quadratics
logarithms
probability
trigonometry
conics
hyperbola
modular arithmetic

The algebra test from JHMT.  40 minutes, no calculators.  Point values in brackets.

1)[5] If the quadratic $(a-d)x^{2}+ax+(a+d)$ has a double root, compute $\frac{a^{2}}{d^{2}}$.

2)[7] Find $n$ such that

$\sum_{k=1}^{n}\log_{10}\left(1+\frac{1}{k}\right) = 1$.

3)[10] Define a sequence of numbers $a_{1}, a_{2}, \ldots , a_{n}$ by two rules that hold for   $n\geq 1$:

$a_{2n}= \sqrt{a_{2n-1}+1}$

$a_{2n+1}= a_{2n}(a_{2n}+2)$

If $a_{1}= 3$, what is $a_{36}$?

4)[13] If $a$, $b$, and $c$ are randomly chosen (with replacement) from the set $(-1, 0, 1)$, compute the probability that
\begin{eqnarray*}a+b^{2}&=& c^{3}\\ b+c^{2}&=& a^{3}\\ c+a^{2}&=& b^{3}\\ \end{eqnarray*}

5)[14] Find an $x$ such that

$\sum_{k=1}^{\infty}\frac{(k-1)(k+1)}{k^{4}}= \sum_{k=1}^{\infty}\frac{x}{k^{2}}-\sum_{k=1}^{\infty}\left (\frac{x}{k^{2}}\right )^{2}$.

6)[14] Find the third smallest prime factor of $N$, where

$N=21^{0}+21^{1}+21^{3}+21^{6}+21^{10}+21^{15}+21^{21}+21^{28}+21^{36}+21^{45}$.

7)[18] What is the minimum value of $\sin^{2}x+2\sin y\cos x+\cos^{2}y$ for real $x$, $y$?

8)[19] Determine the radius of the circle that is tangent to the $y$-axis and to both arms of the hyperbola $y^{2}=x^{2}+1$.

question 5 should actually read "Find an $x$..." Most people got the answer we wanted (1) but that is actually not the smallest $x$.  So we gave credit to everyone for this.
-JHMT

[quote="Barnacle"] question 5 should actually read "Find an x..." Most people got the answer we wanted (1) but that is actually not the smallest x. So we gave credit to everyone for this.
-JHMT[/quote]

Fixed.  By the way, thanks for an excellent competition, and in particular a great algebra round (I'm only slightly biased by the fact that I got a perfect score on it  :P ).

[hide="clickeroo"]Huh.  I think I'm missing something on #7. The answer is 0, yes?  Or am I not seeing something obvious (which is a very real possibility :blush:).

\[\sin^{2}(x)+2\sin(x)\cos(x)+\cos^{2}(x) = (\sin(x)+\cos(x))^{2}. \]
Given that $x$ and $y$ are both real numbers, so are their trig functions.  So the minimum possible value of the square is 0.  For example, if $x=3\pi/4,$ then $\sin(x) =-\cos(x)$.[/hide]

[hide="but unfortunately . . . "]
There are two different angles, $x$ and $y$.  Both $\sin x$ and $\sin y$ appear, as well as $\cos x$ and $\cos y$, so the expression doesn't factor nicely.
[/hide]

[hide="#7 solution"]
$\sin^{2}x$ and $\cos^{2}x$ are $\geq 0$.  So set them equal to 0.  If $\sin x$ is 0, $\cos x$ is $\pm 1$, and vice versa.  We can have $\cos x = 1$ and $\sin y =-1$, which makes the middle term $-2$.   This makes the whole expression equal to $-2$.  It is easy to see that this is the minimum, since all three terms are minimized.
[/hide]

[hide="comments"]
This is a problem where it is easy to be misled (I was, too, at first), and easy to get lured into an ugly approach and miss the quick solution.
[/hide]

Ah!  Yes, it is much more interesting with the two variables.  That makes more sense; it seemed too easy the other way.

This is a perfect example of what I'm always talking about in class; read the problem carefully to make sure we answer what is asked!  That applies to coaches, too.  :-D

And I agree, that test has great problems.

Best,
Catherine Asaro

[quote="Barnacle"]Alternatively:
$\sin^{2}x+2\sin y\cos x+\cos^{2}y,$
$1-\cos^{2}{x}+2\sin{y}\cos{x}+1-\sin^{2}{y},$
$2-(\cos^{2}{x}+\sin^{2}{y}-2\sin{y}\cos{x}),$
$2-(\cos{x}-\sin{y})^{2}.$
The paranthesized term is at most $2$, giving a minimum of $\fbox{-2}$.[/quote]

Nice.  I knew there had to be a more standard algebraic way of doing the problem.

These problems are fun (I really should stop doing this and go finish my work for tonight :-)).

[hide="21+21=42"]Not that this addition problem has anything to do with the solution, other than being the answer to everything.  But I couldn't resist. ;-)

For problem $\#7$, look at mods for the smallest primes.

The smallest prime is 2, and

$21\equiv 1\pmod2\ \ \Rightarrow\ \ N\equiv 10\cdot1\pmod2\equiv 0\pmod2.$

Thus, 2 is the smallest prime that divides $N$.

Three cannot divide $N$ because $21^{0}=1\ \ \Rightarrow\ \ N = 1+3\cdot M,$ where $M$ is some integer.

Five is much the same as the case with 2:

$21\equiv 1\pmod5\ \ \Rightarrow\ \ N\equiv 0\pmod5.$

So 5 is the next smallest prime.

Seven won't work for the same reason as 3: $N = 1+7\cdot M,$ where$M$ is the proverbial and infamous "some integer."

Now we get to 11:

$21\equiv 10\pmod{11}\equiv (-1)\pmod{11},$ which gives

$N\equiv (1-1-1+1+1-1-1+1+1-1)\pmod{11}\equiv 0\pmod{11}.$

So 11 is the third smallest prime, which is what we want.

[color=blue]Answer is 11[/color]

Nice problem.  I think I will have tomorrow's high school class work it out.[/hide]

That is a very pretty recursion.

If folks liked #7, here is similar type problem (Halo teams who did this in class, you can't give the answer! :-))

Find the sum of the possible values of $\tan(\theta)$, where

$27\sin^{2}(\theta)-11\sin(\theta) = 26.$

Best,
Dr. A


ratio
geometry
rectangle
geometry unsolved

$ P$ is a point on the extension of the diagonal $ AC$ of rectangle $ ABCD$ beyond $ C$, such that $ \angle BPD\equal{}\angle CBP$. Find the ratio $ PB: PC$.


calculus
integration
inequalities
trigonometry
induction
function
real analysis

Prove the following inequality.

\[\int^1_0{e^\sin{\pi}x}dx\geqq{e^\frac{2}{\pi}}\]

$e^\sin{\pi}x} = \displaystyle\sum_{n=1}^{\infty} \frac {sin^n({\pi}x)}{n!}$ , and $ \int^1_0 sin^{n+2}({\pi}x)dx  = \frac {n+1}{n+2} \int^1_0 sin^n({\pi}x)dx$
By induction we prove $ \int^1_0 sin^n({\pi}x)dx \geq (\frac {2}{\pi})^n$ ... and so on ..
 :cool: I would like visit Kyoto !  :cool:

[quote="Diogene"] I would like visit Kyoto !  :cool:[/quote]
me too

There's no need to work as hard as Diogene did. This is just Jensen's inequality. Since the exponential is a convex function and $[0,1]$ has measure 1,

\[\int_0^1e^{f(x)}dx\ge e^{\int_0^1f(x)dx}\].


geometry
3D geometry
modular arithmetic
inequalities
number theory solved
number theory

We have this result:
If a,b and c are integers sucht that a/b+b/c+c/a is an integer, then abc is cub of an integer.

More generally, the result still holds even if we replace 3 by any integer n, that is $ \frac ab + \frac bc + \frac ca = n$. (1)

Since $-x ^3 = (-x)^3$ , it suffices to prove that each prime  which divides $abc $ divides it with an exponant = 0 mod[3].

Let $d = \gcd (a,b,c)$. Note that if (a,b,c) satisfies (1) then so does $\left (\frac ad, \frac bd, \frac cd \right)$. And that $abc$ is a cube iff $\frac {abc}{d^3}$  is a cube. Thus, we may assume that $d = 1$.
Moreover (1) may be rewritten as : $a ^2 c + ab ^2  + bc^2 = nabc$. (2)

Let $p$ be a prime which divides $abc$. Then, from (2) and since $d=1$, it divides exactly two of the three numbers $a,b,c$. Suppose that p divides a and b, but not c (the other cases are similar). Let $a = p^xa'$ and $ b = p^y b'$, where neither $a'$ nor $b'$ is divisible by $p$.

From (2), considering the exponant of $p$ in each of the terms, we deduce that $p ^y$ divides $p ^{2x}$ , thus $y  \leq 2x$.
In another hand, $p ^x$ divides $p ^y$, which leads to $x \leq  y$, and then to $x+y  \geq  2x$.

