Let $ (f_j)$ be a Cauchy sequence in $ L^2$. Then there exists a subsequence $ (f_{j_k})$ such that $ f_{j_k}\to f$ almost everywhere. To show that $ f\in L^2$, fix $ \epsilon > 0$. Now there exists $ j_0\in\mathbb{N}$ such that $ \|f_i - f_j\|_2 < \epsilon$ as $ i,j\geq j_0$. If $ j\geq j_0$ then Fatou's lemma gives
\begin{align*} \int |f_j - f|^2d\mu & = & \int\lim_{k\to\infty }|f_j - f_{j_k}|^2d\mu \\
& \leq & \liminf_{k\to\infty}\int |f_j - f_{j_k}|^2d\mu \\
& = & \liminf_{k\to\infty}|f_j - f_{j_k}|_2^2\leq \epsilon^p. \end{align*}
As $ f_k - f\in L^2$ and $ \| f_j - f\|_2\to 0$ as $ j\to\infty$, we find that $ f = f_j - (f_j - f)\in L^2$ and $ f_j\to f$ in $ L^2$.

If you replace $ 2$ by $ p$ in this proof you see that $ L^p$ is complete for every $ 1\leq p < \infty$. This is Riesz's–Fischer's theorem.

Actually, he said l2, which I did not interpret as $ L^2$ but $ l^2$, but who knows?  Only diophantine.

That's correct. I meant the small l, not the capital L. Thanks. Hope anyone can help me with the small l problem.

That's easier, because the first step is easier.

Outline:

1. Since $ |x^{(n)}_k\minus{}x^{(m)}_k|\le\|x^{(n)}\minus{}x^{(m)}\|_2$ for all $ k,$ a sequence of sequences that is Cauchy in $ \ell^2$ is pointwise Cauchy, and hence has a pointwise limit. Let $ y_k\equal{}\lim_{n\to\infty}x^{(n)}_k.$

2. Show that in fact $ \|x^{(n)}\minus{}y\|_2\to 0.$

3. Show that (2) implies that $ y\in \ell^2.$

But you just prove your sequence is pointwise convergent. That does not imply convergence.

Where exactly in the proof of (2) are you getting stuck? ... I'm just wondering if all your postings are homework ?!?  :o

I'm just trying to prove for all epsilon > 0, there exists x_n in lp, |y - x_n| < epsilon. I stuck because x_n has infinitely many elements.
It's just that I have never found a good proof for this problem. All the proofs so far say that we can choose N big enough, but if we have infinitely many elements for x_n, than how can we choose that N?
Thanks

OK, for every $ \varepsilon > 0$ there is some $ N$ such that for every $ m,n\ge N$ you have $ \sum_{k\ge 0}(x^{(n)}_k \minus{} x_k^{(m)})^2 < \varepsilon$. Now let $ m\to\infty$. Why can you do that or - in other words - why can you evaluate the limit under the summation sign?

Sorry. I got confused with letters. Actually, I think my proof was invalid. Replace
[quote="MeKnowsNothing"]$ \liminf_{k\to\infty}|f_j \minus{} f_{j_k}|_2^2$
[/quote]
by $ \liminf_{k\to\infty}\| f_j \minus{} f_{j_k}\|_2^2$.


The senior class at South High School consists of 250 students. Of
these students, 130 have brown hair, 160 have brown eyes, and
90 have both brown hair and brown eyes. How many members of the
senior class have neither brown hair nor brown eyes?

[quote="Interval"]Very good.

One question:

What lead you to make 90 into - 90?[/quote]
What do you mean? He just used PIE.

PIE = Principle of Inclusion-Exclusion.

Google it up for Christ's sake.

[quote="Interval"]He's got:

130 + 160 - 90 + x = 250

Do you see minus 90 in the equation?

Why - 90 and not + 90?[/quote]

It overlaps.
See a venn diagram for more explanation.
You add 130 and 160 people together. Since 90 people have both, you are ADDING 90 2 TIMES. So you subtract 90.


geometry
geometric transformation
homothety
geometry proposed

i had a pre tst yesterday where i had 3 hours and a half to solve 8 geometry problems. i could only solve 5 including where i think i solved a little more than what was given it´s only a conjectur but i think it´s correct anyway here is the new problem :

Consider a circle with diameter $A_1A_2$. We draw the parallel lines $F_1A_1$ and $F_2A_2$; also, the line $F_1F_2$ is tangent to the circle. Let $B_1$  be the foot of the perpendicular from $F_1$  to to $A_1A_2$ and $B_2$  the foot of the perpendicular from  $F_2$  to $A_1A_2$. Prove that the circle of diameter $B_1B_2$  is tangent to $F_1F_2$.

But $B_1,B_2$ lie on the line $A_1,A_2$! :? How, then, can a circle with diameter $B_1B_2$ be tangent to $A_1A_2$, which is the same line as $B_1B_2$?

Marko, I have edited your post to correct some typos. Hope that I didn't get your intention wrong.

[quote="marko avila"]Consider a circle with diameter $A_1A_2$. We draw the parallel lines $F_1A_1$ and $F_2A_2$; also, the line $F_1F_2$ is tangent to the circle. Let $B_1$  be the foot of the perpendicular from $F_1$  to to $A_1A_2$ and $B_2$  the foot of the perpendicular from  $F_2$  to $A_1A_2$. Prove that the circle of diameter $B_1B_2$  is tangent to $F_1F_2$.[/quote]

Actually, this is more or less trivial:

First, consider the case when the line $F_1F_2$ is not parallel to the line $A_1A_2$. Then, the lines $F_1F_2$ and $A_1A_2$ meet at one point Q. Since $F_1A_1\parallel F_2A_2$, we have $\frac{QF_2}{QF_1}=\frac{QA_2}{QA_1}$, and since $F_1B_1\parallel F_2B_2$ (this is because $F_1B_1\perp A_1A_2$ and $F_2B_2\perp A_1A_2$), we have $\frac{QF_2}{QF_1}=\frac{QB_2}{QB_1}$. Thus, $\frac{QA_2}{QA_1}=\frac{QB_2}{QB_1}$. In other words, $\frac{QB_1}{QA_1}=\frac{QB_2}{QA_2}$. Hence, the homothety with center Q and factor $\frac{QB_1}{QA_1}=\frac{QB_2}{QA_2}$ maps the points $A_1$ and $A_2$ to the points $B_1$ and $B_2$. Thus, this homothety maps the circle with diameter $A_1A_2$ to the circle with diameter $B_1B_2$. On the other hand, this homothety maps the line $F_1F_2$ to itself (since this line passes through the center Q of the homothety). Thus, as the line $F_1F_2$ touches the circle with diameter $A_1A_2$, the homothetic image of this line $F_1F_2$ - this is the same line $F_1F_2$ - must touch the homothetic image of the circle with diameter $A_1A_2$ - this is the circle with diameter $B_1B_2$. In other words, the circle with diameter $B_1B_2$ is tangent to the line $F_1F_2$. This solves the problem for the case when the line $F_1F_2$ is not parallel to the line $A_1A_2$.

In the case when it is, the idea of the proof is the same, but the homothety has to be replaced by a translation.

  Darij

Darij ; many thanks yes the problem is now correct. yes i guess the problem is trivial  :blush:  but  im getting better  :lol:  :?


Find all  integers x,y
 $ y(x\plus{}y)\equal{}x^3 \minus{} 7x^2 \plus{} 11x \minus{} 3$.

hello, at the time i have found
$ x\equal{}1,y\equal{}\minus{}2$, $ x\equal{}1,y\equal{}1$, $ x\equal{}2,y\equal{}\minus{}1$
Sonnhard.

$ x\equal{}1,y\equal{}\minus{}2,x\equal{}1,y\equal{}1,x\equal{}2,y\equal{}\minus{}1$ actualy are http://www.wolframalpha.com solutions[url][/url] :wink:

[quote="Obel1x"]$ x \equal{} 1,y \equal{} \minus{} 2,x \equal{} 1,y \equal{} 1,x \equal{} 2,y \equal{} \minus{} 1$ actualy are http://www.wolframalpha.com solutions[url][/url] :wink:[/quote]
canny :lol:

[quote="aktyw19"][quote="Obel1x"]hope this helps a bit
$ xy \plus{} y^2 \equal{} x^3 \minus{} 3x^2 \minus{} 4x^2 \plus{} 12x \minus{} (x \minus{} 3)$
$ xy \plus{} y^2 \equal{} x^2(x \minus{} 3) \minus{} 4x(x \minus{} 3) \minus{} (x \minus{} 3)$
$ y(x \plus{} y) \equal{} (x \minus{} 3)(x^2 \minus{} 4x \minus{} 1)$[/quote]

What then?[/quote]
how do you mean "what then"?
At least I'm giving something it might help you  :lol: ,not doing SPAM like Dr.Sonnhard did,whose answer let's you down ;)
PS:It's all I could do

I'm posting my solution to the equation. Hope it's correct.


Let  $a,b,c>0$  with  $abc=1$. Prove that  :

$\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\leq\dfrac{(a+b+c)(ab+bc+ca)}{a+b+c+ab+bc+ca}$

We prove the stronger result that $LHS\le\dfrac{3}{2}\le RHS$.

The first follows from $1+a^2\ge2a$, cycling and adding.

The second is $\dfrac{3}{2}\le\dfrac{xy}{x+y}$ where $x=\sum a$, $y=\sum ab$, and reduces to $\left(x-\dfrac{3}{2}\right)\left(y-\dfrac{3}{2}\right)\ge\dfrac{9}{4}$.  But by the AM GM, $x\ge3\sqrt[3]{abc}=3$ and $y\ge3\sqrt[3]{abc}^2=3$, and the result follows.


If positive real numbers $ a,b,c$ satisfied $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}3$ ,Prove that $ (2\minus{}a)(2\minus{}b)(2\minus{}c)\geq \sqrt{abc}$

[quote="tang zy"]If positive real numbers $ a,b,c$ satisfied $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 3$ ,Prove that $ (2 \minus{} a)(2 \minus{} b)(2 \minus{} c)\geq \sqrt {abc}$[/quote]

$ \Leftrightarrow (2\minus{}a)^2(2\minus{}b)^2(2\minus{}c)^2\minus{}abc \ge 0,$

setting $ a\plus{}b\plus{}c\equal{}3u, ab\plus{}bc\plus{}ca\equal{}3v^2, abc\equal{}w^3$, that we have $ v^2  \equal{} \frac{{3u^2  \minus{} 1}}{2}$ and $ \sqrt{3}<u \le 1$ and $ 0<w \le 1$, then ineq becomes

$ {w}^{6}\plus{} \left( \minus{}11\plus{}24u\minus{}18{u}^{2} \right) {w}^{3}\plus{}25\minus{}120u\plus{}81{u}^{4}\plus{}234{u}^{2}\minus{}216{u}^{3}\ge 0,$

and $ f(t)\equal{}t^2\plus{} \left( \minus{}11\plus{}24u\minus{}18{u}^{2} \right)t\plus{}25\minus{}120u\plus{}81{u}^{4}\plus{}234{u}^{2}\minus{}216{u}^{3}$ is decreasing in $ (0,1]$, so we just need to prove that when $ b\equal{}c$.

when $ b\equal{}c$, we have $ a\equal{}\sqrt{3\minus{}2c^3}$, then ineq becomes

$ \left( 7\minus{}2{c}^{2} \right)  \left( 2\minus{}c \right) ^{4}\minus{} \left( 4\left( 2\minus{}c \right) ^{4}\plus{}{c}^{2} \right) \sqrt {3\minus{}2{c}^{2}}\ge 0,$

but

$ \left( 7\minus{}2{c}^{2} \right)^2  \left( 2\minus{}c \right) ^{8}\minus{} \left( 4\left( 2\minus{}c \right) ^{4}\plus{}{c}^{2} \right)^2 (3\minus{}2{c}^{2})$

$ \equal{}\left( 4{c}^{10}\minus{}56{c}^{9}\plus{}336{c}^{8}\minus{}1128{c}^{7}\plus{}2353{c}^{6}\minus{}3270{c}^{5}\plus{}3229{c}^{4}\minus{}2304{c}^{3}\plus{}1152{c}^{2}\minus{}512c\plus{}256\right) \left( c\minus{}1 \right) ^{2}$

$ \ge 0.$

so ineq holds.

[quote="tang zy"]Nice my dear friend. :)
there still a more beautiful prove,I will print it next week.[/quote]
We can prove a patial ineq:
$ (2-a)(2-b)\geq \frac{1+c^{2}}{2}$
$ \Leftrightarrow (2-a)(2-b)\geq \frac{4-a^{2}-b^{2}}{2}$
$ \Leftrightarrow 4-4(a+b)+2ab+a^{2}+b^{2}\geq \0$
$ \Leftrightarrow (a+b-2)^{2}\geq \0$,which is obvious
so $ (2-a)^{2}(2-b)^{2}(2-c)^{2}\geq \frac{1+a^{2}}{2}\frac{1+b^{2}}{2}\frac{1+c^{2}}{2}\geq abc$
This is the same to the origins

[quote="tang zy"][quote="tang zy"]Nice my dear friend. :)
there still a more beautiful prove,I will print it next week.[/quote]
We can prove a patial ineq:
$ (2 \minus{} a)(2 \minus{} b)\geq \frac {1 \plus{} c^{2}}{2}$
$ \Leftrightarrow (2 \minus{} a)(2 \minus{} b)\geq \frac {4 \minus{} a^{2} \minus{} b^{2}}{2}$
$ \Leftrightarrow 4 \minus{} 4(a \plus{} b) \plus{} 2ab \plus{} a^{2} \plus{} b^{2}\geq 0$
$ \Leftrightarrow (a \plus{} b \minus{} 2)^{2}\geq 0$,which is obvious
so $ (2 \minus{} a)^{2}(2 \minus{} b)^{2}(2 \minus{} c)^{2}\geq \frac {1 \plus{} a^{2}}{2}\frac {1 \plus{} b^{2}}{2}\frac {1 \plus{} c^{2}}{2}\geq abc$
This is the same to the origins[/quote]

nice... :)


Find the constant term of $\left( 1+x+\frac{2}{x^{3}}\right)^{8}$.

