This guy has me confused on what he's doing.

Here's 3 different equations and they're all different but I don't know why. Pay attention to the top line of each one. That's the part I don't understand.

1. [img]http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut17ex3c.gif[/img]

He multiplies both sides by 10, but doesn't explain why.
Here's the question the equation is answering: How much 20% alcohol solution and 50% alcohol solution must be mixed to get 12 gallons of 30%?

In another similar question, he gets this answer.

2. [img]http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut17ad1c2.gif[/img]

In the other one he multiplied each side by 10, now he's multiplying each side by 100. Why? He doesn't explain it.
Here's the question for that one: How much 25% antifreeze and 50% antifreeze should be combined to give 40 liters of 30% antifreeze?

And the third one.

3. [img]http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut17ad1d3.gif[/img]

He didn't multiply either side by anything. Not 10 or 100.
Here's the question: You recently came into 20,000 dollars (lucky you!) and you want to place part of your money in a savings account paying 7% per year and part in a certificate of deposit paying 9% per year. If you wish to obtain an overall return of 1700 dollars per year, how much would you place in each investment?

So it shows 3 different ways of answering similar questions, but I have no idea why they're different.

Thanks.

[quote="t0rajir0u"]Notice that each time, after he multiplies, the resulting numbers he has to deal with are all [i]whole numbers[/i].  This is used to make the arithmetic easier; you don't have to add decimals or think about decimal places.  That's the only reason.  That isn't three different ways of answering the same question; it's all the same algebra.[/quote]
Look at the third one, he kept the decimal.
I understand what you're saying, and thought about that, until I saw the third one and it confused me.
So I can multiply 10 to both sides or 100 to both sides or a million to both sides and I'll still get the same answer?

[quote="leroyjenkens"]So I can multiply 10 to both sides or 100 to both sides or a million to both sides and I'll still get the same answer?[/quote]

Yes.  It doesn't change the validity of the equality.  The fact that

$ 2 \plus{} 3 \equal{} 5$

is essentially the same thing as the fact that

$ 20 \plus{} 30 \equal{} 50$,

which is just a multiplication by $ 10$ of the previous equality.

[quote="fishythefish"]And if you don't understand how he got the original equations, maybe it's time to start learning weighted averages. (This is covered in Algebra I.)[/quote]
Now I understand, but I guess I should go ahead and find out more about weighted averages.
I just didn't understand why the equations would be done differently. I noticed he wanted to make them whole numbers in the first and second equation, but didn't bother in the third equation and I just wanted to know if there was any reason for that.


I just learned about Mathcamp. I don't know which one is more advanced: mathcamp or mathpath. I thought that Mathcamp was, and then my friend told me today that he googled it and thought that Mathpath was more difficult. Which one is more difficult?

Mathcamp is for high school students.  Mathpath is for middle school students.

[quote="anirudh"]Are middle school students allowed to go to mathcamp?[/quote]
Yes. I know one of my friends went there last year as an 8th grader.
(well that, of course, is assuming that you do well on the quiz/test.)

Okay. Good. Because I sepent teh time to solve the questions on the quiz. I got like 6 questions already.

Also, about what is the 'cutoff'?


What digit is in the 45th decimal place in the decimal equivalent of 9/14?


[quote="shobber"]Find the range of the function $y=\frac{x-1}{x^2+1}$ if $x>1$.[/quote]

[hide]Is it $\displaystyle \left(-\infty,\frac{1-\sqrt{2}}{2})\cup(\frac{1+\sqrt{2}}{2},+\infty\right)$?[/hide]

perhaps $0 < y \leq \frac{\sqrt 2}{2 \sqrt 2 +4}$

[hide]Since $f(1)=\lim_{x\rightarrow +\infty}f(x)=0$, the function has at least one extreme on $(1,+\infty)$, and it must be a maximum, since the function is always positive. Because the function is continuous, it must take all the values between zero and the maximum. The first derivative is $f'(x)={{-x^2+2x+1}\over{(x^2+1)^2}}$ and it equals zero for $x_{1,2}=1\pm\sqrt{2}$, but we take only the positive solution, $1+\sqrt{2}$. Now we have $f(1+\sqrt{2})={{\sqrt{2}-1}\over 2}$. Hence, the range of the function is $(0,{{\sqrt{2}-1}\over 2}\rbrack$.[/hide]

What i did was to differenciate the function to find the maximum

Sure. Your result is correct, but you didn't simplify at the end: ${{\sqrt{2}}\over{2\sqrt{2}+4}}={{\sqrt{2}}\over{2\sqrt{2}+4}}\cdot{{2\sqrt{2}-4}\over{2\sqrt{2}-4}}={{4-4\sqrt{2}}\over {-8}}={{\sqrt{2}-1}\over 2}$

[hide]Since $\frac{x-1}{x^2+1}\rightarrow 0$ as $n\rightarrow \infty$ and when $x>1$, $\frac{x-1}{x^2+1}$ is positive and continuous, the range will be $(0,M]$ where $M$ is the maximum.

Write $\frac{x-1}{x^2+1}\leq M$, since $x^2+1>0$, multiply to get $Mx^2-x+(M+1)\geq 0$. For this to happen, the discriminant is nonpositive. So $1-4M(M+1)\leq 0$.

Since $M$ is positive, $1-4M(M+1)$ is decreasing, so the largest value of $M$ will be when the discriminant is 0, so $4M^2+4M-1=0\Rightarrow (2M+1)^2=2\Rightarrow M=\frac{-1\pm \sqrt{2}}{2}$.

$M$ is positive, so it must be $\frac{\sqrt{2}-1}{2}$, making the range $\boxed{\left( 0,\frac{\sqrt{2}-1}{2} \right]}$.[/hide]


LaTeX
inequalities
search
inequalities proposed

Find max of $ A\equal{}\frac{abc(a\plus{}b\plus{}c\plus{}3\sqrt{a^2\plus{}b^2\plus{}c^2})}{(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)}$
where $ a;b;c$ are positive real numbers

[quote="chien than"]Find max of $ A \equal{} \frac {abc(a \plus{} b \plus{} c \plus{} 3\sqrt {a^2 \plus{} b^2 \plus{} c^2})}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}$
where $ a;b;c$ are positive real numbers[/quote]
Observe that
$ a \plus{} b \plus{} c \leq \sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})}$.
THus $ A \equal{} \frac {abc(a \plus{} b \plus{} c \plus{} 3\sqrt {a^2 \plus{} b^2 \plus{} c^2})}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)} \leq \frac {abc( \sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})} \plus{} 3\sqrt {a^2 \plus{} b^2 \plus{} c^2})}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)} \equal{} \frac {(3\plus{}\sqrt {3})abc}{(\sqrt {a^2 \plus{} b^2 \plus{} c^2})(ab \plus{} bc \plus{} ac)}$.The rest is just AM-GM.

[quote="chien than"]Find max of $ A \equal{} \frac {abc(a \plus{} b \plus{} c \plus{} 3\sqrt {a^2 \plus{} b^2 \plus{} c^2})}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}$
where $ a;b;c$ are positive real numbers[/quote]
It is an easy problem from Hong-Kong 1997.And the answer is $ \frac{3\plus{}\sqrt{3}}{9}$. :)

[color=darkblue][b]Here ia an entertainment in ... LaTeX and a clear proof of this easy inequality ![/b][/color]

[quote="chien than"][color=darkred]Find max of $ A \equal{} \frac {abc(a \plus{} b \plus{} c \plus{} 3\cdot\sqrt {a^2 \plus{} b^2 \plus{} c^2})}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}$, where $ \{a,b,c\}\subset \mathcal R_ \plus{} ^*\ .$[/color][/quote]

$ \left\|\begin{array}{ccccc} a \plus{} b \plus{} c\le\sqrt {3\cdot \left(a^2 \plus{} b^2 \plus{} c^2\right)} & \Longrightarrow & a \plus{} b \plus{} c \plus{} 3\cdot\sqrt {a^2 \plus{} b^2 \plus{} c^2} & \le & \left(3 \plus{} \sqrt 3\right)\cdot\sqrt {a^2 \plus{} b^2 \plus{} c^2} \\
 \\
\left\|\begin{array}{c} 3\cdot\sqrt [3] {a^2b^2c^2}\le ab \plus{} bc \plus{} ca \\
 \\
3\cdot\sqrt [3] {abc}\le \sqrt {a^2 \plus{} b^2 \plus{} c^2}\end{array}\right\|\bigodot & \implies & 9\cdot abc & \le & (ab \plus{} bc \plus{} ca)\sqrt {a^2 \plus{} b^2 \plus{} c^2}\end{array}\right\| \bigodot$

$ \implies$ $ 9\ \cdot\ abc\left(a \plus{} b \plus{} c \plus{} 3\cdot\sqrt {a^2 \plus{} b^2 \plus{} c^2}\right)\le \left(3 \plus{} \sqrt 3\right)\cdot (ab \plus{} bc \plus{} ca)\left(a^2 \plus{} b^2 \plus{} c^2\right)$ $ \implies$

$ \boxed {\frac {abc\left(a \plus{} b \plus{} c \plus{} 3\cdot\sqrt {a^2 \plus{} b^2 \plus{} c^2}\right)}{(ab \plus{} bc \plus{} ca)\left(a^2 \plus{} b^2 \plus{} c^2\right)}\le\frac {3 \plus{} \sqrt 3}{9}}\ .$

Here are the links:
http://www.mathlinks.ro/viewtopic.php?search_id=17857755&t=70886
http://www.mathlinks.ro/viewtopic.php?search_id=17857755&t=47781
http://www.mathlinks.ro/viewtopic.php?search_id=17857755&t=10575


[hide="Rather easy"]Let $ A\in (O_1)$, $ B\in (O_2)$. On the tangent of $ (O_1)$ at $ A$, let point $ C$ such that $ AC\equal{}r$. On the tangent of $ (O_2)$ at $ B$, let point $ D$ such that $ BD\equal{}r$. The circle $ (O_1;O_1C)$ cuts the circle $ (O_2;O_2D)$ at $ O$. The circle $ (O;r)$ is just the circle needed to draw.[/hide]


I hope I'm not annoying anyone by making post in here.  I feel these are genuinely interesting and hope some of you agree:

I wish to make a conjecture:  Spindle inversion is dense in the vicinity of a non-isolated singularity.  If anyone is interested in this, first look at this thread which describes what I mean by spindle inversion angle:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170058

What I propose above is that in any $ \epsilon$-neighborhood of a non-isolated singularity, I can find a pole with any inversion angle $ 0\leq S_\theta\leq 2\pi$.

I'm wrong about this.  I've been looking at $ f(z)\equal{}Sec(1/z)$ and it looks like the real component alternates between $ \pi/2$ and $ 3\pi/2$ and the imaginary component, $ 0$ and $ \pi$.  Ergo,  not dense.  I'll try and better investigate things in the future before I open my mouth.  I think the whole concept of "spindles" has value and hope the moderators allow me to leave the thread here anyway.

Here's the Mathematica code to demonstrate this where nval is the spindle number starting at $ 3\pi/2$ and going in towards the origin.  You can change both nval and Re[f[z]] to Im[f[z]] to see what I mean:

[code]
f[z_] := Sec[1/z];
Clear[nval]
nval = 1
rd[n_] := 1/((2*n + 1)*(Pi/2));
rd2[n_] := 1/((2*n + 3)*(Pi/2)) - rd[n];
p1 = ParametricPlot3D[
     {Re[z], Im[z], Re[f[z]]} /.
       z -> rd[nval] + r*Exp[I*t],
     {r, 0, rd2[nval]}, {t, 0, 2*Pi},
     BoxRatios -> {1, 1, 2}]
[/code]


trigonometry algebra function domain logarithms ratio Vieta

So I was looking at some of my old post and saw that someone mentioned that you could solve equations in the form $Ax^3-Bx+C$ for $A,B,C>0$ by making the substitution $x=P\cos(\theta)$.
well i the spent some time to come up w/ a general formula for this method and came up with

$R_1=\sqrt{\frac{4B}{3A}}\cos\left(\frac{1}{3}\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)\right)$

$R_2=\sqrt{\frac{4B}{3A}}\cos\left(\frac{1}{3}\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)+\frac{2\pi}{3}\right)$

$R_3=\sqrt{\frac{4B}{3A}}\cos\left(\frac{1}{3}\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)+\frac{4\pi}{3}\right)$

Where $R_1,R_2,R_3$ are the roots of $Ax^3-Bx+C$.

I am wondering if there are any limitations on this formula.

(if you would like a proof just let me know and I will post one)

Well one limitation i see is the restriced domain on $cos^{-1}$... if $-3C\sqrt{3AB} > 2B^2$ then the formula will not work.

And i'd like to see the proof, if it isnt too time-consuming to type out..  :D

yeah i just noticed that and found a way around it.
using the fact that $cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$
using this, some algebra and the quadradic equation we find that we can replace
$\cos^{-1}\left(\frac{-3\sqrt{3AB}}{2B^2}\right)$ with
$-\ln\left(\frac{-3C\sqrt{3AB}}{2B^2}\pm\sqrt{\frac{27AC^2}{4B^3}-1}\right)i$
where we use $\pm$ to make the quantity in the $\ln$ positive.
notice how this value will make u find $\cos(i\beta)$ this is not a problem since now we just plug $i\beta$ into the expression for $\cos(\theta)$ that i stated at the beginning.

