how far did you get in competition last year?

never done it, never heard of it till this year, and never will since my city doesn't do it

 :o

He is trying to say he doesn't like MathCoutns....why?

Beats me.  I like Mathcounts.  Maybe he was pushed too much and hated doing the problems?

I was an alternate for my school.  An eighth-grader (last year) got to nationals, along with one of the moderators here...

I had the state swim meet and didn't like MC that much anyway then so I decided not...to do it.

not saying that i dont like mathcounts itself, just the time spent on it could be used somewhere else...

Made it to nationals and bombed. Like nuked. Without all of the stupid mistakes (like retarded stupid) I could have been up to 80 or more places higher (no exaggeration)

And instead of top 50, you really should say top 57.

[quote="binonunquineist"]Come on, everybody always says they bomb stuff. Its always 'I could have gotten this question right, I could have gotten this other question right, I should have gotten higher a few places, I should have won, blah blah blah if it was like that then we would have a few hundred ties for first place with a perfect score  :lol:[/quote]
I know right. I got some stupid mistakes too, but what happens happens. If you never perform well in the competitions due to pressure then it doesn't matter how much you know.

ya, when i was looking back to posts right around national time and after states, i was so annoyed when people were like, "crap, i got a 45!!!!"

-jorian

There are about 6 different topics on this already including definitely at least 2 with a poll that I remember.

[quote="Fanatic"]NONE of your SOB stories can top this:

with 7th & 8th on the target left to go I'm in 2nd place only to Nathan B (I had binon here earlier,flipped em in my head)

I miss the EASIEST #7 EVER, NOT EXAGGERATING.

I then CHANGE THE ANSWER on #8 a SPLIT SECOND before time was called from right to wrong. NOT EXAGGERATING.

Drop to 5th place.

I'm over it now though.

Edited to add (ETA): I agree with binonquiniest about the "shoulda woulda coulda" thing. I was just posting my sob story to (hopefully) make some of u guys feel better.[/quote]

I feel bad 4 u, i know how it feels.
SUcks for me, i screwed up at schools, didn't get to compete at all

30/16 at chapters
27/16 at states
23/16 at nationals

target round is broken with a ti-84 and knowledge of basic ti-basic.

[quote="bpms"][quote="kyyuanmathcount"][quote="Fanatic"]Indiana:

44, 41, 40, 39, 38 -- unknown.

aww i would've made massachusetts (26/12 split is great tiebreaker), oh well.[/quote]

22/6 is even better at nationals.[/quote]
23/8? How would that stack for the tiebraker?[/quote]

Eh...I'd say we'd both win a good number of em.  But still, 22/6 is better ^_^.

Which ones did you get?  I got 3,4,5

[quote="mathgeniuse^ln(x)"]I don't know...

But, I could've gotten a 26, as you sawwith my rants...on Sprint...but no I didn't...because I suck at addition and multiplication...

I got 1, 2, 4, 5, 6, 7,

I was surprised this year since the first sprint I got wrong was #13...[/quote]

My competition is a flip of practice.  I usually get full on targets and mess up sprints, but the competitions:  30/16, 27/12, 22/6.

I also got the first 10 or so right.

[quote="kyyuanmathcount"][/quote][quote="bpms"][quote="kyyuanmathcount"][quote="Fanatic"]Indiana:

44, 41, 40, 39, 38 -- unknown.

aww i would've made massachusetts (26/12 split is great tiebreaker), oh well.[/quote]

22/6 is even better at nationals.[/quote]
23/8? How would that stack for the tiebraker?[/quote][quote="kyyuanmathcount"]

Eh...I'd say we'd both win a good number of em.  But still, 22/6 is better ^_^.

Which ones did you get?  I got 3,4,5[/quote]
The 23/8 isn't me...
I got a 21/12, If I hadn't tried to bf number 28 I would've had a 23, but I learned to skip hard problems and come back. For number six on the target I forgot that you can switch the price of the item and the change so I got half the answer
:roll:

[quote="bpms"]At our states a 32 got 25th place...[/quote]
At our state a 32 got 3-7th place...

[quote="jhredsox"][quote="frost13"]Since there is all this talk about great tie breaker scores at Nats, her is how my student did at Nats last year :

15/16  Gotta be worst tie breaker score ever.  Especially when only like 8 people got all the Targets and most were in the Countdown.[/quote]

was he/she encountering brain damag during and only for the sprint?

-jorian[/quote]

No, he was just way better at Target than Sprint.  He could do any problem, just wasn't very fast.  Need several different skill sets to reach the top.  He was still 40 something though.  Not bad since he only got 33 at state.

[quote="sonny"][quote="Ubemaya"][quote="mathgeniuse^ln(x)"]
Michigan....hmm...the difference between second and third place written was 3...

42
39
[/quote]

In California the difference between second and third place was 0...

41
41

As a matter of fact, the difference between third and fourth was also 0...

41
41

And the difference between fourth and fifth was 0...

41
41

And the difference between fifth and sixth was 0...

41
41

Seventh and eighth (and maybe ninth) got owned, they only got a 40.[/quote]

Dude are you sure about that? Becasue if I recollect correctly, my friend got a 37/39 (one of the two not so sure) at states in Nor Cal and he got 8th place and my other friend got a the same score and he got 11th place. But then again, I may recollect incorrectly.[/quote]

Are you from Miller?
I think 8th place got a 40, it could have been 39. I'm almost positive that 11th place was Felix (forgot his last name) who got a 38.

[quote="bpms"][/quote][quote="Ubemaya"][/quote][quote="sonny"][quote="Ubemaya"][quote="mathgeniuse^ln(x)"]
Michigan....hmm...the difference between second and third place written was 3...

42
39
[/quote]

In California the difference between second and third place was 0...

41
41

As a matter of fact, the difference between third and fourth was also 0...

41
41

And the difference between fourth and fifth was 0...

41
41

And the difference between fifth and sixth was 0...

41
41

Seventh and eighth (and maybe ninth) got owned, they only got a 40.[/quote]

Dude are you sure about that? Becasue if I recollect correctly, my friend got a 37/39 (one of the two not so sure) at states in Nor Cal and he got 8th place and my other friend got a the same score and he got 11th place. But then again, I may recollect incorrectly.[/quote][quote="Ubemaya"]

Are you from Miller?
I think 8th place got a 40, it could have been 39. I'm almost positive that 11th place was Felix (forgot his last name) who got a 38.[/quote][quote="bpms"]
Would Felix's last name be Nguyen?[/quote]

No it'd be Sung. And I'm not from miller. I think Felix got 12th cause i know my friend got 11th. You can go check out the scores at [url]http://www.mathcounts-ca.org/top25indivdavis.pdf[/url]


geometry
geometric transformation
reflection
linear algebra

Does anyone know how to prove the Lorentz transformations without matrices?

There is no point in learning the Lorentz transformation without knowing Linear Algebra, and a bit of group theory...
Of course there are several methods to derive them without using higher maths, but you wont understand the depths behind it, and the whole point of it...
So I suggest you, take a few courses on higher maths and then you can study SR. :wink:

It is perfectly reasonable to investigate special relativity before a course in linear algebra.

For a derivation, I point you towards Einstein's excellent book for the layman, [i]Relativity[/i].
http://www.bartleby.com/173/a1.html

Never mind, but I want to know more about reletavistic momentum.
if O is the rest frame and O' is the frame moving with velocity v relative to O,
and a particle relative to O is moving with speed u and u' relative to O',
then one can assume that the momentum is conserved and therefore derive the formula for relatavistic mass through the formula for relativistic particle speed.
However, how would one go about proving the formula for relatavistic mass:
$ m\equal{}\frac{m_0}{\sqrt{1\minus{}\frac{v^2}{c^2}}}$
without making those assumptions?

What would you take as the definition of relativistic mass?  I would take the definition offered at [url=http://en.wikipedia.org/wiki/Relativistic_mass]Wikipedia:Relativistic mass[/url]:
[quote="http://en.wikipedia.org/wiki/Relativistic_mass"]The term [b]relativistic mass[/b] is also used, and this is the total quantity of energy in a body (divided by $ c^2$).[/quote]
Then using the energy-momentum equation $ E^2\equal{}(mc^2)^2\plus{}(pc)^2$, we have $ \frac{E}{c^2}\equal{}\gamma m_0$.[/quote]

there is a simple way to prove the lorentz transformation on a very elementary level.
Here it goes,
Consider an object by which a photon or stream of light is allowed to be transmitted at one end allowed to reflect off and return to the original position. Now consider that it is traveling at a velocity near light, call it $ v$. Let the separation between the two ends of the object be $ D$. Now we know the time it takes for that photon to reach the other end will be $ t' \equal{} \frac{s}{c}$ where $ s$ is the distance traveled by the light and $ c$ is the speed of light. Notice this would be the observed time of an observer by who is not in the same reference frame as where the beam of light was shot off at. In this given time period the base of the object traveled a length $ l \equal{} vt'$. Now, due to the nature of light, we know that the observed time it took the light to reach the other end for an observer in a the reference frame of the moving object is going to be different. Call it $ t$. We now know that $ t \equal{} \frac{D}{c}$.
We can now set up a right triangle by which the equality $ s^2 \equal{} l^2 \plus{} D^2$ must be satisfied. Plugging in what we know we get, $ c^2t^2 \equal{} c^2t'^2 \plus{} v^2t^2 \Rightarrow t'^2 \equal{} \frac{t^2}{1\minus{}\frac{v^2}{c^2}}$, where the transformation $ t' \equal{} \gamma t$ is satisfied with $ \gamma \equal{} \frac{1}{\sqrt{1 \minus{} \frac{v^2}{c^2}}}$.

It's a lot easier to understand with a visual, but right now i am a tad bit too lazy to make one.That's the simplest derivation i came across. In fact, i had to prove it for an ib exam.

Same with the energy-momentum equation. This derivation is kind of short and fun.

We know $ E \equal{} \gamma m_0 c^2$ or $ E^2 \equal{} \frac{m_0^2 c^4}{1\minus{}\frac{v^2}{c^2}}$. We also know that $ p \equal{} \gamma m_0 v$ or $ p^2c^2 \equal{} \frac{m_0^2v^2c^2}{1 \minus{} \frac{v^2}{c^2}}$. Now consider the quantity
\[ E^2 \minus{} p^2c^2 \equal{} m_0^2 c^2 \left(\frac{c^2 \minus{} v^2}{1 \minus{} \frac{v^2}{c^2}}\right)\equal{} m_0^2 c^4\]
or
\[ E^2 \equal{} m_0^2c^4 \plus{} p^2c^2.\]


A game is played with two bowl of p and q coins respectively. The two forum members Arne and grobber play alternately. Arne begins.
The person being in action takes one of the two piles away and the other pile is taken apart in two new piles by him. You will lose if you cannot perform a complete move.
Determine p and q so that Arne can force his victory !

I'm not sure abt this:

We can see that if A (let Arne be A and grobber be B) finds a pile of 2 then he can take the other pile and split the pile of 2 coins in 1 and 1 so he beats B. Again, if he finds a pile of 4 he might take the other pile and then divide this one in 1 and 3, so the only thing B can do is take the pile of 1 coin and divide the pile of 3 in 1 and 2, so on his next turn A will have a pile of 2 so he will win. Do you see what I mean? By induction, if at the beginning we have p even or q even then A can force a win. Otherwise (meaning p and q odd) he can't force a win (even more, B can force a win): since p and q are odd, let's assume that he takes the pile of p coins and divides the pile of q coins. q is odd so one of the newly formed piles must have an even number of coins, but now it's B's turn and he found a pile with an even number of coins, so he can force a win (based on the previous sentence). So the condition for A to be able to force a win is this: either p or q must be even. Think it's right?

Yeah, you are right. But do not tell it Arne or he will defeat you !!!


combinatorics solved
combinatorics
graph theory

A graph has 12k points. Each point has 3k+6 edges. For any two points the number of points joined to both is the same. Find k.

It's combinatorics, so let's count sth!  :D This a standard method: I saw a solution like this one (for another problem) and I remembered it. Let t be the number of common vertices for 2 vertices.

Let's fix a vertex a and then count all the pairs (x,y) s.t. a-x-y, where a-x means a is connected to x. First of all, the number of such pairs is (3k+6)(3k+5) because we count the edges going away from a and from each point connected to a, x, we count all the edges going away from it except for a-x.

Then we count all the pairs (a,y): there are (12k-1) such pairs (a is fixed). For each such pair we count the x's for which a-x-y. This number is t. By combining these 2 we get the number of wanted pairs (x,y) s.t. a-x-y to be (12k-1)t.

We have an identity: (12k-1)t = (3k+6)(3k+5), which means that 12k-1 | (3k+6)(3k+5). I haven't continued, but we can't be too far away from the truth...  :D

Let n be the common number of vertices joined to any given pair of vertices M and N.

