Don't respond until after this friday.
What did u guys think of the problems?

Interesting logic there, telling us not to respond until Friday, but posting on Tuesday - perhaps posting on Friday would have saved a few words.

So this post is ironic in multiple ways.

Just asking, is NEML the one with Big Al and his brothers?

[quote="mathemonster"]why is telling you not to respond and wasting words ironic?[/quote]

my post was pretty ironic

[quote="mathemonster"]Don't respond until after this friday.[/quote]

The NEML website is accepting meet one scores until October 30, so you should wait until then.

how'd you guys do?
Also if u were at RIML ( worcester ) how did u do?

6:XXXXXXXXXXX
5:XXXXXXXXXXXX
4:XXXXXXXXXXXXXX
3:XXXXXXXXXXXXXX
2:XXXXX
1:X

Cute questions, but I could do all six in my head, so our score distribution seems about right.

Today's RIML result was pretty bad, but more or less expected given the preparation of our students.  (I don't have anything to do with preparing our RIML team, but I know that the majority received their round assignments [i]yesterday[/i] and their first practice materials [i]today.[/i]  I went as our second coach and was asked during the bus ride, by someone on team A, what "mean" means.  Ouch.)

yea it was trivial. RIML i think i was the only 18, which is pretty weird. Everyone seems to think it was a tough meet though.
a nice problem-
$ 4x^3 \minus{} 3x^2 \plus{} k \equal{} 0,$ the $ 3$ soln's of this equation form an arithmetic sequence find $ k$.
Oh Mr. Yanco, could I go to PUMAC with Western Mass this year? You can put me on whatever team u want.

I'm not coaching Western Mass this year.  You could contact Mrs. Blumberg to see whether she's bringing a team (either as Western Mass or as Evil Geniuses).  You should have her email address from the WMass ARML mailing list, but if you don't have it, send me a PM and I'll give it to you.

We are bringing a Worcester Academy team to PUMaC, but we'll be using our minibuses which only seat 14 students each.  That means either one team of eight or I'd need to find a second minibus driver on our faculty (actually I already have two volunteers but I don't have 16 students interested because of the awkward date).

HMNT on an SAT day is pretty awkward too.

[quote="mathemonster"]a nice problem-
$ 4x^3 \minus{} 3x^2 \plus{} k \equal{} 0,$ the $ 3$ soln's of this equation form an arithmetic sequence find $ k$.[/quote]

<intruding on Massachusetts forum>

Nice if you haven't seen it before, but unfortunately this problem is pretty well-known.

which I hadn't so I had to solve it in like 2 minutes.

My school did badly on this one, I was the only 6.

Yeah, I also thought it was harder than the last one - I was only one to ace at my school (middle school though :))

Last time we had a couple perfects.

Obviously you don't know what NEML stands for and is jealous of Massachusetts' awesomeness

NEML/NJML/NYML/CAML/... all have the same problems.

Online score submission ends November 27.  You should probably wait until Saturday to discuss this.

hey mathemonster, where is that avatar from?

Oh wait a sec...

For the first contest my name said
Surya [u]A[/u]. Lexington, MA

and for the second contest it said my name was
Surya [u]B[/u]. Lexington, MA

So, they might've thought I'm two different people, and my name isn't on the cumulative high scores, either... :(

Is there anyway to fix this? [B is correct]


Ok the problem goes like "A regular nine-sided polygon is inscribed in a circle of raidus 13 as shown. Determine the area of the shaded region."

This is relatively simple i did 360/9 = 40 degree, then 40/2=20degree so now i am forming a 20/90/70degree angle. So to  find the area of this one small triangle i do (.5(13tan(20degree))13) which is using the half*base*height. So now i time that result by 2 to find the area of one of the 9 angles and then multiple that by 9.

end result i get is 9(13tan(20degree)13)

but this program says I am wrong and that the answer is 9(.5(13)squared*sin(40degree)) and i am like WTC?!

[quote="Farcus"] So to  find the area of this one small triangle i do (.5(13tan(20degree))13) which is using the half*base*height.[/quote] [hide] The hypotenuse is $13$, so the legs are $13\sin20$ and $13\cos20$. Finding the total area, we get $9\cdot13^{2}\sin20\cos20=\frac{9\cdot13^{2}\sin40}{2}$, from the identity $2\sin\alpha\cos\alpha=\sin2\alpha$ [/hide]


find all integers x,y such that
y  divides  x²+1    and  x² divides  y³+1

It is a famous problem and very useful,also.

Uhh okay...if you say so. But the answer is still the same.

chess64 how you get that $\frac{1}{a^{3}(ay-1)}=1$?

[quote="chess64"]$b$ has to be an integer[/quote]
 you are wrong.you also lose the solution (x,y)=(3,2)


Why do you say  exercice :lol: ,I thinke that will be more interesting if you are going to ask,how many [color=red][b]problems[/b][/color] we solve every day? :D

Do you mean Math exercise of Physical exercise?

I think he means all kinds of excercises...in which case, an uncountable number.

[quote="probability1.01"]I wonder what your numbers mean. I'm assuming they're not hours, as 20+ hours is pretty ridiculous ;).[/quote]

Yes they probably do because many AOPSers do well over 20  problems a day :P

I exercise 24 hours a day.  We all do, without realizing it, not even controlling it.  Living is an exercise of existence.

I don't know if I exercise 24 hrs a day, with all the work and activities, i don't even know if i'm alive or not.  hehe :P


i don't know how to solve this:
Find the sum and the product of the roots (real and complex) of $ x^3 + 3x^2 + 7x - 11 = 0$.

idk where did your LaTeX go wrong...

but it says $ x^3\plus{}3x^2\plus{}7x\minus{}11\equal{}0$.

sum : $ \minus{}\frac{b}{a}$
product : $ \minus{}\frac{d}{a}$

in $ ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}0$

$ a\equal{}1, b\equal{}3, d\equal{}\minus{}11$

sum : -3
product : 11

product of roots is not $ \frac{\minus{}c}{a}$! the product is $ \frac {c}{a} \equal{} \minus{}11$

...actually, it is $ \minus{}\frac{d}{a}$.

when three roots of polynomial $ ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}0$ are $ \alpha, \beta, \gamma$, $ \frac{c}{a}\equal{}\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha$

...so you're saying $ (x \minus{} 1)^3 \equal{} x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1$? wrong!

$ (x \minus{} 1)^3 \equal{} x^3 \minus{} 3x^2 \plus{} 3x \minus{} 1$, a=1 and d=-1.

therefore, $ \minus{} \frac {d}{a} \equal{} \minus{} \frac { \minus{} 1}{1} \equal{} 1$

EDIT : this is cubic equation, not quadratic.

wait is it different for cubics than quadratics??

yes, it is.

quadratic :

$ \alpha\plus{}\beta\equal{}\minus{}\frac{b}{a}$
$ \alpha\beta\equal{}\frac{c}{a}$

cubic :

$ \alpha\plus{}\beta\plus{}\gamma\equal{}\minus{}\frac{b}{a}$
$ \alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha\equal{}\frac{c}{a}$
$ \alpha\beta\gamma\equal{}\minus{}\frac{d}{a}$

etc.

It's obvious that one root is 1.
Then you can have the equation like this: (x-1)(x^2+4x-11)=0
The roots of x^2+4x-11=0 are also the roots of x^3+3x^2+7x-11=0
So R1+R2=-4  R1*R2=-11
Then plus / times the first root 1
So the sum is -4+1=-3
    the product is -11*1=-11

That's it.

ya i thought u were saying for [u]all[/u] polynomials it was -c/a

[quote="mathemonster"]so your saying in the polynomial $ (x \minus{} 1)^3 \equal{} x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1$ the product of the roots is $ \frac { \minus{} d}{a} \equal{} \minus{} 1$ ? wrong! ( i meant d/a above, i just said c/a because i meant for quadratics) Thats interesting that c/a is the product of the roots taken two at a time- i did not know that[/quote]

:rotfl: I saw your icon and was like, what happened to [b]t0rajir0u[/b]? Be careful what you say with that icon, it makes you look a bit more believable ;)

[quote="JZambrano201"]:rotfl: . I thought it was tOrajirOu too, until I saw TZF's post. I couldn't believe that he would get something like that wrong.[/quote]

lol me too  :rotfl:

lol im undermining his credibility-inyhoo i switched my avatar. Sorry all, I realize my mistake. However, fantasylover did not specify that he was talking only about odd degree polynomials, the sum of the roots of a general polynomial is $ ( \minus{} 1)^n a_0/a_n$


calculus
derivative
trigonometry
real analysis
real analysis unsolved

Hi everyone,

Could you tell me if this is correct, please?

The antiderivative of a constante is x
The derivative of X is 1
The derivative of 2x is 2
The antiderivative of f(x) is 0

Thank you very much

[quote="chocolatelover"]The antiderivative of f(x) is 0[/quote]
I recently graded a test paper which claimed that the antiderivative of $ \cos^3 x$ is $ 0$. Too bad that the student did not check the answer by differentiating $ 0$.

To answer your question: No, in general the antiderivative of f(x) is not zero.
The antiderivative of a constant is that constant times x, plus another constant.
The derivatives are correct.


Can an exchange student (who would be staying for one semester only, without social security number) participate in Putnam? I sent an email to maa to ask but have received no reply yet. I want to know about this because I heard that if you don't have a SSN, your paper will not be marked and will be disposed right away.

If you are registered in the conventional way at a North American college or university, you can participate.

[quote="Soarer"]I heard that if you don't have a SSN, your paper will not be marked and will be disposed right away.[/quote]
This is certainly false and you heard wrong.

There are basically two forms to filled out. Neither requires a SSN.

The first is the team entry form. One form per college is filled out by the college's Putnam supervisor, who is in most cases a faculty member in the Math Dept. The students are identified on this form by name only. (Eventually, there will be an arbitrary code number, assigned later.) At the bottom of that form, one has to swear that the students so named are "regularly enrolled students at the university," and obtain the signature of the chair of the department. This is mailed in October, and the primary peculiarity of the form is that the three members of the team must be named on it.

The second is the outside of the envelope, filled in by the student on the day of the Putnam itself. (Twice, actually - the morning envelope and the afternoon envelope.) This asks for name, mailing address, and expected date of graduation, and requires a signature by which you swear that you have followed the rules. No SSN there, either.

The real "gatekeeper" is the local Putnam supervisor - that faculty member I mentioned earlier. In the case of CSULB, that would be me. I have access to student records, and I'm not going to let a stranger sign up - but if I understand your situation, if you were at my university, I would at least know what courses you're taking and I would hope I'd know who you are. If the local supervisor says you can take it, then you can. If it were my college, I'd say yes, but I may not speak for everyone.

In particular, don't bother asking the MAA or Prof. Klosinski; instead, locate the Putnam supervisor at your current university. Soon, because the entry form is due in mid-October.

There are other situations. For instance, suppose a student at a community college (two-year or junior college), where that community college isn't willing to offer the Putnam itself, wishes to take it at a four-year university to which the student might eventually transfer? My sense of fair play tells me that that is reasonable - and I am the gatekeeper. But it probably wouldn't be appropriate for me to put that student on our 3-person team; again it's a matter of my sense of fair play.

Thanks for the reply.

Do exchange students qualify as "regularly-enrolled" students?

The person who decides that is on your campus. Don't ask us; ask him.


inequalities
calculus
derivative
conics
parabola
inequalities unsolved

let $ x,y,z\in \mathbb{R}_\plus{}$ such that $ x\plus{}y\plus{}z\equal{}1$,

and $ x \leq 2y, x \leq 2z, y\leq 2z, y\leq 2x, z\leq 2x, z\leq 2y$

then find the minimum of

$ xyz$ :o

WLOG $ x \le y \le z$ The conditions can all be rewritten as $ z \le 2z$

Pull x and z apart, while keeping the sum the same. The product clearly decreases. We can do this until either $ z\equal{}2x$, or $ z\equal{}1$. Obviously one of these will happen by the time $ x\equal{}1/5$ since then if x was smaller, we couldn't have $ x\plus{}y\plus{}z\equal{}1$ (think of x,z,2x moving on the number line and visualize :P).

$ z\equal{}1$ does not work, since then $ x\equal{}y\equal{}0$.

Therefore, we can attain the minimum value for the product when $ z\equal{}2x$.

$ 3x\plus{}y\equal{}1$ with $ x\le y \le 2x$ We need to find the minimum of $ 2x^2y$

Now, $ y\equal{}1\minus{}3x$ so the inequality condition can be rewritten as: $ x \le 1\minus{}3x \le 2x$ or $ 1/5 \le x \le 1/4$.

Given this bound on x, we need to find the minimum of $ 2x^2(1\minus{}3x)\equal{}\minus{}6x^3\plus{}2x^2$.

The first derivative $ \minus{}18x^2\plus{}4x$ has 0's at x=2/9 and x=0. Since it's an upside-down parabola, it's positive between the roots, and negative everywhere else. Hence, $ \minus{}6x^3\plus{}2x^2$ is decreasing on
$ (\minus{}\inf,0)$ then has a local min at 0, then increases on (0,2/9) then has a local max. at 2/9, then decreases on $ (2/9,\inf)$. Therefore, it's increasing on [1/5,2/9) and decreasing on (2/9,1/4], so the minimal value in [1/5,1/4] is attained either at 1/5 or 1/4. 1/4 gives a minutely smaller value, so the minimum value is attainable when $ x\equal{}y\equal{}1/4$ and $ z\equal{}1/2$ which gives $ xyz\equal{}1/32$.