[hide="ok here is the proof"]
Before I begin it is important to know that $\cos (3\theta)=4\cos^3(\theta)-3\cos(\theta)$
ok
so we have $Ax^3-Bx+C=0$
let $x=P\cos(\theta)$ we now have
$AP^3\cos^3(\theta)-BP\cos(\theta)=-C$
$P(AP^2\cos^3(\theta)-B\cos(\theta))=-C$ now we get this to look like $\cos(3\theta)$
$P\left(\frac{B}{3}\right)\left(A\left(\frac{3}{B}\right)P^2\cos^3(\theta)-3\cos(\theta)\right)=-C$
now we find the value of $P$ that makes $\frac{3AP^2}{B}=4$ and we have $P=\sqrt{\frac{4B}{3A}}$ we use the positive value since we factored out a $P$ once and if it were negative we would have to change the signes.
we now have,
$\frac{B}{3}\sqrt{\frac{4B}{3A}}\cos(3\theta)=-C$
$\cos(3\theta)=\frac{-3C\sqrt{3AB}}{2B^2}$ (you should be able to get this result on your own.)
now to find $\theta$.
$3\theta=\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)$
from this it is clear that
$\theta=\frac{1}{3}\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)$ or
$\theta=\frac{1}{3}\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)+\frac{2\pi}{3}$ or
$\theta=\frac{1}{3}\cos^{-1}\left(\frac{-3C\sqrt{3AB}}{2B^2}\right)+\frac{4\pi}{3}$

at this point to find the roots we just the values of $\theta$ and $P$ that we just found into $x=P\cos(\theta)$.
once u do this u will have the formula that I have given.
[/hide]

[quote="maokid7"][hide="ok here is the proof"]
Before I begin it is important to know that $\cos (3\theta)=4\cos^3(\theta)-3\cos(\theta)$
ok
so we have $Ax^3-Bx+C=0$
let $x=P\cos(\theta)$ we now have
$AP^3\cos^3(\theta)-BP\cos(\theta)=-C$
$P(AP^2\cos^3(\theta)-B\cos(\theta))=-C$ now we get this to look like $\cos(3\theta)$
$P(\frac{B}{3})(A(\frac{3}{B})P^2\cos^3(\theta)-3\cos(\theta))=-C$
now we find the value of $P$ that makes $\frac{3AP^2}{B}=4$ and we have $P=\sqrt{\frac{4B}{3A}}$ we use the positive value since we factored out a $P$ once and if it were negative we would have to change the signes.
we now have,
$\frac{B}{3}\sqrt{\frac{4B}{3A}}\cos(3\theta)=-C$
$\cos(3\theta)=\frac{-3C\sqrt{3AB}}{2B^2}$ (you should be able to get this result on your own.)
now to find $\theta$.
$3\theta=\cos^{-1}(\frac{-3C\sqrt{3AB}}{2B^2})$ or
${3\theta=\cos^{-1}(\frac{-3C\sqrt{3AB}}{2B^2}}+\frac{2\pi}{3})$  or
$3\theta=\cos^{-1}(\frac{-3C\sqrt{3AB}}{2B^2}+\frac{4\pi}{3})$
from this it is clear that
$\theta=\frac{1}{3}\cos^{-1}(\frac{-3C\sqrt{3AB}}{2B^2})$ or
$\theta=\frac{1}{3}\cos^{-1}(\frac{-3C\sqrt{3AB}}{2B^2}+\frac{2\pi}{3})$ or
$\theta=\frac{1}{3}\cos^{-1}(\frac{-3C\sqrt{3AB}}{2B^2}+\frac{4\pi}{3})$

at this point to find the roots we just the values of $\theta$ and $P$ that we just found into $x=P\cos(\theta)$.
once u do this u will have the formula that I have given.
[/hide][/quote]

ok, i understand everything but why $\frac{2k\pi}{3}$ is added... can you explain?

[quote="maokid7"]it is to make sure that u find all values of $\theta$ that work.
like $\cos(\theta)=\frac{1}{2}$ has more solutions for $0\le \theta< 2\pi$ than just $\theta=\frac{\pi}{9}$
it also hase
$\theta=\frac{\pi}{9}+\frac{2\pi}{3}$ and $\theta=\frac{\pi}{9}+\frac{4\pi}{3}$.[/quote]

ok, but the reason i am asking is becuase its the angle that should be increased, not the ratio inside the inverse cos function. the increasing multiples of $\pi$ should be on the side with theta of the formula and not inside the inverse cosine funtion.

[quote="maokid7"] we can replace
$\cos^{-1}\left(\frac{-3\sqrt{3AB}}{2B^2}\right)$ with
$-\ln\left(\frac{-3C\sqrt{3AB}}{2B^2}\pm\sqrt{\frac{27AC^2}{4B^3}-1}\right)i$
[/quote]
ok this doesnt work can someone check my work please.
[hide="work"]
let $y=\frac{-3C\sqrt{3AB}}{2B^2}$
so $\cos^{-1}(y)=\theta$ so $y=\cos(\theta)$ so we just have to solve for theta.
so we have $y=\frac{e^{i\theta}+e^{-i\theta}}{2}$ leading to
$2y=\frac{\left(e^{i\theta}\right)^2+1}{e^{i\theta}}$ so we let $x=e^{i\theta}$
and just have to solve $x^2-2xy+1=0$ for $x$.
we get $x=\frac{-2y\pm\sqrt{4y^2-4}}{2}$
$x=-y\pm\sqrt{y^2+1}$

$e^{i\theta}=\frac{3C\sqrt{3AB}}{2B^2}\pm\sqrt{\frac{27AC^2}{4B^3}-1}$

$i\theta=\ln\left(\frac{3C\sqrt{3AB}}{2B^2}\pm\sqrt{\frac{27AC^2}{4B^3}-1}\right)$

so

$\theta=-\ln\left(\frac{3C\sqrt{3AB}}{2B^2}\pm\sqrt{\frac{27AC^2}{4B^3}-1}\right)i$

[/hide]

[quote="maokid7"]
I am wondering if there are any limitations on this formula.
[/quote]
This looks correct (Derive gives the same set of solutions, more or less), but the limitation is that you would have to actually evaluate the expression for specific values of $A,B,C$  :P .  I prefer Vieta's substitution when solving the cubic - it's always seemed to me the most elegant way of doing it, and you get solutions in terms of radicals rather than trig and inverse trig functions (though you could always, with some effort, turn those trig functions into radicals).

If $x^3 - px^2 + qx - r$ has roots $a,b,c$ then $a+b+c = p$, $ab+bc+ca = q$ and $abc = r$.

The proof follows from $x^3 - px^2 + qx - r = (x-a)(x-b)(x-c)$.

o that...i already knew that just didnt know it had a name well thx.


A word of length $n$ that consists only of the digits $0$ and $1$, is called a bit-string of length $n$. (For example, $000$ and $01101$ are bit-strings of length 3 and 5.) Consider the sequence $s(1), s(2), ...$ of bit-strings of length $n > 1$ which is obtained as follows :
(1) $s(1)$ is the bit-string $00...01$, consisting of $n - 1$ zeros and a $1$ ;
(2) $s(k+1)$ is obtained as follows :
(a) Remove the digit on the left of $s(k)$. This gives a bit-string $t$ of length $n - 1$.
(b) Examine whether the bit-string $t1$ (length $n$, adding a $1$ after $t$) is already in $\{s(1), s(2), ..., s(k)\}$. If this is the not case, then $s(k+1) = t1$. If this is the case then $s(k+1) = t0$.

For example, if $n = 3$ we get :
$s(1) = 001 \rightarrow s(2) = 011 \rightarrow s(3) = 111 \rightarrow s(4) = 110 \rightarrow s(5) = 101$
$\rightarrow s(6) = 010 \rightarrow s(7) = 100 \rightarrow s(8) = 000 \rightarrow s(9) = 001 \rightarrow ...$

Suppose $N = 2^n$.
Prove that the bit-strings $s(1), s(2), ..., s(N)$ of length $n$ are all different.

You should try this guys ! :)
I can guarantee you, it's a very nice and interesting problem !

I AM trying, but I only have a partial soln

If we assume s(something) to be the string which repeats itself the earliest then it must end in a 0, so s(something)=t0 (I used your notation). We have obtained t0 twice from 2 different strings (t0 is the FIRST repeating string) so t0 has been obtained fro 1t and 0t. Since 0t->t0 and 1t->t0 it means t1 was already in the sequence before 0t and 1t. But t1 could only have been obtained from 1t or 0t and since there are no 1t or 0t before the ones from which we obtained t0 (again because t0 is the FIRST word to be repeated), it means that t1 wasn't obtained from anything, which means that t1 must be the first term of the sequence: 00..01, so t0 (the first string to be repeated) is 00..00.

This means that our sequence has this form: s(1), s(2),.., s(k-1), s(k)=00..00 all distinct and then 00..00, 00..00 etc. You can see this is only a partial soln. We still have to prove that k=2^n=N.

OK Grobber, you're on the right way.
A serious hint for the more difficult part of the solution.

Suppose k < 2^n. This means that there is a string s = bt which is not in S.
(S is the set {s(1), s(2), ..., s(k)}.)
In the above notation, t is a string op length n - 1 and b is 0 or 1.
Now suppose that t0 belongs to S.

Eventually you will reach a contradiction.

Nice problem isn't it ? I really like it :)

OK, I think I've got it (thx for the hint, by the way, I don't think I would have thought of it):

We assume, as Arne said, that bt isn't in S. Further more, we assume that t0 is in S. This means that t1 is also in S (because of the way we built the sequence). t is obviously not 00..0 (0 is taken n-1 times). t1 and t0 can only be obtained from 0t and 1t. Since t1 and t0 are both in S, t1 and t0 aren't 00..01 and 00..00, it means that they were obtained from 0t and 1t (0t and 1t can't be repeated since the only string which is repeated is 00..00), which in turn means that both 1t and 0t are in S, so bt is in S, contradiction.

This means that bt not in S ==> t0 not in S. Now let's assume there is such a bt not in S. We repeat this process until we reach 00..00 not in S, which is a contradiction because we obviously reach 00..00 eventually.

Is it correct Arne?

Seems to be correct ! :D I'll post the official solution later.

Did you enjoy solving it ?

This is the official solution :

Let M be the greatest positive integer for which s1, s2, ..., sM are pairwise different. There are only 2n different strings of length n. Hence 1 \leq M \leq 2n. We will prove that M = 2n. Denote S = {s1, s2, ..., sM}.

We firstly prove that sM = 00...00. Indeed, suppose sM = 0u or 1u (where u is a string of length n - 1). The successor of sM cannot end in 1 because of rule 2(b). So sM+1 = u0. Because of the definition of M, u0 has to be in S, so u0 = si where i > 1. Because of rule 2(b), u1 has to be in S also, so u1 = sj. Suppose j > 1. In this case the predecessors of si and sj exist and we know that {si-1, sj-1} = {0u, 1u}. So sM \in {si-1, sj-1}, which is a contradiction. So we must have j = 1. Hence sj = u1 = 00...01 and sM = 00...0 or 10...0. In the latter case we must have 00...0 \in S. But the only possible predecessor of the string 00...0 is 10...0, which is a contradiction. Hence sM = 00...0 and sM-1 = 10...0. (*)

Now suppose that M < 2n. Then there is a string s of length n which is not in S. Take such a string s and write s = bt where t is a string of length n - 1, and b equals 0 or 1.
Suppose that t0 \in S, t0 = si. Because of rule 2 (and because i > 1) we have t1 \in S, t1 = sj. If j > 1 then {si-1, sj-1} = {0t, 1t} and hence s \in S. If j = 1 then t consists of zeros only. In this case 0t = t0 = 00...0 and it follows from (*) that s \in S, which is a contradiction.
So we must have from s = bt doesn't belong to S that s' = t0 doesn't belong to S. If we apply this principle to s', s", s''', ... we finally find out that 00...0 does not belong to S. This is a contradiction with (*). So s doesn't exist and M = 2n.


Daniel and Bruno are in a station waiting for a train. To entertain themselves, decide to calculate the length of a freight train passing through the station without changing the speed. When the front of the train passes through them (Daniel and Bruno are in the same place), Daniel begins to walk towards the train movement and Bruno begins to move in the opposite direction. They go the same speed and each of them for the moment that intersects with the end of the train. Daniel walked 45 meters and Bruno 30. What is the length of the train?

