$ and $ PV\equal{} t \cdot PB$. Determine $ \cos \angle APB$ in terms of $ s$ and $ t$.

[hide]The power of $ P$ with respect to $ \Omega$ is clearly $ sPA^2\equal{}tPB^2$, or $ PB^2\equal{}\frac{s}{t}PA^2$.  Moreover,

\[ AB^2\equal{}AU^2\plus{}UB^2\equal{}AU^2\plus{}PB^2\minus{}PU^2\equal{}(s\minus{}1)^2PA^2\plus{}PB^2\minus{}s^2PA^2\equal{}PA^2\plus{}PB^2\minus{}2sPA^2\equal{}PA^2\plus{}PB^2\minus{}2tPB^2.\]

Therefore,

\[ \cos\angle APB\equal{}\frac{PA^2\plus{}PB^2\minus{}AB^2}{2PA\cdot PB}\equal{}\frac{2sPA^2}{2PA\cdot PB}\equal{}\frac{2tPB^2}{2PA\cdot PB}\equal{}\sqrt{st}.\][/hide]


MATHCOUNTS
AMC
AMC 8
AMC 10
AIME
Purple Comet

Hi Everyone,

I am wondering what I should do know. I am currently in 8th grade and placed 24th in my chapter. On the AMC 8 I scored 19. I was wondering, how should I go about studying for the AMC 10 (apart from the AOPS books).
Should I be doing AMC 8 practices so I can master them and then move on to AMC 10? Should I be doing Mathcount State Tests and then National Tests?
Should I just be doing the AMC 10 tests?

I really need help!

I know 20 was the cutoff. That is why I wondering what I should do in the long run for the AMC 10. I am going to take the AMC 10 in about 11 months!

Just practice on old AMC10s.
I think that way you will get a feel of how the problems are. Therefore you will be able to expect those kind of problems. :)

[quote="Luminescence"]Just do AMC 10s, and refer to the AoPS wiki for explanations. Mathcounts is nothing like the AMCs - it'll help you get better at math, sure, but it won't help you score high on the AMC 10, especially seeing as you'll be taking it very soon.

Wasn't the [AMC 8] cutoff 20, though?[size=75]

[color=green][modedited to make clear that 20 was the AMC8 cutoff][/color][/size][/quote]

there is a cut-off for AMC 10??  I thought everyone can take the amc 8, 10, and 12, just that they have to be in the grade of the numer of below...

Recommendation:  Not too many AMC 10's out there.  And esp. with the no calc rule, it's gonna be harder to find good practice problems.  So try MATHCOUNTS state and national problems, and then when you master those, move on to the harder AMC 10 problems.

Like the answer to just about every question similar to yours, practice. Practice anything that is around your level (or a little above) and understand the solution. You will get better and better if you just practice. Also just wondering if it should be what should I do now? I'm not very good with grammar.

ernie, the difficulty in the AMC's have not changed as a result of the calculator rule--this was just a measure taken to prevent high-tech vs. low-tech calculator issues and unfairness. I'd say AMC10 should be around MATHCOUNTS National level, too. Handbooks (though the easy problems can get annoying, the hard problems are good) and AHSME's/AMC12's are good practice, as well. There are many other good competitions, too, like Purple Comet, as well.

I would do harder than AMC10 problems. Then, if some unexpectedly hard problem crops up, you can ace it :). And you should always study harder than your goal; just in case. But to be exact, what are you aiming for in AMC10? Perfect score, AIME qualification, >90..?

He scored 19 on AMC 8...

If AMC 10 problems are too challenging, try State level MATHCOUNTS problems which are harder than AMC 8 but definitely easier than AMC 10 and National level MATHCOUNTS.  If those are still much too challenging practice on old AMC 8's and AJHSME's, as well as Chapter and School level MATHCOUNTS tests.

I don't think there is any cut-off to take the AMC10/12. Some schools may issue their own rules, though.

pinkmuskrat, yes - as the tests aren't free, you usually have to get a high score on the AMC 8 for your school to let you take it.


find all pairs of integer m  and n, both greater than 1, such that the following equality holds:
\[ \frac {a^{m \plus{} n} \plus{} b^{m \plus{} n} \plus{} c^{m \plus{} n}}{m \plus{} n} \equal{} \frac {a^{m} \plus{} b^{m} \plus{} c^{m}}{m} \equal{} \frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{n}\]


[size=200][color=red][b]       请教一下余弦定理及其运用。
[/b][/color][/size]
（我懂了．）

大哥，先在国内练练，再到国外来吧，不会走，焉会跑？

大哥，先在国内练练，再到国外来吧，不会走，焉会跑？

（文言中＂走＂即＂跑＂！）

一头雾水中……你们在说什么...

[quote="shobber"]善哉善哉......
我看我们单开出一个帖子讨论中国古典文学罢!
 :rotfl:[/quote]
君之見甚佳  :rotfl:


Noah was walking down Elm St. He then made 2 left turns and 3 right turns and is now facing east. Every turn is at a 90˚ angle. What direction was Noah originally going on Elm St.?

2 left+3 right=1 right.

If one right turn results in facing east, Noah was originally facing $ \boxed{\text{north}}$.

2 lefts, and 3 rights makes it one right.
So if he ends up at east, he was 90 deg to the left, which is NORTH

2 lefts and 3 rights=1 right
he goes  east, bcuz he was 90 degrees to the left, and so the answer is [b]NORTH[/b]


Prove or Disprove

If $ v_1, v_2, \cdots, v_k$ span the subspace $ V$ of $ R^n$ and $ w_1, w_2, \cdots w_l$ span the subspace $ W$ of $ R^n$, then all of the vectors
$ v_1, v_2, \cdots, v_k, w_1, w_2, \ldots, w_l$

span the subspace $ V \plus{} W \equal{} \{ v \plus{} w: v\in V, w\in W\}$.

