AMC
AIME
floor function
trigonometry
modular arithmetic
probability
ceiling function

...
Wasn't it only posted up yesterday?  Or was it the day before?
I haven't as yet found a 3-hour block of time to do this.  If I get to sometime this week, chances are I can't practice violin that day...
Once I do it, I'll post up all of my solutions.

[quote="chess64"]
EDIT: Here's #12 as well.

[hide="12"]
$\triangle ABC$ is the classic 13-14-15 triangle with altitude $AD=12$, $BD=CE=5$, $ED=4$.

$AE=\sqrt{4^2+12^2}=4\sqrt{10}$ by Pythagorean Theorem.

$EF=(5\cdot 9)/AE=\frac{9\sqrt{10}}{8}$ by Power of a Point.

$\sin AED = \frac{12}{4\sqrt{10}}=\frac{3\sqrt{10}}{10}$

$\therefore \sin CEF = \frac{3\sqrt{10}}{10}$

$\Rightarrow [CEF]=\frac{1}{2}\sin (CEF)(EC)(EF)=\frac{15}{16}$

$[AEC]=\frac{1}{2}\cdot 5\cdot 12=30$

$\Rightarrow [AFC]=[AEC]+[EFC]=\frac{495}{16}$

$\therefore \boxed {511}$
[/hide]

Is that right?
[/quote]

[hide="Err..."]
I think you did $\cos{\angle AED}$ instead of sine... everything else looks good to me, though. [/hide]

[quote="chess64"]So, start discussing! :) I'll start with the solution to #5... :D

[hide="5"]
Recall that $x^2+y^2=(x+y)^2-2xy$. Since $25$ is a perfect square and $72\cdot 2$ is a perfect square, we can already tell that this will work out well: $25^2+72^2=(25+72)^2-2(25)(72)=97^2-(5\cdot 12)^2$. Factoring as a difference of squares, we find \[ (97+60)(97-60)=157\cdot 37.  \] To be rigorous, we should verify whether 157 is prime. Since $\lfloor \sqrt{157}\rfloor =12$, the largest prime we need to test is 11. None of the primes 2, 3, 5, 7, or 11, divide 157, so it is prime, and it also our answer.
[/hide]

EDIT: Here's #12 as well.

[hide="12"]
$\triangle ABC$ is the classic 13-14-15 triangle with altitude $AD=12$, $BD=CE=5$, $ED=4$.

$AE=\sqrt{4^2+12^2}=4\sqrt{10}$ by Pythagorean Theorem.

$EF=(5\cdot 9)/AE=\frac{9\sqrt{10}}{8}$ by Power of a Point.

$\sin AED = \frac{12}{4\sqrt{10}}=\frac{3\sqrt{10}}{10}$

$\therefore \sin CEF = \frac{3\sqrt{10}}{10}$

$\Rightarrow [CEF]=\frac{1}{2}\sin (CEF)(EC)(EF)=\frac{15}{16}$

$[AEC]=\frac{1}{2}\cdot 5\cdot 12=30$

$\Rightarrow [AFC]=[AEC]+[EFC]=\frac{495}{16}$

$\therefore \boxed {511}$
[/hide]

Is that right?

EDIT2: And #1

[hide="1"]
$S=f(1)+f(2)+\cdots+f(99)$
$=2(49)+3(33)+5(19)+7(14)+11(9)+13(7)+17(5)+19(5)\\+23(4)+29(3)+31(3)+37(2)+41(2)+43(2)+42(2)$
$\equiv \boxed {358}\pmod {1000}$, I hope.
[/hide]

EDIT3: And #2

[hide="2"]
Let the center of circle $\omega_1$ be $O$ and $\omega_2$ be $P$.

$AB=24\sqrt{2}$

Let $\theta = \angle PBO$.

$\tan \theta = \frac{1}{2}$

$\Rightarrow \tan PBC=\tan 2\theta=\frac{4}{3} \Longleftrightarrow \frac{AC}{AB}=\frac{4}{3}$

$\therefore AC=32\sqrt{2}$

$\Rightarrow [ABC]=\frac{1}{2}(AB)(AC)=\boxed {768}$.
[/hide]

Am I the only one done with this so far? I'll wait until I post my other solutions then...[/quote]

for #1, isn't the last prime 47*2, so the answer is actually [hide]368[/hide]

[quote="chess64"]Sorry, that was a typo. And yes, the answer is [hide]368[/hide][/quote]

Grrrrrr: bruteforcing #1 never works. Esp. when it takes 20 minutes.  :mad:

My answers w/o explanations:
[hide]1: 372
2: 768
3: 870
5: 157
7: 677
9: 550[/hide]

Can you post your solutions to #7 and #9? My answer for #9 has to be wrong, because it's greater than 999.

But for #3:

[hide]
Where $0<A<B<C$, 3-digit hailstone numbers must be in the form $CAB$ or $BAC$. So there are $b=2\binom{9}{3}=168$ of them.

Where $0<A<B<C<D$, 4-digit hailstone numbers must be in the form $DBCA$, $DACB$, $CBDA$, $CADB$, or $BADC$. So there are $a=5\binom{9}{4}=630$ of them.

$\therefore a+b=\boxed {798}$.
[/hide]

Hmmm... :(

#4 was a nice problem. Did anyone get [hide]321[/hide]

Number 9:

[hide]Firstly observe that all the houses must have a person in them, so we reduce the problem to arranging 6 people in 7 houses with at least one house having at least 3 persons in it. We can have 3-3, 3-2-1, 3-1-1-1, 4-2, 4-1-1, 5-1, or 6. There are thus $\binom 72+\binom 71\binom 61\binom 51+\binom 71\binom 62 + \binom 71 \binom 61 + \binom 71 \binom 62 + \binom 71 \binom 61 + \binom 71$ ways, which I probably misadded. :D Can't believe I read a problem better than chess64! :o[/hide]

All the houses must have one person in them?? Hmm... yeah, I misread the problem then :oops:

[hide="9"]
We need the number of solutions to $x_1+x_2+\cdots+x_7=13$ for nonnegative integers $x_i$. Assume that $x_7=4$ (we'll need to multiply by 7 at the end because of this) so $x_1+x_2+\cdots+x_6=9$. Let $y_i=x_i+1$ so then $y_1+y_2+\cdots+y_6=3$ where $y_i$ are nonnegative integers. There are $\binom{8}{5}=56$ solutions, and our answer is $56\cdot 7=\boxed {392}$.
[/hide]

Watch the answer be way off... :(

Chess, I think you forgot when there aren't 4 in one house, but instead 5 or 6 or 7.

Ooops..... :blush: whoever said I read problems correctly?

[quote="chess64"]I'm pretty sure the problem said that "find the probability that at least one house has 4 residents", not "one house has at least 4 residents", but I'm too lazy to check :([/quote]

Oops. I actually meant the problem to be the way solafidefarms interpreted it, but I guess it was pretty unclear. It was the last problem I added, so heh, missed that typo... :blush:

And I think chess64's answer to #3 is correct, but I got a very close answer for #4 (I could very well be wrong, I didn't check my answers thoroughly).

BTW, I think #14 is probably the coolest problem. It's not EXTREMELY difficult, but there is a really nice insight to it ;)

[quote="chess64"]Oh, well that makes it much harder :lol:

[hide="My solution to 4"]
$y=\frac{1}{x}$

$\lceil y\rceil = \lfloor y\rfloor + 1 = \lfloor \frac{1}{x}\rfloor +1$

$\Rightarrow \lfloor x\rfloor = \lfloor \frac{1}{x}\rfloor +1\Longleftrightarrow \lfloor x\rfloor - \lfloor \frac{1}{x}\rfloor=1 \qquad (*)$

$(*)$ is satisfied $\forall x\in (1,2)$ trivially. Now if \[ k-1\le x< k,  \] then \[ \frac{1}{k-1}\ge \frac{1}{x}> \frac{1}{k}.  \] But for $k$ integer, the RHS of the inequality is $<1$ but $>0$. So, $\lfloor \frac{1}{x}\rfloor = 0$. So then from $(*)$ we find that $\lfloor x\rfloor = 1$, so $1\le x<2$ is in fact the only possibility.

$\Rightarrow 1<\log_m n < 2$

$1<\log_m n$ and $\log_m n <2$

$\Rightarrow \log m < \log n \Leftrightarrow m<n$ from the left inequality, which we already knew.

$\Rightarrow \log n < 2\log m = \log m^2 \implies n<m^2$--this is the important thing. So:

\[ \begin{tabular}{cc} \text {Value of } m & \# \text { of possibilities for } n\\\hline 10 & 99-11+1=89\\ 9 & 80-10+1=71\\ 8 & 63-9+1=55\\ 7 & 48-8+1=41\\ 6 & 35-7+1=29\\ 5 & 24-6+1=19\\ 4 & 15-5+1=11\\ 3 & 8-4+1=5\\ 2 & 3-3+1=1 \end{tabular}  \]

$\therefore \boxed {321}$[, unless I made an addition error.]
[/hide][/quote]

Yeah that looks right. I probably added wrong.

Here is what I got (and I am really bad at AIMEs in general, by the way):
[hide]
1. 368
4. 321
5. 157
7. 679
9. 371
10. 020
11. 121
12. 631
13. 830[/hide]
For 12 I was unusually certain of my answer; I used similar triangles and the area of a triangle formula A=(1/2)ab sinC. Where you used the power of a point, I switched to a different method and I can't tell what I did wrong. Also, I'm horrible at LaTeX so it will take me a while to post what I did.

That was horrible. I spent about 1 hr and got 2 problems correct and got about halfway done in about 6 others but couldn't do anything.

I got #1 and #5 -- both of them I  bruteforced, but #5 I knew how to do it the original poster's way too...

yea referring to #4 i got 321 too, can someone show me a fast way to do #1? i missed it somehow  :blush:

my answers were:

[hide]
1) 045 (WRONG)
2) 768
3) 798
4) 321
5) 157
6) 270 (? solution to this ne1?)
7) 679
8) 294 (? solution to this too)
9)
10)
11) 150 (?)
12) 631
13) 830
14)
15)

[/hide]

Did anyone get #8 or #10 or whichever one was 2n choose n?

I got 044 from pure logic... my dad did it another way and also got 044. Did any of you guys get that?

[hide="#6"]
I could very possibly have miscounted this, but I've tried checking over quite a few times...

$P_1(x) = (1+x+x^2+\cdots+x^{100})-(x^2+x^5+\cdots+x^{98}) = \frac{x^{101}-1}{x-1}-\frac{x^{101}-x^2}{x^3-1} = \frac{x^{103}-x^{102}-x-1}{x^3-1} = \frac{(x^{102}-1)(x+1)}{x^3-1}$

$P_2(x) = \frac{x^{101}+1}{x+1}$

$P_3(x) = \frac{x^{68}-1}{x-1}$

So the product becomes

$P_1 \cdot P_2 \cdot P_3 = \frac{(x^{102}-1)(x^{101}+1)(x^{68}-1)}{(x^3-1)(x-1)}$

$x^{102}-1$ and $x^{101}+1$ share $1$ root.
$x^{101}+1$ and $x^{68}-1$ share $1$ root.
$x^{68}-1$ and $x^{102}-1$ share $34$ roots.

$x^{102}-1$ and $x^{101}+1$ and $x^{68}-1$ share $1$ root.

By PIE, the number of distinct roots on top is $102+101+68-1-1-34+1 = 236$.

And the number of distinct roots on the bottom is $3$, all of which appear on top as well, so the final answer is

$236 - 3 = 233$.

[/hide]

[hide="#8"]
Using Euler's Formula, we know $V+F = E+2$.

Let $x$ be the number of triangles, $y$ be the number of squares, and $2x$ be the number of pentagons. We know $3x+y = 37$.

$F = 37$.
$E = \frac{3x+4y+5(2x)}{2} = \frac{13x+4y}{2}$.

So we know $x$ is even.

Then $V = E+2-F = \frac{13x+4y}{2}-35$.

So we test $x = 2, 4, 6, 8, 10, 12$ with the corresponding $y = 31, 25, 19, 13, 7, 1$.

Our values for $V$ are $40, 41, 42, 43, 44, 45$. So the sum is $255$.

As far as I know, these polyhedra all exist... I could be wrong, though... [/hide]

paladin8, would it be possible for you to briefly educate me in polyhedron theory. Actually, it might suffice for someone to post a complete solution to #8. I think I figured out the 6 possible numbers of edges for the polyhedra fitting the requirements, and I know what shapes and in what quantities they are composed of, but I can't figure out how to compute the number of vertices form there. I know it is not necessary to determine the exact configuration of the faces, but what is the shortcut? It's bothering me so much.

Ahhh, Euler's Formula is what I needed. Yeah, and I got the same solution you got. Thanks. I find it strange that I didn't see your post before I wrote my previous message (please ignore it).

[quote="Altheman"]could you post some answers??

[hide="here is what i got"]
1. 368
2. 768
3. 798
4. 330
5. 157
6. 266
7. 679
8. 255
9. 392
10. 088
11. 002
12. 143
13. 655
14. --
15. --
[/hide]

#14 is pretty difficult if you want to prove it, i misinterpreted it when i took the test, but looking at it again conjeccture tells me that they are all radii of the circumcircle, so it is just 6*circumradius, or 340

Oh and thank you very much for the problems! They test was good, maybe a little bit too easy. (Or maybe I just got lots of them wrong :P . But again, thank you for taking the time to make the test![/quote]

I think #4 is wrong (discussed earlier in thread), #6 (solution above), #10 (i need to find the answer to this still, but i checked 88 on my calculator and it doesn't work), #11 (only 2? i don't think so...), #12 (discussed), #13 (pretty sure my answer is right).