It follows (from (2)), that $p^{2x}$  divides $p^y$, so that $2x  \leq y$. Then $ 2x = y$ and $x+y = 3x \equiv 0 \pmod 3$, and we are done.

Pierre.

\leq means $\leq$.
\geq means $\geq$.

Actually, these "\leq"'s and "\geq"'s are relics from the old time when this forum had an automatic mechanism replacing every \leq with an $\leq$ sign and every \geq with an $\geq$ sign. Now, in order to get a $\leq$ sign, you have to write:

[code]$\leq$[/code]

And similarly for $\geq$.

  Darij

Since -x=(-x) , it suffices to prove that each prime which divides abc divides it with an exponant = 0 mod[3].
  i didnt understand this explanation.Why ?

let  a+b+c =n
and a=x/y, b=y/z,c=z/x,
then
a+b+c = 3 (n is integer)
ab+bc+ca = r (r is a rational number)
abc=1
then, a,b,c are roots of
T^3-3T^2 + rT - 1 = 0
(r-3)T = (1-T)^3
=> c/a = (1-c)^3/(1-a)^3
xyz = x^3.(y/x).(z/x) = x^3.(c/a) = s^3 (s is a rational number)
xyz is a cube of a rational number, then xyz is a cube of an integer
number

or you can simply just use the identity (x+y+z)/3>=(xyz) at power 1/3 (sorry..i'm to lazy to write it nicely :) ) with equality when a, b and c are equal. :). so when you apply this taking x=a/b, y=b/c and z=c/a you get what you wanted. :).
i somehow don't agree you can have any number there instead of 3, but i'll have to check it out.  [/code]

[quote]
fdump said:
or you can simply just use the identity (x+y+z)/3>=(xyz) at power 1/3 (sorry..i'm to lazy to write it nicely  ) with equality when a, b and c are equal. . so when you apply this taking x=a/b, y=b/c and z=c/a you get what you wanted. .
[/quote]

This is not always true, since a, b, c are not always positive, they maybe negative.

[quote="marko avila"]well the proof pbornsztein gave look pretty correct but i guess there could be a flaw because he concludes very inelegant and with not much detail. and maybe his inequalities at the end are wrong. Anyway pbornsztein: is your generalization something you conjectured and proved or did you read it????? :P[/quote]

His proof is correct. He considers the case when $gcd(a,b,c)=1$ since the general case can be deduced from. with this, he takes a prime that divides $abc$, so it'll divide $a$, $b$ or $c$, but using the equation, it must divide two of these numbers. so he writes $a=p^xa'$ and $b=p^yb'$. without loss of generality, assume that $x\leq y$. then the equation can be written
\[ p^{2x}a'c+p^{x+2y}a'b'^{2} + p^yb'c = np^{x+y}a'b' \]
Then $p^y$ obviously divides the last three terms, thus it also divides $p^{2x}a'c$, but by definition, it doesn't divide $a'$ nor '$c$, so it must divide $p^{2x}$. this implies that $y\leq 2x$.
On the other hand, since $x\leq y$, $p^{2x}$ divides $p^{x+y}$ and $p^{x+2y}$, so it must divide $p^y$, thus $2x\leq y$.

from the two inequalities, you get $y=2x$, so $abc=p^{3x}a'b'c$ where $p$ doesn't divide $a'b'c$. and since this is true for every prime, then $abc$ is a perfect cube.

[quote="mahmoud"]without loss of generality, assume that $x\leq y$.[/quote]
We can't assume this, but since $p^x$ divides the first, the second and the last term, it must also divide the third one, and $x\leqslant y$.

Anyway, pbornsztein's proof is correct, but I believe that there is only a few possible values of $n$ such that $\frac ab + \frac bc + \frac ca = n$.

[quote="Namdung"]Show that if a, b, c be integers satisfying $\frac ab + \frac bc + \frac ca = 3$ then $abc$ is cube of an integer.[/quote]
Can I solve this problem by AM-GM?
Because $\frac ab + \frac bc + \frac ca \geq 3$, so we can get $a=b=c$, so $abc$ must be a cube of an integer.

As Zhaoli said, a,b,c don't have to be positive, so no AM-GM.


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

Let $R$ be an integral domain and suppose that every prime ideal in $R$ is principal. Then is $R$ a P.I.D.

[quote="harazi"]Are you sure you didn't forget that it is noetherian?[/quote]

nope, I quoted this exercice directly from our book.

A theorem of Cohen states that if the prime ideals of a (commutative, unital) ring are finitely generated, then the ring is noetherian, so adding that the ring is noetherian would have been redundant :).

Wow, that's a very nice result. I didn't know it and it does not seem simple, at first sight. Ok, now I have a proof for the result, using this theorem, but it's quite long: the idea is to show first that the ring is UFD and then to use that any prime ideal has an irreducible element.

Quite a number of problems on the same topic lately. Wouldn't you say?
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=122735[/url]
with the extra remark that if every prime ideal of an UFD is principal then every non-zero prime ideal is maximal. This is because if $P\subset Q$ is an inclusion of non-zero prime ideals, then since $P=(p)$ and $Q=(q)$ for some prime elements $p,q$ we get $q|p$ which means $p=q$ and $P=Q$.

WOW! The above proofs are way to complicated. This is a [i]prime[/i] example of why god created the axiom of choice.

Suppose it is false, and let $S$ denote the set of all ideals in $R$ which are not principle. Then by Zorn's lemma there is a maximal ideal in $S$, say $\mathfrak{A}$. Now show that the maximality  of $\mathfrak{A}$ in $S$ implies that $\mathfrak{A}$ is prime, a contradiction. Therefore, our assumption that $S$ was nonempty can not be true. QED

If you cannot fill in the details above, then take a look at Commutative Algebra by Eisenbud. This is on pages 85 and 724, the former being an exercise, and the latter being the solution.

P.S. This is also how you prove the result of Cohen mentioned above, and this technique can prove many more things.


Euler
number theory proposed
number theory

Find the remainder when $19{}^9{}^2$ is divided by 92.

$92 = 2^2 \times 23$ thus $\varphi (92) = 2 \times 22 = 44$ (where $\varphi$ is the Euler phi function).
Since $19$ and $92$ are coprime, it follows from Euler's theorem that $19^{44} = 1 \mod[92]$ and then $N = 19^{92} = 19^4 \mod [92].$
Since $19^2 = 361 = 85 = -7 \mod [92]$ we easily deduce that $N = 49 \mod [92].$

Pierre.


inequalities
calculus
integration
real analysis
real analysis unsolved

[hide="Hint"]Divide $f$ by $\sqrt{2}$ to replace the bounds $1$ and $2$ with $1/\sqrt{2}$ and $\sqrt{2}$. This should make the product of integrals more symmetric.[/hide]

[hide="Solution"]By the arithmetic-geometric mean inequality, \[\sqrt{\int_{0}^{1}f(x)dx \int_{0}^{1}\frac{1}{f(x)}dx}\le \frac{1}{2}\int_{0}^{1}\left(f(x)+\frac{1}{f(x)}\right)dx\] The integrand on the right is at most $\sqrt{2}+1/\sqrt{2}$.[/hide]


summer program
MathPath
MATHCOUNTS
AwesomeMath
geometry
3D geometry
tetrahedron

Who's going to MathPath? I sent in my solutions on Thursday(because of the deadline) but I have a feeling I didn't make it because most were semi-brute force and holey.

But i realllllly wanna go

[quote="Ignite168"]I'm not going to MathPath, I'm going to...[hide]China :D[/hide][/quote]

Same with me.

[quote="fireemblem13"]wow! :o i didn't know so many AoPS'ers were Chinese, looking at the flags under the names.[/quote]

Look at their names, "Rong"
My last name is a bit weird, it's "Chien" but that's the English pronounciations of "Qian2" and "Jia1 Liu2" (my Chinese first name) becomes Jonathan.

[quote="Ignite168"]I'm not going to MathPath, I'm going to...[hide]China :D[/hide][/quote]
cool!  :lol:

I'm going to mathpath.

If I don't die on the way there.

[quote="Treething"]gj I went to mathpath last year. It doesn't help with much (only my opinion), unless you're planning to do MathCounts next year. Then it helps a lot.[/quote]

Really? I didn't think Mathpath helped me at Mathcounts (hey, look what I got at nationals :blush:) as much as it will on the higher-level competitions (especially proofs)...

[quote="cowpi"]Me too, and samath, and ch1n353ch3s54a1l.[/quote]

Yeah, are you guys going to Mathpath or China?

[quote="Klebian"][quote="fireemblem13"]wow! :o i didn't know so many AoPS'ers were Chinese, looking at the flags under the names.[/quote]

If that's a response to cowpi's post, she meant those 3 are going to mathapth, not china.[/quote]

...cowpi's a guy. anyway, i think he meant MathPath because Samath isn't Chinese.