Let $ z\in V \plus{} W$, then $ z \equal{} v \plus{} w$ for some $ v\in V$ and $ w\in W$, so by hypothesis, we have $ v \equal{} \sum_{i \equal{} 1}^k \alpha_i v_i$ and $ w \equal{} \sum_{j \equal{} 1}^l \beta w_j$, so $ z \equal{} \sum_{i \equal{} 1}^k \alpha_i v_i \plus{} \sum_{j \equal{} 1}^l \beta w_j$ which is a linear combination of the set in question, and thus it is a spanning set.

V+ W is the subspace spanned by the union

From that you also have $ dim(V\plus{}W) \equal{} dim(V) \plus{} dim(W) \minus{} dim (V \cap W)$.


Let $d(a)$ be number of positive divisors of integer $a$. Find all positive integers $k$ for which there exists such $n$, that $d(n^{2}) = kd(n)$

see [url=http://www.kalva.demon.co.uk/imo/isoln/isoln983.html]IMO 1998/3[/url] or http://www.mathlinks.ro/Forum/viewtopic.php?p=124439#p124439


Find all prime numbers such that $ C^{p}_{2p} \minus{} 2$ can be divided by $ p^{4}$.

\[ p^4|C_{ap}^{bp}\minus{}C_a^b,p\not |a(b(a\minus{}b)\] if and only if $ p|B_{p\minus{}3}$ - Wolstenholm's prime.
Known only 2 that primes $ p<10^9$.

[quote="Rust"]
\[ p^4|C_{ap}^{bp} \minus{} C_a^b,p\not |a(b(a \minus{} b)\]
if and only if $ p|B_{p \minus{} 3}$ - Wolstenholm's prime.
Known only 2 that primes $ p < 10^9$.[/quote]
Can we find a way to prove that it's right if and only if p=2?
And how do you prove "only 2 that primes $ p < 10^9$?


How do we convert an inequality into the form $ \sum (something)^2\geq0$? I've seen so many people in the forum do this and but the terms inside that (something) are so weird that i hve always wondered how do they arrive at that result. Is it something that is apparent or is it pure skill? I would also like to learn these techniques(if any)...

Most probably [b]arqady[/b] knows answer :)  I think that, about month ago, same question was asked.

hell_ever, give me an example of an inequality. I'll try to help you.

[b]An example[/b] If $ a,b,c,d$ are nonnegative real number, then

$ 4(a\plus{}b\plus{}c)(8a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} 8d^{2} \plus{} 24ab \plus{} 24ac \plus{} 26ad \plus{} 4bc \plus{} 6bd \plus{} 6cd)^2$

$ \geq 27a[3(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2}) \plus{} 10(ab \plus{} bc \plus{} ac \plus{} bc \plus{} bd \plus{} cd)]^2$.

See: http://www.mathlinks.ro//viewtopic.php?t=116350

Arqady, an example - your fifth post here
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177947[/url]

How did you think of doing what you did??

and Ji Chen, i don't understand how you got the result you posted in your link..

You are talking about S.O.S. (Sum of squares); it has been posted several times in the forum. However the best link is this from hungkhtn:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=80127[/url]

And still, isn't this in the wrong forum?

[quote="delegat"][quote]give me an example of an inequality[/quote]

Your solution to: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=181463[/url][/quote]
[quote="arqady"][quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2}\le \frac 1{2}$.[/quote]
Turned out easy:
$ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2}\le \frac 1{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^6 \minus{} 4a^5b \plus{} 13a^4b^2 \minus{} 2a^4bc \minus{} 6a^3b^3 \minus{} 12a^3b^2c \plus{} 10a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(2c^4 \plus{} 2(a^2 \minus{} 4ab \plus{} b^2)c^2 \plus{} a^4 \minus{} 2a^3b \plus{} 4a^2b^2 \minus{} 2ab^3 \plus{} b^4)\geq0,$
which true because $ (a^2 \minus{} 4ab \plus{} b^2)^2 \minus{} 2(a^4 \minus{} 2a^3b \plus{} 4a^2b^2 \minus{} 2ab^3 \plus{} b^4) \equal{}$
$ \equal{} \minus{} (a \minus{} b)^2(a^2 \plus{} 6ab \plus{} b^2)\leq0.$[/quote]
$ \sum_{sym}(a^6 \minus{} 4a^5b \plus{} 13a^4b^2 \minus{} 2a^4bc \minus{} 6a^3b^3 \minus{} 12a^3b^2c \plus{} 10a^2b^2c^2)\equal{}$
$ \equal{}\sum_{cyc}(2a^6 \minus{} 4a^5b\minus{}4a^5c \plus{} 13a^4b^2 \plus{}13a^4b^2\minus{} 4a^4bc \minus{}$
 $ \minus{}12a^3b^3 \minus{} 12a^3b^2c\minus{}12a^3b^2c \plus{} 20a^2b^2c^2)\equal{}$
$ \equal{}\sum_{cyc}(2a^6\minus{}a^5b\minus{}a^5c)\minus{}3\sum_{cyc}(a^5b\plus{}a^5c\minus{}a^4b^2\minus{}a^4c^2)\plus{}$
$ \plus{}2\sum_{cyc}(a^4b^2\plus{}a^4c^2\minus{}2a^4bc)\plus{}6\sum_{cyc}(a^4b^2\plus{}a^4c^2\minus{}2a^3b^3)\plus{}$
$ \plus{}2\sum_{cyc}(a^4b^2\plus{}a^4c^2\minus{}a^3b^2c\minus{}a^3c^2b)\minus{}10\sum_{cyc}(a^3b^2c\plus{}a^3c^2b\minus{}2a^2b^2c^2)\equal{}$
$ \equal{}\sum_{cyc}(a^6\minus{}a^5b\minus{}ab^5\plus{}b^6)\minus{}3\sum_{cyc}(a^5b\minus{}a^4b^2\minus{}a^2b^4\plus{}ab^5)\plus{}$
$ \plus{}2\sum_{cyc}(c^4a^2\minus{}2c^4ab\plus{}c^4b^2)\plus{}6\sum_{cyc}(a^4b^2\minus{}2a^3b^3\plus{}a^2b^4)\plus{}$
$ \plus{}2\sum_{cyc}(a^4c^2\minus{}a^3c^2b\minus{}b^3c^2a\plus{}b^4c^2)\minus{}10\sum_{cyc}(a^3c^2b\minus{}2a^2b^2c^2\plus{}b^3c^2a)\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)^2(a^4\plus{}a^3b\plus{}a^2b^2\plus{}ab^3\plus{}b^4)\minus{}3\sum_{cyc}(a\minus{}b)^2(a^2\plus{}ab\plus{}b^2)ab\plus{}$
$ \plus{}2\sum_{cyc}c^4(a\minus{}b)^2\plus{}6\sum_{cyc}(a\minus{}b)^2a^2b^2\plus{}$
$ \plus{}2\sum_{cyc}(a\minus{}b)^2(a^2\plus{}ab\plus{}b^2)c^2\minus{}10\sum_{cyc}(a\minus{}b)^2abc^2\equal{}$
$ \equal{}\sum_{cyc}(a \minus{} b)^2(2c^4 \plus{} 2(a^2 \minus{} 4ab \plus{} b^2)c^2 \plus{} a^4 \minus{} 2a^3b \plus{} 4a^2b^2 \minus{} 2ab^3 \plus{} b^4)\geq0.$