#11 is just tedious casework and #13 is cool (see USAMO 2003 #1 for a hint). And for #14, your conjecture is right, try proving it!

EDIT: I think I found the answer to 10, my original one was wrong, but the new one at least works...

[quote="Altheman"]i agree that i did 4 wrong, i counted m=n...

6. wow, i missed that the second one was missing 2 mod 3 powers of x... :blush:

10. i messed this one up, i had 2n=88, or n=44

11. i also messed that up, i found that the sequence could have the lowest perfect squares being 2 numbers, 21^2 and 22^2, but that is not the number of sets...[/quote]

How are people getting $n = 44$ for #10? $\binom{88}{44}$ isn't even divisible by $11$ or $13$...

hm lots of differing answers... could u post the correct answers paladin?

OK, my answers are (could be wrong, but oh well):

1. 368
2. 768
3. 798
4. 321
5. 157
6. 233
7. 679
8. 255
9. 567
10. 131
11. 631 (I might've copied the answers wrong since the answer is the same as for #12... too lazy to solve it again)
12. 631
13. 830
14. 340
15. ??? (err, never solved it... I know it's less than 1000 though...)

[quote="gotztahbeazn"]k does anyone have a solution for #6?
this is what i had:

$P_1(x) \cdot (x-1) = x^{101} - 1$

$P_2(x) \cdot (x+1) = x^{101} + 1$

$P_3(x) \cdot (x-1) = x^{68} - 1$

multiplying these together, you get:

$P_1 \cdot P_2 \cdot P_3 \cdot (x-1)^2(x+1) = (x^{202} - 1)(x^{68}-1)$

since we know $1$ and $-1$ are roots already (this part i wasnt sure on but i read some stuff in AoPS Vol. II and got the idea?), we essentially just have the roots of unity for $202$ and $68$, but since no roots of unity will overlap, you will have $202+68 = 270$ roots....

hmm can i see a real solution to this now?[/quote]

It's up in one of my previous posts. You read $P_1$ incorrectly. The powers $x^2, x^5, \ldots, x^{98}$ are omitted.

[quote="PenguinIntegral"]Does anyone have solutions for the last few?[/quote]

Borderline bad post.  If you're really concerned with the problem, repost it. Revive with care. Let old dogs lie. Don't disturb the bones of the forum.  you know?

Discretion.

We can clean up on the work for 12. All we need is AD=12, BE=9, CE=5.

Then $ [ABE] \equal{} 54$, so $ [CEA] \equal{} 84 \minus{} 54 \equal{} 30$, and $ [CEF] \equal{} 54 \cdot (\frac {5}{4 \cdot \sqrt{10}})^2 \equal{} \frac {135}{16}$, so the answer is $ [CEA] \plus{} [CEF] \equal{} \frac {615}{16}$


Prove that the sum of the primitive roots mod p is congruent to u(p-1) mod p.  u is the mobius function.

unknown source


Solve equation
$ 2(x^2\plus{}1)\equal{}5\sqrt{x^3\plus{}1}$

Clearly there is no soln for x < 0 and by inspection we get that there is no root below 6 as RHS > LHS and for x > 7 as LHS > RHS

So the root must lie between 6 and 7.

How do we find it ?  :maybe:

Who said there's no solution for x < 0?

There are two real solutions: one is approximately -0.8204, the other approximately 5.9383

Courtesy Mathematica, the positive exact root is:
$ \frac {25}{26} + \frac {\sqrt {3}}{48} \sqrt {1619 - 15104 \sqrt [3]{\frac {2}{ - 304967 + 1575 \sqrt {38849}}} + 8 (2^{\frac {2}{3}}) \sqrt [3]{ - 304967 + 1575 \sqrt {38849}}}$
$ + \frac {1}{2} ( \frac {1619}{96} + \frac {236}{3} \sqrt [3]{\frac {2}{ - 304967 + 1575 \sqrt {38849}}} - \frac {4^{\frac {1}{3}}}{24} \sqrt [3]{ - 304967 + 1575 \sqrt {38849}} $ $  + \frac {37275}{32 \sqrt {3} \sqrt {1619 - 15104 \sqrt [3]{\frac {2}{ - 304967 + 1575 \sqrt {38849}}} + 8 (2^{\frac {2}{3}}) \sqrt [3]{ - 304967 + 1575 \sqrt {38849}}}} ) ^{\frac {1}{2}}$
where the everything from the big open parenthesis to the big close parenthesis is raised to the 1/2 power.

Yes, it's an ugly problem, and I won't bother to $ \text{\TeX}$ the negative root.

sorry it should have been no soln for x<-1  :wallbash_red:


Triangles A1A2A3 and B1B2B3 are inscribed in a circle so that
A1A2 - B1B2 # A2A3 - B2B3 . Prove that :
 sinA1 + sinA2 + sinA3 = sinB1 + sinB2 + sinB3 if and only if
(A1A2-B1B2) ^3 +(A2A3-B2B3) ^3 +(A3A1-B3B1) ^3  = 3(A1A2-B1B2)(A2A3-B2B3)(A3A1-B3B1)

Let A1A2=a, A2A3=b, A3A1=c
B1B2=a', B2B3=b', B3B1=c'
by acsinA1=absinA2=bcsinA3=2S
and a'c'sinB1=a'b'sinB2=b'c'sinB3=2S'
sinA1+sinA2+sinA3=sinB1+sinB2+sinB3
<=> 2S(a+b+c)/abc = 2S'(a'+b'+c')/a'b'c'
Since they are inscribed in the same circle, R=R'
so abc/a'b'c'=S/S'
Thus <=>a+b+c=a'+b'+c'

Notice that
(a-a')^3+(b-b')^3+(c-c')^3=3(a-a')(b-b')(c-c')
<=>(a+b+c-a'-b'-c')[(a-a')^2+(b-b')^2+(c-c')^2]=0
<=>a+b+c=a'+b'+c'
this finishes my proof


calculus
integration
function
trigonometry
algebra
polynomial
real analysis

Let $ f$ be a continuous function on $ \left[0,2\pi\right]$ to $ R$ such that $ \int^{2\pi}_{0}\ f\left(x\right)sin\left(nx\right)dx\equal{}0$ for each natural number $ n$, then prove that $ f$ is a zero function

That's fairly badly false. Just let $ f(x)\equal{}1.$ Or $ f(x)\equal{}2\minus{}\cos x\plus{}4\cos 3x.$ Or let $ f$ be any restriction to $ [0,2\pi]$ of a $ 2\pi$-periodic even function.

Answer:
[hide]It is false... $ f(x) \equal{} 1$ satisfies this.  Also $ f(x) \equal{} \cos(x)$.  And many more.[/hide]

it was a question from pugh's real mathematical analysis,the question in the book is like what i wrote,
any other correct formulation of the question?

basil  i had  suggest u  re verify the question again.

If it were functions on $ [0,\pi]$ such that $ \int_0^{\pi}f(x)\sin nx\,dx\equal{}0$ for $ n\equal{}1,2,\dots,$ then it would be true.

Or if it were functions on $ [0,2\pi]$ such that $ \int_0^{2\pi}f(x)e^{\minus{}inx}\,dx\equal{}0$ for all $ n\in \mathbb{Z}.$

Or if it were functions on $ [0,2\pi]$ such that $ \int_0^{2\pi}f(x)\cos nx\,dx\equal{}0$ for $ n\equal{}0,1,2,\dots$ and $ \int_0^{2\pi}f(x)\sin nx\,dx\equal{}0$ for $ n\equal{}1,2,\dots.$

how the last two?
any proof for the first?

We're talking about some of the major, central theorems of Fourier series - in particular, this is the theorem that says that a function is uniquely determined by its Fourier coefficients.

One thing to note: there's no particular gain here from talking about continuous functions $ f$ - we might as well be talking about $ L^2$ functions. This is essentially equivalent to the statement that $ \{\cos nx,\sin nx\}$ is [i]complete[/i] orthogonal system in $ L^2[0,2\pi],$ as is $ \{e^{inx}\}$ for $ n\in \mathbb{Z},$ as is $ \{\sin nx\}$ in $ L^2[0,\pi].$ One of the many things this is equivalent to is that the trigonometric polynomials are dense in $ L^2[0,2\pi]$ and the trigonometric sine polynomials are dense in $ L^2[0,\pi].$ (One of the reasons I shied away from continuous functions is that the trig polynomials are only dense in $ C[0,2\pi]$ for functions for which $ f(0)\equal{}f(2\pi)$ and the trig sine polynomials are only dense in $ C[0,\pi]$ for functions for which $ f(0)\equal{}f(\pi)\equal{}0.$)

I'm not willing to develop the whole theory here and now; try looking it up in some good treatment of Fourier series. Fejér's Theorem might be a name to look for.


[img]http://img23.imageshack.us/img23/1538/43402097.png[/img]

The numbers above and below each square are the measure of their sides... (first square has side $ 11$, second $ 9$... in case you don't understand). Now, how much is the area of the regions in gray minus the area of the regions in black?

[hide]Denote the leftmost white area by $ x$, the middle one by $ y$, the rightmost by $ z$. Then

$ g\equal{}(121\minus{}x)\plus{}(49\minus{}y\minus{}z)$

$ b\equal{}(81\minus{}x\minus{}y)\plus{}(25\minus{}z)$

Hence $ g\minus{}b\equal{}121\plus{}49\minus{}81\minus{}25\equal{}64$[/hide]


geometry
conics
hyperbola
geometric transformation
reflection
angle bisector
geometry unsolved

P and Q are 2 points in the area bounded by 2 rays, e and f, coming out from a point O. Describe how to construct, with a ruler and a compass only, an isosceles triangle ABC, such that his base AB is on the ray e, the point C is on the ray f, P is on AC, and Q on BC.

I think that if you say that "AB is on the ray e, the point C is on the ray f" you must say also that the angle is acute, btw i find a solution using the rectangolar hyperbola.

this was my solution:

lemma : Let a triangle ABC, then the locus of the isogonal conjugate of the perpendicolar axis of BC is a rectangolar hyperbola with center the midpoint of BC and asimptote the internal and external angle bisector of the medial triangle from the midpoint of BC.

let H and K the projection of P and Q on E. So now apply the lemma to the degenetete triangle formed by the sideline PQ, PH and QK we get that the locus of the vertex of all isosceles triangle with base on e and P on a side and Q on the other is a rectangolar hyperbola with center the midpoint of PQ and asimptote perpendicolar and parallel to e. Then C stay of the intersection of this hyperbola with f.

This is a costruction with ruler and a compass of the intersection of a conic C with a line L:

Let F be a focus, l a directrix, and e = the eccentricity.
(1) Let  $ H \equal{} L \cap l$
(2) Take an arbitrary point P with pedal Q on the directrix.
(3) Construct a circle, center P, radius e · PQ.
(4) Through P construct the parallel to L, intersecting the directrix at O.
(5) Through O construct the parallel to FH, intersecting the circle above in X and Y .
(6) The parallels through F to PX and PY intersect the given line L at two points on the conic.

Construct Q', the reflection of Q in ray f.  Then If $\alpha$ is the angle between e and f, we have that $\angle PCQ' = \angle PCO + \angle OCQ' = \angle PCO +\angle OCQ = \angle QCP + 2 \angle PCO = \angle BCA + 2 \angle ACO= \angle BCA + 2 (\angle BAC - \alpha) = 180 - 2 \alpha.$

So to find C, draw the circular arc PQ' (away from O) with angle $180 - 2 \alpha$, and C is the intersection of that arc with f.


We know that $ a^n\equal{}a\cdot a\cdots a$ ntimes.
Also $ a^{m/n}\equal{}\sqrt[n]{a^m}$ since $ (a^{m/n})^n\equal{}a^m$.
What is the meaning of $ a^{\sqrt 2}$.


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

Let f be homomorphism from a nilpotent group G to a group G'. Prove that if the restriction of f to the center of G is injective, then f is injective.

$ f$ is injective, otherwise the kernel of the homorphism would be a non-trivial normal subgroup giving a non-trivial intersection with the center, a contradiction to the assumption that the restriction to the center is injective

This is a property of nilpotent groups.

For a proof let $ G$ be a minimal counterexample with regard to its nilpotency length $ d(G)$.

Since the assertion is true for abelian groups, we must have $ d(G)>1$. Choose a normal subgroup $ N$ of $ G$, which has trivial intersection with the center $ Z(G)$. This yields a new normal subgroup $ M$ of $ G$ defined as the direct product of $ N$ with $ Z(G)$. By induction $ M/Z(G)$ has non-trivial intersection with $ Z(G/Z(G))\equal{}Z_{2}(G)/Z(G)$, i.e. $ \exists 1\not \equal{}x\in N,z\in Z(G):\,xz\in Z_{2}(G)\Longrightarrow x\in Z_{2}(G)$. For arbitrary $ g\in G$ we compute for the commutator: $ [x,g]\equal{}x^{\minus{}1}x^{y}\in N\cap Z(G)\equal{}1$ and hence $ x\in Z(G)$, a contradiction.

A finite nilpotent group is direct product of its sylowgroups. This also holds for the kernel. In p-groups every normal nontrivial subgroup intersects nontrivially with the center.

pls note, there was no restriction to finite groups


function
symmetry
algebra proposed
algebra

Hello!!

find all functions $f:{\Bbb R}\to{\Bbb R}$ such that $f(1)=1$ and \[f(x+y)=3^{y}f(x)+2^{x}f(y)\] for all real numbers $x$ and $y$.