Let a be a vertex.
Let E be the set of vertices which are connected with A, and F the set of the vertices wich are not connected with A (except A).
Thus |E| = 3k+6 and |F| = 9k-7.
For each vertex B in E, each other vertex connected with A and B must also belong to E. Then B is connected to n vertices in E, and 3k+5-n vertices in F.
In another hand, for each vertex C in F, any vertex connected to both A and C must belong to E. It folloxs that C is conneceted to n vertices in E.

The total number of edges between a vertex in E and a vertex in F is then N = (3k+6)(3k+5-n) = (9k-7)n.
Expanding, it leads to :
9k 2 -12kn + 33k + n + 30 = 0, from which we deduce that n = 3m, for some positive integer m.
And then 4m = k+3+ (9k+43)/(12k-1).
Then (9k+43)/(12k-1) is an integer, which leads to 9k+43  \geq 12k-1, that is 1  \leq k  \leq 14.
Direct checking shows that k = 3 only gives an integer.

Conversely, it is not difficult to give a example of a graph satifying the statements of the problem.

Pierre.

"Let A be connected with M and N, and M and N connected with each other. Then number of pairs (x,y) won't be (3k+6)(3k+5). Am I right?"

I don't think so: (x,y) are the ordered pairs, so in that particular case you count both (x,y) and (y,x). I'm pretty sure I'm right here. Try it on some particular cases. I think k=3 can be derived from what I wrote:
12k-1 | (3k+6)(3k+5) iff 12k-1 | (k+2)(3k+5) = 3k^2+11k+10 | 12k^2+44k+40 and 12k-1 | 12k^2-k, so 12k-1 | 45k+40 but 12k-1 | 48k-4, so 12k-1 | 3k-44. If 3k-44>0 then 12k-1 <= 3k-44 so we get a negative k, so 3k-44<0, so 12k-1<=44-3k, so 15k<=45, so k<=3.


calculus USAMTS AMC USA(J)MO USAMO derivative

Are we allowed to use calculus on USAMTS?

Is calculus math?
(Yeah, I think so. Be careful not to make a mistake though: at least for the USAMO, calculus is technically acceptable, but if you make a slight mistake, you'll get docked many points. Eh, might not be as bad for USAMTS.)

definitely allowed, but never necessary.  You may be able to use it, but keep in mind you can always do without - which often may be easier.

I guess I could do it without calculus...
But that would take more work.

well, whatever you want to do.  I have yet to see a problem where calculus is that useful.  For instance, I considered using derivatives on problem 3 of last round, but I realized that they wouldn't get me anywhere easily and I was better off using only inequalities.


calculus
integration
trigonometry
logarithms
function
derivative
calculus computations

Hello, I'm using "Calculus for Biology and Medicine" by Claudia Neuhauser.

We have just started Pure Time Differential Equations, and I am stuck on a question.

The question is

dy/dx  =  x + sin x, where y[size=59]0[/size] = 0 and x0 = 0

This is my attempt:

y(x) =  y(0) + integral { (0 --> x)  (x + sin x) }
       =  0   +  [x^2 - cos x](0 --> x)
       =  (x^2 - cos x)  -  (0^2  - cos 0)
       =  x^2  - cos x - 1

Though the correct answer is

(x^2)/2  - cos x - 1

where did the 1/2 come from infront of the x^2?  Thanks for any help in advance!

$ \frac {d}{dx}x^2 \equal{} 2x\Longrightarrow \frac {d}{dx}\frac {1}{2}x^2 \equal{} x$, yielding $ \int x\ dx \equal{} \frac {1}{2}x^2 \plus{} Const.$

Generally speaking $ \int x^n\ dx \equal{} \frac {1}{n \plus{} 1}x^{n \plus{} 1} \plus{} Const.$ for $ n\neq \minus{} 1.$

Shouldn’t the answer be $ y\equal{}\frac{x^2}{2}\minus{}\cos{x}\color{red}\plus{}\color{black}1$? :what?:

Yeah it should be, stupid mistake on my part, thanks guys!

One last question I just want to make sure i'm doing this right:

this was the second question:

dy/dx = 1/(1+x^2)   where y0 = 1 and x0 = 0

My solution:

y(x) = 1 + integral(0 --> x) 1/(1+u^2)

       = 1 + [ln (1+u^2)](0-->x)

       = 1 + ln (1+x^2) - ln (1 + 0^2)

       = 1 + ln (1+ x^2)

Is that right?

[on a side note, why is it nessesary to change x to u?  I thought that was only used in substitution tom simplify things, but how does that simplify things if you're just changing x to u?]

The change from $ x$ to $ u$ is necessary because the bounds of the integral, as variables, cannot be the same variables as the variable you're integrating with respect to.  Therefore if

$ F'(x) \equal{} f(x)$

then we write

$ F(x) \equal{} \int_{c}^{x} f(t) \, dt$

because we are integrating with respect to $ t$ and plugging in $ x$.

[quote="kunny"]$ \int \frac {1}{1 \plus{} x^2}\ dx\neq \ln (1 \plus{} x^2) \plus{} C.$[/quote]

Isn't the antiderivitive of 1/x   = ln x?  if not, what is it?

Just differentiate $ \ln (1 \plus{} x^2)?$ :)

I see people make this mistake all the time.  The antiderivative of $ \frac{1}{x} \, dx$ is $ \ln x$, but that doesn't say anything about the antiderivative of $ \frac{1}{u} \, dx$, which can be very different.  :)

In any case, if you're not extremely comfortable with the basic integrals you should be doing these problems with an integral table next to you.

... and $ \int \frac1{1\plus{}x^2}\,dx$ is on that basic list. It's the arctangent, and you should memorize that.

OH RIGHT!  Thanks it's been a couple months since my last calc course

k, how about this?

y(x) = 1 + integral(0 --> x) 1/(1+u^2)

= 1 + [arctan u](0-->x)

= 1 + (arctan x) - (arctan 0)

= 1 + arctan x

?

That's right. $ y(0)\equal{}1$ gives your justification.

Hullo everyone!
I missed last week's lecture at uni, when pure time differential equations were handled, so I'm fumbling in the dark a bit here.
I have an additional twist (as is always the case with math) where they want to make it all more difficult.

dx/dt = cos(t-3) when x(3)=1.

Do I find x(0) somehow, or don't I need to?

EDIT: Incidentally, I'm using the very same textbook as forumz.  :lol: Small world.

Your expression for the antiderivative will naturally be in terms of $ t \minus{} 3$. Leave it that way- what happens when you try to evaluate at $ 3$?

If by that you mean solve for x(3), then this is what I got:

x(t)= x(3) + (sin(t-3) - sin(3-3))
     =[u] 1 + sin(t-3)[/u]

How's that? Yes? No? Hope a kindly passing surgeon cuts open my head with a spade and puts a new brain in?

Well, that wasn't so bad! Thanks a lot! :)

Again thanks for the help on my last questions!

I have one more relating to pure-time differential equations:

"Suppose taht the amount of phosphorous in a lake at time t, denoted by P(t), follows the equation

dP/dt = 3t + 1      with P(o) = 0

Find the amount of Phosphorous at t = 10

Here's my attempt:

dP/dt = 3t + 1     P(0) = 0

dp = 3t + 1    dt

integral (dp) = P(0) + integral(0 --> 10)  3u + 1    du

P(t) = 0 + (3/2 U^2 + U) (0 --> 10)

      =  (3/2 10^2 + 10) - 0

      = 160

How does that look?  Right/Wrong?  Thanks in advance!

[quote="Drosophila"]I get 160 too.
A good way to check if you've done something right is to see that the function at the beginning of the equation is the derivative of your integrated function.

dP/dt=[u]3t +1[/u]

(3/2t^2 +t)' = [u]3t + 1[/u][/quote]

Thanks for the prompt response and excellent tip! :)


we have 2n+3 points.
one circle is good such as 3points on circle& n point in circlce & n points out of circle.prove that number of good circle =n(mod2)

The exact number of good circles IS $n$ and this appears already on the forum...
Moreover, I think this one is more a combinatoric problem than a number theoretic one...Please be carefull when you post problems (there are some in inappropriate forums almost everywhere...)

Pierre.


superior algebra
superior algebra unsolved

let p be a prime number,let F be a field and let c be member of F,then (x^p) -c is irreducible in F[x] if and only if (x^p) -c has no root in F. :)

($\Rightarrow$) is clear
($\Leftarrow$) Suppose $f(x)=x^{p}-c$ is reducible, say $f(x)=g(x)h(x)$ where $\deg g<\deg f=p$ and $\deg h>0$. In a bigger field all roots of $f(x)$  is of the form $\omega \alpha$ where $\omega$ is a $p$-th roots of unity and $\alpha$ is a $p$-rooth of $c$.
Now up to sign, the constant term of $g$ say $b$ is a multiplication of $d=\deg g$ number of roots of $f(x)$. So
\[\pm b=\omega \alpha^{d}\]
(Here $\omega$ is the multiplication of $d$ numbers of $p$-root of unity, which of course itself is a $p$-root of unity). It follows that
\[(\pm b)^{p}=\omega^{p}\alpha^{dp}=\alpha^{dp}=c^{d}\]
Since $d<p$ and $p$ prime, then $(d,p)=1$ so we can find $s,t$ so that $sd+tp=1$.
Then
\[c=c^{sd}c^{tp}=(\pm b)^{sp}c^{tp}=\left((\pm b)^{s}c^{t}\right)^{p}\]

So $c$ has a $p$-root in $F$


[hide="1"]

say a job = x
then kevin can do $\frac{x}{37}$ a day.
meaning, Jack can do $\frac{x}{37} * \frac{2}{3} = \frac{2x}{111}$ a day, since Kevin does 50% more than Jack.

using the same method, Mike can do $\frac{x}{37} * \frac{5}{6} = \frac{5x}{222}$ a day.

add them all up, and we get the result $\frac{15x}{222} = \frac{5x}{74}$. this means that they can do $\frac{5}{74}$ of a job together in one day....

SO, it would take them $\frac{74}{5}$ days to finish the same job together, or $14.4$ days.
[/hide]

:? I don't think that answer is correct  :?

1) [hide]$15 \text{ days}$

Since Kevin takes 37 days to finish the job, each day he does $\frac{1}{37}$ of the job. Since Kevin works 50 percent more than Jack no matter the time period, and $\frac{1}{K}$ is how much Kevin works in a day, then we have: $\frac{1}{K}=\frac{3}{2}(\frac{1}{J})$

This means that $\frac{1}{J}=\frac{2}{3K}$. Doing the same for Mike, we find that $\frac{1}{M}=\frac{4}{5K}$

To find the total number of days needed, we have to solve for $T$:

$\frac{1}{T}=\frac{1}{K}+\frac{1}{J}+\frac{1}{M}=\frac{1}{K}+\frac{2}{3K}+\frac{4}{5K}=\frac{37}{15K}$

Subsituting in 37 for $K$, we get that $\frac{1}{T}=\frac{1}{15}$ so $T=15$
[/hide]

[hide="1"]I set K = 3/2J = 5/4M.  If t = 1 / K = 37 then K = 1 / 37.  If the total time is equal to the arbitrary unit of work 1 divided by the total rate which can now be put all in terms of K you get:

        t = 1 / (37/15K) or 15 / (37K)  plugging in for K you get t = 15.

My final answer is 15 days.[/hide]

[quote="miraculouspostmaster"]2.
[hide]Case 1: A's, We have six different possibilities to place the three a's, with 5 letter spaces left over, so
6*2^5 for case 1.
Case 2: B's 5 places for the string, 4 left over.  5*2^4
Eliminating cases where both occur (we can have 2 places for AAABBBB and an extra space, or 2 places for BBBBAAA and an extra space) so 2*2+2*2.
$6*2^5+5*2^4-8=264$
Can anyone verify?[/hide][/quote]

Somehow, I think this answer is wrong...does anyone else have another solution?

shouldn't number 2 be 100%(so 2^8), as 3 As = 5Bs and 4Bs = 4As, so you fulfill either criteria anyways.

wait, never mind, read that wrong.

My attempt at 2
[hide] Let x = AAA and Y = BBBB

first we find the number of ways to get 3 A's without paying attention to the 4 B's

there will be x and 5 other random letters

each of the 5 other letters can be A or B, giving 2^5=32 possibilities.

x can be placed in one of 6 places (before the first, between 1st and 2nd, between 2nd and 3rd, between 3rd and 4th, between 4th and 5th or after 5th) ...
32*6 = 192 times when there are 3 A's in a row

Let's find the number of ways to get 4 B's without paying attention to the 3 A's

there will be y and 4 other random letters.

each fo the 4 can be a or b, giving 2^4=16 possibilities, y can be in 5 spots so we multiply 16*5 to get 80

so we have 272 ways ... but we've counted some twice.