Let ABC be a right-angled triangle with angle B is right and AB=2BC. Let D on side AC such that BC=CD, E on side AB such that AD=AE. Prove that $ AD^2\equal{}AB.BE$

Just a straightforward computation.
If BC=x, AB=2x, then by Pythag $ AC \equal{} \sqrt {5}x$, so $ AD \equal{} (\sqrt {5} \minus{} 1)x$ and $ BE \equal{} (2 \minus{} (\sqrt {5} \minus{} 1))x \equal{} (3 \minus{} \sqrt {5})x$. But it is an identity that $ (3 \minus{} \sqrt {5})x2x \equal{} (\sqrt {5} \minus{} 1)^2x^2$

yes，just computation can finish the work


quadratics
geometry
3D geometry
number theory proposed
number theory

It is like this year Chinese equation.
There was long solution and many cases. And no one nice idea!

[quote="ciprian"]Solve in N: $7^x+1=3^y+5^z$. :D (hope you have a short solution)[/quote]

x=y=z=1 is a solution visibly.OTHER SOLUTIONS?X=y=z=0.DONT know how to show for other solutions FO RTHE MOMENT BEEING;

I think the Chinese problem was much easier. This problem has no nice idea too. In my solution imuseed congruences mod 3,5,7,8,10, 13 and 43 :D. I finished the problem using quadratic reciprocity. :D Not a contest problem i think.

I disagree somewhat. We can proceed logically by using residues mod 3,5 and 7, as well as powers of two. After that, we just need the old idea of using residues mod 9 when dealing with cubes, especially since 3 appears in the problem. My solution is still pretty long, but fairly straightforward.

Can somebody post a full solution. I tried to finish the problem like u wrote, but i just dont see where to use quadratic reciprocals. Please give me at least more information what to look.

[quote="ciprian"]Solve in N: $ 7^x \plus{} 1 \equal{} 3^y \plus{} 5^z$. :D (hope you have a short solution)[/quote]

already posted ! see here:  http://www.mathlinks.ro/viewtopic.php?t=183251


inequalities
symmetry
inequalities unsolved

[quote="shao3wangdi"]If x,y,z>0 and xyz=1 prove that  4(ab+bc+ca)^2-(a+b+c)^2(ab+bc+ca)+5(a+b+c)>=3(a+b+c)(ab+bc+ca)-3[/quote]
I suppose you meant $ a,b,c > 0$, $ abc \equal{} 1$ prove that $ 4(ab \plus{} bc \plus{} ca)^2 \minus{} (a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca) \plus{} 5(a \plus{} b \plus{} c) \le 3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 3$
:)

Let $ 3u \equal{} a \plus{} b \plus{} c, 3v^2 \equal{} ab \plus{} bc \plus{} ca, w^3 \equal{} abc$. Then $ w^3 \equal{} 1$ and we have to prove:
$ \minus{} 36v^4 \plus{} 27u^2v^2 \minus{} 15u \plus{} 27uv^2 \minus{} 3\ge 0 \iff$
This is clearly concave in $ v^2$ so fixing $ u,w^3$ we only have to check for maximum and minimum of $ v^2$, which is reached only when two of $ a,b,c$ are equal.([url=http://www.mathlinks.ro/viewtopic.php?t=278791]Because[/url]) Assume wlog because of symmetry $ a \equal{} b \equal{} t$, then $ c \equal{} t^{ \minus{} 2}$.
Inserting we should prove:
$ 3t^8 \minus{} 7t^6 \plus{} 6t^5 \minus{} 6t^3 \plus{} 3t^2 \plus{} 1 \ge 0 \iff$
$ (t \minus{} 1)^2(3t^6 \plus{} 6t^5 \plus{} 2t^4 \plus{} 4t^3 \plus{} 6t^2 \plus{} 2t \plus{} 1) \ge 0$ Which is true :)