[quote="LadyKn1ght"]Please stop quoting quotes :) .

Just a reminder, I'm still going to host an AIM chat for people going to MathPath at 7:00. And maybe I'll host another chat that you could come to Ubemaya.[/quote]

That's good.

How long did it take you to figure that line out, LadyKn1ght?

"quoting quotes" :rotfl:

Sorry, I wasn't on at 7:00, but if anyone's still intrested in going to an AIM chat I'm online now (my AIM screen name is LadyKn1ghtZ, by the way)

[quote="SplashD"][quote="mathgeniuse^ln(x)"]I know this is random, but LadyKn1ght, why did you add an extra Z at the end of your AIM screenname.[/quote]

Do you have to question everybody's AIM screenname?[/quote]

No, I don't.  Sorry if I am hurting your feelings, but I am just curious.  Not as curious as Pandora though. :lol:

[quote="kyyuanmathcount"]I suggest to you this article.

[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php[/url][/quote]

I hear that, but my school isn't that into math.

I have a Z at the end of my aim screen name because then the capital letters make my initals :) . I took off the Z for my AoPS screen name because I didn't want to put my last inital but I ended up saying it in other places anyway.

[quote="mathgeniuse^ln(x)"][quote="kyyuanmathcount"]I suggest to you this article.

[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php[/url][/quote]

Thankfully, I won't be in the Calculus Trap since I am two years ahead of grade level, and that is how far our school goes up to.  Alan Huang on the other hand, will run into it.[/quote]

Are there exceptions to the Calculus trap?

What do you mean by exceptions?

Oh, by the way MathPath does some stuff about Calculus.

[quote="robinhe"][quote="tz"]P.S. Robinhe, does Huron weigh honor cources?
[/quote]

Yes, and every school should.

If you take Geometry AC, then you get more GPA than if you took Geometry.

And AP classes give more GPA than AC classes.[/quote]

Sorry about being off topic, but I have just come across some info that will disprove these statements...possibly.

So I do not know.

I know one thing.  It should be weighed to distinguish people from the normal to the not normal.


What is the greatest common factor of 5!, 10!, and 15!?

[quote="Mathmaticious"]Um.. shouldn't it just be 5! (120)?  :huh:[/quote]

Yes, you are right.

$ 5! \equal{} 5 \times 4 \times 3 \times 2 \times 1$

$ 10! \equal{} 10 \times 9 \times 8 \times ... \times 5 \times 4 \times 3 \times 2 \times 1$

$ 15! \equal{} 15 \times 14 \times 13 \times ... \times 5 \times 4 \times 3 \times 2 \times 1$

Thus, the answer is $ 5 \times 4 \times 3 \times 2 \times 1$, or $ \boxed{120}$.

Any factorial is a factor of a larger factorial. FOr n > k,

$ n!\equal{}n \cdot n\minus{}1 \cdot n\minus{}2 \cdot \hdots \cdot k \cdot k\minus{}1 \cdot \hdots \cdot 1$
$ n!\equal{}k! \cdot n \cdot n\minus{}1 \cdot n\minus{}2 \cdot \hdots \cdot k\plus{}1$.


Given an $m$ by $n$ grid of squares, how many rectangles can be drawn on the grid whose sides overlap with the sides of the squares?

Come up with an algebraic way, and a combinatorial way of representing this.

[hide]In an $m$ by $n$ grid, there are $m+1$ points horizontally and $n+1$ points vertically. Any non linear way of selecting 2 points creates a rectangle.

Combinatorically (real word??) , we simply need to choose two distinct x-coordinates and y-coordinates. Thus, we have $\boxed{\dbinom{m+1}{2}\dbinom{n+1}{2}}$

Algebraically, you have $m+1$ choices for the x-coordinate of the 1st point, and $n+1$ choices for the y-coordinate. For the second point, you have $m$ choices for the x-coordinate and $n$ choices for the y-coordinate. Thus we have $(m+1)(n+1)mn$. But wait! If we look at a rectangle, and label its vertices clockwise $A,B,C,D$, we see that we have counted $(A,C), (C,A), (B,D)$ and  $(D,B)$ all as different rectangles. We have described each rectangle 4 times, so we divide by 4. $\boxed{\frac{(m+1)(n+1)mn}{4}}$[/hide]

[hide]$\binom{n+1}{2}\binom{m+1}{2}$
You have to choose 2 from n+1 choices for (horizontal/vertical endpoints of rectangle) and same goes for m.
[/hide]


topology
function
real analysis
real analysis unsolved

If we can define the convergence of sequence points in a space $ X$, then can we define a topology on $ X$ suchthat the above convergence is equivalentto  the convergence of topology that we defined? Is this topology is unique?

Restated a little: Is it possible to have two different topologies with the same convergent sequences (and the same limits)?

An example: Let $ S$ be an uncountable set, and let $ X$ be the space of all functions from $ S$ to $ \mathbb{R}$. We construct two topologies on $ X$:
1- The box topology: The basic open sets are of the form $ \{f: |f(x)\minus{}g(x)|<\epsilon\text{ for all }x\}$, where $ g$ is another element of $ X$. A sequence converges if and only if it converges uniformly.
2- The "countable box" topology: The basic open sets are of the form $ \{f: |f(x)\minus{}g(x)|<\epsilon\text{ for all }x\in C\}$, where $ C$ is a countable subset of $ S$. The other values are unrestricted. Here, it's clear that a uniformly convergent sequence of functions converges; we want to show the converse.
Let $ f_{n}$ be a sequence that doesn't converge uniformly to $ f$. There is some $ \epsilon$ such that $ |f_{k}(x_{k})\minus{}f(x_{k})|\ge\epsilon$ for infinitely many $ k$ (and associated $ x_{k}$). Pick one of these $ x_{k}$ for each such $ k$; the set of these $ x_{k}$ is countable, so the associated $ \epsilon$-box around $ f$ is open. Since the $ f_{n}$ miss this box infinitely often, they don't converge to $ f$.


function algebra polynomial geometry Support AMC USA(J)MO

Put all of your MCAS rants here! Be sure to be appropriate as well.

Mine:

"What does the 3 represent in the above equation?"
(where the equation was $ y \equal{} 3x \plus{} 10$ and the question was about silver chains)

I dunno... an arbitrary constant created by solving a differential equation?

(A table with only two values given)

Am I supposed to assume it's linear? It could only be defined at those two values, for all we know! It could be an exponential function or a polynomial of degree 194087213956231897562138961895.

Wow, I never got that, but since I'm in 7th grade, that would probably be too hard for us.
:(
And yeah, I really hated those two value graphs I got in elementary, didn't get one in MS though.
All the problems I got were like:
Sally has a garden that is 7' by 3', what is the total area that she can plant flowers in? Be sure to support your answers using work, thoughts, etc.
Still, during MCAS, I just try to be a nice little boy and give the stereotypical responses.

The way I wrote my open response was being dumb. I answered like I have never seen this problem and said pretty elementary things. At the time when I couldn't say simple things I jumped to really difficult stuff. I wonder if they were surprised.

Others things-hopefully my essay will be graded by nice people who are nice and will give me a nice grade-that's advanced of course.
-I uber-phailed the history social studies and blah blah blah MCAS since it tested you on everything, ugh. I know for sure I got 2 wrong on both day 1 and day 2. I think my open responses were good.......but the test was unfair.

MCAS is a waste of time and a waste of time to rant about.

Yes it is...I enjoy not being subjected to them :P

I sat for about an hour today, playing calc games. And that was [i]before[/i] they released us early. Yes, MCAS is a waste of time (especially the math one).

An aside: Would they take off points specifically for the phrase "THIS IS SPARTAAA!!!" appearing in random places in the open responses?

[quote="CatalystOfNostalgia"]UGH I'M SO MAD I FAILED MY MCAS

UGH[/quote]

I was like, wait, you didn't even take MCAS, then I realized that 9th graders also had to take MCAS too! :D

When I took my math MCAS, I finished like an hour early, and then left the testing room. Instead of going back to class, I hung out with my friends in a different class with the excuse that I was "still taking the MCAS".

We left school and got ice cream.

MCAS is awesome.

Same with us... except that we were in a math classroom and I had forgotten to bring my book. So what did I do? I read an algebra textbook. :ninja:

Did I forget to mention that this was the math MCAS?

Imagine how mad my teacher was at that point... :stretcher:

lol..i put the solution to USAMO #1 in a margin

Really? I'm sure the MCAS graders were like, yeah yeah, I don't understand this, so I'll assume it's right and give it a 4/4. Next. :!:

Teehee, it was so dumb.
I spent the whole time writing an inductive proof.  :rotfl:

I had 10th grade math MCAS, did my best to use calculus as much as possible. managed a partial derivative for which I expect to be failed. still finished in about 20-30 minutes per section. MCAS is just sad.