[quote="gauss2"]very easy

f(x)=2^x-3^x[/quote]

but $f(1)\neq 1$   :P  , and one has to find ALL the functions satisfying the given property.

your solution is wrong! if you do correct calculations you get no contradiction :roll:
one can show that the only solution is $f(x)=c(3^{x}-2^{x})$ and $c=1$ in our case :wink:

Hello, I hope that I didn't overlook any link to a solution, so I post mine.
And since I don't know how to post it on a link, I write it down here:

$f(x+y) = 3^{y}f(x)+2^{x}f(y)$

$y=0$ gives $f(0)=0$ as we have seen before. With the symmetry on the left side of the equation one gets $3^{y}f(x)+2^{x}f(y)=3^{x}f(y)+2^{y}f(x) \Leftrightarrow (3^{y}-2^{y})f(x) = (3^{x}-2^{x})f(y)$
Let $y=1$, then $f(x) = 3^{x}-2^{x}$. So it is the only possible solution. Inserting in the equation gives that it is really a solution. q.e.d.


limit
Asymptote
function
real analysis
real analysis theorems

If $ \lim_{x \to 1^ \minus{} } \frac {f(x) \minus{} f(1)}{x \minus{} 1} \equal{} \minus{} \infty$

Do we have: The tangent line at $ 1$ is "Up"??

Is there a rule??

Thanks  :)

[quote="mlok"][url=http://en.wikipedia.org/wiki/Asymptote]Vertical asymptote[/url] - can't touch this.[/quote]

Sorry, Ill edit it  :lol:

Do you see now what I mean?

Most people don't think of tangents as oriented lines. So my answer is "Who cares?"

But knowing how the tangeant line is, is important to draw The graph of a function  :wink:

Is there anything wich is not clear in my question?

 :maybe:  :maybe:

Please help
 :D

$ \|\begin{array}{c}
......\ \mathrm {vertical\ asymptotes\ in\ }x_0\ .....................\\
 \\
f'_l(x_0) = \|\begin{array}{cccc} \infty & \mathrm {means} & \diagup\ \boxed {\blacksquare} & \mathrm {increasing} \\
 \\
- \infty & \mathrm {means} & \diagdown\ \boxed {\blacksquare} & \mathrm {decreasing}\end{array} \\
 \\
f'_r(x_0) = \|\begin{array}{cccc} \infty & \mathrm {means} & \boxed {\blacksquare}\ \diagup & \mathrm {increasing} \\
 \\
- \infty & \mathrm {means} & \boxed {\blacksquare}\ \diagdown & \mathrm {decreasing}\end{array}\end{array}$

l = left ; r = right .

Ok, thank you very much Virgil !

 :)  :wink:


Denote $f(x)=(m-1)x^2-2(m+1)x+(m+1),\ x\in R$. Prove that:
$\boxed {\ \left(\forall\right)\ x\in [-1,1],\ |f(x)|\le 4\Longleftrightarrow |f(-1)|\le 4\ }$

f(-1)=m-1+2m+2+m+1=4m+2 so m=(f(-1)-2)/4 then, maximiun value of f ,is when
(2m-2)x-2(m+1)=0 so therefore (m+1)/(m-1)=x. replacing you can get the solution  :D


The surface area of a cylindrical aluminum can is measure of how much aluminum the can requires. If the can has a radius r and a height h, its surface area A and its volume V are given by the equations:

A=2(pi)r^2 + 2(pi)rh   and V= (pi)r^2h

A) The volume, V, of a 12 oz cola can is 355cm^3. A cola can is approximately cylindrical. Express the cola can's surface area A as a function of its radius r, where r is measured in centimeters. Simlify your answer. (Hint: Your function should only contain two variables, A and r).

B) The manufacturers wish to use the least amount of aluminum (in centimeters squared) necessary to make hte 12 oz cola can. Use your answer in part A to find the minimum amount of aluminum needed. State the values of r and h that minimize the amount of aluminum used.

[quote="golfmaster925"]$ A=2\pi r^{2} + 2\pi rh$   and $ V= \pi r^{2}h$[/quote]

[hide="A"]
Since $ V=355\text{cm}^{3}$ then $ 355=\pi r^{3}h \Rightarrow h=\frac{355}{\pi r^{2}}$.

Then we plug $ h$ into the Area equation and get $ \boxed{A=2\pi r^{2}+\frac{710}{r}}$
[/hide]

[hide="B"]
${ A'=4\pi r -\frac{710}{r^{2}} \Rightarrow 0=2\pi r-\frac{355}{r^2}} \Rightarrow 0=2\pi r^{3}-355 \Rightarrow r=+\sqrt[3]{\frac{355}{2\pi}} \Rightarrow \boxed{r=3.84\text{cm}}$

Take this value of $ r$ and plug it into the Volume equation which is set equal to 355. $ 355=\pi\sqrt[3]{\frac{355}{2\pi}}h \Rightarrow h=\frac{355}{\pi}\cdot\sqrt[3]{\frac{2\pi}{355}} \Rightarrow \boxed{h=29.45\text{cm}}$
[/hide]

[quote="golfmaster925"]Thanks for your help. Can you explain the process of why you did what you did. Thanks.[/quote]
Your Welcome!  :lol:

The process for A was a simple plug, solve for variable, and substitute. The reason you solve for $ h$ instead of of $ r$ is because you want the Area equation to be expressed in terms of $ r$ so you want to remove $ h$.

B was a little more complicated. I took the derivative of my first answer in order to find the relative minima and maxima (of which there was no maxima) to find the lowest point in the graph of $ A$. That would be the radius required to obtain the smallest possible $ A$. Taking the exact value of the radius into the equation with only two variables ($ r,h$) allowed me to solve for the corresponding $ h$ to that $ r$.


MATHCOUNTS
Ring Theory
geometry
3D geometry

What did you get?

I got (as of now)

CHAPTER ROUND:
MATHCOUNTS slinky
MATHCOUNTS hand-powered flashlight
3 MATHCOUNTS pencils (non-mechanical)
MATHCOUNTS lunchbox
MATHCOUNTS trophy (4th place)

STATE ROUND:
MATHCOUNTS mechanical pencil
MATHCOUNTS puzzle pen
MATHCOUNTS t-shirt
2 generic pencils (non-mechanical)
3 MATHCOUNTS trophies (2nd place indiv, 2nd place team member, top 7th grader)
1 book (on Fantastical Engineering - theme-park)

i got 4 MC shirts, lol, i was last one to get the shirts and they had left overs =]

I got

CHAPTER:
3 trophies (1st written, 1st team, 2nd cd)

STATE:
2 trophies (4th written, 1st cd)
1 book

What??? Mathcounts gives other free stuff at chapter and states? Omg, I only got trophies and nothing else :(

james4l is in a very competitive chapter/state that gives out prizes.

Last year, Sprint was sponsoring MATHCOUNTS (chapter) so we got a cheap plastic MATHCOUNTS glass with another cheap MATHCOUNTS bracelet.  I know Darryl Wu got a laptop from MATHCOUNTS.

Chapter:

T-shirt
2 Trophies (2nd individual, 3rd team)

State:

T-shirt
2 Trophies (7th individual, 1st countdown)

do we get calculators at nationals james4l (or anybody else who knows the answ)?
if we do, what type of calcs?

YOU GUYS ARE SOOOOOOO LUCKY!!! OMG... You guys get to have ti-inspire ca+!!!!! NOOOOOOOOOOOOOOO!!!!!! Lucky lucky lucky lucky!!! No No No!!! If you google it, then you'll see the awesomeness of it! I heard that you could download movies and songs on it!! NOOOOOOOOOOOOOOO  :noo:  :noo:  :noo:

NC is a really poor state i think
all i got is...

CHAPTERS
2 Trophies (9th individual, 2th team)
certificate for participating in Chapters
1 "medium"-quality T-Shirt
some snacks during break (which i rejected - since they were pure junk food)
2 tiny slices of pizza
a bottle of water (which disappeared during the countdown round -- unopened)
invitation to State Competition

STATE
3 Trophies (3th Countdown, 2nd Individual, 2nd Team -- all smeared already)
certificate for participating in States
1 "poor"-quality T-Shirt
some snacks during break (which i rejected AGAIN)
low-quality buffet lunch
a bottle of water (which luckily didnt disappear)
three MAHTCOUNTS pencils (which i found out too late that wasn't supposed to take those)
invitation to National Competition

I hope I win something more during Nats... :)

Chou, arnt u taking "not making nats" a little too far. lol. Plus, u have next year to do well. Plus plus, it's not considered "lucky," it's considered "deserved."

Haha, i didn't make nats either :(
But idc, i have 4 yrs of AMC left soo... :)[/quote]

This is an abbreviated list, I'm too lazy to list everything  :D
It seems most other chapters and states give more indiv awards and free stuff, probably because it's harder to get the stuff, my state is tiny, i basically have:

Chapter:
7 trophies (3 team (2nd, 1st, 1st), 1 cd (1st), honor mention, 2 indiv (1st, 1st))
State:
9 trophies (3 team (3rd, 2nd, 3rd), 3 indiv (4th, 2nd, 1st), 3 cd (1st, 1st, 1st))
No other stuff (unless you count pencils/meals/etc)
Nats:
3 plaques
yearbooks
certificates
big plaque for going 3 yrs
2 84's, an nspire
Microsoft Excel 2007
3 hats (one cowboy)
lots of pins
pencils, pads
lots of other sponsor materials I've forgotten/misplaced?

chapter
3 trophies
FOOD
100+1000=1100 in scholarships

state
3 tshirts
3 trophies and a medal
3 name tags(lulz :P)
1 reading light which i lost
about a billion pencils(most of which i lost)
MORE FOOD
500+some college scholarship to be determined

Omg, I feel really jealous of you guys.

I have nothing but trophies and some participation medals from states :(

Then how many individual and team trophies are awarded? Do they also give Countdown trophies to all the Countdown participants?

Chapter:
MC shirt
MC Medal(2nd team)
2 trophies(1st individual, 4th countdown)
3 MC pencils (non-mechanical)

State:
3 pencils (generic, non-mechanical)
MC medal(countdown quarterfinalist)
2 trophies (5th team, 4th individual)
Information booklet for Nats (It's free stuff)

My team was really bad... I'm in elementary school, do I had to gather some people that weren't very good... Their school scores were 12, 14, and 16  :( But at least my individual score and our team round score got 5th at state.

Wow... an elementary student made Nats... I didnt know they were allowed to take Mathcounts.

Hmmm, I think elementary students aren't allowed to participate officially...

Yes, they are.  8th grade and under can participate officially.  However, you can only participate officially 3 times, so if you participate in 5th, 6th, and 7th, you cannot participate in 8th.

well, most elementary schools are k-5, but ours is k6. MathCOUNTS is a middle school competition, and most middle schools are 6-8, so I'm allowed to participate. I can also do the AMC's and I made to AIME. I can't discuss anything else though... :whistling:

[quote="gauss1181"][quote="Ihatepie"]Remember I only did Mc for one year in 8th grade:

Nationals:
Plaque
19th Place Team Trophy
76th Place Individual Trophy
N/A Countdown Trophy
Ti-NSpire
Microsoft Excel 2008
Mathcounts Yearbook
Mathcounts Hat
Some other random assorted things
And best of all....[hide]a box![/hide]

.[/quote]

Huh? At Nats, they give individual/team trophies even for those who didn't place in the top 25% individuals or top 40% at teams??? :maybe:[/quote]Nope. As you probably realized, most of the post from there on was pretty much a joke. They only give the top (3?) teams trophies, and the top 12 individuals. They also don't give a N/A Countdown Trophy  :P

Are we getting Ti Nspires in Nats this year?
 :ninja:

The MC website https://mathcounts.org/Page.aspx?pid=287 talks about eligible students...

Eligible Participants

Students enrolled in the 6th, 7th or 8th grade are eligible to participate in MATHCOUNTS® competitions. Students taking middle school mathematics classes who are not full‑time 6th, 7th or 8th graders are not eligible. Participation in MATHCOUNTS competitions is limited to three years for each student though there is no limit to the number of years a student may participate in the school‑based coaching phase.

[quote="james4l"]Too bad. I really don't like TI-nspires (plus the fact that I already have one). Instead I would rather have a TI-89.[/quote]

My TI-89 Titanium can bash almost all Target and Team round problems in ~30 seconds, which includes typing in. :D

[quote="connaissance"]I explained that my elementary school is K-6, albeit badly, remy1140!! I'm in sixth grade![/quote][/quote]

Yeah, I'm just making it sure, since there were some discussions about elligibility.

GL at nats everyone!

Hm.. They should make MC Rubik's Cubes!!  :blush:  :lol:


calculus
integration
logarithms
trigonometry
calculus computations

[b][u]without[/u][/b] reference to sums of [i]any [/i]known series, solve the integral... :D

$ \int_{0}^{1}\;\;\frac{1}{2x}\;\ln\left(\frac{2+2x}{2-2x}\right)\;\;\textbf dx$

[ PS: answer is hidden below... but don't look quite yet... :P ]

[hide]
$ \text{answer}: \;\;\;\quad \boxed{\frac{\pi^{2}}{8}}$
[/hide]

[quote="pardesi"]is there 2 both on the numerator and denominator like that  :o[/quote]

yes...  :huh: but why is that upsetting you ?

Misan, why did you wrote that and not $ \int^{1}_{0}\frac{1}{2x}\ln\left(\frac{1+x}{1-x}\right)\,dx$ ?

Anyway:

$ I = \frac{1}{2}\int^{1}_{0}\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)\,dx = \frac{1}{2}\left(\left[\ln x \ln\left(\frac{1+x}{1-x}\right)\right]^{1}_{0}-2\int^{1}_{0}\frac{\ln x}{1-x^{2}}\,dx\right)$

$ =-\int^{1}_{0}\frac{\ln x}{1-x^{2}}\,dx =-\int^{\pi/2}_{0}\frac{\ln(\sin t)}{\cos t}\,dt$.