One of those cute problems in which some of the unknowns cancel out from the equations, and remain forever unknown!
[hide]Denote by $ v$ the speed of the train, $ L$ the length of the train, and $ u$ the (common) speed of the boys. Given that ultimately the train bypasses Daniel, it follows that $ v > u$. Then Daniel walks for a time $ t_D \equal{} \frac {L} {v\minus{}u}$, while Bruno walks for a time $ t_B \equal{} \frac {L} {v\plus{}u}$. This yields the system of equations $ \frac {uL} {v\minus{}u} \equal{} 45$, $ \frac {uL} {v\plus{}u} \equal{} 30$, so $ \frac {2} {L} \equal{} \frac {v\plus{}u} {uL} \minus{} \frac {v\minus{}u} {uL} \equal{} \frac {1} {30} \minus{} \frac {1} {45}$, whence $ L \equal{} 180$. The only other relation one can extract is $ v \equal{} 5u$, which leaves $ u$ and $ t_D$, $ t_B$ unknown.[/hide]

ok, but anyone knows how to solve this question ?

the answer is 180 meters.

how to do ?  :o

Has this question other simple method to resolve ?

The post that wasn't yours nor mine gave a nice solution.  :wink:


Harvard college AMC USA(J)MO USAMO articles

I was just wondering how exactly does high schools rank you? Do they look at all of your grades starting from freshman year? and how important is your rank?  :huh:

i think its just your (weighted) gpa. and then they put you in order, starting from the person with the highest gpa (in your graduating class) to the person with the lowest. i would ASSUME that it begins at freshman year, although in the high schools that start in 10th grade, i am sure that the gpa calculation also starts in tenth grade. for transfer students, i am not sure.

but what do i know? i go to an unranked system.

Yes, usually it does start with the freshman year. They take those grades, and then according to whatever system you have, they will add on the weight.

I heard of some schools having valedictorians/rankings only based on unweighted and others only based on weighted, and some have 2 valedictorians.

My suggestion: talk to your guidance counselor, he/she will have the most information about your personal situation.

colleges actually care about class rank a decent amount.  admissions officers say that schools who do not have class rank (like mine) put the top 5 (depending on size of school) or so students at a disadvantage.

Last year, my HS ranked by straight, unweighted GPA. We had 7 valedictorians. That's not even that much, though. Some other schools in WA had 30+, at which point the title loses some value. My HS used to do it by weighted, as in AP/Honors were on a 5.0 scale and everything else was on 4.0 but they changed it last year. I'm not sure how they'll do it in the future. I'd say class rank is relatively important, but it defiintely depends on what school you go to.

Our high school ranks for high school classes you take even before yor freshman year.  And then our valedictorian is just anyone with abvoe a 4.0 gpa, with some type of weighting system for AP classes.  So yea, it's different for different schools.

the high school i'll go to doesn't rank. i dont think that colleges should put that much emphasis on rank. it really isn't a good indicator of performance. they should be more interested in grades, classes, and extracurriculars.

my school uses a system where a 100 in an AP class equals a 5.87 gpa.  :lol:

My school is like... :roll: ... Super noncompetitive or something. We have a 10 point weighted scale - it's rather complicated. I never bother to look at how my average turns out so every semester they just tell me my rank and I'm okay with it. Every course is weighted - we have academic, honors, seminar, and sometimes AP.

[quote]Our school (and our school district) has a regular 4.0 unweighted GPA. (4.0=A, 3.0=B, etc.) All classes contribute to this GPA.

But then we have this thing called "Academic Average" that we use to rank.
For TAG (talented and gifted) Pre-AP, AP classes a 100=6.0, 99=5.9,...90=5.0, etc.
For regular classes, we have 100=5.0, 99=4.9, etc...
Its average for our core classes (math, science, social studies, english, foreign language)[/quote]

the high school i'm going to has a similar system. 4.0 scale for remedial classes, 5.0 for accelerated, and 6.0 for honors. the highest grade you can get is 6.33 for an A+ in an honors class.(AP's are honors classes)

I thought a 5.87 was high but 6.33!  How on earth would people fail in a system like that  :?:

[quote]I thought a 5.87 was high but 6.33! How on earth would people fail in a system like that [/quote]

well, that's the weighted GPA. unweighted is still 4.0 scale. but, yes, a D in an honors class is a 3.67, believe it or not. that's the equivalent of an A- in a remedial class.

[quote="L_Li"]My school reports both weighted and unweighted GPA, which are calculated purely by letter grades, so i've formed a habit of aiming for 89.5's  :lol:

Only the weighted GPA is used for rank, though. my school's pretty competitive for a public school, and there are a good 20 or so people in my class of 379 with a 4.0 unweighted, so weighted is the only way to prevent having more than 5% of the class all in first place.[/quote]

That's like the same as my school. A flat 90 in all your classes = 4.0 (sometimes less because some teachers like to round 89's up). This makes it a LOT easier to get a 4.0; I had a couple of 91's and 92's that would've been A-'s. We have at least 10-20 4.0's in my class right now (7 valedictorians in class of '05, all 4.0's) but I'm not sure whether they will use the weighted GPA system anymore. In any case, I'd rather they not since it puts immense pressure to take every AP class possible (a couple of students in my class are aiming at 13+).

i dont think pressure is as big of an issue as everyone makes it out to be.  my friend and I are aiming to take 17 AP's and we're pretty relaxed about it.  in our school grades aren't weighted at all, so I don't think the "pressure" of wanting to take more AP's necessarily comes from weighting grades to be valedictorian.  neither my friend nor me plans to be valedictorian, but if we take less AP's we would have a better chance of being valedictorian!

ffdb, you may not understand it because you do not have that type of personality, but there are many people who want to be top 10, top 5%, top 2, valedictorian, etc. And when you include weighting into determining the valedictorian, it causes those people to feel more pressured to take more classes.

My response to that: Oh well. If someone's taking more classes than you, doing better in them than you, and not stressing out as much as you, then you're going to have to settle for not being the valedictorian.

Personally, I want my rank to be in the single digits (which my guidance counselors have told me they think it is, it isn't officially calculated for another few weeks), and I guess I did feel a bit of pressure when, in my first semester of highschool, I got a 3.5 GPA. This is because I knew that seniors were graduating valedictorians with like 5.3s and stuff like that. But it didn't worry me too much because since then I've decided to work a bit harder (getting 4.0's so far) and taking college classes which weight (although I took the classes mainly for enjoyment, the weight was a big extra), and I now have like a 4.3'ish.

But, having a system where a valedictorian is determined by unweighted GPA is pretty stupid, IMO. The person who can handle the most work deserves to get the top award, and in many schools there are GREAT differences between the regular program and the IB/AP program, and if the school doesn't acknowledge that, then there will be a good amount of undeserving valedictorians.

Valedictorian means #1, and sometimes people have to face that they cannot always be that.

i agree with you completely.  my school doesn't make any sense.  they think they're being fair to everyone by not weighting to pick valedictorian.  AP class vs. regular class in my school is certainly a big difference, and it's just not right what they're doing.  this year's valedictorian was of course not stupid or anything, she was actually pretty smart with 7 or 8 AP's and a GPA of 97.  but, there was another student who took 15 AP's with a GPA of only 1 point less than hers, 96, and he wasn't even salutatorian.  very very silly in my opinion.

it might not be perfect, but it's certainly better than giving someone who is in a significantly large amount of hard classes like AP's a rank of 5th while another who has only taken a few a rank of 1st.  a couple years ago, the valedictorian in my school could not make the ivy league and the next 9 or so in ranking were all able to make ivy league and other top tier schools including Harvard and MIT.  of course getting into college is not solely based on grades and this is not the best way to judge it, it still reinforces a point.

[quote="h_s_potter2002"]There is a problem with weighted ranks though.  Let's say you have 2 people, person A and person B who both go to a school where they weight grades.  Person A decided to only take 6 AP classes, while Person B decided to take 6 AP classes, and an art class that wasn't weighted.  And let's say they both got all A's.  Then A would have a higher GPA that B, even though person B took the extra class.  So weighted GPA isn't all great.

But I'm kind of glad I go to a school where anyone who gets above a 4.0 is a valedictorian.[/quote]

Well, in most cases, the top contenders will all take the same number of classes regardless of AP, honors or whatever.  Therefore person A would most likely have taken another class also.  Even if this scenario did happen, I wouldn't consider it as a major problem with weighted ranks.  Statistically, if weighted ranks picked the best person 95% of the time, you can't really complain about that other 5%.

It's probably because I'm neither the valedictorian nor salutatorian at my school, but I think these titles are only prestigious if you truly earn them at good high schools.

I think an anecdote my Cal AB teacher told me sums it all up.

Two students were deadlocked in the race for valedictorian. Student Adecided to take some classes over the summer so he could skip the introductory classes and take harder classes as substitutes. Both he and his fellow competitor had 5.0 weighted GPAs. However, these classes taken over the summer by Student A were unweighted. Thus, he had a string of 4.0s instead of 5.0s. He graduated salutatorian; he was unable to catch up.

Stuff like this can almost punish students for wanting to challenge themselves harder.

....  and Student A was probably more prepared for college, and him making salutatorian instead of valedictorian most likely did NOT affect any acceptances/denials, while the colleges see the classes in the summer and understand the work put into them when one didn't have to.

It all really does work out in the end.

This seems like a College Confidential thread.....

I think some school in WA had like 44 valedictorians...A USAMO qualifier pointed out in an article about the high school that there was flagrant grade inflation....but, he was one of the valedictorians....  :roll:

My school has a sort of unweighted GPA, that seems mostly based on a 100-point scale, but they somehow convert it to a 4-point scale.  It includes all high school classes, no matter when you take them.  It usually sucks to have unweighted GPA, but maybe it'll look more impressive to colleges if I have the highest unweighted GPA AND the highest AP count.

I wonder how long they'll let me speek at the graduation ceremony before they drag me off the stage.

[quote="h78reg"][quote]my rank started back in middle school, bc i took several high school classes.[/quote]

that's not very fair. unless you graduate early, now you have to get good grades for a longer amount of time then other kids.[/quote]

Actually it is fair because gpa is an average so he's just taking more classes, which means he can do poorly on one with less impact, but overall it doesn't make a difference.  In my school several students take 8 classes (the normal is 7).  The one advantage is that he will spend his later years in higher classes so if his school has weighted gpa then he will have more high level classes which will increase his gpa.

My school does class rank in a very unfair way, I think. They carried out the weighted GPA to about six decimal places. The result of this is that a difference between and 86 and a 92 will result in rank being extremely skewed. My friend Jacob is top of our class and his weighted GPA is 97.4. My weighted GPA is 93.8 and I'm 53rd in my class. the people in the next eleven ranks above me also have 93.8s, but they have 93.8081s or 93.8087s. I think weighted GPas can be good, but I think what my school has done is overkill. The difference between me and number 52 is about six points on an english test we had last year.

I have a questoin: Since valedectorian can't be calculated until end of senior year, does it even affect college applications? And for class rank do they take into account how you are doing in the beginning of senior year?

[quote="h78reg"]4.0 scale for remedial classes, 5.0 for accelerated, and 6.0 for honors. the highest grade you can get is 6.33 for an A+ in an honors class.(AP's are honors classes)[/quote]

 :!:  :!: Wow!  My school is apparently 'old-school'- normal (non-AP, non-honors) classes are on 4.0 scale, AP and honors classes are on 5.0 .  In addition, A-, B-, etc count for less than A (I think that A- == 3.75).  Unfortunately, there are no bonuses for A+ .  :(

School rank is calculated with the weighted GPA, culumative from 9th grade.

My school, or at least the class of 2007, is freakishly competitive: my current cumulative weighted GPA is 4.495 (all A's except for 2 A- 's) and my class rank is 7th of 83.  For comparison, they gave us a list of all graduates with information including class ranks and GPA; the average GPA of the person ranked #1 is 4.51.


Fie:

$S^`=\{ (x,y,z,t) \ \&\mbox{\mid}x+y-z-t=0\}$             $\ \& \mbox{si}$
$S^=\{ (x,y,z,t) \ \&\mbox{\mid}x-y-z+t=0\}$

doua subspatii in $R^4$ ; determinati intersectia lor si demonstrati identitatea:

$S^`+S^=R^4$


When 39,500 is divided by an integer n, the quotient is 123 and the remainder is 17. What is n? :mad:

If the quotient is 123 and the remainder is 17, we can set up the equation $ 123n \plus{} 17 \equal{} 39500 \implies n \equal{} \boxed{321}$.


Can anyone please posst the link to the 2008 USAPho Quarterfinal Exam solutions, as I've been searchinf for it desperately for the past few days.

I've never seen the solutions to any round actually posted anywhere (except the IPhO).  All I've ever seen is the option to purchase old tests from AAPT (the group that puts out the test).  Do you have a link to the actual test itself because I would love to see it.


I'm stuck in this problem from kalva, please help!
For each non-empty subset of {1, 2, 3, 4, 5, 6, 7} arrange the members in decreasing order with alternating signs and take the sum. For example, for the subset {5} we get 5. For {6, 3, 1} we get 6 - 3 + 1 = 4. Find the sum of all the resulting numbers.

Sorry, i've found the solution in kalva, thanks for paying attention.
( I don't notice the link answer in the bottom of page, so i make this mistake, sorry)


Please could anyone explain to me what is meant by the expression:

(-) as in:     1 (-) n

The full formula I'm looking at is:

1 (-) n [Root Symbol] s/c

(where s/c is 's divided c', beneath the root symbol)

Thanks in advance.

Anyone?

This must be really easy for you guys on here.

I've never seen such a thing before...
Do you have a picture¿

Cheers for the interest, ZetaX.

I've got the answer I require now.