[quote="shao3wangdi"]If x,y,z>0 and xyz=1 prove that  4(ab+bc+ca)^2-(a+b+c)^2(ab+bc+ca)+5(a+b+c)>=3(a+b+c)(ab+bc+ca)-3[/quote]

$ abc\equal{}1$

$ p\equal{} a\plus{}b\plus{}c$ , $ q \equal{} ab\plus{}bc\plus{}ca$

$ f(q)\equal{}4q^2 \minus{}(3\plus{}p)pq\plus{}(5p\plus{}3) \le 0$ , $ ( f(q)$ - convex  by $ q  )$

$ p^2 \ge 3q\ge 9$ and by Shur : $ q \le q_1\equal{}\frac{1}{4}(p^2\plus{}\frac{1}{p})$;

we must prove that $ f(3) \le0$ and $ f(q_1) \le 0$

$ f(3) \equal{} \minus{}(p\minus{}3)(3p\plus{}13) \le 0$ , ( $ p \ge 3$)

$ f(q_1)\equal{} \minus{}3p^3\plus{}29p\minus{}15\plus{}81\frac{1}{p^2} \le 0$ , ( $ p \ge 3$)

[quote="shao3wangdi"]And  now I have another Inequality,can you help me?Let  $ a,b,c>0$  ,$ abc\equal{}1$ and prove that $ 4(a\plus{}b\plus{}c)\plus{}a^3(b\plus{}c)\plus{}b^3(c\plus{}a)\plus{}c^3(a\plus{}b)\ge 3ab(a\plus{}b)\plus{}3bc(b\plus{}c)\plus{}3ca(c\plus{}a)$[/quote]
 Latexed

[quote="shao3wangdi"]And  now I have another Inequality,can you help me?Let  a,b,c>0  ,abc=1 and prove that 4(a+b+c)+a^3(b+c)+b^3(c+a)+c^3(a+b)>=3ab(a+b)+3bc(b+c)+3ca(c+a)[/quote]
It's equivalent to: $ 12u \plus{} 27u^2v^2 \minus{} 18v^4\minus{}3uw^3 \ge 27uv^2\minus{} 9$, so you can use the same method as I've shown you in the first post ;)


The roots of the equation $ 2\sqrt {x} \plus{} 2x^{ \minus{} \frac {1}{2}} \equal{} 5$ can be found by solving:

$ \textbf{(A)}\ 16x^2 \minus{} 92x \plus{} 1 \equal{} 0 \qquad \textbf{(B)}\ 4x^2 \minus{} 25x \plus{} 4 \equal{} 0 \qquad \textbf{(C)}\ 4x^2 \minus{} 17x \plus{} 4 \equal{} 0 \\
\textbf{(D)}\ 2x^2 \minus{} 21x \plus{} 2 \equal{} 0 \qquad \textbf{(E)}\ 4x^2 \minus{} 25x \minus{} 4 \equal{} 0$

[hide="Click for solution"]
$ 2\sqrt{x}\plus{}\frac{2}{\sqrt{x}}\equal{}5$.  Squaring, we obtain $ 4x\plus{}8\plus{}\frac{4}{x}\equal{}25$.  Multiply both sides of the equation by $ x$ and we get $ 4x^2\minus{}17x\plus{}4\equal{}0$, or $ \boxed{\textbf{(C)}}$.
[/hide]

I suppose in the problem there was a plus sign between $ 2\sqrt{x}$ and $ \dfrac{2}{\sqrt{x}}$?

hello, i have an other solution, multipliying the given equation by $ \sqrt{x}$ and we get $ 2x\plus{}2\equal{}5\sqrt{x}$, setting
$ \sqrt{x}\equal{}t$ we have the equation $ 2t^2\minus{}5t\plus{}2\equal{}0$ with solutions $ t_1\equal{}2$ and $ t_2\equal{}\frac{1}{2}$, so the equation above has the solutions $ x_1\equal{}4$ or $ x_2\equal{}\frac{1}{4}$. I have found the solutions of our equation without squaring both sides.
Sonnhard.


calculus
integration
trigonometry
logarithms
calculus computations

\[ \int\frac{1 \plus{} sinx}{1 \plus{} sin2x} dx\]

Hint:

$ I \equal{} \int\frac {1 \plus{} sinx}{1 \plus{} sin2x} dx$

$ I \equal{} \int\frac {1}{1 \plus{} sin2x}dx \plus{} \int\frac {sinx}{1 \plus{} sin2x}dx$

Let $ tanx \equal{} t$

$ Hence,dx \equal{} \frac {dt}{1 \plus{} t^2}$ and

$ sin2x \equal{} \frac {2t}{1 \plus{} t^2} , sinx \equal{} \frac {t}{\sqrt {1 \plus{} t^2}}$

$ I \equal{} \int\frac {1}{1 \plus{} \frac {2t}{1 \plus{} t^2}} \frac {dt}{1 \plus{} t^2} \plus{} \int\frac {\frac {t}{\sqrt {1 \plus{} t^2}}}{1 \plus{} \frac {2t}{1 \plus{} t^2}}\frac {dt}{1 \plus{} t^2}$

[quote="tariq1440"]$ I\equal{}\int\frac {1 \plus{} \sin x}{1 \plus{} \sin 2x} dx$ .[/quote]
[b][u]Elementary level[/u].[/b] Denote $ W\equal{}\int\frac {1}{1\plus{}\sin 2x}\ \mathrm {dx}$ , $ U\equal{}\int \frac {\sin x}{1\plus{}\sin 2x}\ \mathrm{dx}$ , $ V\equal{}\int\frac {\cos x}{1\plus{}\sin 2x}\ \mathrm {dx}$ . Se observa ca $ I\equal{}W\plus{}U$ .