Now I don't see how to solve this last integral: perhaps someone can provide an hint.

I'm not sure if this could exactly help, but..

If we we're to express the integral in a symmetric form as below:

$ -2I = \int^{\pi/2}_{0}\frac{ \sin x (\ln (\sin x))+\cos x (\ln (\cos x))}{\sin x+\cos x}$

And then use symetrry properties to change it into integrals with upper bound pi/4, which is less than 1, could we expand the taylor series of the natural log and then integrate term by term? Long shot lol

Correction and addition in the third block of my last post.
Read $ -\int_{0}^{1}\sum_{k=0}^{\infty}x^{2k}ln(x)dx=-\sum_{k=0}^{\infty}\int_{0}^{1}x^{2k}ln(x)dx$ (because the last series is absolutely convergent).


Please help me prove these identities! I really need help on this!

1. Prove the identity:
1-sina            cos^3a
--------   =     --------
seca              1+sina

2.prove the identity:
cos(a+b)
-------------  =  cosa-tanb
cos(a)sin(b)

I have no idea how to type problems like this on here, so sorry if it's confusing! I had to use a and b since I had no idea how to do the angle symbols. (theta and the other one...)
thank you for help!!!

[hide="1"]
Prove $\frac{1-\sin{\theta}}{\sec{\theta}}=\frac{\cos^{3}{\theta}}{1+\sin{\theta}}$.

$\implies (1-\sin{\theta})(1+\sin{\theta})=\cos^{2}{\theta}$

$\implies 1-\sin^{2}{\theta}=\cos^{2}{\theta}$,

which is true by the Pythagorean Identity.[/hide]

thanks!
I had a bit of a hard time understanding that though... anyone else?

$cos(a+b) = cosacosb-sinasinb$
$\implies \frac{cos(a+b)}{cosasinb}= cotb-tana$

Something different, right? Are you sure that you're right?

I seem to agree with bibobeo:

$\frac{cos(a+b)}{cos(a)cos(b)}= cot(b)-tan(a)$

[quote="Dagny.Taggart"]Please help me prove these identities! I really need help on this!

1. Prove the identity:
1-sina            cos^3a
--------   =     --------
seca              1+sina

2.prove the identity:
cos(a+b)
-------------  =  cosa-tanb
cos(a)sin(b)
[/quote]

Here ya go   :) :
[code]
$(1) \ \ \frac{1 - \sin(a)}{\sec(a)} = \frac{ \cos^3 (a)}{1+ \sin(a)}$

$(2) \ \ \frac{\cos(a+b)}{\cos(a) \cdot \sin(b)}= \cos(a)-\tan(b)$[/code]

$(1) \ \ \frac{1-\sin(a)}{\sec(a)}= \frac{ \cos^{3}(a)}{1+\sin(a)}$

$(2) \ \ \frac{\cos(a+b)}{\cos(a) \cdot \sin(b)}= \cos(a)-\tan(b)$

#1 has been proven,
[hide="here is #2"]
$\frac{\cos(a+b)}{\cos(a) \cdot \sin(b)}= \frac{ \cos(a) \cos(b)-\sin(a) \sin(b)}{ \cos a \sin b}= \frac{ \cos b}{\sin b}-\frac{ \sin a}{\cos b}= \boxed{ \cot b-\tan a}$

Did you type in wrong the right-hand side of the question?[/hide]


probability
geometry
rectangle
analytic geometry
linear algebra
matrix
MATHCOUNTS

A rectangle XYZW has XY = 6, and XW = 8.  If a point M is selected randomly in the interior of this rectangle, what is the probability that triangle XMW has an area equal to or greater than 16 square units?  Express your answer as a common fraction.

Give this rectangle with coordinates X=(0,0); Y=(6,0); W=(0,8); Z=(6,8); M=(a,b). Then the area of the triangle XMW is

1     |0,0,1|
-  * |0,8,1|
2     |a,b,1|

Which is 4a. If you want the area to be greater than or equal to 16, a has to be greater than or equal to 4, so there is a 1/3 chance.

I think I did that right...

[hide]If the are has to be greater than or equal to 16, then the height has to be greater than or equal to 4.

8*4=32
48-32=16
Therefore the probability is 16/48, or 1/3.[/hide]

[hide]Draw the rectangle. You'll notice that the triangle has to have a height of at least 4, and the rectangle has a total length 6. Thus, the triangle has height 4-6. (6-4)/6=1/3

Answer: 1/3[/hide]

[quote="GoBraves"][quote="bob123"][hide]M must be on or above y=2, 6-2/6=4/6=2/3[/hide][/quote]

The area of a triangle is actually 1/2bh, not bh (Im sure you know this, I forget this a lot too) . So M must be on or above 4.[/quote]

Woops heh heh

[hide]
$A\ge(6)\bigl eft\frac{1}{2}\bigr ight(D)$ where D is the distance from M to XW
SO,

$16\ge3D\\
\\
\frac{16}{3}\ge D$

$8\ge\frac{24}{3}$
So $\frac{D}{8}\ge\frac{\frac{16}{8}}{\frac{24}{8}}=\frac{2}{3}$
Finally,
$1-\frac{2}{3}=\frac{1}{3}$
[/hide]

About the matrix method: It works, but I think it is too slow for a mathcounts problem, especially one this easy.

[hide]Self confusion on this one.  Took me too long to figure it out.  The answer is 1/3.[/hide]


The ratio of boys to girls in Mr. Brown's math class is $ 2: 3$.  If there are $ 30$ students in the class, how many more girls than boys are in the class?

\[ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 10
\]

hello
no. of girls=$ 18$ whereas boys=$ 12$
thus the answer=$ 6$
thank u

Wow so this was problem 16 on the 1985 AMC 8. Feels like there was some misplacement going on because 15 was so much harder. :O

But yeah a shortcut for these problems is just to do $30/(2+3) = 6$.


trigonometry
combinatorics unsolved
combinatorics

Let $ f \, : \, X \rightarrow Y$ and let $ C, B \subset Y$. Prove that $ f^{ \minus{} 1} (C \cap D) \equal{} f^{ \minus{} 1}(C) \cap f^{ \minus{} 1}(D)$. Let $ A,B \subset X$. Must $ f(A \cap B) \equal{} f(A) \cap f(B)$ be true?

No, it must not. Take $ f(x) \equal{}\sin x$, $ A\equal{}[0,100\pi]$, $ B\equal{}[333\pi,666\pi]$.

Okay, how about $ f^{\minus{}1}(C\cap D) \equal{} f^{\minus{}1}(C)\cap f^{\minus{}1}(D)$?

$ f(A \cap B) \subseteq f(A) \cap f(B)$ always holds. $ f$ injective guarantees equality.

[quote="zhoraster"]This is really easy. If $ f(x)\in C$ and $ f(x)\in D$, then $ f(x)\in C\cap D$. And vice versa.[/quote]

I did something like that, so that's okay. Thanks!


Hi, all!
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Can a mod delete the advertisement, please?

Why not? Probably, someone'll bw interested in this product, in fact it's a help for the imagination. And now, there is a product for Maple, is available now!
http://www.maplesoft.com/products/thirdparty/nvizx/index.aspx


linear algebra
matrix
quadratics
algebra
polynomial
Ring Theory
superior algebra

Prove that there is no ring epimorphism $ f: \mathcal{M}_2(\mathbb{R})\to\mathcal{M}_3(\mathbb{R})$, where $ \mathbb{R}$ is the field of real numbers.

Now try this version:

Given a field $ k$ and two positive integers $ m,n$, there is a non-zero (unitary) ring homomorphism $ \mathcal{M}_n(k)\to\mathcal{M}_m(k)$ iff $ n$ divides $ m$.

(note that the homomorphism is no longer required to be onto)

[quote="LydianRain"]Every element on the left side satisfies some quadratic polynomial, but not every element on the right does.[/quote]
Actually a ring homo is not necessarily $ \mathbb{R}$-linear, so this argument doesn't work.
(you also need $ f(\mathbb{R} I_2)\subseteq  \mathbb{R}I_3$, which follows by  $ \mathbb{R}I_3\equal{}Z(\mathcal{M}_3(\mathbb{R}))$ and $ f$ epi)

Now assume we have a ring homo $ f: \mathcal{M}_n(k)\to\mathcal{M}_m(k)$ with $ f(I_n) \equal{} I_m$.
For each $ 1\le i,j\le n$ define $ e_{ij}\in \mathcal{M}_n(k)$ with $ (i,j)$-entry $ 1$ and $ 0$ otherwise and set $ f_{ij}: \equal{} f(e_{ij})$.
Then $ e_{ij}e_{kl} \equal{} \delta_{jk}e_{il}$ gives also $ f_{ij}f_{kl} \equal{} \delta_{jk}f_{il}$ for all $ 1\le i,j,k,l\le n$.
Now $ 0 \equal{} f_{ii}^2 \minus{} f_{ii} \equal{} f_{ii}(f_{ii} \minus{} I_m)$ gives $ f_{ii}$ diagonalizable with only eigenvalues $ 0,1$ for all $ 1\le i\le n$.
Then $ f_{ii}f_{jj} \equal{} 0 \equal{} f_{jj}f_{ii}$ for all $ i\not \equal{} j$ gives $ f_{11},f_{22},\ldots,f_{nn}$ simultaneously diagonalizable,
and after composing $ f$ with some conjugation in $ \mathcal{M}_m(k)$,
we may assume $ f_{11},f_{22},\ldots,f_{nn}$ are diagonal matrices with only entries $ 0,1$.

For each $ 1\le r\le m$ we take $ E_r\in\mathcal{M}_m(k)$ with $ (r,r)$-entry $ 1$ and $ 0$ otherwise.
Then for each $ 1\le i\le n$ we find some subset $ J_i\subseteq \{1,2,\ldots,m\}$ with $ f_{ii} \equal{} \sum_{r\in J_i}E_r$.
Then $ f_{ii}f_{jj} \equal{} 0$ gives $ J_i\cap J_j \equal{} \emptyset$ for all $ i\not \equal{} j$ and
$ I_n \equal{} e_{11} \plus{} e_{22} \plus{} \cdots \plus{} e_{nn}$ gives $ I_m \equal{} f(I_n) \equal{} f_{11} \plus{} f_{22} \plus{} \cdots \plus{} f_{nn}$,
which gives $ \{1,2,\ldots, m\} \equal{} \bigcup_{i \equal{} 1}^n J_i$ and $ m \equal{} |J_1| \plus{} |J_2| \plus{} \cdots \plus{} |J_n|$.

But $ f_{ii} \equal{} f_{ij}f_{ji}$ and $ f_{ij} \equal{} f_{ij}f_{jj}$ gives $ |J_i| \equal{} \text{rk} f_{ii}\le \text{rk} f_{ij}\le \text{rk} f_{jj} \equal{} |J_j|$ for all $ i,j$,
so $ |J_1| \equal{} |J_2| \equal{} \cdots \equal{} |J_n|$ and $ m \equal{} n|J_1|$ and $ n|m$.

And now if $ f: \mathcal{M}_2(\mathbb{R})\to\mathcal{M}_3(\mathbb{R})$ is a ring epi, then for some $ A\in \mathcal{M}_2(\mathbb{R})$
we get $ I_m \equal{} f(A) \equal{} f(A\cdot I_n) \equal{} f(A)f(I_n) \equal{} I_mf(I_n) \equal{} f(I_n)$.
This gives the contradiction $ 2|3$.


parameterization
probability
probability and stats

A teacher asks her class, "How many children are there in your family, including yourself?" The mean response is 3 children. How close should this be to the cencus figure for the average number of children per family? Explain your answer.

Not really sure how to explain this correctly. I was thinking about something along the lines of...

A classroom is very small sample size compared to the census figure of the USA, meaning that it will probably be vulnerable to a lot of variance. One would say that there average children per household probably is between 1-4, so we can't be that much off.

What do you guys all think?

You'd also have to consider possible biases. For example, historically, poor people had more kids then rich people. If you take a private school which is presumably attended by rich people, you're numbers might be skewed.

Suppose that there are two types of families: those with three children and those with two children.  A classroom, filled with a representative sample of the child population (and you two are both going wrong in thinking that this assumption is what you should be thinking about), is polled, and the mean number of children in the family of the students is computed.  How does this number compare to the mean number of children in a uniformly selected family from the nation as a whole?  (Further hint: the important idea here remains valid even in the case when every single child in the nation is in this classroom.)

This funny statistical effect is exploited by colleges, which report the mean over classes of class size, not the mean over students of class size, thereby arriving at a number that does not represent the student experience.

Okay, so I understand the response by 0714446459923. Not only is sample size a problem, as I had indicated, but also the possibility that the particular sample could be exposed to major bias as school classrooms vary immensely (one such of variance due to wealth).

I do not quite understand your point Joel. Could you further iterate?

[quote="ProblemSets"] but also the possibility that the particular sample could be exposed to major bias as school classrooms vary immensely (one such of variance due to wealth).[/quote]
You're still missing the point. Yes, there's bias here, but it's not the bias you're thinking of. The bias is present in the very act of sampling, and would still be there inside the most homogeneous and egalitarian community that you could imagine.

[hide="Click only if you have given up completely"]Let's do it this way: there are 50 families in the community, of which 30 of them have 2 school-age children (we ignore all others completely) and 20 of them have 3 school-age children. That's 120 children, and 120 children divided by 50 families is 2.4 children per family.