[quote="Yongyi781"]AH-HA! So it WAS alkjash who was standing with me for the perfect math SAT! I feel so honored. :)[/quote]

Hmmmm...
I'm not sure about that...
I think we are in different regions so we went to different places for the awards thing.
I met perfect there though!

No, no, I'm definitely sure I met you. I was originally invited to Baltimore, but my mom decided it was too far away, so we decided to go to Acton instead. And as for perfect, I didn't even know him at the time, so I didn't even look for him. :|

I now feel even more honored! :lol:

Eh, whatever...
The ceremony was kind of lame...
I had like 8 papers in my hand...

Was this the CTY thing? That was so stupid last year...

The ceremony was stupid...I didn't mean CTY in general was stupid...though I barely knew I was doing it when I was. Huh.

MCAS was soo fun!
I got a few hours of extra to time to stare at the clock, look out the window, sleep, and even twirl my pencil :)

As for the one that alkjash wrote an inductive proof on... I think it was like

Pic #1: One triangle
Pic #2: One triangle with three triangles on the 2nd row
Pic #3: One triangle with three triangles on the 2nd row with five triangles on the 3rd row.
Pic #4: One triangle with three triangles on the 2nd row with five triangles on the 3rd row and seven triangles on the 4th row.....

Find the number of triangles in the 15th row (or something like that).
And find a formula to find the number of triangles on the xth row.

I was too lazy to write a inductive proof and I was afraid that maybe they wouldn't be able to understand it, so I drew a lot of squares and explained my way from there.... I better get some extra points for my amazingly drawn squares XD

the principal at my school is now annoyed at me. he says that partial derivatives on MCAS is a way to fail. watch me have a 5 on BC Calc test and fail math MCAS.... in the same year lol.


algebra
function
domain
Gauss
superior algebra
superior algebra unsolved

Let P is a prime ideal in commutative ring R. Show that P[X] is prime in R[X]

HINT reduce to  P = 0, i.e. D domain => D[x] domain.
Compare to the classical proof of Gauss' Lemma, e.g.
http://google.com/groups?selm=y8zhegzo759.fsf%40nestle.ai.mit.edu


modular arithmetic
floor function
number theory proposed
number theory

Among the numbers $ 11n \plus{} 10^{10}$, where $ 1 \le n \le 10^{10}$ is an integer, how many are squares?

[hide="Solution"]
Notice that $ \sqrt{11n\plus{}10^{10}}\leq \sqrt{11\cdot 10^{10}\plus{}10^{10}} \equal{} 2\sqrt{3}\cdot 10^5$. If we write $ \sqrt{11n\plus{}10^{10}}$ as $ 10^5\plus{}k$, we have $ 1\leq k \leq (2\sqrt{3}\minus{}1)10^5$. Equating $ 11n\plus{}10^{10}$ with $ \left(10^5\plus{}k\right)^2 \equal{} k(k\plus{}2\cdot 10^5) \plus{} 10^{10}$, we find that $ k(k\plus{}2\cdot 10^5)\equal{}11n$, or that $ 11\mid k(k\plus{}2\cdot 10^5)$. If we have $ 11\mid k$, we have $ k\equiv 0 \pmod{11}$; if we have $ 11\mid (k\plus{}2\cdot 10^5)$, we have $ k\plus{}2\cdot 10^5 \equiv 0 \pmod{11} \Leftrightarrow k \equiv 2 \pmod{11}$. These cases are mutually exclusive. Thus, the total number of squares is the number of solutions to the following Diophantine equations:

[list]
[*]$ k\equiv 0 \pmod{11}$, for $ 1\leq k\leq (2\sqrt{3}\minus{}1)10^5$. This has $ \lfloor \frac{(2\sqrt{3}\minus{}1)10^5}{11}\rfloor \equal{} 22400$ solutions.
[*]$ k\equiv 2 \pmod{11}$, for $ 1\leq k\leq (2\sqrt{3}\minus{}1)10^5$. This has $ \lfloor \frac{(2\sqrt{3}\minus{}1)10^5\plus{}9}{11}\rfloor \equal{} 22401$ solutions.[/list]

Therefore, in total there are $ 22400\plus{}22401\equal{}44801$ squares of the form $ 11n\plus{}10^{10}$ where $ 1\leq n\leq 10^{10}$ is an integer.
[/hide]


function
calculus
integration
real analysis
real analysis unsolved

Let $ F : [0; 1]\rightarrow \mathbb{R}$ be a continuously differentiable function. Find all continuously differentiable functions $ G : [0; 1] \rightarrow \mathbb{R}$ , such that for each continuously differentiable function
$ f : [0; 1] \rightarrow \mathbb{R}$ , such that $ f(0) \equal{} f(1) \equal{} 0$,
\[ \int_0^1(F(x)f(x)\plus{}G(x)f'(x))dx\equal{}0\]
holds.

if $ \int_{0}^{t}F(z)dz \minus{} G(t) \equal{} const$ then everything ois OK, indeed just try to integrate $ \int_{0}^{1}F(t)f(t)dt$ by parts..
after you need to choose some compactly supported function $ f$ on [0,1] and so on.. I mean there is some standart procedure,

It suffices to prove that the solution is unique (up to a constant).

If $ G_{0}$ and $ G_{1}$ are solutions, then by putting $ G \equal{} G_{0} \minus{} G_{1}$ we get $ \int_{0}^{1} G f' \equal{} 0$ for every $ f \in C^{1}$ such that $ f(0)\equal{}f(1)\equal{}0$. It is easy to prove that this is equivalent to $ \int_{0}^{1} G f \equal{} 0$ for $ f$ such that $ \int_{0}^{1} f \equal{} 0$. Now we can write $ G\equal{}\bar{G}\plus{}C$, $ \int_{0}^{1} \bar{G} \equal{} 0$, so for any $ f$ such that $ \int f \equal{} 0$: $ 0 \equal{} \int_{0}^{1} G f \equal{} \int_{0}^{1} \bar{G} f \plus{} C\int_{0}^{1}  f \equal{}\int_{0}^{1} \bar{G} f$, and by setting $ f\equal{}\bar{G}$ you get $ \int_{0}^{1} \bar{G}^2 \equal{} 0 \Rightarrow \bar{G}\equal{}0$, so $ G \equiv C$.

I dont have that much time, so i'll just write sketches of the proofs i have. I solved integrated by parts twice to construct a continous function $ u$ depending on $ F$ and $ G$ such that $ \int_0^1 x^k u(x) dx \equal{} 0$ for all natural numbers $ k$ and this is a quite known problem. For this i used that $ x^k\minus{}x$ satisfies the conditions to be $ f$.


in how many ways can a 3x3 grid be filled with the first 9 natural nos. such that the sum of  the nos. along a particular row or column is not a multiple of 5


quadratics
calculus
symmetry
analytic geometry
calculus computations

I have a problem with something.  I have points, (20,9.1), (40,7.8), (60,7.1), (80,7.1), (100,7.8), and (120,9.1).  These are the only information given to me.  How would I be abel to figure the vertex of this quadratic line with teh tools of calculus?

Masoud Zargar

It doesn't take any calculus at all; just look at the symmetry and see that it must be at $x=70$. The $y$-coordinate is a little trickier, as we're interpolating with inexact values. Still, it's pretty easy to see that that value is about 7.0.

Well, it is obvious that $x=70$, however, how would I be able to figure out the y-coordinate without graphing?

Masoud Zargar

$(7.8-y)=a\cdot30^2$
$(7.1-y)=a\cdot10^2$
$\frac{7.8-y}{7.1-y}=9$
$.7=8\cdot(7.1-y)$
$y=7.0$

All of this is of course to the number of significant digits we have, which is small.


For each nonzero integer $ n$ find all functions $ f: \mathbb{R} \minus{} \{\minus{}3,0 \} \rightarrow \mathbb{R}$ satisfying:

$ f(x\plus{}3)\plus{}f \left( \minus{}\frac{9}{x} \right)\equal{}\frac{(1\minus{}n)(x^2\plus{}3x\minus{}9)}{9n(x\plus{}3)}\plus{}\frac{2}{n}$ for all $ x \not\equal{} 0,\minus{}3.$

Furthermore, for each fixed $ n$ find all integers $ x$ for which $ f(x)$ is an integer.