Let z = a or b

thus we need to find the # of ways we can have xyz ...

there are 3! = 6 permutations of xyz. If z = a or b, then we get twice as many, so that's 12.

272-12 = 260[/hide]

All the solutions to #2 have been terribly ignorant toward overlapping.  The answers were 256 and 260 and there are only $2^8=256$ strings!  I think it will be much easier to subtract out the exceptions.  Every time a sequence changes from A to B or from B to A, I split the string there and form different groups of letters.  I figured the number of exceptions is 2+14+36+40+16+1=109.  My guess is 256-109=147.


[quote="kabi"]Check this::
1.http://www.wonderfulinfo.com/amazing/unbelv/madeira_airport/index.htm
2.http://www.wonderfulinfo.com/amazing/unbelv/largestpool/index.htm
3.http://www.wonderfulinfo.com/amazing/unbelv/mmmbt/index.htm
4.http://www.wonderfulinfo.com/amazing/unbelv/kansaiairport/index.htm

for more go to::
http://www.wonderfulinfo.com/amazing/unbelv/index.htm
 :)[/quote]

Hey! Nice links!


geometry conics parabola inequalities symmetry

[quote="Interval"]A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway.  If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?[/quote]
[hide]
The largest area comes from when the 3 sides are equal.
$\frac{2000^{2}}{3^{2}}m^{2}$ is about $444444.44m^{2}$
[/hide]

[quote="Interval"]We have two different answers.  So, which is it?

Thanks![/quote]
cincodemayo5590 is right. the answer is 500,000 sq. m

[quote="Interval"]A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway.  If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?[/quote]

[hide]$l+2w=2000$ so $l=2000-2w$

We want to maximize the area which is $l\cdot w$.  Plug the equation for $l$ back into $l\cdot w$ to get $(2000-2w)w=-2w^{2}+2000w=A$.  The graph of this is a porabola and the maximum point is when $w=\frac{-b}{2a}=\frac{-2000}{4}=500$.  This means $l=1000$ so the area is $500,000$.[/hide]

[quote="Interval"]A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway.  If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?[/quote]
Similar to cincodemayo's solution, but different.

[hide]$l+2w=2000$

$\frac{l+2w}{2}\geq\sqrt{2wl}$

Substitute the equation into the inequality to get $1000\geq\sqrt{2wl}$.

$1000000\geq2wl$

$500000\geq wl$

We know $wl$ is the area, so its greatest possible value is $\boxed{500000}$.[/hide]

[quote="Interval"]There are too many answers to this question.

I don't know which one to accept.[/quote]

You don't need to "accept" only one solution; if there is more than one [i]valid[/i] solution which yields the correct answer, then they are all correct. For most problems, there are many methods for solving.

[quote="Interval"]There are too many answers to this question.

I don't know which one to accept.[/quote]

yeah...i agree with cincodemayo5590.  its not about the answer, its about how to get it and the more ways you know how to get to the answer, the more you learn

[quote="cincodemayo5590"][quote="Quevvy"][quote="Interval"]Okay but HOW did you come to that conclusion based on the word problem information?

How did you find the answer?[/quote]
Since the plot is a square (to maximize the area), and you are only covering 3 sides, so the length of each side is $\frac{2000}{3}m$ Since it's a square, the area is $\left(\frac{2000}{3}\right)^{2}=\frac{2000^{2}}{3^{2}}=444444.\overline{4}m^{2}$

Also, it helps if you draw a picture...[/quote]

For this problem, it isn't necessarily a square; that maximizes area for when you need to cover [i]all[/i] sides.

[hide="Solution"]First, $l+2w=2000$. Then, apply AM-GM:

$\frac{l+2w}{2}\geq \sqrt{l \cdot 2w}$
$\implies A=l\cdot w \leq \left( \frac{l+2w}{2 \sqrt 2}\right)^{2}=\left( \frac{2000}{2 \sqrt2}\right)^{2}= \left( 500 \sqrt 2 \right)^{2}= \boxed{500,000 \text{ m}^{2}}$.[/hide][/quote]
i see my mistake.
however, if it isn't a square, i solve it differently but get a same answer.
$2x+y$=2000
$y=2000-2x$
$A=xy$
$A(x)=-2x^{2}+2000x$
the vertex of the parabola is on the axis of symmetry, which is at $x=\frac{-b}{2a}$
$b=2000$
$a=-2$
$x=\frac{-2000}{-4}$
$x=500$
$A(x)=500000$ $m^{2}$


Find the sum

$ \frac{a\plus{}1}{2}\plus{}\frac{a\plus{}3}{4}\plus{}\frac{a\plus{}7}{8}\plus{}\cdots\plus{}\frac{a\plus{}2^n\minus{}1}{2^n}$

[hide="Answer"] $ \frac{(a\minus{}1)(2^n\minus{}1)}{2^n}\plus{}n$[/hide]

I think you meant $ a\minus{}1$ , but I got your solution, thanks a bunch man, really helpful


linear algebra
matrix
algebra
polynomial
search
algorithm
Gauss

Let $ A,B$ br $ n\times n$ matrices .

We know that if $ A$ is similar to $ B$ then they have the same characteristic polynomial . Is the converse also true ? Namely

if $ A$ and $ B$ have the same characteristic polynomial , then they are similar ?

No. $ \begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $ \begin{pmatrix}1&0\\0&1\end{pmatrix}$ have the same characteristic polynomial but are not similar.

Actually , how should we prove that they are not similar ? Is it we have to prove that there does not exist any invertible matrix $ P$ such that $ A\equal{}PBP^{\minus{}1}$ ?

This particular example is easy, since the identity matrix is only similar to itself. After all, $ P^{ \minus{} 1}IP \equal{} I$ for all invertible $ P.$

But let's make a more general argument.

If $ A$ is similar to $ B$ and $ f(A) \equal{} 0$ for some polynomial $ f,$ then $ f(B) \equal{} 0.$ This follows since $ B \equal{} P^{ \minus{} 1}AP$ implies $ f(B) \equal{} P^{ \minus{} 1}f(A)P.$

In the example I gave, the second matrix (the identity) satisfies the polynomial $ x \minus{} 1,$ and in fact that is its minimal polynomial. The first matrix does not satisfy $ x \minus{} 1.$ Rather, its minimal polynomial is $ (x \minus{} 1)^2.$

So: are two matrices of the same dimension and the same minimal polynomial necessarily similar?

[hide="Hint."]But would having the same minimal polynomial necessarily imply the same characteristic polynomial?[/hide]

So, let's phrase it this way: suppose $ A$ and $ B$ have both the same characteristic polynomial and the same minimal polynomial. Are they similar?

[hide="Hint."]Try looking at the last post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1221281370&t=31816]this topic.[/url][/hide]

[quote]But would having the same minimal polynomial necessarily imply the same characteristic polynomial?[/quote]
I think no .They might not have the same characteristic polynomial

for example

$ \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$ and $ \begin{pmatrix}1&0&0\\0&2&0\\0&0&2\end{pmatrix}$

both have minimum polynomial $ (x\minus{}1)(x\minus{}2)$ and same dimension but different characteristic polynomial

Actually , I just know that two similar matrices must have same characteristic polynomial,same minimum polynomial,same rank,same determinant,same trace,same eigenvalue. Does this mean that if all of the properties are satisfied , then we can say that the two matrices are similar ?

Still unanswered?

[quote="shyong"]Actually , I just know that two similar matrices must have same characteristic polynomial,same minimum polynomial,same rank,same determinant,same trace,same eigenvalue. Does this mean that if all of the properties are satisfied , then we can say that the two matrices are similar ?[/quote]

Nope, take $ \left(\begin{array}{cccc}1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1\end{array}\right)$ and $ \left(\begin{array}{cccc}1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\end{array}\right)$. Both of these matrices have characteristic polynomial $ \left(X \minus{} 1\right)^4$, minimal polynomial $ \left(X \minus{} 1\right)^2$, rank $ 4$, determinant $ 1$, trace $ 4$, eigenvalues $ 1$, $ 1$, $ 1$, $ 1$, albeit with different geometrical multiplicities. But if you really want, I can even give you an instance where the geometrical multiplicities are equal, and still the matrices are not similar. The structure of the blocks in the Jordan decomposition is too complicated to be fully determined by a few parameters.

Something more interesting: Over which fields $ K$ does the following assertion [b](*)[/b]  hold?

[b](*)[/b] If $ A$ and $ B$ are two matrices from $ \text{M}_n\left(K\right)$, then $ A$ and $ B$ are similar if and only if for every polynomial $ P\in K\left[X\right]$ we have $ \text{Rank }\left(P\left(A\right)\right) \equal{} \text{Rank }\left(P\left(B\right)\right)$ ?

At the moment I can just prove that [b](*)[/b] holds over algebraically closed fields $ K$. Probably the general case is going to use the rational canonical form and similar methods, where I don't really have practice in...

  darij

Hi,

May I know where can I find the relevant readings regarding similar matrices having the same characteristic and minimal polynomials? Besides, I would like to know more about Jordan canonical form. I just learned this topic and I find quite difficult to understand the relationships between characterisric polynomials, minimal polynomials and Jordan form. Hope someone could help me out. Thanks.

Hi Darij, we prove that
$ A$ and $ B$ are similar if for every polynomial $ P\in K\left[X\right]$ we have $ \text{Rank }\left(P\left(A\right)\right) \equal{} \text{Rank }\left(P\left(B\right)\right)$.
Let $ p_1(x)\cdots{p_k(x)}$ and $ q_1(x)\cdots{q_l(x)}$ the square free polynomials associated to the decompositions in irreducible unitary factors in $ K$ of the charac- polynomials of $ A$ and $ B$. $ rank(p_i(A))\equal{}rank(p_i(B))<n$; then $ p_i\equal{}q_j$. We deduce easily that $ k\equal{}l$ and $ p_1(x)\cdots{p_k(x)}\equal{}q_1(x)\cdots{q_k(x)}$.
Let $ L$ be the $ K$-algebraic closure; $ a_1,\cdots,a_s$ are the (simple) $ p_1$-roots in $ L$. Let $ dim(ker((A\minus{}a_iI)^t))\equal{}d_t$ ($ d_t$ is independent of $ i$); then $ dim(ker((p_1(A)^t))\equal{}sd_t$. Here $ dim(ker((p_1(A)^t))\equal{}dim(ker((p_1(B)^t))$; then $ dim(ker((A\minus{}a_iI)^t))\equal{}dim(ker((B\minus{}a_iI)^t))$. We conclude by the Jordan theorem.

Loup Blanc, you use the fact that if two matrices $ A$ and $ B$ from $ \text{M}_n\left(K\right)$ are similar in $ \text{M}_n\left(L\right)$, where $ L$ is a field containing $ K$, then they are similar in $ \text{M}_n\left(K\right)$ as well. I can only prove this for $ K$ being infinite. Do you have any swift argument for all $ K$ ?

  darij

Hi Darij,
you are right; I have used this result; it's a consequence of the rational canonical form (Frobenius normal form) algorithm:
1) $ A,B$ are similar iff they have the same rational canonical form over $ K$.
2) The $ A$-rational canonical form over $ K$ is given by the $ A$-Jordan form over $ L$ and the square free polynomial associated to the decomposition in irreducible over $ K$ of the $ A$-minimal polynomial.
      Let us recall that this algorithm is very fast (in $ O(n^3)$ as the Gauss method !).


http://www.logicmazes.com/theseus.html
I'm level 14. put more such similar games please  :)  (like sokoban...)

I agree this is a very nice and hard game. :)

I think that the trick to this game is realizing that there is usually only one or at the very most two moves that you can make at any given time without quickly dying. That said, I don't have enough patience to do much of this. I'm at level 4 right now, and kind of stalled. :rotfl:

[quote="nutz_for2.718281828"]I can't beat level 2. A litle help please?[/quote]

you delay


A bicycle tire has a diameter of 27 inches. How many feet will it travel in four complete revolutions? Express your answer in terms of $ \pi$.

[hide]Ok, the circumference is $ 27 \pi$ inches.  Multiplied by $ 4$ and divided by $ 12$(because the answer has to be in feet), this equals $ 9 \pi$ inches.[/hide]


The best I can do is : http://en.wikipedia.org/wiki/List_of_winners_of_the_Mathcounts_competition

I don't know if you can find anything like that online; try talking to some people in your area (your coach, older team members, MC vets)?

[quote="pinkmuskrat"]Question- Is there any way, and how, can we find past national participants lists? Is there any website where you can find these? Thanks[/quote]

I have the list of winners from each state since 2002 if you really want them.