[hide="Solution"]We'll use trinomial expansion

$(p+q+r)^{n}=\sum_{i+j+k=n}\frac{n!}{i!j!k!}p^{i}q^{j}r^{k}$

Hence we must find all pairs $i,j$ such that

$\frac{8!}{i!j!(8-i-j)!}1^{i}x^{j}\left(\frac{2}{x^{3}}\right)^{8-i-j}$

doesn't depend on $x$

The exponent of $x$ is $j-3(8-i-j)=j-24+3i+3j=3i+4j-24$. Equating it with zero we get $4j=24-3i\iff j=6-{3i\over 4}$. Since both $i,j$ are non-negative integers not greater than $8$, $i$ is divisible by $4$, hence can only be $0,4,8$. For those values we get $j=6,3,0$ respectively.

Therefore we have three constant terms:

$\frac{8!}{0!6!2!}1^{0}x^{6}\left(\frac{2}{x^{3}}\right)^{2}=\frac{8\cdot 7}{2}\cdot 2^{2}=112$

$\frac{8!}{4!3!1!}1^{4}x^{3}\left(\frac{2}{x^{3}}\right)^{1}=\frac{8\cdot 7\cdot 6\cdot 5}{6}\cdot 2^{1}=560$

$\frac{8!}{8!0!0!}1^{8}x^{0}\left(\frac{2}{x^{3}}\right)^{0}=1$

and they add up to $112+560+1=\boxed{673}$[/hide]

[hide]There are three ways we can get a constant term.

Case 1.  We take the $1$ from each of the eight factors, resulting in $1^{8}=1$.

Case 2.  We take the $x$ from three factors and the $\frac{2}{x^{3}}$ from one factor.  This case results in $\binom{8}{3}\binom{5}{1}x^{3}\cdot\frac{2}{x^{3}}=560$.

Case 3.  We take the $x$ from six factors and the $\frac{2}{x^{3}}$ from two factors.  This case results in $\binom{8}{6}\binom{2}{2}x^{6}\left(\frac{2}{x^{3}}\right)^{2}=112$.

The constant term is $1+560+112=673$.[/hide]


modular arithmetic
number theory proposed
number theory

Find all $ (x,y)\in N$ such that
$ x\plus{}y\equal{}(x\minus{}y)^n$
Source :[hide="http://mathnfriend.org/forum/viewtopic.php?p=4049"] [/hide]
2)Find all $ (x,y)\in N$
$ 2^{x}\plus{}y^2\equal{}\frac{(x\plus{}y)^3}{3}$

[quote="TTsphn"]Find all $ (x,y)\in N$ such that
$ x \plus{} y \equal{} (x \minus{} y)^n$
Source :[hide="http://mathnfriend.org/forum/viewtopic.php?p=4049"] [/hide]
[/quote]

$ n>1$ and $ x\minus{}y$ divides $ x\plus{}y$ and so $ x\plus{}y\equal{}k(x\minus{}y)$ and $ x(k\minus{}1)\equal{}y(k\plus{}1)$ and so :
Either $ k$ is even and $ x\equal{}\alpha(k\plus{}1)$ and $ y\equal{}\alpha(k\minus{}1)$. Then $ 2\alpha k\equal{}(2\alpha)^n$ and $ k\equal{}(2\alpha)^{n\minus{}1}$
Either $ k$ is odd and $ x\equal{}\alpha\frac{k\plus{}1}{2}$ and $ y\equal{}\alpha\frac{k\minus{}1}{2}$. Then $ \alpha k\equal{}(\alpha)^n$ and $ k\equal{}(\alpha)^{n\minus{}1}$

And the solutions are :
$ n>1$
$ x\equal{}\frac{\alpha^n\plus{}\alpha}{2}$ and $ y\equal{}\frac{\alpha^n\minus{}\alpha}{2}$ $ \forall\alpha\in\mathbb{N}\minus{}\{1\}$

[quote="TTsphn"]
2)Find all $ (x,y)\in N$
$ 2^{x} \plus{} y^2 \equal{} \frac {(x \plus{} y)^3}{3}$[/quote]

$ x\equal{}0$ gives $ 3\plus{}3y^2\equal{}y^3$, which hasn't solutions.

If $ x \geq 1$, then $ y \equiv (x\plus{}y) \pmod 2 \Rightarrow 2|x$. So $ x\equal{}2a$

$ 3|(x\plus{}y)^3 \Rightarrow 3|(x\plus{}y)$, so $ 9|RHS \Rightarrow 9 | (2^{2a}\plus{}y^2)$, but 3 divides a sum of squares iff it divides every square, but $ 3 \not | 2^x$

So there shouldn't be any solution to the proposed equation.


geometry perimeter Harvard college MATHCOUNTS induction rectangle

I could've sworn that I put a <15 option.
Can a mod add it in?  [size=75][color=red]<done.  -- rcv>[/color][/size]

hate to stereotype about AoPSers, but it's probably ok without

jorian

[quote="ProtestanT"]i haven't gotten it back yet...got any idea how i can find out?[/quote]
Try emailing the coordinator.
If not, then just vote your estimate.

Who voted 46? :blush:

I got <15. Missed all targets! :blush:

Around 29.

I got around 17 on sprint and am sure that I got 6 on target.

37.

23 on sprint and 7 on target.  I won my chapter (Manhattan).

@meenamathgirl: That was our nat'l cutoff last year.

well my estimate is 44 since we discussed almost all the problems after the competition and i didnt have more than 2 wrongs that i knew of
so my estimate is
sprint: 28/30
target: 16/16
sum: 44/46

i also feel that it is acurate because i got 5th written in my region, and the top two written got perfects.

The test didn't seem taht hard except ofr last 5 questions.

I think that if you could do #28 and #29, you are good.

Oh, if anyone cares I did it for practice and got 46/46.

wow, even in ubemaya's chapter there were only 2 44's

jorian

Eh, 45.  STUPID DUMB MISTAKE.  For "how many cubic feet are in a cubic yard?" I put 9. -__-  Oh well, I shouldn't complain, so woot.

Oh you got that wrong.  That was dumb, but you ownde.

i got first at my chapter withe a 28 :wink:  :rotfl:

northern kentucky

anyway- anyone know of some good study material

Yep. look at one of the stickies in this forum, and there is a site with a lot of past mathcounts tests. :D

I won my chapter with.... 29

21/30 sprint
4/8 Target

I made a few stupid mistakes.

Anyone out there know who the strong competitiors are? I know last yr's nj champ was in 8th grade last yr - so he's out .. any other "contenders"??  If you're reading this link, how was your chapter's score this yr?

@everyone, I live in Central Wisconsin, and there were only 8 schools at chapters :rotfl:

[quote="kstan013"]This year there were 43 people at chapters from 8 schools, and we sent 2 teams and 2 individuals[/quote]

that means that not all teams had 8 ppl. is it possible to have more than 8 in a team?

The top score at our chapter was a 27. I got a 20 and came 10th.

[quote="kstan013"][quote=".:Logic:."]OMG, 8 SCHOOLS??!!

 :wacko:

Our chapter has 40+ schools...

How many school teams get to go to state at your chapter??[/quote]This year there were 43 people at chapters from 8 schools, and we sent 2 teams and 2 individuals[/quote]
Yeah.. okay. :)

We sent top 5 teams.... :P

[quote="Walk Around The River"][quote=".:Logic:."]OMG, 8 SCHOOLS??!!

 :wacko:

Our chapter has 40+ schools...

How many school teams get to go to state at your chapter??[/quote]

we had like 11 or 12, and we live in the capital of florida.  i think other chapters might be bigger than ours.  florida seems to have a lot of chapters.

2 teams and 4 individuals go to state, but this year all top 4 individuals were on our team, so just 2 teams or 8 people.[/quote]
Our chapter has 10 schools give our take 2.

wow... 20's and first place?
Difference between our chapter and yours ==== difference between √-1 (i) and 1

My chapter had about 100 people. 2nd got 36, t-4th got 33, 9th got 25.

Top 3 teams and top 12 indiv's not on the top 3 teams make state. My guess is the cutoff was 24-23, maybe lower.

I had 2nd written with a score of 31.  There were like 20 or so schools in the chapter, but only top 4 ind. and top 3 teams go to state.


There are $N$ people at a party. When they enter they each put their hats into a pile. When it is time to leave the party they each randomly select a hat to take from the pile. what is the probability that no one takes their own hat?

[hide]
There are $N!$ ways in which the hats can be distributed. The number of ways where no one takes their own hat is equal to the number of derangements of $N$, so the desired probability is the quotient of the two, I think.
[/hide]

[hide="The relevant expression"] $\frac{!n}{n!}$ [/hide] [hide="Which is somewhere around..."] $\frac{1}{e}$ [/hide] [hide="And is exactly..."] $\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+-... \pm \frac{1}{n!}$ [/hide]

Okay, so that's just application of derrangement formula (I am realizing it must be easily provable with PIE)

i remember doing this using PIE but as of right now i dont remember  how or why i used PIE. :(

You use PIE to count the complement of the number of derangements: the number of arrangements with at least one number in the correct spot.

There are $n$ ways to choose which number is in the correct spot, and $(n-1)!$ ways to arrange the rest. But this double counts the number of arrangements with $2$ numbers in the correct spot, which is equal to $\binom{n}{2}(n-2)!$. You use PIE to correctly count the number of arrangements with at least one number in the correct spot.

The total number of arrangements is equal $n!$, so the number of derangements is equal to \[n!-\binom{n}{1}(n-1)!+\binom{n}{2}(n-2)!-\binom{n}{3}(n-3)!+\cdots+(-1)^{n}\binom{n}{n}0!\]


How many different three-letter monograms can be formed using the letters A, E, I, O, and U if each letter can be used more than once?

Each digit of the monogram has 5 choices (A,E,I,O,U), so there are $ 5^3 \equal{} 125$ possible monograms.


Subtract the first equation from the second and find permutations of solutions. This seems like the way to go; I'll come back and post what I find later since I am in class right now  :lol:

i dont quite get how to find all solutions of
\[a^2+d^2=2c^2\]

ITs equivalent to $(a-d)^2+(a+d)^2=(2c)^2$


I went to the one in Iowa Falls. I got pretty bad scores: 40 on the target and 74 on the sprint. Our two-man team got 50.

Having never practiced a GPML test or gone to another meet this year, I'd say it was an okay start.

hmm...I went to one in january too, except it was a ISU...our team got I think 1st or second for junior GPML...and our 3-person team round (me, alex, lucy) got 2nd in junior for team. I guess it was a good expierence for my first time.


Romeo runs northwest at $13\sqrt{2}$ mph and Juliet drives east at 71 mph. (They broke up.) How far apart are they after twelve minutes?

I'd call this a Sprint round question, maybe number 20.

[hide="Sprint Question, Sprint answer"]
 $17.00$ miles[/hide]

[hide=":blush:...sorry"] i guess i tried to do it too fast [/hide]

[hide]After 12 minutes, or $\frac{1}{5}$  of an hour, Romeo has gone northwest $\frac{13\sqrt{2}}{5}$ miles, and Juliet has gone east for $\frac{71}{5}$ miles.
Romeo has gone northwest whcih means that he is $\frac{13}{5}$ miles north of Juilet and $\frac{13}{5}+\frac{71}{5}=\frac{84}{5}$ west of her. That means that they are $\sqrt{(\frac{13}{5})^{2}+(\frac{84}{5})^{2}}=\sqrt{\frac{169}{25}+\frac{7056}{25}}= \sqrt{\frac{7225}{25}}=\sqrt{289}=\boxed{17}$[/hide]

[hide]The legs of the right triangle are (13rt(2)/rt(2))=13 and (71+13)=84. Pythagorean triple, 13-84-85, and divide 85 by 5, since they only traveled for 1/5 of an hour gives you 17.[/hide]

That's a fairly nice problem.  How did you come up with the numbers to make it work out nicely?

[hide]If Romeo is going due northwest, he makes a $45^\circ$ angle with north.  12 minutes is $\frac{12}{60}=\frac15$ of an hour, so he has gone $\frac{13\sqrt2}{5}$ miles northwest, which means (45-45-90 right triangle) he has gone $\frac{13}{5}$ miles north and $\frac{13}{5}$ miles west.

Juliet has gone $\frac{71}{5}$ miles east.  The distance between them east and west is therefore $\frac{84}{5}$ miles.  The distance north and south is $\frac{13}{5}$ miles east.

In case you do not know that 13, 84, 85 is a right triangle, you can figure it out like so:
$\left(\frac{84}{5}\right)^2+\left(\frac{13}{5}\right)^2=\frac{7056+169}{25}=\frac{7225}{25}=289$
$\sqrt{289}=\boxed{\boxed{17\text{ miles apart}}}$[/hide]

[quote="Iron Fist"]Romeo runs northwest at $13\sqrt{2}$ mph[/quote] Wow, that's a 3 minute, 16 second mile.  :huh: :D

He's really athletic I guess - explains why Juliet fell in love in him :?

Or maybe he was a genetically created human with super human speed, and that's why they broke up...

or maybe he was just sprinting really really fast to get away from her? but for 12 minutes???

it's an unfair world.
 how come Juliet got to keep the car?

Haven't you read the book? She's filthy rich :D.

[hide]
In twelve minutes Romeo would have run $\frac{13\sqrt{2}}{5}$ miles. In twelve minutes Juliet would have run $\frac{71}{5}$ miles. The direction in which Romeo is running makes a 45 degree angle with the north. This means Romeo is $\frac{13}{5}$ miles north of Juliet and $\frac{84}{5}$ miles west of Juliet.
Hence the distance between Romeo and Juliet is $\frac{85}{5}=\boxed{17}.$[/hide]

[hide]  OK.
So if Romeo goes 13root2 miles per hour, he goes 13/5root2 miles in 12 minutes.
If Juliet goes 71 mph, she goes 71/5 miles in 12 minutes.
Then, make a right triangle. The height will be 13/5root2/root2, which is 13/5 (It helps to graph this on a plane). Then the length is 13/5+71/5, or 84/5. The distance apart is 85/5, or 17.
[/hide]
I'm a beast. Weird that I came in 112th.
Has anyone noticed that Daesun is REALLY good at writing problems?