It means calculate everything after the (-) first, then take the 1 away from the result.

I would have thought that the correct layout for this formula should be:

1 - (n [root symbol]s/c)

The formula is for working out a % Depreciation Rate for Fixed Assets in business.

It's come up in a Bookkeeping course I'm studying, and it just didn't make sense.

I think maybe it's a typo! lol

Thanks anyway. I'll leave you to your higher maths.


.( Wolstenholme’s theorem )Prove that  if
     $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{p-1}$. Is expressed as a fraction ,where $p\geq 5$ is a prime ,then $p^2$ divises the numberator.

Ok, after this post : http://www.mathlinks.ro/Forum/viewtopic.php?p=513197#p513197, and the following one by [b]t0rajir0u[/b], I wanted to post this exercise :
Let $p$ be a prime, $p \ge 3$.
1) Show that the sum of elements of $(\mathbb{Z}/p\mathbb{Z})^*$ is zero.
We let $1 + \frac 12 + \ldots + \frac 1{p-1} = \frac{A_p}{(p-1)!}$. Show that $A_p$ is divisible by $p$.
2) Show that the binomial coefficient $\binom {2p}p$ is congruent to 2 modulo $p^2$.

:)

1) just "join" the k-th term with the (p-k)-th term to obtain that it is divisible by $p$. it is much more harder to prove that it is divisible by $p^2$.

2) it follows from the wonderful identity
${{2p}\choose p}={p\choose 0}^2+{p\choose 1}^2+...+{p\choose p}^2$

I think this problem can uses Larange theorem

What is Larange theorem¿
When it means Lagrange, then I just know some other things under that name...

If $p$ is prime and $\sum_{k=1}^{p}\frac{1}{k}=\frac{r}{ps}$ then $r-s$ divise by $p^3$


MATHCOUNTS
modular arithmetic
AMC
AIME
AMC 8
AMC 10

[quote="gauss1181"]whats the highest possible score for AMC 8?[/quote]
25 out of 25

[quote="Harini"]I'm planning to take the AMC 8, and I want to score 24+. I was wondering if anyone made any mock AMC 8's so I could practice. Also, does anyone know if there are past AMC 8 exams on AOPS? And are there other problems that are of similar difficulty so I can practice? Thanks.[/quote]

im taking the AMC 8 this year as well. (8th grade)there's no better way to practice than doing old tests--you can order them through AMC. [url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/2007-Pub,Revision.pdf[/url]

i would recommend chapter (maybe even school) mathcounts competitions for material practice..but each different type of competition requires a different strategy of timing and pacing yourself. the time allotted for both competitions varies greatly.

Here's number 23 from the AMC 8 brochure; it's incredibly easy:
A box contains gold coins. If the gold coins are equally divided among six people, 4 coins are left over. If the coins are equally divided among 5 people, 3 coins are left over. If the box holds the smallest amount of coins that meets these 2 conditions, how many coins are left when equally divided among 7 people?
[hide]
$ x\equiv 4 \pmod{6}$
$ x\equiv 3 \pmod{5}$
$ x\plus{}2\equiv 0 \pmod{6}$
$ x\plus{}2\equiv 0 \pmod{5}$
$ x\plus{}2\equal{}3\cdot2\cdot5\equal{}30\implies x\equal{}28\implies\boxed{0}$

 [/hide]

MOEMS problems are generally good practice too...
but I agree that the school target and chapter sprint+target would be best.

You can buy old tests on the AMC website.

If someone can score like 22 or 23 at least in practice, then why wouldn't that person practice something harder or more challenging? Like AMC 10, or something

There's a BIG jump from the 8 to the 10/12.
I almost got perfect on AMC 8 (23, 2 silly mistakes on 7 and 13), but got owned in the 10. AMC 8 is just basically an introduction, while the 10/12 actually go a bit deeper in to problem solving.

Yes; slightly deeper. I got owned on the 10 too. =D Wow you actually still remember what problems were which.

I don't think it's necessarily such a big jump, but there are just some I had no idea how to do on the 10, especially the harder ones.

Oh, ok. I checked the Olympiad Resources and there aren't any AMC 8 problems posted. So I guess the only way to practice is through buying the old contests through the link... thanks!

I guess I'll make an AMC 8 Mock sometime later.

I didn't find the jump from the AMC 8 to 10 particularly big; In 8th grade, I got a 24 on the AMC 8, and then a 138 on the AMC 10 (two missed questions). I did, however, get completely owned on the AIME that year, though.


Find all $f: R \to R$ such that $f(x) \geq 0$ for all $x$ and

$f(x+y)+f(x-y)-2f(x)-2y^{2}=0.$

Let $g(x)=f(x)-x^{2}-f(0)$, then $g(x+y)+g(x-y)=2g(x),g(0)=0$.
It is known, that if g(x) is continiosly then $g(x)=ax$, therefore $f(x)=x^{2}+ax+b$.


inequalities
limit
integration
real analysis
real analysis unsolved

Let $ n$ be a positive integer. Prove that
$ \sum^n_{i\equal{}1}\sqrt{\frac{i}{n\plus{}1\minus{}i}} \le \frac{n \pi}{2}.$
Can $ \frac{\pi}{2}$ be replaced by a smaller constant?

No, a smaller constant is impossible.
\[{ \lim_{n\to \infty}\frac {1}{n}\sum_{i = 1}^n \sqrt {\frac {i}{n + 1 - i}} = \lim_{n\to \infty}(\sum_{i = 0}^{n - 1}\frac {1}{n}\sqrt {\frac {i}{n - i}} + \frac {1}{\sqrt {n}}\sum_{i = 0}^{n - 1}\frac {1}{n}\sqrt {\frac {1}{1 - \frac {i}{n}}}}) = \int_0^1 \sqrt {\frac {x}{1 - x}}dx = \frac {\pi}{2}\]


Given $3 = \sqrt{a}+\frac{1}{\sqrt{a}}$ where $a \neq 0$, find $a-\frac{1}{a}$.

[quote="Hoborobo"][hide]$3^{2}=(\sqrt{a}+\frac{1}{\sqrt{a}})^{2}$
$9=a+2+\frac1a$
$49=a^{2}+2+\frac1{a^{2}}$
$45=a^{2}-2+\frac1{a^{2}}$
$\sqrt{45}=a-\frac1a$[/hide]

amirite?[/quote]
[hide]How did you get from
$9=a+2+\frac1a$
to
$49=a^{2}+2+\frac1{a^{2}}$
Squaring doesn't do that. ANyways, solving for a will definitely work but notice that
$a-\frac{1}{a}=\frac{a^{2}-1}{a}$...
[/hide]
I have a strong feeling that this can be done with only algebraic manipulations, though it seems pretty tough...

[quote="Hoborobo"][hide]$3^{2}=(\sqrt{a}+\frac{1}{\sqrt{a}})^{2}$
$9=a+2+\frac1a$
$49=a^{2}+2+\frac1{a^{2}}$
$45=a^{2}-2+\frac1{a^{2}}$
$\sqrt{45}=a-\frac1a$[/hide]

amirite?[/quote]
That answer's right it's just not simplified. However, your work doesn't seem to match your answer :rotfl: . @bpms, yes there's a pretty nice algebraic manipulation for this problem. Think manipulations of squares.

[quote="k8reindeer"][hide][quote="bpms"]
How did you get from
$9=a+2+\frac1a$
to
$49=a^{2}+2+\frac1{a^{2}}$
Squaring doesn't do that. I have a strong feeling that this can be done with only algebraic manipulations, though it seems pretty tough...[/quote]
his answer is sound.
$9=a+2+\frac1a$
$7=a+\frac1a$
square both sides
$49=a^{2}+2+\frac1{a^{2}}$
[/hide][/quote]
Oh never mind, I wasn't thinking :|

[quote="NeverOddOrEven"][quote="Hoborobo"][hide]$3^{2}=(\sqrt{a}+\frac{1}{\sqrt{a}})^{2}$
$9=a+2+\frac1a$
$49=a^{2}+2+\frac1{a^{2}}$
$45=a^{2}-2+\frac1{a^{2}}$
$\sqrt{45}=a-\frac1a$[/hide]

amirite?[/quote]
That answer's right it's just not simplified. However, [b]your work is totally wrong[/b] :rotfl:.[/quote]

How so?

oh nvm his work didn't seem to work the way i saw it at first but nvm.


algorithm
game
combinatorics
game strategy
IMO Shortlist

Two players play alternately on a $ 5 \times 5$ board. The first player always enters a $ 1$ into an empty square and the second player always enters a $ 0$ into an empty square. When the board is full, the sum of the numbers in each of the nine $ 3 \times 3$ squares is calculated and the first player's score is the largest such sum. What is the largest score the first player can make, regardless of the responses of the second player?

[hide="Official Solution"]We first prove that the second player can keep $ A\le 6$. Place 10 dominoes on the board as shown in the diargam. Whenever the first player marks a 1 on a cell of a domino, the second player marks a 0 on the other cell of the same domino if it is still empty. Since any $ 3\times 3$ square contains 3 complete dominoes, there are at least 3 0's inside. It follows that $ A\le 6$.

We now prove that the first player can make $ A\equal{}6$. Label the board $ (a,b,c,d,e)\times (1,2,3,4,5)$. Start by marking a 1 on c3. By symmetry, we may assume that second player's response is made on row 4 or 5. Disregard these two rows from now on. The first player then marks a 1 on c2. If the second player does not mark a 0 on c1, the first player will mark a 1 there in the next turn. Then either $ (a,b,c)\times(1,2,3)$ or $ (c,d,e)\times (1,2,3)$ contains 3 1's and no 0's. The first player can put three more 1's in that $ 3\times3$ square. If the second player's second move is on c1, the first player can guarantee 2 1's in $ (b)\times (1,2,3)$ as well as 3 1's in $ (a,d)\times(1,2,3)$. Hence either $ (a,b,c)\times(1,2,3)$ or $ (b,c,d)\times(1,2,3)$ will contain at least 6 1's. It follows that we can have $ A\equal{}6$.[/hide]

I have a slightly different solution...

As in the first solution, a tiling shows that the second player can keep the score at $6$, so it remains to prove that the first player can guarantee a score of $6$. The numbers in the unit squares below represent the number of $3\times3$ squares that they belong to. After the game is over, define $V$ to be the sum over all squares of the values of the squares times the numbers on the squares (i.e. each square contributes its value if it has a $1$ on it and $0$ otherwise). This is clearly equal to the sum of the sums of the $1$'s in the $3\times3$ squares.

Let the first player play on C3 and the second player play, without loss of generality, in the first two rows. Now let the first player play on C4. If the second player does not play on C5, then at least one of the $3\times3$ squares with centers B4 and D4 will be void of $0$'s, so the first player can get a score of $6$ by repeatedly playing in that square. Now assume the second player does play on C5, which has a value of $3$. If the second player's first move was not on C2 (we assumed it was in the first two rows, so this means the first move had value at most $4$), then by the greedy algorithm the first player can guarantee
\begin{align*}
V\ge9(1)&+4(0)+6(1)+3(0)+6(1)+6(0)+6(1)+4(0)+4(1)+4(0)\\
&+3(1)+3(0)+3(1)+2(0)+2(1)+2(0)+2(1)+2(0)+2(1)\\
&+2(0)+2(1)+1(0)+1(1)+1(0)+1(1)=47>9\cdot5,
\end{align*}which means at least one $3\times3$ square has to have $6$ ones at the end. So assume the second player's first move was on C2, and let the first player play on B3. If the second player doesn't move on D3, then again a greedy algorithm will guarantee that the first player has $6$ ones in some square since here the second player's move has value at most $4$ so
\begin{align*}
V\ge9(1)&+6(0)+6(1)+3(0)+6(1)+4(0)+6(1)+4(0)+4(1)+4(0)\\
&+3(1)+3(0)+3(1)+2(0)+2(1)+2(0)+2(1)+2(0)+2(1)\\
&+2(0)+2(1)+1(0)+1(1)+1(0)+1(1)=47>9\cdot5,
\end{align*}But if the second player moves on D3, then the $3\times3$ square with center B4 will have at least $6$ ones if the first player keeps playing in that square, so we are done.

This one was a bit messy for C1
The part that maximum is smaller than 7 was too easy but the proof for 6 was nasty


algebra
polynomial
quadratics
trigonometry
calculus
integration
quadratic formula

Decompose the polynomial:
\[ \ x^{2m}-2|a|^{m}x^{m}cos(m\theta)+a^{2m} \]
in a product of $m$ polynomials of second degree with real coefficients.

We use the quadratic formula to find that the polynomial is zero when \[x^{m}= |a|^{m}(\cos ( m\theta ) \pm i \sin( m\theta))\] which clearly occurs when $x = |a| \left[ \cos \pm \left( \theta+\frac{ 2k\pi }{ m }\right)+i\sin \pm \left( \theta+\frac{ 2k\pi }{ m }\right) \right]$, for all integral $k$.