Now we go to the school (lumping all the grades together) and ask how many (school-age) children are in each child's family. We find 120 children. Of those, 60 are from 2-child families and 60 are from 3-child families. If we average that number over the 120 children, we find an average of 2.5 - which doesn't match the population average of 2.4.
[/hide]

By the way, my riff on the same family of ideas is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=326508]here.[/url]

A specific computational variant of this: suppose in the population, the number of children in a household is given by a Poisson distribution with mean $ \lambda.$ Note that this includes a considerable number ($ e^{\minus{}\lambda}$ to be specific) households with no children. What then will be the average number of children averaged over a class, and how will that number be distributed within the class?

I claim this as an answer to my previous question, with the Poisson distribution. The population per-family average of children per family is $ \lambda.$ The classroom-weighted average number of children per family is $ \lambda\plus{}1.$

Can anyone verify this?

A plausibility check: suppose $ \lambda$ is very small, something like $ 0.01.$ Then most families are childless, and almost all of the families that do have children have one. The school population is entirely taken from those families with children, so the classroom-weighted average has to be at least $ 1.$ Since there do exist, however rare, families with more than one child, that classroom average will be slightly greater than $ 1.$ Having it be $ 1.01$ is plausible.

So, in a Poisson distribution with parameter $ \lambda$, this means that $ e^{-\lambda}\frac{\lambda^n}{n!}$ of all families have $ n$ children, right?  In this case, the average family has
\[ e^{-\lambda} \sum_{n \geq 0} n \cdot \frac{\lambda^n}{n!} = \lambda e^{-\lambda} \sum_{n \geq 0} \frac{\lambda^n}{n!} = \lambda\]
children, which is as it should be.  On the other hand, suppose there are $ F$ families and $ F \lambda$ children in total.  Then there are $ n \cdot e^{-\lambda}\frac{\lambda^n}{n!} F$ children in $ n$-child families and so the average child has
\begin{align*}
\frac{1}{\lambda}\sum_{n \geq 1} n^2 \cdot e^{-\lambda}\frac{\lambda^n}{n!} & = e^{-\lambda} \sum_{n \geq 0} (n + 1) \frac{\lambda^n}{n!} \\
& =  e^{-\lambda} \left(\sum_{n \geq 0} \frac{\lambda^{n + 1}}{n!} + \sum_{n \geq 0} \frac{\lambda^{n}}{n!} \right) \\
& = e^{-\lambda}\left((\lambda + 1)e^\lambda\right) \\
& = \lambda + 1
\end{align*}
children in its family, as Kent claimed.

Is this a reasonable model for the true distribution, do you think?

JBL, i don't really understand your model...

What i got:
Suppose that there are n families and the number of children in each is Poisson($ \lambda$): $ X_{1}$, ... $ X_{n}$

We have 2 new random variables:
a) We choose at random one family and we measure the number of children in this family, $ Y$
b) We choose at random one child and we measure the number of children in the family of this child, $ Z$
(if there are no children then obviously $ Z\equal{}0$ - we can't even choose at random)

Then:
\[ E(Y)\equal{}\lambda\]
\[ E(Z)\equal{}\left(1\minus{}\frac{1}{n}\right)\left(1\minus{}\exp(\minus{}n\lambda)\right)\plus{}\lambda\]

And if $ n\to\infty$ then $ E(Z)\equal{}1\plus{}E(Y)$.

I don't really speak probability and the question in my first sentence was a genuine question, i.e., I wasn't sure my description of the Poisson model was right.  Then my final question was about whether this model is a realistic one for the number of children in families (again, because I don't really know anything about it).


I have found a mistake made by the mod in the game
In #606
[quote="DarkKnight"][color=orange]Accused: G-Unit[/color]
[color=red]Aye (8): Sly Si, Peter VDD, mark, pianoforte, jli, mithsapprentice, samath, supernova, falconwing64[/color]
[color=green]Nay (3): LadyKn1ght, jmadsen, mwhahahaha[/color]

[color=blue]No Lynch (1): blueywinkle [/color]

[b]Edit: 1 more vote and he's dead[/b][/quote]
There have been 9 votes of aye and G-unit was lynched, but the mod has counted 8.
I think there is no problem now as g-unit has been lynched.
Just to remain the mod not to make a similar mistake.

Things of this nature would be best presented by PM to one of the game moderators or myself. Thank you for the concern, however. Locked.


calculus
integration
trigonometry
calculus computations

Evaluate the Integral

$\int\frac{x}{(1+x^{2})\arctan5x}\,dx$

i was thinking let $5x=\tan(\theta)$ or $x=\tan(\theta)$
but neither of those seem to help.


search
MATHCOUNTS
AMC
AMC 8
FTW
AMC 10
AIME

I have nothing to do this summer so I decided to prepare for MathCounts. What is the best way to study? Do you just do problems? Or do you study concepts? How do you make yourself faster?

Maybe my advice doesn't mean much (after all, I pretty much FAILED states), but you should do test after test after test ,assuming you want to get to nationals, not just make states. Soon you'll get the idea of what most Mathcounts problems are and what type of concepts they cover. Then you study those concepts. I found that AOPS volume 1 was a really helpful tool. You can also white out the answers on AMC 8's and use those. One last note: the 2008 mathcounts state was really really REALLY hard, but the 09 state was really really REALLY easy (yet I still failed). What Im getting at is that dont be so nervous about a test that you make stupid mistakes. That what a lot of people in my state did. They ended up placing much worse because they thought the test was going to be a lot harder than it actually was.

Also, if you seriously want to make it far in mathcounts, you really cant practice half-heartedly. Put in all your effort or none. I heard of some people who studied 6 hours a day for nationals! (that might be overkill, though)

[quote="Thunder365"]
Also, if you seriously want to make it far in mathcounts, you really cant practice half-heartedly. Put in all your effort or none. I heard of some people who studied 6 hours a day for nationals! (that might be overkill, though)[/quote]

12 for me.

Really? On average? :wink: You might want to give some reference to what 12 hours means...like school takes up at least 8 hours per day of school. 4 hours of sleep at most?

If you're really really good you can probably get an average of 3 hours per day for one year (summer, winter, and spring breaks, and weekends, pretty much).

12 hours a day on weekends leading up to the competition.

study a lot, do old MC tests, and make sure to correct any wrong answers you made. the third thing was the main mistake i made while studying for states, which i failed. also, dont go overkill with the tests, i learned the hard way that cramming 4 tests in one day is not a good idea for me :wink: . good luck with MC!

@ mewtwo55555 and math154...

I [i]REALLY[/i] need to know how you guys studied for Mathcounts, since you both made CD at nats.

Um well from 5th grade and back, I didn't really have too much of a competitive or math environment in College Station, Texas other than a few random competitions (MOEMS, Mathfax, MathLeague, and the UIL competitions)...but I did get quite a few basic math tricks from those. Then I moved here to Missouri, got really lucky, and landed in Ladue Middle School. I was still pretty horrible, but there was a relatively competitive environment here, and in particular, Mewto55555 and tinytim motivated me a lot. Still though, I wasted a lot of time on my calculator in 6th grade, which made me okay at programming; I also learned some math from MATHCOUNTS and local competition materials, but not too much (for reference, I got a 15 AMC8, ~30 school and chapter MATHCOUNTS, placing 19th at chapter, 40 MathLeague, and I failed in most of the regional competitions (mainly MCTM)). After failing compared to a lot of people, I did a lot of math from these Chinese textbooks in the summer, which helped but was kind of inefficient since I had to ask my parents to translate some of it...fortunately, a friend of my parents told them about AoPS around winter break 2008, so I read through AoPS 1, skipping most of the problems in the later chapters because of lack of time. I also did a lot of MATHCOUNTS chapter and state tests (I think, at least) which I could find online. Apparently I looked on the inside cover on January 21 :P and found out about this website. However, I was pretty much inactive except maybe I played FTW (I don't remember for sure, but I'm pretty sure). So after doing a lot of tests, I started spamming the MATHCOUNTS forum ( :oops: just look at my first pages of posts) around the time I got 5th at states (I managed to get first countdown with some luck against Mewto55555). My scores for other competitions that year were 23 AMC8, 42 school and chapter MATHCOUNTS, 30 state MATHCOUNTS, 40 MathLeague; I also pretty much bombed MCTM and didn't get to do most other regional competitions. The rest of the school year after states I basically didn't do anything...in the summer, I finished most of the AoPS books except for AoPS 2 and the Intermediate books. I did some of the Chinese textbooks too...in the fall I didn't do much math, and got a 24 AMC8. In the winter I did most of AoPS 2, all the AMC10s, several old AHSMEs from the 80s, and then after break I did AIMEs, cramming MATHCOUNTS tests before their respective competitions were held. In the spring I got to take the 10A and 12B, which I got 144 and 115.5 on, and a 9 on the AIME (also bombed USAMO :blush: ). I got 45 school and chapter MATHCOUNTS, 44 state MATHCOUNTS, and 40 national MATHCOUNTS. Note that I didn't mention any countdown preparation...mainly because I didn't do much. This is why I didn't do very well in countdown.

Additionally, if you make Masters' Round, make sure you play poker the night before. Also if you annoy someone who has AA, you will win Masters' Round.

I was being serious.

Anyway, I suggest doing lots of problems. If you find you have trouble with certain types of problems, ask on AoPS or read up in the corresponding book.

I'd type more, but Mathcounts depresses me now.

WHOA Mewtwo55555 is from Missouri??? This is new to me.....
Thanks for all the info stuff. You could turn what you wrote into a memoir lol

there are lots of ways in which you can study or get better at Mathcounts. One way that hasnt been mentioned is reading throgh the Mathcounts marathon posted on this site. It actually very useful, TONS of problems and ways how to solve them.

I found those sources very useful, hundreds of pages of internet review. Last year, i was in 7th, i had a 12 or something at chapter, coming into 56th out of 63 people. It was my first year at MC and i didnt car much either. But then, i just had an urge to do well, and i did review, like a lot, and got a 41 in chapter and 38 at states. I didnt make nats, but the score is a HUGE improvement for me though. Remember "Commitment Leads to action and action leads to your dreams coming closer."

[url]http://www.agmath.com/57427.html[/url]

What Art of Problem Solving books do you suggest I should use? I have Volumes 1 and 2 and Intro to Number Theory, and Counting and Probability.

How should I work on these? 3/4 chapters a day until I finish the book? I think that I will have to do Volume 1 over again because I needed to use the solutions manual a lot. The concepts aren't very hard to master, but many of the word problems are very difficult for me.

Also, should I finish all the books and then do old competitions or work on both things simultaneously?

Do Volume 1 very thoroughly for sure, and think on the concepts and problems deeply. Volume 2 helped my problem solving a lot too (which is really all that matters in MATHCOUNTS, since concepts shouldn't be too advanced), even if most of the concepts were above MATHCOUNTS level. If you feel like you need more problems on a certain specific topic, you can search those problems on AoPS or maybe Google.

Do the Intro books which cover your weaknesses. If you are at least okay at a subject, you can improve quicker by doing problems in that subject. I made the mistake of doing all of the Intro books...while it helped, I probably could've been more efficient in learning the topics and applying them to problems.

You can do problems both while or after doing books; it just depends on who you are. I personally like to finish books before doing other problems, and it's what I did.

As for rates of work, do as much as you feel like. Never force yourself to do more than you can or want to.

As people now have a full summer to work, going through the entire intro series may not be a bad idea, and then saving Volume 1 for reviewing later.

Well, I suppose that's true, but if you're going to do that you probably shouldn't do them and Volume 1 all in a month like I unfortunately did. :| Take the time to absorb all of the materials.

[quote="PowerOfPi"]This might sound weird, but I can't sleep the night before the competition. In the morning, I can't have breakfast. It took me 2+ hours to fall asleep the night before State. I didn't do problems in the morning either. I felt that I did bad and got easy questions wrong during State, and I think that is from the night/morning before. Does anybody know how to stop that? I know it is from being nervous.[/quote]

Pray!! That happened to me too! I couldn't sleep. I was so nervous. You have to practice. And it's okay if you don't get at least 9 hrs of sleep. Just keep doing problems during the morning and night.

[quote]Pray!![/quote]
What if he's atheist... :wink:

Anyway, i wouldnt worry about the competition at all. Worrying about it would just make you even more nervous. Also, dont expect a lot, because if you dont meet it, u just get depressed. Just be like this: its just a little competition for middle schoolers. i have a lot more to do later in life. MC is just a start.....and stuff like that. In fact even if you dont make nats, its actually not that bad. You'll probably want to work harder for AMCs.

[quote="AIME15"]Practicing through the night will only make you tired :wink:[/quote]

Well not the whole night. You probably only need like 8 hrs 30 mins max. But practicing in the morning is the main thing.

@hidn: Well, you can pray even if you aren't atheist. Not to anyone particularly but just say like, "Oh please please let me do good in this competition!"

[quote="jonathanchou711"][quote="AIME15"]Practicing through the night will only make you tired :wink:[/quote]

Well not the whole night. You probably only need like 8 hrs 30 mins max. But practicing in the morning is the main thing.

@hidn: Well, you can pray even if you aren't atheist. Not to anyone particularly but just say like, "Oh please please let me do [b]well[/b] in this competition!"[/quote]

God likes good grammar  :wink:

The night before prepping is pretty much useless. You should just do whatever it takes to ensure on the morning of the test you are ready and have confidence. If you have confidence in yourself, you will do much better. My form of inspiring confidence consisted of doing a test the morning of the competition, and thinking to myself thoughts along the lines of, "I can do this, they won't be able to throw a problem at me that I cannot do, so I just need to do my best."

[quote="Mewto55555"]
You should just do whatever it takes to ensure on the morning of the test you are ready and have confidence. If you have confidence in yourself, you will do much better. My form of inspiring confidence consisted of doing a test the morning of the competition.[/quote]

What kind of problems you should use to test yourself in the morning?
Easy problems or hard problems?