[quote="moldovan"]For each nonzero integer $ n$ find all functions $ f: \mathbb{R} \minus{} \{ \minus{} 3,0 \} \rightarrow \mathbb{R}$ satisfying:

$ f(x \plus{} 3) \plus{} f \left( \minus{} \frac {9}{x} \right) \equal{} \frac {(1 \minus{} n)(x^2 \plus{} 3x \minus{} 9)}{9n(x \plus{} 3)} \plus{} \frac {2}{n}$ for all $ x \not \equal{} 0, \minus{} 3.$[/quote]
If the problem statement is correct, no solution exists since the equation must be true for $ x\equal{}3$ but then $ f(\minus{}\frac 9x)\equal{}f(\minus{}3)$ is not defined.
So I think we must understand that $ f: \mathbb R\backslash\{0,\plus{}3\}\to\mathbb R$

With this modification, this equation may be written $ f(x)\plus{}f(\frac{\minus{}9}{x\minus{}3})\equal{}\frac{n\minus{}1}{n}(\minus{}\frac{x\minus{}3}{9}\plus{}\frac 1x)\plus{}\frac 2n$ $ \forall x\notin\{0,3\}$ and so :

Let then $ g(x)\equal{}\frac{\minus{}9}{x\minus{}3}$ (notice that we have $ g(g(g(x)))\equal{}x$ $ \forall x\notin\{0,3\}$

The equation is $ f(x)\plus{}f(g(x))\equal{}\frac{n\minus{}1}{n}(\frac{1}{g(x)}\plus{}\frac{1}{x})\plus{}\frac 2n$ $ \forall x\notin\{0,3\}$

So :
$ f(x)\plus{}f(g(x))\equal{}\frac{n\minus{}1}{n}(\frac{1}{g(x)}\plus{}\frac{1}{x})\plus{}\frac 2n$ $ \forall x\notin\{0,3\}$

$ f(g(x))\plus{}f(g(g(x)))\equal{}\frac{n\minus{}1}{n}(\frac{1}{g(g(x))}\plus{}\frac{1}{g(x)})\plus{}\frac 2n$ $ \forall x\notin\{0,3\}$

$ f(g(g(x)))\plus{}f(x)\equal{}\frac{n\minus{}1}{n}(\frac{1}{x}\plus{}\frac{1}{g(g(x))})\plus{}\frac 2n$ $ \forall x\notin\{0,3\}$

And so, (adding the first and the third and subtracting the second) : $ \boxed{f(x)\equal{}\frac{n\minus{}1}{nx}\plus{}\frac 1n}$ $ \forall x\notin\{0,3\}$
And it is immediate to verify that this necessary condition is also sufficient.

[quote="moldovan"]Furthermore, for each fixed $ n$ find all integers $ x$ for which $ f(x)$ is an integer.[/quote]

$ f(x)\equal{}\frac{x\plus{}n\minus{}1}{nx}$ can be an integer only :
For $ n\notin\{\minus{}2,1\}$ : if $ x\in\{\minus{}(n\minus{}1),1\}$
For $ n\equal{}1$ : $ \forall x\notin\{0,3\}$
For $ n\equal{}\minus{}2$ : if $ x\equal{}1$


Find roots:
$ x^4\plus{}3x^3\minus{}6x^2\minus{}11x\plus{}15\equal{}0$

Not at all hard if you know how to do these.

[hide="Hint"]
See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1083977#1083977]this[/url].$ c' \equal{} \minus{} 8$
[/hide]
[hide="Solution"]
$ \equal{} (x^2 \plus{} x \minus{} 3)(x^2 \plus{} 5x \minus{} 5)$
$ \implies \boxed{x \equal{} \big( \minus{} 1\pm \sqrt {6}\big),\big(\frac { \minus{} 1\pm \sqrt {13}}{2}\big)}$
[/hide]


modular arithmetic
superior algebra
superior algebra unsolved

Problem. Let $G$ be a finite group such that $3 \nmid |G|$, and let $s$ be the number of equivalence classes formed by the elements of $G$. Prove that
\[
s \equiv |G| \pmod{3}.
\]

I have a short and elegant solution to this problem, but I cannot remember its source (it was from some paper that I read quite a long time ago). Perhaps someone can provide a reference.  :)

--Vesselin

Yup. Or, if you want, the number of irreducible complex characters.  :)

I've heard about characters before, but the first time I read a definition and stuff like that was $5-10$ minutes ago. I found this article about characters and group representations on the Internet, and found something that might be useful, but I'm sure you'll be understanding if it turns out to be total bogus :D.

If we write the character table for our group, the order of the group is the sum of the squares of the characters computed in $1_G$. These squares are all $1\pmod 3$, so the order is congruent to the number of characters in the character table.

Does it have even a remote chance of being correct? :D

But some of the $\chi (1)$ may be divisible by 3. Your observation simply states that the problem is equivalent to showing that the number of multiples of 3 among the $\chi (1)$ (or, in other words, the dimensions of the irreducible representation of $G$) is divisible by 3, but this doesn't prove anything.

My one-line solution, however, doesn't even mention characters; I think it would be more convenient to interpret $s$ as the number of equivalence classes rather than as the number of irreducible characters.

Didn't think about that  :blush:  :blush:

Just a hint. Use the fact that the number of solutions to the equation $[x,y]=1$ is equal to $s|G|$. That is, $G^2$ contains precisely $s|G|$ ordered pairs of commuting elements (Frobenius). It should now be easy.  :)

--Vesselin


Find the general term for $ (x_n)_{n\geq 1}$

$ x_{n\plus{}1}\equal{}\frac{\minus{}x_n\plus{}\sqrt{3(1\minus{}x_n^2)}}{2}$ ,$ \forall n\geq 1$, and with $ x_1 \in [\minus{}1,1]$

[hide="A hint"]$ x_1\equal{}\sin \alpha$[/hide].


The square of $ 5\minus{}\sqrt{y^2\minus{}25}$ is:

$ \textbf{(A)}\ y^2\minus{}5\sqrt{y^2\minus{}25} \qquad
\textbf{(B)}\ \minus{}y^2 \qquad
\textbf{(C)}\ y^2 \\
\textbf{(D)}\ (5\minus{}y)^2 \qquad
\textbf{(E)}\ y^2\minus{}10\sqrt{y^2\minus{}25}$

[hide="Click for solution"]
$ (5\minus{}\sqrt{y^2\minus{}25})^2\equal{}25\minus{}10\sqrt{y^2\minus{}25}\plus{}y^2\minus{}25\equal{}y^2\minus{}10\sqrt{y^2\minus{}25}$, or $ \boxed{\textbf{(E)}}$.
[/hide]


calculus
derivative
calculus computations

Consider a variation of Newton's method in which only one derivative is needed, that is
$ x_{n \plus{} 1} \equal{} x_n \minus{} \frac {f(x_n)}{f'(x_n)}$
Find $ C$ and $ s$ such that $ e_{n \plus{} 1} \equal{} Ce_n^s$.
If you have got a solution, please let me know
 miwalin@163.com

[quote="kunny"]What's $ e_n$?[/quote]
$ e_n\equal{}x_n\minus{}\alpha$, where $ f(\alpha)\equal{}0$.

[quote="Lousin Garckz"]any solution to this problem?[/quote]
It has been posted one year ago, up till now, I haven't heard any comment.


inequalities
vector
linear algebra
matrix

If $ R = \left(\begin{array}{cc}R_{11} & R _{12} \\
0 & R_{22} & \end{array}\right)$, where $ R_{11}\in C^{p\times p}$,   $ R_{22}\in C^{q\times q}$, is normal, then $ R_{11}, R_{22}$ are normal.

By Schur unitary triangulization theorem, $ R_{11}$ and $ R_{22}$ can be made upper triangular by unitary matrices $ U_{11}$ and $ U_{22}$.

So

$ \left(\begin{array}{cc}U^{\ast}_{11}& 0\\ 0 & U^{\ast}_{22}&\end{array}\right)\left(\begin{array}{cc}R_{11}& R_{12}\\ 0 & R_{22}&\end{array}\right) \left(\begin{array}{cc}U_{11}& 0\\ 0 & U_{22}&\end{array}\right) =\\
\left(\begin{array}{cc}U^{\ast}_{11}R_{11}U_{11}& \ast\\ 0 & U^{\ast}_{22}R_{22}U_{22}&\end{array}\right)=\\
\left(\begin{array}{cc}D_{11}& \ast\\ 0 & D_{22}&\end{array}\right)=D$

$ D_{11}$ and $ D_{22}$ are both upper triangular with eigenvalues on the diagonal and $ \left(\begin{array}{cc}U_{11}& 0\\ 0 & U_{22}&\end{array}\right)$ is unitary matrix.

Therefore $ A$ is unitarily equivalent to $ D$.

Further on for unitarily equivalent matrices we have
$ \sum^{n}_{i,j=1}|a_{ij}|^2=\sum^{n}_{i,j=1}|d_{ij}|^2$
and for normal matrices we have
$ \sum^{n}_{i,j=1}|a_{ij}|^2=\sum^{n}_{i=1}|\lambda_{i}|^2$, where $ \lambda_i$ denotes the eigenvalue of $ A$.

Now, since $ A$ and $ D$ are unitarily equivalent and $ D$ has eigenvalues of $ A$ on the diagonal, $ D$ must be diagonal ($ \sum^{n}_{i,j=1}|d_{ij}|^2= \sum^{n}_{i=1}|\lambda_{i}|^2+ \sum^{n}_{i<j}|d_{ij}|^2=\sum^{n}_{i=1}|\lambda_{i}|^2$).