Hence

\begin{eqnarray*}x^{2m}-2|a|^{m}x^{m}\cos ( m \theta )+a^{2m}& = & \prod_{k=0}^{m-1}\left[ x-|a| \mbox{ cis}\left( \theta+\frac{ 2k\pi }{ m }\right) \right] \left[ x-|a| \mbox{ cis}-\left( \theta+\frac{ 2k\pi }{ m }\right) \right] \\ & = & \prod_{k=0}^{m-1}\left( x^{2}-2 |a| \cos \left( \theta+\frac{2k\pi}{m}\right) x+|a|^{2}\right) \end{eqnarray*}

Q.E.D.


I would like to know if there is a seperate quota for class 12 students in the RMO also ,like it is there in the INMO where 6 seats are reserved for class 12 students.
Has anybody seen this years RMO paper. how much is likely to be the cutoff for general students and for class 12 students {if any} i have answered abuot 3 to 4 questions , and i am in class 12 , do i stand anywhere near clearing the examination?

please inform me as soon as possible.

You need atleast 4 full questions and an attempt at a fifth.
 Doesn't make sense to me to have a cutoff just for 12th.
 If they intend a cut-off it should be the same for 11th
 and 12th as they are almost at te same level of mathematics.
 there isn't that much one learns in twelfth that will help him do much better than he did the previous year.

As far as I know a guy from 12th with 5 problems has qualified.
I do not know anyone with lesser problems.

P.S. I got six and qualified.

well i dont know about your region .. but results for delhi region are out .. within 2 weeks!!!!
im in...
generally its believed tat abt 6 from class 12 are slected but the number may exceed depending upon their performance... n yeah they mit hav a sepert benchmark...
i dont know abt qualifying marks in other regions coz the trend mit be different,,, here in delhi for most ppl... 2- 2.5 questions is the required benchmark.. though it mit be higher for class 12th students... but as far as i ve seen if youve done 3-4 ... "prorply"... ul be defn in...
gud luk :)

the results are out in mumbai region and I think that probapbly 5 to 5.5 problems is the cutoff here for class 12 people because I and two of my friends managed to get 4 problems right and had possibly gone in the right direction for the fifth one.  yet we could not qualify

[quote]I got six and qualified[/quote]

Good.I suppose i can see u in next year's camp.

Udi

last year only one from entire u.p (agra) got thru. he got 6 correct. yet in nsec and nsep 12 and 16 got in.i got selected in nsec

well acc to me this yrs delhi RMO paper was quite easy => the cutoff is/should have been higher. one of my friends got all but one correct he stood first.

anant - where did you get the results of this year's nsec from?

In RMO 2006(Delhi) 35 students were selected for INMO. I myself got selected. The maximum no. of students of class 12 is [b]6[/b] among these selections as told by the officials. The cut-off marks were around 21 out of 70

rmo results maharashtra were declared quite late on 16 jan...i got 43 out of 100,think i had done 4.5 probs....last year i had qualified wid same marks n was 17th..this year i m 5 th.....think the ques ppr levels are being increased substantially

Could you please put up the questions in Maharashtra RMO 2007?

are the results of the rmo 07 xam available on d net? i dont think our school has received d results? do v get to know our marks even if we didnt qualify???


geometry
ratio
geometric transformation
homothety
rectangle
Asymptote
similar triangles

[hide]Extend $ PA$ over $ A$ to $ P'$ such that $ AP'\equal{}AP$. Since $ \angle CAP\equal{}90^\circ$, the triangle $ \triangle CPP'$ is isosceles, hence similar to isosceles $ \triangle OPA$ (which it shares an angle with). Therefore the line $ PD$, by homothety, images the midpoint of $ OA$ into the midpoint of $ CP'$ - call that last point $ E$. We conclude that $ PE$ and $ CA$ are the medians of $ \triangle CPP'$, hence $ AD: DC\equal{}1: 2$[/hide]

Nice proof! :)

I was trying to prove $ AMD\sim COD$, where $ M$ is the midpoint of $ AO$. Can anyone come up with a proof of this?

[quote="i_like_pie"]Triangle $ ABC$ is inscribed in circle $ O$ with diameters $ AB$ and $ CP$. The line connecting $ P$ with the midpoint of $ AO$ intersects $ AC$ at $ D$.

Determine, with proof, the ratio $ AD: DC$.[/quote]
Nice solution, Farenhajt!  I have a different one.
[hide="Different Soluion"]
Draw QR parallel to AC through M (so that Q is on AP and R is on PC).  Notice that triangles AQM and APB are similar as are OMR and OAC.  So, we have that $ \frac {MR}{AC} \equal{} \frac {OM}{OA} \equal{} \frac {1}{2}$, so $ 2MR \equal{} AC$.  Also, $ \frac {QM}{PB} \equal{} \frac {AM}{AB} \equal{} \frac {\frac {AO}{2}}{2AO} \equal{} \frac {1}{4}$, so $ 4QM \equal{} PB$.  Also, notice that triangles PMQ and PDA are similar as are triangles PMR and PDC.  So, $ \frac {QM}{AD} \equal{} \frac {PM}{PD} \equal{} \frac {MR}{CD}$, so $ \frac {AD}{CD} \equal{} \frac {QM}{MR} \equal{} \frac {\frac {PB}{4}}{\frac {AC}{2}} \equal{} 2(\frac {PB}{AC})$.  Since ACBP is a rectangle, $ AC \equal{} PB$, and thus $ \frac {AD}{CD} \equal{} \boxed{\frac {1}{2}}$.
[/hide]

[quote="i_like_pie"]I was trying to prove $ AMD\sim COD$, where $ M$ is the midpoint of $ AO$. Can anyone come up with a proof of this?[/quote]

I'm pretty sure that it is not true.

[asy]import cse5; pair O=MP("O",D((0,0))), A=MP("A",D(3*dir(180))), B=MP("B", D(3*dir(0))), C=MP("C", D(3*dir(130))), P=MP("P",D(3*dir(310))), M=MP("M",D((A+O)/2)), D=MP("D",D((2*A+C)/3)); D(CR(O,3)); D(A--B--C--cycle); D(C--P--D--O); D(A--P);[/asy]

[hide="Solution to problem"]Mass points! (Menelaus Theorem)

Place a mass of 2 on C. Then there is a mass of 2 on P, so a mass of 2+2=4 on O. But AM=MO, so there is a mass of 4 on A. Since there is a mass of 4 on A, 2 on C, $ 4 AD \equal{} 2 DC$, so $ AD/DC \equal{} 1/2$.[/hide]

EDIT: Mistake fixed.

[quote="vishalarul"][quote="i_like_pie"]I was trying to prove $ AMD\sim COD$, where $ M$ is the midpoint of $ AO$. Can anyone come up with a proof of this?[/quote]

I'm pretty sure that it is not true.[/quote]
After proving $ AD: DC \equal{} 1: 2$, the facts that $ AM: CO \equal{} 1: 2$ and $ \angle A\equal{}\angle C$ can be used to show $ \triangle AMD\sim\triangle COD$.

[quote="vishalarul"][b]Place a mass of 2 on C[/b]. Then there is a mass of 2 on B, [b]so a mass of 2+2=4 on C[/b].[/quote]
:huh:

This is pretty trivial because you just have $ \triangle APC$, $ O$ is the midpoint of $ CP$, $ M$ is the midpoint of $ AO$, and you extend $ PM$ to hit $ AC$ at $ D$. This is a simple ratio/concurrency problem...which can be done with either mass points or ceva/menelaus [you would have to extend CM].

[hide="straightfoward, compare areas solution"]$ E: \equal{} PD \cap AB$. From "ratio of bases $ \implies$ ratio of areas", compared to the $ [ACP]$.

$ [APB] \equal{} [ACP]$ because $ APBC$ is a rectangle, (angles subtend an arc of a diameter)
$ \implies EBP \equal{} \frac {3}{4}[ACP]$.

$ [ADE] \equal{} \frac {1}{9}[EBP]$, from similar triangles, $ \implies \equal{} \frac {1}{12}[ACP]$.

$ [AEC] \equal{} \frac {1}{4}[ACP]$.

Thus, $ [EDC] \equal{} (\frac {1}{4} \minus{} \frac {1}{12})[ACP] \equal{} \frac {1}{6}[ACP]$.

$ AD: DC \equal{} [ADE]: [EDC] \equal{} 1: 2$.[/hide]

Nice proof! Pretty difficult to come up with, and yet so simple!

[quote="Mandate Of Heaven"]$ E: \equal{} PD \cup AB$[/quote]
I think you meant $ E \equal{} PD\cap AB$.

Nice problem.  Here is my solution which I don't think anyone has mentioned yet.
[hide="pardon the image"][img]11092[/img][/hide]
[hide="solution"]First of all, $ \angle PAC \equal{} \angle ACB \equal{} 90^{\circ}$, so $ ACBP$ is a rectangle and $ AC \equal{} PB$.
Now we draw point $ N$ as the midpoint of $ OB$, so $ AM: MN \equal{} 1: 2$, or $ AM: AN \equal{} 1: 3$.
Draw $ NC$.  Now, we have $ \angle ONC \equal{} \angle OMP$, because they intersept $ AC$ and $ PB$, which are equal.  Thus, $ MP \parallel{} NC$, which means $ MD \parallel{} NC$.  Thus, $ \triangle AMD$ is similar to $ \triangle ANC$.
Similarity preserves ratios, so since $ AM: AN \equal{} 1: 3$, we must have $ AD: AC \equal{} 1: 3$, or $ AD: DC \equal{} 1: 2$.
[/hide]

[quote="archimedes1"]$ MD \parallel{} OC$[/quote]
$ P\equal{}MD\cap OC$

[quote="i_like_pie"][quote="archimedes1"]$ MD \parallel{} OC$[/quote]
$ P \equal{} MD\cap OC$[/quote]
Oops.  I miswrote one of the sides, edited.

[quote="i_like_pie"][quote="archimedes1"]Thus, $ \triangle AMD$ is similar to $ \triangle AOC$.[/quote]
Not trying to be annoying or anything, but... $ \triangle AMD\sim\triangle ANC$[/quote]
lol... You know what I mean  :oops:


I just heard that I got into Caltech and I am really, really excited. But they also said that they will be checking final grades at the end of the year. So I have three questions.

1. I have like a 92 GPA (on a scale of 0-100 obviously), which I guess is not extraordinarly excellent. What (minimum) GPA do you think they would expect me to maintain for the third and fourth quarters?

2. If I get a good score on the USAMO, would that count as a buffer?

3. I have some teachers that try to make me work as hard as possible and will give us much worse grades than we deserve so that we don't "slack off" senior year. Does anyone else have teachers like that, and what are colleges' stances towards them?

I don't plan on completely slacking off these two quarters, but I most likely will slack off a little bit. I just want to know where my absolute limits are.

well, you have already obviously proved yourself to the college, so if you were to slack off a little it would be ok. just keep doing what you've been doing for the last 11.5 years and you'll be fine, even if your grade slips a few points. just don't stop working or anything drastic like that.
and if you were to do really poorly for some reason, i doubt a college wants to hear the excuse that you had a hard and crazy teacher.

Yeah, "keep up the good work" is the basic idea. Just try to emphasize LEARNING as much as you can from your current high school classes, and that should be all right.

tokenadult the problem with me is that i feel that if i want to learn i have to fail some classes (well not fail, but not get all A's, not study much and not doing homework; in my case not math classes but sciences especially) and dedicate some time to real math or the things i like in order to learn something... what do you think.. ihave tried to explain this to my teachers but they say I have to be "all around" and they dont see that (at least in my school) school doesn't make you all around at all.. they present like a false picture of what mathematics, science and humanities and arts are.. and by doing math i learn the correct attitude toward a field and i become more "all around" becuase i respect what a scientist, writer, or artist does..
but also i skip many classes and homeworks because not doing just math, but im very dedicated to tennis and to music... and to be succesful in these fields that i love i have to ignore school..

and especially in this semester, although i already got into a good universities, some of my grades are low because im dedicating much time to math...

what do you think tokenadult or what do you recommend? do you think mit will see my grades this semester?
if i finish with all B's except math, will that bother?

thanks!

[quote="manuel"]tokenadult the problem with me is that i feel that if i want to learn i have to fail some classes (well not fail, but not get all A's, not study much and not doing homework; in my case not math classes but sciences especially) and dedicate some time to real math or the things i like in order to learn something.[/quote]

Oh, I know exactly what you mean. That was the story of my life in high school. I read, read, read about the subjects that interested me, and did schoolwork only sporadically. Much of the assigned schoolwork in my high school was quite pointless. I think Richard Rusczyk (who is a little younger than I am, and grew up in a different part of the country) was a little bit more shrewd in high school, because he maintained high grades in high school (VERY high, in his case), got into a great college, and then had credentials to get a great job after college.

You have been admitted to a great college, and if you pursue your main area of interest--mathematics--diligently, and do enough to keep up with the required homework in those other classes, I expect you will keep your offer of admission. Getting a C might be dangerous, but getting two or three Bs should be reasonable as long as you use your time to pursue learning you can demonstrate to other people. High school is over soon for you--don't give up completely when you are this close to the finish line, but move down the path that fits your future.

[quote="Philip_Leszczynski"]I just heard that I got into Caltech and I am really, really excited. But they also said that they will be checking final grades at the end of the year. So I have three questions.

1. I have like a 92 GPA (on a scale of 0-100 obviously), which I guess is not extraordinarly excellent. What (minimum) GPA do you think they would expect me to maintain for the third and fourth quarters?[/quote]

Somewhere between 75-80.  Don't sweat it, and don't get Cs.

[quote]2. If I get a good score on the USAMO, would that count as a buffer?[/quote]

Yes, but it won't be a problem in the first place.  Don't sweat it.

[quote] 3. I have some teachers that try to make me work as hard as possible and will give us much worse grades than we deserve so that we don't "slack off" senior year. Does anyone else have teachers like that, and what are colleges' stances towards them?

I don't plan on completely slacking off these two quarters, but I most likely will slack off a little bit. I just want to know where my absolute limits are.[/quote]

Those kind of teachers are misanthropic idiots.  Don't get Cs and you'll be fine.

thanks tokenadult... you understand me! :)


Prove that if   3x^2 +x= 4y^2 + y, than x-y is a perfect square

$ \Longrightarrow (x\minus{}y)(3(x\plus{}y)\plus{}1)\equal{}y^2$
We prove that $ \gcd(x\minus{}y,3(x\plus{}y)\plus{}1)\equal{}1$
Let $ d\equal{}\gcd(x\minus{}y,3(x\plus{}y)\plus{}1)$ then $ d|y,d|x\minus{}y$ so $ d|x$
$ \Rightarrow d|3(x\plus{}y)$ so $ d|1$
$ \Rightarrow d\equal{}1$
Imply that $ |x\minus{}y|$ is a perfect square.


Well, I'm taking the AMC8 and AMC10 next year since I'm in 7th grade. I also want to take the AMC12 but my school wont let me take all 3.

Anyone know how I can take it somewhere else?

The AMC 12B is offered at some colleges/universities each year. Check back next year for a list.


abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

If this is valid I would like to see a proof of it if that is possible.

Let M and N be modules that are isomorphic.  Prove that $End(M\oplus N)$ is isomorphic to ${L_{2}}(End(M))$

I think I fixed everything.  This is the 2 x 2 matrix L.

Could you please explain what $N(A)$ stands for?

Let $M$ be a $R$-module. By the universal properties of direct sum and product, there's an isomorphism of $R$-modules

$\text{End}(M^{n}) = \text{Hom}(M^{n},M^{n}) \cong \prod_{n}\text{Hom}(M,M^{n}) \cong \prod_{n}\prod_{n}\text{Hom}(M,M) \nolinebreak \cong \nolinebreak \text{End}(M)^{n \times n}$

The isomorphism is given by

$\alpha : \text{End}(M)^{n \times n}\to \text{End}(M^{n}) , (f_{ij})_{1 \leq i,j \leq n}\mapsto \left((x_{k})_{1 \leq k \leq n}\mapsto \left(\sum_{m=1}^{n}f_{km}(x_{m}) \right)_{1 \leq k \leq n}\right)$

An easy computation shows, that $\alpha$ is multiplicative. So, $\text{End}(M)^{n \times n}$ and $\text{End}(M^{n})$ are isomorphic als $R$-algebras.


algebra
polynomial
superior algebra
superior algebra unsolved

Given a field $ k$ and a ring $ R$ such that $ R$ is finitely generated by elements $ \theta_1,...\theta_n$ with coefficients in $ k$, then $ R$ is the quotient $ k[x_1,...x_n]/I$ for some ideal $ I$.

Is that true?

Yep.

Well, you should state the relationship between $ k$ and $ R$ a little more clearly.  Specifically, you want $ R$ to be finitely generated as a $ k$-algebra.  If you think about it, it's exactly the same as saying that $ R$ is the image of some assignment homomorphism $ f: k[x_1,\ldots x_n] \to R$.


[quote="mdk"]Find $ \sum_{j\equiv 0(mod 3)}^{}C(n,j)$[/quote]
[hide="hint"]Third Roots of Unity[/hide]

Im new to roots of unity, but heres my try:
[hide]
Let $ \omega$ be a third root of unity not equal to 1 such that $ \omega^{3}\equal{}1$ and $ \omega^{2}\plus{}\omega\plus{}1 \equal{} 0$

$ (1\plus{}1)^{n}\equal{}\binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n}{3}\plus{}\binom{n}{4}\plus{}\binom{n}{5}\plus{}...$
$ (1\plus{}\omega)^{n}\equal{}\binom{n}{0}\plus{}\omega\binom{n}{1}\plus{}\omega^{2}\binom{n}{2}\plus{}\binom{n}{3}\plus{}\omega\binom{n}{4}\plus{}\omega^{2}\binom{n}{5}\plus{}...$
$ (1\plus{}\omega^{2})^{n}\equal{}\binom{n}{0}\plus{}\omega^{2}\binom{n}{1}\plus{}\omega\binom{n}{2}\plus{}\binom{n}{3}\plus{}\omega^{2}\binom{n}{4}\plus{}\omega\binom{n}{5}\plus{}...$

So $ (1\plus{}1)^{n}\plus{}(1\plus{}\omega)^{n}\plus{}(1\plus{}\omega^{2})^{n}$
$ \equal{} 3\binom{n}{0}\plus{}(\omega^{2}\plus{}\omega\plus{}1)\binom{n}{1}\plus{}(\omega^{2}\plus{}\omega\plus{}1)\binom{n}{2}\plus{}3\binom{n}{3}\plus{}(\omega^{2}\plus{}\omega\plus{}1)\binom{n}{4}\plus{}(\omega^{2}\plus{}\omega\plus{}1)\binom{n}{5}\plus{}...$
$ \equal{}3\binom{n}{0}\plus{}3\binom{n}{3}\plus{}3\binom{n}{6}\plus{}...$

Divide this by three,
$ \sum_{j\equiv0(mod3)}\binom{n}{j}\equal{}\frac{2^{n}\plus{}(1\plus{}\omega)^{n}\plus{}(1\plus{}\omega^{2})^{n}}{3}$
$ \equal{}\frac{2^{n}\plus{}(\minus{}\omega)^{n}\plus{}(\minus{}\omega^{2})^{n}}{3}$
This is all I can simplify, though...
[/hide]


calculus
integration
trigonometry
calculus computations

$ \int \frac{x \cos x}{x \sin x \plus{} \cos x}\ dx$.

$ \frac{d}{dx}(x \sin x \plus{} \cos x) \equal{} \sin x \plus{} x \cos x \minus{} \sin x \equal{} x \cos x$. Therefore,

$ \int\frac{x\cos x}{x\sin x\plus{}\cos x}\ dx \equal{} \ln|x \sin x \plus{} \cos x| \plus{} C$.


i am in a card board box.  somebody puts four different drinks into a different cardboard box: generic lemon-lime soda, generic diet lemon-lime soda, generic lime-lemon soda, and diet generic lime-lemon soda.  I cannot see, taste, smell, touch or hear(?) the soda. somehow, i know exactly what order they are in.  how can i tell?

[hide]Someone must have told you already[/hide]

umm... [hide] is the answer that the sodas are the same, the 1st and 4th, and the 2nd and 3rd. (these are the ones that are the same) and the order can be determined by how heavy they make you? (no taste required)
[/hide].

IME, I think that diet sodas float (in ice water, anyway) while non-diet sodas sink.

mustafa was right  :D  it was a stupidly easy one.


Ring Theory
superior algebra
superior algebra unsolved

Let $ R\neq 0$ be a commutative ring. Let $ R^\star$ denote the set of units of $ R$.

(1) Show that $ R^\star \equal{}R\setminus \bigcup_{M \mathrm{~maximal~ideal}} M$

(2) Show that $ R$ is local if and only if $ R\setminus R^\star$ is an ideal.

Explainatory sketches:

(1) Use that any ideal $ I \subsetneq R$ is contained in a maximal ideal. Especially, $ (a)$ and thus $ a$ is in $ \bigcup_{M \mathrm{~maximal~ideal}} M$ iff $ (a) \neq R$. And the latter happens iff $ a$ is not a unit.

(2) $ R$ is local iff there is only one maximal ideal iff the union of all maximal ideals is a maximal ideal iff the union of all maximal ideals is an ideal.

Does this mean that (2) can be generalized to say that $ R$ is local if and only if $ R\setminus R^\star$ is a maximal ideal?

Yes, exactly, in the local ring $ (R,m)$ $ R\minus{}m$ are units.


algebra
polynomial
function
domain
superior algebra
superior algebra unsolved

I'm studying for my exam and have some questions which I am completely lost with. They're from another year's exam.
I'll post the questions here instead making topics for each one :\

Consider the homomorphism $ f: \mathbb{Q}[x] \to \mathbb{Q}$ given by $ f(p(x)) \equal{} p(1)$. Does the Universal Property of Quotients allow f to induce a homomorphism $ \tilde f : \mathbb{Q}[x] / (x^2\plus{}1) \to \mathbb{Q}$?
I got no since there isn't a homomorphism from $ \mathbb{Q}[i] \to \mathbb{Q}$.
Same question but $ (x^2\plus{}1)$ becomes $ (x^2\minus{}1)$

4a) Let $ \alpha$ be a root of the polynomial $ x^3\plus{}x\plus{}1$ over $ \mathbb{Q}$ and let $ K \equal{} \mathbb{Q}(\alpha)$. Computer the degree $ [K: \mathbb{Q}]$.

6) Let $ K$ be the field $ \mathbb{C}(x)$ of rational functions of x with complex coefficients, and let $ L$ be the fraction field of the domain $ \mathbb{C}[x,y]/(y^2\minus{}x^3)$.
a) Find an element $ \alpha \in L$ such that $ \alpha^2\equal{}x$.  (I don't know if its right but I just trial and errored to find $ \left( \frac { x^2}{y} \right)^2 \equal{} x$ So I took that to be my $ \alpha$. If thats right is there a method to find it?
b) Find an isomorphism from $ K$ to $ L$.

9) Let $ I$ be a nonzero ideal of $ R\equal{}\mathbb{Z}[\sqrt{2}]$. Prove that $ R/I$ is finite.

10) Let $ L/K$ be an extension of fields with $ [L: K]\equal{}2$. Let $ f(x) \in K[x]$ be an irreducible polynomial of degree $ d$ with coefficients in $ K$. Prove that either $ f(x)$ is irreducible over $ L$, or else $ f(x)$ splits into two irreducible factors in $ L[x]$ of the same degree.

Please respond if you know any of these I am stressed over these questions. I can't do them at all :P
Thanks

4a) $ \mathbb{Q}(\alpha)\cong\mathbb{Q}[x]/(x^3\plus{}x\plus{}1)$. A basic theorem of commutative algebra says the degree is $ 3$ because thats the degree of the (monic) polynomial $ x^3\plus{}x\plus{}1$. (You can find it in one of the chapters of Eisenbud's book on commutative algebra, roughly speaking it's due to the Caley-Hamilton theorem).

6b) Here's a geometric interpretation (I'm not sure its right, I'm just throwing this out there): the curve $ y^2\minus{}x^3\equal{}0$ has a singularity at the origin and is smooth everywhere else (http://www.wolframalpha.com/input/?i=plot+y^2-x^3%3D0). If we delete that point we have an open set. Then projection onto the $ y$ axis gives a birational equivalence between the curve $ y^2\minus{}x^3\equal{}0$ with the affine line. This induces an isomorphism on their function fields which are $ Frac(\mathbb{C}[x,y]/(y^2\minus{}x^3))$ and $ \mathbb{C}(x)$ respectively.

@LKNsngth,

why is 4a related to Cayley-Hamilton? It's degree three because $ x^3\plus{}x\plus{}1$ is irreducible thus $ 1,x,x^2$ forms a basis.

9) Edit: Deleted Idea - pretty sure it wont work

@Soarer: Oh oops... I was confused sorry; we need Caley hamilton to prove a sort of a converse to what I said: namely that if $ R$ is a ring and $ J$ is an ideal of $ R[x]$, $ R[x]/J$ is finitely generated if and only if $ J$ contains a monic polynomial. We need Caley Hamilton for showing that $ R[x]/J$ is finitely generated, $ J$ contains a monic polynomial (we basically consider multiplication by $ x \plus{} J$ as an endomorphism to conclude that $ x \plus{} J$ satisfies a monic polynomial). Sorry about that, I really wasn't thinking when I posted that >.<

6a, actually a simpler way is to look at $ y^2 \equal{} x^3 \Rightarrow \left(\frac {y}{x}\right)^2 \equal{} x$.
The point is the ring is not integrally closed.

9. This is something true for Dedekind domains as well (norm of ideals)
A trick, pick $ a \plus{} b\sqrt {2} \in I$, $ a\plus{}b\sqrt{2} \neq 0$. Its norm $ a^2 \minus{} 2b^2$ is also in $ I$. But it is easy to show that $ \mathbb{Z}[\sqrt {2}]/(a^2 \minus{} 2b^2)$ is finite. $ \mathbb{Z}[\sqrt {2}]/I$ is a quotient of it, thus also finite.


They are given a circle and a point in the exterior of it.
Construct the tangent segments from the point to the circle, with ruler [u]only[/u]   :)

[asy]path unitcircle=E..N..W..S..cycle;
path halfcircle=N..E..S;
draw (unitcircle);
pair A = sqrt(3)/2 * N + 1/2 * W;
pair B = 1/2 * N + sqrt(3)/2 * W;
pair C = cos(0.4) * S + sin(0.4) * W;
pair X = 3.5*E;

pair AP = intersectionpoint (X--A, halfcircle);
pair BP = intersectionpoint (X--B, halfcircle);
pair CP = intersectionpoint (X--C, halfcircle);
pair Y =   intersectionpoint (A--BP, AP--B);
pair Z =   intersectionpoint (C--BP, CP--B);
pair TANG[] =   intersectionpoints ((2*Y-Z)--(2*Z-Y), unitcircle);

draw(X--A);
draw(X--B);
draw(X--C);
draw(A--BP);
draw(B--AP);
draw(B--CP);
draw(C--BP);
draw(TANG[0]--TANG[1]);

dot(A,red); dot(AP, red); dot(B,red); dot(BP,red);
dot(C,red); dot(CP,red); dot(Y, red); dot(Z, red);

draw(X--(1.2 * TANG[0] - 0.2 * X),blue);
draw(X--(1.2 * TANG[1] - 0.2 *X), blue);

dot (TANG[0]);
dot (TANG[1]);
dot(X);[/asy]


function
parameterization
analytic geometry
vector

What is the fastest way to compute the perpendicular distance between two different skew lines (lines on parallel planes) in 3d??

[hide="Outline of approach"]
The minimum distance between two [u]skew[/u] lines is the distance between the unique pair of parallel planes for which each contains one line.

Say the lines are parametrized as $ (l_x (t), l_y (t), l_z (t))$ and $ (m_x (t), m_y (t), m_z (t))$, where each component is a linear function of the parameter $ t$.

Now, let the equation of one plane be $ ax \plus{} by \plus{} cz \equal{} 1$ and $ ax \plus{} by \plus{} cz \equal{} d$ for the other. Plug in the coordinates of each line into its respective plane's equation to solve for $ a,b,c,d$, knowing that equality must arise for all $ t$. This gives the equations of the two planes.

Finally, find the perpendicular distance between the two planes, which is a fairly routine procedure (vectors help here).

** -[i]It is important that the lines are skew! Iff they are parallel, you will get the same plane for both: ie, d=1[/i]
[/hide]
[hide="Example"]
Say the lines are $ (t \plus{} 1,t,t)$ and $ (t,2t,t \plus{} 1)$, and let their respective planes be $ ax \plus{} by \plus{} cz \equal{} 1$ and $ ax \plus{} by \plus{} cz \equal{} d$.

From the first line, we have $ a(t \plus{} 1) \plus{} bt \plus{} ct \equal{} 1 \implies (a \plus{} b \plus{} c)t \equal{} 1 \minus{} a$. Since a multiple of $ t$ is a constant for all $ t$, that multiple must be $ 0$. Thus, $ a \plus{} b \plus{} c \equal{} 0$. Then, $ 1 \minus{} a \equal{} 0 \implies a \equal{} 1$. Therefore, $ 1 \plus{} b \plus{} c \equal{} 0 \implies c \equal{} \minus{} (b \plus{} 1)$.

Now, we move on to the second equation with the second point**. $ t \plus{} 2bt \minus{} (b \plus{} 1)(t \plus{} 1) \equal{} d \implies bt \equal{} b \plus{} d \plus{} 1$. Again, this must hold for all $ t$, so we must have $ b \equal{} 0$. This gives $ b \plus{} d \plus{} 1 \equal{} d \plus{} 1 \equal{} 0 \implies d \equal{} \minus{} 1$. Finally, recall that $ c \equal{} \minus{} (b \plus{} 1) \equal{} \minus{} 1$.

Thus, we must find the distance between the two planes $ x \minus{} z \equal{} 1$ and $ x \minus{} z \equal{} \minus{} 1$, which is routine, and we get $ D \equal{} \sqrt {2}$.

** - [i]If the two lines are parallel, the second equation will degenerate to something without $ t$ when you plug the points in[/i][/hide]

[hide="similar to the Zuton Force way"]
I just cut straight to vectors.

Take the cross product of the two vectors that represent the lines. This gives a perpendicular vector.
This vector is the normal vector to the plane containing one of the lines.

Dump it into point normal form for a plane using a point from the first line.

$ a(x\minus{}x_0)\plus{}b(y\minus{}y_0)\plus{}c(z\minus{}z_0) \equal{} 0$

From here, you have the equation of one plane, so select a point on the other line and use the formula for distance from a point to a plane

$ d \equal{} \frac{ax_0 \plus{} by_0 \plus{} cz_0 \plus{} d}{\sqrt{a^2\plus{}b^2\plus{}c^2}}$

tada!
[/hide]

Let $ \mathcal{L}_1$ pass through $ P_1$ and have a direction vector of $ v_1$.  Let $ \mathcal{L}_2$ pass through $ P_2$ and have a direction vector of $ v_2$. Then the vector $ v\equal{}v_1\, x \,v_2$ is perpendicular to both vectors. Now project $ P_1\minus{}P_2$ onto $ v$ to get the distance. The formula is easily given with dot products \[ \frac{ |(P_1\minus{}P_2)\cdot v|}{|v|}\]

For a sort of proof, consider the planes through the lines perpendicular to $ v$. The distance between the two lines is the distance from these two planes. If we take two points in these planes, and project it onto the vector perpendicular to these planes, we're done.


[quote="dondigo"]Point $P$ is inside a triangle $ABC$ and:
$\frac{PB}{BC}=\frac{PA}{AC}$.
Point $D$ is a point of intersection of line $PC$ and segment $AB$.
My problem is whether $AD$=$DB$[/quote]

Not necessarily. The equation $\frac{PB}{BC}=\frac{PA}{AC}$ just means that the point P lies on the C-Apollonius circle of triangle ABC.

  Darij

[quote="dondigo"]What are the properties of this Appolonius circle? And why the letter C - Appolonius?[/quote]

Here are the basics: http://www.cut-the-knot.org/Curriculum/Geometry/ApollonianCircle.shtml . I could tell you more, but I don't want to help you with a current olympiad problem...

  Darij


In the diagram, $ AB \equal{} BC \equal{} AC \equal{} AD \equal{} x \plus{} 8$. Furthermore, $ CD \equal{} 3x \minus{} 9$. What is the sum of the smallest and largest possible integer values of $ x$?

[asy]size(4cm,4cm);

pair A,B,C,D;

A=(0,0);
B=(-0.1,1.15);
C=(1,0.7);
D=(1.25,-0.2);

draw(A--B--C--D--A);

draw(A--C);

label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);

label("$x+8$",midpoint(A--D),S);
label("$3x-9$",midpoint(D--C),E);[/asy]

Here, we could incorporate the triangle inequality theorem. Since no side of any triangle can be $ 0$ or negative, $ 3x\minus{}9 > 0$, and thus, $ x > 3$. Thus, the smallest integer solution is $ 4$. Also, $ \overline{CD}$ can't be longer than the sum of $ \overline{AC}$ and $ \overline{AD}$. Since $ AC\equal{}AD\equal{}x\plus{}8$, We can set this equation: $ 3x\minus{}9 < 2(x\plus{}8)$; thus, $ x < 25$. Hence, the largest integer solution is $ 24$. Adding $ 4$ and $ 24$, we arrive at our answer of $ \fbox{28}$.

For more information about the triangle inequality theorem:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Triangle_inequality[/url]


MIT college Harvard Yale Princeton Stanford AMC

Join Date: Dec 2008
Posts: 4

question about MIT
MIT doesn't seem to like my school very much. We have about 2 people go to Harvard every year, along with a lot of Yale and Princeton and Stanford acceptances, but nobody has got into MIT for a long, long time. I'm just wondering if it's just a coincidence that they don't take people from my school, or if they really do blacklist schools and refuse to take people. This sucks because I applied EA and was deferred, but i really really really really really really want to go to MIT =(

p.s. [url=http://en.wikipedia.org/wiki/Elizabeth_Shin]this girl[/url] graduated from my school

p.p.s. jian li was also from my school (: lmao

I really doubt MIT blacklists schools.  It's more likely that we're simply looking for different things than other schools; although MIT has done a lot to branch out in recent years, we are still very much a tech school and primarily interested in math and science aptitude.

Hello,

I go to a math and science academy in CT, and we also didn't have any early acceptances into MIT this year, so don't think that MIT just pulls people out of math schools.

Good luck during RD!

My school didn't admit anyone from MIT in recent memory (more than a decade), but a couple of kids go to schools like Stanford and Harvard each year. They don't "blacklist" schools; I got in.

I am a junior (applying fall 2009, hopeful class of 2014), but I can tell you that this admissions process is highly flawed.  So if you are mad that you didnt get in, just use that as motivation and maybe you will get into MIT graduate school (they wont blacklist colleges  :P )

One person that got into MIT from my school greatly "exaggerated" his activites (I know this because he told everybody and he would seem like the type to do it).  I do not think that that is fair to some other people that ended up getting rejected.  This is one of the big issues with the college admissions system; unfamiliarity with applicants and exaggerations/maybe even lies cause more problems than, lets say, what school you go to (as Narcissa implies).

So basically, what I am trying to say is that dont get down just becuase you didnt get in and dont complain and make silly excuses.  Eventually, the person that was better will be better in 20 years with hard work and motivation.  I am pretty sure the person that I talked about^, if he continues his path, he will never truly be sucessful.

And no, they dont blacklist any schools (maybe quotas).  The girl's link that you posted should have no relationship to your situation...going to the same school that she went to doesnt mean you live the same kind of life.

EDIT: @TZF: When I meant flawed, I dont mean flawed from the college's perspective.  They clearly try to do the best with all the information that they have; I meant that the information that the colleges receive can be highly flawed (exaggerated or maybe even lies).  I elaborated more in my second post.

I say this to everyone who complains about college admissions, and so I'll say it again here. The admissions process isn't nearly as flawed as people make it out to be.

The principle objective is to create a new class with the ideal makeup (in terms of interests, viewpoints, experience, background, etc). Frankly, top colleges do that very well. They find the right amount of work to put into decisions so that the cost of doing an additional amount of effort will outweigh the benefit of the expected improvement in the class.

MIT and other top colleges declare openly that they will look at grades and academic achievements in the context of the high school. Obviously they have some information about different high schools that they document. While they probably won't entirely block out a school, it's very possible (and, as I argue, entirely fair) for them to have inclinations towards or against any particular school.

Ultimately, the duty of admissions is first towards the students of the college, and second towards the students that apply and do not get in. It is much, much more important for the college to care about getting the right class composition than for the college to care about that one smart kid that got left behind. After all, that smart kid that got left behind won't be investing four years of his/her life and two hundred thousand dollars in them, and will probably end up at another fine institution, and make their way in life.

While this can be hard to accept for a student who spends years aspiring toward attending a particular institution, it is ultimately the fairer way of looking at things.

TZF has really hit it on the nose here. In the same way as the USAMO selection process, at some point you have to accept that no system will be perfectly fair, and there will always be people who theoretically "should" get in but don't.

To elaborate on my point, I am just stating what I know from my own experiences.  I dont think its totally flawed (atleast not from the college's perspective), but if I was personally choosing (which I have no authority nor the credentials) based on my personal experience among the applicants from my school, my list would be totally different.  This is because I am more familar with the students at my school than a college will ever be.  Of course, this is unrealistic to think that colleges can become this familiar with a student.

Obviously, a college like MIT will try to make the best possible decision from what they know, but sometimes thats not always 20/20.  When I mean that it is flawed, I mean its flawed because people can lie all the time about what they do, which kinda sucks because some people actually do do those activities.

Eventually, it matters what you do in college, not where you go.  A truly brilliant and hard working person can succeed anywhere at any college.

Actually, I do not think the above analogy works because the USAMO is based on scores, which are totally objective.  Of course, how you determine which score is better is subjective, but its mainly objective.  The college application process is mostly subjective...as aside from tests and GPA, I believe most top colleges distinguish between the ECs/awards/essays/recs more...as everyone that applies has great scores (MIT actually does a great job as they have some data tables with acceptance percentage based on a variety of factors to see what they incorporate into their selection process-however, correlation =/ causation)

@Narcissa: you say you were deferred?  Were you accepted later?  And deferred =/ rejected, as like 10% or so eventually get in.

I've been told that MIT has a fairly liberal initial screening process by which they give you a 0-5 index for quantitative things (gpa, in context, sat's, etc) and a 0-5 index for non-quantitative things, and then divide the resulting square of data points into two regions. In the first region, they decide who to accept and who to defer. In the second region, they look to see if anything interesting was missed out that might make them a defer, or even an accept.

If true, this would not lead to any particular changes in strategies for people applying. Still, I think it's a good model of what might go on inside the black-box at MIT or any other competitive college. Some people have the impression that every little hair matters, because it might be the difference between them and someone else. As if they have 15 people in a conference room with both applications on the projector, racking their brains over which candidate to accept. Of course, details can make the small difference in impression that makes the difference, but it's important to get a general feel of how the applications may be considered.


Let $ N$ be a positive integer with $ d$ digits, all of which are the same and are not equal to $ 0$. If $ 11\mid N$, determine $ d\mod 2$.