Let a,b,c be positive numbers such that $ abc=1$. Prove that $ \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$


modular arithmetic
algebra unsolved
algebra

Find all solutions in $ Z$:

$ y^3\minus{}2x^2\equal{}1388$

[quote="r3d30m3j"][hide]$ \pmod{8}$[/hide][/quote]

thanks.
you are right.


Lance Armstrong cycled 519 hours, 45 minutes and 18 seconds to win his six Tour de France titles.  If he cycled non-stop for this amount of time, how many days would this be?  Express your answer as a decimal to the nearest hundredth.

1 day = 24 hours so 519 hours = 21 days 15 hours

1 day = 24*60*10*(6seconds)= 14400 * (6 seconds)

15hours 45minutes 18 seconds = 15*600 (6seconds) + 45*10 (6seconds) + 3 (6seconds)
                                          = (9000+450+3)(6seconds)
                                          =9453 * (6seconds)

Therefore no. of days = 21 + 9453/14400

9453/144 = 65 + 97/144 so it has 66 hundredths rounded off.

No. of days = 21.66 approximately


it is 5000A.D. and Mars is colonized. The Amulet of Hope has kept Mars thriving and colonized. It has been stolen and shattered into 100 peices. the cities of the planet are falling under seige by the evil Quantomites. You are to vanquish the Quantomite King and restore the Amulet of Hope to Mars. you set of from a village with an armory, a training ground, and a medicine shop.

What weapon do you choose?

spear+shield(2)
sword+shield(4)
club+shield(4)
shield(2)

grrr, you took the spear!!!! sword+shield

um :maybe: , there are no more swords left wrightboy. choose another weapon

club+shield

Sorry, I didn't realize what the (4)s and (2)s meant.  My bad.

club+shield(2)
razor shield( spins on your arm and cuts open your opponent, had to make people choose this option and you also have an open spot for a sweet weapon.)(2) oh, and ?(as in an unknown key)+A=secret message

[color=white][hide]oh, about the message. you get a free choice of spear, sword, club, or pm me and spend 1000 special coins(which only james4l has) to get a lightsaber. but, you have to trade in your primary weapon: spear, club, or nothing to get one of these.[/hide][/color]

and your stats are 20HP, (5,5,5) in (strength,speed,defense skill) defense skill is one thing, not two. spear is 5 strength(the amount of damage you do), sword is 5 strength, club is 5 strength, and razor shield is 5 strength.

Hey I pm'ed you too  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:

[hide]
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[/hide]

Richard_Min: you play Mozart and the guard falls asleep. you get the other half, but realize you are ina teleport zone :o . you teleport back to town, thankfully.

Butler: you slash at its legs. -5 damage to monkey. you try to restore electricity, but the walls are rubber, you realize :rotfl: .

Mister Geek: you hold it up. You lose concentration and it falls on spikes. -5 damage to monkey. It retaliates and knocksyou flying. -10 damage to you.

@everyone: If someone could prod eqjj, that would be great because I need him later for the story.

am i dead? if so get butler to electrocute me to restart my hart like those defibrillators

if not move in with the lightsaber and stab it

Mister Geek, you have 10 health. Also, RTFP!!!!!! NO ELECTRICITY!

I throw my razor shield at it like a discus. Then I stab it in the eyes to blind it.

eqjj: you answer the riddle. the gaurd looks at you blankly, then flees into a spike pit. you open the door. inside you see jjfun1 holding a dagger. He thinks you are the guard.

1.persuade him
2.fight him(HARD)
3.Run like heck.

Special OPTION: sit down and cry.

Butler: you move forward. suddenly bars come down in front of you. You ask Mister geek, who was right next to you, "are we in a cage?". He replies, "No, but the monkey is."

Mister Geek: you spot an opening in the wall. go through?

Richard_Min: You have not connected the two sides yet! You find nothing.

I tell it, "If you give us safe passage and helpp us through this labryinth, then we will heal you and set you free."

agree with butler if the answer is not hen leave through the hole

Butler: It can't help you. it is trapped.

Mister Geek: You escape and find your self in a room with k00lperson. you guys leave. You have to ask for your reward.

Richard_Min: it suddenly feels extremely heavy...

eqjj: you persuade him. he agrees to leave, but he must show you something. Do you follow him?

ask for reward

anyway then go get another quest

I start cutting the bars with my lightsaber... BTW, lightsaber isn't electric. It runs on a crystal inside its handle.


Does there exist a convex pentagon such that for any of its inner angles, the angle bisector contains one of the diagonals?


geometry rectangle perimeter quadratics ratio calculus limit

The image below partially shows an area $R$ of a rectangle $ABCD$ with length $2 \cdot r$ and $b$, $|AB| = |CD| = 2 \cdot r$ and $|BC| = |AD| = b$, $b > r$. From the rectangle $R$ a semi-circle with diameter $CD$ is cut off in order to obtain the area $R$. The area $R \in \mathbb{R}^+$ is given. Then please determine all ordered pairs $(r,b) \in \mathbb{R}^+ \times \mathbb{R}^+$ with $r < b$ such that the figure with area $R$ has minimum perimeter.

This is solved in exactly the same way as the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22662]pentagon whose points form rectangle, equilateral triangle[/url], with or without calculus.

Denote $2r = AB = CD,\ b = BC = DA$. Let $p, S$ be the perimeter and area of the curvilinear quadrilateral $ABCD$. Finding the minimum perimeter at a given area is equivalent to finding the maximum area at a given perimeter.

$p = 2b + 2r + \pi r,\ \ S = 2rb - \frac{\pi r^2}{2}$

$b = \frac{p - (\pi + 2)r}{2},\ \ S = \frac{pr - (\pi + 2)r^2}{2} - \frac{\pi r^2}{2}$

$(1 + \frac 3 4 \pi )r^2 - \frac p 2 r + S = 0$

This quadratic equation for $r$ has a real solution, iff the discriminant is greater than or equal to zero:

$D = \frac{p^2}{4} - (4 + 3 \pi)S \ge 0,\ \ S \le \frac{p^2}{4(4 + 3\pi)}$

The area is maximum at equality and the quadratic equation for $r$ becomes

$( 1 + \frac 3 4 \pi ) r^2 - \frac p 2 r + \frac{p^2}{4(4 + 3 \pi)} = 0$

$(4 + 3 \pi)^2 r^2 - 2(4 + 3 \pi) p r + p^2 = 0$

$[(4 + 3 \pi) r - p]^2 = 0$

$p = (4 + 3 \pi)r$

$b = \frac{(4 + 3 \pi)r - (\pi + 2)r}{2} = (1 + \pi)r$

$\frac b r = 1 + \pi$

$\frac{S}{p^2} = \frac{2rb - \frac{\pi r^2}{2}}{(2b + 2r + \pi r)^2} = \frac{4x - \pi}{2(2x + 2 + \pi)^2}$

where $x = \frac b r$. The maximum ratio $\frac{S}{p^2}$ is then

$(\frac{S}{p^2})_{\text{max}} = \frac{4 + 3 \pi}{2(4 + 3 \pi)^2} = \frac{1}{2(4 + 3 \pi)} \approx \frac{1}{27}$ [color=white].[/color] (within 1%)

Check with calculus:

$\frac{\text d}{\text dx}(\frac{S}{p^2}) = \frac{4(2x + 2 + \pi)^2 - 4(4x - \pi)(2x + 2 + \pi)}{2(2x + 2 + \pi)^4} = 0$

$(2x + 2 + \pi) - (4x - \pi) = 0$

$2x = 2 + 2\pi,\ \ x = 1 + \pi$

$\lim_{x \rightarrow 1}(\frac{S}{p^2}) = \lim_{x \rightarrow 1} \frac{4x - \pi}{2(2x + 2 + \pi)^2} = \frac{4 - \pi}{2(4 + \pi)^2} \approx \frac{17 \cdot 20}{2 \cdot 11^2 \cdot 13^2} \approx \frac{1}{120} < (\frac{S}{p^2})_{\text{max}}$

(within 1%)

$\lim_{x \rightarrow \infty}(\frac{S}{p^2}) = \lim_{x \rightarrow \infty} \frac{4x - \pi}{2(2x + 2 + \pi)^2} = 0 < (\frac{S}{p^2})_{\text{max}}$

Thus we have a maximum ratio $\frac{S}{p^2}$ at $x = \frac b r = 1 + \pi$.

If the area $S$ is given, then for $\frac b r = 1 + \pi$,

$S = 2(1 + \pi) r^2 - \frac{\pi r^2}{2} = \frac{4 + 3\pi}{2} r^2$

$r = \sqrt{\frac{2S}{4 + 3\pi}},\ \ b = (1 + \pi) \sqrt{\frac{2S}{4 + 3\pi}}$

which is the only ordered pair $(r,\ b)$, such that the perimeter $p$ is minimum.


search
number theory unsolved
number theory

Let $ p\geq 5$ be a prime number.Prove that for any partition of the set $ P\equal{}\{1,2,3,...,p\minus{}1\}$ in $ 3$ subsets there exists numbers $ x,y,z$ each belonging to a distinct subset,such that $ x\plus{}y\equiv z (mod p)$

What a nice problem! Anybody replied, so I'll post my attempt.

Let $ P$ be partitioned by the sets $ A$, $ B$ and $ C$. Suppose for now that $ A$ has only one element, and call it $ a$. Set $ B$ has the $ n$ elements $ b_1,b_2,...,b_n$ and $ p\minus{}a$ is not in $ B$ (for if it was we could work with set $ C$ instead). Consider the sums $ a\plus{}b_i$ modulo $ p$. None of these sums equals zero or $ a$, so if one of them was not in $ B$ we'd be done. Suppose that all the sums are in $ B$, that is, sor all $ b_i$ we have $ a\plus{}b_i\equiv{b_j} (mod p)$ for some $ b_j$. Note that the $ b_j$'s are all distinct. Adding up all the $ b_i$'s we get $ na\equiv{0} (mod p)$. As $ p$ is a prime it must divide either $ a$ or $ n$, wich is an absurd. So we are done for this special case.

Consider now that all the sets $ A$, $ B$ and $ C$ have more than one element. Using a similar reasoning we find that for each $ a_i$ there must be a $ b_i$ or a $ c_i$ such that $ a_i\plus{}b_i$ is nonzero and is not in $ B$ (or $ a_i\plus{}c_i$ is nonzero and is not in $ C$). If one of these $ a_i\plus{}b_i$ sums were not in $ A$ we'd be done. Assuming that all these sums actually are in $ A$ we must achieve a contradiction. Unfortunately I couldn't make it work. Can someone help me?

It was posted before here http://www.mathlinks.ro/viewtopic.php?search_id=1712261343&t=74320 but recieved no solution.
on the other hand here http://www.mathlinks.ro/viewtopic.php?search_id=1712261343&t=87906 you can find two solutions.


Let F be a field, and x be an indeterminate over F. How to determine all automorphism all automorphism of F(x) leaving F fixed, by describing their values on x?

I think I have posted in a wrong place. I should have posted it in somewhere else.


[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \sum\ \frac {ab}{a^2 \plus{} ab \plus{} bc} \geq\ 2$[/quote]

Expanding after Cauchy-Schwarz yields $ \sum a^4 \plus{} \sum a^2 b^2 \ge 2\sum a^3 b,$ which is true since $ a^4 \plus{} a^2 b^2 \ge 2a^3 b$ by AM-GM :D

only am-gm :lol:

$ \sum{\frac{ab}{a^2\plus{}ab\plus{}bc}} \equal{} \sum{ \frac{ab(b^2\plus{}c^2\plus{}ac)}{(a^2\plus{}ab\plus{}bc)(b^2\plus{}c^2\plus{}ac)}} \geq 4\frac{\sum{ab^3}\plus{}2abc\sum{a}}{(\sum{a^2}\plus{}\sum{ab})^2} \equal{} \frac{ \sum{ab(a^2\plus{}b^2)} \minus{}\sum{a^3b} \plus{}2abc\sum{a}}{(\sum{a^2}\plus{}\sum{ab})^2} \geq \frac{\sum{ab(a^2\plus{}b^2)}\minus{}\frac{\sum{a^4\plus{}(ab)^2}}{2} \plus{}2abc\sum{a}}{(\sum{a^2}\plus{}\sum{ab})^2}$

using pqr to simplify calculations the ineq is equivalent to:
$ \sum{a^2}(\sum{a^2}\minus{}\sum{ab})^2 \geq 0$ :roll:

[quote="nguoivn"][quote="earldbest"]
Expanding after Cauchy-Schwarz yields $ \sum a^4 \plus{} \sum a^2 b^2 \ge 2\sum a^3 b,$ which is true since $ a^4 \plus{} a^2 b^2 \ge 2a^3 b$ by AM-GM :D[/quote]
Can you detail, thank you  :)[/quote]

by Cauchy-Schwarz,

$ \sum \frac{a^4}{a^3 b \plus{} a^2 bc\plus{} a^3 c} \plus{} \sum \frac{a^2 b^2}{a^3 b \plus{} a^2 b^2 \plus{}ab^2 c}$ $ \ge \frac{(\sum a^2 \plus{} \sum ab)^2}{2\sum a^3 b \plus{} 2\sum a^2 bc \plus{} \sum a^3 c \plus{} \sum a^2 b^2}.$

It suffices to show $ \frac{(\sum a^2 \plus{} \sum ab)^2}{2\sum a^3 b \plus{} 2\sum a^2 bc \plus{} \sum a^3 c \plus{} \sum a^2 b^2} \ge 2.$

Expanding and simplifying yields $ \sum a^{4}\plus{}\sum a^{2}b^{2}\ge 2\sum a^{3}b,$ which is true by AM-GM :lol:

[quote="earldbest"]

by Cauchy-Schwarz,

$ \sum \frac {a^4}{a^3 b \plus{} a^2 bc \plus{} a^3 c} \plus{} \sum \frac {a^2 b^2}{a^3 b \plus{} a^2 b^2 \plus{} ab^2 c}$ $ \ge \frac {(\sum a^2 \plus{} \sum ab)^2}{2\sum a^3 b \plus{} 2\sum a^2 bc \plus{} \sum a^3 c \plus{} \sum a^2 b^2}.$

It suffices to show $ \frac {(\sum a^2 \plus{} \sum ab)^2}{2\sum a^3 b \plus{} 2\sum a^2 bc \plus{} \sum a^3 c \plus{} \sum a^2 b^2} \ge 2.$

Expanding and simplifying yields $ \sum a^{4} \plus{} \sum a^{2}b^{2}\ge 2\sum a^{3}b,$ which is true by AM-GM :lol:[/quote]
Very nice solutions, earldbest and anas  :lol: . And here is my full proof:

[quote="manlio"][quote="earldbest"]...by Cauchy-Schwarz,

$ \sum \frac {a^4}{a^3 b \plus{} a^2 bc \plus{} a^3 c} \plus{} \sum \frac {a^2 b^2}{a^3 b \plus{} a^2 b^2 \plus{} ab^2 c}$ $ \ge \frac {(\sum a^2 \plus{} \sum ab)^2}{2\sum a^3 b \plus{} 2\sum a^2 bc \plus{} \sum a^3 c \plus{} \sum a^2 b^2}.$

[/quote]

Sorry, but I don't understand this step. :blush:

Thank you very much.[/quote]

I used [url=http://www.mathlinks.ro/viewtopic.php?t=15125]Cauchy-Schwarz in Engel form/Titu's Lemma[/url] in this step

[hide="edited"]Notice that
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \equal{} \sum\frac {a^{4}}{a^{3}b \plus{} a^{2}bc \plus{} a^{3}c}$
$ \sum \frac {ab}{a^2 \plus{} ab \plus{} bc} \equal{} \sum\frac {a^{2}b^{2}}{a^{3}b \plus{} a^{2}b^{2} \plus{} ab^{2}c}$

$ \frac {a^{4}}{a^{3}b \plus{} a^{2}bc \plus{} a^{3}c} \plus{} \frac {b^{4}}{b^{3}c \plus{} ab^{2}c \plus{} b^{3}a} \plus{}
\frac {c^{4}}{c^{3}a \plus{} ab^{2}c \plus{} c^{3}b}$ $ \plus{} \frac {a^{2}b^{2}}{a^{3}b \plus{} a^{2}b^{2} \plus{} ab^{2}c} \plus{} \frac {b^{2}c^{2}}{b^{3}c \plus{} b^{2}c^{2} \plus{} abc^{2}} \plus{} \frac {c^{2}a^{2}}{c^{3}a \plus{} c^{2}a^{2} \plus{} a ^{2} bc}$ $ \ge RHS.$

I hope this is clearer  :D

[/hide]

Sorry, but I don't think you apply it in correct form.

Please, explain better any step.

Very sorry, earldbest :blush:

Your solution is correct and nice.

I get confused, very sorry. :blush:

Dear anas,

in your nice solution you need not to use p,q,r.

If you expand $ (\sum a^2\plus{}\sum ab)^2\equal{} \sum a^4 \plus{}3\sum a^2b^2\plus{}4abc\sum a\plus{}2\sum ab(a^2\plus{}b^2)$ and clear the numerator you can rewrite your solution as proving

$ \frac{\sum a^2}{\sum ab} \geq \frac{1}{2} \plus{}2\frac{(\sum a^2)^2}{(\sum a^2\plus{}\sum ab)^2}$  (1)

putting $ k\equal{}\frac{\sum a^2}{\sum ab} \geq 1$ we see that (1) can be rewritten after an easy computation as

$ (k\minus{}1)(2k^2\plus{}k\plus{}1) \geq 0$ :)

[quote="manlio"]Dear anas,

in your nice solution you need not to use p,q,r.

If you expand $ (\sum a^2 \plus{} \sum ab)^2 \equal{} \sum a^4 \plus{} 3\sum a^2b^2 \plus{} 4abc\sum a \plus{} 2\sum ab(a^2 \plus{} b^2)$ and clear the numerator you can rewrite your solution as proving

$ \frac {\sum a^2}{\sum ab} \geq \frac {1}{2} \plus{} 2\frac {(\sum a^2)^2}{(\sum a^2 \plus{} \sum ab)^2}$  (1)

putting $ k \equal{} \frac {\sum a^2}{\sum ab} \geq 1$ we see that (1) can be rewritten after an easy computation as

$ (k \minus{} 1)(2k^2 \plus{} k \plus{} 1) \geq 0$ :)[/quote]

Sorry, I made a terrible mistake as I forgot to multiply by 4!!!

It should be $ \sum \frac{ab}{a^2\plus{}ab\plus{}bc}\geq 2\minus{}4(\frac{\sum a^2}{(\sum a^2\plus{}\sum ab)})^2$

so the inequality becomes

$ (\sum a^2\plus{}\sum ab)^2 \geq 4(\sum a^2)(\sum ab)$

which is true by AM-GM.


For all $ n\in N$, $ F(n)$ is divisible by at least one of $ a_{1},...,a_{m}$. Show that we can choose an $ a_{j},\ j\in \overline{1,m}$ s.t. $ a_{j}|F(n)$ for all $ n\in N$.

Suppose not.
Then for every a_j, there exist r_j s.t. a_j doesn't divide p(r_j)
or p_j^b_j doesn't divide p(r_j) and p_j^b-j | a_j for prime p_i.
Write all these expressions. If  p_i=p_j, delete the equation in which p_i has greatest exponent.
So we get all p_i distinct.
Now, as p(x+ka)=p(x) (mod a) and using Chinese Remainder Theorem.
We can construct X s.t. X=r_j(mod p_j^b_j)
Then P(X) is not divisible by either of a_1..a_m


Do you have to get 60 questions correct to view the videos or do you have to just answer 60 questions?

How do you know how many questions you already answered?

When you click the Alcumus link, the banner says how many problems you have completed. You can also click the Report tab and sum the correct, incorrect, and give up values.


Prove there is no polynomial with real coefficients $ P(x)$ for which $ degP\equal{}998$  and which satysfies for each x from real $ (P(x))^2\equal{}P(x^2\plus{}1)\plus{}1$

[quote="matex (L)(L)(L)"]Prove there is no polynomial with real coefficients $ P(x)$ for which $ degP = 998$  and which satysfies for each x from real $ (P(x))^2 = P(x^2 + 1) + 1$[/quote]
I will solve it with $ deg P = n$ . It is easy to see
\[ P = a_0x^n + a_1x^{n - 1} + ...a_{n - 1} Hence \[ (x^n + Q(x))^2 = (x^2 + 1)^n + Q(x^2 + 1) + 1
\]
so that
\[ x^nQ(x) + Q(x)^2 = (x^2 + 1)^n + Q(x^2 + 1) + 1 - x^{2n}
\]

\[ deg LHS = deg RHS
\]

\[ \leftrightarrow n + k = 2(n - 2)
\]

\[ \rightarrow k = n - 2
\]
Repeat that step we will get $ n = 2t$ and $ P(x) = Q(x^2)$ ($ deg Q = t$
So we see that there is no polynomial $ P(x)$ such that the condition
P/s:After added
Sorry for my post ,But I finished my post there because I want you to see my generalization
http://www.mathlinks.ro/viewtopic.php?t=256478
Sorry for my careless post

Allnames is wrong in calculating the degree of both sides.

Here is the proof.

We know $ P(x)^2\equal{}P(\minus{}x)^2$. Since $ P$ is polynomial, we see $ P$ is either even of odd.
1) $ P$ is even.
Suppose $ P$ is odd, $ P(0)\equal{}0$. We define $ a_0\equal{}0,a_n\equal{}a_{n\minus{}1}^2\plus{}1$ and see $ P(a_{2n})\equal{}0,\forall n$, which implies $ P$ has infinitely many roots. Contradiction!

2) There exists polynomial $ g$ such that $ P(x)\equal{}g(x)^2\minus{}1$ and $ g(x)^2\equal{}g(x^2\plus{}1)\plus{}1$.
Since $ P$ is even, we have $ P(x)\equal{}h(x^2)$ with some polynomial $ h$.
$ h(x)^2\equal{}h((x\plus{}1)^2)\plus{}1$, or $ h(x\minus{}1)^2\equal{}h(x^2)\plus{}1$.
Let $ g(x)\equal{}h(x\minus{}1)$, we have $ g(x)^2\equal{}g(x^2\plus{}1)\plus{}1$.
Also $ P(x)\equal{}h(x^2)\equal{}g(x^2\plus{}1)\equal{}g(x)^2\minus{}1$.

So if there exists $ P$ with degree $ 998$ satisfying the equation, by 2), there exists $ g$ with degree $ 499$ still satisfying the equation.

But then $ g$ cannot be even. Contradiction with 1)!

Indeed, it is not hard to show the only $ P$ satisfying the equation is constant function.


Problem 1. Let $x,y,z$ be in $[1,2]$, prove that
1) $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\leq\frac23(x+y+z)$.
2) $(x+y)(y+z)(z+x)\geq8xyz+3(x-y)(y-z)(z-x)$.

Problem 2. Prove that if $x,y,z>0$ then
\[ \frac{x+y+z}{xy+yz+zx}\geq\frac23\left(\frac1{x+y}+\frac1{y+z}+\frac1{z+x}\right). \]

[quote="hitek"]
Problem 2. Prove that if $x,y,z>0$ then
\[ \frac{x+y+z}{xy+yz+zx}\geq\frac23\left(\frac1{x+y}+\frac1{y+z}+\frac1{z+x}\right).  \][/quote]

L-R=

$\frac{(xy+z^2)(x-y)^2+(yz+x^2)(y-z)^2+(xz+y^2)(x-z)^2}{3(x+y)(y+z)(z+x)(xy+yz+xz)} \geq 0$

we have $\sum \frac{xy+yz+zx}{x+y}=\sum z+\sum \frac{xy}{x+y}\leq \sum z+\sum\frac{x+y}{2}=\frac32*\sum x$
so $\frac{x+y+z}{xy+yz+zx}\geq \frac23(\frac{1}{x+y})$

the solution of problem 1 1):
$\sum{x^2}{(y+z)(x+y+z)=\sum{X}{y+z}-\sum{x}{x+y+z}=\sum{x}{y+z}-1}$
let $f=\sum\frac{x}{y+z}$
we know that f is convex.Hence maximum is attained when  $a,b,c$are at the endpoints of an interval.Easy checking gives that this maximum equals \frac53, when x=y=1,z=2


A sequence of integers is defined by $ a_{0}= 0, a_{1}= 2, a_{2}= 9$ and for $ n > 2$
$ a_{n}= a_{n-1}+a_{n-2}-a_{n-3}$
Give a closed form expression for $ a_{n}$.


In the expression $ c\cdot a^b\minus{}d$, the values of $ a$, $ b$, $ c$, and $ d$ are $ 0$, $ 1$, $ 2$, and $ 3$, although not necessarily in that order. What is the maximum possible value of the result?

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 6\qquad
\textbf{(C)}\ 8\qquad
\textbf{(D)}\ 9\qquad
\textbf{(E)}\ 10$

$ c$ and $ a$ are obviously not $ 0$ , so either $ b$ or $ d$ are $ 0$ .

 If $ b$ is $ 0$ then the maximum value is $ 2$.

If $ d$ is $ 0$ then the maximum value is $ 9$

So if I didn't miss anything, the answer is letter $ \fbox {(D)}$


trigonometry
combinatorics proposed
combinatorics

Find in a closed form the following sum $ \sum_{r\equal{}0}^{\infty}\frac{t^{3r}}{(3r)!}$

Hello.

$ f(t) = \Sum_{r= 0 }^{\infty} \frac{t^{3r}}{(3r)!}$

$ f$ is $ C^{\infty}$ function.

So Write down Maclaurin series of f(t), and compare the coefficients and solve the D.Eq.

My solution is $ f(t) = \frac{1}{3} \big(e^t + 2e^{-t/2} \cos[ \frac{\sqrt{3} t}{2} ] \big)$

I'm not sure. and mine is not so beautiful. Does anyone has another elegant solution?

$ f(t) \equal{} \frac {e^t \plus{} e^{\omega t} \plus{} e^{\omega^2 t}}{3}$ where $ \omega$ is a primitive third root of unity.  This is an example of the roots-of-unity filter.  See the discussion, for example, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=242876]here[/url].

hello, you solution is ok, i have also got $ \sum_{r\equal{}0}^\infty\frac{t^{3r}}{(3r)!}\equal{}\frac{1}{3}\left(e^t\plus{}2e^{\minus{}\frac{t}{2}}\cos(\frac{\sqrt{3}t}{2})\right)$
Sonnhard.


The five digits 2, 3, 4, 5, and 6 can be arranged to form 120 different five digit integers. How many are divisible by 11?

I'm pretty sure you use the fact that if a number is divisible by 11, let's say, abcd, then a-b+c-d has to be divisible by 11. Interesting problem...

-rts2007

[hide="solution"] Let the number be abcde. We then have $ (a\plus{}c\plus{}e)\minus{}(b\plus{}d)\equiv0\pmod{11}$. We also know that $ (a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}20$. Thus, $ (a\plus{}c\plus{}e)\minus{}(b\plus{}d)$ must be even and thus must be 0 in this case. Adding the two equations we now have and dividing by two, we get $ a\plus{}c\plus{}e\equal{}10$. It is now easy to see that {a,c,e} must be a permutation of {2,3,5} wihle {b,d} must be a permutation of {4,6}. Thus, there are 3!*2!=12 such numbers.[/hide]


Let $N$ be a positive integer. Two players Alice and Bob take turns to write numbers from the set $\{ 1, 2, \ldots , N \}$ on a blackboard. Alice begins the game by writing $1$ on her first move. If a player has written $n$ on a certain move, the adversary is then allowed to write each $n + 1$ or $2n$ (provided the number does not exceed $N$). The player who writes $N$ wins. We say that $N$ is of type $A$ or type $B$ according to whether Alice or Bob has a winning strategy.

(a) Determine whether $N$ = 2004 is of type $A$ or type $B$.
(b) Find the least $N > 2004$ whose type is different from the type of $2004$.

Shortlist, too, I think ;)
But not from the newest one.

[quote="ZetaX"]But not from the newest one.[/quote]

No, it is from IELL (IMO Extra Long List). Please be so kind and provide evidence next time.  :)

I was sure to have seen this problem in IMO related context, but checking the last ISL's didn't bring it up; thus I thought it's simply older.
Nonetheless, the other three he recently posted are ISL 2004.

[quote="orl"]

No, it is from IELL (IMO Extra Long List). Please be so kind and provide evidence next time.  :)[/quote]

What do you mean? This is shortlist too!  :)

It's indeed ISL 2004, I somehow didn't see it...

Yup! http://www.mathlinks.ro/Forum/post-251895.html#p251895

BTW, what is IMO Extra Long List?  :?

[quote="Saul"]Like the IMO longlist, but longer?[/quote]

No, it was a joke only. That remark is because ZetaX just gave a rough source reference.


[quote="pvthuan"]In triangle, we use standard notations, prove that \[ r_a^2+r_b^2+r_c^2\geq 6r^2+24Rr-p^2.  \][/quote]
$r_a+r_b+r_c=4R+r\ ,\ r_ar_b+r_br_c+r_cr_a=p^2\Longrightarrow$

$r^2_a+r^2_b+r^2_c\ge 6r^2+24Rr-p^2\Longleftrightarrow p^2\le 16R^2-16Rr-5r^2\ .$

It is well-known that $\boxed {p^2\le 4R^2+4Rr+3r^2}\ \ (1)\ .$ But

$4R^2+4Rr+3r^2\le 16R^2-16Rr-5r^2\Longleftrightarrow$

$3R^2-5Rr-2r^2\ge 0\Longleftrightarrow (R-2r)(3R+r)\ge 0$, what is truly.

Therefore, the proposed inequality is weaker then $(1)\ ,$ i.e.

$p^2\le 4R^2+4Rr+3r^2\le 16R^2-16Rr-5r^2\ .$


My name is Joash Boyton, this is my first post.....evidently :roll:
I am 15 yrs old, and am currently  in Year 12, and going for an acadamic certificate...

I am doing Geometry, and I am looking for a website that has all basic postulates, and basically has a large database that you can search, and find what you are looking for...ATM, my first priority, is to try and obtain a list of all Geometric Postulates..

If someone could help, that would be great...and if this is in the wrong part of the forum....my apologies.. :blush:

Thanks,
[url=http;//joashwoodturning.bravehost.com]Joash[/url] :lol:

Euclidean geometry really doesn't have that many postulates / axioms, if I understand your question correctly.

http://en.wikipedia.org/wiki/Euclidean_geometry#Axiomatic_approach


"How old are you?"

"What do you think?"

"May I take a guess?"

"Sure"

"All right.  Please divide your age by three.  What is the remainder?"

"It's two."

"If you divide your age by five, what is the remainder?"

"None."

"How about dividing your age by seven?"

"The remainder is one."

"I see . . . you are fifty years old, aren't you?"

"Amazing! You are right!"

  By knowing the remainders when a person divides his age by three, five, and seven, you can determine his age.  How is it done?

(Taken word for word)

(But don't forget that you can't be older then 120!)


I was wondering if there is some result which generalises Zsigmondys theorem.
For example that in every linear recurrent sequence (except of course some finite cases) there are always new primes that show up as divisors in each term...
It seems like such a result would need much elaboration. Maybe someone has seen an article with any extensions of this theorem :?:

See [url=http://citeseer.ist.psu.edu/cache/papers/cs/25481/http:zSzzSzwww.math.u-bordeaux.frzSz~yurizSzpublzSzpreprszSzprim.pdf/existence-of-primitive-divisors.pdf]Existence of primitive divisors of Lucas and Lehmer numbers[/url] by Yu Bilu, G. Hanrot, P.M. Voutier.  This is probably the most important paper in this area.

Hope this helps.   :)

Very nice, thanks a lot!
quite impressive  :)

You are very welcome.  For further interest, you can get the book, [url=http://books.google.com/books?id=LmfonVHe7MMC&pg=PA279&lpg=PA279&dq=%2289j+11016%22&source=web&ots=g8H2Ok90Ja&sig=ZJrw1xFxGynLa-_xwzEaIm5k-ik#PPP1,M1]Recurrence Sequences[/url].  This will tell about the history of the problem, starting with Zsigmondy and leading to the above mentioned paper and, in addition, many other remarkable facts about the more general-recurrence sequences (not just Lehmer and Lucas).  :)

A number of generalization is described in the paper:

[url=http://arxiv.org/abs/math/0412079]Primes Generated by Recurrence Sequences[/url]


trigonometry
geometry
inradius
circumcircle
inequalities proposed
inequalities

Let $ ABC$ be a triangle.Prove that :
$ \frac{a^2}{b^2\plus{}c^2}\plus{}\frac{b^2}{a^2\plus{}c^2}\plus{}\frac{c^2}{a^2\plus{}b^2} \ge 2\minus{}\frac {r}{R}$,where r and R are  its inradius and circumradius,respectively.

There are problems from my brother: [b]evarist[/b]
 $ \frac{14\minus{}\frac{4r}{R}}{9\minus{}\frac{2r}{R}}\le\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\le 2\minus{}\frac{r}{R}\le \frac{a^{2}}{b^{2}\plus{}c^{2}}\plus{}\frac{b^{2}}{c^{2}\plus{}a^{2}}\plus{}\frac{c^{2}}{a^{2}\plus{}b^{2}}\le 6(1\minus{}\frac{r}{R})^{2}$
Posted here: http://mathfriends.org/forum/viewtopic.php?t=299

My proof:
From $ sin^2 \frac{A}{2} \le \frac{a^2}{b^2\plus{}c^2}$ which is easy to prove result:
$ \sum_{cyc} sin^2 \frac{A}{2} \le \sum_{cyc} \frac{a^2}{b^2\plus{}c^2}$
but $ \sum_{cyc} sin^2 \frac{A}{2}\equal{}2\minus{}\frac {r}{R}$,which is well kown identity. :lol:
 NguyenDungTN it's too nice and hard the sequence of Evarist!


integration
calculus
calculus computations

[quote="maokid7"]Find $f(x)$ if
$f''(x)+x^{2}f'(x)+2xf(x)=1$ and $f(0)=f'(0)=1$

i got bored in class and made this up, but i don't really know how to solve it.  :blush:[/quote]

It may written:

$y''+(x^{2}y)'=1$ and $y(0)=y'(0)=1$ wich implies the easy inhomogen equation: $y'+x^{2}y=x+1$ Results:
$y=e^{-\frac{x^{3}}{3}}\left({1+\int{(x+1)e^{\frac{x^{3}}{3}}}dx}\right)$


Can anyone teach me how to solve these problems? I'm not sure of how to solve them. I think I'm having trouble knowing what to use like sine, cosine, and tangent. =/

[img]http://img205.exs.cx/img205/9126/trigonometricratios7bb.png[/img]

use the mnemonic SOHCAHTOA to help you with trig ratios. Sine is Opposite over Hypotenuse; Cosine is Adjacent over Hypotenuse; and Tangent is Opposite over Adjacent. Therefore, applying these to your triangles, we get:

sine 37=6(the opposite side to the angle)/ x(the hypotenuse)
therefore: x=6/sin37

similarly, to find out y, since we the side opposite and we want to find the side adjacent, we should use tangent (remember: TOA)
tan37=6/y
therefore y=6/tan37

for the other two right-angled triangles, just apply the same principles. Using SOHCAHTOA will help a lot.

Cheers!

I think I know how to do this now. woot I just didn't know what trigonometric ratio to use for the problems. Thanks anyways. :D


greatest common divisor
number theory
least common multiple

G(a,b)= GCF {a,b}, L(a,b)= LCM {a,b}
If G(a,b)=1 and L(a,b)= 24

Find all possible pairs (a,b)

sory :blush:

:?:

No problem condition is given, just two definitions.

[hide] Note that L(a,b)*G(a,b)=ab.  Plug in G(a,b)=1, so we have that L(a,b)=ab.  Thus, we can find all of the "factor pairs" of 24 and see which ones have a GCF of 1.  We know that the factor pairs of 24 are (1,24), (2,12), and (3,8).  GCF(2,12)=2, so that doesn't work, but the rest work.  However, (a,b) is not the same as (b,a) so we also have to include (24,1) and (8,3).  Thus, our solutions are (24,1), (3,8), (1,24), and (8,3).  [/hide]


Can someone explain the functions of the compunds given below...

1. In the presence of silver catalyst, ethylene reacts with oxygen to give P. Compound P on heating with acidified water gives Q. 1H NMR spectrum of P has only 1 signal while that of Q contains 2 signals. Identify P and Q.

2. Phenol ---(HSbF6)(halogen)----> A
A gives 2 equal intensity peaks at 172 and 174 in the highest m/z region of its mass spectrum. It gives 3 isomeric mononitro derivatives on nitration under mild conditions. A=??

3. What transformation will citric acid undergo when treated with warm conc. H2SO4 at 45-50 C?

After warming citric acid with sulphuric acid, anisole is added to the reaction vessel and A($ C_{12}H_{12}O_5$) is obtained.

On heating with Ac2O, A forms an anhydride.
A has 2 replaceable H+ (basicity is 2)
1 mole of A  reacts with 1 mole of bromine to gine an addition product.

A=??

Ok then...   :!:

1. Ethylene, also known as ethene (from its IUPAC Nomenclature name) is partially oxidised with a silver catalyst to form [b]epoxyethane[/b].

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/2/2d/Ethylene-oxide-from-xtal-3D-balls.png/150px-Ethylene-oxide-from-xtal-3D-balls.png[/img]

It reacts with water (not necessarily acidified water) to form [b]ethan-1,2-diol[/b].

2. Not sure what at the stuff in the arrow is supposed to be i.e. (HSbF6)(halogen) :?:

3. Usually most things will [b]condensate[/b] with concentrated with $ {H_2}S{O_4}$, but when you require heat, it becomes a [b]hydrogenation[/b] reaction.

Since A can react with bromine to give an addition product, we know and deduce that it is an electrophile and it undergoes an electrophilic addition.

However, I'm not really quite sure about the wording of the question since $ {Ac_2}O$ is itself an anhydride, so not sure how an anhydride with an anhydride forms another anhydride!

Sorry.  :oops:

[quote="pizza1512"]
2. Not sure what at the stuff in the arrow is supposed to be i.e. (HSbF6)(halogen) :?:
[/quote]

It means reacting with "magic" acid or "super" acid in presence of a halogen.

[quote="Pizza1512"]3. Usually most things will condensate with concentrated with $ H_2SO_4$, but when you require heat, it becomes a hydrogenation reaction.[/quote]

You should review your vocabulary: instead of "hydrogenation" it should be "dehydration".


geometry
analytic geometry
graphing lines
slope

Find equation of the line which is equidistant from parallel lines $ 9x\plus{}6y\minus{}7\equal{}0$ and $ 3x\plus{}2y\plus{}6\equal{}0$. Please give its solution. Its answer is $ 18x\plus{}12y\plus{}11\equal{}0$. Thanks a lot.