So $ U_{11}$ and $ U_{22}$ unitarily diagonalize $ R_{11}$ and $ R_{22}$ and by the spectral theorem for normal matrices: $ M$ is normal if and only if it is unitarily diagonalizable, $ R_{11}$ and $ R_{22}$ are normal.

I hope, I am not wrong.

Am I doing something wrong, or is the problem easier than what we usually expect from miwalin? Since $ R$ is normal, we have $ RR^{\ast} \equal{} R^{\ast}R$. But $ R \equal{} \left(\begin{array}{cc}R_{11} & R_{12} \\
0 & R_{22}\end{array}\right)$ implies $ R^{\ast} \equal{} \left(\begin{array}{cc}R_{11}^{\ast} & 0 \\
R_{12}^{\ast} & R_{22}^{\ast}\end{array}\right)$. Thus, using the standard rules for multiplying block matrices, we get

$ RR^{\ast} \equal{} \left(\begin{array}{cc}R_{11}R_{11}^{\ast} \plus{} R_{12}R_{12}^{\ast} & ... \\
... & ...\end{array}\right)$

and

$ R^{\ast}R \equal{} \left(\begin{array}{cc}R_{11}^{\ast}R_{11} & ... \\
... & ...\end{array}\right)$,

where $ ...$'s stand for entries we don't care about. This already yields $ R_{11}R_{11}^{\ast} \plus{} R_{12}R_{12}^{\ast} \equal{} R_{11}^{\ast}R_{11}$, so that $ R_{12}R_{12}^{\ast} \equal{} R_{11}^{\ast}R_{11} \minus{} R_{11}R_{11}^{\ast}$, and thus

$ \mathrm{Tr}\left(R_{12}R_{12}^{\ast}\right) \equal{} \mathrm{Tr}\left(R_{11}^{\ast}R_{11} \minus{} R_{11}R_{11}^{\ast}\right) \equal{} 0$.

But $ \mathrm{Tr}\left(R_{12}R_{12}^{\ast}\right)$ is the sum of the squares of the lengths of the row-vectors of the matrix $ R_{12}$, and thus can only be $ 0$ if $ R_{12} \equal{} 0$. Now we are left with a block-diagonal matrix, and the rest is easy...

[color=blue][b]Generalization.[/b] Let
$ R \equal{} \left(\begin{array}{ccccc}R_{1,1} & R_{1,2} & R_{1,3} & \hdots & R_{1,n} \\
0 & R_{2,2} & R_{2,3} & \hdots & R_{2,n} \\
0 & 0 & R_{3,3} & \hdots & R_{3,n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \hdots & R_{n,n}\end{array}\right)$
be a block-triangular matrix over $ \mathbb{C}$ (this means, in particular, that $ R_{i,i}$ is a square matrix for every $ i$). Prove that if $ R$ is a normal matrix, then every $ R_{i,i}$ is a normal matrix, and $ R_{i,j} \equal{} 0$ for any $ i < j$.[/color]

  darij

[quote="darij grinberg"]Am I doing something wrong, or is the problem easier than what we usually expect from miwalin? Since $ R$ is normal, we have $ RR^{\ast} \equal{} R^{\ast}R$. But $ R \equal{} \left(\begin{array}{cc}R_{11} & R_{12} \\
0 & R_{22}\end{array}\right)$ implies $ R^{\ast} \equal{} \left(\begin{array}{cc}R_{11}^{\ast} & 0 \\
R_{12}^{\ast} & R_{22}^{\ast}\end{array}\right)$. Thus, using the standard rules for multiplying block matrices, we get

$ RR^{\ast} \equal{} \left(\begin{array}{cc}R_{11}R_{11}^{\ast} \plus{} R_{12}R_{12}^{\ast} & ... \\
... & ...\end{array}\right)$

and

$ R^{\ast}R \equal{} \left(\begin{array}{cc}R_{11}^{\ast}R_{11} & ... \\
... & ...\end{array}\right)$,

where $ ...$'s stand for entries we don't care about. This already yields $ R_{11}R_{11}^{\ast} \plus{} R_{12}R_{12}^{\ast} \equal{} R_{11}^{\ast}R_{11}$, so that $ R_{12}R_{12}^{\ast} \equal{} R_{11}^{\ast}R_{11} \minus{} R_{11}R_{11}^{\ast}$, and thus

$ \mathrm{Tr}\left(R_{12}R_{12}^{\ast}\right) \equal{} \mathrm{Tr}\left(R_{11}^{\ast}R_{11} \minus{} R_{11}R_{11}^{\ast}\right) \equal{} 0$.

But $ \mathrm{Tr}\left(R_{12}R_{12}^{\ast}\right)$ is the sum of the squares of the lengths of the row-vectors of the matrix $ R_{12}$, and thus can only be $ 0$ if $ R_{12} \equal{} 0$. Now we are left with a block-diagonal matrix, and the rest is easy...
  darij[/quote]
  Solution of you the same of me . Very Nice !


inequalities
geometry
trigonometry
function
geometry unsolved

Do you know any proof for Ono's inequality that states:
For any acute triangle with area $ A$

$ 27(b^2\plus{}c^2\minus{}a^2)^2(c^2\plus{}a^2\minus{}b^2)^2(a^2\plus{}b^2\minus{}c^2)^2  \le (4A)^6$

$ \cot A \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{4A} \ , \ \cot B \equal{} \frac {a^2 \plus{} c^2 \minus{} b^2}{4A} \ , \ \cot C \equal{} \frac {a^2 \plus{} b^2 \minus{} c^2}{4A}$

Then the inequality becomes:

$ \cot A \cdot \cot B \cdot \cot C \le \frac {_1}{^9}\sqrt {3} \ \Longleftrightarrow \ \tan A \plus{} \tan B \plus{} \tan C \ge 3\sqrt {3}$

Which follows from Jensen's inequality for the fuction $ \tan (\varphi) , \varphi \in \left(0, \frac {\pi}{2} \right)$

Thanks! However, is there any purely geometric proof?

Area $ A=\frac{\sqrt{xy+yz+zx}}{2} $, $ a^2=y+z $ and the others sides, for an acute-angled triangle. Replace and you will get something obvious. Greetings! :-)


function
calculus
derivative
real analysis
real analysis unsolved

If We have a function f with this Condition:
1.f is continues in all point of interval [a,b] .
2. for all x in this interval we have f´(x)>=0 .
3.The set of points which f´(x)=0 ,and f`(x) does not exist(Critical points) is countable .

Does neccessairly we can say f is increasing?

Well wherever $ f'$ is defined and positive, it is increasing.  If it is not defined at some point, then the function isn't decreasing there.  Same with if $ f'\equal{}0$.

Well, it is not that simple ;) (actually, what Jrav wrote makes little sense, if any). Let me restate the question to avoid any misinterpretations:

Suppose that $ f$ is continuous on $ [a,b]$ and, outside a countable subset of $ [a,b]$, the derivative $ f'$ exists and is strictly positive. Is it necessarily true that $ f(b)>f(a)$?

I believe that the problem has already appeared on the forum but, as far as I remember, it was left unsolved. Let's see if somebody can solve it this time. :)


Why is the hydrogen written second to nitrogen in $NH_{3}$, is it acceptable to write $H_{3}N$?

the notation is that acids are written then bases , except for metals that is


I'm studying for my matric exam in math this monday. PLEASE help someone!!!
I am in the middle of an equation and I have the answer but I don't understand how to get from step 2 to step 3!?
Here is the sum in full:

find(without a calculator) the value of:
1 sin^2(-100) / cos190sin40cos400
2 = (-sin80)^2 / (-cos10)sin40cos40
3 = sin^280 / (-sin80)(1/2)(sin80)
4 = -2

My calculator gives the same answer as the memo but I still don't understand how sin40 times cos40 can give (1/2)sin80??

Can anyone please help?

$ \sin 80\equal{}2\sin 40\cos 40$ by the [url=http://mathworld.wolfram.com/Double-AngleFormulas.html]Double Angle Formula[/url] for Sin, therefore, $ \dfrac{\sin 80}{2}\equal{}\sin 40\cos 40$

thanks a lot TonyP, I only realize now how incredibly easy that actually is, can't believe I didn't think of that. Guess I'll have to keep practising...
 :dry:


algebra
polynomial
abstract algebra
Galois Theory
group theory
superior algebra
superior algebra unsolved

if we add the 15th root of unity to the field of rational numbers. why do we get a Galois extention!? It is a separable extention, the problem is whether it is normal.
I thought this way: if $ Z_{15}$ is the 15th root of unity and a root of $ X^{15} \minus{} 1 \equal{} 0$, then  every power of $ Z_{15}$ is also a solution. So adding $ Z_{15}$ is equivalent to adding all the roots of the polynomial $ X^{15} \minus{} 1$. Now the minimal polynomial of $ Z_{15}$ divides  $ X^{15} \minus{} 1$, so we can find all roots of the minimal polynomial  by just adding $ Z_{15}$.
Is this correct? Is there another way?