Hi, sorry for reviving-
Can anyone post the 2009 one? I lost my copy. Thanks

i know i'm reviving this but i didn't feel like making a new thread just for it.

does anyone have any of the past top 57's? well actually, just 2006 and 2005. thanks :)


How many ways are there to arrange the letters of the word BA$ _1$N$ _1$A$ _2$N$ _2$A$ _3$, in which the three A's and the two N's are considered different?

The correct answer is $ 6! \equal{} 720$, but I got $ \frac {6!}{3! * 2!} \equal{} 60$

This is a trick question!

No it's not :P

If you read carefully, and notice the weird "A$ _1$" stuff, you can't miss this one.

Yes! Because they are considered DIFFERENT!!!

[hide=sol]They are differnet so $6!=6\cdot5\cdot4\cdot3\cdot2=720$.[/hide]


Find all functions $ f: \Bbb{R}_{0}^{ \plus{} }\rightarrow \Bbb{R}_{0}^{ \plus{} }$ with the following properties:

(a) We have $ f\left( xf\left( y\right) \right) \equal{}f\left( x\plus{}y\right)$ for all x and y belong $ \Bbb{R}_{0}^{ \plus{} }.$
(b) We have $ f\left(2\right) \equal{} 0.$

I suppose $ \mathbb{R}_{0}^{\plus{}} \equal{} \mathbb{R}_{\geq 0}$.

Filling in $ x\equal{}0$, $ y\equal{}2$ gives us
$ f(0) \equal{} 0$.

Filling in $ y\equal{}0$ gives $ f(xf(0)) \equal{} f(x)$, so $ f(0)\equal{}f(x)$ for all $ x$, so the only function that might satisfies both equations is

$ f(x) \equal{} 0 \;\forall x \in \mathbb{R}_{\geq 0}$.

This function indeed satisfies them.

[quote="angelstt"]Find all functions $ f: \Bbb{R}_{0}^{ \plus{} }\rightarrow \Bbb{R}_{0}^{ \plus{} }$ with the following properties:

(a) We have $ f\left( xf\left( y\right) \right) \equal{} f\left( x \plus{} y\right)$ for all x and y belong $ \Bbb{R}_{0}^{ \plus{} }.$
(b) We have $ f\left(2\right) \equal{} 0.$[/quote]

Set $ x\equal{}0$ in $ (a)$ and you get $ f(y)\equal{}f(0)$ $ \forall y$ and so $ f(x)$ is a constant function. Then $ (b)$ gives us the unique solution $ \boxed{f(x)\equal{}0}$ $ \forall x$

 :blush:  oops, too late :)


Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{ab+4bc+ca}{b^{2}+c^{2}}+\frac{bc+4ca+ab}{c^{2}+a^{2}}+\frac{ca+4ab+bc}{a^{2}+b^{2}}\ge 4.\]

Sorry to ask this, but are you sure? It seems to me that the right hand side should be $9$...

Actually, playing around with it for the triplet $(4.13,3.81,2.97)$
I find a value of $8.83$...
This is not a typical inequality I guess... I'll get back later

EDIT: Sorry, $f(4.13,3.81,2.97)=8.82$
also, I found:
$f(.02,4.56,6.38)=4.05$ so I would propose:
\[9\geq \frac{ab+4bc+ca}{b^{2}+c^{2}}+\frac{bc+4ca+ab}{c^{2}+a^{2}}+\frac{ca+4ab+bc}{a^{2}+b^{2}}\ge 4\]

It is problem 6 on page 302 of Algebraic Inequalities by Vasile Cirtoaje.

Oh... I created it from another problem of VASC (on a Vietnamese book)
\[\sum \frac{ab-bc+ca}{b^{2}+c^{2}}\ge \frac{3}{2}\]


geometry inequalities trigonometry calculus integration probability algebra

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729749#p2729749]Problem 1[/url]

Let $ ABC$ be a triangle with $ \left|AB\right| \equal{} 14$, $ \left|BC\right| \equal{} 12$, $ \left|AC\right| \equal{} 10$. Let $ D$ be a point on $ \left[AC\right]$ and $ E$ be a point on $ \left[BC\right]$ such that $ \left|AD\right| \equal{} 4$ and $ Area\left(ABC\right) \equal{} 2Area\left(CDE\right)$. Find $ Area\left(ABE\right)$.

A) $ 4\sqrt {6}$  B) $ 6\sqrt {2}$ C) $ 3\sqrt {6}$ D) $ 4\sqrt {2}$ E) $ 4\sqrt {5}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729750#p2729750]Problem 2[/url]

How many ordered integer pairs $ \left(x,y\right)$ are there such that $ xy \equal{} 4\left(y^{2} \plus{} x\right)$?

A) 0  B) 3  C) 7  D) 14  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729752#p2729752]Problem 3[/url]

Four boxes with ball capacity 3, 5, 7 and 8 are given. In how many ways can 19 same balls be put into these boxes?

A) 34  B) 35  C) 36  D) 40  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476030#p1476030]Problem 4[/url]

If inequality $ \frac {\sin ^{3} x}{\cos x} \plus{} \frac {\cos ^{3} x}{\sin x} \ge k$ is hold for every $ x\in \left(0,\frac {\pi }{2} \right)$, what is the largest possible value of $ k$?

A) $ \frac {1}{2}$  B) $ \frac {3}{4}$  C) 1  D) $ \frac {3}{2}$  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729754#p2729754]Problem 5[/url]

Let $ ABC$ be an isosceles triangle with $ \left|AB\right| \equal{} \left|AC\right| \equal{} 10$ and $ \left|BC\right| \equal{} 12$. $ P$ and $ R$ are points on $ \left[BC\right]$ such that $ \left|BP\right| \equal{} \left|RC\right| \equal{} 3$. $ S$ and $ T$ are midpoints of $ \left[AB\right]$ and $ \left[AC\right]$, respectively. If $ M$ and $ N$ are the foot of perpendiculars from $ S$ and $ R$ to $ PT$, then find $ \left|MN\right|$.

A) $ \frac {9\sqrt {13} }{26}$ B) $ \frac {12 \minus{} 2\sqrt {13} }{13}$  C) $ \frac {5\sqrt {13} \plus{} 20}{13}$  D) $ 15\sqrt {3}$ E) $ \frac {10\sqrt {13} }{13}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476028#p1476028]Problem 6[/url]

If $ a,b,c\in {\rm Z}$ and
\[ \begin{array}{l} {x\equiv a\, \, \, \left(mod\, 14\right)} \\
{x\equiv b\, \, \, \left(mod\, 15\right)} \\
{x\equiv c\, \, \, \left(mod\, 16\right)} \end{array}
\]
, the number of integral solutions of the congruence system on the interval $ 0\le x < 2000$ cannot be

A) 0  B) 1  C) 2  D) 3  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476031#p1476031]Problem 7[/url]

Six cards with numbers 1, 1, 3, 4, 4, 5 are given. We are drawing 3 cards from 6 given cards one by one and are forming a three-digit number with the numbers over the cards drawn according to the drawing order. Find the probability that this three-digit number is a multiple of 3. (The card drawn is not put back)

A) $ \frac {1}{5}$  B) $ \frac {2}{5}$  C) $ \frac {3}{7}$  D) $ \frac {1}{2}$  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476035#p1476035]Problem 8[/url]

If the polynomial $ P\left(x\right)$ satisfies $ 2P\left(x\right) \equal{} P\left(x \plus{} 3\right) \plus{} P\left(x \minus{} 3\right)$ for every real number $ x$, degree of $ P\left(x\right)$ will be at most

A) 0  B) 1  C) 2  D) 3  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729755#p2729755]Problem 9[/url]

Find the area of inscribed convex octagon, if the length of four sides is 2, and length of other four sides is $ 6\sqrt {2}$.

A) 120  B) $ 24 \plus{} 68\sqrt {2}$  C) $ 88\sqrt {2}$ D) 124  E) $ 72\sqrt {3}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729756#p2729756]Problem 10[/url]

Let $ n$ be the greatest common divisor of integers $ a,b,c$. If it is known that
\[ \begin{array}{l} {x \plus{} 2y \plus{} 3z \equal{} a} \\
{2x \plus{} y \minus{} 2z \equal{} b} \\
{3x \plus{} y \plus{} 5z \equal{} c} \end{array}
\]
has a solution in integers, what is the smallest possible value of positive number $ n$?

A) 7  B) 14  C) 28  D) 56  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476036#p1476036]Problem 11[/url]

* * * *
 * * *
  * *
   *

Place all numbers from 1 to 10 to the boxes such that every number except the uppermost is equal to the difference between the two numbers on its top. The number in the lower box is at most

A) 1  B) 2  C) 3  D) 4  E) 5

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476037#p1476037]Problem 12[/url]
\[ \begin{array}{c} {x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 21} \\
{x \plus{} y \plus{} z \plus{} xyz \equal{} \minus{} 3} \\
{x^{2} yz \plus{} y^{2} xz \plus{} z^{2} xy \equal{} \minus{} 40} \end{array}
\]
The number of real triples $ \left(x,y,z\right)$ satisfying above system is

A) 0  B) 3  C) 6  D) 12  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729757#p2729757]Problem 13[/url]

Square $ BDEC$ with center $ F$ is constructed to the out of triangle $ ABC$ such that $ \angle A \equal{} 90{}^\circ$, $ \left|AB\right| \equal{} \sqrt {12}$, $ \left|AC\right| \equal{} 2$. If $ \left[AF\right]\bigcap \left[BC\right] \equal{} \left\{G\right\}$ , then $ \left|BG\right|$ will be

A) $ 6 \minus{} 2\sqrt {3}$ B) $ 2\sqrt {3} \minus{} 1$ C) $ 2 \plus{} \sqrt {3}$ D) $ 4 \minus{} \sqrt {3}$ E) $ 5 \minus{} 2\sqrt {2}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476038#p1476038]Problem 14[/url]

Find the sum of squares of the digits of the least positive integer having 72 positive divisors.

A) 41  B) 65  C) 110  D) 123  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476039#p1476039]Problem 15[/url]

2 squares are painted in blue and 2 squares are painted in red on a $ 3\times 3$ board in such a way that two square with same color is neither at same row nor at same column. In how many different ways can these four squares be painted?

A) 198  B) 288  C) 396  D) 576  E) 792

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476043#p1476043]Problem 16[/url]

How many pairs of rational number $ \left(x,y\right)$ are there satisfying the equation $ y \equal{} \sqrt {x^{2} \plus{} \frac {1}{1999} }$?

A) 0  B) 2  C) 3  D) 4  E) Infinitely many

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729759#p2729759]Problem 17[/url]

In a regular pyramid with top point $ T$ and equilateral base $ ABC$, let $ P$, $ Q$, $ R$, $ S$ be the midpoints of $ \left[AB\right]$, $ \left[BC\right]$, $ \left[CT\right]$ and $ \left[TA\right]$, respectively. If $ \left|AB\right| \equal{} 6$ and the altitude of pyramid is equal to $ 2\sqrt {15}$, then area of $ PQRS$ will be

A) $ 4\sqrt {15}$ B) $ 8\sqrt {2}$ C) $ 8\sqrt {3}$ D) $ 6\sqrt {5}$ E) $ 9\sqrt {2}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476047#p1476047]Problem 18[/url]

Let $ t_{k} \left(n\right)$ show the sum of $ k^{th}$ power of digits of positive number $ n$. For which $ k$, the condition that $ t_{k} \left(n\right)$ is a multiple of 3 does not imply the condition that $ n$ is a multiple of 3?

A) 3  B) 6  C) 9  D) 15  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476046#p1476046]Problem 19[/url]

$ k$ black pieces are placed on $ k$ consecutive squares of top row starting from upper left of a $ 2\times 5$ board. We are placing white pieces on empty squares one by one in arbitrary order. Two squares is said to adjacent if they have common vertex. When a white piece is placed on a square, the pieces on adjacent squares change their color. For which $ k$, when all the squares are filled, it is possible that color of every piece is white?

A) 0  B) 1  C) 2  D) 3  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476048#p1476048]Problem 20[/url]

How many pairs of real numbers $ \left(x,y\right)$ are there such that $ x^{4} \minus{} 2^{ \minus{} y^{2} } x^{2} \minus{} \left\| x^{2} \right\| \plus{} 1 \equal{} 0$, where $ \left\| a\right\|$ denotes the greatest integer not exceeding $ a$.

A) 0  B) 1  C) 2  D) 4  E) Infinitely many

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729760#p2729760]Problem 21[/url]

$ ABC$ is a triangle with $ \angle BAC \equal{} 10{}^\circ$, $ \angle ABC \equal{} 150{}^\circ$. Let $ X$ be a point on $ \left[AC\right]$ such that $ \left|AX\right| \equal{} \left|BC\right|$. Find $ \angle BXC$.