If you use hard problems and you didn't do well, will that hurt your confidence in the real test?

Do you grade yourself for the morning test?

If the problem is kind of easy and you made many stupid mistakes, will that hurt your confidence in the real test?

It all depends on YOU. I did 08 nationals the morning before. I wasn't expecting to do too well, so it wouldn't hurt my confidence too much. I always grade every test I do. For one thing, I wanted to see my score. If it was a low score, I figure "Hey, this wasn't the real test, I should be glad I got my low score on this test as opposed to that one". If its a high score, then my confidence gets a boost. Also, you want to see where you are making mistakes. If today is a day when my arithmetic is bad, I would slow down on the test to check that. If I'm misreading a bunch, I'll go back and read twice (which was the case for me!).


calculus
derivative
analytic geometry
graphing lines
slope
quadratics
function

Find the equation of all lines through the origin that are tangent to the curve y= x^3-9x^2-16x

Please explain how you get it or AIM me. Info in sig. thank you. I've been wokring on it with a friend, but it's hard.

My teacher saays it's more than one line, but i don't know. If u could please explain to me how to do it I would appreciate it. I'm prolly missing something other than taking the derivative and using that slope in the point slope formula. Please help. thank you.

A line passing through a point $(a,b)$ with slope $m$ has equation $y-b=m(x-a)$. For a tangent line we have $b=y(a)=a^{3}-9a^{2}-16a$ and $m=y'(a)=3a^{2}-18a-16$. Hence the equation $y-(a^{3}-9a^{2}-16a)=(3a^{2}-18a-16)(x-a)$.
Let's see if this line passes through the origin by plugging zeroes for $x$ and $y$ into the equation:
$-(a^{3}-9a^{2}-16a)=(3a^{2}-18a-16)(-a)$
It remains to solve this equation. It may look scary, but if you simplify it a bit, you'll get $a(\text{quadratic})=0$, which is not so hard to solve.

So i have solved the quation and got a=0 and a= 9/2. Could you please explain to me why we did this though? I really want to learn how to do it. I understand why you took the derivative and put that equivalent to the slope. I am assuming you changed the function variable from x to a because of preference. Now that I have solved the a values how do I go about writing the equations for them? y-0=16(x) and y-0=-92.25(x-4.5)? Are both those equations tangent to the original? How would I go about checking that?

I wanted to give names to the coordinates of the point where the line touches the curve. It would be a bad idea to call them $(x,y)$, because I'd need $x$ and $y$ for the equation of the line. So I used $(a,b)$. Could have been $(x_{0},y_{0})$ or something else.

As long as an equation is of the form $y-(a^{3}-9a^{2}-16a)=(3a^{2}-18a-16)(x-a)$ with some $a$, it describes a line tangent to the curve. This is because of the way in which the equation was derived.

There may be something wrong in your computations, because your second line does not pass through $(0,0)$.

I'm still getting 9/2 as the other root. I'm unsure of how to write the equation though. I'm getting y=-36.25x+163.125 But i'm pretty sure that's incorrect. I plugged in y-0=-36.25 ( the result of putting 9/2 into the equation 3a^2-18a-16) times (x-4.5) so the equation in whole is y=-36.25(x-4.5). If you could offer a little more help I'd appreciate it. Or even more of an explanation, thank you.

[quote="blaze8"]Find the equation of all lines through the origin that are tangent to the curve $y= x^{3}-9x^{2}-16x.$[/quote]

We can set the desired equation of the tangent line as $y=mx.$
$x^{3}-9x^{2}-16x=mx\Longleftrightarrow x\{x^{2}-9x-(m+16)\}=0.$
This will have the double root. Clearly the condition cab be obtained by $m+16=0$ or $Discriminat: 9^{2}+4(m+16)=0,$ yielding $m=-16,-\frac{145}{4}.$

Thus the tangent line are $y=-16x$ and $y=-\frac{145}{4}x.$

In the first message you posted, you plugged in the point (0,0) for x and y. I don't think you can do that, you can only plug values into a and b. If you have y-b=m(x-a), which is the equation you have in simplist form, you cannot plug any values into x and y, only a and b.


when you solved a geometry problem,how can you sure that your solution is right for each situation?
are you check it for every question that you solved or not?
please advice me about that. :D
i have doubt in every geometry things and theorems,what should i do?(even the equation of triangles) :o

[quote="Arne"]What? :huh: Someone please remove this.[/quote]
what is your problem man? if you don't want help me,don't but you shouldn't be so rude like that.ok!thanks. :D

ok
u should thing that it is over of a furmulla or teorem :)
and be able to prove the first thing & something like that.
(such the prove of nagel point u should use anoyher theorem. :ninja:


$ tg(k\plus{}1)x \minus{}tgkx \equal{}\frac{sinx}{coskx cos(k\plus{}1)x}$, therefore $ \sum_{k\equal{}1}^n \frac{1}{coskxcos(k\plus{}1)x}\equal{}\frac{tg(n\plus{}1)x\minus{}tgx}{sinx}$. Therefore $ tg(n\plus{}1)x\equal{}tgx,x\not \equal{}k\pi\to x\equal{}\frac{m \pi}{n}, n\not |m.$


Let $ N$ be the set of positive integers. Find all functions $ f(x): N \rightarrow N$ such that:
$ f(f(m)+f(n)) = m+n \forall m,n \in N$

This was posted many times, hope someone will give you a link..

Same idea http://www.mathlinks.ro/Forum/viewtopic.php?t=148143  :wink:


Maybe we should change the count to 0 or 50 game to this game:

If your first name starts from A-L, then subtract a number.
If not, then add a number.

Same rules as the other one except different, more even teams. :D

So let's start with....

[b]25[/b]

[b]31[/b]

A-L/M-Z would have been more even.

howd it go from 25 to 31?

And A-M is 13 letters, N-Z is 13 letters.

So I assume you posted 26, so I'll do 25.

[b]26[/b]

I was just fooling with you because you said to add a number as opposed to add 1.

Additionally, there are more A-M names than N-Z names, and the split happens at about A-L and M-Z. That's why all the schools divide everything that way.

23 :D

The message is too small. Please make the message longer before submitting.

19
The message is too small. Please make the message longer before submitting.

This game isn't fair either...

[b]12[/b]

11

Yay!!! Girls won in boy vs. girls!!! check out boy vs. girl round two!!!

11

This game is hopeless and very slow, at least compared to b vs g 2.

10... i like boys vs. girls a lot better... but i'll still post anyways :D

YEAH!!!! A-L WON!!! congrats to everyone who partcipated!!!!

Aww dang it. I missed it. Oh well, at least my team won.  :lol:


trigonometry
inequalities unsolved
inequalities

prove that for all $0 \leq \alpha \leq 2$ we have

$\sum \left( \sin A \right)^{\alpha} \leq \sum \left(\cos\left(\frac {A}{2}\right)\right)^{\alpha}$

thank you very much

sorry, (i have so many mistakes in typing)

it is $\sum \left(sin A \right)$

please! is there anyone can help me?
i just have the solution for the cases
i)  for $0\leq \alpha \leq 1$
ii) for $1\leq \alpha \leq 2$ and the traingle ABC is acute
 :(


The positive three-digit integer $ N$ has a ones digit of $ 0$. What is the probability that $ N$ is divisible by $ 4$? Express your answer as a common fraction.

An integer is divisible by 4 if and only if a number formed from the last two digits is divisible by 4. If the units digit is 0, all the numbers with even tens digits are divisible by 4 (00, 20, 40, 60, 80), and all the numbers with odd tens digits are not (10, 30, 50, 70, 90). Since there are the same number of even digits as odd digits, there is a $ \boxed{1/2}$ probability that $ N$ is divisible by 4.


trigonometry
induction
algebra theorems
algebra

I just used DeMoivre's Theorem on one of old AHSME problem. But I wonder, can anyone prove this theorem so it makes easier to understand?

Thank you very much.

What theorem do you mean?
\[(\cos t+i\sin t)^n=\cos nt+i\sin nt.\]
Right?

I think that's the one.

It was one about complex numbers' relationships. Is there more than one DeMoivre's Theorem? :?

Let $f(t)=\cos t+i\sin t$,then $f'(t)=-\sin t+i\cos t=i(\cos t+i\sin t)$ so we have $f'(t)=if(t)\Longleftrightarrow f(t)=Ce^{it}$.
Since $f(0)=1$,we obtain $C=1$.Therefore we obtain Euler's formula $e^{it}=\cos t+i\sin t$.
Then $e^{int}=(e^{it})^n\Longleftrightarrow (\cos t+i\sin t)^n=\cos nt+i\sin nt\ (n=0,\pm1,\pm2,\cdots)$.

Proof:
you have $cos(x)+i \cdot sin(x) = e^{ix}$ and from $e^{inx}=(e^{ix})^n$ follows $cos(nx)+i \cdot sin(nx)=(cos(x)+i \cdot sin(x))^n$

Edit: kunny was faster...

[quote="Silverfalcon"] can anyone prove this theorem so it makes easier to understand?[/quote]

To Silverfalcon:
You can use induction.

kunny

I personally liked the induction proof, here's my paper on it that I wrote a while for [url=http://onlinemathcircle.com/]OnlineMathCircle[/url]

http://onlinemathcircle.com/wp-content/uploads/2011/04/17-induction.pdf

Here is the proof written in Japanese.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1747440#p1747440[/url]


algebra
function
domain
inequalities
integration
geometry
3D geometry

Before I ask a question, check the following please. Suppose we are seeking conditions on $ q > 1$ and $ \alpha\in\mathbf R$ under which the problem (call it PRB)
\[ \{\begin{aligned} - \Delta u & \ge |u|^q\rho^\alpha(x) & & \text{ in } \Omega, \\
u & = 0 & & \text{ on } \partial\Omega \end{aligned}
\]
has only trivial solutions in
\[ X = \{ u \in H_0^1(\Omega) \mid |u|^q\rho^\alpha \in L^1(\Omega) \}.
\]
Here $ \rho(x) : = \mathrm{dist}(x,\partial\Omega)$ and $ \Omega$ is a bounded smooth domain in $ \mathbf R^n$.

Assume that $ \Omega = B$, the open unit ball, and let $ \Gamma(q,\alpha) : = q + \alpha + 1 > 0$. Then the function
\[ u(x) = \gamma (1 - |x|)^s, \quad \max(1 - \frac n2, - \frac {\alpha + 1}q,0) < s < 1,
\]
is a solution of PRB for some small $ \gamma\in(0,1)$ and some $ s$ close to 1.

[hide="Proof"]
The function $ u$ belongs to $ X$ and is a weak solution to the following problem:
\[ \{\begin{aligned} - \Delta u & = \gamma s ( 1 - s + \frac {n - 1}{|x|} (1 - |x|) ) (1 - |x|)^{s - 2} & & \text{ in } \Omega, \\
u & = 0 & & \text{ on } \partial\Omega. \end{aligned}
\]
Therefore the inequality $ - \Delta u \ge |u|^q\rho^\alpha$ is equivalent to
\[ \gamma^{q - 1} (1 - |x|)^{\Gamma(q,\alpha) - (q - 1)(1 - s)} \le s ( 1 - s + \frac {n - 1}{|x|} (1 - |x|) ).
\]
Take $ s$ sufficiently close to 1 so that $ 0 < \Gamma(q,\alpha) - (q - 1)(1 - s)$. Now we can easily find necessary $ \gamma$.
[/hide]

So far, so good. :) What is the question then?

Ok. Let $ u \in X$ be a solution of PRB, then for any nonnegative $ \varphi \in H_0^1(\Omega) \cap H^2(\Omega)$
\[ \int_\Omega |u|^q\rho^\alpha\varphi \le \minus{} \int_\Omega u \Delta\varphi.
\]
By Holder's inequality,
\[ \minus{} \int_\Omega u \Delta\varphi \le \frac 1q \int_\Omega |u|^q\rho^\alpha\varphi \plus{} \frac 1p \int_{\{\varphi\neq0\}} \frac {|\Delta\varphi|^p}{(\varphi\rho^\alpha)^{p \minus{} 1}},
\]
where $ p \equal{} q/(q \minus{} 1)$. Therefore
\[ \int_\Omega |u|^q\rho^\alpha\varphi \le \int_{\{\varphi\neq0\}} \frac {|\Delta\varphi|^p}{(\varphi\rho^\alpha)^{p \minus{} 1}} \equal{} : I
\]
for any nonnegative $ \varphi \in H_0^1(\Omega) \cap H^2(\Omega)$.

For any $ \epsilon > 0$, let $ \varphi \equal{} \varphi_\epsilon \in H_0^1(\Omega) \cap H^2(\Omega)$, $ 0 \le \varphi_\epsilon \le 1$, be a function such that

1) $ \varphi_\epsilon(x) \equal{} 1$ for $ |x| \le 1 \minus{} \epsilon$;

2) There is a constant $ C > 0$ s.t. $ \varphi_\epsilon(x) \ge C \rho(x)$ for any $ x\in\Omega$ and $ |\Delta \varphi_\epsilon| \le C\epsilon^{ \minus{} 2}$, for any $ \epsilon > 0$.