Rename $ 9x \plus{} 6y \minus{} 7$ to $ 3x \plus{} 2y \minus{} \frac {7}{3}$. Since the average of $ \minus{} \frac {7}{3}$ and $ 6$ is $ \frac {11}{6}$, we have $ 3x \plus{} 2y \plus{} \frac {11}{6} \equal{} 0 \equal{} 18x \plus{} 12y \plus{} 11$

Hello, Thanks , but still it is a problem. When I solved it I renamed it and used the formula i.e. $ |c1\minus{}c2|/ \sqrt{a^2\minus{}b^2}$. and I could not get the solution. Also I didn't understand why you took average of $ \minus{}7/3$ and $ 6$. Please explain it. Thank you so much.

[quote="ninni"]Hello, Thanks , but still it is a problem. When I solved it I renamed it and used the formula i.e. $ |c1 \minus{} c2|/ \sqrt {a^2 \minus{} b^2}$. and I could not get the solution. Also I didn't understand why you took average of $ \minus{} 7/3$ and $ 6$. Please explain it. Thank you so much.[/quote]
He/She took the average of $ \minus{}\frac{7}{3}$ and $ 6$ because you asked for the line that is equadistant from those two lines. Since they have the same slope the difference is in the y-intercept. By taking the average of the two y-intercepts it is possible to find the equation of the line

Suppose $ L_!\equal{}3x\plus{}2y\minus{}\frac{7}{3}\\ \ L_2\equal{}3x\plus{}2y\plus{}6$
Let them cut the Y-axis at A & B, respy. and the reqd line  cuts it at C,
The perp from B to $ L_1$ cuts reqd line at H....and $ L_1$ at l....
Apply similarity of triangles to ABL & ACH.....

[quote="SuperNerd123"][quote="ninni"]Hello, Thanks , but still it is a problem. When I solved it I renamed it and used the formula i.e. $ |c1 \minus{} c2|/ \sqrt {a^2 \minus{} b^2}$. and I could not get the solution. Also I didn't understand why you took average of $ \minus{} 7/3$ and $ 6$. Please explain it. Thank you so much.[/quote]
He/She took the average of $ \minus{} \frac {7}{3}$ and $ 6$ because you asked for the line that is equadistant from those two lines. Since they have the same slope the difference is in the y-intercept. By taking the average of the two y-intercepts it is possible to find the equation of the line[/quote]

ok I understand that we should take average of y-intercepts. But I think y-intercept of $ 3x\plus{}2y\minus{}\frac{7}{3}$ is $ \frac{\minus{}(\minus{}7)}{3}/2 \equal{} \frac{7}{6}$ and y-intercept of  $ 3x\plus{}2y\plus{}6$ is $ \frac{\minus{}6}{2} \equal{} \minus{}3$. Please help again.Thanks.

Ok the 2 equations are:
$ 3x \plus{} 2y \minus{} \frac {7}{3} \equal{} 0$

$ 3x \plus{} 2y \plus{} 6 \equal{} 0$
Putting both of these in $ y \equal{} mx \plus{} b$ form we have:
$ y \equal{} \minus{} \frac {3}{2}x \plus{} \frac {7}{6}$

$ y \equal{} \minus{} \frac {3}{2}x \minus{} 3$
The average is $ \frac {\frac {7}{6} \minus{} 3}{2} \equal{} \minus{} \frac {11}{12}$
This means the equation is $ \boxed{ y\equal{}\minus{} \frac {3}{2}x \minus{} \frac {11}{12}}$
Ihatepie said the answer was $ 18x \plus{} 12y \plus{} 11\equal{}0$
If you put this into $ y \equal{} mx \plus{} b$ from then you get $ \boxed{y\equal{} \minus{} \frac {3}{2}x \minus{} \frac {11}{12}}$
Which is the answer

[quote="SuperNerd123"]Ok the 2 equations are:
$ 3x \plus{} 2y \minus{} \frac {7}{3} \equal{} 0$

$ 3x \plus{} 2y \plus{} 6 \equal{} 0$
Putting both of these in $ y \equal{} mx \plus{} b$ form we have:
$ y \equal{} \minus{} \frac {3}{2}x \plus{} \frac {7}{6}$

$ y \equal{} \minus{} \frac {3}{2}x \minus{} 3$
The average is $ \frac {\frac {7}{6} \minus{} 3}{2} \equal{} \minus{} \frac {11}{12}$
This means the equation is $ \boxed{ \minus{} \frac {3}{2}x \minus{} \frac {11}{12}}$
Ihatepie said the answer was $ 18x \plus{} 12y \plus{} 11$
If you put this into $ y \equal{} mx \plus{} b$ from then you get $ \boxed{ \minus{} \frac {3}{2}x \minus{} \frac {11}{12}}$
Which is the answer[/quote]

oh thanks. I was doing the calculation mistake. :-)


I was working on the Broken Necklace problem, and got stuck on this problem. The program that I wrote managed to get the first test case, a string of 29 characters, but timed out on the second one, which was "rrr." I have no idea why it would time out on the "rrr," but I thought I would create an conditional statement to handle the cases where all the beads were the same color, only I don't know how to code something like "if all the elements of an array are equal"  :blush: Could someone help me fix the problem? Any help would greatly be appreciated  :yup:

I included the problem and the program I wrote below

[code]
Broken Necklace

You have a necklace of N red, white, or blue beads (3<=N<=350) some of which are red, others blue, and others white, arranged at random. Here are two examples for n=29:

                1 2                               1 2
            r b b r                           b r r b
          r         b                       b         b
         r           r                     b           r
        r             r                   w             r
       b               r                 w               w
      b                 b               r                 r
      b                 b               b                 b
      b                 b               r                 b
       r               r                 b               r
        b             r                   r             r
         b           r                     r           r
           r       r                         r       b
             r b r                             r r w
            Figure A                         Figure B
                        r red bead
                        b blue bead
                        w white bead

The beads considered first and second in the text that follows have been marked in the picture.

The configuration in Figure A may be represented as a string of b's and r's, where b represents a blue bead and r represents a red one, as follows: brbrrrbbbrrrrrbrrbbrbbbbrrrrb .

Suppose you are to break the necklace at some point, lay it out straight, and then collect beads of the same color from one end until you reach a bead of a different color, and do the same for the other end (which might not be of the same color as the beads collected before this).

Determine the point where the necklace should be broken so that the most number of beads can be collected.
Example

For example, for the necklace in Figure A, 8 beads can be collected, with the breaking point either between bead 9 and bead 10 or else between bead 24 and bead 25.

In some necklaces, white beads had been included as shown in Figure B above. When collecting beads, a white bead that is encountered may be treated as either red or blue and then painted with the desired color. The string that represents this configuration will include the three symbols r, b and w.

Write a program to determine the largest number of beads that can be collected from a supplied necklace.
PROGRAM NAME: beads
INPUT FORMAT
Line 1: 	N, the number of beads
Line 2: 	a string of N characters, each of which is r, b, or w
SAMPLE INPUT (file beads.in)

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

OUTPUT FORMAT
A single line containing the maximum of number of beads that can be collected from the supplied necklace.
SAMPLE OUTPUT (file beads.out)

11

OUTPUT EXPLANATION
Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.

wwwbbrwrbrbrrbrbrwrwwrbwrwrrb wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
                       ****** *****
[/code]

[code]
import java.io.*;
import java.util.*;

public class beads {
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new FileReader("beads.in"));
        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("beads.out")));

        int n = Integer.parseInt(f.readLine());
        String s = f.readLine();
        char beadNecklace[] = s.toCharArray();
        int beadCount;
        int beadCounterL=1;
        int beadCounterR=1;
        int beadMax = 2;

        for (int j=0; j<n; j++)
            {
                int k = (j+1)%n;
                char leftMark = beadNecklace[j];
                char rightMark = beadNecklace[k];

                while(beadNecklace[j] == beadNecklace[(int)((j-beadCounterL) - Math.floor((double)(j-beadCounterL)/n) * n)] || beadNecklace[(int)((j-beadCounterL) - \
Math.floor((double)(j-beadCounterL)/n) * n)] == 'w'  && beadCounterL<n)
                    {
                        beadCounterL += 1;
                    }

                while(beadNecklace[k] == beadNecklace[(k+beadCounterR)%n] || beadNecklace[(k+beadCounterR)%n] == 'w' && beadCounterR<n)
                    {
                        beadCounterR += 1;
                    }
                if(beadCounterL==n-1)
                    {
                        beadCount = n;
                    }
                else
                    {
                        beadCount = beadCounterL + beadCounterR;

                        if(beadCount>beadMax)
                            {
                                beadMax = beadCount;
                            }
                    }

                beadCounterL = 1;
                beadCounterR = 1;

            }

        out.write(beadMax + "\n");

        out.close();                                  // close the output file
        System.exit(0);                               // don't omit this!

    }
}
[/code]

Theres a few problems here.

You've already noticed the program times out on RRR, so you just need to trace through your code to see why that happens. Each of your while loops are of the form while (bead matches or bead is white and counter < n).
In java (or anything), && has higher precedence than ||, so thats being read as while (bead matches or [bead is white and counter < n]). So if the bead matches, its always going to return true regardless of how high the counter is.

Adding some brackets semi-fixes that problem, but not quite. Again, in the RRR case, when will it stop? It'll stop when the left and right counters both hit n, so it will return 6 instead of 3.

Finally, you'll need to consider the case where the end beads are white :)

(By the way, just out of style, I think I'd prefer something like this:
[code](j-beadCounterL+n)%n[/code]
rather than
[code](int)((j-beadCounterL) - Math.floor((double)(j-beadCounterL)/n) * n)[/code], but I guess your way seems to work fine).

EDIT: Seems like I was too slow this time :(.

All right, first of all (I'm following nebula42's example now :) ) if you can't get your program to work, try running it on your computer with the buggy input. It helps to add some debugging information to your program to see what it is doing - commands to print values of your variables to screen in each pass of the main "for" cycle etc.:
[code]
for (int j = 0; j < n; j ++) {
	System.out.println("Beginning of the FOR cycle.");
	while (something) {
		beadCounterL += 1;
		System.out.print("Value of beadCounterL: ");
		System.out.println(beadCounterL);
	}
	System.out.println("WHILE cycle finished");
	...
[/code]

If you still don't feel like finding the problem in your program on your own,
[hide="click here"]
The reason why your program times out is that if all the beads are of the same color, both your while cycles will go on forever and your program will never stop! Although you did include the "beadCounter < n" check, it doesn't do what you intended because of the priority of logical operators. Your condition looks like this:
[code]
while (something || something_else && beadCounterL < n) {
	beadCounterL += 1;
}
[/code]
but as long as "something" evaluates to true, your cycle will continue to run forever. What you need to do is to include this pair of brackets:
[code]
while ((something || something_else) && beadCounterL < n) {
	beadCounterL += 1;
}
[/code]
[/hide]
Now if you want to know how to check if all the elements of an array are equal, this is what I would do:
[code]
boolean areEqual = true;
for (int i = 0; i < n; i ++) {
	if (array[i] != array[0]) {
		areEqual = false;
	}
}
[/code]
You can probably easily see how it works.

One final note to simplify your code:
[code]
// This is too complicated:
// while(beadNecklace[j] == beadNecklace[(int)((j-beadCounterL) - Math.floor((double)(j-beadCounterL)/n) * n)] || ...

// Try this instead:
while(beadNecklace[j] == beadNecklace[(j-beadCounterL + n) % n] || ...
[/code]

Ohhh, thanks for pointing those things out  :lol: I'll try the printing intermediate output to check for these sort of problems.

I'll try to limit my questions from now on  :P

[quote="bus"]
Now if you want to know how to check if all the elements of an array are equal, this is what I would do:
[code]
boolean areEqual = true;
for (int i = 0; i < n; i ++) {
	if (array[i] != array[0]) {
		areEqual = false;
	}
}
[/code]
You can probably easily see how it works.[/quote]

But that takes the same amount of time whether the arrays differ in the first or the last element. Personally I'd either go [code]
boolean areEqual = true;
for (int i = 0; i < n && areEqual; i ++) {
	if (array[i] != array[0]) {
		areEqual = false;
	}
}
[/code]

or

[code]
boolean areEqual = true;
for (int i = 0; i < n; i ++) {
	if (array[i] != array[0]) {
		areEqual = false;
		break;
	}
}
[/code]

The second is simpler, but as far as I know, break has a higher overhead than simply checking to make sure areEqual every time. It's only about .01 second for large arrays though, so it's not that important.

Asking questions is not a bad thing, my point is just that you'll get better at coding when you can figure out this kind of problems yourself :wink:.


Hey all,

As a matter of curiosity, who do you reckon the smartest member on ML.

Bomb :)

Myself of course :D

No, I think if you look in the Olympiad/College section, you'll quickly pick out a few who are certainly in the top. (Myth, harazi, peter scholze, ... and many others I forget right now)

I don't know the AoPS instructors personally, but I see absolutely no reason why they should be the smartest persons on this site... :?

i think its either:
MCrawford	  or
rrusczyk	  or
DPatrick	  or
nsato

This topic is bound to get flamy - thus it will be locked.


The roots of the polynom $ 3x^{3}\minus{}4x^{2}\plus{}3x\plus{}1$  are $ a,b$ and $ c$. Find $ a^{2}\plus{}b^{2}\plus{}c^{2}$.

[hide]You just use Newton's and Vieta's factorization ,than Vieta's formula.[/hide]

from vieta formulas:
$ a\plus{}b\plus{}c\equal{}4/3 < \equal{} > (a \plus{} b \plus{}c)^2 \equal{} 16/9$(1)
also from these formulas $ ab \plus{} bc \plus{} ca \equal{} 1$
so (1)=> $ a^2 \plus{}b^2 \plus{}c^2\equal{}16/9 \minus{}2\equal{}\minus{}2/9$


Jim's stride measures $ 2\frac{1}{2}$ feet, and Jeffrey's stride measures
$ 2\frac{2}{3}$ feet. There are 5280 feet in a mile. If Jim and Jeffrey each
walk one mile, what is the ratio of the number of strides Jim takes to the number of strides Jeffrey takes? Express your answer as a common fraction.

first we figure out how many strides each of the people take

jim takes $ \frac{5280}{\frac{5}{2}}$ strides, or $ 2112$ strides
Jeffrey takes $ \frac{5280}{\frac{8}{3}}$ strides, or $ 1980$ strides

the ratio of strides Jim takes to Jeffrey takes is $ \frac{2112}{1980} \equal{} \frac{16}{15}$

Or just note that the ratio of the length of their strides are 15/16, so its the reciprocal or 16/15.


1) Show that there exist 3 consecutive integers such that each is divisible by the 1999th power of an integer.

2) Show that there are 1999 consecutive integers, each of which is divisible by the cube of an integer.

[hide="1"] Find a number which is $-1\mod{5^{1999}}, 0\mod{3^{1999}}, 1\mod{2^{1999}}$.  Then add and subtract one from it to get the two other numbers.[/hide]

[hide="2"] Similar method...take the first 1999 primes and do essentially the same thing. [/hide]


vector
function
linear algebra
linear algebra unsolved

Prove that: $ \|A\cdot B\|_{op}\leq \|A\|_{op}\cdot\|B\|_{op}$, where $ A,\; B$ are matrixes.

Which norm? There are many different choices that have this property.

It is Operator norm: $ \|A\|_{op}\equal{}\sup\{\frac{\|Av\|}{\|v\|}: \; v\in V\;\textbf {with} \;v\ne 0\}$ and we should prove that: $ \|A\cdot B\|_{op}\leq \|A\|_{op}\cdot\|B\|_{op}$.

Of course, you didn't specify which vector norm it's based on- but it doesn't matter, because the proof is the same for all operator norms. $ \frac{\|ABv\|}{\|v\|}\equal{}\frac{\|ABv\|}{\|Bv\|}\cdot\frac{\|Bv\|}{\|v\|}$ (if $ ABv\neq 0$), and $ \sup fg\le \sup f\cdot \sup g$ for nonnegative functions $ f,g$.


when is the polynomial P(X)=(X-a1)(X-a2)...(X-an) irreducible?
(a1,...,an are different integers)

You will need to tell us exactly what you call "irreducible" ...

By definition iff $n=1$ ;)
But isn't there just anything missing¿ (e.g. a $\pm 1$)

Suppose for contradiction that $f(x)=(x-a_{1})^{2}(x-a_{2})^{2}\dots(x-a_{n})^{2}+1=p(x)q(x)$. Then $p(a_{i})q(a_{i})=1$ and if for some $i,j$, $p(a_{i})p(a_{j})=-1$ then $p(x)$ has a real root and this is contradiction (because $f$ does not have real roots.). If that $p(a_{i})=1,q(a_{i})=1$. So $(x-a_{1})(x-a_{2})\dots(x-a_{n})|p-1,q-1$. Thus $p(x)=q(x)=(x-a_{1})(x-a_{2})\dots (x-a_{n})+1$ and this is contradiction. for $p(a_{i})=q(a_{i})=-1$ we have a similiar contradiction.


trigonometry
calculus
calculus computations

If $f=f(x),g=g(y)$  please show me how to solve the following:

$g(y)\frac{d^{2}f}{dx^{2}}+f(x)\frac{d^{2}g}{dy^{2}}+1 =0$

Can't separate the variables and I can't solve.Can it be solved ?

Thank you in advance.

Should there be $f(x)$ instead of $f(y)$?

It looks like you are trying to solve $\Delta u(x,y)+1=0$ (with some boundary conditions?) using separation of variables. It may be a good idea to introduce $v(x,y)=u(x,y)+(x^{2}+y^{2})/4$, because then $\Delta v(x,y)=0$.

Yeah ,I edited it.
So,it is allowed to use that substitution and PDE becomes
Laplace equation?

I would at least give it a try... the only possible drawback is that the boundary conditions for $v$ are different.


a) notam x'=yx si obtinem $tx^2y'=yx^2$ , de unde y=ct si x=0. astfel x'=ctx si din conditia initiala c=1. obtinem x'=tx, adica $x= e^{\frac{t^2}{2}+c'}$ si folosind iar conditia initiala avem  $x= e^{\frac{t^2}{2}-\frac{1}{2}}$


Prove that for $x,y,z \in \mathbb{R}$ and $x^{2}+y^{2}+z^{2}= 2$:

\[x+y+z \leq 2+xyz\]

[quote="Jan"]Prove that for $x,y,z \in \mathbb{R}$ and $x^{2}+y^{2}+z^{2}= 2$:
\[x+y+z \leq 2+xyz \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399

i have a feeling that you can mass expand

$\left[\frac{1}{2}*\left(\sum x^{2}\right)\left(\sum x\right)-xyz\right]^{2}\ge 2\cdot\left(\frac{1}{2}*\sum x^{2}\right)^{3}$

$\left[\sum x^{3}-xyz+(\sum x)(\sum xy)\right]^{2}\ge\left(\sum x^{2}\right)^{3}$
but that is a boring way...i know


If a regular polygon is inserted at vertex A and shares sides AB and AC with the 2 regular pentagons, how many sides will this new polygon have?

[hide]Let's solve thus problem in general, with $n$ representing the number of sides in the given regular polygons and $x$ representing the number of sides in the inserted regular polygon.

The degree measure of each angle in a regular polygon with $n$ sides is: $\frac{180(n-2)}{n}$

Since we have two of the same polygons and need to insert a third that shares adjacent sides, we have:

$\frac{180(n-2)}{n}+\frac{180(n-2)}{n}+\frac{180(x-2)}{x}=360$
$\frac{360(n-2)}{n}+\frac{180(x-2)}{x}=360$
$\frac{360(n-2)}{n}-360=-\frac{180(x-2)}{x}$
$\frac{360(n-2)-360n}{n}=-\frac{180(x-2)}{x}$
$\frac{720}{n}=\frac{180x-360}{x}$
$720x=n(180x-360)$
$x=\frac{2n}{n-4}$

Letting $n=5$ as stated in the problem yields $x=\boxed{10}$
[/hide]


Find all functions continuous $ f: R\rightarrow R$ such that :
$ f(f(x))\equal{}f(x)\minus{}\frac{x}{5}$

I tried but I couldn't
I will continue trying
bye

[quote="conan6032"]Find all functions continuous $ f: R\rightarrow R$ such that :
$ f(f(x)) \equal{} f(x) \minus{} \frac {x}{5}$[/quote]
Nobody can't solve it???  :(


Hello, this topic is 2 discuss INO's

Since I gave INPhO and INChO i'll tell u sumthing.

Bansal topper is getting 75/80 in INPho and and 75/100 in INChO (my marks r not worth telling).Post ur marks  as well as that of people u kno. :lol:

Doublepost ..spam :P
better continue the discussion in GD

betw ,i m getting 54 in INPhO

rank 1 bansal u mean shubham tulsiani???

u ppl r from which batches in bansal............???

No one frm Bansal.INChO paper out.Post answers

InPhO also out. I am gonna try it now. Will post answers in 3 hours.  :)

Its less than 3 hours but I am done with the paper.Since this paper was either you know it or you dont type. And sadly it was easier than JEE .

My answers:

1.  c
2.  b
3.  b
4.  d
5.  c
6.  d
7.  ?
8.  d
9.  ?
10 ?
11 c
12 d
13 ?
14 ?
15 ?
16 a
17 ?
18 a
19 ?
20 ?
21 c
22 ?
23 c
24 ?
25 c
26 a
27 a
28 ?
29 ?
30 c
31 d
32 ?
33 c
34 a
35 c
36 c
38 c

B

1.c
2 ?
3 ?
4. b
5. e
6. e
7. c
8. a
9. a
10.b
11.e
12.b
13.a
14.b
15.b
16.c
17.d
18.e
19.a
20 c

C

1. $ P_0(1 \minus{} \frac {V}{V_0})$

2. $ P_0(1 \minus{} \frac {V}{V_0})\frac {V}{R}$

3.$ \frac {P_0}{r}(1 \minus{} 2\frac {V}{V_0})$

4.$ \frac {P_0V_0}{4R}$

5.
6.$ \frac {R}{\gamma \minus{} 1}$

7.$ \frac {R}{\gamma \minus{} 1} \plus{} \frac {P_0}{R}(1 \minus{} \frac {V}{V_0})(1 \minus{} 2\frac {V}{V_0})$

8. $ \frac {3}{2}$

9.$ Put \gamma \equal{} \frac {3}{2} in 7.$

10.

Now I will look into the questions I couldnt do.

HEY GUYS...

How did you ppl calc your marks without the keys....

can anyone post the right and complete keys if possible...

and on the above post. is your 4th ans correct? i got b

There are few mistakes in my solutions ofc. I am not sure of the 4th one. Also 6th one outta be a. And I found answer to 9th and 28th.

no chem lovers here???
i m getting 50 or so.. any chance???????????

I dunno about the chem cut off.
But i will post the correct keys after open test(8 feb).

chemrok said d cutoff will be 50-55.. agree/ disagree??


[quote="chatriyan"]yesterday i went for a long running around the local lakes. as the running was quite long, i runned at various stages. in the first stage i runned half of the overall distance, stage two saw half of the remaining distance plus 35 meters covered. stage three covered three-quarters of the remaining distance. stage four completed half of the remaining distance plus 75 metres. stage five completed the journey with a final burst of 150 metres. how far did i run in total ?

give me how u solved in steps[/quote]
[hide="If this is for homework, you should try it yourself again before asking people here."]
Go backwards.
1. Stage 5 was 150 meters, so start with that.
2. 150 was left after running Stage 4 (1/2 of what was left plus 75), so 150=x-(x/2+75)
3. Now for math: 150=x/2-75, so 225=x/2, so x=450.
4. Now you have 450.
5. Stage 3 was 3/4 of the distance left, so 450=y/4 (I'm using new variables as to avoid confusion)
6. More math: 450=y/4, so 1800=y.
7. Now you have 1800.
8. 1800 was left after running Stage 2 (1/2 of what was left plus 35), so 1800=z-(z/2+35) (Again, new variable).
9. Even more math: 1800=z/2-35, so 1835=z/2, so 3670=z.
10. Now you have 3670.
11. 3670 was half of the total, so 3670=a/2 (again, new variable)
12. MATH!!: 3670=a/2, so 7340=a
13. That is your answer.[/hide]
Based on your grammar, I'd say you just didn't feel like doing this, but oh well.
FYI, it's "ran", not "runned". JSYK.

But... As of Now.... U dont have rites to comment on Me.....

Just keep ur Dick Out

1) Edit your post if it's < 2 days
2) Usually, it's beneficial to new members to be commented on by older members so they can see what their doing wrong. This happened a lot to me at the beginning.
3) Language like that is going to get you banned.

dude, someone is wasting their precious time in order to help you with whatever this may be. It could be your hw, but who knows? Plus, it really isnt a brainteaser/puzzle. So you should
1: thank him without any mouthing back
2: watch your language
and
3: apologize to erabel. He wasnt writing anything bad abt you :mad:

take a chill pill, and then chillax

you are seriously harshing my mellow

i feel sad... your comment was mean... apologize... and no language, if you don't mind.

I think he just wanted a free answer, so just made a random account to ask.

If you want free answers, you should really go to Yahoo Answers.....


Helo
take two numbers $ \alpha$ & $ \beta$ from $ \mathbb{R}\minus{}\{\frac{\pi}{2}\plus{}k\pi$.

we have: $ P(x)\equal{}ax^2\plus{}ax\plus{}b$

suppose that: $ P(tan\alpha)\equal{}P(tan\beta)\equal{}0$

prove this: $ P(tan(\alpha\plus{}\beta))\equal{}\frac{b}{cos^2(\alpha\plus{}\beta)}$

$ \beta \equal{} 0 \implies 0 \equal{} \frac{b}{\cos ^2 \alpha} \implies b\equal{}0$  :huh:


The problem makes the tacit assumption that [color=blue]a piece of gum (of any color) costs an integer number of cents[/color]. Once agreeing on that, since $ 129 \equal{} 10\cdot 12 \plus{} 9$, it means that we need that many red gumballs (costing $ 3$ cents more), so that to offset the remainder $ 9$ of $ 129$ when divided by $ 12$. The available values are $ 3$, $ 7$ and $ 11$, but we are told there were more white gumballs than red, so the only value fitting is $ 3$.
Therefore there are $ 9$ white gumballs, each costing $ 10$ cents, and $ 3$ red gumballs, each costing $ 13$ cents.

[hide="What I got"]Let $ a=\text{\# of white gumballs}$ and $ b=\text{cost of white gumballs in cents}$. We have:
\[ a \cdot b+(12-a)(b+3)=129 \\
ab+12b+36-ab-3a=129 \\
12b-3a=93 \\
4b-a=31 \\
4b=31+a\]
But since $ a$ is the number of white gumballs, it is a positive integer and $ 6<a \le 12$. After substituting this back in, we have:
\[ 37<4b \le 43\]
Assuming $ b$ is an integer, we see that the only possible value of $ b$ is $ 10$, which we plug back in and find everything else:
\[ 4b=31+a \\
40=31+a \\
a=9\]
And since there must be a total of 12 gumballs, the number of red gumballs is $ 12-9=3$ and their cost is $ 10+3=13$ cents.

Finally, we have $ \boxed{\text{3 red gumballs at 13 cents each and 9 white gumballs at 10 cents each}}$.[/hide]

[quote="spaceguy524"]Anna bought 12 pieces of gum consisting of only red gumballs and white gumballs. The total cost was $ \$1.29$. The red gumballs each cost 3 cents more than each white gumball, and she bought fewer red gumballs than white gumballs. How many white gumballs did she buy?

I need help on this problem. Please post how you got your solution and show your work, so I can understand it. ($ \text{\LaTeX}$ would be nice too.)[/quote]
you could just solve it algebraicly.':ddr:


>
>

Chris and Pat have taken 40 courses together.
There are 100 points maximum available in each course per student.

For each course the grading system is:

A  90 -> 100
B  80 -> 89
C  70 -> 79

A  4.0
B  3.0
C  2.0

After the completion of these 40 courses (but no other courses than these 40 have
been taken by either student), Chris has a grade point average of 3.375, and a grade
point average of 2.500 is held by Pat.

Is it possible for Pat to have a greater total number of points across all 40 courses
as compared to the total points for Chris?

[quote="minime1235813"]No they cant. Because: A = Average of points (given), P = All points added up (were finding this) N = Number of items (40)
A = P/N
For Chris, 3.375 = P/40; P = 135 so thats 135 points added up
For Pat, 2.5 = P/40; P = 100
135 > 100, so No.[/quote]

[quote="Arrange your tan"]
Chris and Pat have taken 40 courses together.
There are 100 points maximum available in each course per student.
[/quote]

No, you don't have the right understanding.  With 40 courses apiece
and a maximum of 100 points per course, the students have their
totals out of $ 40\cdot100 \equal{} 4000$ points apiece maximum.

* * * Open Hint/Discussion:

This situation can vary, but for relative simplicity, suppose that
Chris gets either an 80 or a 90 in the 40 courses, and that Pat
gets either an 89 or a 79 in the 40 courses.

Let x and y be appropriate positive integers for the next inequalities.

Then, for Chris's points to be less than Pat's points, you might have:

(x)(90) + (40 - x)(80) < (y)(89) + (40 - y)(79) . . . (*)

Look at the values for the letter grades.  Chris gets the upper two
(4 and 3), and Pat gets the lower two (3 and 2), *if* you settle for
those grade limitations they are given as discussed as above.

(x)(4) + (40 - x)(3) > (y)(3) + (40 - y)(2)

This becomes:

$ y < x \plus{} 40$ . . . (Though this is some upper limit, I did not make use of this in
 . . . . . . . . . . . . . . my solution in a direct way.)

This becomes:

$ y \ge x \plus{} 5$.

The values of x and y can vary.

End of the Open Hint/Discussion * * *

For my choice of an answer:
(The answer can vary.)
[hide]
Let x = 15
Let y = 20

Using (*) from above, x = 15 gets substituted on the left-hand side and
y = 20 gets substituted on the right-hand side:

[(15)(90) + (25)(80)] points versus [(20)(89) + (20)(79)] points

3350 points versus 3360 points

So Pat has more total points than Chris, but he/she has a lower grade
point average than Chris.

Grade point average work for Chris:

(15)4 + (25)3 = 60 + 75 = 135

135/40 = 3.375

-------------------------------------------------------------

Grade point average work for Pat:

(20)3 + (20)2 = 60 + 40 = 100

100/40 = 2.500
[/hide]


What is the sum of all primes $p$ for which $p+1=2x^2$ and $p^2+1=2y^2$  and $(x,y)\in\mathbb{Z}$

am i missing something in this question.  It seems kinda open ended and vague

I have a partial solution:

Squaring the equation $p+1 = 2x^2$ yields $p^2 + 2p + 1 = 4x^4$, so that $p^2 + 1 = 4x^4 - 2p$.  But $p^2 + 1 = 2y^2$, so $2y^2 = 4x^4 - 2p$, or $2y^2 + 2p = 4x^4$.

Now, subtracting 2 from both sides of $p+1 = 2x^2$ and squaring both sides yields $p^2 - 2p + 1 = 4x^4 - 8x^2 + 4$.  But again, $p^2 + 1 = 2y^2$, so our equation becomes $2y^2 - 2p = 4x^4 - 8x^2 + 4$.

Adding $2y^2 + 2p = 4x^4$ to $2y^2 - 2p = 4x^4 - 8x^2 + 4$ yields $4y^2 = 8x^2 - 8x^2 + 4$, or $y^2 = 2x^4 - 2x^2 + 1$.  But the only integral solutions to this equation are (x,y) = (-2, 5), (-1,1), (0, 1), (1,1), (2,5).  The corresponding values of $p$ are 7, 1, -1, 1, and 7.  Of these, the only prime is 7, so the sum is 7.

However, I don't know how to prove that the five above solutions are the only integral ones.  Can anyone do this?


integration
trigonometry
symmetry
calculus
calculus computations

Ascertain a recurrence for $ a_n \equal{} \int_0^{\infty }\frac {x^{n \minus{} 1}\cdot\arctan x}{(x^2 \plus{} 1)^{n}}\ \mathrm {dx}$ , where $ n\in \mathcal N\ .$

Re-shaping this: let $ t=\arctan x.$ Then $ dx=\sec^2t\,dt,(x^2+1)=\sec^2t,x=\tan t.$

$ \int_0^{\frac{\pi}2}\frac{t\tan^{n-1}t\sec^2t}{\sec^{2n}t}\,dt=
\int_0^{\frac{\pi}2}t\sin^{n-1}t\cos^{n-1}t\,dt.$

Now by a symmetry argument, this is

${ \frac{\pi}4\int_0^{\frac{\pi}2}\sin^{n-1}t\cos^{n-1}t\,dt=
\frac{\pi}{2^{n+2}}\int_0^{\pi}\sin^{n-1}\theta\,d\theta}$

Recursions are easy to write for this last form.

The integral can be calculated, without using any recurrence relation, as follows:

Let $ I$ be the value of your integral.

Using the substitution $ x \equal{} 1/y$ we get, after  straight forward calculations, that

$ I \equal{} \frac {\pi}{2}\int\limits_{0}^{\infty}\frac {x^{n \minus{} 1}}{(1 \plus{} x^{2})^{n}}dx \equal{} \frac {\pi}{4}\beta(n/2,n/2)$.

Cant we apply Wali's formula to the expression Mr Merryfield got?
i.e. $ sin^{n\minus{}1}t . cos^{n\minus{}1}t$


Two vectors (2D) S and E define arc with center in Z(0,0). How can I find if some vector D starting in Z(0,0) intersect the arc(S,E,Z).

inline bool PVector2::BetweenVectors( const PVector2 &vecStart, const PVector2 &vecEnd, DIR direction ) const
{
	float angle1 = Angle( vecStart );
	float angle2 = vecEnd.Angle( vecStart );
	if ( direction == DIR::CCW )
	{
		if ( angle1 < 0 ) angle1 += PC_TWO_PI;
		if ( angle2 < 0 ) angle2 += PC_TWO_PI;
		return angle2 > angle1;
	}
	else
	{
		if ( angle1 > 0 ) angle1 -= PC_TWO_PI;
		if ( angle2 > 0 ) angle2 -= PC_TWO_PI;
		return angle2 < angle1;
	}
}


Find the values of $ \sin 18^\circ ,\ \cos 18^\circ$.

We have $ \sin{54^0} = \cos{36^0}$

$ \Longleftrightarrow \sin{3.18^0} = \cos{2.18^0}$

$ \Longleftrightarrow 3\sin{18^0} - 4\sin^3{18^0} = 1 - 2\sin^2{18^0}$

Let $ t = \sin{18^0}$ we get equation: $ 3t - 4t^3 = 1 - 2t^2$

$ \Longleftrightarrow 4t^3 - 2t^2 - 3t + 1 = 0$

$ \Longleftrightarrow (t - 1)(4t^2 + 2t - 1) = 0$

$ \Longleftrightarrow$ $  \|{ \begin {array}{c} t = 1 \\
t = \frac { - 1 + \sqrt {5}}{4} \\
t = \frac { - 1 - \sqrt {5}}{4} \end {array}}$

But $ t = \sin{18^0} \Longrightarrow 0 < t < 1$, hence this equation have only root $ t = \frac { - 1 + \sqrt {5}}{4}$

Your solution is important as well., tanhnam2902. :lol:

You can type $ 18^\circ$ as this.
[code]$18^\circ$[/code]


Gauss calculus integration trigonometry search

Consider triangle ABC,the continues charges distribution lies on the side of BC with the linear charge density such as m,prove that the direction of electric force in the point A lies on the bisector of the viewing angle BAC

[size=150][color=blue]
MOD EDIT:TOO BIG AND COLOURFUL[/color][/size]

COnsider a cylrndrical gaussian surface of lenght = lenght of the side BC....

Nopw use gauss law to find the Electric field at A......

I don't know gauss law,I want a simple solution  :(  :(  :(

The Gauss Law solution works only when the charge distribution is infinite.

You can obviosly do this by integration, but I think that it can be done using only trigonometry. I'll try to put up a picture.

very very thanks if you can prove that by integration or trigonometry writte it,I don't success with both of two way  :(

actually [url=http://www.mathlinks.ro/viewtopic.php?search_id=1666719383&t=45864]here[/url] is a detailled discussion with some wonderful methods on this problme

That's a really nice solution

You might want to try the sequel:
Using this result, or otherwise calculate the electric field in a plane perpendicular to a long charged rod that contains one of the end points of the rod.


Ross Mathematics Program
induction
number theory unsolved
number theory

We call a set of natural numbers { x1 , ... xn }, a " complete set" if for all i we have: xi | x1+ x2 + ... + xn.

 Prove that for every set of natural numbers: A, there exists a complete set :B such that A is a subset of B. :?

This solution is wrong. So I deleted it.

Everything is OK, BUT sets $S(n,i)$ are not mutually disjoint. As I understood your  solution is based on the fact that
\[1+\sum_{i=2}^{n-1}\sum_{j\in S(n,i)}j=n!\cdot (n+1)/2.\]

let n >= 2 be a positive integer, the set
$S_n = \{1,2,...,n,\frac{n(n+1)}{2}\} \cup \{\frac{(n+1)!}{k!}-\frac{(n+1)!}{(k+1)!}:k=1,2,...,n-2\}$
is a complete set

for any set A that contains only natural numbers, let m be the largest member of A, let n be any positive integer such that n >= m and n >= 2, then A is a subset of S_n

Pete_Ross I solved this problem,and my solution is similiar.
But I have seen a solution with induction


I've done all the Mathcounts tests from the 1998-1999 year to 2004, and I've run out of practice tests! Does anyone know a website where I can print/buy old competition sets from before 1998?

I have all the nationals rounds, but I got them from my nationals coach, so I couldn't tell you where to get them. I'm sure you could buy them from MathCounts, somehow.

Can anyone scan them and send them? (if its not a violation). I can't get a hold of them anywhere and my nationals coach has only 1 year.

[quote="mysmartmouth"][quote="09husbri"]If somebody proves to me that it's not a violation, I'd be happy to send you some tests.[/quote]

Let's write a letter to MathCounts for permission; I would also like to scan and distirbute some old MathCounts tests.  It's not like they trying to make money; they are a nonprofit organization.  But maybe they need money to support all the free stuff they give us at Chapter and State, not to mention the trophies and plaques....[/quote]
I know but I dont have 100+ dollars to spend on old tests.

I wouldn't mind buying competitions prior to 1999, but they are not even on the website for purchase. If they aren't even available to buy, I don't see why MathCounts would mind distributing old sets for free.

I believe that the only way to procure older tests is to get them from experienced coaches and such.  Since I'm assuming you've gotten to nats, your state coach or other team members may have old tests.  I don't think your actually supposed to distribute these at all, but I don't know for sure.

This is the first year at nats for all of my teammates and me. My coach only has 1 year of nats (i did them). I don't think its a copyright violation.

[quote="chess64"]Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Scan them! Okay sorry. :blush: [size=184][b]Please scan them![/b][/size]   :D[/quote]

I will NOT scan them w/o written consent from MATHCOUNTS or AoPS.  So if you want me to scan them for you, sure, but first, forward me permission from MathCounts!  You can email me at mysmartmouth@gmail.com

Rounds from before 1990 iare not sold by MathCounts, so there is no copyright violation.  They are "out of print" and there is no liability.  It is like classic books like Moby Dick.  Every publishing Co. prints classics, b/c there is no copyright, b/c the original author or his/her co's/heirs do not print the books anymore.

If you have to buy them off MathCounts, it's illegal, but I don't mind if you guys scan it.

[quote="Press alt f4"]I have some random problem sets that m,y coach gave me if you want those...[/quote]Like not part of a set? Sure.

[quote="Press alt f4"]I have some random problem sets that m,y coach gave me if you want those...[/quote]

I'll take them plz

There are lots of problems from old nationals in the All-Time Greatest MATHCOUNTS Problems book  . . .  you can buy it on the mathcounts website. A lot of those problems are also on the elias saab website.

Mathcounts stops printing competition books after about five years. I don't think it's illegal to scan the really old ones. You can also get them from people on the problem writing committee (our coach got them from Harold Reiter). Good luck finding them.

[quote="Press alt f4"]I have some random problem sets that m,y coach gave me if you want those...[/quote]

Sure! How can I get them?

I cant find most of them now but I have a few still I gave the rest back to my coach a few days ago.

[quote="Phelpedo"]I found 95-96 sprint...[/quote]WHERE????

I have ones as old as 1989... to 2004. (hehe... my teacher had some, my brother had some, and I had some :D ) I also have them on CD.

Wasn't 2004 unusually hard though? (compared to the previous few years anyway). Maybe we're seeing the beginning of a new trend.

[quote="kingof21penguins"]1990 is the hardest year but my coach has all nats from 1989 to 2004[/quote]

1990 was the easiest! It was the only year not to have problems in the all-time greatest book. Besides, I got a 24 on the national sprint that year.

heh my coach has everything from 1981 or something to 2003

I think a few people are working on creating complete Nationals sets for EVERY year.  I will post more information here (as will countless other poeple, I'm sure) if and when they become available.


geometry
3D geometry
analytic geometry
combinatorics proposed
combinatorics

1000 points reside in a 10x10x10 grid. Some of the points are colored red, and the rest are black. Find, with proof, the maximum number of dots that can be colored red such that no three red points lie in the same vertical, horizontal, or sideways line (they can be in the same diagonal line).

Let $ 10 \equal{} n$.  Look down any face of the cube and observe that along each of $ n^2$ lines you may see at most $ 2$ dots, so $ 2n^2$ is immediately an upper bound.  In fact, it's not difficult to achieve: coordinatize the points as $ (x, y, z)$ where each coordinate is taken modulo $ n$, and color red exactly those points such that $ x \equal{} y \plus{} z$ or $ x \equal{} y \plus{} z \plus{} 1$.  (As you look through the slices $ z \equal{} 0, z \equal{} 1, \ldots, z \equal{} n \minus{} 1$, this gives you two "rotating diagonals.")

Much harder follow-up: compute the number of ways to color $ 2n^2$ dots in an $ n \times n \times n$ cubic grid so that every axis-parallel line contains exactly two colored dots.  (Probably we want the answer in the form of some kind of generating function.)


Current top 30.
We usually take a team of 15, two alternates, and anyone who wants to pay their way entirely.

Name	Vestavia		Grissom			Total	School
Seva Tchernov	10		11			210	Vestavia
Johnny Hu	9		10			190	Grissom
Evan Zhao	10		9			190	Vestavia
Wei Yan	10		9			190	Vestavia
Wayne Zhu	10		8			180	Vestavia
Ryan Muller	10		7			170	Vestavia
John Huang	10		6			160	Vestavia
Kevin Hu	9		7			160	Vestavia
Zhenya Kaliberova	8		8			160	Vestavia
Kurt von Laven	8		7			150	Randolph
Chris Thomas	8		7			150	Vestavia
Liangsha Xu	5		10			150	Vestavia
Alan Chou	6		8			140	Grissom
Kevin Chou	7		7			140	Grissom
John Reinhardt	6		7			130	Vestavia
Steven Huang	6		6			120	Randolph
Ali Cem Randa	6		6			120	Vestavia
Amy Wu	5		7			120	Vestavia
Alex Crichton	5		6			110	Vestavia
Nathan Bishop	7		4			110	Vestavia
Parker Robbins	5		6			110	Vestavia
Nikhil Goil	5		5			100	Grissom
Anne Weng	5		5			100	Vestavia
William Desmond	3		7			100	Vestavia
Yichen Feng	4		6			100	Vestavia
Matthew Nubbe	-		9			90	Huntsville
Erin Schmale	6		3			90	Randolph
Carter Spires	3		6			90	Vestavia
Steve Kim	4		4			80	Grissom
Sanket Shah	3		5			80	Randolph

Eventually I'll make a website for this.... (maybe)

yay! steven and sanket went to mathcounts with me. =)

so who's on the alabama arml team this year?


Which of the following belongs to this precondition: The velocity vectors connecting position A to the adjacent positions appear to have the same magnitude and direction. Therefore, the acceleration is zero—and so is the net force.

A. A car is moving along a straight road at a constant speed.
B. A car is moving along a straight road while slowing down.
C. A car is moving along a straight road while speeding up.
D. A hockey puck slides along a smooth (i.e., frictionless) icy surface.
E. A hockey puck slides along a rough concrete surface.
F. A cockroach is speeding up from rest.
G. A rock is thrown horizontally; air resistance is negligible.
H. A rock is thrown horizontally; air resistance is substantial.
I. A rock is dropped vertically; air resistance is negligible.
J. A rock is dropped vertically; air resistance is substantial.

I'll start you off:
A and D both meet the pre-condition.  Why?  Because there is no change in the velocity vector.


[quote="kstan013"]hey everyone i'm starting a cool club and you can only be in it if you're cool so post here anyone who's cool and if you're not cool then BACK OFF[/quote]

The old thread got locked.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142080

put quotes around "kstan013" to make the quote box :wink:

[code][quote="kstan013"]blahblahblah[/quote][/code]
[quote="kstan013"]blahblahblah[/quote]

Dude, 1=2 turned traitor to the cool club.
And what's to stop me from putting "Proud member of the cool club" without pming u for the translation?  :D

IT'S JUST A STUPID CLUB!!! DO NOT PM ME ASKING IF I CAN BE IN THE CLUB!!!!!!


geometry
modular arithmetic
number theory

this is in number theory but i dont know what it means.i read the definitions but still i dont get it.
$ \overline{k}\equal{}$???
$ \overline{0}\equal{}$?
$ \overline{1}\equal{}$?
$ \overline{2}\equal{}$?

I think it depends on the number of digits.  I have a Vedic Arithmetic book that uses these, I think they're called "vinculums."
First note that in a normal base ten numbering system, for a number you'll have something like ...1000A+100B+10c+1D.  In our traditional system, we always use + signs.  However, theree is another way of writing numbers that makes a digit "negative."  For example,
$ AB\overline{C}D\equal{}1000A\plus{}100B\minus{}10C\plus{}D$.
This system can be used for simplification.  For example, we can write
$ 1993$ as $ 200\overline{7}$.
There are probably details in your book about the cool ways you can use vinculums.

It has a lot of different meanings depending. Could mean a repeating number, in physics it's used to indicate sig figs.

True, but judging from the context of the post...
binomial_4eva, could you give us some more details on how this is used?

[quote="binomial_4eva"]ok, it is in the first chapters of number theory book. so its to do with number theory, i think.
sorry about my first post, i know it is confusing because they also use it in geometry to represent angles i think.
but ok in this book this is written as $ \overline{k}\equal{}\{m\in\mathbb{Z}\mbox{such that }m\equiv k\pmod n\}$ i dont know what that means.[/quote]

That denotes a residue or congruence class. For example, in modulo 3 the classes are:

$ \overline{0}_{3}\equal{}\{...,\minus{}6,\minus{}3, 0, 3, 6, ...\}$
$ \overline{1}_{3}\equal{}\{...,\minus{}5,\minus{}2, 1, 4, 7, ...\}$
$ \overline{2}_{3}\equal{}\{...,\minus{}4,\minus{}1, 2, 5, 8, ...\}$

As you can see, $ \overline{k}_{n}$ is the set of all integers that are congruent to $ k\pmod{n}$, jsut as your book says.

Ew. When was that book published? The standard notation for congruence classes would be $ [0]_{3}, [1]_{3}, [2]_{3}$ instead of $ \overline0_{3},\overline1_{3},\overline2_{3}$.


I have a question,
Given a positive integer $a, a>1, a$ being a non square ,does there always exist an odd  prime $p$,such that $a$ is a primitive root in the multiplicative group $mod(p)$ ?

Thanks
Cheers
Bhargav

Artin's conjecture, it's open.
It was proven that there are at most two $a$ such that it is wrong.


[quote="tarquin"][hide]Let $b$ be the common difference. Then

$3(a_1 + 7b) = 5(a_1 + 12b)$

$b = -\frac 2{39} a_1$

We know enough information to find $S_{20}$ and $S_{21}$ in terms of $a_1$ so we have

$S_{20} = \frac {10}{39} a_1$ and $S_{21} = -\frac {21}{78} a_1$ so clearly $S_{20} > S_{21}$[/hide][/quote]
Nice solution, tarquin.
I wonder if there is any other solutions?


$F(x) = 20$, $f(7) = 24$.
the forumla is $a (b^x)$ so solve $f(x) = 20$ and $f(7) = 24$ using $a(b^x)$

[size=75][color=green]Learn $\LaTeX$! - mod[/color][/size]

i'm trying to learn latex....lol

[quote="Marius_P"]Hy, i've been trying to solve this problem, and I have a question: is it a*b^x or (a*b)^x???[/quote]
It's a*b^x.

its y=a times b^x power

y= a times b raised to the x power


advanced fields
advanced fields unsolved

Give an example of an uncountable numberfield,naturally,except for $ \mathbb{R},\mathbb{C}$.

number fields are finite extensions of $ \mathbb{Q}$, in particular they are countable. so you use another definition of number fields?

"number fields are finite extensions of $ \mathbb Q$"?If so ,how can we get the real number field $ \mathbb R$? :)

As completion of $ \mathbb Q$¿
To rephrase -oo-:
Number Fields are _defined_ to be finite algebraic extensions of $ \mathbb Q$. Thus your question makes no sense.

"Number field" in the traditional sense means just "subfield of $ \mathbb C$", cf. [url=http://eom.springer.de/n/n067910.htm]EOM[/url]. In modern usage, "number field" usually means "algebraic number field" in the sense given above; note that this definition allows for "abstract" extensions of $ \mathbb Q$ with no preferred embedding into $ \mathbb C$.

In the traditional sense, the question makes perfect sense: can we write down explicitly any uncountable subfields of $ \mathbb C$ besides $ \mathbb R$ and $ \mathbb C$ itself? owk

Not that it is very explicite, but it's closely related to http://www.mathlinks.ro/viewtopic.php?t=177829 .


linear algebra
matrix
algebra
polynomial
vector

Question from Sheldon Axler's book:

Find a linear transform $ T$ acting on $ R^4$ with no (real) eigenvalues.

I think we can do this using the characteristic polynomial, but in the book we have not yet discussed the characteristic polynomial. How do we go about finding such a transform without the characteristic polynomial?

You're going to need something equivalent to the characteristic polynomial, at least for this one matrix.

I'd go with $ T\equal{}\begin{bmatrix}0&1&0&0\\\minus{}1&0&0&0\\0&0&0&1\\0&0&\minus{}1&0\end{bmatrix}$ as a simple example. The determinant of $ T\plus{}\lambda I$ is $ (\lambda^2\plus{}1)^2$, which is positive for all real $ \lambda$.

How do we work backwards from our polynomial $ (\lambda^{2}\plus{}1)^{2}$ to construct our desired matrix $ T$?

Also, I am pretty sure there is a way to find such a $ T$ without this method, beacuse Axler put this question 3 chapters before the characteristic polynomial is discussed, it is no accident. It is the second last question of the chapter 'eigenvalues and eigenvectors'.

If you prefer to avoid the concept of characteristic polynomial, you can show directly that jmerry's matrix $ T$ has no real eigenvalues.  Namely, suppose that $ Tx \equal{} \lambda x$ and see what you get.

Ok, thanks Rabi B, you pretty much answered my question. Its quite obvious actually. Consider the equation $ Tx \equal{}\lambda x$ for an arbitrary matrix with entries $ a_ij$ and constrain the entries $ a_ij$ such that the solution must be that $ x\equal{}0$. This is messy but we can try before hand to take a matrix which just permutes the co-ordinates and maybe adds minus signs and see if we can constrain our $ x$. We can.

Notice that in order to have an $ n\times n$ real matrix with no real eigenvectors, you must have $ n$ be an even number. This has a simple reason: an odd degree real polynomial must have at least one real root.

For the example I used, it doesn't have to be messy. Look at the sum of the squares of the entries in the vector; multiplying by $ T\minus{}\lambda I$ multiplies that by $ \lambda^2\plus{}1$.

I don't understand what your saying here? I was saying that finding a $ T$ such that $ T(x)\equal{}\lambda x$ constrains $ x$ to be $ 0$ is messy unless we guess that we can just have a permutation matrix for $ T$ with some minuses thrown in.

... which is why I didn't say anything about a more generic $ T$. Mine [i]is[/i] a "permutation matrix with some minuses thrown in". It's also an orthogonal matrix, which enables the other clean method I mentioned.

What I meant is that I was hoping that you would explain this a bit more to me:

[quote="jmerry"] For the example I used, it doesn't have to be messy. Look at the sum of the squares of the entries in the vector; multiplying by T-\lambda I multiplies that by \lambda^2+1.[/quote]

why are we looking at the sum of squares of the entries in our vector?