Anyone want to have some sort of meetup? Friday night? Saturday night?

Wow, we're talking about this already?  ARML's not for another two months!

And yeah, it would probably be on a Friday night.

[quote="CatalystOfNostalgia"]Most teams leave by Saturday night, don't they?[/quote]Some do. I forgot about that - mine stays until Sunday morning.

A lot of teams leave before sunday as far as I know.  For instance, my team leaves around 4 on saturday. So friday night will be better for a meeting.

Waah I'm not going to Penn State :what?: . See you guys next year, as I assume Georgia will be a failure and not be repeated.

PS We always left immediately after the contest, and I am surprised to hear that some teams don't.

Well, for the Ohio team, it takes like 9 hours for us to get from Penn State to Cincinnati, the last stop.  So, if we left at 4 pm, Cincinnati people would get back at 1 am Sunday, so it just works better to leave Sunday morning.

Friday night seems like the best option then.

I started the thread early so people who are going can trickle in over the next two months (this forum isn't exactly super-active).

Well, this is something I would consider.  You just need to explicitly state a place and time.  Something like "Findlay Commons" isn't good enough, because there are several possible places to gather there.

I'd try to get there. now, question- Its U Penn or Penn State? I know theres a big difference, but I'm not sure which ARML goes to.

It's Penn State. That's State College, PA, not Philadelphia.

sounds cool- pity I can't eat at Chinese restaurants cause of a shellfish allergy :(

I'm going, and I would be happy to meet-up

I know nothing about the schedule at ARML though so...

And, I would love to go to a Chinese restaurant  :lol:

I'd possibly go if i can make it. I'm usually practicing with my team for the song contest(and at the first round btw everyone cheer for Lehigh Valley Fire's song :) ).  But it sounds good; chinese rocks as well

gah, I can't go until next year!!! I will have restless leg syndrome for 13 months!!!

I'm going to Penn State (On AAST, which won the song last year  :P )

I went last year... going for my second time at age 11!!!

actually managed to get one of the Individuals last year =D

Anyway, I THINK we leave right after the contest, so like Friday

night would be best for a meeting, but I dunno if I can do that...

I think my Dad is coming with me (......)

I'll ask him if people start showing interest about this.......

[quote="CatalystOfNostalgia"]We should set up a thing where like 500 people go into some Chinese restaurant (I believe there are at least two) at the same time. I totally want to take a video of the reactions of the people behind the counter.[/quote]Yes. Very yes.

So Friday night, then?

Friday night is certainly the optimal time... hopefully before the song contest prelims of course

And someone definitely bring a video camera to the Chinese place :P

Yeah- I may not be able to eat there, but I'd go just to see that reaction!

But then, there are a few questions left to answer:

[b]Which Chinese restaurant?[/b]
CoN said that there are at least two.  Which one will it be, and where is it?

[b]When will this occur?[/b]
"Before the song contest prelims" is too general.  A time in the form "HH:MM" is preferred.

8:00ish is probably best - as far as I can remember the CT team was free by then last year.

oh man...I didn't think my idea was ACTUALLY going to be implemented, sadly I can't even go to ARML this year.

[quote="worthawholebean"]On second though, I'm not sure this is going to work as we won't have many people. Let's just meet at a restaurant - does anyone remember them well?[/quote]

There is a pizza one really nearby as well as a smallish supermarket down the hill from the college.

I say let's just all try to meet at 8:00 at the pizza place.

Sounds good! I'll definitely be there if I'm not choreographing embarrassing dances  for the Song Contest  :P

Uhhh, does anyone know the name of the pizza place? And about how far is it from the Commons? (Sorry, I have terrible navigation skills  :( )

It's one of the first places you see when you arrive downtown from the campus. I don't remember the name.

You go down the hill take a RIGHT, then cross the street.

8:00 at pizza place- got it and can eat at those- yay! cyall there

[quote="worthawholebean"]Anyway, I'm the boy with orange hair.[/quote]

[img]http://upload.wikimedia.org/wikipedia/en/6/64/IchigoAnimeEp113.jpg[/img]

??

Umm.... also...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205630

YOINK!

[quote="lingomaniac88"][quote="worthawholebean"]Anyway, I'm the boy with orange hair.[/quote]
How do you know you're going to be the only one?
(I don't have orange hair, BTW.  I'm just bringing up a point.)[/quote]

how many other non-japanese people with orange hair do you know?!?!?!  :huh:

way to leave us hanging at the pizza place wawb


complex numbers
superior algebra
superior algebra solved


Let $ a,b,$ and $ c$ be three angles such that $ a\plus{}b\plus{}c\equal{}\pi$. If we are given that $ \tan{a}\tan{b}\equal{}\csc{\frac{\pi}{3}}$, than determine $ \frac{\cos{a}\cos{b}}{\cos{c}}$

[hide]set cosc=180-a-b and try expanding[/hide]

[quote="erdogankerem123"]Let $ a,b,$ and $ c$ be three angles such that $ a \plus{} b \plus{} c \equal{} \pi$. If we are given that $ \tan{a}\tan{b} \equal{} \csc{\frac {\pi}{3}}$, than determine $ \frac {\cos{a}\cos{b}}{\cos{c}}$[/quote][hide]$ \tan\theta\equal{}\frac{\sin\theta}{\cos\theta}$

$ \cos(\alpha\plus{}\beta)\equal{}\cos\alpha\cos\beta\minus{}\sin\alpha\sin\beta$

$ \cos(\pi\minus{}\theta)\equal{}\minus{}\cos\theta$

$ \tan{a}\tan{b} \equal{} \csc{\frac {\pi}{3}}$

$ \frac{\sin a\sin b}{\cos a\cos b}\equal{}\csc{\frac {\pi}{3}}$

$ \sin a\sin b\equal{}\cos a\cos b\csc{\frac {\pi}{3}}$

$ \sin a\sin b\minus{}\cos a\cos b\equal{}\cos a\cos b\left(\csc{\frac {\pi}{3}}\minus{}1\right)$

$ \minus{}\cos(a\plus{}b)\equal{}\minus{}\cos(\pi\minus{}c)\equal{}\cos c\equal{}\cos a\cos b\left(\csc{\frac {\pi}{3}}\minus{}1\right)$

$ \frac {\cos{a}\cos{b}}{\cos{c}}\equal{}\frac1{\csc{\frac {\pi}{3}}\minus{}1}\equal{}1$[/hide]


Im an 8th grader who took AMC12
Im just wondering if the "floor score" for qualifying into USAMO applies to me as well even thought i took 12, not 10.

(I hope it does)

Yes, it does. It's all good for 10th graders and under.

Funny thing is, if i took the AIME right now, i would get about 0-1 question right. I have 2 weeks (which i plan to sleep 0 hour to study  :lol: ) to prepare. Any thoughts on how well i'll do?
(GOTTA BE SMARTER!)

Well, first of all, you can't get no sleep, or you'll do worse than if you didn't study at all...

"The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value."

Are you absolutely positive that the floor score applies to me?  It clearly says "among students who took the AMC 10A or AMC 10B..."
I really need to know

I don't know where you're getting this quote, but it's out of date.

As usual, if you want the most comprehensive, accurate information, go to the source.  The 2007 USAMO Teachers' Manual is posted on the [url=http://www.unl.edu/amc]AMC website[/url].  (See Chapter 2: Selection Process)

i got it from the official website (near the bottom of the USAMO page)

Whoa...am I reading wrong, or does the manual imply that during the selection of floor-score qualifiers, they are STILL going to count the AMC-12 and AMC-10 equally?! After last year's issues, I would have expected them to have figured out something better by now. A score on the AMC-12 is worth much, much more than the same score on the AMC-10 no matter how you look at it.

Last year they changed the process at the last minute to avoid that and just let everyone who made the floor take it. I believe that if the situation comes up again, they will avoid it again.

That provision was always in there, but it was never invoked until last year.  Then they increased the number of qualifiers to un-invoke it and let everyone who got the floor score take USAMO.

I understand that, but it seems to me that having such a rule at all is just asking for trouble, even if the organizers have managed to dodge it so far. What if there's a 300-way tie at 7 or 8 or whatever? Why not just have a more useful rule in place in case of emergency? Anything would be better than counting the 12 and the 10 equally.

I agree, actually.  I have often wondered why they don't just say that they'll raise the AIME floor if too many people qualify.  That would very heavily favor AMC12 takers.  Or give priority to younger students--there was a year or two (2000 and/or 2001) where a floor-like system was used, with a floor of 8 for 10th graders and 7 for 9th grade and below.


This problem must be posted before.

Let $ S$ and $ 2p$ be the area and perimeter of the triangle $ ABC$, respectively. Prove that

$ \tan\dfrac{A}{2}\cdot\tan\dfrac{B}{2}\cdot\tan\dfrac{C}{2} \equal{} \dfrac{S}{p^2}$

Inscribe a circle in the triangle. Then use two-tangent theorem and area forumla for the triangle. So we have that x+y=a, x+z=b, z+y=c. Therefore x+y+z=p. And r(x+y+z)=s. Also tan A/2=r/x etc. so we can set up an equation.

$ \frac{r}{y}\cdot\frac{r}{x}\cdot\frac{r}{z}\equal{}\frac{r(x\plus{}y\plus{}z)}{(x\plus{}y\plus{}z)^2}$

$ \frac{r^3}{xyz}\equal{}\frac{r}{x\plus{}y\plus{}z}$

$ \frac{r^2}{xyz}\equal{}\frac{1}{x\plus{}y\plus{}z}$

Now I can't seem to get any further.

[hide]$ \tan\dfrac{A}{2}\cdot \tan\dfrac{B}{2}\cdot \tan \dfrac{C}{2} = \dfrac{\sin \displaystyle{\dfrac{A}{2} \cdot \sin\dfrac{B}{2}\cdot \sin \dfrac{C}{2}}}{\displaystyle{\cos \dfrac{A}{2} \cdot \cos \dfrac{B}{2} \cdot \cos \dfrac{C}{2}}}$

[color=darkblue]Lemma 1[/color]:

$ 1 - \cos A = \dfrac{2\cdot(p - b)\cdot(p - c)}{bc}$
Prove it.

Hence,

$ \sin\dfrac{A}{2} = \sqrt {\dfrac{1 - \cos A}{2}} = \sqrt {\dfrac{ (p - b)\cdot (p - c)}{bc}}$;
$ \sin\dfrac{B}{2} = \sqrt {\dfrac{ (p - a)\cdot (p - c)}{ac}}$;
$ \sin\dfrac{C}{2} = \sqrt {\dfrac{ (p - a)\cdot (p - b)}{ab}}$.

Therefore,

$ \sin \dfrac{A}{2} \cdot \sin\dfrac{B}{2}\cdot \sin \dfrac{C}{2} = \dfrac{(p - a)\cdot(p - b)\cdot(p - c)}{abc}$

But

$ S = \sqrt {p\cdot(p - a)\cdot(p - b)\cdot(p - c)}\Longrightarrow (p - a)\cdot(p - b)\cdot(p - c) = \dfrac{S^2}{p}$

Thus,

$ \sin \dfrac{A}{2} \cdot \sin\dfrac{B}{2}\cdot \sin \dfrac{C}{2} = \dfrac{S^2}{p\cdot abc} (*)$

[color=darkblue]Lemma 2[/color]:

$ 1 + \cos A = \dfrac{2p\cdot(p - a)}{bc}$
Prove it.

Hence,

$ \cos\dfrac{A}{2} = \sqrt {\dfrac{1 + \cos A}{2}} = \sqrt {\dfrac{p\cdot (p - a)}{bc}}$;
$ \cos\dfrac{B}{2} = \sqrt {\dfrac{p\cdot (p - b)}{ac}}$;
$ \cos\dfrac{C}{2} = \sqrt {\dfrac{p\cdot (p - c)}{ab}}$.

Therefore,

$ \cos\dfrac{A}{2}\cdot \cos\dfrac{B}{2}\cdot \cos\dfrac{C}{2} = \dfrac{p}{abc}\cdot\sqrt {p\cdot(p - a)\cdot(p - b)\cdot(p - c)}$

We know that $ S = \sqrt {p\cdot(p - a)\cdot(p - b)\cdot(p - c)}$

Thus

$ \cos\dfrac{A}{2}\cdot \cos\dfrac{B}{2}\cdot \cos\dfrac{C}{2} = \dfrac{p\cdot S}{abc}(**)$

And from $ (*)$ and $ (**)$

$ \tan\dfrac{A}{2}\cdot \tan\dfrac{B}{2}\cdot \tan \dfrac{C}{2} = \dfrac{S}{p^2}$[/hide]

[quote="sonny"]Inscribe a circle in the triangle. Then use two-tangent theorem and area forumla for the triangle. So we have that x+y=a, x+z=b, z+y=c. Therefore x+y+z=p. And r(x+y+z)=s. Also tan A/2=r/x etc. so we can set up an equation.

$ \frac {r}{y}\cdot\frac {r}{x}\cdot\frac {r}{z} \equal{} \frac {r(x \plus{} y \plus{} z)}{(x \plus{} y \plus{} z)^2}$

$ \frac {r^3}{xyz} \equal{} \frac {r}{x \plus{} y \plus{} z}$

$ \frac {r^2}{xyz} \equal{} \frac {1}{x \plus{} y \plus{} z}$

Now I can't seem to get any further.[/quote]
sonny, you had it!
[hide="Follow Up"]
By Heron's, the area of the triangle is \[ \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\equal{}\sqrt{(x\plus{}y\plus{}z\minus{}x\minus{}y)(x\plus{}y\plus{}z\minus{}y\minus{}z)(x\plus{}y\plus{}z\minus{}x\minus{}z)(x\plus{}y\plus{}z)}\equal{}\sqrt{xyz(x\plus{}y\plus{}z)}\] \[ [ABC]^2\equal{}xyz(x\plus{}y\plus{}z)\] \[ \frac{xyz}{x\plus{}y\plus{}z}\equal{}\frac{[ABC]^2}{(x\plus{}y\plus{}z)^2}\equal{}(\frac{[ABC]}{s})^2\equal{}r^2\implies \frac{r^2}{xyz}\equal{}\frac{1}{x\plus{}y\plus{}z}\] [/hide]

[quote="The QuattoMaster 6000"][quote="sonny"]Inscribe a circle in the triangle. Then use two-tangent theorem and area forumla for the triangle. So we have that x+y=a, x+z=b, z+y=c. Therefore x+y+z=p. And r(x+y+z)=s. Also tan A/2=r/x etc. so we can set up an equation.

$ \frac {r}{y}\cdot\frac {r}{x}\cdot\frac {r}{z} \equal{} \frac {r(x \plus{} y \plus{} z)}{(x \plus{} y \plus{} z)^2}$

$ \frac {r^3}{xyz} \equal{} \frac {r}{x \plus{} y \plus{} z}$

$ \frac {r^2}{xyz} \equal{} \frac {1}{x \plus{} y \plus{} z}$

Now I can't seem to get any further.[/quote]
sonny, you had it!
[hide="Follow Up"]
By Heron's, the area of the triangle is
\[ \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)} \equal{} \sqrt {(x \plus{} y \plus{} z \minus{} x \minus{} y)(x \plus{} y \plus{} z \minus{} y \minus{} z)(x \plus{} y \plus{} z \minus{} x \minus{} z)(x \plus{} y \plus{} z)} \equal{} \sqrt {xyz(x \plus{} y \plus{} z)}
\]

\[ [ABC]^2 \equal{} xyz(x \plus{} y \plus{} z)
\]

\[ \frac {xyz}{x \plus{} y \plus{} z} \equal{} \frac {[ABC]^2}{(x \plus{} y \plus{} z)^2} \equal{} (\frac {[ABC]}{s})^2 \equal{} r^2\implies \frac {r^2}{xyz} \equal{} \frac {1}{x \plus{} y \plus{} z}
\]
[/hide][/quote]

Ah, using Heron's never came to mind, but thanks!


If $ G$ is the barycenter of the triangle $ ABC$, prove that $ \overline{AB}^2\plus{}\overline{BC}^2\plus{}\overline{CA}^2\equal{}3\cdot(\overline{GA}^2\plus{}\overline{GB}^2\plus{}\overline{GC}^2)$

[hide="Hint"] Stewart Theorem[/hide]

Let $ A=a$, $ B=b$, $ C=c$, $ G=g=\frac{1}{3}*(a+b+c)$ where $ a,b,c$ are complex numbers.

Then \[ \sum |AB|^2-3|GA|^2=\sum (a-b)(\overline{a}-\overline{b})-3(g-a)(\overline{g}-\overline{a})\]
\[ \sum (a\overline{a} -a\overline{b}-b\overline{a}+b\overline{b})
-3
(g\overline{g} -g\overline{a}-a\overline{g}+a\overline{a})\]

\[ \sum[ (-a\overline{a} )+(-a\overline{b}-b\overline{a})+3a*\overline{g} ]-3\sum (g\overline{g} -g\overline{a})
\]
\[ \sum a*( \overline{a}+\overline{b}+\overline{c})-
a\overline{a}-a\overline{b}-a\overline{c}=0\]

which proves the identity.

Is a barycenter, just the centroid of the triangle?

Yep, if centroid means the point of intersection of the medians.


[quote="diophantient"]I lost my feedback from last year.
Is there anyway I can reaccess the photos of grader feedback?[/quote]
I also lost my feedback for Year 17 and Year 18. However, the Year 16 feedback is still there... I wonder why.

Yes, I can see my Year 16, Year 17, and Year 18 feedback now.

Is there a date on when Year 19 Round 1 will be released?
Can I submit my Entry Form with my set of solutions for Round 1?
And, if I'm going to use LaTeX, I need only to submit the tex file and the pdf file, and any images (even though I won't have any since I'll use Asymptote), right?

The date for Round 1 release is not yet set - it will be within the next couple weeks.

You can send your Entry Form with Round 1.

Those are indeed the files to send if you use LaTeX.


Ring Theory
function
algebra
polynomial
number theory
prime factorization
complex numbers

[quote="sharkman"]Use synthetic division:

f(x) = x^3 - 7x^2 - x + 87 ;   Show 5 + 2i is a zero.

The term 5 + 2i confuses me due to the letter i next to 2.[/quote]
Consider the UFD $ \mathbb{Z} i$
We have that under this ring, unique prime factorization exists, as well as a unique factorization domain. Incidentally, there exists a Norm function that is commutative such that $ N(x) \cdot N(y)\equal{}N(xy)$ and $ N(x)\equal{}|x|^{2}$. So we have that a gaussian integer is prime iff the norm function is prime, since we aren't guaranteed that a "prime" number if the field $ \mathbb{Z}$ is actually prime.

This is some interesting math, and this is the basis of it (Fundamental Theorem of Arithmetic).

[quote="sharkman"]Use synthetic division:

f(x) = x^3 - 7x^2 - x + 87 ;   Show 5 + 2i is a zero.

The term 5 + 2i confuses me due to the letter i next to 2.[/quote]

Short: i² = -1. So just work with i as if it was any real number and substitute -1 for any i² you come across.

Just make sure that you remember that not all $ \mathbb{Z} \sqrt{d}$ are UFD.

We know that for a real-coefficient polynomial, if it has an imaginary zero, then it must have its conjugate also as a zero, which implies it has a factor of degree 2 with real coefficient obtained by
$ (x \minus{} a)^2 \plus{} b^2$ for a zero $ a \plus{} bi$
and then you divide the polynomial by this quardratic and check its remainder.

(Directly dividing it by the imaginary number is...kind of tedious, because of the repeated foiling)

P.S. What's UFD and all the other stuff?

A discussion of algebraic number theory is hardly appropriate for introducing someone to imaginary numbers.  :|  [b]diophantient[/b], your discussion isn't even relevant to his problem.

[b]sharkman:[/b]  $ i$ is called an "imaginary number."  It is one of the roots of the equation $ x^2 \plus{} 1 \equal{} 0$, which has no solution in the ordinary ("real") numbers.  However, it is such a useful mathematical tool that it is used even though it is not "real."

The rules for working with numbers of the form $ a \plus{} bi$ (called "complex numbers"), where $ a, b$ are real, are basically the same as the rules for working with ordinary numbers, except that whenever we calculate $ i^2$ we write it as $ \minus{}1$.  For example,

$ (2i) \cdot (3i) \equal{} \minus{}6$
$ i(1 \plus{} 2i) \equal{} \minus{}2 \plus{} i$

The complex numbers are very useful because it turns out that any polynomial of dimension $ n$ will have [b]exactly[/b] $ n$ complex roots (counting multiplicity).  Anyway, for your problem, all you have to do is divide by $ (x \minus{} (5 \plus{} 2i))$.  This is equivalent to evaluating

$ f(5 \plus{} 2i) \equal{} (5 \plus{} 2i)^3 \minus{} 7(5 \plus{} 2i)^2 \minus{} (5 \plus{} 2i) \plus{} 87$

and showing that it is $ 0$.

I think the point of his post was to be annoying to sharkman who posted lots of stupid problems a few months ago.

and since this post isn't going to be total spam

Look at this:
http://www.purplemath.com/modules/synthdiv.htm


conics
parabola
analytic geometry
symmetry
rotation

45. On an ordinary coordinate system with perpendicular x and y axes, the graph of $x=y^{2}$ is a parabola. If the angle between the positive x and y axes is changed to 60 degrees, but the parabola's orientation is unchanged (i.e., vertex at the origin, axis of symmetry on the x-axis), what is its equation with respect to the new pair of axes?

A) $4x=3y^{2}-2y$
B) $4x=2y^{2}-3y$
C) $2x=y^{2}\sqrt{3}-y$
D) $2x=y^{2}\sqrt{3}-2y$
E) $x\sqrt{3}=3y^{2}-2y$

Observation:
[hide]
It seems clear we'll need to use polar coordinates, but I'm not really sure how to apply them. I tried a few variations of $\theta+30^{o}$ but none really worked.
[/hide]

The correct answer from the key:
[hide]
A) $4x=3y^{2}-2y$

But I don't know how to get it yet.
[/hide]

Take three points on the parabola that are easy to transfer over, like $(0,0)$, $(-1, 1),$ and $(1, 1)$ and find the final value for these (like the first one won't change). Then just find the appropriate parabola for this either by plugging in (that's kinda cheating though) or, you could use matrices to rotate each x value, y value, etc. check out aops2


geometry
3D geometry
sphere
function
topology
advanced fields
advanced fields unsolved

Let S^2 be the unit sphere {(x,y,z) in R^3: x^2 +y^2 + z^2 = 1} in R^2.
Let d be the arc length metric along great circles, ie. the distance between P,Q in S^2 is the angle POQ subtended by P,Q at the origin 0 = (0,0,0) in R^3
(Another way to describe this is as follows. The plane through P,Q and O intersects S^2 in a circle C. Then d(P,Q) is the arc length distance between P and Q in C.)

Now let g:S^2 -> S^2 be a map such that d(g(P),g(Q)) >= d(P,Q) for all P, Q in S^2.
Fix P_1, Q_1 in S^2.
How do I show that there is a subsequence (g_i)^n such that g^(n_i) (P_1) and g^(n_i) (Q_1) are convergent?

Thank you for your help,

LJB

A partial solution, which probably isn't the style you're looking for:

First, we show that $g$ is actually an isometry, which makes it continuous. Somewhere on the forum, there was a thread discussing this problem (If $X$ is a compact metric space, and $f$ is a map from $X$ to itself such that $d(f(x),f(y))\ge d(x,y)$, then $f$ is an isometry), but I couldn't find the thread.

Since the functions are continuous and the spaces are compact, it should be asy to find convergent subsequences of any sequence. It may not even be necessary to show that $g$ is continuous.


Let  $ A,B$are $ n$x$ n$ Hermitian matrices and positive definite.
Prove that :
1) $ AB^{\minus{}2}A\plus{}BA^{\minus{}2}B \geq 2I_n$
2) $ A^{\frac{1}{2}}(B\plus{}C)^{\minus{}1}A^{\frac{1}{2}}\plus{}B^{\frac{1}{2}}(A\plus{}C)^{\minus{}1}B^{\frac{1}{2}}\plus{}C^{\frac{1}{2}}(A\plus{}B)^{\minus{}1}C^{\frac{1}{2}} \geq \frac{3I_n}{2}$
 Here $ I_n$ is identity matrix.
 "$ \geq$" is semi-positive definite.
Thanks ! :roll:

It is false ! Sorry everyone ! :blush:
But we have :
1) $ Tr( AB^{ \minus{} 2}A) \plus{} Tr(BA^{ \minus{} 2}B) \geq 2n$
2)$ Tr(A^{\frac {1}{2}}(B \plus{} C)^{ \minus{} 1}A^{\frac {1}{2}}) \plus{} Tr(B^{\frac {1}{2}}(A \plus{} C)^{ \minus{} 1}B^{\frac {1}{2}}) \plus{} Tr(C^{\frac {1}{2}}(A \plus{} B)^{ \minus{} 1}C^{\frac {1}{2}}) \geq\frac {3n}{2}$
( is true  :P )
http://www.mathlinks.ro/Forum/viewtopic.php?t=272660


Is there any way to send in the entry form for the USAMTS by e-mail?  If not, can I send in the entry form separately from the solutions for round 1?  I would like to send in my solutions by e-mail so that they have the possibility of being published.  Also should I send in the LaTeX source file as a text file or otherwise in my e-mail?

The entry form requires hand-written signatures, so it must be sent in by mail or fax, to the address (if by mail) or phone # (if by fax) shown on the form.

It is perfectly acceptable to send in the form(s) by mail, but the solutions by email; in fact, this is the preferred method.

For your solutions, you can send them as attachments to an email message.  Send the .tex file (the input file), the .pdf file (the output file), and any pictures you might be including.

More details here:
[url]http://www.artofproblemsolving.com/USAMTS/AoPS_U_AbPart.php[/url]


It would help a lot if the question had said that it was using the LINES as the paths and not the squares.  I've missed these types of questions soo much because of that. Or you could add dots for points e f and g
that would also help

[quote="GameBot"]How many 9 step paths are there from $ \textrm{E}$ to $ \textrm{G}$ which pass through $ \textrm{F}$?[asy]size(4cm,4cm);int w=6;int h=5;int i;pen p=fontsize(9);for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("G", (w-1,0), SE, p);label("E", (0,h-1), NW, p);label("F", (3,3), NE, p);[/asy][/quote]

You have $ \binom{4}{1}$ ways to get from $ E$ to $ F$.  Then, you have $ \binom{5}{2}$ ways to get from $ F$ to $ G$.  Multiplying reveals $ \boxed{40}$.

But I don't get the reason for 5 choose 2.

The reason for $\binom{5}{2}$ is because we can move downwards twice, so we can pick two out of the five steps to get from $F$ to $G$.

please hide your answers, and i am also pretty sure your answer is incorrect

[hide = Solution]This is the number of ways from E-F, multiplied by the number of ways from F-G. From E-F is just $\binom{4}{1} = 4$, and from F-G is just $\binom{5}{2} = 10$.
Multiplied gives $10 \cdot 4 = \boxed{40}$. [/hide]

[quote=mathwinner1]How many 9-step paths are there from $E$ to $G$?[/quote]

[hide=Sol]That would be $\binom{9}{5} = 126$[/hide]

Woah, a lot of different answers here! Here is my approach on this question.

To get from [b]E[/b] to [b]G[/b] within $9$ steps means that one can only get closer to [b]G[/b] with each move (bare minimum is $4$ down and $5$ right, $4+5=9$).

Let us first compute the number of possibilities to get from [b]E[/b] to [b]F[/b]. One must go $1$ down and $3$ right.

$$\frac{(1+3)!}{1!3!} = \frac{4!}{1!3!} = \frac{24}{6} = 4.$$

Now let us compute the number of possibilities to get from [b]F[/b] to [b]G[/b]. One must go $2$ right and $3$ down.

$$\frac{(2+3)!}{2!3!}=\frac{5!}{2!3!}=\frac{120}{2 \cdot 6}=\frac{120}{12}=10.$$

Multiplying the two possibilities, we get $4 \cdot 10 = \boxed{40}$.

[quote=vectorfunction]40 should technically be incorrect, as it is not stated within the problem that the paths must be down and over.[/quote]

It does not state down or right, however there is no way to make it from E to G in 9 steps if you go up or left.


It's called a bump, not an UPPP!

If you want word problems, check out a few of them that made your thread go DOWN!!!!

They are all really nice word problems made by committees.

[quote="AIME15"]It's called a bump, not an UPPP![/quote]

no

And there are no real world problems my dude. smh

There are some real world problems on the AMC 10. Check it out on the Wiki or on the Contests section (you can find the button by scrolling up, clicking on 'Contests' then click on 'USA', then click on 'AMC 10').

[quote="TheProblem856"][quote="Rajiv"][quote="TheProblem856"]I need some probability word problems...They should "real world" problems. For example, an Ice Cream can have 3 scoops of different ice cream...There are 21 different flavors, how many ways can...etc.
someone hook me up with some problems and the answers to them.

I need at least 1 of each:
Combination
Linear
Circular Permutation
Permutation with Repetition

THX[/quote]
What is probability that a selection of four cards from a pack of cards will contain [b]atleast[/b] two black cards or an honour [b]instead[/b].

This question satisfies most of your demands.[/quote]

Thanks...Is this a combination problem? Whats an honour?[/quote]
Don't know honours, never played?
The king, queen, jack and the ace are the honours.

In the US, they're called face cards, and an Ace is just...well...an Ace :)


Let m and n be given positive integers such that mn divides m^2 + n^2 + m. Prove that m is a square of an integer.

We have that $mn$  divides $m^2+n^2+m$
Suppose $p^{2k+1}$ divides $m$, then $p^{2k+1}$ divides $n^2$, so $p^{k+1}$ divides $n$.
Hence $p^{2k+2}$ divides $mn$, but is a factor of $m^2+n^2$, too.
We conclude that $p^{2k+2}$ will divide $m$.

I have other solution
$ Let gcd(m,n) \equal{} d$ then $ m \equal{} da,n \equal{} db$
$ gcd(a,b) \equal{} 1$
................
This is my result $ m^{2}\plus{}n^{2}\plus{}m \equal{} 3mn$
Then there exist infty (m,n) satisfy the equation(You can file all )


Solve the following equation:
4x^5-7=m^2
4x^13-7=n^2
where x,m,n are integers.

I think the only solution is ( x, m, n )=( 2, 11, 181 ), but I don't know how to prove it.Can anyone help me? Many thanks！ :huh:

You can subtract the equations and then factor it.

$ 4x^5(x^4\plus{}1)(x^2\plus{}1)(x\plus{}1)(x\minus{}1) \equal{} (m\plus{}n)(m\minus{}n)$

Maybe that'll help you.

The problem asked for [b]real[/b] $ x$.
There are couple of ways to show that $ x$ is still an integer. Then multiply both equations and try to bound the result between consecutive squares.

Or if you feel lucky, try to solve $ 4x^5\minus{}7\equal{}y^2$ in integers, without the second equation (or for the second one without the first).

Zetax siad "multiply both equations and bound the result does help" Would you say it more clearly?

ZeTaX meant

$ (4x^9 \minus{} 3.5x^4 \minus{} 1)^2 < (4x^5 \minus{} 7)(4x^{13} \minus{} 7) < (4x^9 \minus{} 3.5x^4)^2 \text{ } \forall x \ge 4$

(it is not difficult to see $ x$ is even)

The intermediate term is $ (mn)^2$, which implies $ 4x^9 \minus{} 3.5x^4 \minus{} 1 < mn < 4x^9 \minus{} 3.5x^4$. So there is an integer between two consecutive integers: contradiction.


hi
how do you pronounce this in English? :)
thanks


i came by solving a symmetry problem...

$\text { Factor } a^3 + b^3 + c^3 - 3abc$

and in the solution it said

$ = (a+b+c)(a^2+b^2+c^2) - \displaystyle \sum_{\sigma} (a^2b + b^2a) - 3abc$

what does it mean by $\displaystyle \sum_{\sigma}$?

This is a really cool factoring, also mentioned in the first chapter of Math Olympiad Treasures. The key to that one is making a polynomial with a, b, and c as roots.

[hide="The Factoring"]
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ [/hide]

well that one was the answer that the book was leaning towards...hmm


Normally when learning about algorithms and bubble sorts and searches, you speak
of using ONE dummy variable.  Could a search/sort algorithm be made faster, on average, by including two, or maybe three dummy variables in the flowchart, instead of the expected one dummy variable?

Could you give an example of the dummy variables of which you speak?


Calculati primitivele urmatoarelor functii
a)$f:\mathbb{R}\to\mathbb{R}$, $f(x)=\frac1{2x^3+3}$
b)$f:\left(0,\frac\pi2\right)\to\mathbb{R}$,

$f(x)=\frac{2\sin x+3\cos x}{2\cos x+3\sin x}$

c) $f: (0,\infty)\to\mathbb{R} $

$f(x)=\frac{x^2+1}{(x^2-1)\sqrt{x^4-1}}$

1. descompunere in factori ireductibili si apoi in fractii
2.se determina a si b din relatia a(2cosx +3sinx)+b(3cosx-2sinx)=2sinx+3cosx
3.trebuie sa avem x>1. Substitutia $t^2=\frac{x^2+1}{x^2-1}$


I decided to start a new one cause the other one was getting very cluttered.
New rules:
Everybody starts out with $ \$1000$. You can get a job to earn more. The only person you can work for until someone makes another factory is me. You need $ \$10,000$ to make factory. If you take a job at my factory, then you get paid 1000 dollars per day that you are alive. Every day, $ \$100$ is deducted from your account for daily food expenses. Also, I might pick random people per week that get sick and have to pay $ \$200$ to live. And, you can get health insurance for $ \$100$ a week to save you from being sick.  You may not work for more than one factory .More rules will be introduced as the game progresses. sign ups start now.

i sign up and pay for life insurance
EDIT: and work at your factory
p. s.
when are the life insurance fees?

[b]Join[/b] is factory worker seriously the only job.....

:lol: dude, don't count the teeth of a gifted horse! its the only job you got!

Teeth of a gifted horse...

So basically we live in a factory, get money, and eat, and get sick, and that's all?

[quote="Poincare"]I decided to start a new one cause the other one was getting very cluttered.
New rules:
Everybody starts out with $ \$1000$. You can get a job to earn more. The only person you can work for until someone makes another factory is me. You need $ \$10,000$ to make factory. If you take a job at my factory, then you get paid 1000 dollars per day that you are alive. Every day, $ \$100$ is deducted from your account for daily food expenses. Also, I might pick random people per week that get sick and have to pay $ \$200$ to live. And, you can get health insurance for $ \$100$ a week to save you from being sick.  You may not work for more than one factory .[i][size=200][color=red][b]More rules will be introduced as the game progresses[/b][/color][/size][/i]. sign ups start now.[/quote]

I join and I get a job at that factory.

Why thank you!
[color=white]I hope that good behavior can get me promoted and I earn more.[/color]

[b]I join and get a job at the factory....
[/b]
wow that's a lot of money to earn for a factory worker! but then, cost of living is very high as well. lol.

yay?
i now have 2700 dollars (assuming you give me my health insurance now)
so...

Yeah, i also have 2700 bucks...

This is quite boring, and no, i wouldn't say its better than aops city.

sorry guys game not started yet. but this post signifies that it starts.
Chaching! all you guys you work at my factory get $ \$1000$! But, $ \$100$ reduction for food prices. I will accept late sign ups. And, for the weekly sick thing, there's gonna be a sick person every month :P so, for this month itssss........ AIME 15! $ \$200$ off you account!

So you pick the person who complains to be sick...

I don't think this game is very fair.

so now food is free?
oh well... did you make sure i have health insurance?

it is.
Stevenmeow: $ \$2700$
AIME15:
1=2:$ \$2700$
james4l:$ \$2700$
chenshsi:$ \$2700$
crazychinesemaniac:$ \$2700$
ak909090: $ \$2700$

ok, i'm announcing a lottery. But, it has special rules, more coming up tomorrow (gotta go to bed). PEOPLE, PLEASE SIGN UP.


abstract algebra
superior algebra
superior algebra unsolved

Consider 6 groups :abelian, Solvable, finite generated solvable, nilpotent, supersolvable and polycyclcic. Help me to Find the relationship among them. If group A is B, please prove. If group A isnot group B, please get an axample to prove it.

If we restrict ourselves to finitely generated groups, we can consider the following arrangement of classes of groups:
cyclic < abelian < nilpotent < supersolvable < polycyclic < solvable < finitely generated group.

Source: http://en.wikipedia.org/wiki/Solvable_group

I can't understand this diagram clearly, because If A is abelian, so that we can't have A is finitely generated.

[quote="abdul"]If we restrict ourselves to finitely generated groups
[/quote]

@dhdung This is only considering finitely generated groups. That's why we can say, if A is abelian, it is finitely generated. Because all of them are.

ohh, I see.
In the case that A is an arbitary group. How are their relationship?


ratio
geometry
geometric transformation
rotation
analytic geometry
Pythagorean Theorem
geometry open

Is there a way of defining lengths and angles that isn't circular?

All the definitions of angle I have seen define it as the ratio of two lengths, usually the arclength between two straight lines and the length of the lines.

However, length is defined using the pythagorean theorem, which uses a right triangle, which isn't defined if we haven't defined a right angle yet.  Also, nearly all proofs of the pythagorean theorem rely on either angles or the concept of rotating triangles (in which case we have to define rotation, which will be hard without using the concept of angles.)

Is there any way around this conundrum?  They're both intuitive concepts we know from the real world, so much so that pinning down an exact definition is actually rather difficult.

If we use coordinates as our basis for geometry, we may define a point to be an ordered pair $(x,y)$ and the distance from $(x_{1},y_{1})$ to $(x_{2},y_{2})$ as $\sqrt{\left( x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}.$ This is just a definition and may be made arbitrarily as long as it gives a unique real number $AB$ for all points $A$ and $B$. We then define angle by arctangent series and proceed to show that our system satisfies the postulates of standard Euclidean geometry. The Pythagorean theorem is only necessary if we proceed in the other direction, making synthetic geometry the ultimate basis, taking length as an undefined term, either defining angle in terms of length or leaving it undefined also, and adopting synthetic postulates. From these foundations the Pythagorean theorem is proved, coordinates are defined in terms of length and perpendicularity, and the distance formula appears as a theorem rather than a definition.


in the older entries in my blog, where we could put the mood we were in, it says things like "mood: icon_important" and "mood: challenge" and it show those little icons.  how can I fix that?

also, when will i be able to change my blog's background again?

You can change your blog background even now. Look at ccy's blog, or Richard's blog how they look for example.

I will restore the moods in the near future also. For now you can use "Currently".


[quote="calc rulz"]I emailed my solutions but didn't include all that info (name, ID, round #, etc) in the actual document (just in the subject line), will my solutions still be valid?[/quote]

Hopefully we'll catch that when we file the paper, but in the future, make sure you get that info on every single page.


calculus
integration
calculus computations

$ \int_0^\infty\,\frac{\textbf dx}{1+\left(e^x-1-x\right)^2\right)}$

 :|

Heh misan. I am still relatively new here but if i EVER had a calculus question I would definitely ask you. This integral must indeed be...well for lack of an elegant word, hard :P


integration
calculus
calculus computations

Hello everyone:

Could someone please point out my mistake in solving this differential equation?

Thank you.

---

1. Solve the differential equation $ x^2 dy = -2dx$.

---

$ x^2 dy = -2dx$

$ dy = \frac{-2dx}{x^2}$

Since: $ u = x^2, du = 2x dx$, therefore: $ -\frac{du}{x} = -2dx$.

$ y = \int \frac{-du}{u^2x}
= \int \frac{1}{ux} \times du
= -ln|ux| + C
= -ln|x^3| + C$

The answer provided is
[hide]$ y = \frac{2}{x} + C$[/hide]

Your substitution $ u\equal{}x^2$ isn't right. It should be $ u \equal{} \minus{}\frac{1}{x^2}$

But an easier way to do all this is
$ dy \equal{} \minus{}2\frac{dx}{x^2}$

Integrate both sides to their respective,
$ y \equal{}2\frac{1}{x} \plus{}C$

Also you don't need this to be a $ \text{DE}$

Just put it in the form  $ \frac{dy}{dx} \equal{} \frac{\minus{}2}{x^2}$
and integrate normally,

$ \int {\minus{}2x^{\minus{}2}}dx$ and use $ \int {x^n} \equal{} \frac{x^{n\plus{}1}}{n\plus{}1}$

[color=green] The quote removed by moderator. It never makes sense to quote the previous post in full.[/color]

Thank you for your reply, Euclid's Turtle.

I did not get this integration initially as I was not introduced to  $ \int {x^n} \equal{} \frac {x^{n \plus{} 1}}{n \plus{} 1}$; somehow, the book I am using does not mention it.

Would you care to explain this integration rule?

Well, this integration rule follows from the power rule, which states that $ \frac {d}{dx}x^n \equal{} nx^{n \minus{} 1}$...

In other words, $ \frac {d}{dx}\left[\frac {x^{n \plus{} 1}}{n \plus{} 1}\right] \equal{} x^n \Longleftrightarrow \int x^n dx \equal{} \frac {x^{n \plus{} 1}}{n \plus{} 1} \plus{} C, \text{ given } n\not \equal{} \minus{} 1$

Yes :)

Also to try out the power rule, try to figure out for which $ p$,  $ \int_1^\infty {\frac {1}{x^p} dx}$ converges. :)


inequalities
number theory unsolved
number theory

Find the number of sequences $ \mathop {\{\mathop a\nolimits_n \} }\nolimits_{n \equal{} 1}^\infty$ of integers ,which satisfy $ \mathop a\nolimits_n \ne \minus{} 1$ and $ \mathop a\nolimits_{n \plus{} 2} \equal{} \frac {{\mathop a\nolimits_n \plus{} 2006}}{{\mathop a\nolimits_{n \plus{} 1} \plus{} 1}}$ for every positive integer $ n$

[b] Step 1. [/b] $ a_1 a_2 \geq 2006$

[b] Proof. [/b] Assume that $ a_1 a_2 < 2006$.
This implies $ a_1 < \frac{2006}{a_2}$

According to the recurrence, $ a_3 \equal{} \frac{a_1 \plus{} 2006}{a_2 \plus{} 1} < \frac{ \frac{2006}{a_2} \plus{} 2006}{a_2 \plus{} 1} \equal{} \frac{2006}{a_2}$
This implies $ a_2 a_3 < 2006$
By repeating this process, we get $ a_n a_{n\plus{}1} < 2006$ for all n.

Assume $ a_n a_{n\plus{}1} < 2006$. We have $ a_{n\plus{}1} < \frac{2006}{a_n}$
This yields $ a_{n\plus{}2} \equal{} \frac{a_n \plus{} 2006}{a_{n\plus{}1} \plus{} 1} > \frac{a_n \plus{} 2006}{\frac{2006}{a_n} \plus{} 1} \equal{} a_n$.

In other words, we have $ a_{n\plus{}2} > a_n$ for every n.
Since all $ a_i$ are positive integers, there exists an integer k s.t. $ a_k > 2006$
This is contradictory to $ a_n a_{n\plus{}1} < 2006$.

[b] Step 2. [/b] $ a_1 a_2 \leq 2006$
[b] Proof. [/b] Proceed as the same to Claim 1, except reversing the inequalities for each step.

[b] Step 3 [/b] If such an integer sequence $ a_n$ exists, then $ a_1 a_2 \equal{} 2006$
[b] Proof. [/b] Trivial.

[b] Step 4 [/b] If $ a_1 a_2 \equal{} 2006$, $ a_{2n\plus{}1} \equal{} a_1$ and $ a_{2n\plus{}2} \equal{} a_2$.
[b] Proof. [/b] A trivial inductive step kills it.

By Step 3 and 4, we have that all sequences are representable in the form $ a_{2n\plus{}1} \equal{} a_1$, $ a_{2n} \equal{} \frac{2006}{a_1}$ ($ a_1$ is a divisor of 2006)[/b]


im not sure if this is the right forum to ask this but yeah:
im taking the AP physics B exam in May and I was wondering what study tool is there to prepare for it. i don't need a textbook, just a reference book that has practice AP tests and test hints, ectr.

thanks!

You mean an AP review book?  There are several well-known brands - Barron's, Princeton Review, 5 Steps to a 5 - but I don't know which one has the best AP Physics B book.  You'll have to look up reviews online.

Just from personal experience Barron's had a harder test that the actual AP test. And it had explanations I liked.


The number 25 can be written as the sum of two or five consecutive whole numbers. What is the positive difference between the products of the numbers in each set?

Two consecutive numbers that add up to 25 is 12 and 13.  Their product is 156.

Five consecutive numbers that add up to 25 is 3,4,5,6,7.  Their product is 2520.

The positive difference of the numbers is 2520-156=2364.


Prove that the real numbers $x_1,x_2,...,x_n$ exist such that at least $n-1$ of them are irrational (for every natural number like $n$ which is bigger than one)and:

$0<\frac{\sqrt{2}}{2}(x_1+x_2+x_3+...+x_n-n)<2\sqrt{n}-1$

Let $S$ be an arbitrary real number. You can always find $x_1,...,x_{n-1}\in\mathbb{R}\backslash\mathbb{Q}$ and  $x_n\in\mathbb{R}$ such that
     $x_1+...+x_n=S$.
Indeed, if $S$ is rational, take
     $x_1=x_2=...=x_{n-1}=\sqrt{2}$
and
     $x_n=S-(n-1)\sqrt{2}$.
If $S$ is irrational, take
     $x_1=x_2=...=x_n=\frac S{n}$.

Now to solve your problem it suffices to choose $S$ s.t.
     $0<\frac{\sqrt{2}}{2}(S-n)<2\sqrt{n}-1$.


A palindrome is a number that reads the same forwards and backwards. The sum of a particular set of three consecutive positive integers is a three-digit palindrome. If the sum is less than 220, what is the greatest possible value for the largest of the three integers in the set?

[hide] Since the three integers are consecutive, you need the greatest palindrome less than 220 that is divisible by 3. That is 171. $ 56 \plus{} 57 \plus{} 58 \equal{} 171$, so the greatest possible value for the largest of the three integers in the set is 58.[/hide]


topology
advanced fields
advanced fields unsolved

1) What's an example of a topological space $ X$ which is limit-point compact but not countably compact?
2) What's an example of a topological space $ X$ which is limit-point compact but not sequentially compact?

I can't seem to find suited counterexamples.

I'll look these up when I get near my counterexamples book (pretty far away now). Things to note about these properties.

For 1 you can't have anything that satisfies $ T_1$, because limit-point compact and countably compact are equivalent there.

Limit-point and sequentially compact are equivalent in a metric space, so don't look in spaces that are metric-like. There may be an even weaker condition that makes these two equivalent that I'm not thinking of right now.

If no one has answered in a few hours, I'll be back at my room and can look these up.

Sorry. I got back last night and completely forgot.

1. is pretty easy actually. The odd-even topology works. Let $ X\equal{}\mathbb{N}$ and to topology is the one generated by the two sets $ \{\{2k\}_{k\in\mathbb{N}}, \{2k\plus{}1\}_{k\in\mathbb{N}}\}$.

2. you can also use odd-even topology. If you want a different one, take $ X\equal{}(0,1)$ and define $ U_n\equal{}(0, 1\minus{}1/n)$, for $ n\equal{}2,3,\ldots$. Now $ \mathfrak{T}\equal{}\{\emptyset, X, U_n\}$ forms a topology, sometimes called the Nested Interval Topology. This is limit point compact but not sequentially compact. This example works for 1. as well.


Given $ f(x)\equal{}5x\plus{}2$ and $ g(x)\equal{}3x^2\minus{}4x$, evaluate $ f(f(1))\minus{}g(g(2))$.

[hide="Solution"]
Just plug and find the values, then plug again.
\begin{align*}f(f(1))-g(g(2))&=f(5\cdot1+2)-g(3\cdot2^2-4\cdot2)\\&=f(5+2)-g(3\cdot4-8)\\&=f(7)-g(4)\\&=(7\cdot5+2)-(3\cdot4^2-4\cdot4)\\&=(35+2)-(3\cdot16-16)\\&=37-32\\&=\boxed{5}\end{align*} Not much explaining I can do here :|[/hide]

Since we are composing a function within itself, we replace the function itself for the $ x$ in the outer iteration. Plugging in, we get $ 5(5(1) \plus{} 2) \plus{} 2 \minus{} (3(3(2)^2 \minus{} 4(2))^2 \minus{} 4(3(2)^2 \minus{} 4(2))$, which simplies to $ 5(7) \plus{} 2 \minus{} 3(4)^2 \plus{} 4(4) \equal{} 35 \plus{} 2 \minus{} 48 \plus{} 16 \equal{} 37 \minus{} 48 \plus{} 16 \equal{} \minus{} 11 \plus{} 16 \equal{} \boxed{5}$.


Tina randomly selects two distinct numbers from the set
$ \{1,2,3,4,5\}$, and Sergio randomly selects a number from the set
$ \{1,2,\ldots,10\}$. What is the probability that Sergio's number is larger than
the sum of the two numbers chosen by Tina?

[hide="Solution"]First, pick Tina's numbers.  The sums $ 9$, $ 8$, $ 4$, and $ 3$ can be picked in just one way, and $ 7$, $ 6$, and $ 5$ can be picked in two ways.  The probability for each of Sergio's choices (from $ 1$ to $ 10$) is $ \frac{1}{10}$.  If we take all of his selections in decreasing order, the probability that his choice is greater is $ \left (\frac{1}{10} \right ) \left (1\plus{}\frac{9}{10}\plus{}\frac{8}{10}\plus{}\frac{6}{10}\plus{}\frac{4}{20}\plus{}\frac{2}{10}\plus{}\frac{1}{10}\plus{}0\plus{}0\plus{}0\right )\equal{}\boxed{\frac{2}{5}}$.[/hide]

I have a question on the Wiki Solution [url=http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_10A_Problems/Problem_24]here[/url]

Why do you not "flip" and count (5,4) when you count (4,5)? Aren't they different?

Order doesn't matter. It's the same reason you don't flip any of the others.

Why not write $\frac{4}{20}$ as $\frac{2}{10}$ because other numbers have the denominator 10?  :huh:

I think that was a typo -- it was meant to be $\frac{4}{10}$, not $\frac{4}{20}$.

Seriously? I got this problem right after I read this thread.
Well, solved it anyways.
[hide=Solution]
I used casework.
There are ten cases, and they add up to 2/5.[/hide]

[hide=my solution](probably doesn't generalize)

Kinda hard to explain, but since (2+4)/2=3, and 3 is the mean of Tina's set, the most common sum that Tina would get is probably 6. There are 4 numbers after 6 in Sergio's set, so the answer is 4/10, or [b]2/5[/b].
[/hide]

Alcumus overcomplicated things. [hide=Sol]
Avg in first set is 3, so sum avg is 6. We can get 7,8,9,10 out of 10 so 4/10=2/5.
[/hide]

That does not always work
Consider the following example:
Choose 2 numbers from the set $(1,2,3,4,5,6,7,8,9, 382910103837)$
Then choose 1 number from the set $(87584930263, 87483947338, 84354657689)$
By your logic, sum avg first set is less than all 3 numbers in the second set. Therefore the probability that the two numbers chosen from the first set sum to less than the number chosen from the second set is 1.
Except the biggest number in the first set is greater than all 3 numbers in the second set. Therefore, the probability mentioned above must be less than 1.


Find the sum of the solutions of $8^{x^2+3x+10}=4^{x^2-x}$

Big hint: [hide]think powers of two...[/hide]

[hide]this is equivalent to
$ 3(x^2 + 3x+ 10 ) = 2(X^2 -x) $

Thus we get, $ x^2 + 11x +30 = 0 $

Thus the sum of the roots is $ -11 $ by properties of quadratics[/hide]

[i]<remember hide!!>[/i]

I'm not quite sure this is the right level for this forum. I'm considering brining it to the basics.

I think that it's alright. The hardest part is making the base equal on both sides.


Role PM's have been sent.  This game will work like any other mafia game except for the following exceptions [b](Please Read)[/b]:

- Night kills as well as day lynches will be revealed to the public.
- There are some roles that alter the number of votes placed on somebody, so don't think that anybody's cheating.
- Every night, somebody's role (chosen at random) will be blocked.
- Absolutely NO EDITING OF POSTS or POSTING AFTER ONE'S DEATH.  If you wish to do one of those two things, please PM me with what you would like to edit/post and I will give you permission if I see fit.
- Everybody must post EVERY 72 HOURS or they will be mod-killed.
- You [i]may[/i] quote your exact role PM, as they are easy to fake.
- Good luck and have fun!  :wink:
-There are roles that alter the number of votes on a player. If I lynch a player with too few votes or don't lynch a player with enough votes, there is most likely a reason. Because of this, I have decided that [u]it would make things much easier if everybody got exactly one vote.
[/u]
EDIT: 6 votes to lynch for day 1 (remember, this does not necessarily imply exactly 6 people voting).

Alive:
- ajai
- 142857
- Ars
- hunter34
- Peter
- pianoforte
- Hamster1800
- CheeseIsGood
- perfect
- DonkeyKong
- mnmath
- BoesFX

[url]http://en.wikipedia.org/wiki/Great_Expectations[/url]

........reading the website..........

Wait, you guys haven't read Great Expectations?  It's a required reading book, isn't it?

*Resumes normal mod status*

Eh? I never even [i]heared[/i] of Great Expectations.  :|
I thought the title meant that it would be an all secret game so we'd have great expectations on weird roles to come out. :?

[quote="Peter"]heared[/quote]Not an English major, eh?

No... math major... like most people on a math site I guess?

All I know is that it is a classic by Dickens...

We should have a Lord of the Flies Theme sometime! Someone do that.

Can I have the list of players?

btw, I have never read Dickens's books.

Mafia Game 28 Players List (12 Players):
- ajai
- 142857
- Ars
- hunter34
- Peter
- pianoforte
- Hamster1800
- CheeseIsGood
- perfect
- DonkeyKong
- mnmath
- BoesFX

1=2 just shut up you are not in the game
OMGUS vote: BoesFX
an Miyomiyo though I have read many of Charles Dickens books I never read this.
In addition not all of us are from America or England, as a result not all of us will have read English classics

[b]Vote:Peter[/b] for using a first name as a user name.

Nobody answered my question :(
How many votes do we get? As many as we like (like 26) or one (like 27)?

I [b]AM[/b] Magwitch. Of course, I don't trust CheeseIsGood after he claims like that :D

Maybe the mafias isn't good and we townies are really lucky :lol:

Heh, I did do pretty badly. I might have been able to think of a better claim, but I knew I was trapped anyway. Here are Hamster's PMs

[quote="Hamster1800"]So it's night, and we lynched our other teammate. Unfortunately, he had a good ability, but the good news is nobody suspects us right now.

Before we decide on who to kill, what is your ability? I take one less vote to lynch :(

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

I don't know about if he had another ability, but his was that he gets double votes.

My role PM doesn't say anything about kills, so I guess it's just you, but I can help decide, I guess.

I say Peter is an obvious choice to kill, but it seems really unfair to him since it'd be nice to let him play another day... Another benefit is that when people notice that Peter wasn't killed, they'll probably lynch him because they think that he can only survive the night if he's mafia.

kill either perfect or donkeykong, I guess, or an inactive to keep the game alive. In any case, leave Peter alive and we'll have an easy day 2.

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

I'm not sure who to kill of those three, any of them will be fine.

Also, don't accuse Peter immediately, let the others bring it up first. Otherwise we might slip up and make it look like we conversed through the night, hence mafia / mason. [/quote]

Ah I'm pretty good at predicting who the angel protects :)

So yeah, really I think that had not piano gotten me on night 1, I would have lasted a while. Of course, without a night kill, I don't think I would really have had a chance, but meh. For example, when you look at day 2, it's clear that piano knew I was mafia even before he announced it ("I specifically wanted Hamster to post his innocents.") But nobody else did (as far as I can tell)

Also, I was thinking of claiming angel and that god was a liar right before getting lynched in hopes of evoking a counterclaim. Turns out we didn't need it and got the angel anyway... Out of curiousity, what was the role name? Had I claimed, I would have claimed Clara for the reason of
[quote="Wikipedia"]
He also has a secret called Clara. She is secret because Herbert knows his mother would say she is below his "station." She's actually a sweet, fairy-like girl who takes care of her dying drunk of a father.
[/quote]

Et Voilá.
[hide="All Roles"]
[hide="hunter34"]You are [b]Miss Havisham[/b]. Your daytime votes count double.

You are on the [u]evil[/u] team.
Your goal is to reduce the number of players on the good team to the number of players on the evil team. If you do so, your team wins. Every night (but not day) you may converse via PM with the members of you team (who are listed below).

The players on your team are [i]CheeseIsGood[/i] and [i]Hamster1800[/i].

Good luck and have fun! [/hide][hide="CheeseIsGood"]You are [b]Estella[/b]. Your daytime votes do not count. Every night, you may choose one player to kill.

You are on the [u]evil[/u] team.
Your goal is to reduce the number of players on the good team to the number of players on the evil team. If you do so, your team wins. Every night (but not day) you may converse via PM with the members of you team (who are listed below).

The players on your team are [i]hunter34[/i] and [i]Hamster1800[/i].

Good luck and have fun! [/hide][hide="Hamster1800"]You are [b]Drummle[/b]. You take one less vote to lynch than everybody else.

You are on the [u]evil[/u] team.
Your goal is to reduce the number of players on the good team to the number of players on the evil team. If you do so, your team wins. Every night (but not day) you may converse via PM with the members of you team (who are listed below).

The players on your team are [i]CheeseIsGood[/i] and [i]hunter34[/i].

Good luck and have fun! [/hide][hide="142857"]You are [b]Pip[/b]. Every night, you may target one player. If the player is...
(a) Miss Havisham or Drummle, he/she dies.
(b) Estella, you die.
Once during the game you may PM me during the day with a player's username. They are unlynchable for the day.
Use your power cautiously.

You are on the good team.
Your goal is to reduce the number of players on the evil team to zero. If you do so, your team wins.

Every night, you may converse via PM with your team member [i]Ars[/i].

Good luck and have fun! [/hide][hide="Ars"]You are [b]Herbert[/b]. You have no power, except that you gain the power of you team member listed below once he/she dies.

You are on the [u]good[/u] team.
Your goal is to reduce the number of players on the evil team to zero. If you do so, your team wins.

Every night, you may converse via PM with your team member [i]142857[/i]. Note that this is the player whose role you will recieve once he/she dies.

Good luck and have fun! [/hide][hide="pianoforte"]You are [b]Mr. Jaggers[/b]. Every night, you may target one player. You will get a result PM telling you if they are on the good team or the evil team.

You are on the [u]good[/u] team.
Your goal is to reduce the number of players on the evil team to zero. If you do so, your team wins.

Good luck and have fun! [/hide][hide="BoesFX"]You are [b]Magwitch[/b]. Every night, you may kill one player who you think is on the evil team.

You are on the [u]good[/u] team.
Your goal is to reduce the number of players on the evil team to zero. If you do so, your team wins.

Good luck and have fun! [/hide][hide="perfect628"]You are [b]Joe[/b]. Every night, you may target one player. They are saved from any killing role (even if they are on the evil team).

You are on the [u]good[/u] team.
Your goal is to reduce the number of players on the evil team to zero. If you do so, your team wins.

Good luck and have fun! [/hide][hide="Peter, mnmath, ajai, DonkeyKong"]You are a [b]civilian[/b]. You have no particular power besides the outsanding ability to twiddle your thumbs for extended periods of time. If you wish to do so, you may send me a night PM just to feel like you are getting in on some of the action. Unfortunately, I will not count it for anything and will probably just reply saying "Shut up; I have better things to do with my time".

You are on the [u]good[/u] team.
Your goal is to reduce the number of players on the evil team to zero. If you do so, your team wins.

Good luck and have fun! [/hide]
[/hide][hide="Night Actions"][hide="Night 1"][hide="CheeseIsGood (Mafia)"]BoesFX[/hide][hide="perfect628 (Angel)"]BoesFX[/hide][hide="pianoforte (Inspector)"]Hamster1800[/hide][hide="142857 (Special Pip role)"]Nobody[/hide][hide="BoesFX (Vigilante)"]Nobody[/hide][hide="Random Role Block"]142857[/hide]
[/hide][hide="Night 2"][hide="CheeseIsGood (Mafia)"]perfect628[/hide][hide="perfect628 (Angel)"]pianoforte[/hide][hide="pianoforte (Inspector)"]Peter[/hide][hide="142857 (Special Pip role)"]Nobody[/hide][hide="BoesFX (Vigilante)"]Nobody[/hide][hide="Random Role Block"]pianoforte[/hide]
[/hide][/hide]

miyomiyo, why was I role-blocked? Or did I just miss it on the roles?...

And, yeah the roles were a bit unbalanced.

Hamster, I wouldn't have inspected DK, I was pretty certain he was just an innocent not acting competently.

Actually, I think it was the fact that your avatar is black and flashy why I inspected you. No one else was interesting.

[quote="Hamster1800"]Also, I was thinking of claiming angel and that god was a liar right before getting lynched in hopes of evoking a counterclaim. Turns out we didn't need it and got the angel anyway... Out of curiousity, what was the role name? Had I claimed, I would have claimed Clara for the reason of
[quote="Wikipedia"]
He also has a secret called Clara. She is secret because Herbert knows his mother would say she is below his "station." She's actually a sweet, fairy-like girl who takes care of her dying drunk of a father.
[/quote][/quote]

That would have been a better move on your part. There's no way I could have counterclaimed, for exactly those reasons. Though, a miller-doctor claim sounds a bit suspicious...

I wouldn't have claimed miller-doctor, just that god lied. My friend was once a god where he just lied to the inspector every night regardless of who he chose to inspect.

Hmm, I've never seen god as a role...*researching*

A couple of flaws with this game
-Town was REDICULOUSLY overpowered, with masons, one of whom could make someone unlynchable and had that scumkilling ability, with a cop, a doc, and a vig
-Scum was even further underpowered. One of them took one less vote to kill, and that business with only one of them being able to kill is not good at all.
-You revealed role PMs of those lynched. This tells town more than they need to know.
-You made it clear from the beginning of the game that lynches might be a bit weird. This is not necessary. Town should expect some minor surprises..
-That random roleblocking is nonsense, seriously. It can seriously cripple either side if it happens at the wrong time
-Quoting role PMs. This rule should not exist. It makes it harder for the scum to fakeclaim if they have to rewrite an entire role PM, and have no idea what roles are not present in the game. In a themed game like this one, you should provide the scum with plausible rolenames that are not possessed by others in the game.
-You should have posted the townie PM. At least then scum has [i]something[/i] to go by.

Still, the roles were pretty good. Yes it was unbalanced, but the theme was done fairly nicely. Limiting everyone to one vote was good. And trying to enforce a post every 72 hours thing was also good.

This is the kind of game that I want to see more of AoPS mafia. With a bit more standardization we can probably have better games than we have had in the past few...

I realized the horrible setup on day one when I thought, "Who's the 4th mafia again?" and saw that there wasn't one.  I'll edit the All Roles post soon.  And piano, check the very first post of the game.

[quote="Hamster1800"]God=moderator in case that wasn't sarcasm...[/quote]

Well, it wasn't, no one calls mods gods. What you're referring to is an insane cop.

[quote="Hamster1800"]Actually, yes people do call mods gods, just perhaps not online. Whenever I play with the school's math team, he is called god and not the mod.[/quote]Same here, but that's not used online.


Express the following sum as a common fraction:
\[ \frac{1}{1\cdot 2} \plus{} \frac{1}{2\cdot 3} \plus{} \frac{1}{3\cdot 4} \plus{} \frac{1}{4\cdot 5} \plus{} \ldots \plus{} \frac{1}{9\cdot 10}\]

First we need to simplify each fraction.

$ \frac{1}{2} \plus{} \frac{1}{6} \plus{} \frac{1}{12} \plus{} \frac{1}{20} \plus{} \frac{1}{30} \plus{} \frac{1}{42} \plus{} \frac{1}{56} \plus{} \frac{1}{72} \plus{} \frac{1}{90}$

Now we add these together, to get an answer of $ \boxed{\frac{9}{10}}$.

I think there should be some less time-consuming way to solve this, but I can't think of it.

[quote="isabella2296"]First we need to simplify each fraction.

$ \frac {1}{2} \plus{} \frac {1}{6} \plus{} \frac {1}{12} \plus{} \frac {1}{20} \plus{} \frac {1}{30} \plus{} \frac {1}{42} \plus{} \frac {1}{56} \plus{} \frac {1}{72} \plus{} \frac {1}{90}$

Now we add these together, to get an answer of $ \boxed{\frac {9}{10}}$.

I think there should be some less time-consuming way to solve this, but I can't think of it.[/quote]

Formula:

$ \frac {1}{1\cdot2} \plus{} \frac {1}{2\cdot3} \plus{} \frac {1}{3\cdot4} \plus{} .... \plus{} \frac {1}{(n \minus{} 1)\cdot n} \equal{} \frac {n \minus{} 1}{n}$


Let $a,b,c\geq 0$ be real numbers such that $a+b+c\geq abc$. Prove the following inequality :
$\sqrt{\frac{a^{2}}{1+5bc}}+\sqrt{\frac{b^{2}}{1+5ac}}+\sqrt{\frac{c^{2}}{1+5ab}}\geq \frac{a+b+c}{4}$.

Well, no one interested ?
I'm looking for a nice classical solution using CS, AM-GM,... homogenization or elegant substitution. My solution uses a (quite) technical result...
Someone?

indeed, very nice :) Thank you ! I didn't think about Holder using three terms!

[quote="andyciup"]Use Holder and it`s a one-liner:
\[\ds \sum\frac{a}{\sqrt{1+5bc}}\cdot\sum\frac{a}{\sqrt{1+5bc}}\cdot\sum a(1+5bc)\geq (a+b+c)^{3}\]
Good enough? :)[/quote]

I guess you need $\sum a(1+5bc)=\sum a+15 abc\leq 16\sum a$ so that
\[\ds \sum\frac{a}{\sqrt{1+5bc}}\cdot\sum\frac{a}{\sqrt{1+5bc}}\cdot\sum a(1+5bc)\geq  \ds \sum\frac{a}{\sqrt{1+5bc}}\cdot\sum\frac{a}{\sqrt{1+5bc}}\cdot\sum 16 \sum a\geq (a+b+c)^{3}\]
$\text{The above is }\iff \text{ to: }$ \[\left(\sum\frac{a}{\sqrt{1+5bc}}\right)^{2}\geq \frac{(\sum a)^{2}}{16}\iff \sum\frac{a}{\sqrt{1+5bc}}\geq \frac{a+b+c}{4}\]

What was your "technhical solution Crux? And what was the method of findint this, Andyciup?


An acute angle of a right triangle measures $3x+27$ degrees.  Given that $x$ is an integer, how many possible values are there for $x$?

[hide]3x+27<90; x<21. So there are 20 possible values for x.  :) [/hide]

Cj--nothing preventing x from being negative. (Or zero)

Here is a hint

[hide]
\[0<3x+27<90\]
\[-27<3x<63\]

You should be able to go from here.

Remember that $x$ does not necessarily have to be greater than $0$ just because the angle is greater than $0$.

[/hide]

[hide]
 :P [/hide]
[hide]$0<3x+27<90$

$-27<3x<63$

$-9<x<21$

$x$ can be from $-8$ to $20$

there are $29$ possible x values.[/hide]


You're in the middle of a locked room. There is a pipe which is 3 meters long and 2cms wide and also cylindyric. there is a pingpong ball in the pipe. You have a 10 meter long rope, sunglasses, cowboy hat, a knife , and your brain.
you have to take the table tennis ball out of it :lol:
It doesn't have an answer like "I'll go out and take it."
Think about it.
try to answer the poll and it will help.
my answer was brain
one week I'll wait and than answer.

er...so if I don't choose brain doesn't it REALLY mean I won't use my brain?
any vote involves brain.

[quote="Hikaru79"]What... exactly is the question? Locked room, pipe... now what? Are we trying to get out of the room? I'm totally confused  :?[/quote]Yeah, I don't get what you're asking either...

edited...
sorry i missed the biggest part of the question :lol:  :blush:  :P

read the message where i wrote about "edit"
try to do it again.
though you cannot.

Where I live ping-pong balls wouldn't fit into such a pipe unless they were crushed.

[quote="DusT"]What about using your mouth, not brain???
Couldn't you just blow it??? And then it would get out! If you wouldn't be able to get it out this way, the only way would be to get a blacksmith and to destroy the pipe  :D[/quote]

You still used your brain, it's impossible not to.

Your brain is in a pipe. How do you get it out?
   	10 meter long rope
  	sunglasses
  	cowboy hat
  	knife
  	ping-pong ball

There, fixed.

sklauiskjgiewa;k sd.... *keels over dead*

So how am I supposed to do anything without my brain? :P

Here's how I see it.  Assuming the room is big enough, you could tie the pipe to the rope and since its 10 meters long, then swinging it around in circles as fast as you can will put quite a bit of pressure on the ball, probably enough to force it out.  Although I suppose that took my brain  :?

It is probably ok, but I would like to remind you that you used your hands too (you didn't tie the rope to the pipe with your brain ;) ) , so I don't see why you can't use your mouth???? I don't think this is a "only-one-solution-puzzle", so maybe the proposer could enlighten us and tell us the official solution :)


In $ \triangle{ABC}$, we have $ AB\equal{}1$ and $ AC\equal{}2$. Side $ BC$ and the median from $ A$ to $ BC$ have the same length. What is $ BC$?

$ \textbf{(A)}\ \frac{1\plus{}\sqrt2}{2} \qquad
\textbf{(B)}\ \frac{1\plus{}\sqrt3}{2} \qquad
\textbf{(C)}\ \sqrt2 \qquad
\textbf{(D)}\ \frac{3}{2} \qquad
\textbf{(E)}\ \sqrt3$

stewarts...too bad i forgot it lol.

but im pretty sure its something like a^2m+b^2n=dad+man?

[hide]

Let the midpoint of $ BC$ be $ M$. We can also place a point $ D$ such that $ CD\parallel{}AB$ and $ AC\parallel{}BD$. Since $ ABCD$ is a parallelogram, $ AM\equal{}MD\equal{}BC\equal{}x$. In a parallelogram, the sum of the squares of the sides is equal to the sum of the squares of the diagonal, so:
\[ 2(2^2\plus{}1^2)\equal{}(2x)^2\plus{}x^2\equal{}5x^2\]
Hence, $ BC\equal{}x\equal{}\sqrt2\Longrightarrow\mathrm{ (C) \ }$

[/hide]

yea so just use stewarts like i said

let 1/2BC=1/2median=x

So median=2x and BC=2x

From stewarts, you know that (AB^2)(x)+(AC^2)(x)=(2x)(2x)(2x)+(x)(x)(2x)...hard to explain without picture, but if you draw it out and you know what stewarts is, its pretty easy to understand

so x+4x=8x^3+2x^3, 5x=10x^3, x isnt 0, so 5=10x^2, x^2=1/2, $ x \equal{} \frac {\sqrt {2}}{2}$, so $ BC\equal{}2x\equal{}\sqrt{2}$

Alternatively

 [geogebra]c9aed04c2cb9a6260e9ac80749ff526362aefe52[/geogebra]

$ \alpha \equal{} \pi \minus{} \beta$ So from cosine law

$ 1^2 \equal{} x^2 \plus{} \left(\frac{1}{2}x\right)^2 \minus{} 2(x)\left(\frac{1}{2}x\right)\cos(\alpha) \qquad (1)$
$ 2^2 \equal{} x^2 \plus{} \left(\frac{1}{2}x\right)^2 \plus{} 2(x)\left(\frac{1}{2}x\right)\cos(\alpha) \qquad (2)$

$ 5 \equal{} \frac{5}{2}x^2 \qquad (1)\plus{}(2)$

$ x\equal{}\pm \sqrt{2} \Rightarrow x \equal{} \sqrt{2}$

[b]Solution:[/b] Letting $\overline{BC}$ and the median be of length $2x$. By stewart's we have that $2x^3 + 8x^3 = x + 4x \Rightarrow x = \frac{\sqrt{2}}{2} \Rightarrow 2x = \boxed{\sqrt{2}}$.


here are some problems from my class.

find the percent of increase.

old 12
new 18

old 23
new 45

and also

3(x+1)=15

5*=625

9(x+1)=8(x+2)

$ 3(x\plus{}1)\equal{}15$
$ 3x\plus{}3\equal{}15$
Then we subtract three from both sides
$ 3x\equal{}12$
We divide both sides by three
$ x\equal{}4$

Solutions to all the problems.

[hide="1)"]the increase in the price is $ 6$, which is over a $ 12$ old price, so $ \frac {6}{12} \equal{} \boxed{50\%}$[/hide]

[hide="2)"]the increase in the price is $ 22$, which is over a $ 23$ old price, so $ \frac {22}{23}\approx\boxed{97\%}$[/hide]

[hide="3)"]$ 3(x \plus{} 1) \equal{} 15\implies x \plus{} 1 \equal{} 5\implies x \equal{} \boxed{4}$[/hide]

[hide="4)"]$ 5^n \equal{} 625$. Taking the base 5 logarithm of both sides gives $ \log_5{5^n} \equal{} \log_5{625}\implies n \equal{} \boxed{4}$[/hide]

[hide="5)"]expanding gives $ 9x \plus{} 9 \equal{} 8x \plus{} 16\implies x \equal{} \boxed{7}$[/hide]

If that's what you wanted :maybe:


I have an equation that looks very simple, yet every time I try to apply trig identities, I end up going around in circles.  Any help solving it would be greatly appreciated.

a*SIN(:theta:) + b*COS(:theta:) = c, where :theta: is unknown and a,b,c are constants.

[quote="shobber"][hide="hint"]
$a\cdot\sin{\theta}+b\cdot\cos{\theta}=\sqrt{a^2+b^2}\sin{(\theta+\varphi)}$ where $\tan{\varphi}=\frac{b}{a}$[/hide][/quote]

Well, that allowed me to solve it, but where did it come from?  I wasn't able to find this trig identity anywhere (not even in the sticky at the top of the forum).

Also, how do I make the math in my posts look nicely formated?

Thanks,
Steve

Consider the equation $a\sin{\theta} + b\cos{\theta} = c$. Does this look like any trigonometric identity you've seen before? If you can recognize, this looks similar to the identity $\sin({x+y}) = \sin{x}\cos{y} + \sin{y}\cos{x}$. How are we going to be able to transform that equation into this form, though?

The key is using the Pythagorean identity. We know that we need to find some $x$ such that it satisfies $(\frac{a}{x})^2+(\frac{b}{x})^2 = 1$. Through a little work, we can see that $x$ must equal $\sqrt{a^2+b^2}$.

Going back to our original equation, we divide both sides by $x$, and find $\sin{\beta}\cos{\theta} + \sin{\theta}\cos{\beta} = c\sqrt{a^2+b^2}$ for some angle $\beta$. However, since we know that know $\sin{\beta} = \frac{b}{\sqrt{a^2+b^2}}$ and $\cos{\beta}=\frac{a}{\sqrt{a^2+b^2}}$, we know that $\tan{\beta} = \frac{b}{a}$, and then we are done.

Hope that helps! :)

I think we should add this to the Trig identities at the top of the forum


find area
find ar(DEF)
Find Area of DEF If
ar(ADE) = 5 sq units
ar(EBF) = 4 sq units
ar(DFC) = 3 sq units

FIND ar(DEF) [img]http://img5.imageshack.us/img5/7826/74271509.jpg[/img]

if $ |FC| \equal{} a$  $ \implies$   $ |DC| \equal{} \frac {6}{a}$
if $ |BF| \equal{} b$  $ \implies$      $ |EB| \equal{} \frac {8}{b}$  $ \implies$   $ |AE| \equal{} \frac {6}{a} \minus{} \frac {8}{b}$
$ \frac {6b \minus{} 8a}{ab}.(a \plus{} b) \equal{} 10$  $ \implies$
$ (3b \minus{} 4a)(a \plus{} b) \equal{} 5ab$  $ \implies$
$ 3ab \minus{} 4a^{2} \minus{} 4ab \plus{} 3b^{2} \equal{} 5ab$  $ \implies$
$ 3b^{2} \equal{} 6ab \plus{} 4a^{2}$  $ \implies$   $ \frac {3b^{2}}{a^{2}} \equal{} \frac {6b}{a} \plus{} 4$  ,$ \frac {b}{a} \equal{} x$  $ \implies$
$ 3x^{2} \equal{} 6x \plus{} 4$  $ \implies$  $ 3x^{2} \minus{} 6x \minus{} 4 \equal{} 0$ $ \implies$  $ \frac {b}{a} \equal{} x \equal{} 1 \plus{} \frac {\sqrt {21}}{3}$
$ A(ABCD) \equal{} 6 \plus{} \frac {6b}{a}$  $ \implies$
$ A(DEF) \equal{} \frac {6b}{a} \minus{} 6 \equal{} 6.(\frac {b}{a} \minus{} 1) \equal{} 2\sqrt {21}$   :wink:


For $ a \plus{} b \plus{} c \equal{} 1$ $ such that a,b,c > 0$
   Prove that
$ \frac {\sqrt {a^{2} \plus{} abc}}{ab \plus{} c} \plus{} \frac {\sqrt {b^{2} \plus{} abc}}{bc \plus{} a} \plus{} \frac {\sqrt {c^{2} \plus{} abc}}{ca \plus{} b}\leq\frac {1}{2. \sqrt {abc}}$


Hey, I'm new here, I'm a high school student, from Skopje that's the capital of Macedonia.I'm not that good at, math's mostly because I don't study it that much, I do excellent in other subjects but me and math's haven't been actually been good friends.
I currently have a C in the subject, all my other 10 classes are A. but now my C is in jeopardy, we did a semester test I didn't do to good, probably goanna get an F.
But I did mange to write the test down. We get about 7 tasks I don't know exactly how you call them, pardon me, to do in 45 min.
Now I do fell a bit ashamed to ask for help, but I'll do, if any one can solve one of the tasks I put up here, I will be deeply great full, I need to see how they are solved the process, I can't study form books, I only look at the examples and then try to figure out a pattern, but most of this tasks aren't very well explained or aren't even in my book, my teacher wants to give out different tasks, cause she think we should be more creative.
so I beg out of any one who can help, if you have the time, please do I can't offer much but my gratitude.
 here are the tasks, there 10 of them if you could do one it will be a big help.

Because i can write them here i can't use superscript and stuff like that i will attach a word file.

THANKS UPFORNT.[/u]

[quote="namek"]i can't use superscript and stuff like that[/quote]

$LaTeX!!$

I'm thinking this is the test that he has  already taken - the one he believed he failed.

[quote="namek"]...we did a semester test I didn't do to good, probably goanna get an F.
But I did mange to write the test down....[/quote]

which means that this is the test he failed.

Continuing from my last post, not all people know Latex.

Anyway...

#4 Simplify:

$\frac{(1- \cos A)^2 + \sin^2 A}{(1 + \cos A)^2 + \sin^2 A}$.

I'm not sure how far you want it to expand but I'll go far as I want to. :P

Numerator: $(1 - \cos A)^2 + \sin^2 A = 1 - 2 \cos A + \cos^2 A + \sin^2 A = 2 - 2 \cos A$

Denominator: $(1 + \cos A)^2 + \sin^2 A = 1 + 2 \cos + \cos^2 A + \sin^2 A = 2 + 2 \cos A$

$\frac{2 - 2 \cos A}{2 + 2 \cos A} = \frac{1 - \cos A}{1 + \cos A}$

I'm not sure if you're supposed to stop there but if not:

$\frac{1 - \cos A}{1 + \cos A} \cdot \frac{1 - \cos A}{1 - \cos A} = \frac{(1 - \cos A)^2}{1 - \cos^2 A} \Rightarrow \frac{(1 - \cos A)^2}{\sin^2 A}$

You can expand the upper part $(1 - \cos A)^2$ but I just didn't do it since I don't know how far you want it to be simplified.

#8

$\cos^2 X - \sin^2 X = \sin X$

Rewrite $\cos^2 X$ and we get:

$1 - \sin^2 X - \sin^2 X = \sin X\\
0 = 2 \sin^2 X + \sin X - 1\\
0 = (2 - \sin X - 1)(\sin X + 1)$

$2 \sin x - 1 = 0\\
2 \sin x = 1\\
\sin x = \frac{1}{2}\\
x = \frac{\pi}{6}, \frac{5 \pi}{6}$

$\sin x + 1 = 0\\
\sin x = -1\\
x = \frac{3 \pi}{2}$

Since the period is not specified, I assume we have to add $+2n\pi$ in our solution.

So, the final solution is $\boxed {\frac{3 \pi}{2} + 2n \pi , \frac{\pi}{6} + 2n \pi, \frac{5 \pi}{6} + 2n \pi}$.

Does tg denote the tangent function and does ctg denote cotangent?

If so here is number 2
Rewrite in terms of sin and cos
$\frac{\cos^2A}{\sin^2A} - \cos^2A = \frac{\cos^2A}{\sin^2A}(\cos^2A)$
Common denominator
$\frac{\cos^2A - \cos^2A(\sin^2A)}{\sin^2A} = \frac{\cos^4A}{\sin^2A}$
The denominators cancel
$\cos^2A - \cos^2A(\sin^2A) = \cos^4A$
$\cos^2A(1 - \sin^2A) = \cos^4A$
$\cos^2A(\cos^2A) = \cos^4A$
$\cos^4A = \cos^4A$

Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,Thanks,

No i did take the test, but on Tuesday i'll be questioned to fix my greade, i just like to thank everybody for there help, this means a lot to me guys , thanks thanks.

Happy to know this forum was helpful to you.

This thread is going to be moved up - This is not Beginner Level Trig.

don't think anyone has answered #5 yet so here it is:
$\frac{\tan{A}+\sin{A}}{\tan{A}-\sin{A}}=\frac{\frac{\sin{A}+\sin{A}\cos{A}}{\cos{A}}}{\frac{\sin{A}-\sin{A}\cos{A}}{\cos{A}}}$
$=\frac{\sin{A}(1+\cos{A})}{\sin{A}(1-cos{A})}=\frac{1+\cos{A}}{1-\cos{A}}=\frac{1}{\tan^2({\frac{A}{2}})}$

assuming #9 is: $\frac{\sin{x}+\cos{x}}{2}=1$ (even if it isn't you can use the same idea...)
$\sin{x}+\cos{x}=2 \Rightarrow (\sin{x}+\cos{x})^2=4\Rightarrow \sin{(2x)}+1=4\Rightarrow \sin{(2x)}=3\Rightarrow$ there is no answer, because $-1\le \sin{\theta}\le 1$.
assuming the problem was $\sin{x}+\cos{x}=1$ then $(\sin{x}+\cos{x})^2=1\Rightarrow \sin{(2x)}=0\Rightarrow \boxed{x=\frac{\pi}{2}+k\pi}$

[quote="xxreddevilzxx"]i think that $\cos{2A}$ would be simpler.[/quote]
you mean for #1??
the problem was not asking to simplify, rather it was asking to prove the identity... ;)

you people are great, i`m so sorry to take time out of you schedule, so you could help me, and you don`t even know me.

god bless you.
i love you all ( in a none  prison movie kind a way ).


[hide]
Let $a = 2^x-4$ and $b = 4^x-2$.

We have $a^3+b^3 = (a+b)^3 \Rightarrow 3a^2b+3ab^2 = 3ab(a+b) = 0$

So

$a = 0 \Rightarrow 2^x-4 = 0 \Rightarrow x = 2$.
$b = 0 \Rightarrow 4^x-2 = 0 \Rightarrow x = \frac{1}{2}$.
$a+b = 0 \Rightarrow 2^x-4 = 4^x-2 \Rightarrow (2^x)^2-2^x+2 = (2^x-2)(2^x+1) = 0 \Rightarrow x = 1$.

The sum of our solutions is $2+\frac{1}{2}+1 = \frac{7}{2}$.[/hide]


I play Ruy Lopez as White, and as Black I play Modern Defense. I don't play the Sicilian because it has razor-sharp lines and one wrong move could spell disaster.

I play the Ruy Lopez as white and the Scandinavian as black.

Which variation of the Scandinavian Defence do you play after 3.Nc3?

I play Qa5 (as taught by my tutor). I usually go off of instinct after those few of moves.

Heh I used to be decent at chess.

White- Italian :D
Black- French :D

White- Depends
Black- Usually French

I dont play Chess as much these day because of the intensity of math.  :lol:

White: Reti system or Ruy Lopez

Black: Pirc defense.

this is funny...

BTW, what are your ratings?

White: Queen's Gambit

Black (against e4): King Pawn's game
Black (against d4, c4, or Nf3): Dutch (2. f5)

Well, used to be ranked #9 in chess for my grade, got up to #4 though :P, but got sick
and now i'm #11 =(
WHITE: Ruy Lopez Exchange Variation with 0-0 d4
BLACK: Depends on what he plays :P Semi-Tarrash if d4, I will play the Fried Liver out if he goes there and i will pwnz00r him!


trigonometry
geometry
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

There is a triangle with the sides 39.7 and 48.1 and the included angle 85.4 degrees.  Find the area of the triangle.  I know I can use a combination of the Law of Cosines and Heron's formula, but I heard of another easier way of doing it.

[quote="basketball9"]i THINk it somewhat depends on where the sides are located in relation to the angle.[/quote]

The included angle is the angle between the sides.

EDIT: Also, this should be in HS Basics, at least.

The included angle is the angle between the two sides

The area formula that is being mentioned is $ \frac12ab\sin{\theta}$.

Note: This only works if its the included angle.

hello, this site will also help you
http://en.wikipedia.org/wiki/Triangle
Sonnhard.

If we have side $ a\equal{}39.7$ and side $ b\equal{}48.1$ and $ \theta \equal{} 85.4 ^ \circ$ We use the formula, $ \dfrac{a \cdot b \cdot sin \theta}{2}\equal{}\dfrac{39.7 \cdot 48.1 \cdot sin 85.4}{2}$ which is about $ \dfrac {39.7 \cdot 48.1 \cdot 1}{2}$ $ \equal{} \boxed{1909.57}$


calculate
$ I\equal{}\int_0^1 \frac{x^7\plus{}1}{x^6\plus{}1}dx$


limit
calculus
derivative
integration
real analysis
real analysis solved

Find
\[\lim_{\lambda \rightarrow 0} \sum_{j = 1}^\infty (4 \lambda^2 j^2 - 1) e^{-2 \lambda^2 j^2}.\]

For $\lambda\neq 0$ the series is a derivative of $f(x)=-\sum_{j\geq 1}\lambda e^{-2\lambda^2 j^2}$ and $f(x)=F'(x)$, where $F(x)=\frac{1}{4}\sum_{j\geq 1}\frac{e^{-2\lambda^2 j^2}}{j^2}$.

The limit is $\frac12.$

Let $f(x)=(4x^2-1)e^{-2x^2}.$ Note that $\int_0^{\infty}f(x)\,dx=0.$

The sum is $\sum_{j=1}^{\infty}f(j\lambda)$

We would have to multiply this by $\lambda$ for it to be a Riemann sum for the integral of $f.$ The error in such a Riemann sum would have something to do with the derivative of $f.$

Let's try this: (Using the Mean Value Theorem):

$f(j\lambda)=\frac1{\lambda}\int_{(j-1)\lambda}^{j\lambda} f(x)\,dx + \frac{\lambda}2\,f'(\xi_j)$ where $\xi_j\in((j-1)\lambda,j\lambda).$

Sum that up over $j$ and we have a Riemann sum for

$\frac12\int_0^{\infty}f'(x)\,dx = -\frac12\,f(0)=\frac12.$


Okay, so today, in history class, taking a big history test, there was this question:

Name three animals brought to the New World by Spain.

I thought of horses and cattle right on the spot. However, since time was running out, and I had to finish other questions first, I just randomly put Slaves as the 3rd answer. Afterwards, I had no idea why I did so, but I'm pretty sure I thought that human beings are in biological standards still animals?

Slaves were considered animals during the era of segregation in the 20th century, and part of a mentioned reason I remember reading about was that they were below human standards.

Anyways, 1) Do you think that should be a valid answer? By the way, I know that I would not have gotten it right anyway.
              2) Do you think human beings such as slaves are animals? Not that I am saying life today is devoid of slavery; it's not.
              3) Also, what would you say is the best definition for an animal?

That's the least politically correct answer you could possibly think of. Doesn't matter how well you can justify it; it's still a pretty messed up answer.

If your teacher interprets your interpretation of slaves as animals not as you meant, there is no way you got that correct. However, if (s)he is smart enough to recognize humans as mammals and therefore as animals, then you might have gotten it correct. "Might" because (s)he still might not accept slaves because they were not introduced as a new animal altogether, they were introduced in the presence of other humans.

I, personally, would say your answer is incorrect because of the aforementioned reason. If your teacher thinks otherwise, you got lucky.

@ Lazarus - It's only a "messed up answer" to those who interpret it non-scientifically.

No, it's wrong. Humans are all the same animal, regardless of whether or not they are slaves. So when slaves were brought to the New World, the Spanish were not introducing a new animal. Humans were already there.

Also, there's nothing wrong with stating that humans are animals. It's when you make a distinction between humans based on some other trait (they are slaves, etc) and refer to some as animals and others as not that you're doing something (as Lazarus says) "pretty messed up."

When you get your test back, can you tell us what your teacher said?

Slaves are the same species as humans, so scientifically, they could be animals as long as humans are animals.

Humans belong in the animal kingdom, but that doesn't mean that humans are animals. The same could be said for slaves.

It could be said that an animal is a movable creature that can affect its surroundings through physical actions and has the same or similar (consistent) lifestyle/life as all the other creatures of that species. However, if I defined it this way, household pets and their wild counterparts are not animals if they're both classified as the same species.

If this definition was refined further (such as "An animal is a movable creature that can affect its surroundings through physical actions and has a similar lifestyle/life as all the other creatures of that species living in the same community"), it would eventually become clearer that humans today can be excluded called "animals." (Humans living as close together as possible or having as many similar characteristics as possible still won't necesarily have a similar life. If the definition was refined to the point where it states "An animal is a movable creature that can affect its surroundings through physical actions and has the same life as all the other creatures with the same life," then the last part of the definition becomes completely useless.)
This doesn't necessarily mean that slaves or humans from the past can be excluded from being called animals, because some slaves could have similar lives or lifestyles, and same with some humans in the past. However modern humans have lifestyles that change very rapidly, and go in all sorts of directions such that no defining characteristics other than "having the same life" can generally show people with similar lives.

That's what sets humans apart from most animals, because most animals of certain kinds have repetitive or similar lives. If animals with similar lives are set apart, there would be specific characteristics (such as species and/or subspecies) that can define which animals would have which types of lives. It's extremely hard to do the same with humans because many lives change and changing lives are much more complex.

From a religious point of view (creationist point of view), God put a soul in us and set us apart from the animals by allowing us to have the ability to communicate with him. Another characteristic that sets us apart from the animals is that God allowed the first man to name all the animals that he created. That kind of puts authority on us as the namegivers of the animals. Other characteristics we have that most animals don't have include thieving, lying, deceiving, trickery, and murder for gains other than personal.

With both ways I described I supported that humans are not animals, and similarly slaves would not be easily classified as animals.

And like the post a bit above said, even if humans are classified as animals, they already existed in the Americas, so they're not new from Spain.

I want to know if you got the answer right or not on the test when you get it back :D

humans are definitely animals, but slaves aren't a type of human. slaves aren't separated from the rest of humankind biologically. however, africans might have been acceptable as a subspecies of human, though i doubt subspecies would count for the test.

however, given that this is a math forum, and math is logical and proof-oriented, the argument that "god put souls in humans" is unscientific; i don't think that qualifies as an argument. math has even stricter requirements for validity than all other areas of science. in science, many supporting examples would be considered sufficient proof, but anyone who tried to use examples as a proof in math would get a low USAMO score, so religious arguments probably won't work in this case.

For all of you that wanted to know, it was counted wrong. There weren't any comments. It was just right or wrong grading I guess. =0

lol. i admire your willingness to put such a controversial answer though ^^

A good piece of background reading on the "Columbian exchange" is Jared Diamond's book [i]Guns, Germs, and Steel[/i].

Almost all of the major domesticated animals of the world were domesticated in Eurasia, and the Spanish brought most of them to the New World: cattle, swine, horses, donkeys, sheep, goats. The Americans were seriously lacking in domestic animals. There were the Andean camel species: llama and alpaca - but only in the Andes. There were domestic dogs. And there were a few small food animals, such as guinea pigs and turkeys. And that's it. No hunting buffalo from horseback. No lard in [i]frijoles refritos[/i] - in fact, no fried food at all. No side of beef cooking on the barbeque.

There were humans in the Americas, of course. Many millions of them, speaking hundreds of languages, living on the shores of the Arctic and in Tierra del Fuego and in nearly every place in between. And they did have a decent assortment of food crops in some of those places: corn, beans, squash, chile peppers, potatoes, and so on.