Then we have
\[ I \le C' \int_{B \setminus B_{1 \minus{} \epsilon}} \frac {\epsilon^{ \minus{} 2p}}{\rho^{p \plus{} \alpha(p \minus{} 1) \minus{} 1}}.
\]
Suppose that $ p \plus{} \alpha(p \minus{} 1) \minus{} 1 < 0$ or equivalently that $ \alpha < \minus{} 1$, then
\[ I \le C' \int_{B \setminus B_{1 \minus{} \epsilon}} \epsilon^{ \minus{} p \minus{} (2p \plus{} \alpha(p \minus{} 1) \minus{} 1)} \equal{} C' \int_{B \setminus B_{1 \minus{} \epsilon}} \epsilon^{ \minus{} p \minus{} \frac {\Gamma(q,\alpha)}{q \minus{} 1}} \equal{} \epsilon^{\minus{} p \minus{} \frac {\Gamma(q,\alpha)}{q \minus{} 1} \plus{} n} O(1).
\]
We deduce that if $ \alpha < \minus{} 1$ and $ \minus{} p \minus{} \frac {\Gamma(q,\alpha)}{q \minus{} 1} \plus{} n > 0$, then our problem, PRB, has only trivial solution [because as $ \epsilon\to0$, $ \int_\Omega |u|^q\rho^\alpha \varphi_\epsilon \to \int_\Omega |u|^q\rho^\alpha$ and $ \epsilon^{ \minus{} p \minus{} \frac {\Gamma(q,\alpha)}{q \minus{} 1} \plus{} n} \to 0$.] In particular, if $ (q,\alpha)$ and $ n$ are so that $ 0 < \Gamma(q,\alpha) < (n \minus{} p)(q \minus{} 1)$ (e.g. for $ n$ large enough) and $ \alpha < \minus{} 1$, then PRB has no nontrivial solutions in $ X$.

But my previous post shows that PRB has a nontrivial solution whenever $ \Gamma(q,\alpha) > 0$. Where is a mistake?

The volume of the $ \varepsilon$-shell near the unit sphere is $ \varepsilon$, not $ \varepsilon^n$! :roll:


Let a sequence of integers {$ a_{n}$},n in N be given,defined by
$ a_0\equal{}1,a_{n}\equal{}a_{n\minus{}1}\plus{}a_{[n/3]}$
for all n in $ N*$.
Show that for all primes $ p<\equal{}13$,there are infinitely many integer number $ k$ such that $ a_{k}$ is divided by $ p$

for $ p\equal{}3$
for $ r\in\{0,1,2\}: \ a_{3n\plus{}r}\equal{}a_{3n\minus{}1}\plus{}(r\plus{}1)a_n$
$ u_{n}\equal{}a_{3n\minus{}1}$ gives $ u_{n\plus{}1}\equal{}a_{3n\plus{}2}\equal{}a_{3n\minus{}1}\plus{}3a_{n}\equal{}u_{n}\plus{}3a_n$ then $ u_n\equal{}u_1\plus{}3\sum_{k\equal{}1}^{n\minus{}1}a_k\equal{}3\sum_{k\equal{}0}^{n\minus{}1}a_k$
then $ r\in\{0,1,2\}: \ a_{3n\plus{}r}\equal{}3\sum_{k\equal{}0}^{n\minus{}1}a_k\plus{}(r\plus{}1)a_n$ and $ \forall h\in\mathbb{N}: \ 3|a_{3h\plus{}2}$

For $ p\equal{}2$:

Suppose there are only finitely many even integers in the sequence. Let $ N$ be the maximum integer such that $ a_N$ is even. Then $ a_{N\plus{}1}$, $ a_{3N\plus{}2}$ and $ a_{3N\plus{}5}$ must be odd. (NB: The sequence $ (a_n)_{n\,\equal{}\,0}^\infty$ is strictly increasing.)

Now $ a_{3N\plus{}2}\equal{}3(a_0\plus{}\ldots\plus{}a_N)\ \Rightarrow\ (a_0\plus{}\ldots\plus{}a_N)$ is odd (since the LHS is odd). But then

$ a_{3N\plus{}5}\equal{}3[(a_0\plus{}\ldots\plus{}a_N)\plus{}a_{N\plus{}1}]$

from which we see that $ a_{3N\plus{}5}$ and $ a_{N\plus{}1}$ cannot both be odd after all. Contradiction.

Hence the sequence must contain infinitely many even integers.

For all $ p\leq 13$. First find some number in the sequence divisible by p. Let it be $ a_{n}$. Working modulo p now. We have $ a_{3n\plus{}2}\equal{}a_{3n\plus{}1}\equal{}a_{3n}\equal{}a_{3n\minus{}1}\equal{}m$. So the numbers in the sequence from $ a_{3(3n\minus{}1)\minus{}1}\equal{}a_{9n\minus{}4}$ to $ a_{9n\plus{}8}$ form arithmetic progression(modulo p) with difference m and length 12. So in this progression there is at least one element divisible by p. And so for each $ a_{n}$ divisible by p there is greater number in the sequence divisible by p. The problem is solved.


[quote="the desk of asmallgoat"]The Game of Piles:
directions for a turn:
-add 6 to smallest pile
-add 3 to second smallest
-add 1 to the other 3 piles
-if there's a tie for which pile gets 6 or 3 the pile who's two neighbouring piles are larger than the other's neighbours gets the 6 or 3

The piles are sitting in a circle going in alphabetical order. There are piles A-E, and you start with n in A, n-1 in B, n-2 in C, etc.

A turn is done all at the same time, sort of like in the game of life.

Will the piles ever be equal?
[/quote]

Notice that this is actually a pretty easy game as asmallgoat told me(most of the time.) The real point is to go really fast(see how I put it in MC forum because of this.)

Also, I actually have the answers(it's written on her desk.) Oh and usually you don't start with n,n-1,n-2 etc. in you piles. It's rather random.
[hide="the ones asmallgoat told me to tell you are"]
70-70-69-72-71
[/hide]

When you say its random, do you mean the values that start out are random or do you mean that the order of the values are random?

[quote="CircleSquared"]If there is a tie, it doesn't matter which one receives some addition, the numbers turn out the same any way.  They're just in different order.

@ZhangPeiJin:
It probably means the values that start out are random.  Numbers in different orders don't really matter.[/quote]

Well it actually sort of matters when it comes to order. Read the real instructions. When I say random I mean numbers and order. The two I supply here(the n,n-1, etc and asmallgoat's favorite one) aren't random; take them in a circle the way they appear. I just mean if you were to actually want to play the game...which unfortunately is what asmallgoat does instead of doing school(sometimes.) I don't see how it would be addicting but asmallgoat makes conjectures and tries to "prove"(rather go through and find a pattern) them.


modular arithmetic
number theory unsolved
number theory

Is there an elementary proof for korselt's criterion: if $ a$ is a positive integer then $ n$ divides $ a^n \minus{} a$ iff it is squarefree and $ p \minus{} 1|n \minus{} 1$ for every prime factor $ p$ of $ n$?. I tried to prove it, but i couldn't do it... :blush:

[hide="if case"]
For each prime $ p$ dividing $ n$, either $ a$ is divisible by $ p$, or $ a^{p\minus{}1}\equiv 1\pmod{p}\Rightarrow a^{n\minus{}1}\equiv 1\pmod{p}$. Either way, $ p$ divides $ a^n\minus{}a$.

As $ a^n\minus{}a$ is divisible by every prime dividing $ n$, it's divisible by the product of those primes. For squarefree $ n$, this product is $ n$.
[/hide]
[hide="only if case"]
To show $ n$ is squarefree, let $ p$ be a prime dividing $ n$. Then $ (1\plus{}p)^n\minus{}(1\plus{}p)$ is divisible by $ n$, but
\[ (1\plus{}p)^n\minus{}(1\plus{}p)\equiv (1\plus{}np)\minus{}(1\plus{}p)\equiv (n\minus{}1)p\pmod{p^2}\]
$ p$ can't divide both $ n$ and $ n\minus{}1$, so $ (1\plus{}p)^n\minus{}(1\plus{}p)$ is not divisible by $ p^2$. As a result, $ n$ is not divisible by $ p^2$.

It remains to show that if $ p$ is a prime dividing $ n$, $ p\minus{}1$ divides $ n\minus{}1$. To do this, let $ g$ be a primitive root modulo $ p$. As $ g^n\minus{}g$ is divisible by $ n$, it's divisible by $ p$.

So $ g^{n\minus{}1}\equiv 1\pmod{p}$ (of course $ g$ is not divisible by $ p$). As $ g$ is a primitive root, $ p\minus{}1$ (the order of $ g$ modulo $ p$) must divide $ n\minus{}1$.
[/hide]


inequalities
trigonometry
inequalities unsolved

Prove that in any acute-angled triangle the following inequalities holds:

  $ \frac {4\minus{}\sqrt{2}}{2}\leq\frac {a}{b \plus{} c}\sin A \plus{} \frac {b}{c \plus{} a}\sin B \plus{} \frac {c}{a \plus{} b}\sin C < 2$.

[quote="Marius Mainea"]Prove that in any acute-angled triangle the following inequalities holds:

  $ \frac {3\sqrt {3}}{4}\leq\frac {a}{b \plus{} c}\sin A \plus{} \frac {b}{c \plus{} a}\sin B \plus{} \frac {c}{a \plus{} b}\sin C$.[/quote]
I think, it's wrong. Try $ a\equal{}b\equal{}\sin46^{\circ}$ and $ c\equal{}\sin88^{\circ}.$ :wink:

But I guess this inequality is true: $ \frac {a}{b \plus{} c}\sin A \plus{} \frac {b}{c \plus{} a}\sin B \plus{} \frac {c}{a \plus{} b}\sin C<2$.
 Because $ \sin A, \sin B, \sin C < 1$ the inequality becomes: $ \sum \frac{a}{b\plus{}c} <2$.
Its also known the following result: [i]If $ 0<a<b$ and $ x>0$, then $ \frac{a}{b} < \frac{a\plus{}x}{b\plus{}x}$[/i].
Now, because $ a<b\plus{}c$ we get that $ \frac{a}{b\plus{}c}<\frac{2a}{a\plus{}b\plus{}c}$.
Conclusion: $ \sum \frac{a}{b\plus{}c}< \sum \frac{2a}{a\plus{}b\plus{}c}\equal{}2$


Prove that:

$B^3(x) = \sum_{n=0}^\infty (2n+1){2n \choose n}x^n$ where $B(x) = \frac{1}{\sqrt{1-4x}}$

I think the proof is well known but I could not solve it.


Hi everybody!Maybe this is not the right section but as you can easily notice I'm new here... :?
I need to have basic knowledges on diophantine equations.
What are diophantic equations?
Does it exist a special program used to solve them?
(as I learn fast would you mind telling me some hard problems whose solutions are based on them :D )
Ciao! :)
Giuglia

I think this will be useful to you:
[url=http://mathworld.wolfram.com/DiophantineEquation.html]mathworld.wolfram.com/DiophantineEquation.html[/url]
Ciao!


let $ a,b,c,d>0$ and $ a\plus{}b\plus{}c\plus{}d\equal{}4$ then prove that

 \[ \frac{1\plus{}ab}{1\plus{}b^2c^2}\plus{}\frac{1\plus{}bc}{1\plus{}c^2d^2}\plus{}\frac{1\plus{}cd}{1\plus{}d^2a^2}\plus{}\frac{1\plus{}da}{1\plus{}a^2b^2}\geq4\]

[quote="M.A"]let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ then prove that
\[ \frac {1 \plus{} ab}{1 \plus{} b^2c^2} \plus{} \frac {1 \plus{} bc}{1 \plus{} c^2d^2} \plus{} \frac {1 \plus{} cd}{1 \plus{} d^2a^2} \plus{} \frac {1 \plus{} da}{1 \plus{} a^2b^2}\geq4
\]
[/quote]

$ \frac{1 \plus{} ab}{1 \plus{} b^2c^2}\equal{}1 \plus{} ab \minus{} \frac {b^2c^2 \plus{} b^3c^2a}{1 \plus{} b^2c^2}\geq 1 \plus{} ab \minus{} {b^2c^2(1\plus{}ba)}{2bc}$
$ LHS\geq4 \plus{} \sum_{cyc} ab \minus{} \frac {\sum_{cyc} ab \plus{} \sum_{cyc}b^2ca }{2}\geq4$ we need proe this.
$ \sum_{cyc} ab\geq  \sum_{cyc}b^2ca$ .
its to easy for me

$ a\equal{}b\equal{}c\equal{}d\equal{}1$

[quote="rachid"][quote="Enjar0"][quote="M.A"]let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ then prove that
\[ \frac {1 \plus{} ab}{1 \plus{} b^2c^2} \plus{} \frac {1 \plus{} bc}{1 \plus{} c^2d^2} \plus{} \frac {1 \plus{} cd}{1 \plus{} d^2a^2} \plus{} \frac {1 \plus{} da}{1 \plus{} a^2b^2}\geq4
\]
[/quote]

$ \frac {1 \plus{} ab}{1 \plus{} b^2c^2} \equal{} 1 \plus{} ab \minus{} \frac {b^2c^2 \plus{} b^3c^2a}{1 \plus{} b^2c^2}\geq 1 \plus{} ab \minus{} {b^2c^2(1 \plus{} ba)}{2bc}$
[/quote]
Are you sure Enjar0 ???.[/quote]

  $ \frac {1 \plus{} ab}{1 \plus{} b^2c^2} \equal{} 1 \plus{} ab \minus{} \frac {b^2c^2 \plus{} b^3c^2a}{1 \plus{} b^2c^2}\geq 1 \plus{} ab \minus{} \frac{b^2c^2(1 \plus{} ba)}{2bc}$


Prove that for any odd prime number p, there doesnot exist integer number A such that:
1. $A^3 - A$ is not a multiple of p
2. $\sum_{i=1}^{p-1}\frac{{p \choose i}}{p}.A^i$ is a multiple of p

Hi nmtruong1986,

If I understood your question correctly, then I guess there is something wrong in your post.

1st part :
For example, let $p = 5$ and $A = 2$. Then, $A^3 - A = 2^3 - 2 = 6$ which is not a multiple of $p$.

2nd part:
For example, let $p = 3$ and $A = 2$. Then, $LHS = 6$ which is a multiple of $p$.