Consider the action of the Galois group on a polynomial that has a linear factor.

You need to be slightly more specific when talking about 15th roots of unity, since some are primitive and others are not.

You are, however, basically correct.  If $ \zeta\equal{}e^{\frac{2 \pi i}{15}}$ then the roots of $ X^{15}\minus{}1$ are exactly the powers of $ \zeta$, so $ \mathbb{Q}(\zeta)$ is the splitting field of this polynomial.

On the other hand, if you add the "wrong" root of unity, you will still get a Galois extension, just one for $ X^3\minus{}1$ or $ X^5\minus{}1$.

SO in all cases we still have a Galois extention, thank you.
t0rajir0u: what do you mean exactly? THe first thing I think of is that the Galois group is not acting on a transitive way on the roots. But what did you want to say?

Another question: if we add $ (-3)^{\frac{1}{6}}$ then we get a Galois extention  over $ \mathbb{Q}$ of degree 6: the possible groups are in this case the cylic group or the dihedral group of order 6. I think it is the seconde one: $ \mathbb{Q}((3)^\frac{1}{3})$ over $ \mathbb{Q}$  is not a normal extention, if I can prove it is a subfield of $ \mathbb{Q}((-3)^\frac{1}{6})$ then we finished, because all subfield of an Abelian extention are normal. But I can't prove it, can anyone help?

The Galois group is cyclic of order $ 6$, with generator $ \alpha \to \zeta \alpha$, where $ \alpha^6\equal{}\minus{}3$ and $ \zeta$ is a primitive 6th root of unity.  $ \sqrt[3]{3} \notin \mathbb{Q}(\alpha)$.

[quote="LydianRain"]The Galois group is cyclic of order $ 6$, with generator $ \alpha \to \zeta \alpha$, where $ \alpha^6 \equal{} \minus{} 3$ and $ \zeta$ is a primitive 6th root of unity.  $ \sqrt [3]{3} \notin \mathbb{Q}(\alpha)$.[/quote]
this can't be true. I think you make a mistake. Because $ \sqrt [3]{3}$=$ \minus{}\sqrt [3]{\minus{}3}$. There are three subfields of degree 3 over $ \mathbb{Q}$ and one of degree 2 over  $ \mathbb{Q}$ namely $ \mathbb{Q}(\zeta_{3})$

You are right, I did make a mistake, $ \sqrt[3]{3} \in \mathbb{Q}(\alpha)$.  (Be careful with this kind of notation, what we really mean to say is that there is a subfield isomorphic to $ \mathbb{Q}(\sqrt[3]{3})$)

As you say, that means that the Galois group is the dihedral group of order $ 6$.

The mistake with the "generator" I provided is that $ \zeta$ also changes under this automorphism, so that $ \alpha \to \zeta \alpha$ has order $ 2$.  That's another way to see the Galois group, as we can explicitly construct three automorphisms of order $ 2$.


Euler
geometry
circumcircle
linear algebra

Prove that the altitudes of a triangle are concurrent using vectors.

Use the Euler line: if the origin is the circumcenter, the orthocenter of $ABC$ is $A+B+C$.
Existence of the circumcenter is easy, of course.


There are $ 9$ black ball and $ 7$ white ball in a box
Take $ 1$ ball from the box, and without return the first ball into the box, take again $ 1$ ball.
Without return the first and second ball, take $ 1$ ball again, and without return the first, second, and third ball, take $ 1$ ball again
How many ways such that the first ball and second ball is same with the third and fourth ball?

Example:
1st ball: Black
2nd ball: White
3rd ball: White
4th ball: Black

That is one ways


To find the height of a pole, a surveyor moves 60 feet away from the base of the pole and then, from an eye-level height of 6 feet.  measures the angle of elevation to the top of the pole to be 26 degrees.  Find the height of the pole to the nearest foot.

Basic trig.  $ \tan{26^{\circ}}\equal{}\frac{h}{60}\Rightarrow h\equal{}29.26$.  Add in the extra 6 feet and we get $ h\approx 35\mathrm{ft}$.


Express the sum $ \frac{8}{9}$ and $ \frac{9}{8}$ as a repeating decimal.

idk if therz an ezer way... but i'd add the fractions

9/8 + 8/9 = 17/72 (2 lazy 2 use latex)

and then i'd divide 17 by 72

.236111111111111111111111111111111111111

my bad...  :blush:

thats what went wrong  :rotfl:


The number $ 1$ is written on a black board, using the operations $ A$ and $ B$ find a sequence of operations to get to $ \dfrac{2009}{2^{2009}}$.

Operation $ A$: Divides the number by $ 2$.
Operation $ B$: Subtract the number to $ 1$.

Examples:

If you use the operation $ A$ on $ \dfrac{5}{8}$ you get $ \dfrac{5}{8}\cdot \dfrac{1}{2} \equal{} \dfrac{5}{16}$.

If you use the operation $ B$ on $ \dfrac{7}{16}$ you get $ 1\minus{}\dfrac{7}{16}\equal{}\dfrac{9}{16}$.

[hide="My answer"]
$ 3A,B,2A,B,A,B,5A,B,1998A$
[/hide]

[hide="We start by working"]!!!SDRAWKCAB

Note that $ \frac{2048}{2048}\minus{}\frac{2009}{2048}\equal{}\frac{39}{2048}$, and $ \frac{64}{64}\minus{}\frac{39}{64}\equal{}\frac{25}{64}$, and $ \frac{32}{32}\minus{}\frac{25}{32}\equal{}\frac{7}{32}$, and $ \frac{8}{8}\minus{}\frac{7}{8}\equal{}\frac{1}{8}$. Hence we can do AAABAABABAAAAAB(lots of A's) to get $ \frac{2009}{2^{2009}}$.[/hide]

In other words, I agree with you.

Can anyone generalize? I was looking at the kinds of numbers that are attainable.

$ \frac {1}{2^n}\implies\frac {2^{n} \minus{} 1}{2^n}\implies\frac {2^n \minus{} 1}{2^{n \plus{} 1}}\implies \frac {2^{n \plus{} 1} \minus{} 2^{n} \plus{} 1}{2^{n \plus{} 1}}$, etc.

I think with induction it would be easy to show that $ \frac {2^{n} \minus{} 2^{n \minus{} 1} \plus{} 2^{n \minus{} 2} \plus{} \cdots\pm 1}{2^n}$ is attainable.

The numerator of that is $ \sum_{k \equal{} 0}^{n}( \minus{} 2)^k \equal{} \frac {( \minus{} 2)^{n \plus{} 1} \minus{} 1}{ \minus{} 3}$, so we can prove that any number of the form $ \frac {( \minus{} 1)^n\cdot 2^{n \plus{} 1} \plus{} 1}{3\cdot 2^n}$ can be attained, which is a cool result.

Edit: Actually I'm slightly mistaken (that sum is incorrect), but it's salvageable. I'll prove it after I've finished attempting to study for my history test...

[quote="1=2"][hide="We start by working"]!!!SDRAWKCAB

Note that $ \frac {2048}{2048} \minus{} \frac {2009}{2048} \equal{} \frac {39}{2048}$, and $ \frac {64}{64} \minus{} \frac {39}{64} \equal{} \frac {25}{64}$, and $ \frac {32}{32} \minus{} \frac {25}{32} \equal{} \frac {7}{32}$, and $ \frac {8}{8} \minus{} \frac {7}{8} \equal{} \frac {1}{8}$. Hence we can do AAABAABABAAAAAB(lots of A's) to get $ \frac {2009}{2^{2009}}$.[/hide]

In other words, I agree with you.[/quote]

I'm glad that we got the same answer, I'm hoping for a gold on the test  :lol: .

@grn_trtle: that is pretty cool, if you do prove it, I would like to see the proof.

This is almost the same answer I wrote when I did the test.
grn_trtle is right. You always have to start with Operation A and so you will get a number of the form $ \dfrac{1}{2^n}$ Then doing operation B gives you:

$ 1 \minus{} \dfrac{1}{2^n} \equal{} \dfrac{2^n \minus{} 1}{2^n}$ If you repeat this process you will get what grn_turtle said, the only difference will be the signs.

Now notice that $ 2009 \equal{} 2^ {11} \minus{} 2^6 \plus{} 2^5 \minus{} 2^3 \plus{} 1$ which is easy to get with the strategy I described.

(Starting off with number 1) Operation A 3 times you will have $ \dfrac{1}{2^3}$
Operation B once and you will get $ \dfrac{2^3 \minus{} 1}{2^3}$
Operation A twice to get $ \dfrac{2^3 \minus{} 1}{2^5}$ and then Operation B to get
$ \dfrac{2^5 \minus{} 2^3 \plus{} 1}{2^5}$. Now Operation A and then B and you will have:

$ \dfrac{2^6 \minus{} 2^5 \plus{} 2^3 \minus{} 1}{2^6}$  Finally just do operation A 5 times and then operation B once and you will have :

$ \dfrac{2^ {11} \minus{} 2^6 \plus{} 2^5 \minus{} 2^3 \plus{} 1}{2^ {11}} \equal{} \dfrac{2009}{2^{11}}$

Now you just have to apply operation A 1998 times to get the desired result.