EDIT:  Forgot to hide.

[hide]For a number to be divisible by 11, the sum of the "odd-position" digits (ones, hundreds, ten thousands, etc.) must equal the sum of the "even-position" digits (tens, thousands, hundred thousands, etc.), or the difference must be divisible by 11. If all the digits are the same, the only way this is possible is for there to be an even number of digits (because 11 is not a digit).  Therefore, $ d$ must be $ 0 \pmod{2}$.[/hide]


function
real analysis
real analysis unsolved

If $ f_n$ is a sequence of characteristic function of sets in $ \mathbb{X}$(sigma algebra)
and if $ f_n$ converges to in $ L_p$.

Show that $ f$ is (almost equal to) the characteristic function of a set in $ \mathbb{X}$

-------------------------------------------------
Sorry for posting same question twice...

I don't know how to delete this...
----------------------------------------------------

Extract a pointwise convergent subsequence..


limit
real analysis
real analysis unsolved

Ascertain $ \lim_{n\to\infty }\ \sum_{k \equal{} 1}^n\ \frac {1}{n \plus{} k \plus{} k^2}$ ([b]D[/b]istrictual [b]M[/b]athematical [b]O[/b]lympiad).

Zero, of course  :sleeping: (dominated convergence theorem with $ \frac 1{k^2}$ as a majorant if you know it, splitting the range of the summation into $ k<n^{2/3}$ and $ k\ge n^{2/3}$ if you don't). A slightly better question: find a simple equivalent for this sum as $ n\to\infty$. ;)


inequalities
logarithms
inequalities proposed

Help me :

Prove that : $ (1\plus{}\frac{1}{n})^n < 3\minus{}\frac{3}{n\plus{}1}$ $ ( n \in N*)$

False for $ n\equal{}1$, $ n\equal{}2$, ... and trivial for large $ n$ since LHS's sup is $ e$, while RHS's sup is $ 3$.

[quote="Duy_VN"]Help me :

Prove that : $ (1 \plus{} \frac {1}{n})^n < 3 \minus{} \frac {3}{n \plus{} 1}$ $ ( n \in N*)$[/quote]
I think it must be
$ (1 \plus{} \frac {1}{n})^n > 3 \minus{} \frac {3}{n \plus{} 1}$  with $ n \ge 3$ ;)


Hello,

A paper on a website mentions that

The equation x^3 + y^3 = 3*a*x*y is equivalent to:

x = 3*a*t/(1+t^3)
y = 3*a*t^2/(1+t^3)

I am not able to really understand the process of arriving at this particular solution?

Are there any hints / guidelines for such conversions?

Thanks for any help.

And where does it come from? I don't see it after reading the given equation

Ok, another hint: Let $y = tx$.  Substitute this into the equation $x^3 + y^3 = 3axy$ and see what you get.


geometry
3D geometry
algebra unsolved
algebra

I am part the way solving the cubic equation x^3+3x^2+6x+8=0 using ferro/tartaglia/cardan method.

completeing the cube: X^2+3X=-4 where X=x+1
let X=u-v which, after some manipulation gives the solution u^3=-2 (+/-) sqrt(5) and v=1/u

I am usure where the three solutions come from at this point for (u,v). I will be grateful fro any help.

hello, i hope this will help you
http://www.sosmath.com/algebra/factor/fac11/fac11.html
Sonnhard.

See also post #8 in
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=235713


Here is a cute classical problem. Let $T$ be a tree on $n$ vertices. Assume that we may write a positive integer on each of the $n-1$ edges of $T$ in such a way that the resulting $\binom{n}{2}$ sums on the $\binom{n}{2}$ paths of $T$ form a permutation of $1,2,\ldots,\binom{n}{2}$. Show that $n$ is either of the form $n=k^2$, or of the form $n=k^2+2$.

[b]Open Question. [/b] For which $n$ is this possible?

--Vesselin

See here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=odd+even&t=22106

Pierre.


[quote="Inspired By Nature"]Sorry about the messy diagram.
XYZ is equilateral with side length 2. ZV and YW are medians and PQRS is a square. Find the ratio of the area of PQRS to XYZ.
[hide=" :D "]Its been a while since I posted anything. :) [/hide][/quote]

[hide]Call the intersection of the medians O. In XYZ medians are altitudes. call the intersection of the median connecting to X and bisecting YZ  A. The altitudes are of length $\sqrt{3}$. Then by the pythagorean theorem $\overline{OY}^{2}=(\sqrt{3}-\overline{OY})^{2}+1$ and $OY=\frac{2\sqrt{3}}{3}$. In triangle OYZ, the intersection of OA and PS iS B. We have OA=$\frac{\sqrt{3}}{3}$. OYA is similar to OPB. Call the square length s. then $\frac{\frac{s}{2}}{\frac{\sqrt{3}}{3}-s}=\frac{1}{\frac{\sqrt{3}}{3}}$. Solving, $s=\frac{4\sqrt{3}-2}{11}$. Then the area of the square is $\frac{52-16\sqrt{3}}{121}$ and the area of the triangle is $\sqrt{3}$. Divide to get $\frac{52-16\sqrt{3}}{121\sqrt{3}}$[/hide]

it is a bit messy. Also, i got $\frac{52-16\sqrt{3}}{121\sqrt{3}}$.
I think you made an error. :maybe:

[color=darkblue][b]Lemma.[/b] Let $ABC$ be a triangle and let $MNPQ$ be a rctangle so that $N\in [AB]$, $P\in [AC]$, $\{M,Q\}\subset [BC]\ .$ Denote $MQ=NP=x$ and $MN=PQ=y\ .$ Then there is the relation $\boxed{\ \frac{x}{a}+\frac{y}{h_{a}}=1\ }\ .$ Particularly, $x=y$ ($MNPQ$- square) $\Longrightarrow$ $\boxed{\ \frac{1}{x}=\frac{1}{a}+\frac{1}{h_{a}}\ }$, i.e. $x=\frac{2S}{a+h_{a}}\ .$

[b]Remark.[/b] In the our case, for the equilateral triangle $XYZ$ suppose w.l.o.g. $YZ=1$. Thus, $OY=OZ=\frac{\sqrt 3}{3}$, $h_{o}=\frac{\sqrt 3}{6}$, $2S=\frac{\sqrt 3}{6}$ and $x=\frac{\frac{\sqrt 3}{6}}{1+\frac{\sqrt 3}{6}}$ $\Longrightarrow$ $x=\frac{2\sqrt 3-1}{11}\ .$ Therefore, $\rho =\frac{x^{2}}{[XYZ]}=$ $\frac{\left(\frac{2\sqrt 3-1}{11}\right)^{2}}{\frac{\sqrt 3}{4}}$ $\Longrightarrow$ $\rho=\frac{52-16\sqrt 3}{121\sqrt 3}\ .$[/color]

Thanks guys, its always good to get different approaches. :)


modular arithmetic
number theory theorems
number theory

For given coprime $a,b$, is $\varphi(b)$ the [i]minimum[/i] of $\{n|n>0, a^{n}\equiv 1\pmod{b}\}$?

When also $a$ is given, then it's trivially wrong: $a=1$ and any $b$.
But when you allow $a$ to run through all coprime residue classes, then it's still wrong: $b=15$.

And I remember answering this question in the past...


Let x_0,y_0 and z_0 be three real numbers , pozitive <>0, which are known. we have the series:
x_n,y_n and z_n defined like so:
x_(n+1)=(x_n+y_n+z_n)/3
y_(n+1)=(x_n*y_n+y_n*z_n+z_n*x_n)/(x_n+y_n+z_n)
z_(n+1)=3*x_n*y_n*z_n/(x_n*y_n+y_n*z_n+z_n*x_n).

prove that these sequences converge and find out it's limits.

I only managed to see that x_(n+1)*y_(n+1)*z_(n+1)=x_n*y_n*z_n=x_0*y_0*z_0 which everybody I guess would have seen it!

hope for a solution!
bye bye

I'll just consider x_0>y_0>z_0 (the other cases should work out) (hopefully). x_1 will  decrease. z_n*y_n will be, for all n, (constant)/(sum of x y and z). If x decreases, y and z should get bigger. Thus, x, y, and z get closer together. Hmmm....this can probably be rigorized with AM-GM-HM and that sort of stuff.

Whoa...actually, by AM-HM, x_n:ge:z_n for n>0.

This (much wanted  :!:  :!: ) problem just appeared on the imo-problem mailing list. Here is my answer :

Lemma 1 : (a+b+c) >= 3(ab+bc+ca)

expands to a+b+c >= ab+bc+ca, well known (using rearrangement for instance)

Lemma 2 : (ab+bc+ca) >= 3abc(a+b+c)

expands to ab + bc + ca >= abc + abc + abc or 1/a + 1/b + /c >= 1/ab+1/bc+1/ca (lemma1)

By lemma 1 and 2, x_n >= y_n >= z_n for n >= 1.

Also min(a,b,c) <= _mean_(a,b,c) <= max(a,b,c) where _mean_ is either arithmetic (x_n) or harmonic mean (z_n).

So x_n is decreasing and z_n is increasing for n > 1, so they are convergent towards L1 and L2 > 0.

y_n = x_0*y_0*z_0 / (x_n*z_n) is also convergent towards L3.

L1 = (L1+L2+L3)/3  with L1>=L2>=L3 is only possible if L1=L2=L3 = (x0y0z0)^1/3


aki la version  pdf  del examen selectivo para la IMO 2006

Tomado el sab 20 de mayo, en las instalaciones
de Campus de la Univ. Catolica del Peru, Lima (San Miguel)

PD: Cualquier correccion de edicion hacerla llegar..

Carlos Bravo ;)
Lima - PERU

:? Mi única observación son las tildes.

$Tipe$


geometry
3D geometry
Vieta
algebra
polynomial
calculus
integration

2006, S28

The sum of three numbers is 4, the sum of their squares is 10, the sum of their cubes is 22. What is the sum of their fourth powers?

2006, S30

Three circles of radius 1, 2 and 3 are all tangent to each other. A smaller circle is tangent to all 3 circles. The radius of the smaller circle can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$?

I've only read the first problem.  It is trivial using Vieta's Theorem and some well known identities.  I will leave the rest for you.

JavaMan, perhaps you might want to change the $x,y,z$ to $a,b,c$ on the 2nd line of your solution :D

Thanks, changed it.

[hide="2"]
This is a start, I'm thinking of a nicer solution.

Notice by connecting the centers you get a 3-4-5 triangle with area 6. Connect the centers of the circles, with the center of the middle circle. The 3-4-5 triangle is divided into 3 areas which sum to 6.

Use herons a few times,
\[\sqrt{2r(3+r)}+\sqrt{3r(4+r)}+\sqrt{6r(5+r)}= 6 \implies r=6/23, \boxed{6+23=29}\]
[/hide]

you can always be cool and use descartes' circle theorem for #2

I see, sure makes things easier, but was their intended solution to use descartes' circle theorem?

Do you know if they sell these books by any chance.

Damn i didn't think of using Heron's for the 2nd problem  :blush: . Thanks people. The Australian MC books are available but pretty expensive: http://www.amt.edu.au/ and you go to the book shop. But they're pretty expensive compared with the American MC solutions and problems. However, the questions are quite interesting. In fact, it's my favourite math competition  :P .

2006 S23:
A tetrehedron PQRS has angles QRP, SRP, and QRS at one vertex. It is given that angle QRP is 30 degrees and angle SRP is 40 degrees. If QRS is an integral number of degrees, how many possible values are there for QRS?

A. 59
B. 69
C. 290
D. 10
E. 70

2006 S24:

For how many positive integers $a$ does there exist a real number $x$ so that $\sqrt{a-x}+\sqrt{a+x}= a$?

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

2006 S25:

Eight points are marked on a circle. One of these points is P. Chord connect some or all of these points to one another, such that different numbers of chords are joined to the seven points other than P. What is the smallest possible number of chords connected to P?

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

2006 S26 (integer answer question):

A teacher asks all the students in a class to write down sums of different 2-digit numbers on the board. She says that no matter what, there is a sum that will be written down thrice. What is the smallest number of students in the class for her to be correct?

JavaMan, what does $\sigma_{2}$ and $\sigma_{3}$ refer to in the solution of S28?

And how do you get that $x^{3}-4x^{2}+3x+2 = 4$? What does $x$ represent?

There's another thing: Calculators can't be used in the Australian Mathematics Competition, so is there an elegant way to solve the surds obtained from using Heron's Formula in S30?

More problems:

2006, S27

In a regular polygon there are two diagonals such that the angle between them is $50^{o}$. What is the smallest number of sides of the polygon for which this is possible?

2006, S22

The vertices of a cube are each labelled with one of the integers 1, 2, 3, ..., 8. A [i]face-sum[/i] is the sum of the labels of the four vertices on the face of a cube. What is the maximum number of equal face-sums in any of these labellings?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

2006, S14

In a triangle $PQR$, $\sin P = \frac{1}{3}$ and $\sin Q = \frac{1}{4}$. How many different values can the size of $R$ have?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

2006, S15

How many different pairs of 2-digit numbers multiply to give a 3-digit number with all digits the same?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9