Man, I need LaTeX

[quote="jclarkemathL314159"]
Has anyone noticed that Daesun is REALLY good at writing problems?

Man, I need LaTeX[/quote]

unfortunate side effect from studying too much

[quote="math92"][quote="jclarkemathL314159"]Man, I need LaTeX[/quote]
actually, its like built in to the posting reply thing.  Just put dollar signs around what you want to be latexed.  If i want to write 2/3 in latex i would do \frac{2}{3} surronded by dollar signs which looks like this:
$\frac{2}{3}$[/quote]

$\displaystyle{\frac{2}{3}}$ looks a lot better.

that would be    $\$$ \displaystyle\frac{2}{3}  $\$$


geometry
inequalities
geometric transformation
rotation
parallelogram
geometry proposed

A square is divided in four parts by two perpendicular lines, in such a way that three of these parts have areas equal to 1. Show that the square has area equal to 4.

Let the side of the square be $ a$.
Then let the sides of one of the parts(the part that is beetwen the other two) be $ x,y$.

We know the following

$ xy\equal{}1$
$ (a\minus{}y)x\equal{}ax\minus{}xy\equal{}1$
$ (a\minus{}x)y\equal{}ay\minus{}xy\equal{}1$

If we look at the last two (and  put in the first one)

$ ax\minus{}1\equal{}1$
$ ay\minus{}1\equal{}1$

It follows that $ x\equal{}y$. And from here $ x\equal{}y\equal{}1$

$ a\minus{}1\equal{}1$ so $ a\equal{}2$ so the area is indeed $ 4$.

@Wichking. the perpendicular lines are not necessarily parallel to sides of square. :maybe:

[img]http://i130.photobucket.com/albums/p279/ITAQWeb/geo-1.png[/img]

Hmmm... Name the points as in the figure above. O is the center of the square. GH // NM and IJ // PQ. Y and Z are the lines' intercepts.
WLOG, assume that E lies in the left section of the square (in the pentagon OGAI).
Denote [polygon] be the area of the polygon.
It's easy to see that [MEPB] <= [HYPB] <= [MEQC]. Equality holds when E lies on GH. Do this similarly to the remain parts.
Then, we can show that there are 3 parts of the square having equal areas iif E is the center, which means the area of the square is 4.

That solution seem to lose some generality, supose that Q lie on the segment [AD].  Then it's to hard to prove the inequality [EYHM]=<[YHCQ].

It's an god solution, anyway. Those geometric problemas are very hard, it seems that is no way to prove...

We name the lines a and b. a and b are perpendicular, and each divides the square in two parts, one of which has area 2. This means that line b is a 90º rotation of a about the center of the square, O. We draw the other 90º rotation, c, and the 180º rotation, d. This divides the square in 9 parts.

Because each line is a rotation about O of the other lines, there are two groups of four parts that are rotations of the other ones, and therefore have the same area. Let the area of those parts be r and s, and the area of the central part be t. a and b divide the square in four parts of area r, r+s, r+s and r+2s+t. Three of them can only be equal if s=0, which means that b and c are the same line, and so are a and d, and therefore the central part has area 0. All four parts have area r=1.

Label $P_1P_2P_3P_4$ the given square and $M$ is a point inside it. A line through $M$ cuts $\overline{P_1P_2},\overline{P_3P_4}$ at $P,T$ and another line through $M$ perpendicular to $PT$ cuts $\overline{P_2P_3},\overline{P_4P_1}$ at $S,U.$ $O$ is the center of $P_1P_2P_3P_4.$ Parallel through $O$ to $PT$ cuts $\overline{P_1P_2},\overline{P_3P_4}$ at $E,G$ and parallel through $O$ to $SU$ cuts $\overline{P_2P_3},\overline{P_4P_1}$ at $F,H.$ Let $X \equiv EG \cap SU,$ $Y \equiv FH \cap TP.$  Denote $S,$ $S_A,$ $S_B,$ $S_C,$ $S_D,$ $S_R$ the areas of $P_1P_2P_3P_4,$ $UHYM,$ $PEOY,$ $SFOX,$ $TGXM,$ $OXMY,$ respectively.

WLOG assume that $O$ lies inside the quadrilateral $MPP_2S.$ Clearly, the lines $EG$ and $FH$ divide $P_1P_2P_3P_4$ into four quadrilaterals with equal areas. Therefore, we have

$\frac{_1}{^4}S-S_B+S_A=\frac{_1}{^4}S-S_A-S_C-S_R=\frac{_1}{^4}S-S_D+S_C=1.$

From these expressions we obtain $S_A+S_D=S_B+S_C$ $\Longrightarrow$ parallelograms $PEGT$ y $SFHU$ have equal areas. But since they have equal altitudes, then it follows that $PE=FS$ $\Longrightarrow$ $OXMY$ is a square. Thus, $S_A=S_C$ and $S_B=S_D.$ Again, using the expressions found previously, we obtain $S_A=S_B=S_C=S_D=0$ $\Longrightarrow$ $O \equiv M$ and the conclusion follows.


analytic geometry
function
combinatorics solved
combinatorics

Find the minimum natural number k satisfy: For any system of unit radii circles in the plane, such that no two circles have at least two common points, we can color each of these circles with one of k colors sothat there are no two tangent circles colored the same color

We will prove that $k=4$ is the desired minimum, even if the set of circles is infinite.

First, note that a system $S$ of circles which satisfies the statement of the problem is clearly countable : Since no two circles have a common interior point, we may choose a point with both coordinates rational in each of the circles and these points are pairwise distincts. Thus we defined an injective function from $S$ onto $Q^2$. Using that $Q^2$ is countable, we deduce that $S$ is countable too.

It follows that there are only a countable number of directions defined by the pair of centers of the circles from $S$. Therefore, we may select an orthogonal system of coordinates such that the centers of the circles have pairwise distincts x-coordinate.
Now let these centers be $O_n(x_n,y_n)$ where $n = ...,-2,-1,0,1,2,....$ and such that $x_i < x_j$ if and only if $i<j$.

Let's $G$ be the graph whose vertices are the $O_i$'s, two joined by an edge if and only if their respective associated circles are tangents.
The problem is to find the chromatic number of this graph.
But, it is easy to prove that each $O_i$ has at most 6 neighbours, and among these neighbours there are at most three which have an x-coordinate smaller than $x_i$ (resp. bigger than $x_i$).
Now we color the circles, which is equivalent to color the $O_i$'s, with colors $c_1,c_2,c_3,c_4$ as follows :
Give $O_1,O_2,O_3$ colors $c_1,c_2,c_3$ respectively.
For $i \geq 3$ suppose that $O_1,...,O_i$ has been colored properly. Thus, from above at most three of the neighbours of $O_{i+1}$ are already colored. Thus, at least one color is still available to color $O_{i+1}$ properly. Give $O_{i+1}$ one these available colors.
Thus, we color all the $O_i$'s for $i \geq 0$.

Then color all the $O_i$ for $i<0$ as above.

Thus, the chromatic number is at most 4.

Conversely : Suppose that we only have 3 colors.
Let a diamond be two equilateral triangles of side-length 2 sharing a side. In all the following, each vertiex of a diamond will be the center of circle with radius 1. Thus, we will only talk about vertices (we'll only have to take care that the points are at least 2 apart so that the circles do not overlap).
Clearly, the two opposite vertices of a diamond, that is those which are not on the common side, must have the same color.

Sorry, but the following would be much clear with a figure....

Let $ \gamma$ be a circle with radius $R$. Let $n$ be a positive integer.
Choose $R$ so that $2n \theta + 2 \beta = 2 \pi$ where $\beta = arcsin( \frac {1} {R})$ and $ \theta = arcsin( \frac {\sqrt{3}} {R}$.
It follows that we may select $n$ points on $ \Gamma$, say $A_1,...,A_n$ such that the distance $A_iA_{i+1} = 2 \sqrt {3}$ for $i=1,...,n-1$ and $A_nA_1 = 2$.

Let $d_i$ be the diamond with opposite vertices $A_i$ and $A_{i+1}$.
Now choose $n$ to be sufficientely large so that any two of the diamonds $d_1,...,d_{n-1}$ have vertices with mutual distance at least 2 (I mean except for the $A_i$'s obviously, I'm talking about the non-opposite vertices of the diamonds. This needs $R$ and then $n$ to be sufficientely large).
From above each of the $A_i$'s must have the same color. Since $A_nA_1$ we then have two neighbours with the same color. A contradiction.
Thus the chromatic number is at least 4, and we are done.

Pierre.


geometry
cyclic quadrilateral
similar triangles
number theory unsolved
number theory

Let $a,b,c,d \in N$ and $a^2+ab+b^2=c^2+cd+d^2$. Prove that $a+b+c+d$ is a composite number. ;)

if $(a,b)= (c,d)$ it is trivial
$3(a+b+c+d)(a+b-c-d)=3((a+b)^2-(c+d)^2)=3(ab-cd)=(c-d)^2-(a-b)^2=(c+a-d-b)(c+b-d-a)$
if $a+b+c+d=p$ is prime then
$p|(c+a-d-b)$ or $p|(c+b-d-a)$
because $-p<c+a-d-b<p$ and $-p< c+b-d-a <p $ and $(a,b)\neq (c,d)$ we have contadiction

I guess the "beautiful" idea you are talking about is the geometric approach (like in IMO 2001 question 6)?

(Take a cyclic quadrilateral $PQRS$ with $\angle{SPR} = \angle{SQR} = 120^{\circ}$ and $PS = a$, $PR = b$, $QR = c$ and $QS = d$. Then you probably need to use Ptolemy or similar triangles or something, haven't finished it since the idea is so well-known.)


calculus
integration
inequalities
real analysis
real analysis unsolved

show that any positive rational $x$ can be written as $x=a_{1}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+..$ where  the $a_{k}$ are integers and $0\leq a_{k}\leq{k-1}$ for all $k>1$

i am not sure if this qn belongs to analysis, moderators feel free to move it to the appropriate forum  :)

In fact, every real number $x$ can be written as $x = \sum_{n=1}^{\infty}\frac{a_{n}}{n!}$, where $a_{n}$ are integers with $0\leq a_{n}< n$ for all $n>1$, and this representation is unique if $x$ is irrational.

the proof is easy. Let $[x]$ be the integral part of $x$. then we define $a_{n}$ and $x_{n}$ as follows:

$a_{1}= x_{1}= [x]$,
$a_{n}= [n!\left( x-x_{n-1}\right)]$, $x_{n}= x_{n-1}+\frac{a_{n}}{n!}$ for $n\geq 2$

you can check that $a_{n}$ defined in this way satisfies

$0\leq a_{n}< n$ for all $n>1$, and
$0 \leq x-x_{n}< \frac{1}{n!}$.

the last inequality yields that the sequence $x_{n}$ converges to $x$ as $n\to \infty$.


$1)$ In a sequence of $(4n+1)$ terms,the first $(2n+1)$  terms are in A.P whose common difference is $2$ and the last $(2n+1)$ terms are in G.P whose common ratio is $0.5$.If the middle terms of the A.P and G.P are equal,then the middle term of the sequence is

$a) \frac{n\cdot 2^{n+1}}{2^{n-1}}$

$b) \frac{n\cdot 2^{n+1}}{2^{2n}-1}$

$c) n\cdot 2^n$

$d)\text{none of these}$

$2)$ In the sequence $1,2,2,4,4,4,4,8,8,8,8,8,8,8,8,\cdots,$ where $n$ consecutive terms have the value $n$,find the $1025th term$

your answer for the first one is right thoiugh i didn't understand your solution properly.....

[quote="sen"][hide]We have sequence

the first $2n+1$ terms in AP:

$a, a+2, a+4, a+6 \cdots a+4n$

the last $2n+1$ terms in GP:

$a+4n, \frac{a+4n}{2^1}, \frac{a+4n}{2^2} \cdots \frac{a+4n}{2^{2n}}$

We know that $a+2n = \frac{a+4n}{2^n}$

$\implies 2^n(a+2n)=a+4n$

$2^na-a=4n-2^n2n$

$a(2^n-1)=4n-2^n2n$

$a=\frac{4n-2^n2n}{2^n-1}$ hmm what did i do wrong?[/hide][/quote]

Nothing at all :lol:


trigonometry
inequalities
calculus
geometry solved
geometry

Prove that in any triangle the following inequality holds!

2*p2>=27Rr.

well .. mitzah I like your problems... :) it's the second one for which I think I have a solution...

well... we know that (a+b+c)^3>=27*a*b*c and 2p=a+b+c so we get that: 8*p^3>=27*a*b*c and we can rewrite this the following way :
2*p2>=27*(abc)/(4S) * S/p and we know that  4RS=abc  and rp=S so we have what we were looking for...:)
THE END.

that's xactly the solution that i got but for abt half an hour i was lost in the wacky world of trig.but in the end the solution turned out to be straightforward am-gm

I think I have another solution. You can rewrite the ineq like this
   2*(p/R)*(p/R)>=27r/R
   But we have 4cos(A/2)cos(B/2)cos(C/2)=p/R
   4sin(A/2)sin(B/2)sin(C/2)=r/R
   So the ineq becomes after some more calculs where we use that cos (A/2)^2=p(p-a)/bc and sin(A/2)=.....
     p*p*p/abc>=27/8
     But we have p=a+b+c/2 so we will have to proove that (a+b+c)^3>=27abc which is true
      Hope it is true!


limit
search
inequalities
real analysis
real analysis solved

Hi!

Although I have a solution to the following problem, it's not the solution one is meant to find.

Let $ (g_n) $ be a strictly monotonically increasing divergent sequence. Furthermore let there be a sequence $ (f_n) $, so that $ \lim_{n \to \infty} \frac{f_{n+1} -  f_n}{g_{n+1} -  g_n} = a$. Prove that  $ \lim_{n \to \infty} \frac{f_n}{g_n} = a $.