Kindly clarify this.

Thanks,
Sambit

What he means is: for an odd prime $p$ there's no number $A$ which satisfies [b]both[/b] conditions (1. and 2.) at the same time. In other words, prove that if $p$ is an odd prime and $p|\displaystyle \sum_{i=1}^{p-1}\frac{\binom pi}pA^i$, then $p|A^3-A$.

that statement is wrong. a counterexample is $p=7$, $A=18$.

Peter


So...how easy is it to get partial credit on the USAMO if you don't really know the solution to a problem? That is, how does one get a 1 or 2 out of 7 on a problem? Could someone provide some insight on how the grading system works? Thanks.

I think it's fairly hard to get a few points for just writing something. If your method can lead to a solution, you can get partial credit, but otherwise you won't. Having said that though, most of the time when people tell me their score, it isn't 0(mod 7).

Hehe, so you're saying that most people you hear from gets partial credit?

-interesting_move

So does anyone have any tips for how to squeeze a few partial credit points from the graders?

I just wrote how however far I had gotten on each problem, working towards a solution, and I got a total of 6.


Help!!! I cant the basic probability concepts down in head...I have a head like a rock !!! Can someone help me understand the basics,e.g. If I have 10 ritz crackers, 13 wheat thins and 15 triscuits. If I randomly pick 6 crackers out of the 38, how many different ways can I do this ???

If someone wants 3 ritzes at a given time..and I have to serve all of the crackers, how many different ways can I do this ?

Whats the probability of getting all of the same variety out of the 6 that I pick ???

I need some basic rules that I can apply for unordered, ordered, etc....

In dire straits !!!!

EDIT: I wrote this in a hurry so [i]all[/i] of these answers are wrong. I am keeping this post for discussion purposes. The correct answers are found below... somewhere...

[quote="girl58"]Help!!! I cant the basic probability concepts down in head...I have a head like a rock !!! Can someone help me understand the basics,e.g. If I have 10 ritz crackers, 13 wheat thins and 15 triscuits. If I randomly pick 6 crackers out of the 38, how many different ways can I do this ???[/quote]

This isn't even a probability problem. SInce you have more than 6 of all types of crackers, the answer would be $3^6=729$

[quote="girl58"]If someone wants 3 ritzes at a given time..and I have to serve all of the crackers, how many different ways can I do this ?[/quote]

Since you already have 3 of the six crackers set, you need to find the possibility for the 3 remaining crackers: $3^3=27$. Sorry if this is a little confusing.

NOTE: For all of the above problems I am finding the number of permutations, not combinations.

[quote="girl58"]Whats the probability of getting all of the same variety out of the 6 that I pick ???[/quote]

An easy way to do this is to find the probability of getting 6 ritzes, 6 wheatthins and 6 triscuits (my fave crackers). Since none of the possibilities overlap, you can simply add them. the possibility of getting 6 ritzes is $10/38*9/37*8/36*7/35*6/34*5/33$  :approx: $0.0009128$ (Do you see why?) Similarly, the probability of getting 6 wheatthins is $13/38*12/37*11/36*10/35*9/34*8/33$  :approx: $0.007459$ and the probability of getting 6 triscuits is $15/38*14/37*13/36*12/35*11/34*10/33$  :approx: $0.0217555$. Adding, you get the solution: $0.0301273$.

For the second question, I thought that you had to serve all 38 of the crackers 3 at a time. The number wouldn't be too small, and 38 isn't even divisible by 3.

For the first question you can use partitions.

Partition it like  _ _ | _ _ | _ _
Then, since we have to have at least 1 of each, we take out the middle:
_ | _ _ | _
And there are $\frac {6}{2!4!} = 15$ ways to pick 2 places for the separators.
Sorry if you don't understand, but it's another way to find how many combinations of different Ritzes you can make.

236factorial considering your name you really should understand this stuff better  :D

Sorry I messed up. There's only $10$ ways.

Now that I think about it, I think the answer to the first question is $28$. You can find it by adding up different combinations (Ritz = A, Wheatthins = B, Triscuit = C):

[code]Only A      : 1 (AAAAAA)
Only B      : 1
Only C      : 1
A and B     : 5 (AAAAAB, AAAABB ... ABBBBB)
A and C     : 5
B and C     : 5
A, B, and C :10 (4 A's = 1 combo, 3 A's = 2 combos, ... 1 A = 4 combos)
[/code]

Add them up and you get $28$.

This can be generalized by the formula $(x+1)(x+2)/2$, where x is the number of crackers you are serving, provided that $A>x$, $B>x$, and $C>x$.

I did it where you have to pick at least one of each. Oops

For #3, there are 3 feasible combinations (AAAAAA, BBBBBB, and CCCCCC) out of 28 possible (as demonstrated above). Therefore the answer is $3/28$ ( :approx: $0.107143$)

I don't get question #2.


I know this must have been asked before, but i want an updated one

I am on aops a total of maybe 30-35 minutes a day usually. I go on and off, with a computer handy, and check once every few hours for a few minutes.

I used to spend a lot of time here every day, but I got busy with college... :(

The poll options are rather strange, don't you think? I mean, it should've been something like 15-30min, 30-60min, 1-2h etc. Why 26-50min, 50-60min?  :huh:

On topic: I log in frequently and briefly to see if there's anything new. In total, I'd say about 20 minutes.

Not as much as I would like to; not as much as I should.

i get on all the time because it is my homepage
i shouldn't be doing that though
i have to get my priorities in order
gotta get packed

Maybe like 30 minutes a day? I dunno

[quote="now a ranger"][quote="math92"]I used to go on a lot...like 2 hours a day. Now i go on 15 min a day.[/quote]

I have a feeling that will be how my life is next year as a freshman...

My cousin got on about 3 hours a day, and he was in 7th grade...

now he's in 9th, and he doesn't get on like ever...[/quote]

Why would going to high school change your time on AOPS?

i was the opposite. I have always liked math, but I suddenly became super interested and enjoying it when I got into highschool.

[quote="arkhammedos"]sorry but AOPS is an abreviation of what ?[/quote]

Art of Problem Solving

[quote="Xevarion"][quote="Elemennop"]Why would going to high school change your time on AOPS?[/quote]
Same reason college changed mine: suddenly there's a lot more work that's not contest math.[/quote]

Yea but there's a difference between the workload at princeton and the workload at a public high school...

I go on for like 15 minutes a day but it might seem like more, because I leave my computer on

About 30 minutes per day.
Pool is really strange. :maybe:

I dont know about the pool, but the poll is strange :P . I get on around an hour a day, much of wic is spent reading peoples solutions and duplicating them

About an hour if I have leisure time. Otherwise, I won't browse it or just a few minutes.

I dont spend a lot of time on it, but mostly every day I get on AoPS for 26-50 min

I'd say 50-60. I'm the contrary of some people, it's my freshman year and I've spent about thirty times more time on this site so far this year than my seventh and eighth grade years combined. Sudden interest in math = more time on forums.

I'm on AoPS for like 2 hours but I usually just leave it on while I'm doing something else

I usually go on like 3-5 hours a day, depending on how much projects/homework there are.

I go for like 2 to 3 hours to answer like 3 questions.  :rotfl:

I have a life..... but part of it is on AOPS :lol:

50-60 roughly I guess. Like most, I leave it open when I do stuff.

1-2 hours.. but I do other things
So, I am really on for about 30-40 min

probably anywhere from 0 minutes - 1.5 hours a day. Usually around 40-55 minutes on school days, which is less than I should, less than I want to, and more than I have time for.


how many ways can you tile a $2$ by $n$ board with $1$ by $2$ and/or $2$ by $2$ tiles

n.b. a recurrence relation solution would be preferable...

[hide]If there are $n$ collumns, you can fit $n$ 2x1's vertically. Horizontally it's $2(n-1) = 2n-2$. 2x2's can only fit in $n-1$ times. Add together and get $4n-3$.[/hide]

i dont think thats quite right. try 3
it should be
[hide]the fibonacci sequence, because by adding an extra row, we are adding all the ways to make it previously, with 2 less rows to change[/hide]

tarquin, you can use both 1by2 and 2by2 tiles on the same board


Let $ x,y\geq 0$.Prove that:
$ (x^2\plus{}y^2)^2\plus{}x^3\plus{}y^3\plus{}4xy^2\plus{}x^2\plus{}y^2\geq 3x^3y\plus{}5xy^2\plus{}xy$

I'm quite surprised that I see no solution yet... :(

Where are you, arqady?

[quote="Inequalities Master"]I'm very sorry, the initial ineuality should be:
$ (x^2 \plus{} y^2)^2 \plus{} x^3 \plus{} y^3 \plus{} 4xy^2 \plus{} x^2 \plus{} y^2\geq 3x^3y \plus{} 5x^2y \plus{} xy$

so arqady's solution helps only if $ y\geq x$
Sorry again for the mistake when posting.[/quote]
My solution :)


Gene and Anne have one coin and continue to flip it until one of them wins this game.
Gene wins if the sequence “Head-Head” appears before the sequence “Tail-Head”.
Anne wins if the sequence “Tail-Head” appears before the sequence “Head-Head”.
What is Anne’s probability of winning this game?

A.1/4   	B.1/3	 C.1/2	  D.2/3	      E.3/4

[hide=" click to reveal hidden content"]

This looks like $ \frac{3}{4}$ to me. We could use a tree diagram, which I think checks.

[/hide]


logarithms
real analysis
real analysis unsolved

Let $\sum_{i \ge 0} a_i x^i$ be a power series with radius of convergence $R \in (0,\infty)$, and assume that $a_i \ge 0$ for all $i$. Let $P_N(x) = \sum_{i=0}^N a_i x^i$ be the N-th partial sum.
Determine all $x > 0$ for which $\limsup_{N \to \infty} \frac{1}{N} \ln P_N(x)$ exists, and find the limit for these $x$.

No takers? It's been out for two weeks.

Hint: the $\limsup$ exists for [b]all[/b] $x > 0$.

The interesting case is $x\ge R.$ By the root test, for all $\epsilon>0,\,\exists\,M$ such that $a_k<\frac{M}{(R-\epsilon)^k}.$

Hence $P_N(x)\le M\sum_{k=0}^N\left(\frac x{R-\epsilon}\right)^k\le C \left(\frac x{R-\epsilon}\right)^{N+1}$ (geometric sum).

Take the logarithm, divide by $N,$ and let $N\to\infty$, and we get

$\limsup_{N\to\infty}\ln P_n(x)\le \ln\left(\frac xR\right).$

I would assume that this is the limit as well, although I haven't quite proved that yet.

The case $x<R$ is fairly simple, with the limit being zero in that case.

[quote="xirti"][quote="Kent Merryfield"]The interesting case is $x\ge R.$ By the root test, for all $\epsilon>0,\,\exists\,M$ such that $a_k<\frac{M}{(R-\epsilon)^k}.$[/quote]

What is the root test? Can you explain? Thank you![/quote]
$R^{-1} = \limsup_{n \to \infty} \left( |a_n| \right)^{1/n}$

Thank you! I kenw this formula but I didn't know that it is called root test :blush:  Stupid me  :D


Given triangle $ ABC$ and incenter $ (I)$ touch $ BC, CA, AB$ at $ D, E, F$. Through an arbitrary point $ P$ draw the lines parallel with $ BC, CA, AB$ and intersect the altitude from $ A, B, C$ at $ M, N, P$ resp.Prove that $ DM, EN, FP$ are concurrent.

Something is wrong. See the sketch (I replaced the initial arbitrary point $ P$ with $ X$ to avoid repeating)

Sorry here is correct problem
Given triangle $ ABC$ an arbitrary point $ P, DEF$ is pedal triangle of $ P$, through  $ P$ draw the lines parallel with $ BC, CA, AB$ and intersect the altitude from $ A, B, C$ at $ M, N, P$ resp. Prove that $ DM, EN, FP$ are concurrent.
 I was wrong with incenter. :)

I think there is a typo. We must say as $ Q,$ the intersection point of the altitude through $ C,$ from the line through $ P$ and parallel to $ AB.$ So, the concurrency lines are $ DM,$ $ EN,$ $ FQ$ $ ($ instead of $ FP$ $ ).$

A genaralization of this problem $ ($ [b][size=100]The paracevian perspector theorem[/size][/b] $ ),$ has already been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=126138

Kostas Vittas.

[quote="gemath"][quote="vittasko"]I think there is a typo. $ Q,$ is the intersection point of the altitude through $ C,$ from the line through $ P$ and parallel to $ AB.$

A genaralization of this problem $ ($ [b][size=100]The paracevian perspector theorem[/size][/b] $ ),$ has already been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=126138

Kostas Vittas.[/quote]
Dear vittasko I don't understand what do you mean point $ Q$? I think the problem I proposed is true
[quote="gemath"]
Given triangle $ ABC$ an arbitrary point $ P, DEF$ is pedal triangle of $ P$, through  $ P$ draw the lines parallel with $ BC, CA, AB$ and intersect the altitude from $ A, B, C$ at $ M, N, P$ resp. Prove that $ DM, EN, FP$ are concurrent.
[/quote]
Anyway thank you very much for your link, I have never known them before. :)[/quote]
Sorry my friends I made a big mistake when proposed this problem, thank you vittasko very much to show it for me, the correct problem and nice solution is in the vittasko's link and I recorrect my problem here

Given triangle $ ABC$ an arbitrary point $ P, DEF$ is pedal triangle of $ P$, through  $ P$ draw the lines parallel with $ BC, CA, AB$ and intersect the altitude from $ A, B, C$ at $ M, N, Q$ resp. Prove that $ DM, EN, FQ$ are concurrent.

Thank you for your interrest. :oops: