trigonometry
analytic geometry
graphing lines
slope
function

[quote="cdymdcool"]Many racetracks have banked turns, which allow the cars to travel faster around the curves than if the curves were flat.  Actually, cars could also make turns on these banked curves if there were no friciton at all.  Explain.[/quote]
The force acting on the car is $ \frac {mv^2}{r}$ horizontally. The force can be split into two components, $ \frac {mv^2}{r} \sin A$ and $ \frac {mv^2}{r} \cos A$ where $ A$ is the angle of inclination. The component force of gravitation acting along the slope is $ mg \sin A$. It is easy to attain a velocity $ v$ such that $ \frac {mv^2}{r} \cos A$ balances the force $ mg \sin A$. Hence there is no need of friction as such.

Hope this explains it all!

[hide="Note:"]The force $ \frac{mv^2}{R}$ is a pseudo force arising due to the centripetal acceleration, the centrifugal force.[/hide]

[quote="cdymdcool"]Many racetracks have banked turns, which allow the cars to travel faster around the curves than if the curves were flat.  Actually, cars could also make turns on these banked curves if there were no friciton at all.  Explain.[/quote]

I had told Rajiv that I wasn't going to bother with this thread, but I'm going to have to rescind that decision because the thread bothers me.

It bothers me because the premise is false. It is not that there is no friction, but rather that banking the turns of a road allows one to take advantage of what friction there is.

Consider a car as if it were resting on an inclined plane then you can write the following equations by inspection:

\begin{eqnarray*}\sum F_t\,\colon \mu N - mg\sin\theta & = & 0 \\ \sum F_n\,\colon N - mg\cos\theta & = & 0\end{eqnarray*}

Solving these simultaneously,

\[ \frac{\mu N}{N} = \frac{mg \sin\theta}{mg \cos\theta} \implies \mu =\tan\theta\]

Now, we know that on a level road, $ \mu mg =\frac{m v_{max}^{2}}{R}$.

It follows that $ \mu =\frac{v_{max}^{2}}{gR}$.

However, from the sum of forces analysis we just did for the inclined plane $ \mu =\tan\theta$

Equating the two expressions:

\[ \boxed{\tan\theta =\frac{v_{max}^{2}}{gR}}\]

[quote="Dr. No"]
$ \boxed{\tan\theta \equal{} \frac {v_{max}^{2}}{gR}}$
[/quote]
 :oops: Sorry to bother you!

But since the equation of the answer does not have any $ \mu$ term, doesn't that mean the $ \theta$ does not depend upon $ \mu$. I'm confused...

[quote="Rajiv"][quote="Dr. No"]
$ \boxed{\tan\theta \equal{} \frac {v_{max}^{2}}{gR}}$
[/quote]
 :oops: Sorry to bother you![/quote]
No worries, my friend.  :wink:

[quote="Rajiv"]But since the equation of the answer does not have any $ \mu$ term, doesn't that mean the $ \theta$ does not depend upon $ \mu$. I'm confused...[/quote]

The final expression doesn't explicitly depend upon $ \mu$, you're right, but if you'll look at how I derived it, $ \tan\theta$ is $ \mu$ as a function of $ \theta$, i.e. $ \mu(\theta) \equal{}\tan\theta$.

Just because $ \mu$ doesn't pop up explicitly in a sum of forces equation doesn't mean that there is no friction present. Friction by definition is the (shear) force of contact between two bodies in contact with one another. From that definition whenever there is such a force in play you have friction.

Static friction is a function of the materials that are in contact with one another. As such there is a maximum angle at which one body will rest upon another without sliding off; this angle is called the angle of repose.

\[ \text{Angle of repose (AOR)} \equal{} \tan^{\minus{}1}(\mu_s)\]

By the way, I'm going to be away until Tuesday. If you all have any more questions I can answer, I'll try to answer them then. In other words, if I don't reply immediately, I'm not necessarily ignoring anyone.


A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents.  A subcommittee of 3 members is randomly chosen.  What is the probability that the subcommittee consists of three Republicans?

5/143
I hope i'm not making the Mcain Obama mistake...

So there are $ 13*12*11$ ways of choosing 3 members from the total members, and $ 5*4*3$ ways of choosing 3 republicans, so it would be $ 5*4*3/13*12*11$ i think...

Or maybe it's this:
there are 13C3 combinations for choosing 3 members out of the total, 13 and 5C3 ways of choosing 3 members out of 5 republicans, so it's $ 5C3/13C3$ and then reduce to get $ 5/13$...

I'm unsure now after watching that alcumus video...

The latter is correct. We don't care about the order they are chosen, so combinations is the way to go. I believe the answer is indeed $ 5/13$.


trigonometry
geometry
geometric transformation
reflection
LaTeX
analytic geometry
vector

no calculators

1. what is $\tan 75^{\circ}$?

2. simplify $\cos{12}\cos{24}\cos{48}\cos{96}$. those are in degrees.

3. what is $\cos^{2}10+\cos^{2}50-\sin 40\sin 80$. again those are in degrees.

4. suppose $\frac{\cos{3x}}{\cos x}=\frac{1}{3}$ for some angle $x$, $0\leq x\leq\frac{\pi}{2}$. determine $\frac{\sin{3x}}{\sin x}$ for the same $x$.

5. find $Im((\cos 12+i\sin 12+\cos 48+i\sin 48)^{6})$. angles are in degrees.

6. evaluate $\sin 10\sin 20\sin 30...\sin 170$. angles are in degrees.

i would like to know how to do these.

[hide="answers"]

1. $2+\sqrt{3}$
2. $-1/16$
3. $3/4$
4. $7/3$
5. 0
6. $9/65536$
[/hide]

[hide="The awesomeness of a trig identity"]
6.Using $\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$, we immediately find that the answer is $\frac{9}{65536}$[/hide]

[hide="4."]
this is a nice problem, using the triple angle formulas, we get

cos 3x/cosx=(cos^2x+3)/4=1/3

sin3x/sinx=-1/4(sin^2x-3)

from here it is trivial using sin^2x+cos^2x=1[/hide]

This kind of is boring because it's bashed. Probably a more elegant way.

[hide="3."]Work $\cos^{2}10+\cos^{2}50-\sin 40\sin 80$ in two parts.

$\cos^{2}10+\cos^{2}50 = \cos^{2}(30-20)+\cos^{2}(30+20) = 2\left(\frac{3}{4}\cos^{2}20+\frac{1}{4}\sin^{2}20\right) = \frac{1}{2}+\cos^{2}20$

Then, $\sin 40\sin 80 = \frac{1}{2}(\cos40-\cos120)$ by product to sum formula.
And use double angle formula to get $\cos40 = 2\cos^{2}20-1$, and pluggin in, $\frac{1}{2}(\cos40-\cos120) = \cos^{2}20-\frac{1}{4}$.

Subtracting yields the desired answer of $\frac{3}{4}$.[/hide]

[hide="5"]
$(cis12+cis48)^{6}$
$cis^{6}12+6cis^{5}12cis 48+15cis^{4}12cis^{2}48+20cis^{3}12cis^{3}48+15cis^{2}12cis^{4}48+6cis12 cis^{5}48+cis^{6}48$
$cis72+6cis108+48+15cis144+20cis120+15cis216+6cis252+cis288$
$(cis72+cis288)+6(cis108+cis252)+15(cis144+cis216)+20cis180$

For each pair, the sum of the angles add up to 360. Thus, they are reflections of each other across the real axis, so the imaginary parts becomes 0. The term $20cis180$ is also on the real axis, so its imaginary part is 0. Thus the total imaginary part is $\boxed{0}$

btw, does anybody know how to use cis in latex?

EDIT- Using my calculator, I found out that cis12+cis48 when written in polar coordinates has an angle of 30 degrees, so taking it to the sixth power would give it an angle of 180 degrees, placing it on the real axis. Does anybody know how to prove that cis12+cis48 is a 30 degree number without a calculator?

EDIT 2- I realized that adding any two complex numbers with equal absolute values, will give a comple numbers whose angle is the average of the original numbers' angles. Or you could also do what 10000th User did.
[/hide]

[quote="tjhance"][hide="1"]
Use the trig identity: $\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
$\tan 75=\sqrt{1-\cos150}{1+\cos150}$
$\cos150$ is the additive inverse of $\cos30$, which u can find to be $\frac{\sqrt{3}}{2}$, making $\cos150=\frac{-\sqrt{3}}{2}$
$\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}$
After some simplification, that gives you $\boxed{2+\sqrt{3}}$
[/hide]

[hide="4"]
Use the identity: $\cos(x+y)=\cos x\cos y-sin x\sin y$
as well as the identity: $\sin(x+y)=\sin x\cos y+\sin y\cos x$
$\cos3x=\cos 2x\cos x-\sin 2x\sin x$
$=(\cos^{2}x-\sin^{2}x)\cos x-(2\sin x\cos x)\sin x$
$=\cos^{3}x-\sin^{2}x\cos x-2\sin^{2}x\cos x$
$=\cos^{3}x-\cos x(\sin^{2}x+2\sin^{2}x)$
$=\cos^{3}x-\cos x(3\sin^{2}x)$
Now I use the identity: $\sin^{2}x+\cos^{2}x=1$
$=\cos^{3}x-\cos x(3-3\cos^{2}x)$
$=\cos^{3}x-3\cos x+3\cos^{3}x$
$=4\cos^{3}x-3\cos x$

$\frac{\cos 3x}{\cos x}=\frac{4\cos^{3}x-3\cos x}{\cos x}=4\cos^{2}x-3$
$4\cos^{2}x-3=\frac{1}{3}$
$4\cos^{2}x=\frac{10}{3}$
$\cos^{2}=\frac{10}{12}=\frac{5}{6}$
$\sin^{2}x+\cos^{2}x=1$
$\sin^{2}x+\frac{5}{6}=1$
$\sin^{2}x=\frac{1}{6}$

We can derive a similar identity as the one above: $\sin 3x=3\sin x-4\sin^{3}x$
$\frac{\sin 3x}{\sin x}=\frac{3\sin x-4\sin^{3}x}{\sin x}=3-4\sin^{2}x=3-4\left(\frac{1}{6}\right)=\boxed{\frac{7}{3}}$

[/hide][/quote]
[hide="another way to do number 1"]
$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a \tan b}$[/hide]

[i]Mod edit: let's not call other people's solutions 'ugly'...[/i]
EDIT: Sorry, I meant a way to do it with a less ugly identity.

[quote="bpms"][quote="tjhance"][hide="1"]
Use the trig identity: $\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
$\tan 75=\sqrt{1-\cos150}{1+\cos150}$
$\cos150$ is the additive inverse of $\cos30$, which u can find to be $\frac{\sqrt{3}}{2}$, making $\cos150=\frac{-\sqrt{3}}{2}$
$\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}$
After some simplification, that gives you $\boxed{2+\sqrt{3}}$
[/hide]

[hide="4"]
Use the identity: $\cos(x+y)=\cos x\cos y-sin x\sin y$
as well as the identity: $\sin(x+y)=\sin x\cos y+\sin y\cos x$
$\cos3x=\cos 2x\cos x-\sin 2x\sin x$
$=(\cos^{2}x-\sin^{2}x)\cos x-(2\sin x\cos x)\sin x$
$=\cos^{3}x-\sin^{2}x\cos x-2\sin^{2}x\cos x$
$=\cos^{3}x-\cos x(\sin^{2}x+2\sin^{2}x)$
$=\cos^{3}x-\cos x(3\sin^{2}x)$
Now I use the identity: $\sin^{2}x+\cos^{2}x=1$
$=\cos^{3}x-\cos x(3-3\cos^{2}x)$
$=\cos^{3}x-3\cos x+3\cos^{3}x$
$=4\cos^{3}x-3\cos x$

$\frac{\cos 3x}{\cos x}=\frac{4\cos^{3}x-3\cos x}{\cos x}=4\cos^{2}x-3$
$4\cos^{2}x-3=\frac{1}{3}$
$4\cos^{2}x=\frac{10}{3}$
$\cos^{2}=\frac{10}{12}=\frac{5}{6}$
$\sin^{2}x+\cos^{2}x=1$
$\sin^{2}x+\frac{5}{6}=1$
$\sin^{2}x=\frac{1}{6}$

We can derive a similar identity as the one above: $\sin 3x=3\sin x-4\sin^{3}x$
$\frac{\sin 3x}{\sin x}=\frac{3\sin x-4\sin^{3}x}{\sin x}=3-4\sin^{2}x=3-4\left(\frac{1}{6}\right)=\boxed{\frac{7}{3}}$

[/hide][/quote]
[hide="another way to do number 1"]
$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a \tan b}$[/hide]

[i]Mod edit: let's not call other people's solutions 'ugly'...[/i][/quote]
Like bpms said (for #1 I mean):
[hide="1"]$\tan 75=\tan (45+30)=\frac{\tan 45+\tan 30}{1-\tan 45\tan 30}=\frac{1+\frac1{\sqrt3}}{1-\frac1{\sqrt3}}=2+\sqrt3$[/hide]

[hide="2"]Note that
$\cos 96=-\cos 84=-\sin 6$,
$\sin 96=\sin 84=\cos 6$, and
$\sin 12=2\sin 6\cos 6$,

we have

$\begin{aligned}\cos 12\cos 24\cos 48\cos 96&=-\sin 6\cos 12\cos 24\cos 48\\ &=-\frac{\sin 12}{2\cos 6}\cos 12\cos 24\cos 48\\ &=-\frac{\sin 24}{2^{2}\cos 6}\cos 24\cos 48\\ &=-\frac{\sin 48}{2^{3}\cos 6}\cos 48\\ &=-\frac{\sin 96}{2^{4}\cos 6}=-\frac1{16}\end{aligned}$[/hide]

[hide="3"]
$\cos 10=\sin 80$
$\cos 50=\sin 40$

Use product to sum formulas for sin-cos and sin-sin, and sum to products for sin-sin,

$\begin{aligned}\cos^{2}10+\cos^{2}50-\sin 40\sin 80&=\sin 80\cos 10+\sin 40\cos 50-\sin 40\sin 80\\ &=\frac12\left(\sin 90+\sin 70+\sin 90+\sin (-10)+\cos 120-\cos (-40)\right)\\ &=\frac12\left(1+1-\frac12+(\sin 70-\sin 10-\cos 40)\right)\\ &=\frac34+\frac12(2\cos 40\sin 30-\cos 40)=\frac34 \end{aligned}$[/hide]

EDIT: tjhance,
[quote="tjhance"]Does anybody know how to prove that cis12+cis48 is a 30 degree number without a calculator?[/quote]Well you can prove that $\sin 12+\sin 48=2\sin 30\cos 18$ and $\cos 12+\cos 48=2\cos 30\cos 18$ from sum to product formula for sin-sin and cos-cos. #5 would be $\mathrm{Im}\left(((2\cos 18)(\cos 30+i\sin 30))^{6}\right)$... :)

Uh....I'm guessing bpms had originally posted this solution for #1, but I don't see it, so i guess I'll show you the way I did it (yeah, I actually have this formula memorized, lol)[hide]
$\tan{70}=\tan{(30+45)}=\frac{\tan{30}+\tan{45}}{1-\tan{30}\tan{45}}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=2+\sqrt{3}$[/hide]
Edit: Yeah, my bad, I was on a crappy comp at my school and nothing showed up when I tried to look at your post. Otherwise, I wouldn't have posted.

[quote="aahmadi"]Uh....I'm guessing bpms had originally posted this solution for #1, but I don't see it, so i guess I'll show you the way I did it (yeah, I actually have this formula memorized, lol)
$\tan{70}=\tan{(30+45)}=\frac{\tan{30}+\tan{45}}{1-\tan{30}\tan{45}}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=2+\sqrt{3}$[/quote]
I didn't post the whole solution, just the hint.
That formula is easy to derive anyways.

Elemennop wrote:For #5, just

[hide]

draw the two vectors, and it's obvious that the sum of the two vectors bisects the angle formed by the two individual vectors, so it makes 30degrees with x axis..

[/hide]

Yes I realized that if you read my EDIT

though not in those words

[quote="iamagenius"]no calculators

i would like to know how to do these.

[/quote]

learn the double angle, half angle, sum and difference formulas


binomial coefficients
combinatorics unsolved
combinatorics

[quote="brettburns83"]1) Prove Pascal's formula
(nCk) = (n-1 Ck) + (n-1 Ck-1)

by substituting the values of the binomial coefficients as given by (nCk)=n!/k!(n-k)!

I've gotten as far as:

n(n-1)...(n-k+1)/ k(k-1)...1= (n-1)(n-2)...(n-k)/k(k-1)...1 + (n-1)(n-2)...(n-k+1)/(k-1)...1

Not sure if I'm doing this right, when I plug random values in it works but I don't see how I can get to a point where I prove they equal each other.[/quote]
mmmmmmmmm...
as far as I know it has a real easy proof with combinatorics:
we want to choose $k$ persons from $n$ persons
we do that by $C(n,k)$
in the other hand we can choose these persons in two ways:
$i)$we choose person $x$
$ii)$we dont choose person $x$
now we calculate these two:
$i)$we want $k-1$ other persons from $n-1$ persons so we can do that in $C(n-1,k-1)$ ways
$ii)$we want $k$ persons from $n-1$ persons so we can do that in $C(n-1,k)$ different ways
so we have:
$C(n,k)=C(n-1,k)+C(n-1,k-1)$

or you can prove it your way:
$C(n-1,k)+C(n-1,k-1)= \frac{(n-1)!}{k!(n-1-k)!}+ \frac{(n-1)!}{(k-1)!(n-1-k+1)!}= \frac{(n-k)(n-1)!+k(n-1)!}{k!(n-k)!}= \frac{(n-1)!(n-k+k)}{k!(n-k)!}= \frac{(n-1)!n}{k!(n-k)!}= \frac{n!}{k!(n-k)!}=C(n,k)$
now our proof is completed
 ;)
did you get it??

Thanks for the help I just got hung up on the next step with what to do with all of the + and - k's and 1's. I follow what you did thought thanks


induction
linear algebra
linear algebra unsolved

Induction on $n$ that $A^{n}=\frac{7^{n}-5^{n}}{2}A+C$ and $B_{n}=\frac{7^{n}-5^{n}}{2}B+C$, Where $C$ is a matrix.


trigonometry
calculus
LaTeX
trig identities

one of the questions for the aops calculus diagnostic test is

$ (cos pi/12 \plus{}i(sin pi/12))^9$

but according to this formula i found on the aops wiki

$ e^{iA}\equal{}sin A \plus{} cos A$

i do not get the correct answer which is $ \minus{}\frac{\sqrt{2}}{2}\plus{}i\frac{\sqrt{2}}{2}$

can someone show me a proof to the answer because i am very confused

[quote="peregrinefalcon88"]one of the questions for the aops calculus diagnostic test is

$ (cos pi/12 + i(sin pi/12))^9$

but according to this formula i found on the aops wiki

$ e^{iA} = sin A + cos A$

i do not get the correct answer which is $ - \frac {\sqrt {2}}{2} + i\frac {\sqrt {2}}{2}$

can someone show me a proof to the answer because i am very confused[/quote]

A $ \text{\LaTeX}$ note (backslashes are your friend), put a backslash behind pi,cos and sin to get and use the \frac to do division.:

\[ (\cos \dfrac{\pi}{12} + i(\sin \dfrac{\pi}{12}))^9\]

This is the best I can possibly make it look in $ \text{\LaTeX}$:

\[ \left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^9\]

[code]$\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^9$[/code]

Anyway, using $ e^{i\theta}=\text{cis}\,\theta$ ("cis" is a common abbreviation) is not completely necessary, because you can just use De Moivre's theorem (even though the theorem follows immediately from that formula).

[b]Theorem.[/b] (De Moivre). For any $ x\in\mathbb{R}$, $ (\cos\theta+i\sin\theta)^x=\cos x\theta+i\sin x\theta$.

Thanks guys that was the only problem i couldn't solve but i still think i need to learn more trig. I printed some pages from the aops wiki (trig identities and the laws of sin, cos, tan) but do you guys know any other sources that would help?

Plane Trigonometry by S.L. Loney

Volume 2


vector
geometry
parallelogram
real analysis
real analysis theorems

Suppose $ V$ is a normed real vector space that satisfies the parallelogram law: for all $ x$ and $ y,$ $ \|x\plus{}y\|^2\plus{}\|x\minus{}y\|^2\equal{}2\|x\|^2\plus{}2\|y\|^2.$

Show that $ \langle x,y\rangle\equal{}\frac14\left(\|x\plus{}y\|^2\minus{}\|x\minus{}y\|^2\right)$ satisfies the axioms of an inner product on $ V.$

====

And if you can do that, what would you do if $ V$ were a complex vector space?

[quote="Kent Merryfield"]Suppose $ V$ is a normed real vector space that satisfies the parallelogram law: for all $ x$ and $ y,$ $ \|x \plus{} y\|^2 \plus{} \|x \minus{} y\|^2 \equal{} 2\|x\|^2 \plus{} 2\|y\|^2.$

Show that $ \langle x,y\rangle \equal{} \frac14\left(\|x \plus{} y\|^2 \minus{} \|x \minus{} y\|^2\right)$ satisfies the axioms of an inner product on $ V.$

====

And if you can do that, what would you do if $ V$ were a complex vector space?[/quote]
This is the Jordan–von Neumann theorem. You can find an outline of how to prove it [url=http://www.math.uci.edu/~toikhber/TEACH/F05/M260A/NOTES/PARAL/paral.pdf]here[/url] (pdf file).

Thanks for the useful and succinct reference. I stated this in the middle of a class without having checked beforehand that I knew the proof and discovered fairly quickly that if I didn't have that proof prepared, I wasn't going to find it on the fly.


Let $ ABC$ be a triangle and the points $ M\in(AB),\ N\in(AC),\ P\in BN\cap CM$.
Prove that $ [AMPN]\equal{}[BPC]\Longleftrightarrow \frac{MA}{MB}\equal{}\frac{NC}{NA}$ .

Does $ M\in(AB),\ N\in(AC),\ P\in BN\cap CM$ mean M is on AB, N is on AC, and P is the intersection of BN and CM?

Yes, it does mean that.
Are you sure this is right? I have
$ [AMPN] \equal{} [ABC] \minus{} [BMC] \minus{} [BNC] \plus{} [BPC] \equal{}
\newline [ABC](1 \minus{} \frac {CN}{AC} \minus{} \frac {BM}{AB}) \plus{} [BPC]$
So $ [AMPN] \equal{} [BPC] \implies \frac {CN}{AC} \plus{} \frac {BM}{AB} \equal{} 1$, which I don't think is equivalent to what you have.

It is right because $ \frac{CN}{AC}\plus{}\frac{BM}{AB}\equal{}1\ \Longleftrightarrow \frac{CN}{AC}\equal{}\frac{AM}{AB}\ \Longleftrightarrow \frac{CN}{NA}\equal{}\frac{AM}{MB}$.  :P
Thanks.

Oh, wow, I can't believe I missed that  :D


AMC AIME USA(J)MO USAMO function USAMTS ARML

TX are such "awesome" letters for initials.

ne ways,

im indian, barely 5ft,kind of skinny and in 8th grade right now.

edit: im failish at math, ive only taken 1 AIME in my life (ive looked at others)  in which i got a 5 :(.

i play the piano,violin,and i swim ocasionally because my mom wants me to get taller,also b/c its fun.

i have a bunch of board games and im not sure which to bring.  any suggestions?

I'm most likely going so:
I am a sophmore from Gunn High School, and I am bad at math. I am terrible at piano as well, but overly enthusiastic about it. I will be sure to bring a CD player and a lot of CDs (since I don't have one of those certain Apple products with a lot of memory). I kinda play chess, and I am somewhat bad at it. In general I fail?

I'll be there.
Shri Ganeshram from Wichita Falls, TX.
I'm pretty decent at UIL stuff, but I'm a noob at AIME.
I'm Indian, 9th grade, bout 6 feet tall. I make games for fun.
I'll play chess, but I'm not too good. I will definitely bring my phone and that certain touch version of an apple device who has an SSD of 32gb.

Anyone like basketball?

I'm Chinese, I love Metallica and Nirvana, I like staying up late (doing math, what else), I'm very disorganized, and I can get around 8-12's on AIMEs.

Oh yeah, freshman from AAST.

I also kind of like chess but fail at it at the same time.

Hi.

My name is Sam.. Korean.. from Plano, TX which is about a 20 minute drive from the UTD site. I also personally know the director, Dr. Andreescu, so..

First time in AMSP.. I'm a 7th grader, but I'm 5' 7" and could pass for a freshman?? ha..

Um.. Anyone play tennis? Because I'm definitely bringing my tennis racket, and despite all the math, I have to train for my summer excellence circuit tournaments.. And really, I'm maybe too involved in tennis, but maybe @ camp, no time?

Also bringing a portable game system to do it in bed instead of sleeping after lights off.. The DS. Which is suckish compared to a PSP, but it does the job: keep you busy and somewhat entertained.

Um, everyone is talking about chess so.. I hate chess. I am very epic phail at it, too. But for the board games, I'm good with the games like monopoly and stuff life that where you go around and around and get money and stuff and land on random spots. Like Life, too.

I wouldn't prefer to talk about my mathematical performance, let's just say I epic phailed at the AIME last year + this year. But still above average: average tends to range 2.5 ~ 3.5 ... (from Wikipedia. haha..)

Looking forward to the camp.. Maybe some of us will room together??

I'm most probably going, as well.

My name is Sona Chowdhary.
From Florence, SC.
9th grade, 15, 5'2-ish

I'm not that great at math but enjoy it, nonetheless. I used to play chess a lot but haven't for the past year or two... I play the piano, take karate, and play tennis.

Sjyoo: Perhaps we could play tennis together.

You'll find out my name if you are going there: Cooler initials than T.X!

Finishing 9th, 4'7"
Glasses (transitions)
Will probably ask you if you have Aops or not.....
Asian(chinese)
Going to AMSP in TX
Taken 1 AIME, 1 AMC 10, and 1 AMC 8

failed MC chapter..
Lacks common sense.
Likes to Jog (Track/XC team) and like to play cards...I'm decent - know the rules of many board games.. (Chess, checkers)
Clueless at times.
Will probably bring GBA-SP...
Meh I dont know how else to describe myself.. being 4'7" Should make me easy enough to find/recognize.

I'm some sort of generic person, have a name, attend school, and am between three and seven feet tall.

More importantly, I guess, I've taken three AMC 12s, one AMC 10, two AIMEs, and no USAMOs. The last number will not increase this year. My USAMO index expressed as a function of years since 2008 is monotonously decreasing, my AMC scores expressed as a similar function are monotonously decreasing, and my AIME scores expressed as a similar function are constant. Bronze on USAMTS, first at states this year. I suck at math.

Also I'm going to division B ARML in case anyone else is.

15 Finishing 9th 5 7'' chinese
Plays basketball, soccer, tennis,
Plays violin a lot
Taken AMC 8, 10, 12, and AIME
Movie/ Music fanatic
:) I'll see you guys there

hadasah

initials: ML
I wonder who else is a girl...
I'm and ABC, horrible at sports, though I do ballet. about 5ft5. will finish 8th grade, 13 right now, but 14 at camp.
I also suck at chess, and I only personally know one of the people going to AMSP
I'm living in China right now , which complicates a lot of things like mail and transportation and forgetting things...
I think that's all....

At least you know one person....
I don't know anyone, in both Cali and TX, and I'm going to both!
BTW, Texas is not the stereotypical "wild west"... In fact its similar to the north, except its more spread out, flat, and people say "y'all"

Btw, I know Dallas pretty well...
I go there like twice a month.

[quote="serialk11r"]I'm most likely going so:
 I am bad at math. I am terrible at piano as well?[/quote]

Yea right...

Anyways, I'm currently a sophomore in AL and fail at math and piano more than serialk11r, seeing as I only practice an hour a day, and didn't start to really study math hard til last year  (so no usamo for me).

Anyways, I liked running cross country, even though I was the slowest person on the team, and track, but didn't have time to run this school year. I'm guessing there won't be an indoor track nearby, but it would be awesome if there were. Anyone else like to run?

I fail at chess since I haven't played in 4 years. I also like to play ping-pong.

Oh, and I have been 6ft tall for about 3 years.

Does anyone know if there is like a fitness center at Univ. of Dallas that we can use? It says it has like tennis courts, basketball courts and such but im not sure if it actually has a fitness facility with like weights or something.

Well the thing is the only math I've ever done was:
12/2007-4/2008: bunch of AIMEs, AMCs.
3/2009-4/2009: handful of olympiad questions.

For piano, I am pretty horrible because I had a pretty bad teacher until last year. Now I practice a lot but I almost feel like I am stupid for doing so because there are so many little middle schoolers better than me.

then you clearly improve 2347293x times faster than I do.

[quote="mathtyro"]Anyone from NJ? I might need to share a flight with someone...[/quote]

Im from NJ too, but im only 14.  What part of NJ do u live in?  im from Princeton Jct.

I'm 14, in 9th grade, and I live in Oklahoma.
I went to AMSP in 2007, but not in 2008.
3 AMC 10s, 2 AMC 12s, 1 AIME.
I also do Chem Olympiad and made the National test this year, but that didn't go very well...
Other than math, I like fencing, reading, and writing. And I jump rope when I need to calm down.

[quote="FightTheTide"]Hey everyone. I'm Jay, and I'll be on staff as an assistant this year. I've gone to AMSP for the past two years, and I live in Dallas, so let me know if you have any questions.[/quote]

YESSSSS! ^

hi i'm russell
half korean
i play the cello, swim, do whatever
i listen to a lot of music, and watch a lot of movies

lookin forward to it

Is anyone else going on the Chicago O Hare Airport?

[quote="FightTheTide"]Hey everyone. I'm Jay, and I'll be on staff as an assistant this year. I've gone to AMSP for the past two years, and I live in Dallas, so let me know if you have any questions.[/quote]

Wait how do you get invited as an assistant? Do they just call you and ask you to be an assistant?

i think u can contact them and tell them u want to be an assistant.

My name is Eric and I'm from San Antonio, TX. I'm 5 foot 2ish, or something like that, and I'm probably bringing a bunch of books like Forgotten Realms, or other junk. I'm Chinese and I also play the piano and violin. BTW, I'm 12 and I'm finishing 7th grade..............I play chess and um..........................I also play tennis, but I'm not that good....................... :D .................. :ninja:  :ninja: ..........................Anyone else bringing books? Maybe we could share.........I'm probably bringing some games (if I can find them, my dad is crazy about not wanting me playing games).....That was probably terrible grammar (points to the kind of sentence in parentheses before).............yep, that's it, oh wait!!!!, I did MathCounts this year, got 6th in chapter (Bexar chapter), and I have no idea what I got in state, I haven't done any AMC's yet, though I have looked at old AMC 10s, and I think I did okay.....I'm probably going to be the worst one at math at AMSP, so yeah, I'll probably be relying on you guys for help

Hi, my name is [censored for potentially dangerous content]. I live in [censored for potentially dangerous content]. I'm finishing 6th grade (OMG! that got through!). To see my transcript, look at my sig. Specific, isn't it? I may or may not be going to AMSP (that is all I could say before revealing too much). I am looking forward to see everyone. I play violin (and do a lot of competitions), am good at chess (2nd best K6 in state... yay?), did spelling bee (did decently well, though would have done better had I read a medical reference book), did scifair (1st in state... yay again?), does a lot of math (duh), likes model rocketry, cares for the environment (created a countywide EARTH club :P ), does WAY too many activities... yeah...

Oh, and my personality is like this:[censored for potentially dangerous content].

darnit. I hate those stupid letter readers.

hm just a suggestion but
if you meet anyone from delaware named edward li, RUN FOR YOUR LIFE. HE IS THE MOST PERVERTED KID YOU WILL EVER MEET.
:)

I'm not so sure about the "most" part, but okay.

[quote="FFfreak127"]I also play tennis, but I'm not that good....................... :D .................. :ninja:  :ninja: ..........................[/quote]

Finally! Someone who plays tennis!!! Please bring your racket, or I have a spare.  :lol: I've been looking ages to find someone who plays tennis, and it's OK if you're not that good..

Hi guys, I think this is my second intro...? Anyways, my name is Sam and I'm a local from Plano, TX, which is about 20 minute drive away from UTD.. I'm 5' 8", but I'm in 7th grade so whaaaat... (does race affect height? and if it does, I'm Korean btw.) lol. As said, my biggest hobby besides math, which shouldn't really be a "hobby" although many enjoy it, is tennis. Besides tennis, I also swim competitively (swim team) and play some b-ball.. (height advantage? lol) Those are really my only sports.. Um.. what else to say.. Uh.. right now, it's almost 8:00 AM here ONE DAY before AMSP! YAY! Lookin forward to it.. Still not decided on which courses, but oh well.. :D :D :D.. Um.. yea.. that's all for now, and see you here in Dallas!! IT'S GONNA B FUN!

I swim competitively and that's my only sport...

Unfortunately, I can't bring my violin (wah), so I'm not practicing when I'm there. I'm also not bring any athletic gear except my swim trunks and racing suit (why not?). So excited... Tomorrow is AMSP day!!

I can't wait... bringing a lot of books (one for Mr. Andreescu to sign), but I'm not bringing any non-math books... i figured I could check them the at the library. I'm not going to bring my chess set, unfortunately, so could someone plz bring theirs? I want to get higher than 1564 after AMSP...

See you there!! I wonder who I'll be sharing a room with... any other sixth graders? seventh graders? ppl who I would not want to share a room with?

~AMSPS (AMSP Student)

how are the dorms and the rooms? How many people are in a room etc. can someone please give me more info on this, i would like to know.

it explained this in the packet... and a whole bunch of other places... haven't you read that stuff?

We are going to live in apartment-like dorms with similarly-aged people. There are 4 bedrooms in each dorm, there will be 6 to a dorm. Each person shares a room with someone else, and there is a college student or someone similar in the leftover bedroom. there will be 2 bathrooms, a washer/dryer, and  kitchen in each dorm.

seriously... read the stuff that they give you...

[quote="asdfjklasdfjkl"]yea...i play tennis
im in B14's Champs
btw whats "summer excellence circuit"??[/quote]

I was in B12's Champs, then I aged up and got pulled down to ZATs.. And now I'm working back up to CMZ's.. Although I'm really at the bottom of champs.. You'll be surprised that I was a champ  :blush: ..

Anyways, summer excellences are what you would call a "special tourney"... I will send you message further on this..

Yay MORE TENNIS PEOPLE! I hope you are crazy about tennis as I am... Tennis is basically ALL I DO in my free time... I'll basically be playing tennis all the recreational time there :D .. Anyone else tennis?

What other recreational sports are popular (I know tennis isn't "POPULAR" but.. yea..) ??

My name is Rolland He and I'm from Troy, Michigan.
I'm currently 15 and will be a sophomore in the fall.
I play piano and violin, although I'm pretty bad at both.
I used to play chess and I was halfway-decent, but years of not playing it has probably caused by skills to falter by quite a bit.
I like to play tennis (I majorly suck, but I still play it) and sometimes swim...
Science is cool and I've been doing Sci-oly for a couple of years now.
I really like computer programming and am somewhat of a computer junkie  :lol:
I've AMC'd 5 times, AIME'd 4 times, USAMO'd once (although I did much worse than I should have). I am hoping to make MOP next year.
Looking forward to meet some new people!

YAY FOR SWIMMING!!!i swim competively too...(i hav to skip junoir olympics for this... :( ) and um...you dont HAVE to read the stuff they give you...i found all that out when i got there last year...learn as you go???i guess???

erm, gatorlover, you seem a little hyper...

To identify me, look for a sorta-tall to-be-7th-grader  wearing an MC Nats shirt, jean shorts, NASA hat, and green backpack. I'll also be kinda perverted, very obnoxious at times, very quiet at times (best of both worlds: interpersonal and intrapersonal), and may be bouncing around and saying "This place is amazing!" in awe of a math summer camp (I did at at MC Nats and SciBowl nats...). Overall, i'll be a really eccentric person...

ummm.... you didn't go to scibowlnats..... anyway, hope every1 is having fun at amsp.


invariant
vector
abstract algebra
induction
linear algebra
linear algebra unsolved

assume that $X$ is a space on $\mathbb{C}$ with finite dimension and  $L(X)$ denote the set of all operator on $X$ .suppose that $S$ is semigroup of $L(X)$,prove that $S$ is simultaneously triangulable iff for every $A,B,C \in S$ we have $tr(ABC)=tr(CBA)$.
this is nice theorem that one of its consequence is Kaplansky theorem :if $S$ is semigroup of all n*n matrice such that trace is constant on $S$ then $S$ is simultaneously triangulable.
and the other consequence:if all member of $S$ are idempotent then $S$ is simultaneously triangulable.

Suppose we have a semigroup $ S \subset L(X) $, such that for every $ A,B,C \in S $
we have $ tr(ABC) = tr(CBA) $. We have to prove that $ S $ is simultaneously triangulable.
We have that $ tr(ABC) = tr(CBA) = tr(BAC) $, so $ tr((AB - BA)C) = 0 $. Because of $ S $
is a semigroup, we have that $ tr((AB - BA)DC) = 0 $ for every $ A,B,C,D $ in $ S $, so
$ tr(C(AB - BA)D) = 0 $ for every $ A,B,C,D $ in $ S $. Now denote by $ T $ the span
over $ \mathbb{C} $ of the operators of the form $ C(AB - BA)D $, where $ A,B,C,D $ in $ S $.
I claim that $ T $ is a semi-algebra, i.e. product of two elements of $ T $ is also
in $ T $. Indeed, $ C'(A'B' - B'A')D' C''(A''B'' - B''A'')D'' =
(C'A'B'D'C'')(A''B'' - B''A'')D'' - (C'B'A'D'C'')(A''B'' - B''A'')D'' $.
We know that $ tr(X) = 0 $ for every $ X \in T $. Because of $ T $ is semi-algebra,
we have that also $ tr(X^{k}) = 0 $ for every $ k \in \mathbb{N} $. Therefore
we have that every element of $ T $ is nilpotent.
Now let's solve the following problem:
If we have a semigroup $ S \subset L(X) $ such that if we define $ T $ as above,
then every element of $ T $ is nilpotent. Then $ S $ is simultaneously triangulable.

It is clear that the original problem follows from this problem.
Also for the new problem it is enough to prove that there exists a common eigenvector
for all elements of $ T $. Indeed, the property of the new problem remains the
same if we take a quotient by a subspace, which is invariant under the action of
all elements in $ S $. Let's now solve this problem.
Suppose we have that every element of $ T $ is nilpotent. Let us prove that in general,
if we have a semi-algebra $ R $ of nilpotents, acting on the complex vector space
$ V $,  then there exists a vector, lying in kernel of each element of $ R $.
Indeed, if we have an invariant subspace $ W \subset V $, then if there exists
some $ w \in W $, such that $ Rw \neq \{ 0 \} $, then $ Rw \subset W $
is invariant subspace of all elements of $ R $, and $ Rw $ is stricly smaller than
$ W $, because it cannot contain $ w $. Therefore if we take the smallest
$ R $ -invariant subspace of $ V $, then every element of it lies in the kernel of
all operators from $ R $. Now, if we take our $ T $ and apply this to it,
we will get that there exists a vector $ v \in X $ such that
$ C(AB - BA)Dv = 0 $, for every $ A,B,C,D $ in $ S $.
Denote by $ Y \subset X $ the span of all vectors of the form $ Dv $, where $ D \in S $.
Now there are some cases:
if $ Y \subsetneq X $, then $ Y $ is $ S $ -invariant, therefore by induction on the dimension
of $ X $, we will get a common eigenvector.
If $ Y = X $, then we have that $ C(AB - BA) = 0 $ for every $ A,B,C \in S $.
In this case, if for every $ A,B \in S $ we have $ AB = BA $, than every two
elements of $ S $ commute, therefore there exists a common eigenvector.
Now, if there exist $ A,B \in S $, such that $ AB-BA \neq 0 $, then there exists
$ w \in X $, such that $ u := (AB-BA)w \neq 0 $, and we have that $ Cu = 0 $ for
every $ C \in S $, therefore $ u $ is a common eigenvector.

I hope it is correct.


1. In an $ ABC$ triangle such that $ m(\angle B)>m(\angle C)$, the internal and external bisectors of vertice $ A$ intersects $ BC$ respectively at points $ D$ and $ E$. $ P$ is a variable point on $ EA$ such that $ A$ is on $ [EP]$. $ DP$ intersects $ AC$ at $ M$ and $ ME$ intersects $ AD$ at $ Q$. Prove that all $ PQ$ lines have a common point as $ P$ varies.

2. A graph has $ 30$ vertices, $ 105$ edges and $ 4822$ unordered edge pairs whose endpoints are disjoint. Find the maximal possible difference of degrees of two vertices in this graph.

3.The equation $ x^3\minus{}ax^2\plus{}bx\minus{}c\equal{}0$ has three (not necessarily different) positive real roots. Find the minimal possible value of $ \frac{1\plus{}a\plus{}b\plus{}c}{3\plus{}2a\plus{}b}\minus{}\frac{c}{b}$.

4.The sequence $ (x_n)$ is defined as; $ x_1\equal{}a$, $ x_2\equal{}b$ and for all positive integer $ n$, $ x_{n\plus{}2}\equal{}2008x_{n\plus{}1}\minus{}x_n$. Prove that there are some positive integers $ a,b$ such that $ 1\plus{}2006x_{n\plus{}1}x_n$ is a perfect square for all positive integer $ n$.

5.$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD\equal{}\frac{BD^2}{AB\plus{}AD}\equal{}\frac{CD^2}{AC\plus{}AD}$. $ E$ is a point such that $ D$ is on $ [AE]$ and $ CD\equal{}\frac{DE^2}{CD\plus{}CE}$. Prove that $ AE\equal{}AB\plus{}AC$.

6. There are $ n$ voters and $ m$ candidates. Every voter makes a certain arrangement list of all candidates (there is one person in every place $ 1,2,...m$) and votes for the first $ k$ people in his/her list. The candidates with most votes are selected and say them winners. A poll profile is all of this $ n$ lists.
If $ a$ is a candidate, $ R$ and $ R'$ are two poll profiles. $ R'$ is $ a\minus{}good$ for $ R$ if and only if for every voter; the people which in a worse position than $ a$ in $ R$ is also in a worse position than $ a$ in $ R'$. We say positive integer $ k$ is monotone if and only if for every $ R$ poll profile and every winner $ a$ for $ R$ poll profile is also a winner for all $ a\minus{}good$ $ R'$ poll profiles. Prove that $ k$ is monotone if and only if $ k>\frac{m(n\minus{}1)}{n}$.


What is the sum of the first two positive integers $ n$ such that $ 360 \cdot n$ has exactly 72 positive divisors.

EDIT: is this like more of intermediate?

[hide="EDIT Hint Now  :oops: "]
$ 360 = 2^{3}\cdot 3^{2}\cdot 5 \Rightarrow 4 \cdot 3 \cdot 2 = 24 \ \text{factors}$
[/hide]

No both answers posted (by AstroPhys and hammertime) are wrong.

Say its $ 2^{2}=4=n$, okay then $ 360n=2^{5}*3^{2}*5^{1}$, which has $ 6*3*2=36$,not $ 72$ factors.

[quote="hammertime"][hide]Or you could realize that n must have 3 factors that are non-existent in the prime factorization of 360. This would be 7^2, which equals 49. 4 + 9 = 13.[/hide][/quote]

Read the question man. Hes asking for the sum of the first two numbers, not the sum of the digits of the first number that comes up.

Plus, 49 isnt the first two numbers that come up...i beleve 28 works...

also astro, sure 2^2 has 3 factors,but when added on to 2^3,it has 2^5-6 factors.

Wait, what? 360*49 = 17640. The prime factorization of 17640 is 2^3 * 3^2 * 7^2 * 5. 4*3*3*2 = 72. Therefore 17640 has 72 factors. Or am I still not understanding what this problem is asking for?
[hide]Wait, is it 49 + 121 = 170? [/hide]

Thats still wrong.

Yes, you are right 7^2 and 11^2 work but thats not the only way (definedly not the smallest) to get 72 factors.

28 works also, as mentioned by spell it out 11.  There are a lot smaller ways to get 72 factors for n.

Look your numbers do give 72 factors,but they are not the smallest two numbers.  Try 28*360 thats 2^5*3^2*5*7which has 6*3*2*2=72 factors.

[quote="hammertime"]Oh, right I see now. That makes things tougher.
So it's
[hide] 28 + 44 = 72 [/hide][/quote]

How did you get that?  Im sure theres a smaller answer (brute forced a smaller answer)

Does anyone want to post a non guessing solution to this?


vector
calculus
integration
absolute value
calculus computations

hello, your series does converge.
Sonnhard.

Are you sure it doesn't converge? If you take the absolute value of each term, then the series would approach infinity, so it clearly is not absolutely convergent. If you were to graph the points on the complex plane, you would see points spiralling into the origin. If you were to add the vectors of those points, it seems that it should converge. I believe that it is conditionally convergent, but I can't prove it.

See what you can do with [url=http://en.wikipedia.org/wiki/Abel%27s_test]Abel's test[/url].

Isn't the series equal to $ \sum_{k \equal{} 1}^\infty ( \minus{} 1)^k\sum_{n \equal{} k^2}^{(k \plus{} 1)^2 \minus{} 1}{1\over n} \equal{} \sum_{k \equal{} 1}^\infty ( \minus{} 1)^k\left({1\over k^2} \plus{} \ldots \plus{} {1\over (k \plus{} 1)^2 \minus{} 1}\right)$?

$ b_{k \plus{} 1}\le b_k$ where $ b_k \equal{} {1\over k^2} \plus{} \ldots \plus{} {1\over (k \plus{} 1)^2 \minus{} 1}$ (I have proved it by comparison with the integral of $ 1/x^2$) and $ b_k\to0$ (also by comparison with the integral). It follows that the series is Leibnitz.

[u]By the way problem 11384 of Amer.Math.Monthly October 2008 is quite similar. See also problem of Amer.Math.Monthly Vol.85, No.3, March 1978, 207-209 wich is more difficult. [/u]

Abel's test isn't applicable because it only works for values |z|<1 (converges) or |z|>1 (diverges). In this case z=-1, so the test fails.

I'm talking about the first result in the Wikipedia article called Abel's test, not the second.

are u saying that the partial sums of the series (-1 )^[n^1/2] are bounded?

I was hoping that that would be the case, but now I can see that this can't work.  Anyway, Arrigo-Sacchi is right.


been working on this for an hour. seems like the answer is RIGHT THERE, but i can't get part (c).
A girl with mass $ m_{1}$ and a sled with mass $ m_{2}$ are on the frictionless ice of a frozen lake, a distance d apart but connected by a rope of negligible mass. The girl exerts a horizontal force with magnitude $ F$ on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? (c) How far from the girl’s initial position do they meet? State your answers in terms of the given variables.

Thanks!

Solution

I'm not so sure about my solutions to parts (a) and (b) but I will give them a go anyway. Part (c), for is much more fun, as I find kinematics much easier than dynamics.

(a) Let the origin be set at the initial position of the girl, with the displacement of the sled being positive. The rope is slack before the girl exerts a force on it. Then she pulls it taught such that the force is $ F$. The rope will hence, in turn, exert a force of $ F$ on the sled. Thus by Newton's second law of motion:

$ a_{2}\equal{}\frac{F}{m_{2}}$ where $ a_{2}$ is the acceleration of the sled.

(b) By Newton's third law of motion, the sled will exert an equal, but opposite force on the girl (I hope!). Therefore:

$ a_{1}\equal{}\minus{}\frac{F}{m_{1}}$ where $ a_{1}$ is the acceleration of the girl. The negative sign can be removed if we are interested in the magnitude.

(c) I am going to use $ a_{1}$ and $ a_{2}$ in this question, in case the accelerations in the previous parts are wrong.

For this part, we make use of the equation $ s \equal{} ut\plus{}\frac{1}{2}a t^{2}$ which gives the displacement $ s$ of a mass from it's starting position, where it has initial velocity $ u$ when it undergoes uniform acceleration $ a$ for a time $ t$.

For the girl, the starting position is at the origin and the initial velocity is $ 0$.

Hence: $ s_{1}\equal{}\frac{1}{2}a_{1}t^{2}$

$ \Rightarrow t^{2}\equal{}\frac{2s_{1}}{a_{1}}$

However, the inital displacement of the sled is $ d$, and so to get the displacement of the sled from the origin, we must add this displecement, to the displacement of the sled from $ d$.

Hence $ s_{2}\equal{}\frac{1}{2}a_{2}t^{2}\plus{}d$

When the sled and the girl meet, the vaules of $ s_{1}$ and $ s_{2}$ are the same (as are the values for $ t$). Lets say the displacement of the point at which they meet is $ x$.

$ \Rightarrow t^{2}\equal{}\frac{2x}{a_{1}}$

Substituting this into the second equation:

$ x \equal{}\frac{1}{2}a_{2}\frac{2x}{a_{1}}\plus{}d$

$ \Rightarrow a_{1}x \equal{} a_{2}x\plus{}a_{1}d$

$ \Rightarrow x(a_{1}\minus{}a_{2}) \equal{} d$

$ \Rightarrow x \equal{}\frac{a_{1}d}{a_{1}\minus{}a_{2}}$

Now if parts (a) and (b) are correct then:

$ x \equal{}\frac{\frac{F}{m_{1}}d}{\frac{F}{m_{1}}\plus{}\frac{F}{m_{2}}}$

(A plus appears since the direction of $ a_{2}$ is negative)

$ \Rightarrow x \equal{}\frac{\frac{d}{m_{1}}}{\frac{1}{m_{1}}\plus{}\frac{1}{m_{2}}}$

$ \Rightarrow x \equal{}\frac{m_{2}d}{m_{1}\plus{}m_{2}}$

If the first parts are correct, then it is quite interesting to note that the point at which they meet is the same regardless of the force applied to the rope.

Now lets go further and find the time at which they meet:

$ t^{2}\equal{}\frac{2x}{a_{1}}\equal{}\frac{\frac{2m_{2}d}{m_{1}\plus{}m_{2}}}{\frac{F}{m_{1}}}\equal{}\frac{2m_{1}m_{2}d}{F(m_{1}\plus{}m_{2})}$

$ \Rightarrow t \equal{}\sqrt{\frac{2m_{1}m_{2}d}{F(m_{1}\plus{}m_{2})}}$ so the time taken for them to meet (assuming the first parts correct) is inversely proporional to the square root of the force applied. This would be plausible, as you would expect them to come together more quickly if a greater force is exerted.

Thanks I got the same answer yesterday night =)


two cyndrical cans have the same volume. The height of the first can is triple the other. If the radius of the narrower can is 12 units how many units are in the length of the radius of the wider can? Express your answer in siplest radical form.

I know the answer but not the solution

Thanks so much ( i hope this is not annoying anyone cause it is my third question in two days)

Answer
[hide]12 square roots of 3[/hide]

[hide]h is the height of the shorter can, 3h is the height of the taller.

Volume of narrower/taller can =$12^2*3h\pi$
Volume of wider/shorter can=$r^2*h\pi$

Their volumes are the same, so $12^2*3h\pi=r^2*h\pi$
cancelling, $12^2*3=r^2$
square rooting, $r=12\sqrt{3}$[/hide]


function
floor function
algebra
algebra unsolved

Let $ \left\lfloor x\right\rfloor$ denote the floor function (the largest integer less than or equal to $ x$).  Find the average value of the quantity $ \left\lfloor 2x^{3}\minus{}2\left\lfloor x^{3}\right\rfloor\right\rfloor$ on the interval $ (\minus{}3/2,3/2)$.

[b]Lemma 1[/b]: $ \lfloor a + n\rfloor = \lfloor a\rfloor + n\ \forall n\in \mathbb Z$
[b]Lemma 2[/b]: $ \lfloor a\rfloor + \lfloor - a\rfloor = - 1\ \forall a\not \in \mathbb Z$

Suppose $ x^3\not \in \mathbb Z$

$ \lfloor 2x^{3} - 2\lfloor x^{3}\rfloor\rfloor\ \ + \ \ \lfloor 2( - x)^{3} - 2\lfloor ( - x)^{3}\rfloor\rfloor =$
$ = (\lfloor 2x^3\rfloor - 2\lfloor x^{3}\rfloor )\ \ + \ \ (\lfloor - 2x^3\rfloor - 2\lfloor - x^{3}\rfloor ) =$
$ = (\lfloor 2x^3\rfloor + \lfloor - 2x^{3}\rfloor)\ \ - \ \ 2(\lfloor x^{3}\rfloor + \lfloor - x^{3}\rfloor) =$
$ = (-1) - 2( - 1) = 1$

[b]Lemma 3[/b]: The average value of $ f(x)$ in the interval $ ( - l,l)$ is the same as:
Average value of $ \frac {f(x) + f( - x)}{2}$ in the interval $ [0,l)$

Let $ f(x) = \lfloor 2x^{3} - 2\lfloor x^{3}\rfloor\rfloor$

But $ \frac {f(x) + f( - x)}{2} = \frac {1}{2}\ \forall \ x^3\not \in \mathbb Z$

However the values of the above expression for $ x^3\in \mathbb Z$ contribute infinitely small amount in the average. Hence negligible.

Answer: $ \boxed{\frac {1}{2}}$

 :lol:

Let $ y\equal{}x^3$, you interest $ f(y)\equal{}[2y]\minus{}2[y]$. Obviosly $ f(y\plus{}m)\equal{}f(y)$ - periodic function with period 1.
Therefore sufficintly consider $ f(y)\in Z$ in $ [0,1)$. $ f(y)\equal{}0,0\le y<\frac 12, f(y)\equal{}1,\frac 12\le y<1$.

[quote="Rust"]Let $ y \equal{} x^3$, you interest $ f(y) \equal{} [2y] \minus{} 2[y]$. Obviosly $ f(y \plus{} m) \equal{} f(y)$ - periodic function with period 1.
Therefore sufficintly consider $ f(y)\in Z$ in $ [0,1)$. $ f(y) \equal{} 0,0\le y < \frac 12, f(y) \equal{} 1,\frac 12\le y < 1$.[/quote]
I think it'll be more calculations than it appears as, we need to consider $ x$ in the intervals $ [0,0.5),[0.5,1),[1,\sqrt [3]{1.5}),[\sqrt [3]{1.5},\sqrt [3]{2}),[\sqrt [3]{2},\sqrt [3]{2.5}),[\sqrt [3]{2.5},\sqrt [3]{3}),[\sqrt [3]{3},1.5)$ and weigh them by the length of the intervals and all that divided by total length of the interval...
and take in the negatives...

If you consider symmetric intervfl $ (\minus{}a,a)$, then $ f(x)\plus{}f(\minus{}x)\equal{}1$, therefore average value is $ \frac 12$ - without calculation.

[quote="Rust"]If you consider symmetric intervfl $ ( \minus{} a,a)$, then $ f(x) \plus{} f( \minus{} x) \equal{} 1$, therefore average value is $ \frac 12$ - without calculation.[/quote]
Have you read my 1st post totally? :wink:


1. The real valuve of x which satisfies
x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3=3^3 is k. Find K.

2. Find the four walues of x for which (x-3)^4+(x-5)^4=-8.

Thank you sooo much for helping me. :oops:  :mad:

I can easily see that -2 would work in the first equation, since if you subtract the 3^3 from both side and plug in , all terms will cancel with a corresponding term that it the negative of it, except for the middle one, which is 0. I guess you could expand the whole thing and then factor our x+2 to get the others.

For the next equation, I saw that 4+i is one solution, because by symmetry, you would want the real parts of both of the numbers raised to the four to be opposites. Thus, 4-i is another solution, and you could factor to find the other two.

First Question: dgreenb801's right, (x+2) is a factor, but cubing all those expressions is mad long. So use a substitution, y=(x+2). The equation becomes $ (y\minus{}2)^3\plus{}(y\minus{}1)^3\plus{}y^3\plus{}(y\plus{}1)^3\plus{}(y\plus{}2)^3\plus{}(y\plus{}3)^3\equal{}27$
Now use the fact that $ (y\minus{}a)^3\plus{}(y\plus{}a)^3\equal{}2y^3\plus{}6a^2y$. We get: $ (2y^3\plus{}24y)\plus{}(2y^3\plus{}6y)\plus{}y^3\plus{}(y^3\plus{}9y^2\plus{}27y\plus{}27)\equal{}27\rightarrow y(6y^2\plus{}9y\plus{}57)\equal{}0$. So y=0 is the only real solution, and thus x=-2 is the only real solution, but the other ones are easily attained by solving for y.

Second Question: to avoid factoring a fourth degree polynomial, let y=x-4. Then the equation becomes $ 2y^4\plus{}12y^2\plus{}2\equal{}\minus{}8$, which we can solve to get $ y^2\equal{}\minus{}1$ or $ \minus{}5$. Hence, $ y\equal{}\pm i$ or $ \pm\sqrt 5i$, and so $ x\equal{}4\pm i$ or $ 4\pm\sqrt 5i$.


integration trigonometry calculus blogs logarithms function algebra

$ \int log_3x dx$

$ \int_{0}^{pi}sin^2xdx$

$ \int(x^2)/(x^2\minus{}4x\plus{}5)dx$

2) $ \int_{0}^{\pi}\sin^{2}x\ dx$

Recall that $ \cos(2x) \equal{} 1 \minus{} 2\sin^{2}x\Rightarrow\sin^{2}x \equal{} \frac {1}{2}\left(1 \minus{} \cos(2x)\right)$.  Then our integral is now

$ \frac {1}{2}\int_{0}^{\pi}1 \minus{} \cos(2x)\ dx \equal{} \frac {1}{2}\left(x \minus{} \frac {1}{2}\sin(2x)\right)\bigg|_{0}^{\pi} \equal{} \frac {1}{2}\left(\pi \minus{} 0 \minus{} 0 \plus{} 0\right) \equal{} \frac{\pi}{2}$

While it's true that these don't belong here (for future reference, try http://www.artofproblemsolving.com/Forum/index.php?f=296 ) I'll provide hints for the other two that JRav hasn't done.

1) I'll assume that's a $ \log_3 x$. What do you know about using $ \log$ in an integral? Isn't it usually much easier to have $ \ln$ to work with? Perhaps there's some way to convert bases.

3) As far as I can see from looking at this one, it's probably going to entail a partial-fractions approach combined with a trig substitution.

1.
[hide]$ \frac {\ln x}{\ln 3}$
Take out the $ \ln 3$
Integrating $ \ln x$ is easy with integrating by parts
$ \int {\ln x dx}$
= $ x \ln x \minus{} \int{ x * \frac {1}{x} dx}$
= $ x \ln x \minus{} \int {1 dx}$
= $ x \ln x \minus{} x \plus{} C$
Then divide by $ \ln 3$
= $ \frac {x \ln x \minus{} x}{\ln 3} \plus{} C$
[/hide]

@JRav
#2 should be $ \frac {\pi}{2}$. You forgot to divide. :)

3. Looks like some form of arctangent so complete the square

Please use the Calculus Computations and Tutorials Forum in the future.

moved.

How would one figure out such partial fraction decompositions?  I know how to do partial fractions, just how does one do more strange ones like that, or $ \frac{1}{x^{6}\plus{}1}$?

It's a routine algorithm.
Step 0: If it's an improper rational function, find the polynomial part by long division. Keep going with the fractional part.
Step 1: Factor the denominator completely. This is the hard part in general.
Step 2: Write down the generic form; if one of the factors of the denominator is $ (x\minus{}1)^2$, we get terms of $ \frac{a}{x\minus{}1}$ and $ \frac{b}{(x\minus{}1)^2}$. Make sure that the number of unknowns is equal to the degree of the denominator.
Step 3: Solve the resulting system of linear equations matching the generic form with the original fraction. The linear algebra can often be simplified by setting $ x$ equal to a root and looking at the resulting equations.

You don't have to consider partial fraction decomposition in order to compute $ \int\frac{1}{1\plus{}x^6}\,dx$.


Let $ a,b,c \ge 0$ such that $ a\plus{}b\plus{}c \ge abc$
Then prove that $ a^2\plus{}b^2\plus{}c^2 \ge abc$

$ a^2\plus{}b^2\plus{}c^2 \geq abc \sqrt{\frac{a\plus{}b\plus{}c}{abc}}\geq abc$

Nice, there's another solution with absurd.

Actually the following stronger inequality holds:

$ a^2\plus{}b^2\plus{}c^2 \ge \sqrt{3} abc$


Determine all linear transformations $ T: V \rightarrow V$ such that $ T \equal{} T^2$.
Hints:
[hide]Note that $ x \equal{} T(x) \plus{} (x \minus{} T(x))$ for every $ x \in V$, and show that $ V \equal{} \{y: T(y) \equal{} y\} \oplus N(T)$[/hide]

I understand that any transformation T that we are looking for will posess the property described in the hint, but I was wondering if anyone could give me a start to this problem as I'm at a loss as to how to approach it.

Thanks!

Recall that for an arbitrary vector $ a$ we have $ x \equal{} \text{Proj}_{a} x \plus{} (x \minus{} \text{Proj}_{a}(x))$ where the first term is parallel to $ a$ and the second term is perpendicular.

Another way of thinking about the problem:  $ T$ only has eigenvalues $ 0$ and $ 1$, so it either takes its eigenvectors to zero or preserves them.  What are the possible sizes of each respective eigenspace?  What happens if you take the eigenvectors as a basis?


trigonometry
algebra
polynomial
calculus
Galois Theory
algebra unsolved

Solve the equation $ \sin x \plus{} \sin 2x \plus{} \sin 3x \equal{} 1$

You can see at
http://www.mathlinks.ro/viewtopic.php?t=224264

This problem have posted.  You will get solution from way find sum: $ S \equal{} \sin{x} \plus{} \sin{2x} \plus{} \sin{3x} \plus{} ... \plus{} \sin{nx}, n\in N*$.

I'm very sorry but I dont understand all these posts. It seems that everybody knows how to solve the equation but carefully avoid giving the answer and I'm unable to find the solution from all these messages (even after reading [url]http://www.mathlinks.ro/viewtopic.php?t=224264[/url], as suggested).

I's very easy to get a $ 6^{th}$ power polynomial (using either $ \tan\frac x2$, either $ \tan x$, for example), but none of the obtained polynomial seem to me easy to solve.

Would it be possible to have somebody giving some explicit answer, please ?  :oops:

Thanks in advance.

With $ t\equal{}tan(x/2)$ we obtain the equation:
$ p(t)\equal{}t^{6}\minus{}4t^{5}\plus{}3t^{4}\plus{}16t^{3}\plus{}3t^{2}\minus{}12t\plus{}1\equal{}0$
The Galois group of $ p$ is $ S_6$ that is not solvable ; then we cannot solve with radicals this equation.

i think it can be solved in this way.pardon me if i am wrong.$ \sin x \plus{} \sin 2x \plus{} \sin 3x \equal{} sin x \plus{} \ 2sin x cos x \plus{} \ 3 sin x \minus{} 4(sin x)^{3}$.now rearrange the terms and try

Let me give a synthesis :

The equation to solve is $ \sin(x)\plus{}\sin(2x)\plus{}\sin(3x)\equal{}1$
Obviously, any quick worksheet usage gives the two solutions (0.171668975... and 1.243579362... modulus $ 2\pi$) so the problem is not to show the two values but to find a formula which gives them.

Math pro found the solution by finding a degree 3 polynomial ([url]http://www.mathlinks.ro/viewtopic.php?p=1245946#1245946[/url]) but avoided to give us this polynomial. :(

Dr Sonnhard Graubner found a solution by finding a degree 6 polynomial ([url]http://www.mathlinks.ro/viewtopic.php?p=1246039#1246039[/url]) but omitted to say us how to solve it (except with calculus method, since with such method we can immediatly find the solution from the original equation).

Thanhnam2902 found a solution by replacing the original equation with $ \cos(\frac x2)\minus{}\cos(\frac{7x}{2})\equal{}1$ whose solution, according to him is much more simple ([url]http://www.mathlinks.ro/viewtopic.php?p=1246873#1246873[/url]). But, unfortunately, he did not give us this obvious solution. :(

Thanhnam2902 and Dr Sonnhard Graubner again, found a general solution of $ \sin{x} \plus{} \sin{2x} \plus{} \sin{3x} \plus{} ... \plus{} \sin{nx}\equal{}1$ ([url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1517349#1517349[/url]) but, once again, forgot to give us the solution in the "basic" case where $ n\equal{}3$ :(

Would one of you, guys, be kind enough to give us a full solution, and not just something like "the solution can easily be found thru ...".

I'm about to think that none of you got the solution. :oops:

[quote="loup blanc"]pco read my preceding post.
The Galois theory says that you can't obtain explicitly the 2 solutions.
Evariste is well-known to have brought instruments giving negative results.
Be not disappointed too much ; the mathematics are often like that.[/quote]

This does not mean there is no explicit solution using $ \arcsin, \arctan, \exp, W, ...$

And all the posters I named seemed to have the solution, so ... .

"This does not mean there is no explicit solution using........"
You are right pco but it's very far from probable.


If a,b,c are the sides of a triangle and x,y,z arbitrary real numbers prove that

(ax) 2 +(by) 2 +(cz) 2  \geq xy(a 2 +b 2 -c 2)+yz(b 2 +c 2 -a 2 )+zx(c 2 +a 2 -b 2 )

First of all, there is a simple proof usind the second degree equation.
 But I prefer the following:
   Write it as sum a^2(x-y)(x-z)>=0. Consider u=x-y, v=y-z, w=z-x. Among u,v,w two have the same sign. Let them be u and v. The inequality becomes (u+v)(a^2u+c^2v)>=b^2uv. Since u and v have the same sign, we can take u,v>=0. From Cauchy (u+v)(c^2v+a^2u)>=(csqrt(uv)+asqrt(uv))^2=uv(a+c)^2>=b^2uv.

You notice that a^2+b^2-c^2 = 2abcosC.
  Moreover, we also have this useful inequality
     If ABC is a triangle, x,y,z, are real then
  \sumxycosC <=1/2( \sum x^2).
   And we have done . :D

Using vectors, $LHS-RHS = |z(A-B) + x (B-C) + y (C-A)|^2 \geq 0$

********************************************************

Amazingly, this inequality can also be derived from the inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558

by taking u = 0, v = x-z, w=y-x and M=A, then

$(u+v+w)(uMA^2 + vMB^2 + w MC^2) \geq vwa^2 + wub^2 + uvc^2$ translates into:

$(y-z)((x-z)c^2 + (y-x)b^2) \geq (x-z)(y-x)a^2$ which is equivalent to what we want to prove.


abstract algebra
algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Can someone find a counterexample when $R$ is a finitely generated ring, but is not a quotient of a Noetherian ring.

(If $R$ is a $k$-algebra the statement is correct)

Being a quotient of a Noetherian ring is the same as being Noetherian. The statement that every finitely generated algebra over a Noetherian ring is again Noetherian is known as Hilbert's basis theorem. It may be reduced to the fact that the polynomial ring in one indeterminate over a Noetherian ring is again Noetherian. owk

uhm, i'm not following. $R$ was not at all assumed to be finitely generated over $k$ (a field) or any other Noetherian ring. It can be the most arbitrary f.g. ring you wish :)

So what do you mean by "finitely generated" in an absolute sense? owk

the additive group is finitely generated and the ring structure is respected (not sure how to formalize this)

example: $\bf Z$

Then, the ring in question is f.g. as a $\mathbb Z$-algebra (which is equivalent to the most natural sense of being finitely generated: containing a finite subset $E$ such that there is no proper subring containing $E$). owk


Let $ ABC$ is a triangle, let $ (O;R)$ is circum circle of $ ABC$ triangle, let $ (I;r)$ is inscribed circle of $ ABC$ triangle. Let $ AB \equal{} c, CA \equal{} b, BC \equal{} a$. Prove that: $ \frac {a^3}{b.r \plus{} c.R} \plus{} \frac {b^3}{c.r \plus{} a.R} \plus{} \frac {c^3}{a.r \plus{} b.R}\ge\frac {a^2 \plus{} b^2 \plus{} c^2}{r \plus{} R}$

$ \boxed{
\begin{array}{cc}
\frac {a^3}{br + cR} + \frac {b^3}{cr + aR} + \frac {c^3}{ar + bR}\ge\frac {a^2 + b^2 + c^2}{r + R}
\\\\
\frac {a^4}{abr + acR} + \frac {b^4}{bcr + abR} + \frac {c^4}{car + bcR}\ge\frac {a^2 + b^2 + c^2}{r + R}
\\ \\ %Error. "SMALL" is a bad command.
{\mathrm{ By\ Cauchy's \ Lemma: }}
\\\\
\frac {a^4}{abr + acR} + \frac {b^4}{bcr + abR} + \frac {c^4}{car + bcR}\ge \frac{(a^2+b^2+c^2)^2}{abr + acR+bcr + abR+car + bcR}\\\\
\frac{(a^2+b^2+c^2)^2}{(ab+bc+ca)(r+R)}\ge\frac {a^2 + b^2 + c^2}{r + R}
\\\\a^2+b^2+c^2 \ge ab+bc+ca
\end{array}
}$

[quote="Ahiles"]$ \boxed{ \begin{array}{cc} \frac {a^3}{br + cR} + \frac {b^3}{cr + aR} + \frac {c^3}{ar + bR}\ge\frac {a^2 + b^2 + c^2}{r + R} \\
 \\
\frac {a^4}{abr + acR} + \frac {b^4}{bcr + abR} + \frac {c^4}{car + bcR}\ge\frac {a^2 + b^2 + c^2}{r + R} \\
 \\
%Error. "SMALL" is a bad command.
{\mathrm{ By\ Cauchy's \ Lemma: }} \\
 \\
\frac {a^4}{abr + acR} + \frac {b^4}{bcr + abR} + \frac {c^4}{car + bcR}\ge \frac {(a^2 + b^2 + c^2)^2}{abr + acR + bcr + abR + car + bcR} \\
 \\
\frac {(a^2 + b^2 + c^2)^2}{(ab + bc + ca)(r + R)}\ge\frac {a^2 + b^2 + c^2}{r + R} \\
 \\
a^2 + b^2 + c^2 \ge ab + bc + ca \end{array} }$[/quote]

Thank you very much.


Let $a$ and $b$ be positive real numbers, prove that

$\dfrac{a+\sqrt{ab}+b}{3} \leq \sqrt{(\dfrac{a^{2/3}+b^{2/3}}{2})^{3}}$

Hi detective!
   I use AM-GM according to your hint already! But I think my solution is a little bit ugly and I hope that s.o else can give a nice sols for you.
  Anyway , here is mine.
Suppose  $a^{\frac23}+b^{\frac23}=2$
And we put $a=x^6$ $b=y^6$ So we will have a new puzzle :
  $x^4+y^4=2$  Prove that: $x^6+y^6+x^3y^3 \leq3$
We have :
$x^6+y^6+x^3y^3=x^2(2-y^4)+x^3y^3+y^2(2-x^4)=(x^2+y^2)(2-x^2y^2)+x^3y^3= \sqrt{2(x^2y^2+1)}(2-x^2y^2)+x^3y^3 \leq \frac12 (2+x^2y^2+1)(2-x^2y^2)+x^3y^3 = \frac12 (6-x^2y^2-x^4y^4)+x^3y^3= 3+x^3y^3-\frac12 (x^2y^2+x^4y^4) \leq 3+x^3y^3-\frac12(2\sqrt{x^6y^6})=3$
  Right?

wrong. see carefully what you have done.

I actually don't see, how to use AM-GM sophisticatedly. Here's my solution:

[hide]
Let $x=\sqrt[6]{a}$ and $y=\sqrt[6]{b}$
Then the inequality is equivalent to
\[ x^{12}+27x^8y^4+27x^4y^8+y^{12}\geq 24x^6y^6 + 16x^9y^3 + 16x^3y^9  \]
Dividing this inequality by $y^{12}$, we obtain
\[ \left(\frac xy\right)^{12} - 16\left(\frac xy\right)^9 + 27\left(\frac xy\right)^8 - 24\left(\frac xy\right)^6 + 27\left(\frac xy\right)^4 - 16\left(\frac xy\right)^3 +1 \geq 0  \]
Now, let $t = \frac{x}{y}$, we obtain
\[ t^{12} -16t^9 + 27t^8 - 24t^6 +27t^4 - 16t^3 + 1 \geq 0  \]
This can be factorized to:
\[ \left( t-1 \right)^4\cdot\left(t^8 + 4t^7 + 10t^6 + 4t^5 - 2t^4+ 4t^3 + 10t^2 + 4t + 1 \right) \geq 0  \]

The latter bracket is obviously greater than $0$, hence this inequality is proven. We get equality iff $t=1$ which is equivalent to $a=b$.

[/hide]

BTW, the ineq posted in the contest was
$(\frac{\sqrt{a}+\sqrt{b}}{2})^2\leq\frac{a+\sqrt[3] {a^2b}+\sqrt[3] {ab^2}+b}{4}\leq\frac{a+\sqrt{ab}+b}{3} \leq \sqrt{(\frac{a^{2/3}+b^{2/3}}{2})^{3}}$

 :)

[quote="Yimin Ge"][quote="silouan"]
[...]
which is equivalent to $(\frac{x^3+y^3}{2})^2\leq (\frac{x^2+y^2}{2})^3$

where $a=x^3$  and $b=y^3$  which is true by the power means
[/quote]

I'm afraid it is not, the inequality is exactly the other way round ;)[/quote]

 I agree with you. Indeed:
$(\frac{x^3+y^3}{2})^2\geq (\frac{x^2+y^2}{2})^3$
(Prove it isn't hard at all. Just let LHS-RHS and factor $(x-y)^2$ out


[size=150][b]Congratulations on receiving Nobel prize in physics of 3 Japanese![/b][/size]

[url]http://nobelprize.org/nobel_prizes/physics/laureates/2008/index.html[/url]

[url]http://www.washingtonpost.com/wp-dyn/content/article/2008/10/07/AR2008100700196.html[/url]


Can you give a link? I don't see anything wrong with it.

There was nothing wrong with the factoring itself.  The thing was, the question was asking for the sum of the prime factors of $ 2^{15} \minus{} 1$.  Factoring $ 2^{15} \minus{} 1$ into $ (2^{15/2} \plus{} 1)(2^{15/2} \minus{} 1)$ doesn't really help with solving the problem.

For anyone else who would like to help, the link is this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=320917

It's right, but I think what your classmates were saying was that it wouldn't do anything to help solve the question. Why dont you edit your post and put the question down? We can help you solve it.


please help in introduction to tensors or produce a link   :)  :)  :)


Show that for every $n \geq 5$ we have that the intenger part of $\frac{3^{n}-1}{2^{n}}$ is equal the intenger part of $\frac{3^{n}-2}{2^{n}}$.

It is equavalent to $2^{n}\not |(3^{n}-1)$ and obviosly for n>2.

yes, I knew that equivalence, but I am trying to do it by another way, because I want that to solve a problem.


Solve in integers the following equation:
x^n +y^n =1, where n is a natural number.

Well I guess I'll take the credit for it, a 'Guest' solved it, which is anonymous, so I'll take credit for it unless the solver identifies himself.
By the way, registration is free.


İff $tan11x=24$ and $tan19x=34$ then find $tan5x$

Use the formula to determine $tan(8x) = tan(19x-11x)$ then to determine $tan(3x) = tan(11x-8x)$ then to deduce $tan(5x) = tan(8x-3x)$.

Pierre.

It is not hard to chek, that exist not x, satisfied your conditions.

How can you check that?by a computer or by hand?
could you post your way?

You can find in this way $tg 2x$ and $tg(3x-x)$. they are rational. But $tg2x\not =\frac{2tgx}{1-tg^{2}x}$.
I think exist not x, suth that $tg(ax)=b,tg(cx)=d$, were a,b,c,d are integer $\not =0$.

Thanks a lot.But I think your proposition is not true for $b=d$
but if $b\not=d$,I have no idea.

It is not true for $b=d=\pm 1$.
Exactly my statement: if a,b,c,d integer, $|b|+|d|>1$, $|a|\not =|c|$, then exist not x, suth that $tg(ax)=b, \ tg(cx)=d$.


MATHCOUNTS
analytic geometry
AMC
AIME
AMC 10

I am considering creating an AMC mock contest. It would be very similar in difficulty to the AMC 12, and would follow the same rules (i.e. 75 minutes, 25 questions, 6 points for a correct answer and 2.5 for a skipped question, any SAT acceptable calculator may be used, etc.) If there are enough interested people I will go ahead and write this contest. Please let me know if you are interested!

[b]UPDATE[/b]--I have written the contest, making some problems on my own and taking some from a variety of sources. It is a little easier than recent AMC 12s, so I'll say that it's approximately "AMC 11" difficulty. All I need to do now is to check the problems over for mistakes. I would like this contest to be taken under real AMC conditions, so hopefully we can agree on a date/time, and I will post the problems exactly then. After 75 minutes are up, PMs should be sent to me with your answers. So, let me know what times will work for you and we will see if we can find one that works for everyone so that the contest can be taken by everyone at the same time.

[b]UPDATE(2)[/b]--Will a weekday next week at some time between 2:30 and 5:30 Eastern work for everyone? If so, let me know which weekdays would be preferable.

[b]UPDATE(3)[/b]--The contest will take place on Friday, July 23 at 3:00 PM Eastern time. Full details on the contest are available [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=100659#100659]here[/url].

I would, but I won't be able to do it til the end of August

Crud, I'd like to be in it, but I probably can't.

I am.

But can you explain the problems after you're done with them?  That would really help me out.

Yeah, we could definitely discuss the problems after the contest is over.

I'll join and I hope it can help me on creating second SC.  Just to clear things up, are you going to clarify the scores on the public or just post the winner or something like that? And is this for beginner or intermediater (I know the level, though).

I am planning on creating a scoreboard of high scores. I will also PM each participant with the correct answers after the contest.

As for difficulty--some of the questions are AMC 10 level and others are AMC 12 level. That is why I am calling this an "AMC 11".

No solutions necessary, since this contest is a simulation of a real AMC contest. Answers must be PMed immediately after the contest is over, however.

Can everyone post dates and times (Eastern time, please) that they would be available?

[i]Edit[/i]: To clarify--I am hoping that everyone can agree on a time to take the contest, so that everyone will take the test at the same time.

I'm pretty sure I can do it from 8:00 AM EST - 7:00 PM EST, but NOT ON SAT/SUN because I am [i]very[/i] busy on those days.  I'll probably forget, but I'll do it just in case.  Is a TI-89 SAT acceptable?

[quote="JSRosen3"]TI-89s are fine, as is [url=http://www.collegeboard.com/counselors/hs/sat/registration/calc.html]any calculator that is acceptable for the SAT[/url].[/quote]

Oh.  OK.  It seems like MathCounts acceptable calculator = SAT acceptable calculator.

I'll do it, sure. Fresh practice  questions from anywhere are welcomed to me.

I will also join, though don't expect me to score high at all.

I am!! Are you going to post bad scores?
Im still a big beginner at this stuff, so it'd be kind of embarrassing to see my bad score up. Anyway, I need more practice, so this would be perfect  :lol:

Don't worry...only scores of 100 or higher will be posted (or the top 10, if there are not 10 people who scored at least 100). Your individual score will be PMed to you. If you want to gauge the difficulty of the problems, go to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14236]this[/url] thread and scroll down a bit. There are 5 sample problems posted there.

Umm...I'll join.  I'm bored so I guess I need the practice.

I didn't want to start another Mock AMC thread and this one looked like the most appropriate one.

I think that we should try to have like a Mock AMC every month.  It is a lot of fun and really good practice.  It doesn't have to be set in stone that on exactly the nth day of the month we will have it but just around that time.  I know that it takes a lot of hard to put together one of these.  However, if we alternate who puts it together that should lighten the load a bit.  We could also have 2 or more ppl collaborate to put it together.  What do you guys think?  Do you think that we would be able to do it?

Good Idea, I think I will make the next AMC.

We should space these out a bit though, so that we can encourage wider participation...many AoPS-ers are away for the summer. However, I like the idea of more of these mock contests. The Mock AIME this Saturday is going to be awesome!  ;)

A Mock AMC every month thats a great idea.  Only one thing could we try to have one that is acctually about the difficulty of the AMC 10?  The last 2 people have been saying were AMC 12 like.

Ok, I finished writing the contest. How many people are interested in taking a mock AMC again?

I would.  Just make sure to give some time between the Mock AIME and your Mock AMC.

Guess I'll have to miss the next couple. I'm going to be 12 hours away until September 4th.


$ a,b,c,d$ are positive natural numbers such that $ ab\equal{}cd$

then prove that :$ a^2\plus{}b^2\plus{}c^2\plus{}d^2$ is not prime.

[hide="Hidden by mod"]$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} (a \minus{} b)^2 \plus{} (c \plus{} d)^2 \equal{} (a \plus{} b)^2 \plus{} (c \minus{} d)^2$

Suppose $ a < c\le d < b$; then $ d \minus{} c$ is smaller than any of the other squared terms, so that we have two different representations of $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$ as the sum of two squares. However, it is well-known that this implies that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$ is composite.[/hide]

[hide="Hidden by mod"]$ \frac {a}{c} \equal{} \frac {b}{d} \equal{} \frac {x}{y}$ where $ gcd(x,y) \equal{} 1$
Let $ u,v \in N$ such that $ a \equal{} ux,c \equal{} uy,b \equal{} vx,d \equal{} vy$
Therefore $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} (u^2 \plus{} v^2)(x^2 \plus{} y^2)$ is not prime  :lol:[/hide]

[quote="apollo"]$ \frac {a}{c} \equal{} \frac {b}{d} \equal{} \frac {x}{y}$ where $ gcd(x,y) \equal{} 1$
Let $ u,v \in N$ such that $ a \equal{} ux,c \equal{} uy,b \equal{} vx,d \equal{} vy$
Therefore $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} (u^2 \plus{} v^2)(x^2 \plus{} y^2)$ is not prime  :lol:[/quote]

nice

Let $ w\equal{}a^2$ etc. so $ wx\equal{}yz$. Let $ p\equal{}w\plus{}x\plus{}y\plus{}z\equal{}a^2\plus{}b^2\plus{}c^2\plus{}d^2$ and we must show $ p$ not prime.

Suppose it were. Then $ yp\equal{}wy\plus{}xy\plus{}y^2\plus{}wx\equal{}(y\plus{}w)(y\plus{}x)$, so $ p$ would have to divide either $ y\plus{}w$ or $ y\plus{}x$. However, $ 0<y\plus{}w<p$ and $ 0<y\plus{}x<p$, which gives a contradiction.

[quote="apollo"]$ \frac {a}{c} \equal{} \frac {b}{d}$[/quote]
I think you mean $ \frac {a}{c} \equal{} \frac {d}{b}$ since we have $ ab\equal{}cd$.
Anyway, I don't understand why you need this passage... it seems that to suppose (is this the right verb? :blush: ) $ gcd(x,y)\equal{}1$ it's however enough to proceed. Correct?

Goodbye
GioMott

[quote="t0rajir0u"][quote="FOURRIER"][quote="bertram"]However, it is well-known that this implies that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$ is composite.[/quote]

Can you prove this lema or give a hint?[/quote]

Alternately:  if a number has two representations as a sum of squares, it cannot be the norm of a [url=http://mathworld.wolfram.com/GaussianPrime.html]Gaussian prime[/url].[/quote]

Does this proves the lema?
I don't know
Is there an elementary proof of it?

This lemma is from Engel's book, too. ;)

[hide="Proof"]Suppose that positive integers $ a,b,c,d$ satisfy $ a^2\plus{}b^2\equal{}c^2\plus{}d^2\equal{}n$. Then we have $ n^2 \equal{} (ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2 \equal{} (ac\minus{}bd)^2\plus{}(ad\plus{}bc)^2$ and $ n|(ac\plus{}bd)(ad\plus{}bc) \equal{} (a^2\plus{}b^2)(c^2\plus{}d^2)$. If $ n$ is prime, then WLOG $ n|ac\plus{}bd$; then $ n\equal{}ac\plus{}bd\implies ad\minus{}bc \equal{} 0 \implies \frac{a}{c} \equal{} \frac{b}{d} \implies (a,b) \equal{} (c,d)$.

Thus, if there are two different representations of $ n$ as the sum of two squares, then $ n$ has to be composite.[/hide]

I think this problem is also from More Mathematical Morsels, with the proof a la Apollo.

Hi , the same proof as Apollo' in a more "natural" way :
suppose $ab = cd $
let's show that $ a^2 + b^2 + c^2+ d^2 $ is not prime .
let $ \delta = a \wedge c $ and $a'$ , $c' $ integers satisfying $a = \delta a'$ and $ c = \delta c' $ , we know then that $a' \wedge c' = 1 $
Then we see that $ ab = cd \Longrightarrow a'b = c'd $ the last equality implies (due to GAUSS theorem) that $ a' | d$ then there exists an integer $ \lambda$ such that $ d = \lambda a' $ that implies (by reporting it in the equality $a'b = c'd$ ) $ b= \lambda c' $ .
Under these facts , we can write :
$a^2+b^2+c^2+d^2 = \delta ^2 a'^2 + \lambda^2 c'^2 + \delta^2 c'^2 + \lambda^2 a'^2 $

which is obviously equal to : $ (\delta^2 + \lambda^2) (a'^2 + c'^2) $
thus $a^2 + b^2 + c^2 + d^2$ is "rather" composite than prime :-)


Inequality
IMO
IMO 1984
three variable inequality
calculus

$x,y,z\geq 0$ and $x+y+z=1$
Prove that
$xy+yz+zx-2xyz\leq\frac{7}{27}$

Very easy. This inequality is equivalent with $xy+yz+zx\leq\frac{7+54xyz}{27}$,but this last one follows immediately from Schur written as $xy+yz+zx\leq\frac{9xyz+1}{4}$ and from AM-GM $xyz\leq\frac{(x+y+z)^3}{27}$.So the conclusion follows. ;)

Not [i]that[/i] new...

http://www.mathlinks.ro/Forum/viewtopic.php?t=35433
http://www.mathlinks.ro/Forum/viewtopic.php?t=33007
http://www.mathlinks.ro/Forum/viewtopic.php?t=40857 post #59 problem 15
http://www.kalva.demon.co.uk/imo/isoln/isoln841.html

  darij

What is the official solution to this IMO 84 inequality ?


What would be a good linear algebra book for self study?

What level are you looking for?  Mostly computational or mostly proofs?  I used Linear Algebra by Friedberg, Insel, and Spence and I thought that was a good book.  You may also want to try Linear Algebra Done Right.

Both of those are proof-based books, as my experience with linear algebra has been mostly proof-based.

i would recommend
[i]T. S. Blyth, E. F. Robertson: Basic linear algebra, Springer, 2002
T. S. Blyth, E. F. Robertson: Further linear algebra, Springer, 2002[/i]

i used those for my linear algebra studies (and sometimes i "consulted" [i]Gilbert Strang's  Linear algebra and its applications[/i])

I would recommend [i]"Linear Algebra and Its Applications" Third Edition, Update.[/i] by David C. Lay.

Apropos Linear Algebra Done Right: What is the usefulness of developing things without the use of determinants? Is it because determinants aren't 'nice' to compute for high dimension?

[url=http://www.axler.net/DwD.pdf]the author can probably tell you better than i can.[/url]


I was just wondering, who is planning to go to the NC School of Math and Science?  Or at least planning to compete for a spot?  It seems to be the right thing for a smart person to do...  But I know some people who doesn't want to go there.  Sooo.  I was wondering what the Math Geniuses on the AoPs posts thought.....

(I know this was posted forever ago, but anyway)

I'm applying/applied to NCSSM. Particularly at AoPS, I'm nothing amazing, but I'm out of courses at my school... so hopefully I'll get in. And since I don't live in Wake County, I'm not in the valley of death. There seems to be something against NCSSM here, but I don't see what's wrong with it... whatever it is it can't be worth not having math for 2 years. lol


logarithms
number theory proposed
number theory

[quote="surshu"]Find all pairs of unequal integers m, n such that mˇn = nˇm.[/quote]
 is it $ m^{n}=n^{m}$? Right?

If so , it is equivalent to $ \frac{\ln n}{n}=\frac{\ln m}{m}$.
[quote="surshu"]yes, i dont know how to type standard sign for that,  ô without o[/quote]

See here http://www.mathlinks.ro/LaTeX/AoPS_L_About.php  :wink:


Let $ ABCD$ be convex quadrilateral with $ \angle C\equal{}\angle D\equal{}90{}^\circ$. The circle $ K$ passing through $ A$ and $ B$ is tangent to $ CD$ at $ C$. Let $ E$ be the intersection of  $ K$ and $ \left[AD\right]$. If $ \left|BC\right|\equal{}20$, $ \left|AD\right|\equal{}16$, then $ \left|CE\right|$ is

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\  6\sqrt{2} \qquad\textbf{(C)}\  4\sqrt{5} \qquad\textbf{(D)}\  7\sqrt{2} \qquad\textbf{(E)}\ 10$

[hide="Solution"]Let $ O$ be the center of the circle $ K$. So it is on $ BC$ and $ OA \equal{} 10$. Let $ F$ be the foot of the perpendicular from $ A$ to $ BC$. $ AD \equal{} CF \equal{} 16 \Rightarrow OF \equal{} 6 \Rightarrow AF \equal{} CD \equal{} 8$.
Power of $ D$,  $ CD^2\equal{}AD \cdot DE \Rightarrow DE \equal{} 4$. In $ \triangle CDE$, $ CD \equal{} 8, DE\equal{}4$, so $ CE \equal{} 4\sqrt 5$.
$ \boxed C$[/hide]


MATHCOUNTS
analytic geometry
graphing lines
slope
conics
parabola

past Mathcounts State problem:

A merchant has determined that if the selling price of a peach is 15 cents, he will sell 400 each day.  He has also determined that for each cent the price is increased, he will sell 16 fewer peaches each day.  At what price, in cents, whould he sell each peach in order to maximize his daily income from peaches.

[hide]Trial and error until the max. 20[/hide]
another solution
[hide]On a coordinate plane, we can see that this is like a line. x is 15 and y is 400. Then x is 16 and y is 384. The slope of the line is -16. The line is 16x + y = 640. We want the maximum value of xy. y = 640 - 16x. Substitue that in and we get x(640-16x) which is 640x - 16x^2. The max value for x is the average of the roots or -b/(2a) which comes out 20.[/hide]

This was a sprint round question, so my guess is you wouldn't want to keep doing out the multiplication, but if you are quick, maybe that way is fine :)

Can anyone do this without guess and check?  That would be useful, especially if this was a Target and the answer wasn't a whole number, or even a rational # making guess and check impossible.

hint for anyone who is seriously stumped:
( frost13 is my coach )
[hide]think graphing[/hide]

I've actually done this sprint before and I did it the first way I said. [hide]I actually jumped on the number 20, then did 21 and 19 before writing down the answer, so I guess I was lucky. =D[/hide]

You have to make an equation, but how I have no clue.

try writing an equation for the amount of money made from the sales like this

income = (#peaches)(cost per peach)

then define a variable for how mnay times you are going to raise price / sell 16 less peaches

I was thinking in that direction, but when I tried what I thought it was, it didn't work.  But yours is a little different, so it would probably work.

[hide](400-16x)(15+x)=money he will make. 60000+160x-16x2, so you want the maximum value of 160x-16x2, which is 5, because if you were to add 6 cents, then you will add 160 and - (6+5)*16, which is 176, so he should sell his peaches at 20 cents.[/hide]

[hide]bob, right equation(expression) right answer.  But how did you get the answer so all can repeat your process?  If no response, others can try stting bob's expression equal to y to get the equation of a parabola.  Where is the maximum point on this parabola????? [/hide]

Don't you need to complete the square or something like that?

Did I do it right in my second solution?
[hide]Maximum point on the porabola is (5, 60400).[/hide]

I belive that you're right about the maximum point.

[hide]
tarquin, your only mistake comes in using the mistake bob made that i missed last time. (15)(400) = 6,000 not 60,000 :)

So we are getting the right point(pretty much) but how are we all getting it???
[/hide]

shouldn't you hide the above comment, Frost? :?

I meant that I had a 2nd solution in my first post and I was wondering if I had done it correctly there. Yeah you complete the square or use -b/(2a) to find where x is in the maximum value because in a porabola, the maximum/minimum spot for y is always directly in the middle.


A fixed circle of radius $ a$ is centered at $ (0,a)$. For every $ \theta$ (angle between $ OQ$ and the horizontal axis), the point $ P$ is the midpoint of the segment $ QR$, where $ Q$ is a point on the circle, $ R$ is a point on the line $ y\equal{}2a$ and the points $ O,Q,R$ are collinear. The curve traced out by $ P$ for $ 0<\theta<\pi$ is called the longbow curve. Find the parametric equations for this longbow curve.

The equation of the circle is $ x^{2} \plus{} (y\minus{}a)^{2} \equal{} a^{2}$. The line $ OQR$ has equation $ y\equal{}x\tan\theta$. We can find the coordinates of $ Q$ in terms of $ a$ and $ \theta$ by solving between these. Then solving between the line $ OQR$ and $ y\equal{}2a$ will give the coordinates of $ R$. Then we find the midpoint by finding the arithmetic means of the $ x$ and $ y$ cordinates of $ Q$ and $ R$.


geometry
trigonometry
ratio
geometric transformation
homothety
geometric series

Draw a regular pentagon with side length $2$ and its diagonals. Then in the pentagon that has vertices as the pairwise intersection of these diagonals, draw its diagonals. Iterate this process infinitely. What is the sum of the areas of these pentagons, including the original?

Eventually, with some trig manipulation, you will get that the area of any regular pentagon is equal to

$A= \frac{5{s^{2}}}{4}\cdot \cot \frac{\pi}{5}$
$A = \frac{s^{2}{\sqrt{25+10\sqrt{5}}}}{4}$

Clearly, the area of the first pentagon is ${\sqrt{25+10\sqrt{5}}}$

Now comes the tricky part.
Each diagonal has a sidelength of $\frac{s}{\frac{1+\sqrt{5}}{2}}$, or $\frac{2s}{1+\sqrt{5}}$
due to the fact that a regular pentagon generates many sets of congruent isoceles triangles. (proof of this is not given, but may be derived easily)
Now we need to find the sidelength of each subsequent pentagon.

We have the diagonal length.
This is equal to the sum of the side length of the small pentagon, and two other equal, extra sections. Call the sidelength of the small pentagon $a$ and the extra sections $b$
We know that $a+b = \text{Sidelength of the Large Pentagon}$, and that $a+2b = \text{Diagonal of the Large Pentagon}$

Remember, all we want is a.

$b = \text{Sidelength}-a$
$b = \frac{\text{Diagonal}-a}{2}$

$\text{Sidelength}-a = \frac{\text{Diagonal}-a}{2}$
$2S-2a = D$
$2a = 2S-D$
$a = S-\frac{1}{2}D$
$a = S-\frac{1}{2}\cdot \frac{2S}{1+\sqrt{5}}$
$a = S-\frac{S}{1+\sqrt{5}}$
$a = \frac{S(1+\sqrt{5})-S}{1+\sqrt{5}}$
$a = \frac{S\sqrt{5}}{1+\sqrt{5}}$
We're almost done.
We now need to set up a sum of the areas.

$\text{Total Area}= s^{2}\left(\frac{{\sqrt{25+10\sqrt{5}}}}{4}\right)+{s_{2}}^{2}\left(\frac{{\sqrt{25+10\sqrt{5}}}}{4}\right) \ldots$

There is a constant factor of $\left(\frac{{\sqrt{25+10\sqrt{5}}}}{4}\right)$, so factor that out:

$\text{Area}= \left (\frac{{\sqrt{25+10\sqrt{5}}}}{4}\right) \cdot \left(s^{2}+{s_{2}}^{2}+{s_{3}}^{2}\ldots\right)$
Where the subscript refers to the sidelength of the respective iteration.
----------------------------------------------------------------------------------
Let $Q = \frac{\sqrt{5}}{1+\sqrt{5}}$
The sidelengths form a geometric series, but we want the SQUARE of successive sidelengths, so we get a series like this:
$s^{2}+(sQ)^{2}+(s{Q^{2}})^{2}+(s{Q^{3}})^{2}+\ldots+(s{Q^{N}})^{2}$
Factor out an s^2, and we get that the sum is equal to
$s^{2}(1+Q^{2}+Q^{4}+\ldots+Q^{2n})$
That factor on the right looks like a geometric series to me.
Let's evaluate.
First Term: $Q^{0}= 1$
Common ratio: $Q^{2}= \frac{15-5\sqrt{5}}{8}$

Sum that puppy!
$\frac{1}{1-\frac{15-5\sqrt{5}}{8}}$
$\frac{1}{\frac{5\sqrt{5}-7}{8}}$
$\frac{8}{5\sqrt{5}-7}$

Now we're finished.
Our total area looks like this:
$\text{Area}= \left (\frac{{\sqrt{25+10\sqrt{5}}}}{4}\right) \cdot s^{2}\cdot \frac{8}{5\sqrt{5}-7}$

Substitute 2 in for s, and the answer works out to be [hide="Answer"]
$\left (\frac{8{\sqrt{25+10\sqrt{5}}}}{5\sqrt{5}-7}\right)$, which is roughly equal to $\boxed{13.17007}$[/hide]

Is this long solution correct?

uhhh...it is long

i think you can make it a lot shorter if you don't derive the ratios...just use the trig values

it is a geometric series

initial area is

5 tan72

[label ABCDE, and A'B'C'D'E', so the the labels represent a homothety with negative ratio about the center]

the length ratio is DE/D'E'=BE/BD'=2sin 54/sec 36

this squared is the area ratio, so our answer is

$\frac{5\tan 36}{1-(2\sin 54\cos36)^{-2}}=\frac{5\tan 36}{1-4\sin^{-4}54}=$some messy stuff

...

your solution has the area as cot 72...it should be tan 72, etc


topology
advanced fields
advanced fields solved

Let $x_{0}$ and $-x_{0}$ be two antipodal point of $\mathbb{S}^{2}$. Let $\gamma_{1}$ be a path from $-x_{0}$ to $x_{0}$ along the great circle of $\mathbb{S}^{2}$ and $\gamma_{2}$ be a path from $x_{0}$ to $-x_{0}$ along the great circle of $\mathbb{S}^{2}$ such that $\gamma_{1}\cap \gamma_{2}=\{x_{0},-x_{0}\}$. (Thus $\gamma_{1}: x_{0}\simeq-x_{0}$ and $\gamma_{2}:-x_{0}\simeq x_{0}$ are paths in $\mathbb{S}^{2}$.) Let $p: \mathbb{S}^{2}\to\mathbb{RP}^{2}$ be a covering map. Let $\omega=p\circ \gamma_{2}$.
1. Prove that $\omega$ is a loop in $\mathbb{RP}^{2}$.
2. Prove that $\omega$ is not nullhomotopic.
3. Prove that $\omega\omega$ is nullhomotopic.

3/ is immediate because the fondamental group of $\mathbb R \mathbb P^{2}$ has $2$ elements, and 1/ is obvious. If $w$ were nullhomotopic, there would exist an homotopy $H: [0;1] \times [0;1] \to \mathbb R \mathbb P^{2}$ such that $H_{0}= p \circ \gamma_{2}$. And there is a unique lifting $\hat H$ of $H$ satisfying $\hat H (t,0) = \gamma_{2}(t)$. This is impossible because $\hat H (t,1)=\gamma_{2}(1)$ and $\hat H (t,0)=\gamma_{2}(0)$  for all $0 \leq t \leq 1$, and $\hat H (.,1)$ must be constant. (why did you introduce $\gamma_{1}$)

[quote="alekk"]3/ is immediate because the fondamental group of $ \mathbb R \mathbb P^{2}$ has $ 2$ elements, and 1/ is obvious. If $ w$ were nullhomotopic, there would exist an homotopy $ H: [0;1] \times [0;1] \to \mathbb R \mathbb P^{2}$ such that $ H_{0} \equal{} p \circ \gamma_{2}$. And there is a unique lifting $ \hat H$ of $ H$ satisfying $ \hat H (t,0) \equal{} \gamma_{2}(t)$. This is impossible because $ \hat H (t,1) \equal{} \gamma_{2}(1)$ and $ \hat H (t,0) \equal{} \gamma_{2}(0)$  for all $ 0 \leq t \leq 1$, and $ \hat H (.,1)$ must be constant. (why did you introduce $ \gamma_{1}$)[/quote]

Given the low level of the exercise, I doubt he's supposed to know that $ \pi_1(\mathbb{RP}^2) \equal{} \mathbb{Z}/2$.


summer program
MathZoom
email
Alcumus
AMC
AIME
USA(J)MO

Greetings, fellow LCRF members. Feel free to use this forum to talk about math, LCRF, or anything else MathZoom-related.

Yay! I have the first comment. I don't want to leave MathZoom. It's been the best 2.5 weeks of my life. I want to go next year if I don't make MOP.

2nd post! *slinks off into distance* :ninja:

can i join the club....add a C for cool or an A at the end for awesome? =)

I agree... mathzoom was pretty awesome... I was there all 3 weeks =D

[quote="CountdownKing"]Who from MathZoom is going to do WOOT this year?[/quote]I am! :lol:

Yay!

I am too, if I somehow manage to convince my dad...

calc84maniac, I have a feeling you have my email address listed wrong because I didn't get "This be spam" even though a brief glance at Cynthia's thing showed my screen name.

lol, i'm not sure who's who. I might do woot, though i probably won't have the time to be devoted to it fully. Btw, this is Chris.

Click on my username and my "occupation" will give away my identity.

In order to keep this subforum much cooler than the other subforums, I will add another post. Unforunately it is a consecutive post. Oh well.

Another LCRF member is doing WOOT! Yay! Me!

LCRF is going to have an iTest team. If you are an LCRF member, join by contacting me or Cynthia. If you are a non-LCRF member, you can join too if we don't get enough people.

Aaah, am I the only person who posts?

Hm..Never taken the iTest. What's it like?

No clue. It's getting changed for this year. More questions, more easy questions, and no proof tiebreakers.

Yay, thanks to spam, this subforum now has the most posts on the MathZoom forum. Obviously, this is because this is the best subforum ever!

For anyone who became obsessed over the last week of MathZoom, (or before)
[url]http://www.setgame.com[/url]
and
[url]http://www.gamepixies.com/set/[/url]

yay! in one post i have ruined this subforum by making it...well, a lot wwwwwwwwwwwwiiiiiiiiiiiiiiiiiiiiidddddddddddddeeeeeeeeeerrrrrrrrrrrrrrrrrr. youre welcome

hmm...the LCRF subforum has died...CPR anyone?

Funny that you think the LCRF forum has died. In fact I expect quite an influx of posts in the coming days due to the fact that today and yesterday were both important dates in the history of LCRF. July 20th is one, also.

Yesterday was the first anniversary of Stephen Appreciation Day (SAD) so I hope everyone appreciated Stephen yesterday. Today is the first anniversary of Brendan Appreciation Day (BAD), so I hope everyone appreciates Brendan today. (These are merely anniversaries of the days and not the days themselves, because SAD only occurs on Thursdays in years which are congruent to 8 mod 10, and BAD only occurs the day after SAD.) The next Day Appreciation Year is 2098.

Monday July 20th will be one year after the creation of LCRF: Losers, Castoffs, Rejects, and Failures.

Yay!

hmm...I see no influx of posts...maybe my computer is malfunctioning...

Of course I appreciated Stephen and Brendan on those days. It's kind of sad that none of us are likely to be alive for the next Day Appreciation Year.

hmm...yay!...but I still don't see an influx of posts.

YAY!  ANNIVERSARY!
HI TURAKKKKKKKKKKK
whoa ure rite that is really quite unfortunate...

GO LCRF!

aw the anniversary's over!
earlier i tried to contribute to the influx of posts by adding a consecutive post.  but unfortunately i didnt feel like waiting 10 minutes so yeah...   :lol:

Ah, but the anniversary of Spam Day is July 24th or something, so you can still contribute to our influx of posts.

On a more mathematical note, thank you so much to MathZoom...I spent most of today reviewing my notes from last year. They still help me greatly.

YAY!!!!

hmm...yay!!!
hi cynthiaaaaa!
is two posts an influx?

[quote]influx n. arrival in large numbers: a sudden arrival of a large number of people or things
Microsoft® Encarta® 2007. © 1993-2006 Microsoft Corporation. All rights reserved.[/quote]

two=/=influx :(

hmm...i will try to remember spam day...this forum should expect a post from me every 10 mins or so...maybe...

YYYYYYYYYYYYYAAAAAAAAAAAAAAAAAAAAYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Spam day aniversary!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 :D  :spam:  :D  :spam:  :D
Yayyyyyyyyyyyyyyyy!!!!!!!!!

Too many smilies in your post: 13.
You can have at most 5 smilies in this message. Please remove some smilies before submitting this message.

gahh...
me see no influx of post :(
me sad :( :(

[quote="CountdownKing"]Greetings, fellow LCRF members. Feel free to use this forum to talk about math, LCRF, or anything else MathZoom-related.

Yay! I have the first comment. I don't want to leave MathZoom. It's been the best 2.5 weeks of my life. [b]I want to go next year if I don't make MOP.[/b][/quote]

bleh.

Hmmm...I'd like to say that today is the 2nd anniversary of SAD, but since I'm on the East Coast right now, that's not strictly true. Therefore I instead wish everyone a very happy 2nd anniversary of BAD today.

I would also like to wish everybody happy anniversaries of SAD and BAD. I hope that Stephen and Brendan are well-appreciated.


counting
distinguishability
modular arithmetic
combinatorics unsolved
combinatorics

Let $k$ be a positive integer. Let $S_k = \{(a,b)\mid a,b = 1,2,...,k\}$. Two elements $(a,b)$ and $(c,d)$ are of $S_k$ are said to be indistinguishable if and only if $a - c \in\{-1,0,1\}\pmod{k}$ and $b - d \in \{-1,0, 1\}\pmod{k}$. Let $r_k$ denote the maximum number of pairwise indistinguishable elements in $S_k$. Determine $r_5$, $r_7$ and $r_k$ in general.

Hi!
Maybe you mean "Let r_k denote the maximum number of pairwise [b]distinguishable[/b] elements in S_k".  :?

Now that you brought this up, did anyone manage to get more than $n^2+1$ when $k=2n+1$?

What is your example for $n^2+1$? I see obvious one for $n^2$.

For $k=2n$ we immediately obtain $r_k=n^2$.

I don't have a numerical example, because I used some geometric arguments and didn't want to go through all the trouble of translating those into numbers  :blush:

I noticed that it's equivalent to finding the greatest number of disjoint sets of the form $\{(i,j),(i+1,j),(i,j+1),(i+1,j+1)\}$ that we can find in $\mathbb Z_{2n+1}\times\mathbb Z_{2n+1}$. Then I considered these sets squares on a torus on which we mark the points $(i,j)$, and then played around with $n^2$ squares on such a torus (represented as a square on a piece of paper, of course, with the proper side identifications :)) in such a way that I managed to squeeze another square in.

I'll try to translate it into a numerical example.

[Edit: indeed, the case $k$ even is a piece of cake :D]

But, Grobber, for $k=3$ it is impossible to invent two distinguishable pairs. Or I am missing something again?

$k=3$ is an exception. Here are $5$ pairs for $k=5$ (I prefer to use $0,\ldots k-1$ instead of $1,\ldots,k$): $(0,0),(1,3),(2,1),(3,4),(4,2)$.

Since I'm not good at reading between the lines, you'll have to tell me (please :)) what you meant by that question :D.

I was waiting yesterday your general example. I even stopped to think about it, since I was sure you will complete solution soon.

I will not complete the solution soon, so start thinking :). As for the general example, sorry: too lazy. If only I had the will power to draw a diagram...

Just so that you don't think I'm lying (:D), here are $10$ pairs for $k=7$:

$(0,0),(0,2),(1,4),(2,0),(2,2),(3,5),(4,1),(4,3),(5,6),(6,4)$.

[Edited according to Myth's comment below :)]

I think, you meant $(6,4)$ instead of $(6,3)$ ;)

Ok! I have just geometricaly realized your example. So I am able to construct it for all odd $n$. Thank you! :D
So we need to prove it is impossible to place $n^2+2$ squares.

Indeed :)

Ok, I think this is it (maybe you can find a simpler one? :? ):

First take all the pairs $(2i,2j)$ with $i,j\in\overline{0,n-2}$. Then take $(2n-2,2j+1)$ with $j\in\overline{0,n-2}$, then $(2i-1,2n-2)$ with  $i\in\overline{0,n-2}$ (modulo $k=2n+1$), and, finally, $(2n-3,2n-1)$ and $(2n-1,2n)$.

I guess it wasn't that hard :).

I am sure it is not that hard, but I can't prove estimate! Only obvious estimate $n^2+n/2$ comes to my mind :(


[quote="math92"]The squares of two consecutive integers differ by 1987. What is the sum of these two integers?[/quote]

[hide]x^2= x^2

(x+1)^2=x^2 + 2x + 1

   x^2 + 2x + 1
-  x^2
   --------------
     2x + 1

Thus,

2x + 1 = 1987 (Now we can simply this even more, or observe that x + x+1= 2x +1= 1987 Thus, their sum is 1987.)

2x= 1986

x=993 and x+1= 994

993 + 994 = 1987[/hide]