Along with the large domestic animals, Europeans (led by Spaniards) also brought a host of devastating bacteria and viruses: smallpox, tuberculosis, measles, yellow fever, typhoid fever, and so on. And the diseases did more to destroy American civilizations than did Spanish swords and muskets. There was no corresponding scourge of multiple illnesses spreading from the New World back to Europe. (The one arguable exception is syphilis, and its direction of travel is unclear.) One of Diamond's points is that the very reason Europeans had so many deadly epidemic diseases to spread is that they had been incubating and exchanging them with their herds of domestic animals for thousands of years.


limit
logarithms
real analysis
real analysis unsolved

Show that $\lim n!^{1/n}=+\infty$

Here is my solution, but I want to see some other ones and make sure that mine is good:

Given $M$, we can take $n > n_{0}$ such that
$\frac{M^{n}}{n!}< 1$, and such an $n_{0}$ exists because $\lim \frac{x^{n}}{n!}= 0$ for all $x$.  Thus $n!^{1/n}> M$ for all $n > n_{0}$, which means the limit is infinite.

If $x_{n}=\sqrt[n]{n!}$ then

$\ln x_{n}=y_{n}=\frac{\ln 1+\ln 2+\cdots+\ln n}{n}$.
Then by Cesaro Stoltz lemma we get that

$\lim y_{n}=\lim \ln (n+1)=\infty$, hence $\lim x_{n}=\infty$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120964]this topic.[/url]


[b]Problem:[/b] Construct a bounded set of real numbers with exactly three limit points.

I apologize if this is a repost, but I was not able to find this post when i searched.

Take three convergent sequences with different limits, and splice them together.

This should probably have been posted in Unsolved and Proposed Problems.


$ ({1}\plus{}{\frac{x}{x\plus{}y\plus{}z}})({1}\plus{}{\frac{y}{x\plus{}y\plus{}z}})({1}\plus{}{\frac{z}{x\plus{}y\plus{}z}})\equal{}>64/27$
for x,y,z positive real numbers.
note:i want to find an elegant solution i think we can solve it by brute force :)

It's false
try $ x\equal{}1$ $ y\equal{}2$ $ z\equal{}3$

If you put $ p \equal{} x \plus{} y \plus{} z$ $ q \equal{} zx \plus{} yz \plus{} xy$ $ r \equal{} yxz$ You can rewrite your LHS as
$ 2 \plus{} \frac {pq \plus{} r}{p^3}$

So maybe from there you can find a true inequality

[quote="mehmetcantu"]$ ({1} \plus{} {\frac {x}{x \plus{} y \plus{} z}})({1} \plus{} {\frac {y}{x \plus{} y \plus{} z}})({1} \plus{} {\frac {z}{x \plus{} y \plus{} z}}) \equal{} > 64/27$
for x,y,z positive real numbers.
note:i want to find an elegant solution i think we can solve it by brute force :)[/quote]
It should be $ \le$ and not $ \ge$ ;)

yes mathias you are right :) sorry for typo
and from lambrusus substition it is so easy from a.m g.m to show that
$ \frac{q}{p^2}\plus{}\frac{r}{p^3}<\equal{}1/3\plus{}1/27$

put: $ a\equal{}\frac{x}{x\plus{}y\plus{}z}$ etc...
then:   $ a\plus{}b\plus{}c\equal{}1$
the ineq is:   $ (a\plus{}1)(b\plus{}1)(c\plus{}1)\leq\frac{64}{27}$
we have:
$ (a\plus{}1)(b\plus{}1)(c\plus{}1)\equal{}a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca\plus{}abc\plus{}1$
$ ab\plus{}bc\plus{}ca\leq \frac{1}{3}(a\plus{}b\plus{}c)^2\equal{}\frac{1}{3}$
$ abc\leq \frac{1}{27}$
summation:
$ ab\plus{}bc\plus{}ca\plus{}abc\plus{}a\plus{}b\plus{}c\plus{}1\leq \frac{1}{3}\plus{}\frac{1}{27}\plus{}2\equal{}\frac{64}{27}$

[quote="mehmetcantu"]$ ({1} \plus{} {\frac {x}{x \plus{} y \plus{} z}})({1} \plus{} {\frac {y}{x \plus{} y \plus{} z}})({1} \plus{} {\frac {z}{x \plus{} y \plus{} z}}) \equal{} > 64/27$
for x,y,z positive real numbers.
note:i want to find an elegant solution i think we can solve it by brute force :)[/quote]
One step AM-GM:
$ \prod_{cyc} \left ( 1 \plus{} {\frac {x}{x \plus{} y \plus{} z}} \right ) \le \left ( \frac{\sum_{cyc} \left ( 1 \plus{} {\frac {x}{x \plus{} y \plus{} z}} \right )}{3} \right )^3 \equal{} \frac{64}{27}$


I'm planning on teaching myself some elementary number theory this summer (ideally, something similar to a first semester or year of college) and I'm looking for the best book to study from.  The problem is that most books I've read about or recommended by other people on the forum don't contain detailed solutions to the problems (like Aops excellently does) or even any answers whatsoever!

If anyone has any recommondations, I would greatly appreciate it.

I think Stark's NT book and the one by Niven are both pretty acclaimed.

If you go [url=http://wangsblog.com/jeffrey/?page_id=155]here[/url], there's a section on NT books. Those were the books we were supposed to get for PROMYS; basically, they are like textbooks for a college introduction to number theory course so they're pretty decent. They can be a little high-level and hard to understand at times, but overall I liked them.

I also like the Stark NT book.  It's helped me tremendously in my Intro NT class.  Some of the proofs though are going to be hard to figure out without an instructor; Stark skips some of the intermediate steps which are assumed.  I think its kind of dry, especially if you're going to teach yourself.

The classic book by Hardy and Wright (I think it's called something like Introduction to the Theory of Numbers) is pretty well written and interesting and covers lots of cool topics.  But it doesn't have much in the way of exercises and solutions.

[quote="Basketball4Life"]Thanks for all the suggestions.  My only problem is that I'm looking at all these books but it is hard to tell if they come with detailed solutions or answers to all the problems or whether there are many example problems within the chapter.  I really am looking for something with soulutions becuase I find it very helpful to see how someone else does the problem after I solve it or can't solve it.[/quote]

The Stark has "selected solutions" which are pretty expansive, IMO.


[quote="geniusbliss"]For a real-valued function $ f$,
where, $ a,n$ are natural numbers,
and $ f(a) \equal{} f(2009)$,$ f(a \plus{} 1) \equal{} f(2010)$,$ f(a \plus{} 2) \equal{} f(2011)$,
and , $ f(a \plus{} n) \equal{} \frac {f(n) \minus{} 1}{f(n) \plus{} 1}$.
show that,
$ f(a \plus{} 4n) \equal{} f(n)$[/quote]

Your question is not clear. It is for all $ a,n$ or for given $ a$ and arbitrary $ n$ or something else?

sorry stephencheng $ a$ is a fixed positive integer which satisfies,
$ f(a) \equal{} f(2009) .......$,
and $ n$ is an arbitrary positive integer

hello anyone up for a solution ?
i got it but there could be a better solution with good algebra

well here is my solution
we have
$ f(a \plus{} 4n)\equal{}f(a\plus{}3a\plus{}n)\equal{}\frac{f(3a\plus{}n)\minus{}1}{f(3a\plus{}n)\plus{}1}$
here if u go on using the condition $ f(a \plus{} n) \equal{} \frac {f(n) \minus{} 1}{f(n) \plus{} 1}$ like writing $ f(3a \plus{} n)\equal{}\frac{f(2a\plus{}n) \minus{}1}{f(2a \plus{} n) \plus{}1}$ and similarly if u use the condition and proceed for $ f(2a \plus{} n)$ and continue the procedure till u do for $ f(a\plus{}n)$ you get to prove the sum with easy simplification.


bache ha hamegi salam bebinid khodaei vaze  tarkybiat inja kheili kharabe movafeghid ye marathon rah bendazim. :?:

ghablan boode hamchin chizi mitooni edameye oon shooro koni va dar zemn biyayn marathone jabr rah bendazimmm

ok man jabro payam vali tarkybiato haminja dobare shoroo konim ghablan manzooret male babakghalebie?

ok manam moafegham pas shooroo mikonam
adade tabiyie $n<145$ mafrooz ast . dar har soal shoma mitoonin beporsin ke adade $n$ az adade $a$ koochektar ast ya kheir
javabe soale $i$ om bad az porsidane soale $(i+1)$ om dade mishe . shoma hagh darin $10$ ta soal beporsin (javabe soale $10$ om belafasele dade mishe) . rahi baraye peida kardane $n$ biabid

farz konid $k$ va $t$ do adade sahihe nesbat beham aval va bozorgtar az $1$ bashe.ba shoroo az jaygashte $(1,2,\dots\,n)$ az adade $1,2,\dots,n$ mitavanim jaye do adad ra ba ham avaz konim,agar ekhtelafe anha barabare $k$ ya $t$ basahd. sabet konid mitavan har jaygashti az adade $1,2,\dots,n$ ra ba chand marhale taviz be dast avard agar va tanha agar $n\geq\ k+t-1$.

$27$ noghte motamayez dar safhe darim ke hich se taei hamkhat nistand.$4$ noghte az in majmooe, noghate rashaye yek moraba vahed hastan ke $23$ noghteye digar dar in moraba gharar darand.sabet konid ke se noghteye motamayez  $X,Y,Z$ vojood darand ke $S_{XYZ}\leq\frac{1}{48}$.

[size=150]A game is played on a 2001 x 2001 board as follows. The first player's piece is the policeman, the second player's piece is the robber. Each piece can move one square south, one square east or one square northwest. In addition, the policeman (but not the robber) can move from the bottom right to the top left square in a single move. The policeman starts in the central square, and the robber starts one square diagonally northeast of the policeman. If the policeman moves onto the same square as the robber, then the robber is captured and the first player wins. However, the robber may move onto the same square as the policeman without being captured (and play continues). Show that the robber can avoid capture for at least 10000 moves, but that the policeman can ultimately capture the robber.[/size]

aya $12$ mokaab mostatile $p_{1},p_{2},\dots,p_{12}$ vojood darad ke yalhaye anha movazie mehvarhaye mokhtasat bashand va $p_{i},p_{j}$ noghteye moshtarak nadashte bashand agar va faghat agar $i-j$ be peymaneye $12$ barabare $1$ ya $-1$ bashad .

agha jan olampiad ke faghat jabro hendese nist age mikhaid olampiadi beshid baiad tarkybiatetoon dar darajeye aval ghavi bashe pas in marathono po konid Dge :)

rooye yek khat $2k-1$ parekhatte sefid va $2k-1$ parekhatte siah vojood darad. har parekhat sefid ba haddeaghal $k$ parekhat siah va har parekhatte siah ba haddeaghal $k$ pare khat sefid eshterak darad.sabet konid pare khatte siahi vojood darada ke bahameye parekhathaye sefid eshterak darad va parekhatte sefidi vojood darad ke be hameye parekhathaye sefid eshterak darad.

[quote="sepehr ahmadzadeh"]dar yek jadvale $9*2004$ har yek az adade $1$ ta $2004$ , $9$ bar neveshte shodeand . betorike tafazole har $2$ adadi ke dar yek sotoon gharar darand az $3$ bishtar nist , kamtarin meghdare majmooe adade satre aval ra biabid.[/quote]
lotfan age kasi rahe in soalo midune rahesho benevise,merc

[quote="sepehr ahmadzadeh 2"]javabe in masale $2005004$ esbatesh asoone ba esteghra sabet mishe agar $f(n)$ ra kamtarin meghdar begirim darinsoorat $f(n)=n-1+(1+2+\dots+(n-3)$[/quote]

man halate min ro peyda karde budam vali natunestam esteghra bezanam, mamnun misham age lotf koni esbateto benevisi :oops:

[quote="sepehr ahmadzadeh"]rooye yek daiere $2006$ noghte ba $17$ rang rang shode.mikhahim teedadi vatar rasm konim betorike do sare har vatar hamrang bashad va hich do vatari rooye daiere ya dakhele an noghteye taghato nadashte bashand . haddeaxare tedad in vatarhara biabid.[/quote]
kasi mitune esteghrasho benevise lotfan,merc :blush:

adade satre aval ra $a_{1},\dots,a_{n}$ namgozari mikonim..vazeh ast behtarin halat $9(a_{i}+3)\geq 9i$ yani $a_{i}\geq i-3$ tarif mikonim $f(n)=(1+2+\dots+(n-3))+n-1$ hal ba esteghra sabet mikonim $f(n)$ kamtarin meghdar ast..$2$ halat ra barresi mikonim::
$(i)$ be ezaye har $i ; a_{i}> i-3$ pas $\sum_{i=1}^{n}a_{i}\geq 1+1+1+2+3+\dots+(n-3)+(n-3)=f(n)$
$(ii)$ be ezaye haddeaghal yek $4\leq i \leq n-1 ; a_{i}=i-3$ ke banabar farze esteghra $\sum_{i=1}^{n}a_{i}\geq f(i)+f(n-i)+i(n-i)>f(n)$ pas $f(n)$ kamtarin meghdar ast pas baraye jadvale$9*2004$ haddeaghal majmooe adade satre aval barabare $2005004$ mibashad..

hala in soale ghashang baraye leningerado hal konid :wink:
$22$ noghte dar bazeye $[0,1]$ dade shode ast.dar har marhale mitavanim $2$ noghte ra hazf va bejaye anha noghteye vasate anahara gharar dahim,in amal ra $20$ bar anjam midahim ta nahayatan $2$ noghte baghi bemanad.sabet konid mitavan tori in amaal ra anjam dad ke faseleye beine $2$ noghteye nahaei $0/001$ sahavad..