A) 15  B) 20  C) 25  D) 30  E) 35

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729761#p2729761]Problem 22[/url]

If $ m,n\in Z$, then $ m^{2} \plus{} 3mn \minus{} 4n^{2}$ cannot be

A) 69  B) 76  C) 91  D) 94  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729763#p2729763]Problem 23[/url]

Hour part of a defective digital watch displays only the numbers from 1 to 12. After one minute from $ n: 59$, although it must display $ \left(n \plus{} 1\right): 00$, it displays $ 2n: 00$ (Think in $ mod\, 12$). For example, after $ 7: 59$, it displays $ 2: 00$ instead of $ 8: 00$. If we set the watch to an arbitrary time, what is the probability that hour part displays 4 after exactly one day?

A) $ \frac {1}{12}$  B) $ \frac {1}{4}$  C) $ \frac {1}{3}$  D) $ \frac {1}{2}$  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729765#p2729765]Problem 24[/url]

Polynomial $ f\left(x\right)$ satisfies $ \left(x \minus{} 1\right)f\left(x \plus{} 1\right) \minus{} \left(x \plus{} 2\right)f\left(x\right) \equal{} 0$ for every $ x\in \Re$. If $ f\left(2\right) \equal{} 6$, $ f\left({\tfrac{3}{2}} \right) \equal{} ?$

A) -6  B) 0  C) $ \frac {3}{2}$  D) $ \frac {15}{8}$  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729766#p2729766]Problem 25[/url]

$ \angle BAC \equal{} 80{}^\circ$, $ \left|AB\right| \equal{} \left|AC\right|$, $ K\in \left[AB\right]$, $ L\in \left[AB\right.$, $ \left|AB\right|^{2} \equal{} \left|AK\right|\cdot \left|AL\right|$, $ \left|BL\right| \equal{} \left|BC\right|$, $ \angle KCB \equal{} ?$

A) 20  B) 25  C) 30  D) 35  E) 40

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729768#p2729768]Problem 26[/url]

Let $ x$, $ y$, $ z$ be integers such that
\[ \begin{array}{l} {x \minus{} 3y \plus{} 2z \equal{} 1} \\
{2x \plus{} y \minus{} 5z \equal{} 7} \end{array}
\]
Then $ z$ can be

A) $ 3^{111}$  B) $ 4^{111}$  C) $ 5^{111}$  D) $ 6^{111}$  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476049#p1476049]Problem 27[/url]

Points on a square with side length $ c$ are either painted blue or red. Find the smallest possible value of $ c$ such that how the points are painted, there exist two points with same color having a distance not less than $ \sqrt {5}$.

A) $ \frac {\sqrt {10} }{2}$  B) 2  C) $ \sqrt {5}$  D) $ 2\sqrt {2}$ E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476050#p1476050]Problem 28[/url]

Find the number of functions defined on positive real numbers such that $ f\left(1\right) \equal{} 1$ and for every $ x,y\in \Re$, $ f\left(x^{2} y^{2} \right) \equal{} f\left(x^{4} \plus{} y^{4} \right)$.

A) 0  B) 1  C) 2  D) 4  E) Infinitely many

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729769#p2729769]Problem 29[/url]

The length of the altitude of equilateral triangle $ ABC$ is 3. A3. A circle with radius 2, which is tangent to $ \left[BC\right]$ at its midpoint, meets other two sides. If the circle meets $ AB$ and $ AC$ at $ D$ and $ E$, find the ratio $ \frac {Area\, \left(ABC\right)}{Area\, \left(ADE\right)}$.

A) $ 2\left(5 \plus{} \sqrt {3} \right)$ B) $ 7\sqrt {2}$ C) $ 5\sqrt {3}$ D) $ 2\left(3 \plus{} \sqrt {5} \right)$  E) $ 2\left(\sqrt {3} \plus{} \sqrt {5} \right)$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476052#p1476052]Problem 30[/url]

$ a_{i} \in \left\{0,1,2,3,4\right\}$ for every $ 0\le i\le 9$. If $ 6\sum _{i \equal{} 0}^{9}a_{i} 5^{i} \equiv 1\, \, \left(mod\, 5^{10} \right)$, $ a_{9} \equal{} ?$

A) 0  B) 1  C) 2  D) 3  E) 4

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729773#p2729773]Problem 31[/url]

30 same balls are put into four boxes $ A$, $ B$, $ C$, $ D$ in such a way that sum of number of balls in $ A$ and $ B$ is greater than sum of in $ C$ and $ D$. How many possible ways are there?

A) 2472  B) 2600 C) 2728 D) 2856 E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476053#p1476053]Problem 32[/url]

Let $ \left(a_{n} \right)_{n \equal{} 1}^{\infty }$ be a sequence on real numbers such that $ a{}_{n \plus{} 1} \equal{} a_{n} a_{n \plus{} 2}$ for every $ n\ge 1$. The number of elements in the set $ \left\{a_{n} : n\ge 1\right\}$ cannot be

A) 2  B) 3  C) 4  D) 5  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2729774#p2729774]Problem 33[/url]

$ \left|AC\right| \equal{} \sqrt {2}$, $ B$ is the midpoint of $ \left[AC\right]$, $ E$ is the midpoint of arc $ AB$ of a circle having chord $ \left[AB\right]$, and $ D$ is the point of tangency drawing from $ C$.($ D$ lies on the opposite side of line $ AB$ to $ E$). If $ \left[DE\right]\bigcap \left[AB\right] \equal{} \left\{F\right\}$, $ \left|CF\right| \equal{} ?$

A) $ 5\sqrt {2}$  B) $ 4\sqrt {2}$ C) 8  D) 6  E) $ 4\sqrt {3}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476055#p1476055]Problem 34[/url]

For how many primes $ p$, there exits unique integers $ r$ and $ s$ such that for every integer $ x$ $ x^{3} \minus{} x \plus{} 2\equiv \left(x \minus{} r\right)^{2} \left(x \minus{} s\right)\, \, \left(mod\, p\right)$?

A) 0  B) 1  C) 2  D) 3  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476056#p1476056]Problem 35[/url]

Flights are arranged between 13 countries. For $ k\ge 2$, the sequence $ A_{1} ,A_{2} ,\ldots A_{k}$ is said to a cycle if there exist a flight from $ A_{1}$ to $ A_{2}$, from $ A_{2}$ to $ A_{3}$, $ \ldots$, from $ A_{k \minus{} 1}$ to $ A_{1k}$, and from $ A_{k}$ to $ A_{1}$. What is the smallest possible number of flights such that how the flights are arranged, there exist a cycle?

A) 14  B) 53  C) 66  D) 79  E) 156

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476057#p1476057]Problem 36[/url]
Let $ x_{1} ,x_{2} ,\ldots ,x_{9}$ be real numbers on $ \left[ \minus{} 1,1\right]$. If $ \sum _{i \equal{} 1}^{9}x_{i}^{3} \equal{} 0$, then what is the largest possible value of $ \sum _{i \equal{} 1}^{9}x_{i}$?

A) 1  B) $ \frac {3}{2}$  C) 3  D) $ \frac {9}{2}$  E) None

This is the first round of turkish mathematical olympiads. 2 and half an hour for 36 problems. Every Turkish high-school participates with 6 contestants.
Please look at some of the problems. What are the appropriate forums for these multiple choices? According to your answers,  I will write down the problems at those forums later.

This is your topic:

http://www.mathlinks.ro/viewtopic.php?t=112177

my aim is to answer the problems in the contest. i rewrote all of them in latex. soon i will add them in seperate sub-forums, algebra, geometry, etc.
maybe i must have added this latex form to the  old post. it was a mistake. what is your answer about problem types? which subforums are appropriate?
in general, because the time for the test is limited and they are multiple-choice questions, their level is below from classic olympiad problems. But there are nice problems, eg. see the problems   4,30,36.

Pre-Olympiad for easier, put the hardest in Olympiad. That's only my opinion...

I have added some problems into pre-olympiad section. because i have no permission to edit the first post, i will give them here.

[url=http://www.mathlinks.ro/viewtopic.php?t=272605]Problem 4[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272603]Problem 6[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272606]Problem 7[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272608]Problem 8[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272609]Problem 11[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272610]Problem 12[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272611]Problem 14[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272612]Problem 15[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272613]Problem 16[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272615]Problem 18[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272614]Problem 19[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272616]Problem 20[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272617]Problem 27[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272618]Problem 28[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272619]Problem 30[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272620]Problem 32[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272621]Problem 34[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272622]Problem 35[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=272623]Problem 36[/url]


Given a piece of paper in rectangular shape of dimensions 1cm by 2 cm.
Using this piece of paper to cover a circle with radius r cm, what is the maximum value of r ?
Hint:
[hide]Of course, the answer is not 0.5 . [/hide]


This is not a story! Here we take up a topic and start discussing it in a very 'serious' tone. Jumping from topic to topic is allowed only if there is a logical flow.

I choose 'Cows' as the first topic.

[b]"Cows are the animals with the most stoical personality ...[/b]

they moo with heart and they make mighty food...

and they leave behind heaps of ... er ... fertility.

cow dung is one of the most used natural fuels

Many people in alberta canada have died from mad cow disease. Those poor souls. R.I.P. At least they died doing what they love. Eating large amounts of steak. Them fat asses.

Talking about asses, they are not really famous for an intellectual acumen, but you must give them credit for being hard working. They are the proletariat.

asses are really just a "fancy"/weird word for donkey. Donkeys are/were used to transport goods/people....Aesop's Fables

In some parts of Mexico donkey meat is eaten.

Mexico is a nice and sunny place. IMO was hosted last year by Mexico.

The three-letter abbreviation IMO may have several meanings, depending on context:
In My Opinion
International Mathematical Olympiad
International Maritime Organization
International Meteor Organisation
and more.

Meteors can be very dangerous when you are out in the open.
My advice is to avoid their acquaintance and intimacy until you are safely home.

Making the acquaintance of a meteor in an unpleasant context is highly improbable, but unfortunately it is only finitely improbable.  Thus any device that acts as an Infinite Improbability Engine may very well cause this to happen when turned on.

The fact that "an Infinite Improbability Engine cannot be larger than $3.7 * 10^3 \, m^7$" is an allegory of Maxwell's Equations.

But you also need a fresh cup of tea. It is better to stay at home and have this cup yourself while watching the meteors against the night sky.

A good tea to drink is Chai tea. Chai tea is made in India, but you can make it yourself.

Sure. I can make it myself because i live in india.

But I make it myself too, and I live in USA.

did you know?
you can make the same thing of tea anywhere on earth and it will still taste relatively the same, unless you make it underwater or something.

I can't understand how people kiss underwater

which destroys the very purpose of kissing - to rub lips

what did you write there bubka that got edited to "-"?Naughty boy.....

One day,'-' met "-". "-" wasn't happy. He went and commited suicide. He lived happily ever after.

Guess who "-" is.

"-" is dead. Poor chap -- never found happiness in his life. Only in death did he reach his salvation.

Wait a minute. He could have been sent to the bad place. ;)

but then ._. met up with '-' to discuss the death of "_" who went to the same school as another guy, `~' but looked stupid so no one liked him. Then one day, *__* came along and killed them all

How do you know? Did they tell you about this????

I suspect I forgot the purpose of this thread

Shady has a point. Let us stop this nonsense and get back to the topic - which was a serious piece of writing on a non-serious topic. And [i]don't[/i] digress from the chsen topic too much.

My next chosen topic is: HAIRPINS.
I invite someone to start. Remember - keep your tone serious throughout, even when you write something incredibly funny.

I thought that we could change the topic!(This is on the topic)

mat dear, if you can't be serious, be quiet. (This is most definitely NOT on the topic, and neither is the last one)

Where is the talking mat?

Hair pins: Girls wear them. (or at least girls do).


linear algebra
matrix
calculus
integration
algorithm
superior algebra
superior algebra unsolved

Let $A$ is associative ring in which $\forall a\in A$ $a^2=0$ and $\forall n\in\mathbb N$ $na=0\Rightarrow a=0.$
Prove that $abc=0,$ where $a\in A,$ $b\in A,$ $c\in A.$

From the fact that $(a+b)^2=0$ deduce that $ab=-ba$. Now it is only left to permute the multiplications(our ring is associative) in $abc=-bca=cab=-abc$ which implies that $abc+abc=0$ which together with the condition that our ring is of zero characteristic gives us $abc=0$.

Or, observe that
$0=(a+bc)^2+(ab+c)^2-(b+ca)^2=2abc$, hence $abc=0.$

What does it mean $A^6=0$-we have that $a^3=0$ and surely $a^6=0$ for all $a \in A$.

[quote="enescu"]It means that the product of any 6 elements of $A$ is zero.[/quote]
Yes! :lol:

[quote="arqady"]OK
Let $A$ is associative ring in which $\forall a\in A$ $a^3=0$ and $\forall n\in\mathbb N$ $na=0\Rightarrow a=0.$
Prove that $A^6=0.$ ;)[/quote]

Proceeding in a similar manner to before, we observe that:

$\forall x,y,z\in A: (x+y+z)^{3}-(x+y-z)^{3}-(y+z-x)^{3}-(z+x-y)^{3}=0$

which can be simplified to:

$\forall x,y,z\in A: xyz+xzy+yxz+yzx+zxy+zyx=0$ (Eq 1)

We are asked to show:

$\forall a,b,c,d,e,f \in A: abcdef=0$

We observe that there are $6!=720$ permutations of the variables $a,b,c,d,e,f$ giving rise to $720$ products of the six distinct variables.  We also observe that by partitioning the six variables using four partitions that can be placed in $7\choose 4$ ways, we can come up with constraints on the $720$ permutations.  For example, from the following partition:
$a|b|c|de|f$
we would get the following constraint:

$abcdef+abdecf+acbdef+acdebf+adebcf+adecbf=0$ (Eq 2)

The partitions can be placed in ${7\choose 4}=35$ ways, giving rise to $35$ constraints for each permutation;  but each constraint contains six permutations.  So the total number of constraints similar to Eq 2 is thus $35*720/6=4200$.  This gives us a set of $4200$ equations in $720$ unknowns.  This gives rise to a $4200$x$720$ matrix $M$ where $M_{ij}=1$ if constraint $i$ involves permutation $j$, and $M_{ij}=0$ otherwise.  If we can show that $M$ has rank $720$ when considered as a matrix over $\mathbb Q$, we will be done.  (It suffices to perform computations on $M$ over the rationals because of the condition that $\forall n\in\mathbb N$ $na=0\Rightarrow a=0$.  If the rank is $720$ when the matrix is considered as a matrix over $\mathbb Q$, then some integral multiple of $abcdef$ must equal $0$, and hence $abcdef=0$.)  I wrote a Java program which verified that $M$ does, indeed, have rank $720$, so we are done.

Writing the Java program itself was interesting.  I was concerned that if I used floating point arithmetic, then due to rounding while reducing the matrix some number that should be zero might not actually be zero, leading to a false conclusion that the matrix has a particular rank.  And if I compute the exact numerator and denominator for every rational number in the matrix as I reduce it, the numbers quickly become unmanageable.

What ended up working was to perform the computations on the matrix over the field ${\mathbb Z}_{p}$ for some prime number $p$.  If the matrix has rank $720$ when considered as a matrix over ${\mathbb Z}_{p}$, then it must also have rank $720$ when considered as a matrix over ${\mathbb Q}$ (although the converse doesn't necessarily hold).  Performing computations over ${\mathbb Z}_{p}$ is quite reasonable, and I was able to determine that the matrix has rank $720$ when considered as a matrix over ${\mathbb Z}_{5}$, and hence over ${\mathbb Q}$.  Hence the original result we were asked to show holds.

[quote="arqady"]It is very interesting, but I do not think, that it is possible to consider it the proof.[/quote]

I guess this means you are taking one of the following three positions:

1. You don't consider proofs partly using computers to be valid proofs.  This is perhaps a reasonable position to take but it would put you at odds with the fact that proofs, in part, by computer are quite common in modern mathematics, and are generally accepted by journal reviewers provided the computer algorithm is described in such a manner that it can be readily reproduced.
2. You consider some portion of the non-computer portion of my argument to be fallacious--in which case it would be good to clarify where you think the fallacy is.
3. You ran the algorithm as specified by myself but got a different result--ie you feel I made a programming error.

It would be helpful if you clarified your position on this.

[quote="arqady"]Dear NoYoo-hooForMe!
 I willingly trust, that in your program there is no mistake. In my opinion, your solution indirectly acknowledges that the statement is true, but not more than that! It should be solved in a way a person could understand, rather than a computer.
I will explain: If i am playing against a chess computer program which beats grand masters, I can rely on it practically, but not theoretically. There could be any latent mistake which even the programmers of this program cannot find, which can lead to a 100% defeat, when the opponent is playing in a certain way. Nobody is able to verify that there is no such mistake in the program. I think, that in mathematics, such things are inadmissible. But i am in no way insisting on it at all. I will only say one thing - this problem has a simple enough non computer oriented solution.[/quote]

There is one important difference and one important similarity between a chess program and my solution.  The similarity is that because there are a finite number of possible chess positions, it theoretically is possible to write a perfect computer chess playing program just as my program completely analyzes the problem.  The difference is that, AFAIK, such a chess program is far beyond the capabilities of even the most powerful supercomputer for the forseeable future, whereas my program is well within the capabilities of even my very ordinary laptop.  (As an aside, it would be interesting to note when Moore's Law forecasts a perfect analysis of chess by computer).

I'm not sure that I agree that my solution cannot be understood by a human.  I reduced the problem to the question of computing the rank of a matrix over ${\mathbb Z}_{5}$ using steps that I think a human can readily understand.  I defined the matrix completely (with the exception that I did not define the order of the rows or columns, but that has no effect on the rank).  Computing that rank then requires only a rote algorithm that humans can easily understand and that has been known for at least 200 years.  It was only the actual execution of that rote algorithm that I left to a machine.  This is another important difference with current chess programs--the chess programs may behave in ways that are not readily predictable by humans.  Not so with my program.

If I, or someone else, comes up with a proof that does not require a computer and for which the math is reasonably accessible, I will certainly agree that that is a more elegant solution.  I do not claim that this solution is especially elegant.

[quote="arqady"]I know to prove, that if $a^5=0$ then $A^{15}=0$ for 2 generatives.
 Whether you can check up, what $A^{15}=0$ generally?
 How much I know, it is unsolved problem. ;)[/quote]

I think that case and the general case is probably beyond the limits of my laptop :) .  But I will keep thinking about a simpler solution.

[quote="perfect_radio"]Could you share with us that non-computer proof?[/quote]
If are you mean to:
Let $A$ is associative ring in which $\forall a\in A$ $a^3=0$ and $\forall n\in\mathbb N$   $na=0\Rightarrow a=0.$
Prove that $A^6=0.$ ?
 If so, then a hint: prove that $a^2bc^2=0$ $\forall a,b,c$ from $A$.

[quote="perfect_radio"]I meant the one with $abcdef=0$.[/quote]
OK. Hint: prove that for all $a,b,c$ from $A$ exist $a^2bc^2=0.$ ;)

Sorry, I got confused. $A^6$ does not make much sense too me  :(

Has anybody proved arqady's hint (after almost a year)?


If you took this year's state competition, decide whether it's hard or easier than last year.. If you didn't yet, tell us what you to practice... PLEASE DO NOT DISCUSS THE PROBLEMS!

I don't think it's a good idea for people to give ideas on what to study if they have seen the test.  Maybe I am misreading your intent, can't quite get the gist of the last part of your post.

Never mind on the last part
Just wondering, how many people took their test?

Off-topic:

Good luck to everyone who's taking their state test tomorrow! (I'm one of them, so I'm not going to get a chance to post here until after I finish the competition.)

i didnt take last year's officially but as a practice and i thot it was easier

I took test today and didn't made it..
It was pretty easy except the sprint I guess..

silver, do you know what your score/rank/state was?

I don't know my total score.. For score, I only know I got 6 rights on target...Since I don't know my score, I don't know my rank neither.. I'm from PA

i did better at state then i did at chapter, chapter i had 38, state i had 40....I didn't think it was really that hard

silverfalcon, your in pennsylvania right? did shaunak kishore make nationals?

[quote="Silverfalcon"]Yes. In the first place... He's amazing kid.. I'm just feel proud to see him in distance...[/quote]

And what was Shaunak's score?  46?  Last year, did he not move up in the countdown round by 3 positions or so?

Last year in state he started 8th, but moved up to 1st.  In nats he got 5th or something after moving up a few.

actually, i think he got 7th or 8th in nationals and moved up to about 5th or 4th. i remember him beating this other person who just randomly beeped in and said nonsense. think it was arnav tripithy

i just realized how stupid my last post was. sorry.


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_{0}^{\infty} \frac{1}{1\plus{}x^{3}}\ dx$.

Glad to see you again, AndrewTom. The answer is (3^1/2)*2pi/9.  Accidentally, the answer is as same as the answer of the problem we solved yesterday.

hello, use the identity
$ \frac{1}{x^3\plus{}1}\equal{}\frac{1}{3(x\plus{}1)}\plus{}\frac{1}{3}\frac{2\minus{}x}{x^2\minus{}x\plus{}1}$
Sonnhard.


Hi,

Prove that for $ n \in \mathbb{N}^*$, we have :  $ \left( 1\plus{} \frac {1}{2n} \right)^n \minus{} \left(1\minus{} \frac{1}{2n} \right)^n > 1$.

thanks,

blang.

I am very sorry, it's really obvious  :blush:


[quote="1=2"]Compute

$\sqrt[8]{-527-336i}$

Last time I posted a problem like this, people were too lazy to do it. So if you haven't tried to solve this problem, don't post.[/quote]

Cis form and expand.

[hide="Solution"] We'll just take the square root three times.  Now,

$(a+bi)^{2}= (a^{2}-b^{2})+2abi =-527-336i$

So $b^{2}-a^{2}= (b-a)(b+a) = 527 = 17 \cdot 31$, from which we have $b = \pm 24, a = \pm 7$.  Indeed, $2ab = 2 \cdot 7 \cdot 24 = 336$.  So, [b]one of[/b] the square roots is

$\sqrt{-527-336i}=-7+24i$

If you're familiar with Pythagorean triples, then you know that

$-7+24i = (3+4i)^{2}$

And that

$3+4i = (2+i)^{2}$

So, [b]one of[/b] the eighth roots is

$\sqrt[8]{-527-336i}= 2+i$

And the others can be obtained easily. [/hide]


inequalities
calculus
derivative
limit
function
real analysis
real analysis unsolved

[color=darkred]Prove that for the [b]T. Lalescu[/b]'s sequence $ L_{n}\equal{}\sqrt [n\plus{}1]{(n\plus{}1) !}\minus{}\sqrt [n]{n!}$ exists the "[b]clenching[/b]" $ \left(\frac{n}{n\plus{}1}\right)^{n\plus{}1}< L_{n}<\left(\frac{n}{n\plus{}1}\right)^{n}$.[/color]

[quote][color=darkred]Prove that for the [b]T. Lalescu[/b]'s sequence $ \boxed{\ L_{n}\equal{}\sqrt [n\plus{}1]{(n\plus{}1) !}\minus{}\sqrt [n]{n!}\ }$ exists the "[b]clenching[/b]" $ \boxed{\ \left(\frac{n}{n\plus{}1}\right)^{n\plus{}1}< L_{n}<\left(\frac{n}{n\plus{}1}\right)^{n}\ }$.[/color][/quote]
[b]Proof.[/b]

$ 1\blacktriangleright$ $ \mathrm{L.H.S.\ : }$ ?????????????????????? [b]Here is a nice and dificult problem (if is possible, without derivatives)  ! [/b]??????????????????????

$ 2\blacktriangleright$ $ \mathrm{R.H.S.\ :\ G.M.\le A.M.\ for\ }$ $ x_{k}: \equal{}\left\{\begin{array}{ccc}\frac{\sqrt [n]{n!}}{n}&\mathrm{if}& k\in\overline{1,n}\\ \\ \left(\frac{n}{n\plus{}1}\right)^{n}&\mathrm{if}& k \equal{} n\plus{}1\end{array}\right\|$ $ \implies$ $ \frac{n\cdot\frac{\sqrt [n]{n!}}{n}\plus{}\left(\frac{n}{n\plus{}1}\right)^{n}}{n\plus{}1}>$ $ \sqrt [n\plus{}1]{\left(\frac{\sqrt [n]{n!}}{n}\right)^{n}\cdot\left(\frac{n}{n\plus{}1}\right)^{n}}\equal{}$

$ \frac{\sqrt [n\plus{}1]{(n\plus{}1)!}}{n\plus{}1}\implies$ $ \sqrt [n]{n!}\plus{}\left(\frac{n}{n\plus{}1}\right)^{n}>\sqrt [n\plus{}1]{(n\plus{}1)!}\implies$ $ \sqrt [n\plus{}1]{(n\plus{}1)!}\minus{}\sqrt [n]{n!}<\left(\frac{n}{n\plus{}1}\right)^{n}$ $ \implies$ $ L_{n}<\left(\frac{n}{n\plus{}1}\right)^{n}$.

[b]Remark.[/b] $ \lim_{n\to\infty }\left(\frac{n}{n\plus{}1}\right)^{n}\equal{}\frac{1}{e}\implies\boxed{\ \lim_{n\to\infty}\left(\sqrt [n\plus{}1]{(n\plus{}1)!}\minus{}\sqrt [n]{n!}\right)\equal{}\frac{1}{e}\ }$.

Sorry, I don't an own proof. I have a old intricate proof with derivatives from one former my student.

Ok,thank you.
Does it need to consider the function $ \Gamma$?
Can you post the proof you know?

zhaobin,

You are right: More generally, for proving that the left hand side of Virgil's inequality holds, we have that:

$ \left(\frac{x}{x\plus{}1}\right)^{x\plus{}1}\leq\frac{1}{e}\leq\sqrt[x\plus{}1]{\Gamma(x\plus{}2)}\minus{}\sqrt[x]{\Gamma(x\plus{}1)}$, for all $ x>0$.

The proof of this result is based on the fact that the function

$ f(x)\equal{}\sqrt[x\plus{}1]{\Gamma(x\plus{}2)}\minus{}\sqrt[x]{\Gamma(x\plus{}1)}$ defined on $ (0,\infty)$ is a completely monotonic function which maps the interval $ (0,\infty)$ onto the interval$ (\frac{1}{e}, 1\minus{}e^{\minus{}\gamma})$.

Thus, our function decreases to $ \frac{1}{e}$. The proof of this result is not elementary at all and it is based on a Stieltjes transform technique.

[color=darkblue][b]That's it ![/b] The [b]Didilica[/b]'s proof is similarly, [b]as idea[/b], to the my student's, i.e. $ \left(\frac{n}{n\plus{}1}\right)^{n\plus{}1}<\frac{1}{e}<L_{n}$. Can say that I love it forever ! Sounds it strangely, no?
I tried posting it, to find an elementary proof for $ \frac{1}{e}<L_{n}$. It is one from the most aesthetic in all respects inequalities what I met in the my life !

After many years I didn't manage it elementary. Maybe you have more success what I wish you.[/color]

Thank you,I think that is very amazing result $ f(x)$ is mononic decreases to $ \frac{1}{e}$.
Because  $ 1\minus{}e^{\minus{}\gamma}>\frac{1}{e}$ is not trivial,either.
Can you offer a file or a link for the not elementary proof?


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Let $M=S^{1}\times D^{2}$. Let $M_{k}$ denote the space obtained by deleting $k$ points from the interior of $M$. Calculate the homology of $M_{k}$.

For $k=0$ we have the homotopy type of $S^{1}$ because our space is the solid torus.
For $k=1$ we have the homotopy type of a sphere with two identified points.
For $k>2$ we have a necklace of $k$ spheres.

To get a visual of the last two, look at the solid torus with $k$ points removed from its interior. Imagine that instead of the $k$ points, $k$ small nonintersecting balls are missing. Start expanding these balls so that they cover the interior of the torus. Since we cannot intersect them, they will be separated by membranes that are discs with boundry on the boundry of the solid torus. Shrink these disks to points and you get a necklace of spheres, or if there is only one room (one point removed), you get a sphere with two points identified (on the outside, not on the inside; not that it matters topologically).

You can give these spaces simple CW complex structures and compute homology.


real analysis
calculus
derivative
linear algebra
matrix
function
topology

Suppose that $f \in C^{2}([0, 1])$ and define $\Gamma = \{y = f(x)\}$ in $\mathbb{R}^{2}$, $0 \le x \le 1$. Show that:
$m(\Gamma+\Gamma) > 0 \Leftrightarrow \exists$ an open set $O \in \Gamma+\Gamma \Leftrightarrow f$ is not linear.

($m$ is the Lebesgue measure on $\mathbb{R}^{2}$)

It suffices to prove that $f$ nonlinear $\implies$ the interior of $\Gamma+\Gamma$ is nonempty.

Consider the map $F\colon [0,1]^{2}\to\mathbb R^{2}$ defined by $F(x_{1},x_{2})=(x_{1}+x_{2},f(x_{1})+f(x_{2}))$. Its derivative is \[DF(x_{1},x_{2})=\begin{pmatrix}1 & 1 \\ f'(x_{1}) & f'(x_{2}) \end{pmatrix}\] Since $f$ is nonlinear, there exist $x_{1},x_{2}\in (0,1)$ such that $f'(x_{1})\ne f'(x_{2})$. Hence $\det DF(x_{1},x_{2})\ne 0$, and we can apply the Inverse Function theorem.

Looks like I only used $f\in C^{1}[0,1]$.


while i was studying economics course, I had a problem solving for a T where

e^(-rT)=(0.2a-T)r+1

I really need to solve this.

please~~~~~~~~

Thanks in advance.

Great haloween

hello, do you meant the equation $ e^{\minus{}rT}\equal{}(0,2a\minus{}T)r\plus{}1$? What about the meaning of the variables $ r,T$ and $ a$?
Sonnhard.

He wants to solve for $ T$.  I think Lambert-W function must be employed.


Hello everybody,

1 problem gives me hard time:
A semicircular cross section of a one way tunnel. the single traffic lane is 12feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least 1/2 foot when they are inside the traffic lane, what should be the limit of the height of vehicles that are allowed to use the tunnel?
(Diameter is 20 feet)

I thought that answer must be Radius (10) less 4 (1 side) less 1/2 foot = 5 1/2 feet: But it seems the answer is 7 1/2 feet. has anyone a glue as of how to resolve this problem?

Here's the answer:

This is kind of hard to explain without a diagram, but I'll give it a shot. Draw the circle and the rectangle in it, making the corners of the rectangle tangent to the circle, and having the rectangle centered in retrospect to the diameter of the circle. Draw in a radius that splits the semicircle in half, giving you a quarter circle. Now draw in another radius (10) that is also the diagonal of the rectangle. This creates a right triangle with base six (as it is half of the lane width). Since it is a 3:4:5 right triangle, the height is 8. However, you have to subtract a $1/2$ foot for your answer, as the problem asks you to do so. Therefore your answer is $7$ $1/2$

Ok thank you very much for your info, definitly I would never have had think about this way for resolving it!

Cheers


$ \frac{11}{5}x\minus{}\sqrt{2x\plus{}1}\equal{}3y\minus{}\sqrt{4y\minus{}1}\plus{}2,\quad (x\geq 0; y\geq 1; x\equal{}5k; x,y,k\in Z)$

[quote="EcstaticPotter"]Since 4y-1 cannot be a perfect square 4y-1 must be equal to 2x+1[/quote]

Using this fact,...
[hide]$ 4y \minus{} 1 \equal{} 2x \plus{} 1$

$ 4y \minus{} 2 \equal{} 2x$

Divide both sides by 2:

$ 2y \minus{} 1 \equal{} x$

or $ x \equal{} 2y \minus{} 1$ . . . (*)

Because the square root terms on opposite sides of the equation are equal to
each other, eliminate them and get:

$ \frac{11}{5}x \equal{} 3y \plus{} 2$

Multiply both sides by 5:

$ 11x \equal{} 15y \plus{} 10$ . . . (**)

Option to substitute (*) into (**):

$ 11(2y \minus{} 1) \equal{} 15y \plus{} 10$

$ 22y \minus{} 11 \equal{} 15y \plus{} 10$

$ 7y \equal{} 21$

$ y \equal{} 3$

Using (*),

$ x \equal{} 2(3) \minus{} 1$

$ x \equal{} 5$

The solution, and the only solution given the parameters, is:

$ (5, 3)$

[/hide]


calculus
derivative
logarithms
geometry
function
real analysis
real analysis unsolved

More generally, for $x>1$, we have $\sum_{k\ge 0}\frac{2^k}{x^{2^k}+1}=\frac 1{x-1}$. If we replace $x>1$ with $\frac 1x,\ x<1$ and divide the relation we find by $x$, we get $\sum_{k\ge 0}\frac{2^kx^{2^k-1}}{1+x^{2^k}}=\frac 1{1-x}\ (*)$.

The LHS in $(*)$ is the sum of the derivatives of $\ln(1+x^{2^k})$, i.e. the derivative of the sum $\sum_{k\ge 0}\ln(1+x^{2^k})$, which is the derivative of $\ln(\prod_{k\ge 0}(1+x^{2^k}))=\ln(\sum_{t\ge 0}x^t)=\ln\left(\frac 1{1-x}\right)$, and this is clearly equal to the RHS in $(*)$.

I didn't prove the necessary convergence issues, because everything is pretty clear in that area: all the series of functions converge uniformly in any compact interval included in $(0,1)$.


Dear Mathlinkers,
The following problem is well know:
let ABC a triangle, H the orthocenter of ABC, G the median point of ABC, 0 the circumcircle of ABC, X the second intersection of AG with 0, X' the symmetric of X wrt BC. (cyclically, Y', Z').
It is knowed that H, X', Y' and Z' are cocyclic on the G-Hagge circle noted 1 which is a generalization of the Fuhrmann's circle.
Now, my problem.
With the same hypothesis, let U the point of intersection of XX' with BC, 2 the symmetric of 0 wrt BC, V the point of intersection (near A) of AG with 2, V' the orthogonal projection of V on BC, P the antipode of H on 1, P' the orthogonal projection of P on AH and A' the foot of the A-altitude of ABC.
Prove that   (1) P'V' is parallel to GU    (2) A'G is parallel to V'X'   (3) P', G and X' are collinear.
Sincerely
Jean-Louis

Dear Mathlinkers,
any synthetics ideas?
Sincerely
Jean-Louis

Check out the recent paper by Bradley and Smith at forumgeom.fau.edu/FG2007volume7/FG200730.pdf

This immediately gives (3): P' is the image of X under the similarity taking ABCG to X'Y'Z'G, and thus P',G,X' are collinear.

It probably will also be helpful for doing (1) and (2)....

Dear Hiltigore and Mathlinkers,
I saw this paper and I am ok with you. But all my work goes through synthetic proofs and therefore I don't want use any transformation.
I continue my research.
Thank you.
Sincerely
Jean-Louis

Dear Mathlinkers,
you can see for example
http://perso.orange.fr/jl.ayme/   vol. 5  Le P-cercle de Hagge p. 32.
Sincerely
Jean-Louis

Dear Mathlinkers,
an article concerning the “’Du cercle des huit points au cercle des six pieds or the theorem on the six pedals’’ has been put on my website with interesting examples.
http://perso.orange.fr/jl.ayme    vol. 9  p. 14
You can use Google translator
Sincerely
Jean-Louis

Let $\triangle ABC$ be the medial triangle of $\triangle A_1B_1C_1$，$G_1$ and $H_1$ respectively are the median point and orthocenter of $\triangle A_1B_1C_1$
$D$ is the feet of $A_1$ wrt  $\triangle A_1BC$，$AG$ intersect $BC$ at $M$，$K$ is the symmedian point of $\triangle A_1BC$，$A_1K$ respectively intersect
$AD$ and $BC$ at $L$ and $J$，$AG$ intersect $\odot G_1BC$ at $N$，$S$ is the midpoint of $A_1N$，$LT \perp A_1D$，$T \in A_1D$ .

Then $A_1X'$、$B_1Y'$ and $C_1Z'$ are concurrent at $P$，$P$ is the symmedian point of $\triangle A_1B_1C_1$ . $MB^2=MX \cdot MA_1$ .
$A_1D \cdot A_1H=A_1K \cdot A_1N$，$2SV=A_1X$，$\frac{HV}{SV}=\frac{2H_1X}{A_1X}=\frac{DA'}{DA_1}=\frac{LT}{DT}$，so $\triangle HVS \sim \triangle LTD$，so $\triangle A_1HS \sim \triangle A_1LD$
So $A_1D \cdot A_1H=A_1S \cdot A_1L$，so $A_1L=2A_1K$，so $L \equiv P$，i .e . $P$ is on $AD$ . Cuz $\frac{DP}{AP}=\frac{PT}{PP'}=\frac{DJ}{2MJ}=\frac{DU}{2V'U}=\frac{A'V'}{4MV'}$
So $P'$、$V'$、$G_1$ are collinear，so $\boxed{P'V' \parallel GU}$ . $\triangle XMU$ is homotetive with $\triangle A_1MD$，$G$ is the centroid of $\triangle AA'D$，so $XV' \parallel A_1A'$
So $\angle GA'D=\angle ADA'=\angle A_1A'D=\angle XV'U=\angle X'V'U$，so $\boxed{A'G \parallel V'X'}$  . Let $E$ and $F$ be the midpoints of $VG_1$ and $XG$
Then $P'G \parallel V'E \parallel UF \parallel X'G$，so $\boxed{P',G,X'}$ are collinear . Done .


function
real analysis
real analysis unsolved

There exits a function $ \mathbb{R} \to \mathbb{R}$ such that every point is local maximum or local minimum point?