There comes an instructional commentary with the problem:

First find upper and lower bounds for $ f_n -  f_{n-1} $, then for $ \sum_{k=n_0}^n f_n -  f_{n-1} $.

But I can't find bounds that would be of any help to solve the problem. My prove is based on another theorem that was much easier to prove for me and that contains the theorem mentioned above as a special case.

Nevertheless: Could anybody of you help me to find a solution that is based on the instructional commentary? It's really important to me.

This is just Cesaro-Stoltz. Maybe you could also search the forum? You'll find interesting stuff. It must have been discussed before.

Thank you, grobber! I didn't know the name "Cesaro-Stolz theorem" for this result. After your posting I could find the proof I was looking for.

There is a likely theorem.
It is:Let the sequence $x_{n}$ monotinically tends to $0$.Let $y_{n}$ is a sequence
s.t. $y_{n}{\rightarrow}{0}$ and $\lim_{n{\rightarrow}\infty}{\frac{y_{n+1}-y_{n}}{x_{n+1}-x{n}}}=a$,then
$\lim_{n{\rightarrow}\infty}{\frac{y_{n}}{x_{n}}}=a$.
The Cesaro-Stolz theorem and the theorem above are are similiar to l'Hospitale rules for functions.

Could someone please post a proof for the similar result mentioned by LevonNurbekian?

As Levon(me) posted the result then he can prove it.
Here is the proof.
First let's consider that $|a|<\infty$ and $x_{n}$ is increasing(then $x_{n}\leq{0}$).Now since $\lim_{n{\rightarrow}\infty}{\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}}=a$.Then for every $\epsilon>0$
there exists $N$ s.t.
$a-\epsilon<\frac{y_{N+1}-y_{N}}{x_{N+1}-x_{N}}<a+\epsilon \\
a-\epsilon<\frac{y_{N+2}-y_{N+1}}{x_{N+2}-x_{N+1}}<a+\epsilon \\
\cdots \\
a-\epsilon<\frac{y_{n}-y_{n-1}}{x_{n}-x_{n-1}}<a+\epsilon \\$
for every $n>N$.We can rewrite it in the following form(as $x_{n}$ is increasing)
$(a-\epsilon)(x_{N+1}-x_{N})<y_{N+1}-y_{N}<(a+\epsilon)(x_{N+1}-x_{N}) \\
(a-\epsilon)(x_{N+2}-x_{N+1})<y_{N+2}-y_{N+1}<(a+\epsilon)(x_{N+2}-x_{N+1}) \\
\cdots \\
(a-\epsilon)(x_{n}-x_{n-1})<y_{n}-y_{n-1}<(a+\epsilon)(x_{n}-x_{n-1}) \\$
Now add up all these inequalities and get
$(a-\epsilon)(x_{n}-x_{N})<y_{n}-y_{N}<(a+\epsilon)(x_{n}-x_{N})$ so
when $n{\rightarrow}\infty$ we get $(a-\epsilon)x_{N}{\geq}y_{N}{\geq}(a+\epsilon)x_{N}$ so since $x_{n}\leq{0}$ we have $(a-\epsilon){\leq}\frac{y_{N}}{x_{N}}{\leq}(a+\epsilon)$.We can write the same thing for every $n>N$.That means that $\lim_{n{\rightarrow}\infty}{\frac{y_{n}}{x_{n}}}=a$.

Thank you very much Levon. I was so afraid about this proof that I couldn't think clearly about it. That's my major problem with math...


Prove that if $ z^2\plus{}\frac{1}{z^2} \in \left [ \minus{}2,2 \right ], z \in \mathbb{C}$ then $ z^n\plus{}\frac{1}{z^n} \in \mathbb{C}$.

I dunno. Maybe expand it? Also, might it get better attention in High School  Intermediate?

-rts2007

have to agree with rts2007...
this would get a lot more attention in high school intermediate.

The question is obviously mis-stated -- is the last $ \mathbb C$ supposed to be $ \mathbb  R$ or something?  (I haven't checked that the problem becomes true in this case, though.)

[quote="moldovan"]Prove that if $ z^2 \plus{} \frac {1}{z^2} \in \left [ \minus{} 2,2 \right ], z \in \mathbb{C}$ then $ z^n \plus{} \frac {1}{z^n} \in \mathbb{R}$.[/quote]
Let $ z \equal{} re^{i\theta}$. Then $ r^2\cos2\theta \plus{} (1/r^2)\cos2\theta \plus{} i(r^2\sin2\theta \minus{} (1/r^2)\sin2\theta)\in\mathbb{R}$
$ \implies (r^2 \minus{} 1/r^2)\sin 2\theta \equal{} 0$.
The case $ \theta \equal{} \pi k$ would lead $ \Im[r^ne^{in\theta} \plus{} (1/r^n)e^{ \minus{} in\theta}] \equal{} 0$ and we are done.

$ r^2 \minus{} 1/r^2 \equal{} 0\implies r \equal{} 1$.

Then $ r^ne^{in\theta} \plus{} (1/r^n)e^{ \minus{} in\theta} \equal{} 2\cos n\theta \plus{} i(\sin n\theta \plus{} \sin( \minus{} n\theta))\in\mathbb{R}$

EDIT: Meh, actually we don't have to worry about the angle since from the initial condition we have $ (r^2\plus{}1/r^2)\cos2\theta\in[\minus{}2,2]\implies r^2\plus{}1/r^2\equal{}2\implies r\equal{}1$


$ \tan \frac{\angle DCE}{2} \equal{} \frac{\tan \frac{\pi}{3}}{3} \equal{} \tan \frac{\pi}{6}$ then $ \angle DCE \equal{} \frac{\pi}{3}$ and $ \triangle CDE$ is equilateral.

See that AD*AB = AC^2, hence <ACD = <ABC = 30, consequently AD = CD. Similarly CE = BE, but AD = DE = BE, so the triangle CDE is eqilateral.

Best regards,
sunken rock

[quote="STARS"]Triangle $ ABC$ is isosceles with $ AC \equal{} BC$ and $ \angle{C} \equal{} 120^o$. Points $ D$ and $ E$ are chosen on segment $ AB$ so that $ |AD| \equal{} |DE| \equal{} |EB|$. Find the sizes of the angles of triangle $ CDE$.[/quote]

[color=red][b] A simple Solution[/b][/color]

 If $ M$ is the midpoint of $ AB$ and $ K,L$ are the projections of $ D,E$ to $ AC,BC$ respectively , then $ DK \equal{} \frac {1}{2} AD \equal{} DM$. Thus $ DK \equal{} DM$ , so $ CD$ is the bisector of $ \angle ACM \equal{} 60$. From now on the solution can be done easilly.

[u] Babis[/u]

 [b]Note[/b]

[i]I had proposed this problem in this forum some months ago ![/i]

Call $ \widehat{CDE}\equal{}\widehat{CED}\equal{}\alpha$, $ \widehat{DCE}\equal{}\beta$ and $ \widehat{BCE}\equal{}\widehat{ACD}\equal{}\gamma$

You have
$ \beta \plus{} 2\alpha \equal{} \pi$ and $ \beta \plus{} 2\alpha \equal{} \pi \Rightarrow \alpha \equal{} \gamma \plus{} 30$

Apply the sin theorem on $ \Delta CDE$ $ \Rightarrow \frac{CE}{\sin \alpha}\equal{}\frac{DE}{\sin \beta} \Rightarrow \frac{\sin 2\alpha}{\sin \alpha}\equal{}\frac{DE}{CE}$

and on $ \Delta BCE$ $ \Rightarrow \frac{CE}{\sin 30}\equal{}\frac{EB}{\sin \gamma} \Rightarrow \frac{\sin (\alpha\minus{}30)}{\sin 30}\equal{}\frac{DE}{CE}$

So

$ \frac{\sin (\alpha\minus{}30)}{\sin 30}\equal{}\frac{\sin 2\alpha}{\sin \alpha} \Rightarrow$ $ \sin\alpha\cos30 \minus{} \cos\alpha\sin30 \equal{} \cos\alpha$ $ \Rightarrow \frac{\sqrt 3}{2}\sin\alpha \equal{} \frac{3}{2}\cos\alpha \Rightarrow$ $ \cos^2 \alpha \equal{} \frac{1}{4} \Rightarrow \alpha \equal{} 60$


Let $ (X, d)$ be a metric space. If $ (x_n)$ is a sequence in $ X$ which converges to both $ a \in X$ and $ b \in X$, prove that $ a\equal{}b.$

Let $ \epsilon > 0.$

If n is sufficiently large we have

$ d(a,b) \leq d(a,x_{n}) \plus{} d(x_{n},b) < \frac {\epsilon}{2} \plus{} \frac {\epsilon}{2} \equal{} \epsilon.$

The result follows now from the latter inequality.


inequalities
limit
logarithms
calculus
calculus computations

Let $ H_n:=\displaystyle \sum\limits_{k=1}^{n}\frac{1}{k} $ and $A_n:= \displaystyle H_n-\frac{1}{2^n}\sum\limits_{k=1}^{n}{n \choose k}H_k\; . $ Find $  L:=\displaystyle \lim\limits_{n\to \infty}A_n\; . $ Prove that $ \displaystyle \frac{n}{n+2}\le\displaystyle  (L-A_n)n2^n\le \displaystyle 1\; ,\; \forall n\in {\mathbb N}^* . $


Do you think that the piano is a string instrument or a percussion instrument?
Just wondering

I prefer to call the piano a string instrument, as a child I thought it should have it's own category, but then as I looked in the piano and saw all the strings and dampers and everything, I called it a string instrument.
I just love the tone of a tuned grand piano.

I say string instrument because it's the strings which produce the sound, and it doesn't really matter how you get them to vibrate. For example, violins you use a bow, or you could use your fingers to pluck to string.

According to the Hornbastel Sachs classification, the piano is a simple chordophone with keyboard sounded by hammers. Why do we care so much about the way instruments make sound? Why don't we classify instruments based on the sound they make?

I think that piano has more likeness to a string instrument, as percussion instruments are mainly musical enhancement, and string instruments usually carry the melodies.

But it is built/ sometimes played as a percussion instrument. I don't know.

pianos are basically a combination of the harp and xylophone. technically it's considered percussion, but its structure is a bit of both.


modular arithmetic
algebra unsolved
algebra

[quote="Rust"]Therefore $a_{n+h}=a_{n}$ if and only if $36|h$. [/quote]
''='' here i.e.? Post concrete your solution, please!

$a_{0}\equiv-1 \pmod{3}$,$a_{0}\equiv 1 \pmod{3}$ and $a_{n+2}\equiv a_{n+1}-a_{n}-1 \pmod{3}$ for all $n\geq 0$.
So
$\begin{array}{c|c|c|c|c|c|c|c|c}n&0&1&2&3&4&5&6&7\\\hline a_{n}\pmod{3}&-1&1&1&-1&0&0&-1&1 \end{array}$
and $3|a_{n+h}-a_{n}$ for all $n\geq 0$ forces $6|h$.
And then $a_{n+h}-a_{n}={5^{n+3}\over 6}(5^{h}-1)$ for all $n\geq 0$
using the above formula given by [b]Rust[/b].
The rest is trivial.


On the game show Wheel of Fraction, the spinner wheel consists of a disk divided equally in 6 sectors. Sector one is labeled Bankrupt , sector two is labeled 1000 dollors , sector three is labeled 300 dollors , sector four is lableled 5000 dollors , sector five is lableled 400 dollors  and sector six is lableled  7000 dollors . What is the probability that you will earn exactly 6700 dollors  dollors in your first 4 spins? Express your answer as a common fraction in lowest terms.

I don't understand the solution...
[hide]
It is clear that one needs (4 !)/(6^4)=1/54 after simplification.
[/hide]

Well, the only way to get 6700 dollars is to get a 300,400,1000,5000.

There are 4!=24 ways to get this (as order matters, you could get 300,400,1000,5000 or 5000,400,1000,300, etc.)

There are 6x6x6x6 ways to spin total.

Thus its 24/6^4=4/6^3=4/216=1/54.

^ His explanation is correct, but I find it hard to believe that the solution was just:
"It is clear that one needs (4 !)/(6^4)=1/54 after simplification."

Thanks I understand now! I didn't find the solution very clear and wanted know how it was done.


Find all pairs $ (x;y)$ of real number with $ 0<x<\frac{\pi}{2}$ such that:
         $ \frac{(sinx)^{2y}}{(cosx)^{\frac{y^{2}}{2}}}\plus{}\frac{(cosx)^{2y}}{(sinx)^{\frac{y^{2}}{2}}}\equal{}sin2x$


[quote="ckck"]Twenty percent of the total cost of publishing the second newspaper during the academic year was the cost of the paper, while eighty percent was teh cost of the printing. Over the following summer, the cost of the paper increased by twenty percent, while the cost of printing decreased by twenty percent. By what percent did the total cost of publishing the paper decrease?[/quote]
[hide]1.2*.2+.8*.8=.88
12%[/hide]

[quote="ckck"]Twenty percent of the total cost of publishing the second newspaper during the academic year was the cost of the paper, while eighty percent was teh cost of the printing. Over the following summer, the cost of the paper increased by twenty percent, while the cost of printing decreased by twenty percent. By what percent did the total cost of publishing the paper decrease?[/quote]
[hide="answer"]
paper=20%
printing= 80%
after summer:
paper is 24% of original
printing is 64% of original
add them up to get 88% of the original price after summer
Therefore the total cost decreased by 12%.[/hide]

EDIT: calculation error-did not add 24 and 64 right.


Find all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that for all real $x$ we have
\[f(x)=f\left(x^{2}+\frac{x}{3}+\frac{1}{9}\right). \]

[quote="pohoatza"]could you put all the problems from your TST please?[/quote]
See here http://www.mathlinks.ro/Forum/viewtopic.php?p=808465#808465  :wink:

[quote="Rust"]Let $x=y+\frac{1}{3}, g(y)=f(y+\frac{1}{3})$, then $g(y)=f(y+\frac{1}{3})=f(y^{2}+y+\frac{1}{3})=g(y^{2}+y).$
It is easy to construct sequense $g(y_{i})=g(y_{i+1}), y_{0}=y,y_{i}\to 0$, therefore if $g(y)$ is continiosly in 0, then $g(y)\equiv g(0)$.[/quote]
No easy! Please post concrete your solution!

$g(x)=f(x-{1\over 6})$ gives $g(x)=g(x^{2}+{1\over 4})$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=141283]Done[/url]


Let $k,l,m,n$ be non negative integers.Find the pair of $(k,l,m,n)$ such that  the following equality holds for all $x$ except $0$.

\[\frac{(x+1)^k}{x^l}-1=\frac{(x+1)^m}{x^n}\]

Rewrite it as $[(x+1)^k - x^l]x^n = x^l(x+1)^m$ which can be considered as an equality between polynomials. Thus, it holds for all $x$, and $0$ must have the same multiplicity in both sides.
This means $l=n.$ And, we rewrite it as $(x+1)^k-x^n = (x+1)^m$ which, as above, has to hold for all $x$.
If $k,m \geq 1$ we have a contradiction when $x=-1.$
If $k=0$ then it gives $1-x^n = (x+1)^m$ which , by equalizing degree and coefficients, leads to a direct contradiction.

Then $k >0$ and $m=0$ so that $(x+1)^k = 1+x^n.$
As above, it follows that $k=n=1.$

Thus, the unique solution is $(k,l,m,n) = (1,1,0,1).$

Pierre.

[quote="pbornsztein"]Oops...since they are supposed to be nonzero, there is therefore no solution...

Pierre.[/quote]

Thank you for your reply again ,Pierre.

Sorry for that I was made a typo.I have just edited. Your answer is correct.

kunny


function
real analysis
real analysis unsolved

Does there exist an enumeration of the set of rationals as $ q_{1},q_{2}...$ such that the sum of the squares of the differences of successive terms converges.

I think so. Let $ (q_{i})$ be some enumeration of rationals. Change this enumeration inductively, starting with $ k=1$, by putting some of the rationals $ q_{i}$, $ i>k+1$, between $ q_{k}$ and $ q_{k+1}$. Do it so that the sum of squared differences beginning with $ q_{k}$ and ending with $ q_{k+1}$ is at most $ 2^{-k}$.

More generally, we can achieve $ \sum \varphi(|q_{i+1}-q_{i}|)<\infty$ for any function $ \varphi$ such that $ t^{-1}\varphi(t)\to 0$ as $ t\to 0$.

Yeah, that was essentialy the idea I had although I wasn't that sure of it at first. I didn't even think about that thing at the end though, so thanks!


For those of you who have taken most of the AIME's, which one(s), since 1983, do you think is/are the hardest?

This is off the top of my head, but wasn't 1999 very tough?

It depends on what you mean by "hard". On some of them, the first few were harder, while on some others it's the middle few or last few that were harder.

Anyway, I found the 89 AIME as a whole to be slightly harder than most. And last year's 2003A had one of the hardest geometry problems that I had seen in a while (#15).

I mean in general harder (like the beginning-middle problems, as opposed to the hardest #15's (the difficulties of which don't really matter too much anyways since few people get those). Equivalently, the AIME's that corresponded to the lowest AIME floors for the USAMO.

Actually, I agree with JBL. The 99 AIME was possibly the hardest. Followed by the 2003A. I don't have any statistics to back that up, but that's just my hunch from looking at them.

Really? Wow, that is cool. I got a 7 (shoulda been an 8 if only had i noticed that 2sin7cos7 = sin14) on that 2003A aime (took it recently). I definitely thought that the B was harder from 2003.

I havent done 92-2000A  but of the rest i think 2000B, 2001B, 2002B, 2003B were hardest.  funny how its all Bs

i think difficulty is much too relative to the person to make any general conclusions about which are harder. on the aime usually it's just as simple as 'which ones were you able to figure out in three hours' or what were you thinking like that day.

HDBJA4lyfe

After taking the (2001?) alternate exam officially, I looked at the A exam and I thought it was harder.  Remember, though, difficulty means different things to different people -- I got a 12 (I think) that year, and I thought there were 4 questions on the A exam I might not have gotten in 3 hours.  So 'hard' for me had more to do with the last 5 questions than anything else, although for most people the difficulty of 5-10 is probably more relevant, particularly in deciding how much time you have to work on the last 5.

I thought 1994 was challenging. My opinion about the 2003 A was that it was one of the easiest ever, since the first three were extremely simple. 4,5,6,9,12 were all pretty easy too. So i think it wasnt a hard exam. 2003 B wasnt bad either, but 2002 B was hard.

I haven't seen the past two or three years of AIMEs, but the 1994 had the lowest scores up through at least the late 90s.  The highest score was a 12.

[quote="ml2"]On the 2003A i believe there was a perfect score[/quote]

There were two 15's on the regular AIME last year.

I just took the practice 2003 B and got my best score ever. It might have been luck, but the first 10 problems were very basic.

I also enjoyed the lack of systems and logarithms.

i think 94 was hard and also 92... i did terrible on both :(

i hope this years will be easier cuz if i get a 7 and don't qualify its still better than a 5 and dont qualify... besides it looks better on your transcript, no? :D


Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ such that for all $x$ and $y$,

$f(xf(y)+x)=xy+f(x)$

If there exist $\alpha$ such that $f(\alpha)=-1$, for $y=\alpha$ we get $f(0)=\alpha x+f(x)$. Therefore $f$ is linear and it is easy to check that the only linear solutions are $f(x)=x$ and $f(x)=-x$. So we only need to prove that $f$ takes the value -1.  For $x\neq 0$ and $y=\frac{-1-f(x)}{x}$ we have $f(xf(y)+x)=-1$, and therefore the proof is comlete.


symmetry
number theory unsolved
number theory

find all ordered triple of primes$(p,q,r)$ such that

$p|q^r+1\ \ ,\ \ q|r^p+1\ \ ,\ \ r|p^q+1$

[i][edited][/i]

Just checking: initially I thought there would be cyclic symmetry on the conditions, but it seems there isn't.  Is that right?


AMC AIME LaTeX AoPSwiki geometry algebra polynomial

The results of the Mock AIME 1 for 2007

Many people thought that this test was very difficult and they may have been right.  Anyway, the top 3 scores were as follows:

:winner_first:
jb05 (11)

:winner_second:
sirorange (10)

:winner_third: (tie)
xantos C. Guin (8)
rem (8)

The answers are as follows:

[hide="answers"]
1. 162
2. 026
3. 187
4. 665
5. 605
6. 011
7. 017
8. 137
9. 045
10. 023
11. 927
12. 335
13. 012
14. 090
15. 909

[/hide]

A few things of note:

The most people got problem 3 correct, Followed closely by problem 5.

The average score was about 5.15.

My 1st, 2nd, and 3rd favorite problems were #9, #15, and #14, respectively.

Every problem was solved by somebody except for problem #14.

138 problems were guessed [hide="and..."]0 guesses were correct.  :P [/hide]

I spent a while writing these solutions. If you have any questions, or disagree with my solutions just comment or send me a pm. I would still enjoy to talk about the problems even after the contest is over. I hope that if you took the test and did not do so well, that you can try and learn something. Most of the problems were just application of techniques that a problem solver should learn. I may make another Mock AIME later this year. I will make sure to have sombody check the first few problems to make sure they are easy enough.  :P

EDIT: I don't know what happened with the link, the file is still attached when I try to edit the post, here is the link:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=4604[/url]

Thanks a lot for organizing the AIME and grading. You did a great job! :)

[quote="JSteinhardt"]If I LaTeX my alternate solution to 15 will you put it up?[/quote]

How about if we start putting the mock contest in the AoPSWiki?  Then people can edit solutions whenever and add alternate solutions, etc?

I'll start putting this one in.

[quote="JSteinhardt"]If I LaTeX my alternate solution to 15 will you put it up?[/quote]

If you want to then sure! Anybody else with alternate solutions that are significantly different than my solutions can do the same. If it is a geometry problem, you have to make the diagram though (in a pdf) so I can compile the solutions. Also I would prefer the solution in latex coding (i.e. not already compiled).

[quote="vvr1590"][hide="question"]
So if the vertices of a triangle are at (a, d), (b, e), (c, f), the centroid is at

(1/3(a+b+c), 1/3(d+e+f)) ?

[/hide][/quote]

yes

@Altheman: Okay I LaTeXed it, and it is attached. Have fun!

[quote="vvr1590"]wow the solution to  number 6 is really cool.  Are there places to find other examples like this so i can practice that tactic?[/quote]

Try problem #5

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70988[/url]

Sorry for all the questions, but
[hide]
You say that you use the rational root theorem on

p^3 - 45p + 101/25p^2 -1 = 0,

and get 5 as a possibility.

But the factors of 1 are 1 and -1, so the only possible roots of
the polynomial are 1 and -1.  Where did i go wrong?
[/hide]

All the coefficients have to be integral, so you must multiply by 25 first.

I don't see any solutions... huh?  :huh:

[quote="surge"]I don't see any solutions... huh?  :huh:[/quote]

i edited the post...the solutions are there when I try and edit it, i posted the link too

[quote="matt276eagles"]Here's a solution to #2 that avoids casework:

[hide]$n$ has up to 7 digits.  Let $x_{k}$ be the $k$th digit from the left of $n$.  We have $x_{1}+x_{2}+\cdots+x_{7}=12$, where $0\le x_{i}\le 9$.  Ignoring the restriction $x_{i}\le 9$, there are $\binom{18}{6}=18564$ solutions to this equation.

Suppose some digit is greater than or equal to 10, WLOG $x_{1}\ge 10$.  Let $a=x_{1}-10$.  Then we have $a+x_{2}+x_{3}+\cdots+x_{7}=2$, which has $\binom{8}{6}=28$ solutions.  The answer is $18564-7\cdot 28=18368$, which has a digital sum of $\boxed{026}$.[/hide]

On the solution to #15, you used the fact that the sum of the $n$ th powers of $k$th roots of unity for any $n\not\equiv 0\pmod{k}$ is 0.  Can you show a proof?[/quote]

Let $x^{n}-1=0$, the roots are $r_{k}=\exp \frac{2\pi*i*k}{n}$ for $k=0,1,...,n-1$. Factor out 1, then for the other roots, \[\sum_{q=0}^{n-1}(r_{k})^{q}=0\] for $k=1,2,...,n-1$ \[(r_{k})^{q}=\exp\frac{2\pi*i*k*q}{n}=\left[\exp\frac{2\pi*i*q}{n}\right]^{k}=(r_{q})^{k}\] So the above sum becomes: \[\sum_{q=0}^{n-1}(r_{q})^{k}=0\] for $k=1,2,...,n-1$ as desired.


[quote="kshanmugam"]If you Need online tutoring or Homework help in Math or Statistics Please visit us at

 [b]http://groups.yahoo.com/group/statisticsonlinehelp/
http://groups.yahoo.com/group/mathonlinehelp/[/b][/quote]

It sounds great, at the beginning you may not have to many "costumers", but eventually, if you have enough publicity (and in the web it does not need to be too much) the amount of job may be overwelming.


In matrix form: $ \begin{pmatrix}f_1\\ f_2\end{pmatrix}'\equal{}\begin{pmatrix}4&\minus{}5\\\minus{}2&1\end{pmatrix}\begin{pmatrix}f_1\\ f_2\end{pmatrix}.$

You're going to have to diagonalize (find the eigenvalues and eigenvectors of) the matrix

$ \begin{pmatrix}4&\minus{}5\\\minus{}2&1\end{pmatrix}$

or do something algebraically equivalent to that, such as find the exponential of that matrix, or (better if you had initial conditions), apply Laplace transforms to the system.

General principles:

We have the linear homogeneous constant coefficient system $ y'\equal{}Ay.$ The set of all solutions to this will be a vector space (a subspace of the set of all sufficiently smooth vector valued functions). That is, if we find some solutions, then any linear combination of those will also be a solution.

Suppose we find that $ Av\equal{}\lambda v$ for some $ \lambda\in\mathbb{C}$ and some $ v\ne 0.$ (That is, we have an eigenvalue and and eigenvector. Consider the function $ y\equal{}e^{\lambda t}v.$ Then $ y'\equal{}\lambda e^{\lambda t}v\equal{}Ay$ and we have a solution.

From your diagonalization work, this gives us solutions $ y\equal{}e^{6t}\begin{pmatrix}\minus{}5/2\\1\end{pmatrix}$ and $ y\equal{}e^{\minus{}t}\begin{pmatrix}1\\1\end{pmatrix}.$ Therefore, every function of the form
\[ y\equal{}C_1e^{6t}\begin{pmatrix}\minus{}5/2\\1\end{pmatrix}\plus{}C_2e^{\minus{}t}\begin{pmatrix}1\\1\end{pmatrix}\]
for arbitrary constants $ C_1,C_2$ is a solution.

Furthermore, given any possible initial conditions (for any $ t$), we can solve for $ C_1$ and $ C_2.$ (This is essentially because your matrix $ P$ is invertible.) Therefore, we have found all of the solutions.


inequalities
Functional Analysis
real analysis
real analysis unsolved

[quote="alexilic"]If $a_n$ is nonnegative sequence, and we know:
\[ \sum {a_n x_n^2} < \infty \quad \rightarrow \quad \sum {x_n} < \infty  \]
for every nonnegative sequence $x_n$, prove
\[ \sum {\frac{1}{a_n} < \infty  }\][/quote]
Here's a quick proof, using a bit of functional analysis:
Let $l^2_a = \{ (x_n)_{n \ge 1} \big| \sum_{n\ge 1} a_n x_n^2 < \infty\}$ with norm $\|x\|_a = (\sum_{n\ge 1} a_n x_n^2)^{1/2}$. This is a Hilbert space. The assumption is that $l_a^2 \subset l^1$, i.e. the embedding $J: l_a^2 \to l^1$ is pointwise bounded. By Banach-Steinhaus, the embedding is continuous, i.e. there exists a constant $C$ such that
$\|x\|_{l^1} \le C \|x\|_{l^2_a}$ for all $x$. Written out, this is equivalent to
$\sum_{n \ge 1}|x_n| \le C \left(\sum_{n\ge 1}a_n x_n^2 \right)^{1/2}$
for all sequences for which the right hand side is finite.

Now consider the sequence $x_n = \frac{1}{a_n}$ for $n \le N$, where $N$ is arbitrary, and $x_n = 0$ for all larger $n$. Then
$\sum_{n=1}^N\frac{1}{a_n} \le C \left(\sum_{n=1}^N \frac{1}{a_n} \right)^{1/2}$
and therefore
$\sum_{n=1}^N\frac{1}{a_n} \le C^2$.
Send $N \to \infty$ to conclude the proof.

[i]This is a simple example of how a "reverse  $H"older$ inequality" implies a bound for the reciprocal of a weight. Of course the condition $\sum \frac{1}{a_n} < \infty$ is also sufficient..[/i]

Nice proof, hpe.
There is also a nice related problem on Banach-Steinhaus theorem application:
Let $1 \leq p \leq +\infty$ and $q$-conjugate of $p$,i.e. $\frac{1}{p}+\frac{1}{q}=1$.$(a_n) \in \mathbb{C}^{\mathbb{N}}$ such that for any $(b_n)\in l^p$ the series $\sum{a_nb_n}$ converges. Prove that $(a_n) \in l^q$.


geometry
ratio
quadratics
function
algebra
quadratic formula
algebra unsolved

[tex] (x_{0},y_{0}) [/tex] is non-negative solution of :
(1) [tex]5x^2-2x(p-a-2c)+(p-a)^2=0[/tex]
(2) [tex]5y^2-2y(p-c-2a)+(p-c)^2=0[/tex]

with a,b,c are three sides of a triangel, p=(a+b+c)/2;
Show that:

[tex] \frac{b-c}{b-a} =  \frac{x_{0}-b}{y_{0}-b}  [/tex]

From what did this problem originate?  If the statement is true, you can prove it by direct calculation, but I assume you are looking for a more intelligent approach.

This problem is true. It's from my unsolved geometry problem. I need the shortest solution. Plz tell me more clearly!

How bout this approach?

Replace x with x+b in the first equation, simlarily for y.  We are left to prove then that $x_0 / y_0 = (b-c)/(b-a)$.

Now, replace x and y with k(b-c) and m(b-a).  It remains to prove that the ratio of the roots k and m is 1.

If we expanded the quadratic formula for the expression of k, it is a function in a,b,c, call this f(a,b,c).

It is easy to show that m = f(c,b,a).  If f(a,b,c) = f(c,b,a), we are done.

So we just need to plug $x= k(b-c) + b$ and check that the expression for the positive root of that quadratic in k is symmetric in a,c.

I haven't tried, but it looks like it could work.

I tried but it didn't work (because it's long long way to reach the short proof ^_^). I think it's not simple like that. Please check it again! Thanks!

[quote="Silva"][tex] (x_{0},y_{0}) [/tex] is non-negative solution of :
(1) [tex]5x^2-2x(p-a-2c)+(p-a)^2=0[/tex]
(2) [tex]5y^2-2y(p-c-2a)+(p-c)^2=0[/tex]

with a,b,c are three sides of a triangel, p=(a+b+c)/2;[/quote]

These equations don't have nonnegative solutions, since p - a - 2c and p - c - 2a are negative for any triangle with sides a, b, c. But even if one forgets about the nonnegative condition and checks all possible combinations of solutions, one doesn't get $\frac{b-c}{b-a} =  \frac{x_0-b}{y_0-b}$. I guess you made a typo somewhere. Anyway, what was the geometry problem that made you come up with this question?

  Darij

I can't tell that problem but you are right ^_^. Let me see it again (latest: after my master give solution^_^)


I was wondering if anyone recommends EPGY for University-level courses?  Are the EPGY courses comparable to actual university level courses?  And do colleges generally accept EGPY or at least give the option of a placement test? (I know this depends on the school, but is EPGY reputable?)  In case it's relevant, I will be going into my senior year of high school.

Thanks.

I can't answer your question, but I will also be taking  EPGY in September, taking Multivariable Calculus. When I was in CTY, my TA recommended the course to me (he's in MIT) so I guess it's pretty high-class when it comes to college entrances...


linear algebra
matrix
algebra
polynomial
vector
function
superior algebra

Prove that for any matrix $A$ the dimension of the commutant of the matrix can be written as the sum of squares of some integers whose sum if $n$.

are you sure that it is not just for the case where $u$ is diagonalisable ?
here is a sln for this case:

let $E_1,..,E_r$ be the proper subspace of $u$ and $v$ an endomorphism that
commutes with $u$. Since $u$ and $v$ commute, $E_i$ is $v$-stable. So if you
write the matrix of $v$ in a basis adapted to $E_1,..,E_r$ you clearly see that
it is a matrix with square blocks on the main diagonal, zero everywhere else.
Conclusion eaisly follows and you prove that the dimension of the commutant is
${{{dim(E_1)}^2 + dim(E_2}^2 +..+dim(E_k}^2$)

For the general case and an arbitrary field I don't know how to do this. We cannot
even use Jordan block ?

ps:for a Jordan block of size $n$ it is easy to show that the dimension we are
looking for is $n$ so it is still true.

No, the problem is true for the general case. The case when the matrix is diagonalisable was the first question.

call $P=\prod_{i}(x-a_i)^{d_i}$ the caracteristic polynomial of $u$, where
$Sp(u)=\{a_1,..,a_r\}$ with multiplicity $d_i$.
Call $E_i=ker((u-a_iId)^{d_i})$. Then if $v$ commutes with $u$ then
each $E_i$ is $v$-stable. So if one write the matrix of $u$ in a basis
adapted to $E=E_1+..+E_r$ we obtain a matrix that has square blocks
on the main diagonal.
Then if the field is algebraicly closed we can use jordan blocks in
each $E_i$ and the conclusion should not be very hard, just computations.

in the general case we have to investigate ont the dimension of the commutant of
a nilpotent endomorphism, which could be a hard task. i don't know ?

Can you please explain the proof in case of a Diagonalisable operator.The general case I think is worked out in Jacobson,Lectures in Abstract Algebra Vol -2

working without paper, a friend and me seem to have shown that the commutant of a nilpotent jordan block (a matrix which has n-1 1s right over diagonal and 0s elsewhere) of dimension $n$ is generated by its powers, hence has dimension $n$: from this the thesis will follow.

as for the proof, it's just a matter of calculation: i think it can be shown directly that a matrix commuting with our nilpotent $N$ has to have constant entries on "diagonals" (parallel to the principal one) and has to be upper triangular. (please, don't ask me to post calculations :D)

1) Here $K$ is algebrically closed and $A\in\mathcal{M}_{n}(K)$ is nilpotent. Its Jordan decomposition is $A=diag(J_{k_{1}},\cdots,J_{k_{r}})$ where $J_{k}$ is the Jordan block of dimension $k$ and $k_{1}+\cdots+k_{r}=n$. $\forall{j}, dim(ker(A^{j})=dim(ker(A^{j-1}))+\alpha_{j}$ where $\alpha_{j}=card\{i|k_{i}\geq{j}\}$. Let $\mathcal{C}$ be the subspace $\{B\in\mathcal{M}_{n}(K)|AB=BA\}$. Then $dim(\mathcal{C})=\sum_{i=1}^{n}{\alpha_{i}^{2}}$.
Proof: $K^{n}=F_{1}+F_{2}+\cdots+F_{n}$ (a direct sum) with $F_{1}+\cdots+F_{i}=ker(A^{i})$ and $dim(F_{i})=\alpha_{i}$. We change the order of the basis vectors with respect to this direct sum. Now $A$ is in the form: \[\begin{pmatrix}0&N_{1}&0&\cdots&\cdots\\0&0&N_{2}&0&\cdots\\0&0&0&N_{3}&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{pmatrix}\] where $N_{i}$ is a surjective matrix of rank $\alpha_{i}$. Let $B\in\mathcal{C}$; $ker(A^{j})$ is $B$-stable then $B$ is in the form:\[\begin{pmatrix}B_{1,1}&B_{1,2}&B_{1,3}&\cdots&\cdots\\0&B_{2,2}&B_{2,3}&B_{2,4}&\cdots\\0&0&B_{3,3}&B_{3,4}&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{pmatrix}\] We construct the $B$-columns by recurrence: For $B_{1,1}$ there exist no condition then the degree of freedom is $\alpha_{1}^{2}$. For the $B$-column $j$ (the $B$-columns $1,\cdots,j-1$ are known), the relations are: $N_{1}B_{2,j}=known, N_{2}B_{3,j}=known,\cdots,N_{j-1}B_{j,j}=known$.
The degree of freedom is for $B_{1,j}: \alpha_{1}\alpha_{j}$, for $B_{2,j}: \alpha_{2}\alpha_{j}-\alpha_{1}\alpha_{j}$, $\cdots$, for $B_{j-1,j}: \alpha_{j-1}\alpha_{j}-\alpha_{j-2}\alpha_{j}$, for $B_{j,j}: \alpha_{j}^{2}-\alpha_{j-1}\alpha_{j}$. Thus the total degree of freedom  for the $B$-column $j$ is $\alpha_{j}^{2}$.
2) If $K=\mathbb{R}$ then let $\mathcal{C}=\{B\in\mathcal{M}_{n}(C)|AB=BA\}$ and $\mathcal{D}=\{U\in\mathcal{M}_{n}(R)|AU=UA\}$. Let $B=B_{r}+iB_{c}\in\mathcal{C}$; then $\forall{t}\in\mathbb{R},B_{r}+t.B_{c}\in\mathcal{D}$. Let $B_{1},\cdots,B_{p}$ be a $\mathcal{C}$-basis. Then there exists an inversible square submatrix $S$ of $B_{j,r}+iB_{j,c}$ of rank $p$. We consider the function:$f: t\in\mathbb{R}\to det(S_{t})$ where $S_{t}$ is the $S$ matrix where we have changed $i$ into $t$. $f$ is a non identically $0$ polynomial; thus there exists $t\in\mathbb{R}$ s.t. $f(t)\not={0}$ and $dim(\mathcal{D})\geq{dim(\mathcal{C})}$. Conclusion: $dim(\mathcal{D})={dim(\mathcal{C})}$.

:blush: In my last post, the part 1) is false because the matrix $N_{i}$ is of rank $\alpha_{i+1}$ and isn't surjective but is one to one.
The part 2) is correct.

"Errare humanum est , perseverare diabolicum"
Other method:
Here $K$ is algebrically closed and $A\in\mathcal{M}_{n}(K)$ is nilpotent. Its Jordan decomposition is  $A=diag(J_{k_{1}},\cdots,J_{k_{r}})$ where  $J_{k}$ is the Jordan block of dimension  $k$ , $k_{1}\geq{k_{2}}\geq\cdots$ and $k_{1}+\cdots+k_{r}=n$. Let  $\mathcal{C}$ be the $A$-centralizer.
Proposition 1: $dim(\mathcal{C})=k_{1}+3k_{2}+5k_{3}+\cdots+(2r-1)k_{r}$.
Proof: We seek $B\in\mathcal{C}$ in the block form associated to the $A$-Jordan decomposition: $B=[B_{k_{i}k_{j}}]$.
The relations are: $B_{k_{i}k_{j}}J_{k_{j}}=J_{k_{i}}B_{k_{i}k_{j}}$. It's easy to see that the degree of freedom in $B_{k_{i}k_{j}}$ is $inf(k_{i},k_{j})$. Then the total degree of freedom in  $B$ is (column by column): $(k_{1}+k_{2}+k_{3}+\cdots)+(2k_{2}+k_{3}+k_{4}+\cdots)+(3k_{3}+k_{4}+\cdots)+\cdots=k_{1}+3k_{2}+5k_{3}+\cdots$.
Proposition 2: Let $\alpha_{j}=card\{i|k_{i}\geq{j}\}=dim(ker(A^{j})-dim(ker(A^{j-1}))$. Then $dim(\mathcal{C})=\sum_{j}\alpha_{j}^{2}$.
Proof (by recurrence on $r$): assume that $k_{1}+3k_{2}+5k_{3}+\cdots+(2r-1)k_{r}=\alpha_{1}^{2}+\cdots+\alpha_{s}^{2}$ with $\sum\alpha_{i}=n$. Then $k_{1}+3k_{2}+5k_{3}+\cdots+(2r-1)k_{r}+(2r+1)k_{r+1}=\alpha_{1}^{2}+\cdots+\alpha_{s}^{2}+(2r+1)k_{r+1}$ with $k_{r}\geq{k_{r+1}}$; But $\alpha_{1}=\cdots=\alpha_{k_{r+1}}=r$;
 then the relation becomes:
$k_{1}+3k_{2}+5k_{3}+\cdots+(2r-1)k_{r}+(2r+1)k_{r+1}=(\alpha_{1}+1)^{2}+\cdots+(\alpha_{k_{r+1}}+1)^{2}+\cdots+\alpha_{s}^{2}$
with $(\alpha_{1}+1)+\cdots+(\alpha_{k_{r+1}}+1)+\cdots+\alpha_{s}=n+k_{r+1}$.


Refer back to the thread titled "Take 5 signups" for the rules.

The first topic is...

COUNTRIES THAT CONSUME THE MOST BOTTLED WATER ANNUALLY PER CAPITA

Good Luck! PM me anywhere from 1-5 answers. All at once, please.

...and round one is over the answers are:

Italy
Spain
France
Mexico
Belgium

besttate and perfect628-9
ckck-(-4)

The next category is...

TOP 5 POOREST COUNTRIES PER PERSON

No one has been replying, so take 5 is officially dead.

its cuz its to hard to find the answers, or maybe i just suck with search engines


[quote="ryan22"]Find all solutions and equations to 2 sin (2A) + 1 = 0???

I would be happy if someone can help me in some way

Ryan[/quote]
[hide]
A=-15

just isolate A so you have
sin(2A)=$\frac{-1}{2}$
2A=-30
$\boxed{A=-15}$[/hide]

[quote="Farenhajt"]Wrong.

$2A=-{\pi\over 6}+2n\pi\implies A=-{\pi\over 12}+n\pi, n\in\mathbb{Z}$[/quote]

huh? what's n?

Haven't your teacher told you about the periodicity of trig functions? Hardly, but if (s)he hasn't, look it up in any textbook.

My, why people like to post when they don't know what they're talking about?

[quote="Farenhajt"]Wrong.

$2A=-{\pi\over 6}+2n\pi\implies A=-{\pi\over 12}+n\pi, n\in\mathbb{Z}$[/quote]

Also $2A={\pi+\pi\over 6}+2n\pi$

Sorry but the first "pi" must be before the "pi/6". I'm a beginer in Latex.

I honestly don't know what you guys are talking about... can you please explain...?  :oops:  :(

[quote="Altheman"][quote="Farenhajt"]
My, why people like to post when they don't know what they're talking about?[/quote]
that is a bit harsh...  :mad:

[quote="Farenhajt"]Wrong.

$2A=-{\pi\over 6}+2n\pi\implies A=-{\pi\over 12}+n\pi, n\in\mathbb{Z}$[/quote]

you missed $2A=\frac{7}{6}\pi+2\pi*n \implies A=\frac{7}{12}\pi + \pi*n \, \, \forall n\in \mathbb{Z}$ and $x=-\frac{1}{6}\pi+ \ip*n \, \, \forall n\in\mathbb{Z}$

basically what this is saying is that you have your principle solutions $x=\frac{7}{12}\pi, -\frac{1}{12}\pi$, but you can add any multiple of $\pi$ and that will also be a solution, you ought to ask your teacher about this if you do not understand, this idea lies in the fact that $\sin (x)=\sin (2\pi+x)=...$

notice that in what i did, i added the $2\pi$s in before i divided by 2 to isolate A from 2A[/quote]

ahhhhhh i see...
Thanks! :)

Farenhajt, easy.. No need to get so serious about it.

Now, ch1n353ch3s54a1l's answer was correct in degrees. $30^\circ = \frac{\pi}{12}$ and $\frac{\pi}{12}$ is what's called the radian. This is based on the idea of circle being $2 \pi$ instead of $360^\circ$. Only difference was that ch1n353ch3s54a1l forgot to include the $+2n \pi$ which will include the ALL angles, instead of just one. Remember, the trig functions repeat and that's why it is necessary to include $+ 2n \pi$ unless the specific interval is given.

BTW, thanks to Altheman for explaining this before me.  :D


quadratics LaTeX function algebra \/closed

A major problem I have is that whenever I try to do Ascii Art or make formulas using Ascii (both use a lot of spaces) the spaces get deleted.

Making an Ascii Signature

Whenever I try to put a signature with Ascii art or a lot of spaces in it, the signature on my profile has deleted spaces, but when I go to the signature control panel, It is perfectly fine.

If I wanted to write a quadratic equation

x²+6x+8

That would be fine. But when I write the quadratic or the first part of the cubic formula, this happens.
       ______
-b±√b²-4ac
--------------
        2a

                                  ________________________________________________
First root = -b      1    3/2b^3-9abc+27a²d+√(2b^3-9abc+27a²d)²-4(b²-3ac)^3
                --- - ---   /-------------------------------------------------------
                 3a    3a √                                       2
                                  ________________________________________________
                        1     3/2b^3-9abc+27a²d-√(2b^3-9abc+27a²d)²-4(b²-3ac)^3
                     - ---   /--------------------------------------------------------
                        3a √                                       2

See the screen shot for what it should have looked like. If anyone can fix this, that would be greatly appreciated.

Can't you just add some white text before it?

Yes,  I  do  think  that        the    forum automatically                   makes  multiple      spaces       appear      as only                                     one        space. (Quote this post!)

For your problem with the quadratic equation, what's wrong with $ \text{\LaTeX}$?

Web browsers automatically convert multiple spaces into a single space. This is a function of the web browser, not AoPS.

If you insist on having spaces, use the code tags:
[code]-b ± √b²-4ac
------------
     2a [/code]

For formulas, it's better to use LaTeX.

$ \frac { - b}{3a} - \frac {1}{3a}\sqrt [3]{\frac {2b^3 - 9abc + 27a^2d + \sqrt {2b^3 - 9abc + 27a \& 2d}^2 - 4(b^2 - 3ac)^3}{2}}$


Determine all monotonic and surjective functions $ f:\mathbb{R}\rightarrow\mathbb{Z}$ which satisfy simultaniously:

a) $ f(f(x))=f(x)$,  $ \forall x\in\mathbb{R}$.

b) $ f(2x)-f(x)=f\left(x+\ds\frac{1}{2}\right)$, $ \forall x\in\mathbb{R}$.

[quote="haus"]Determine all monotonic and surjective functions $ f:\mathbb{R}\rightarrow\mathbb{Z}$ which satisfy simultaniously:

a) $ f(f(x)) = f(x)$,  $ \forall x\in\mathbb{R}$.

b) $ f(2x)-f(x) = f\left(x+\ds\frac{1}{2}\right)$, $ \forall x\in\mathbb{R}$.[/quote]

I'm not sure what the definitions of monotonic and surjective are, so please correct me if I'm incorrect.

[hide]Plug in $ 1/2$ into the second equation to get $ f(1/2) = 0$. Plug in $ 1/2$ into the first equation:

$ f(f(1/2)) = f(1/2)$

$ f(0) = 0$

Does this contradict the monotonicity of $ f$, or does this imply that $ f(x) = 0$, $ \forall x\in\mathbb{R}$? If the latter is true, then wouldn't this contradict the surjectiveness of $ f$?

So, I think that there are no such functions that satisfy those conditions.[/hide]

[quote="MellowMelon"]Whoa, where'd those last conclusions come from? All you can get from the first statement is $ f(n) \equal{} n$ for all integers $ n$. (to get this, note that since $ f$ is surjective, you can replace $ f(x)$ with any element in the codomain) EDIT: Ok, I think I see what you were arguing now. Basically, since $ f(0) \equal{} f(1/2) \equal{} 0$, it means that $ f(x) \equal{} 0$ for all $ 0\leq x\leq 1/2$ by the monotonic condition.

Monotonic means that $ a\geq b\Rightarrow f(a)\geq f(b)\vee f(a)\leq f(b)$. Surjective means that every element in the codomain ($ \mathbb{Z}$ here) can be written in the form $ f(x)$ for some $ x$ in the domain.

[hide="Hint for the actual problem"]Conclude that $ f(n) \equal{} n$ (see above), then compute $ f(\frac{n}{2^{k}})$ for all odd $ n$ and positive $ k$. Since the function is monotonic, you will get all values of $ f$.[/hide][/quote]

Thank you for the clarification.

I think I have found another method to solve the problem. I'm probably wrong (as usual)...

[hide]Let $ m$ be any integer. Plug it into the second equation:

$ f(2m)\minus{}f(m) \equal{} f(m\plus{}1/2)$

$ f(m\plus{}1/2) \equal{} m$,

since $ f(n) \equal{} n$ for integers. Since $ f(m) \equal{} m$, this implies that $ f(x) \equal{} m$ for all $ x\in[m,m\plus{}1/2]$ (I will be frequently using this assertion in the steps below, with different values of $ m$). Now, plug in $ m\plus{}3/4$:

$ f(2m\plus{}3/2)\minus{}f(m\plus{}3/4) \equal{} f(m\plus{}5/4)$

$ 2m\plus{}1 \equal{} f(m\plus{}3/4)\plus{}f(m\plus{}5/4)$

But, $ f(m\plus{}5/4)$ must equal $ m\plus{}1$ (since $ m\plus{}5/4\in [m\plus{}1,m\plus{}3/2]$), so:

$ 2m\plus{}1 \equal{} f(m\plus{}3/4)\plus{}(m\plus{}1)$

$ f(m\plus{}3/4) \equal{} m$.

Since $ f(m\plus{}1/2) \equal{} m$, the monotonicity of $ f$ implies that $ f(x) \equal{} m$ for all $ x\in[m\plus{}1/2,m\plus{}3/4]$. Since we already established that this is true for $ x\in [m,m\plus{}1/2]$, we can deduce that it must be true for all $ x\in [m,m\plus{}3/4]$

We can repeat this procedure infinitely (is this valid?), to get that $ f(x) \equal{} m$ for all $ x\in [m,m\plus{}1)$. So, the function must be the floor function.[/hide]

That last step is... well it's correct, but on an actual contest you would probably need to formalize it with induction. Plug in $ m\plus{}1\minus{}\frac{1}{2^{k}}$ and show you get your identity given $ k\minus{}1$.

[quote="MellowMelon"]Monotonic means that $ a\geq b\Rightarrow f(a)\geq f(b)\vee f(a)\leq f(b)$.[/quote]

This is not true, and it's an example of why it's probably best not to use formal logic notation out-of-context.  A function is monotonic if $ (\forall a\forall b\; (a\geq b\implies f(a)\geq f(b)))\vee(\forall a\forall b\; (a\geq b\implies f(a)\leq f(b)))$.  I personally recommend sticking to English as far as statements of quantification and implication are concerned.


Prove that there are many infinitely positive integers $ n$ such that $ \phi(\sigma(n))$ divides $ n$.


[quote="b555"]can we say that  n!+ r is  divisible  by  r  ? :?:[/quote]

$ 105!\plus{}r$ is divisible by $ r \le 105$ because $ r|105!$ for $ 2 \le r \le 105$ as $ n!\equal{}2 \times 3 \times 4 \times ... \times 105$

[quote="b555"]can we say that  n!+ r is  divisible  by  r  ? :?:[/quote]Only when $ r\le n$. You'll immediately get a counterexample if $ r>n\plus{}1$, in general.

[quote="10000th User"][quote="b555"]can we say that  n!+ r is  divisible  by  r  ? :?:[/quote]Only when $ r\le n$. You'll immediately get a counterexample if $ r > n \plus{} 1$, in general.[/quote]
Not really. If $ n\plus{}2$ is composite, you'll be fine. For example $ 100!\plus{}102$ is clearly divisible by 2, and is thus composite.

but what we are discussing  is  100!+ 102  is divisible  by 102  or not.   its obviously  is divisible  by  2.    can u please explain that sir? :?:

[quote="Brut3Forc3"][quote="10000th User"][quote="b555"]can we say that  n!+ r is  divisible  by  r  ? :?:[/quote]Only when $ r\le n$. You'll immediately get a counterexample if $ r > n \plus{} 1$, in general.[/quote]
Not really. If $ n \plus{} 2$ is composite, you'll be fine. For example $ 100! \plus{} 102$ is clearly divisible by 2, and is thus composite.[/quote]I think you meant 102 and not just 2 and your example is still correct. What I meant is that $ n!\plus{}r$ is not divisible by $ r$ for every positive integer $ r\ge n\plus{}1$. Try $ 100!\plus{}101$. This is clearly not divisible by 101. However, there are certain r values that do have such property. And if you see it that far, this set of r values is finite.

How are you going to work this out? Is it impossible because of the ! contains two? :huh:  :?:

Note that 105! contains a factor of 2,3,4,5,...,105. Thus, for $ n!\plus{}r$, this contains a factor of r. For example, $ 105!\plus{}103\equal{}103(105 \cdot 104 \cdot 102 \cdot \hdots \cdot 1\plus{}1)$.


$ general \ rule \equal{} \frac {\# \ of \ letters!}{\# \ of \ repeated \ letters!}$

Example:

Find the number of ways we can arrange the word MISSISSIPPI. (common Mathcounts problem).

$ \frac{11!}{2!4!2!}$

You can read/listen more about this stuff in Alcumus, located on the left underneath "Classroom" and above "For The Win."

[quote="asalate"]So when a problem asks how many different arrangements the letters P E O P L E can be made into, how would one go about solving this?
is there a general rule?
any help would be great![/quote]
I'd recommend to get the Introduction to Counting and Probability book, which explains this more in detail.


geometry email trigonometry AMC AIME MATHCOUNTS inequalities

Hello,

We are excited to announce the first annual NCSSM middle school math contest, to be held at NCSSM on (tentatively, providing there is not too much conflict) April 4, 2009. The contest's difficulty will challenge even the brightest middle schoolers. We will offer power (proof), team (short answer), and individual (short answer) rounds, with the individual round split into three subjects - Algebra, Geometry, and Discrete Math. Each student will take two subjects or a general test, and the top scorers in each subject will qualify for the Final Round, a short-answer test. A combination of Final Round and subject test scores will determine the overall individual winners. The top teams and individuals will receive prizes.

For more information and questions email

jiangty (at) ncssm (dot) edu
or
tenoriop (at) ncssm (dot) edu

For later updates (none yet) see [url]http://ncssmmath.blogspot.com/[/url]

[quote="not_trig"]Hello,
The contest's difficulty will challenge even the brightest middle schoolers. [/quote]
eh?

New website up here: [url]http://math.ncssm.edu/~rash/mach8[/url]. Tips on website design are appreciated :D

This is going to be pretty exciting

dude lol in like 3 years there is going to be like 5000 middle school math competitions

not_trig DO try to not make trivial problems that you can find easily (like AIME 1986 or something)

Well, a) we won't steal AIME problems, and b) AIME problems are likely too hard for anything beyond last few questions on individual/team or tiebreakers.

Also, we are taking problem submissions if you are in high school or college - email them to the above email address (or PM them to me.) Also, give me your name/school so we can recognize you if we use the problem.

See even the middle schoolers agree you should be doing a high school contest.

What is going to be the approximate difficulty of these problems?

Individual tests will range from Chapter Mathcounts (easy end) to medium AIME (hard end). Tiebreakers will range from State Mathcounts (easy end) to hard AIME (or beyond!). Team will range from Chapter Mathcounts Team to medium AIME.

Power round will be challenging but certainly within reach.

(these are all very rough estimates, though)

Dude I will see if I can make some non-fail problems.

The website says we will take either general or two of the three you mentioned. Has anything been changed?

No, the test format will not be changed.

You misunderstood my question. The test website and your post do not have the same information.

Your post says that each student will take all three subjects, while the website says a student can chose either general or two of the three you mentioned. Which is correct?

Oh really? Sorry. This post is outdated - can a mod edit it to say "two out of three subject tests"? Thanks.

"We will offer four different individual tests: Algebra, Geometry, Discrete Math, and General. Students will take either two of the three subjects (Algebra, Geometry, or Discrete Math) tests, or just the General test.

The Algebra and Geometry tests will cover topics similar, but not limited, to those on the Algebra I and II and Geometry subject tests given by the State Math Contest. The Discrete Math test will cover basic concepts in number theory and combinatorics. Finally, the General test will involve a mix of all three subjects."

Is this a middle school contest or high school contest?

Checking with the North Carolina Department of Public Instruction web page about the curriculum, all you are going to test is in the high school math curriculum section.

Probably it is better for you to just follow Mathcounts or AMC 8  Curriculum.

MATHCOUNTS Curriculum
MATHCOUNTS questions are written with the curricula for grades 6-8 in mind. In addition, many
problems are designed to challenge and accelerate student learning, and questions become progressively
more difficult at each level of competition. Possible topics include:
 Algebra  Geometry
 Charts, Graphs & Tables  Logic
 Computation  Measurement
 Consumer Math  Number Theory
 Equations & Inequalities  Probability
 Equivalent Expressions  Statistics
 Estimation & Approximation

The majority of the problems are included in the middle school curriculum; however, some of the harder ones may involve more advanced concepts in lieu of a computational mess. Moreover, I believe that Algebra and Geometry taught at middle schools should not be very different from those courses at high schools (though I may be mistaken, having taken it when I was in a K-12 school.)

The test is meant to be challenging; only the brightest will be able to answer the hardest questions. The easier problems will certainly be doable by everyone. However, we will keep your input in mind as we continue to revise our tests. Thanks!

@ jb05: Yes, thanks!

Yes it has - sorry for the un-updated website. I thought it was changed, but apparently not... it'll be changed to the correct date ASAP.

[quote="dnkywin"][quote="n^4 4"]Why in the world would you change the date to be the same weekend as national mathcounts??[/quote]

Kudos to the above.[/quote]

Sorry about that - there were no other dates that didn't conflict. Unfortunately, we think that the benefit of having more kids come is more valuable than just the 4 on the national MC team. Since this is the first year, we just want as many people as possible to show up, to raise awareness and get feedback. There will still be strong competitors attending, though, as there are many kids who don't make Nats that turn out to be very smart anyways.

Wait I am confused. You're going to make this the hardest middle school competition out there and not even let the top 4 come? Seriously?

This isn't meant to be the "hardest" competition out there. There will simply be some challenging problems. I am sure there are people who are around the same level as the Nats team who didn't make Nats, but didn't realize their full potential at States either.

*shudders* Ew. I failed yesterday. xD

Will yesterday's participants just be sent their scores, or will their answer sheets also be given back? (b/c I really want to go over all the problems I screwed up and find out what stupid mistakes I made... lol)

DAMIEN I DEMAND YOU SHOW US YOUR ANSWER KEYS THAT YOU MADE DURING THE TEST.

[quote="not_trig"]Your brashness may result in displeasure on the part of the contest coordinators. Calm down and maybe you will receive answers.[/quote]

Lol you sound slightly official

P.S. check your PMs

Hi guys!

We're going to be running the 2nd Annual MaCh 8 on December 12, 2009. We promise it will be a lot more organized because we will have more than 2 people preparing it! If you're interested, go to http://math.ncssm.edu/~rash/mach8, fill out the registration form, and send it either to me at jiangty@ncssm.edu, or Bryce Taylor at taylorbw@ncssm.edu. The format and registration form are the same as last year. (The website is still the same as last year's, but it'll be updated soon with problems, results, and acknowledgments.)

Thanks!
Damien Jiang

(P.S. We have a new blog at ncssmmach8.wordpress.com; Wordpress has LaTeX functionality. Yay.)

Hello all,

Unfortunately, December 12 has a conflict with an NCSSM event. So, we are moving the contest date to December 5, 2009. Please tell your teachers to sign up if you're interested!

Thanks,
Damien Jiang

[quote="not_trig"]We promise it will be a lot more organized because we will have more than 2 people preparing it![/quote]
...and I assume you'll make the answers before the competition begins  :wink:

Also, I'll be expecting the final round to be completed before 12:00 AM on the day of the test.

Yes, everything will be done at least a day in advance. This will be far more organized than last year because we'll have about ten times as many people working on it!

We have results! Thanks to all who came, and special thanks to Ms. Stephens, Mr. Batterson, and Mr. Taylor for bringing their students. Also, we'd like to thank the Art of Problem Solving for providing gift certificates as prizes!

Congratulations to the following winners:

Algebra

5. (Tie) Jenny Wang and Evan Liang
4. Damien Gu
3. Pranay Orugunta
1. (Tie) Allan Jiang and Justin Luo

Geometry

4. Allan Jiang
3. Sam Chen
2. Roy Li
1. Justin Luo

Discrete Math

5. Matthew Zheng
4. Vinit Ranjan
3. Roy Li
2. Jenny Wang
1. Evan Liang

Individual Finals

3. Justin Luo (Carnage A)
2. Nick Gentile (Stanford Chargers)
1. Damien Gu (Ligon A)

Team Round

3. Carnage B
2. Ligon A
1. Carnage A

Also, thanks to our student volunteers (in no particular order) - Bryce Taylor, Claire Zeng, Seon Kang, Jessie Duan, Anthony Wu, Nick Woodcock, George Zhao, Alex Yoo, Kathleen Lan, Lilly Park, Christy Vaughn, Winston Liu, Sam Stone, and Cole Finney for their help, and Mr. Philip Rash and Mr. Chris Gann for their support.

GJ GUYS!!!  :D

Who got 5th in geometry?

[quote="professordad"]
Who got 5th in geometry?[/quote]

It was Gordon Ma. (b-flat, can you edit that in? Thanks!)


search
number theory proposed
number theory

[quote="ali666"]a,b,c,d are positive integer such that [img]http://www.mathlinks.ro/Forum/latexrender/pictures/f05ad2501b26bc8d58c4ab048e1516e0.gif[/img] prove taht a+b+c+d is not prime.[/quote]

Let $a+b+c+d=p$, where $p$ - prime. So we have that $a+b+c+d\geq4$ and $p$ - odd. Therefore, one or three of variables are odd. If one we have that one side is odd and other is even. If we have three odd numbers then again one side is odd and other even. We have contradiction, so $a+b+c+d$ cannot be a prime.

[quote="ciprian"]When 3 numbers are odd i think that both sides are odd so where is the contradiction?[/quote]
yeah i agree with him when 3 numbers are  odd then  both sides are odd!!!!!
but when 3 numbers are even one side is odd and the other side is even and it's contradiction.

I gave the same solution as the one in the link of ehsan2004...it is quite natural i think an you don't really need a geometric interpretation ;) Here is a similar problem for who is interested (maybe is also on the forum but i am to lazy to search :P ):

Let $a,b,c,d,e,f$ positive integers and let $S=a+b+c+d+e+f$. If $S$ is a divisor of $ab+bc+ca-de-ef-fd$ and of $abd+def$ than $S$ is a composite number.


let $C(O,R),C'(O',r)$, be two intersected circles.for an arbitrary point$A$,

we have:$AO',AO$, intersect circles$C',C$, at points $B,C$,respectively.

($B,C$ are the first intersections)

with the property of: $\frac{AB}{AC}=\frac{r}{R}$

find the locus of $A$.

[hide]$\frac{AB}{AC}=\frac{r}{R}= \frac{AB+r}{AC+R}= \frac{AO'}{AO}$

Hence the locus is the Apollonius circle which divides the segment $OO'$ at the ratio $\frac{AO'}{AO}= \frac{r}{R}$

(this circle passes through the intersection points of the two circles because then $AO'=r$ and $AO=R$

[color=brown]Note: If the point $A$ is on the exterior of the two circles, then it's ok.
But how do we find the points $B,C$ if $A$ is inside the circles? What is then the "first" intersection??[/color][/hide]


[quote="Treething"]In Treething's latest game, Trouser Snake, each of the four arrow keys are randomly assigned to a direction (up, left, down, or right). Multiple keys assigned to the same direction are not allowed - each arrow key must be assigned to a unique direction.

If Trouser Snake automatically reconfigures the arrow assignations if all four arrow keys corresponds to the correct direction, find the possibility that the up arrow key corresponds to the upwards direction.[/quote]
[hide]5 ways that work out of 23 configurations is $\frac{5}{23}$[/hide]
I almost got tricked by this one :D

[quote="myspacesucks"]Most people who read this will probably almost get tricked at first, but it in the end you should get it.  To get the answer, you need to focus on 4 (the number that the is how many arrow keys there are).  When you think about it, its simple, (4+3)-2/(4+3)+4^2.
It gives you 5/23, the right answer.[/quote]
You are overcomplicating it :roll:
[hide]There are 6 ways normally that have up=up, but one doesn't work. There are 24 ways normally, but one doesn't work. $\frac{5}{23}$[/hide]

the one doesn't work because that's when all of them are in the right direction?

-jorian


probability
expected value
probability and stats

In the World Series, the top teams in the National League and the American League play the best-of-seven series; that is, they play until one team has won four games. (No tie is allowed, so this results in a maximum of seven games.) Suppose the teams are evenly matched, so that the probability that either team wins a given game is 1/2.

A)Use a coin or modeling program to model a World Series, where "heads" represents a win by Team A and "tails" a win by team B. Perform this experiment at least 80 times, keeping track of how many games are needed to decide each series. Estimate the probability that an evenly matched series will end in four games. Do  the same for five, six, and seven games.

B) What is the probability that the series will end in four games? Five games? Six games? Seven games? Compare with your estimates from part (A)

C) Find the expected value of the number of games until the series ends. [Hint: this will be P(four games) X 4 + P(five) X 5 + P(six) + P(seven) X 7]

How do I do this?

Double-posting is rude -- please don't do it.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=281906


Prove that addition of natural numbers is commutative.
Here's my attempt:
Let $ M \equal{} \left\{ {m \in N,m \plus{} y \equal{} y \plus{} m} \right\}$
I. $ 0 \plus{} y \equal{} y \equal{} y \plus{} 0 \Rightarrow 0 \in M$ (considering 0 a natural number)
II.$ \begin{array}{l}
  m \plus{} y \equal{} y \plus{} m \\
  f(m \plus{} y) \equal{} f(y \plus{} m) \equal{} y \plus{} f(m) \\
\end{array}$
Now logic tells that $ f(m \plus{} y) \equal{} f(m) \plus{} y$ or simply $ m \plus{} y \plus{} 1 \equal{} m \plus{} 1 \plus{} y$ and by induction this would show that M=N, but i don't see that in my book's definition of addition. Any help?
Anyway feel free to ignore my way (maybe a dead-end) of demonstration and start from the beginning.

I don't understand your proof; it seems to me that you are assuming commutativity to prove commutativity.

If I'm not mistaken, commutativity of addition is axiomatic.

What is your book's definition of addition?  And what is $ f(x)$?

[b]The Zuton Force:[/b]  I believe he is considering the set $ M$ of natural numbers that commute additively and trying to show that it contains the natural numbers.  Of course, without understanding what axiomatization of arithmetic he's using, this question is empty.

The additive identity property only states it one way - not both. So you can say that y+0=y, but not y=0+y.

[b]The Zuton Force[/b]: Like t0rajir0u said, i'm assuming M is a subset of N where the predicate i want to proof is true and i'm trying to show that M=N by the principle of induction:
If for $ M \subseteq N$ we have:
I) $ 0 \in M$ and
II) $ \forall n \in M,f(n) \in M$ where f(n) = n+1
then $ M\equal{}N$
Commutativity isn't axiomatic. I was cheking the net before and found that in E. Landau's book "Foundation of Analysis", commutativity is proven by using the equivalence that i mentioned in the first post but i can't find that book now.

[b]t0rajir0u[/b]: I'm using Peano's axioms to define the System of natural numbers like this: $ N \equal{} \left\{ {0,f(0),f(f(0),...} \right\}$ where $ f(0) \equal{} 1,f(1) \equal{} 2...f(n) \equal{} n \plus{} 1$. Here's my book's definition of addition:
Addition of two natural numbers:
i) $ n \plus{} 0 \equal{} n, \forall n \in N$
ii) $ n \plus{} f(m) \equal{} f(n\plus{}m), \forall n, m \in N$

[b]worthawholebean[/b]: You're absolutely right. I'm again at the beginning now :(

I am guessing here that this is Peano Arithmetic with $ 0 \in \mathbb{N}$, $ f$ is the successor function, and that you are using this definition of addition:
$ x \plus{} 0 \equal{} x$
$ f(x\plus{}y) \equal{} x \plus{} f(y)$

You can put in the definition that $ 0 \plus{} x \equal{} x$, but this is stronger than you need, and if you're going to do stuff at this level you may as well use the bare minimum. So you need another way to do the base case.

Start by proving that $ x \plus{} 0 \equal{} 0 \plus{} x$ by using induction, and then conclude that $ x \plus{} 1 \equal{} 1 \plus{} x$ by taking $ f$ of both sides. I assume you've already proven $ f(x) \equal{} x \plus{} 1$ and that addition is associative. Then $ m\plus{}x \equal{} x\plus{}m$ implies that
\[ f(m)\plus{}x \equal{} (m \plus{} 1) \plus{} x \equal{} m \plus{} (1 \plus{} x) \equal{} m \plus{} (x \plus{} 1) \equal{} (m \plus{} x) \plus{} 1 \equal{} (x \plus{} m) \plus{} 1 \equal{} x \plus{} (m \plus{} 1) \equal{} x \plus{} f(m)\]
completing the induction.

And I see from the post made while I was typing this that my assumptions are correct.