$ W\equal{}\int \frac {\sin^2x\plus{}\cos^2x}{(\sin x\plus{}\cos x)^2}\ \mathrm {dx}\equal{}$ $ \int\frac {1\plus{}\tan^2x}{(1\plus{}\tan x)^2}\ \mathrm{dx}\equal{}$ $ \int\frac {(1\plus{}\tan x)'}{(1\plus{}\tan x)^2}\ \mathrm {dx}\equal{}\minus{}\frac {1}{1\plus{}\tan x}\plus{}\mathrm C\implies \boxed {\ W\equal{}\minus{}\frac {\cos x}{\sin x\plus{}\cos x}\plus{}\mathrm C\ }$ .

$ V\minus{}U\equal{}\int\frac {\cos x\minus{}\sin x}{1\plus{}\sin 2x}\ \mathrm {dx}\equal{}$ $ \int\frac {(\sin x\plus{}\cos x)'}{(\sin x\plus{}\cos x)^2}\ \mathrm {dx}\equal{}$ $ \minus{}\frac {1}{\sin x\plus{}\cos x}\plus{}\mathrm C\implies \boxed {\ V\minus{}U\equal{}\minus{}\frac {1}{\sin x\plus{}\cos x}\plus{}\mathrm C\ }$ .

$ V\plus{}U\equal{}\int\frac {\sin x\plus{}\cos x}{(\sin x\plus{}\cos x)^2}\ \mathrm{dx}\equal{}\int\frac {1}{\sin x\plus{}\cos x}\ \mathrm {dx}\equal{}\int\frac {\left(x\minus{}\frac {\pi}{4}\right)'}{\sqrt 2\cdot \cos\left(x\minus{}\frac {\pi}{4}\right)}\ \mathrm{dx}$ . Since $ \int\frac {1}{\cos t}\ \mathrm{dt}\equal{}\frac 12\cdot\ln\left|\frac {1\plus{}\sin t}{1\minus{}\sin t}\right|\plus{}\mathrm C$

obtain $ \boxed {\ V\plus{}U\equal{}\frac {\sqrt 2}{4}\cdot\ln\left|\frac {1\plus{}\sin \left(x\minus{}\frac {\pi}{4}\right)}{1\minus{}\sin\left(x\minus{}\frac {\pi}{4}\right) }\right|\plus{}\mathrm C\ }$ a.s.o.


[hide]
If we factor out a $2$ from the set of numbers in the numerator, we get the same set as the denominator. Their LCMs are then the same. And since LCM(2m, 2n) = 2 LCM (m, n), the expression equals 2.
[/hide]

Sweet, I didn't know you could factor something like that. Are there cases where it doesn't work?

Nope, works for any case. You can also do it with GCD.


Find all sets of three consecutive of even integers whose sum is between 168 and 180.

if the middle even integer is $x$, then the smallest is $x-2$ and the biggest is $x+2$.
thus, to satisfy the first number,

\[ x-2 + x + x+2 = 168 \]
\[ 3x = 168 \]
\[ x = 56 \]

so the first set of numbers are ${54, 56, 58}$.

to satisfy the other number,

\[ x-2 + x + x+2 = 180 \]
\[ 3x = 180 \]
\[ x = 60 \]

so the last set of numbers are ${58, 60, 62}$.

the sets between them is ${56,58,60}$.

[quote="buzzer11"]i don't quite understand what you did... :huh:[/quote]

He found the largest possilbe combinations and the smallest, and then saw if there were any between them.


percent geometry LaTeX AMC USA(J)MO USAMO

If a plane is infinetely filled with circles, what percent of the plane is circles? :?:

All of the plane. First large circles are drawn, and then smaller and smaller circles, filling up all of the spaces. With infinitely many circles this is possible.

Edit: Never mind how I did the problem, I did it wrong.

no, but that would be a good question all circles are the same size

it is easy to see that the most space is taken up when all the circles are tangent to each other.  The arrangement should be so that any three centers of mutually tangent circles, when connected, form an equilateral triangle.  The square case obviously doesn't work if you calculate the area in between the four circles.

This gives that the area of the region in between the three circles (assuming that they are all unit, but it doesn't matter) is $ \sqrt {3} \minus{} \frac {\pi}{2}$. This takes up $ \frac {2\sqrt {3} \minus{} \pi}{2\sqrt {3}}*100$% of the area of one equilateral triangle.
Since there are infinitely many, this should be the answer. =P

Edit: Fixed Latex, I hope

Hahahaha, wow, I'm stupid. Yeah, Mathcrazed is right.

[quote="mathcrazed"]it is easy to see that the most space is taken up when all the circles are tangent to each other.  The arrangement should be so that any three centers of mutually tangent circles, when connected, form an equilateral triangle.  The square case obviously doesn't work if you calculate the area in between the four circles.

This gives that the area of the region in between the three circles (assuming that they are all unit, but it doesn't matter) is $ \sqrt {3} - \frac {\pi}{2}$. This takes up $ \frac {2\sqrt {3} - \pi}{2\sqrt {3}*100%}$ of the area of one equilateral triangle.
Since there are infinitely many, this should be the answer. =P[/quote]

Whenever you have a percent symbol (%), you have to backslash it (\%). This is because the percent symbol is a comment indicator in $ \text{\LaTeX}$.

Anyway, the "unparseable" thing was supposed to be
$ \frac {2\sqrt {3} - \pi}{2\sqrt {3}*100}\%$

I fixed it... and it wasn't supposed to be that... 100 isn't in the denominator =P

odd this was like an USAMO problem from last year...

No it wasn't... usamo = can u cover all lattice points.


geometry
3D geometry
sphere
trigonometry
geometry unsolved

I am having problems getting this question started, any help is appreciated.