Any constant function can do the trick.  :wink:

There exits a non constant function, witch has this properties?

Why has this function this properties, you know the solution?

Let c=1 and d=0. Then every rational number is a local maximum, while every irrational is a local minimum. It all follows from the the fact that $ \mathbb{R} \equal{} \mathbb{Q} \cup \mathbb{I}$, dear [b]csabafarkas[/b].

This has, probably, been posted before but, since we have more "newcomers" than "old timers" here now, I'll ask it anyway:

Find (with proof) all continuous functions with this property.

Any constant function still does the trick.  :)

I liked [b]fedja[/b]'s question, but not to spoil the fun of those still trying to solve it on their own, I'll only give a [hide="hint."]
You should have managed to prove that such a function is necessarily [i]locally[/i] constant.

Next, try to use (or, "to prove and to use" :)) the fact that a locally constant continuous function over a connected set is actually constant.[/hide]


inequalities
integration
calculus
complex numbers
real analysis
real analysis unsolved

Let  $f \in L^{1}$, $f>0$. Why is $|\hat{f}(y)| < \hat{f}(0)$ for $y \neq 0$. Here ^ denotes the Fourrier transform.

Why does the inequality have to be strict?

For any $\theta\in\mathbb R$ we have $\hat f(0)-\mathrm{Re}(e^{i\theta}\hat f(y))=\int_{-\infty}^{\infty}(1-\mathrm{Re}(e^{i(xy+\theta)}))f(x)dx$, where the term in parentheses is positive a.e.

Notice the advantage gained by replacing nonlinear $|\cdot |$ with linear $\mathrm{Re}(e^{i\theta}\cdot)$.

Sorry I'm not following, why is $\mathrm{Im}(e^{i\theta}\hat f(y))=0$ then?

and in your integral above, should $xy$ be $-xy$?

thanx

mlok never said that $\mathrm{Im}(e^{i\theta}\hat f(y))=0$ for all $\theta.$ Instead, he proved his real-part identity for all $\theta.$ Now note that you can choose one particular $\theta$ such that $\mathrm{Re}(e^{i\theta}\hat f(y))=|\hat{f}(y)|.$

As for $xy$ versus $-xy,$ that's a matter of convention in defining the Fourier transform. Personally, I'd use $-2\pi xy.$ If you look closely, you'll see that which version of the convention you use makes no difference to the argument.

[quote]Now note that you can choose one particular \theta such that \mathrm{Re}(e^{i\theta}\hat f(y))=|\hat{f}(y)|.
[/quote]

I understood that part. But why could we pick any random $\theta$?

Let's recap. You wanted to prove that $|\hat{f}(y)| < \hat{f}(0)$ under certain assumption.
I proved that $\mathrm{Re}(e^{i\theta}\hat{f}(y))<\hat{f}(0)$ for any $\theta\in [0,2\pi]$.
It remains to use the following: $|\hat{f}(y)|=\max_{\theta\in [0,2\pi]}\mathrm{Re}(e^{i\theta}\hat{f}(y))$.

I got it now. It was the last identity I was not aware of. Everything is clear now. Thanks


A negative element of a set is defined as an element to the left of the 0 on the number line, and a positive element is defined as an element to the right of the 0 on the number line.

Under the natural order of the integers, (Z,<), you have ... -3, -2, -1, 0, 1, 2, 3,...
in which case 3 is a positive element since it is to the right of the 0 on the number line.

However,
when you take an order other than the natural order, such as (Z, >), in which case you obtain:
 ...,3, 2, 1, 0, -1, -2, -3,...

3 is to the left of the 0 on the number line, and therefore a negative element.  :)

Sorry people. I didn't mean to start WWlll.

I just thought this was an interesting point.

*I looked in my math textbook. Would it help the confusion if I restated that as a negative element and not as a negative number? My math book states 3 under (Z, >)as a negative element and all of the negative elements in that set as the set of the "negative integers".

Sorry for any trouble this has caused, and sorry for icantpeeloranges's remarks.  At least he deleted some of them. Knowing him, he doesn't mean the comments as offensively as they seem. Sorry to all.

 :)  :)  :)  :)

[quote="xxreddevilzxx"]what is going on here? What was the originalquestion?[/quote]I didn't post a question, just  a cool thing that we had discussed in our math class, and that is in our textbooks.  :)
            *****I edited my original post, so hopefully it should make more sense*****
It's just a thought, not something to get into a fight over. I'm not trying to offend anyone, and sorry to anyone who has been offended in this.  :)

[i]Mod Note: This thread has been severely modified. Please don't make personal comments. This forum is for healthy discussion and for everybody to learn. Post that don't pertain to the discussion at hand have been deleted, and the original poster has modified the original post so that it conveys the correct meaning.[/i]


parameterization
number theory proposed
number theory

[quote="Bugi"]Determine the biggest possible value of $ m$ for which the equation $ 2005x \plus{} 2007y \equal{} m$ has unique solution in natural numbers.[/quote]
It is easy to see that $ 2 \cdot 2005 \cdot 2007$ has a unique solution. $ 2005x \plus{} 2007y \equal{} 2 \cdot 2005 \cdot 2007$ gives $ 2007 \mid x$ and $ 2005 \mid y$. Let $ x \equal{} 2005x_1$ and $ y \equal{} 2007y_1$ then we have $ x_1\plus{}y_1 \equal{} 2$, which has the only solution in natural numbers $ x\equal{}y\equal{}1$.
Then assume $ m > 2 \cdot 2005 \cdot 2007$ and it has a unique solution:
We easily obtain $ x > 2007$ or $ y > 2005$. (Suppose the contrary: Then $ 2005x \plus{} 2007y \le 2 \cdot 2005 \cdot 2007$...)
If $ x > 2007$ Then $ x_2 \equal{} x \minus{} 2007$, $ y_2 \equal{} y \plus{} 2005$ is also a solution.
If $ y > 2005$ Then $ x_2 \equal{} x \plus{} 2007$, $ y_2 \equal{} y \minus{} 2005$ is also a solution.

So the answer is $ \boxed{2\cdot2005\cdot2007}$


Over the duration of 2006, Herman has purchased a particular beverage on a regular basis. He notes near the end of the year that he only ever bought two at a time, and only ever on Sundays and Thursdays. What was the beverage, and what was the rationale behind his spending habits?

is herman in a cult?
is her religious?
is he buying the beverage for himself?

Could Herman buy the beverage only on Thursdays and Sundays?

Was Herman surprised/did he realize that he only bought the beverage on Thursdays and Sundays?

Was the beverage alcoholic?

but mustafa you can't but alcohol on Sunday can you? :?:

Did he drink the beverage or use it for some other purpose?

Is the beverage sold in packages of 2?

Could Herman buy the beverage only on Thursdays and Sundays? [b]No.[/b]

Was Herman surprised/did he realize that he only bought the beverage on Thursdays and Sundays? [b]Somewhat, but when he thought about it there was a very good reason for it[/b].

Was the beverage alcoholic? [b]No[/b].

Did he drink the beverage or use it for some other purpose? [b]He drank the beverage.[/b]

Is the beverage sold in packages of 2? [b]No. But for his purposes, one wasn't quite enough.[/b]

Is it related to a holiday?

Is he single, or a family man?

are those weekdays weekdays? [b]They are days of the week, though Sunday would probably be considered a weekend as opposed to a weekday[/b]
Did he buy it in a store or a bar? [b]Yes to store, no to bar.[/b]
Is he single, or a family man?[b]Yes to single, no to family.[/b]
Is it related to a holiday? [b]No[/b]

is he dating?
is he a criminal?
do these beverages contain a poisoned substance?
or am i asking questions which don't matter? :rotfl:

is he dating? [b]Irrelevent[/b]
is he a criminal? [b]No[/b]
do these beverages contain a poisoned substance? [b]Too much of it, maybe...[/b]
or am i asking questions which don't matter? [b]...[/b]

is he buying weed killer or something along those lines, and his yard is just a little too big for one squirt thing, and the store's sales are on sunday and thursday?

or maybe, the store has buy one get one free on sundays and thursdays?

No, the Thursday and Sunday bit has to do with his own lifestyle, not some store promotion.

Is it medicaton? [b]Not prescription, and usually not referred to by our society as such![/b]

does he have sandy looks?
is he planning to murder someone?
does he have 5 cats?
will he feed his relative an omlette?

funcia, stop spamming :spam: ! PGA tour tournaments begin on Thursday and end on Sunday. Relavent or Irrelavent?

hey i am serious i mean it if you have sandy looks you are taking arsenic and becoming immune they should be relevant!

does he have sandy looks? [b]IR[/b]
is he planning to murder someone? [b]No[/b]
does he have 5 cats? [b]IR[/b]
will he feed his relative an omlette? [b]IR[/b]
PGA tour tournaments begin on Thursday and end on Sunday. [b]IR[/b]

[hide="hint"]He is a college student.[hide="double hint"]An undiciplined one, at that.[/hide][/hide]

coomon man ig this was like a murder mystery i jus read they'd all be r
ok where is the beverage from?

You can buy two at a time, so you can buy $10000000000000000$ in one day as long as it is two at a time, right?

You can buy two at a time, so you can buy 10000000000000000 in one day as long as it is two at a time, right? [b]You'd die if you drank that many of them.[/b]

[hide="Another BIG hint"]What days are long-term college assignments usually due?[/hide]

does this have to do with the faculty at a college watching herman?
is the beverage a "feel-good" thing?

are these beverages caffeinated? (or extremely caffeinated beverages e.g. mountain dew, red bull, doubleshot espresso, etc. for that matter)

Did he buy the beverages on that day because he needed them on Monday and Friday?

This is hard...

[hide="Solution that would have worked"]
He is Dr. Jekyll/Mr. Hyde and is buying whatever the chemical was that he drank to change between these personalities.

Might have worked without the hint about long-term projects.
[/hide]

1100th post!

This is supposed to be 2006, not 1850's. :|

[quote="funcia"][hide]he bought a beverage with caffeine to cram![/hide][/quote]
Yay! He is a college student and a really hardcore procrastinator. He waits until the last day to do all of his assignments, frequently pulling all-nighters. As his long-term assignments are either due on Monday or Friday, he buys energy drinks on Thursday and Sunday.

The two at a time bit was a little red herring.

see! it works to ask IR questions!  :rotfl:


Here are all Nationals I have.
I don't have anything before 1991.
If anyone is interested in trading with me, please email me at epsrno@gmail.com. Thanks!

Yay! Finally someone has the 2006 test!

Can you send that to me? (check your PM for my email address)

I will trade the 2009 test for anyone who has the 2007 test...

good one...i used to have all of them, sopme one gave them to me, but the the flash drive got erased...

Please pay more attention to the stickies and post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]here[/url] when asking for tests in the future. Thanks.


A regular hexagon of area $H$ is inscribed in a convex polygon of area $P$. Show that $P \leq  \frac{3}{2}H$. When does the equality occur?

in the figure below $ S[A_1B_2B_1] \leq S[B_2'QB_2] ,  B_2B_1'B_1  and B_2B_3B_2'$ are equilateral triangles same with $ B_iB_{i\plus{}1}O$ with area $ \frac{H}{6}\Rightarrow$
$ S[B_1A_1B_2] \plus{} S[B_2A_2B_3]   \leq \frac{H}{6}$
$ S[B_3A_3B_4] \plus{} S[B_4B_4B_5] \leq \frac{H}{6}$
$ S[B_5A_5B_6]  \plus{} S[B_6A_6B_1] \leq \frac{H}{6}$
summing this three and H we get $ P \leq \frac{3}{2}H$
equality holds when $ A_1$ coresponds with the center of $ B_1B_2B_2'$ same for the other $ A_i$

I'm wrong about the equality case and i do not know obtained

can anyone post more precise solution,please

equality holds only when $p$  is an equilateral triangle


Six points A, B , C, D, E and F lie in clockwise order on a circle. The chords AD, BE and CF are concurrent. Let P, Q and R be the midpoints of the chords AD, BE and CF respectively. Two chords AG and AH are drawn so that AG//BE and AH//CF.
Prove that triagnles PQR and DGH are similar.

this is from israel (98, 99 or 00, if i'm not wrong)
i didn't try this much time to give a sintetic solution, but i solved with complex numbers. isn't difficult, try this way! (i'm very busy now, and don't have time to post a solution!  :blush: )


If $a,b,c$ are positive real numbers, and $a^{2}+b^{2}+c^{2}=1$, prove:

$\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\geq \frac{3}{4}(a \sqrt a+b \sqrt b+c \sqrt c)^{2}$