Let Triangle ABC be a right triangle on the unit spere S and with right angle at C. Proove that:

sin(|Tri ABC|) = [sin(a)*sin(b)]/[1+ cos(a)*cos(b)]

where sin(|Tri ABC|) means sin(area of triangle ABC)

What is the analogous result in euclidian geometry?

thanks in advance

Should you say the [u]numerical[/u] area of the sperical triangle?  Areas are expressed in square linear units.  Am I being picky?

I'll solve this a step at a time.

The area of any spherical triangle is equal to the spherical excess (relative to a plane triangle) which is (A+B+C)-pi.
Since C is 90deg, the area is equal to A+B-pi/2.

By substituting the left side of the equation can be expanded and then simplified to sin(A)sin(B)-cos(A)cos(B).

I'll probably try using the cosine rule for side for sides or another formula to state A, B, and C in terms of a, b, and c.
Note: Mathworld has some great stuff at [url=http://mathworld.wolfram.com/SphericalTrigonometry.html]http://mathworld.wolfram.com/SphericalTrigonometry.html[/url]

(Sorry about criticizing you on the units; I just realized myself that spherical triangles are very strange when it comes to their measurements and equations.)

By the way, I hope I'm assuming right when I say that A, B, and C are the spherical angles and a, b, and c are the spherical sides.


vector
linear algebra
matrix
abstract algebra
geometry
Galois Theory
superior algebra

Do you have any books in mind that might deal with abstract algebra for IMC preparation????
Thank you for your interest........

: Linear Algebra
:: Finite-Dimensional Vector Spaces, by Paul Halmos
:: Linear Algebra, by Serge Lang
:: Linear Algebra: Challenging Problems for Students, by Fuzhen Zhang
:: Linear Algebra Problem Book, by Paul Halmos
:: Matrix Analysis, by Horn and Johnson

: General Algebra
:: Algebra, by Thomas Hungerford
:: Abstract Algebra, by Dummit and Foote
:: Algebra, by Birkhoff and MacLane
:: Basic Algebra I, by Jacobson
:: Basic Algebra II, by Jacobson

: Field and Galois theory
:: Lectures in Abstract Algebra: Field and Galois Theory, by Jacobson
:: Field and Galois Theory, by Morandi
:: Galois Theory, by Escofier

: Commutative Ring and Module Theory
:: Commutative Ring Theory, by Matsumura
:: Commutative Algebra: with a View Toward Algebraic Geometry, by Eisenbud
:: Introduction to Commutative Algebra, by Atiyah and MacDonald
:: Commutative Algebra I, by Zariski and Samuel
:: Commutative Algebra II, by Zariski and Samuel
:: Exercises in Classical Ring Theory, by Lam
:: Exercises in Modules and Rings, by Lam

Is my list of algebra books. All are very decent. However, I'm not saying all topics will appear on the IMC...

You can log in http://www.bookilook.com to find some books, and then download files easily. Good luck.


function
induction
number theory
greatest common divisor
group theory
abstract algebra
algebra unsolved

Find all functions $f(x,y):\mathbb{Q}\to\mathbb{Q}$ satisfying the condition:
a)$ f(0,0)=0$
b)$ f(x,y)=f(y,x)$
c)$ f(x,f(y,z))=f(f(x,y),z)$
d)$f(x+z,y+z)=f(x,y)+z \quad  \forall x,y,z\in \mathbb{Q}$

The solutions appear to be $\min(x,y)$ and $\max(x,y)$.

There is only two solution, namely $f(x,y)=\min\{x,y\}$ and $f(x,y)=\max\{x,y\}$. Denote $g(x)=f(x,0)=f(0,x)$
So, the following equalities take place:

$(0)$ $g(0)=0$
$(1)$ $g(g(x))=g(x)$ (Put $y=z=0$ in $c)$)
$(2)$ $g(x)=g(-x)+x$ (Turn $(x,y,z)$ into $(0,-x,x)$ in $d)$)
$(3)$ $g(g(x)-y)+y=g(g(y)-x)+x$ (Let $y=0$ in $c)$)

Let $\mathbb{A}$ be the set of the fixed points of $g$. In a view to $(1)$ we get that $\mathbb{A}=g(\mathbb{Q})$
The key observation is $\mathbb{A}=\mathbb{Q}_+\cup\{0\}$ or $\mathbb{A}=\mathbb{Q}_-\cup\{0\}$ but it is not so obvious, at least it
was for me. Firstly, put $y=-g(x)$ in $(3)$, therefore

$(4)$ $g(g(x)+g(y))=g(x)+g(y)$. Hence, if $a,b\in\mathbb{A}$ then $a+b\in\mathbb{A}$.

Furthermore, putting $y=kg(x)$ for every positive integer less that $n$ we establish $g(ng(x))=ng(x)$ over induction, since $k=1$ is true.
Secondly, $g(\mathbb{Q})\neq\{0\}$, for if there were the equality $(2)$ wouldn't hold for non-zero rational numbers.

Let $h:\mathbb{Q}\mapsto\mathbb{N}$ be a fuction that $h(x)$ is the least positive for which $xh(x)$ is an integer. This is only for easier use.

Assume that there are $a,b\in\mathbb{A}$ different in signs. Then there exist positive integer $N_0\in\mathbb{A}$ and negative integer $M_0\in\mathbb{A}$, since $g(h(a)a)=h(a)a$ and $g(h(b)b)=h(b)b$.

Define $N=\min\{\mathbb{N}\cap\mathbb{A}\}$ and $M=\max\{(-\mathbb{N})\cap\mathbb{A}\}$. Since $g(N)=g(-N)+N$, $M+N\neq0$. Due to $(4)$ $M+N\in\mathbb{A}$ but that contradicts to minimality of $N$ or maximatlity of $M$.
Thus, $g\geq0$ or $g\leq0$. Assume that $g(x)\geq0$ for every rational number $x$.

Put $y=0.5g(x)$ in $(3)$, so $g(0.5g(x))+0.5g(x)=g(g(0.5g(x))-0.5g(x))+0.5g(x)$ or $0.5g(x)=g(0.5g(x)-g(0.5g(x)))$. That implies that

$(5)$ if $a\in\mathbb{A}$ then $0.5a\in\mathbb{A}$.

set $y=h(g(x))g(x)$ in $(3)$. Then $g(-(h(g(x))-1)g(x))+h(g(x))g(x)=g(h(g(x))g(x)-x)+x$. Since $(h(g(x))-1)g(x)\in\mathbb{A}$ then $g(-(h(g(x))-1)g(x))=0$ due to $(2)$. Therefore $g(h(g(x))g(x)-x)=h(g(x))g(x)-x$.
That implies that for every positive integer $n$ there is a fraction in $\mathbb{A}$ with $n$ as denominator written in the lowest terms. Denote $u:\mathbb{N}\mapsto\mathbb{N}$ where $u(n)$ is the least positive integer $r$ such that $\frac{r}{n}\in\mathbb{A}$.
Our goal is to show that $u\equiv 1$. First of all, $u(n)$ is odd, since $\frac{u(n)}{2n}\in\mathbb{A}$.

Let us say that there exist positive integer $k$ not multiple of $u(n)$ such that $\frac{k}{n}\in\mathbb{A}$. Take the least one. Then $k$ is odd. However, $u(n)+k$ is even and $u(n)<\frac{u(n)+k}{2}<k$. Since $(4)$ and $(5)$ hold, $\frac{u(n)+k}{2n}$ is in $\mathbb{A}$.
Contradiction to the minimality of $k$. So, if $\frac{r}{n}\in\mathbb{A}$ then $r$ is multiple of $u(n)$. In a view to $g(-(h(g(x))-1)g(x))+h(g(x))g(x)=g(h(g(x))g(x)-x)+x$ we can conclude that $n$ and $u(n)$ are coprime, just set $x=n^{-1}$.

Since $\frac{u(n\cdot u(n))}{n\cdot u(n)}$ is in $\mathbb{A}$, $u(n)\cdot \frac{u(n\cdot u(n))}{n\cdot u(n)}=\frac{u(n\cdot u(n))}{n}$ is in $\mathbb{A}$. Therefore, $u(n\cdot u(n))$ is a multiple of $u(n)$. But we establisted $\text{GCD}(u(n),n\cdot u(n))=1$, so $u(n)=1$.

That all we need: $\frac{1}{n}\in\mathbb{A}$ for any positive integer $n$, therefore, $\frac{r}{n}\in\mathbb{A}$ for any positive integers $n$, $r$. Hence, $\mathbb{Q}_+\cup\{0\}\subseteq\mathbb{A}\subseteq\mathbb{Q}_+\cup\{0\}$ or $\mathbb{A}=\mathbb{Q}_+\cup\{0\}$. Therefore, $g(\mathbb{Q}_-)={0}$. That implies $g(x)=\max\{x,0\}$.

Change $(x,y,z)$ to $(x-y,0,y)$ in $d)$. So, $f(x,y)=g(x-y)+y=\max\{x-y,0\}+y=\max\{x,y\}$.

Similarly we can show that $f(x,y)=\min\{x,y\}$ in the case $g\leq 0$.

PLEASE, post something elegant. This is so plodding solution that took me about 2 days to solve. I hope that it is correct. :)

Here's another way of looking at the problem:
Let $x*y=f(x,y)$. Then $*$ is a binary operation with the following properties:
(a)$0*0=0$
(b)$*$ is commutative
(c)$*$ is associative
(d)$+$ distributes over $*$

The grubby stuff is probably necessary, but the problem belongs to abstract algebra more than the ordinary theory of functional equations. The additive structure of the rationals is essential (every finite set is contained in a subgroup generated by one element).

To uTOPi_a: I don't think your solution is plodding  ;)
jmerry, you're right. Like this:
Denote $ r(x)=0*x $.We have $S_n = 0*y*2y*....*ny = nr(y)$(proof by induction).Consider $S_{2n}:S_{2n} = 2nr(y)$ and $S_{2n}= nr(y) + r(ny)$. Thus,nr(y)=r(ny) for all n in $N^*$,y in Q
It's easy to verify that:r(x) :!=: 0 ; r(r(x)) = r(x).Thus, there exists z: r(z) = z
If z>0 we have $f(x,y) = max (x,y)$
If z<0 we have $f(x,y) = min (x,y)$


how do you graph something like $ |x \minus{} 1| \plus{} |y \plus{} 1| \equal{} 5$

The process of graphing this takes some time:

$ |x\minus{}1|\plus{}|y\plus{}1|\equal{}5$

Rearrange it:

$ |y\plus{}1|\equal{}\minus{}|x\minus{}1|\plus{}5$

Now pick a value for $ x$, let's say $ 1$:

$ |y\plus{}1|\equal{}\minus{}|1\minus{}1|\plus{}5$

So:

$ |y\plus{}1|\equal{}5$

Now solve for $ y$

$ y\plus{}1\equal{}5$ or $ y\plus{}1\equal{}\minus{}5$

Therefore the possible values of $ y$ when $ x\equal{}1$ are $ 4$ and $ \minus{}6$

If Im correct, the graph should be like a sideways "v" opening to the left.

It's usually a quadrilateral; graph the 4 different parts of it based on the signs of the absolute values.


calculus
integration
calculus computations

hello, we have after a short calculation $ \int\frac{dx}{x\minus{}\sqrt{x^2\plus{}2x\plus{}4}}\,dx\equal{}\minus{}\frac{1}{2}\int\frac{x}{x\plus{}2}\,dx\plus{}\int\frac{\sqrt{x^2\plus{}2x\plus{}4}}{\minus{}2x\minus{}4}\,dx$, the first integral is easy, by second one we need the so called Euler's substitution $ \sqrt{x^2\plus{}2x\plus{}4}\equal{}t\plus{}x$.
Sonnhard.

$ \int \frac {dx} {x \minus{} \sqrt {x^2 \plus{} 2x \plus{} 4}}$
$ \equal{} \int \frac {x \plus{} \sqrt {x^2 \plus{} 2x \plus{} 4}} {(x \minus{} \sqrt {x^2 \plus{} 2x \plus{} 4})(x \plus{} \sqrt {x^2 \plus{} 2x \plus{} 4}) }dx$
$ \equal{} \minus{} \frac {1}{2}\int\frac {x \plus{} \sqrt {x^2 \plus{} 2x \plus{} 4}}{x \plus{} 2}dx$
$ \equal{} [ \minus{} \frac {1}{2}\int\frac {x}{x \plus{} 2}dx] \plus{} [ \minus{} \frac {1}{2}\int\frac {\sqrt {x^2 \plus{} 2x \plus{} 4}}{x \plus{} 2}dx]$
$ \equal{} I_1 \plus{} I_2$
$ I_1 \equal{} \minus{} \frac {1}{2}\int\frac {x}{x \plus{} 2}dx$
$ \equal{} \minus{} \frac {1}{2}\int\frac {(x \plus{} 2) \minus{} 2}{x \plus{} 2}dx$
$ \equal{} \minus{} \frac {1}{2}\int(1 \minus{} \frac {2}{x \plus{} 2})dx$
$ \equal{} \minus{} \frac {1}{2}[x \minus{} 2\ln(x \plus{} 2)] \plus{} C$
$ I_2 \equal{} \minus{} \frac {1}{2}\int\frac {\sqrt {x^2 \plus{} 2x \plus{} 4}}{x \plus{} 2}dx$
$ \equal{} \minus{} \frac {1}{2}\int\frac {\sqrt {(x \plus{} 2)^2 \minus{} 2(x \plus{} 2) \plus{} 4}}{x \plus{} 2}dx$
Who can tell me how to solve $ I_2$?Thank you!

I see It.
$ I_2 \equal{} \minus{} \frac {1}{2}\int\frac {\sqrt {x^2 \plus{} 2x \plus{} 4}}{x \plus{} 2}dx$
Let $ x \plus{} 2 \equal{} t$
$ I_2 \equal{} \minus{} \frac {1}{2}\int\frac {\sqrt {t^2 \minus{} 2t \plus{} 4}}{t}dt$
$ \equal{} \minus{} \frac {1}{2}\int\frac {t^2 \minus{} 2t \plus{} 4}{t\sqrt {t^2 \minus{} 2t \plus{} 4}}dt$
$ \equal{} \minus{} \frac {1}{2}\int\frac {t}{\sqrt {t^2 \minus{} 2t \plus{} 4}}dt \minus{} \frac {1}{2}\int\frac { \minus{} 2}{\sqrt {t^2 \minus{} 2t \plus{} 4}}dt \minus{} \frac {1}{2}\int\frac {4}{t\sqrt {t^2 \minus{} 2t \plus{} 4}}dt$


My school had a snow day on the day of the AMC A. So, I need to decide whether to take the 10B or 12B (I'm a 10th grader)

1. Is it required to take the AMC 10B to qualify for the USAMO through the floor? Or is being a sophomore enough?

    http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml says you would need to take the AMC 10B.

    http://www.unl.edu/amc/f-miscellaneous/faq.shtml (bottom of the page) says being a sophomore is enough.

2. Which is the better option?

3. Last year, I got a 127.5 on my AMC 12B and a 4 on my AIME. But it listed my index as 140, probably due to USAMTS. Does the MAA take the top score for the USAMO Index?

[quote="Bictor717"]You can still read the FAQ. I think 3) addresses your concerns well.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693[/quote]
That's from about four years ago, chances are that some of it (possibly even what he's asking about) is outdated.

I personally am worried about this, since I hadn't noticed, and I'm taking both AMC 12s (despite being able to take the 10).

This is why I'm considering the 10B for security.

When in doubt, read the AMC Teachers' Manuals.  For example, look at last year's [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-USAMO-TM/TM-USAMO,2007.pdf]USAMO Manual[/url].  Page 4 makes it clear that 10th graders or below, whether they took the AMC 10 or 12, can qualify under the floor procedure.  The AMC 10 can still give an advantage, because it is more likely to result in a higher USAMO index, which can help for breaking ties in the floor procedure.

It's not just used for breaking ties... as long as you're sure you'll qualify for AIME (or have already done so through USAMTS) I'd definitely suggest the 10. It's just that it's easy to screw up (I know a few people who did) and fall below 120. If you had 127.5 last year I wouldn't be so worried if I were you.

Basically, I think qualifying AIME by AMC 12 is still more advantageous than AMC 10 because in the next round, you will decide the floor score(and the ones who qualify by AMC 12 is considered first, not by AMC 10). If you qualify through USAMTS, how many score you receive(100?)? And do you know how many scores we need to qualify for USAMO(last years I think they get too high: 210 which means at least 10 in AIME).

Last year, the first year of the USAMO expansion, the index was 197.5 - extremely low compared to previous years. It's very hard to say where it'll fall this year, but you could also just get a 15 and not worry about it.