Does there exist a sequence $c_n\rightarrow 0$ such that for every continuous
function $f$ on the unit circle there are partial sums of
$\sum_{n\geq 1}c_nz^n$ tending to $f$ pointwise ?

Below is the response I got from Alexander Olevskii, who is the best expert on such questions I know:

"In fact, it is easy to prove that no series $\sum_{n\in\mathbb Z} c_n e^{int}$, $c_n=O(1)$, may have two subsequences of partial sums that converge for all $t$  to $f=0$  and  $g=1$ respectively".

Don't ask me how to prove it and why it is easy, though. I don't know... :lol:


analytic geometry
combinatorics proposed
combinatorics

We call a set of points [i]free[/i] if there is no equilateral triangle with the vertices among the points of the set. Prove that every set of $n$ points in the plane contains a [i]free[/i] subset with at least $\sqrt{n}$ elements.

Let $f(n)$ be the minimal size of the largest free subset among any distribution of $n$ points in the plane.
We have to prove that $f(n) \geq \sqrt{n}$ for all $n$.

Note that $f$ is clearly non-decreasing, so that it suffices to prove that $f(n^{2}+1) \geq n+1.$

Now, let's consider a set of $n^{2}+1$ points in the plane. Since it is a finite set of points, we may choose a system of coordinates such that no two of the given $n^{2}+1$ points have the same $x$-coordinate.
Let $M_{1}, \cdots , M_{n^{2}+1}$ be the given points with $x_{1}< \cdots < x_{n^{2}+1}.$

Now, the $y$-coordinates form a sequence of length $n^{2}+1$. But it is well-known that such a sequence contains a non-decreasing subsequence of length $n+1$ or a non-increasing sequence of length $n+1$. Wlog, we may assume that we are in the first case, and let $P_{1}, \cdots ,P_{n+1}$ be some $n+1$ of the initial set of points such that $x_{i}< x_{j}$ and $y_{i}\leq y_{j}$ when $i<j$.
It is a straightforward verification that this set of $n+1$ points cannot contains the three vertices of an equilateral triangle.
Hence result.

Pierre.

Another solution:

Let $F$ be maximal free subset of set $A$ with $n$
elements. Let $|F|=k$, then $|F'| = n-k$, $F' = A\setminus F$.

Connect point $X\in F'$  with unordered pair $\{Y,Z\}\subset F$ iff $XY = XZ=YZ$.

Then each point $X$ has degree at least 1 since it would
contradict maximality of set $F$ and each pair  $\{Y,Z\}$ has
degree at most 2 (by contradiction, if pair is connected with
three points, say $A,B,C$ then all triangles $AYZ$, $BYZ$ and
$CYZ$ are equilateral, so two points from set $\{A,B,C\}$ are
equal). Now we have

\[\;\;\;\;\;2\cdot{k\choose 2}\geq n-k\]

and problem is solved.

Now that you re-opened this old thread with this simple question, here is an immediate formal proof.

For any finite configuration $C$ of points in the plane, denote by $F_C$ a largest free subset of $C$. Then, for $C\subset C'$, we have $|F_C| \leq |F_{C'}|$, since $F_C \subseteq C \subset C'$, thus $F_C$ is a free subset of $C'$.
But we want to prove $f(n) = \min_{|C| = n} |F_C| \leq  \min_{|C'| = n+1} |F_{C'}| = f(n+1)$, and according to the above, $|F_C| \leq |F_{C'}|$ for any $C\subset C'$ with $|C|=n$ and $|C'| = n+1$, so the claim is obvious.


Let $n$ be a positive integer which is not prime. Prove that there exist $k, a_{1},a_{2},...a_{k}>1$ positive integers such that $a_{1}+a_{2}+\cdots+a_{k}=n(\frac1{a_{1}}+\frac1{a_{2}}+\cdots+\frac1{a_{k}})$

Edit: the $a_{i}'s$ have to be grater than 1. Sorry, my mistake :blush:

[quote="felixmann"]Let $n$ be a positive integer which is not prime. Prove that there exist $k, a_{1},a_{2},...a_{k}$ positive integers such that $a_{1}+a_{2}+\cdots+a_{k}=n(\frac1{a_{1}}+\frac1{a_{2}}+\cdots+\frac1{a_{k}})$[/quote]
It is obviosly, that work $k=2n,a_{1}=a_{2}=...=a_{n}=n,a_{n+1}=a_{n+2}=...=a_{2n}=1.$ (n can be prime)


Find the maximum and the minimum of $\frac{1}{P}$ where
$P=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{y}{x}-\frac{z}{y}-\frac{x}{z}$
Where $x,y,z$ are real numbers satisfying:
$x^{3}+y^{3}+z^{3}+3xyz = xy(x+y)+yz(y+z)+zx(z+x)$

[quote="Anh Cuong"]Find the maximum and the minimum of
$P=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{y}{x}-\frac{z}{y}-\frac{x}{y}$
Where $x,y,z$ are real numbers satisfying:
$x^{3}+y^{3}+z^{3}+3xyz = xy(x+y)+yz(y+z)+zx(z+x)$[/quote]

THe equation is just the equality case of Schur's, but that means either one variable is 0 or x=y=z. since there cannot the division by zero, x=y=z and the only value can be 0.

[quote="rem"]But $x,y,z$ are just reals and [b]not necessarily non-negative reals[/b].[/quote]

yeah, sorry. forget what i said.

also, is the expression supposed to be symmetric?

Ahh, yes it is symmetric indeed, sorry for the typo, i fixed it already. Have fun  :) .

if my calculations are correct then P can take any real value...

Can you give the solution, i'm sure it is not.  :wink:

$P=\frac{(x-y)(y-z)(z-x)}{xyz}$ now put $y=x+m, z=x+n$
and we have $P=\frac{(m^{2}+n^{2}-mn)^{3}mn}{(m-n)(m+n)(2m^{3}+n^{3}-2m^{2}n)(2n^{3}+m^{3}-2n^{2}m)}$ and taking $m\rightarrow n$ $P\rightarrow \infty$
$m\rightarrow 0$ $P\rightarrow 0$
So P can take any real value...

Are you sure your calculations are right? If $m \to n$ then in the original statement of the problem it is $y = z = x+m$ and that gives $P=0$...

I got that (m-n) is a double root in the denominator. :wink:

Sorry to anybody has tried this problem, actually it must be finding the max and min of $\frac{1}{P}$ as I edited. :)

because of my above post I still think that it can take any real value...

Not really sure, my answer is that:
$\frac{-2\sqrt{3}}{9}\leq \frac{1}{P}\leq \frac{2\sqrt{3}}{9}$  :wink: .


calculus
integration
logarithms
calculus computations

$ \int _0^1 \frac {x^b \minus{} 1}{\ln x}dx \equal{} ?$

[hide]Set $ f(b) \equal{} \int_0^1\frac{x^b\minus{}1}{\ln x}\,dx$. Then $ \frac{df}{db} \equal{} \int_0^1 \frac{d}{db}\left(\frac{x^b\minus{}1}{\ln x}\right) \,dx \equal{} \int_0^1 x^b \,dx \equal{} \frac{1}{b\plus{}1}$. Therefore,

$ df \equal{} \frac{1}{b\plus{}1} db \equal{}> f(b) \equal{} \ln(b\plus{}1) \plus{} C$.

Since f(0) = 0, we have C = 0, and so $ f(b) \equal{} \ln(b\plus{}1)$.[/hide]

[quote="Carcul"][hide]Set $ f(b) \equal{} \int_0^1\frac {x^b \minus{} 1}{\ln x}\,dx$. Then $ \frac {df}{db} \equal{} \int_0^1 \frac {d}{db}\left(\frac {x^b \minus{} 1}{\ln x}\right) \,dx \equal{} \int_0^1 x^b \,dx \equal{} \frac {1}{b \plus{} 1}$. Therefore,

$ df \equal{} \frac {1}{b \plus{} 1} db \equal{} > f(b) \equal{} \ln(b \plus{} 1) \plus{} C$.

Since f(0) = 0, we have C = 0, and so $ f(b) \equal{} \ln(b \plus{} 1)$.[/hide][/quote]
Thanks to you too Carcul.


algorithm
number theory unsolved
number theory

The positive integers $ a_1, a_2, ..., a_n$ have $ 1$ as their highest common factor. Any two of them, moreover, have the same least common multiple. Prove that there exists an integer $ p$ such that, for any integer $ u$, exactly one of the numbers $ u$ and $ p\minus{}u$ can be expressed in the form $ a_1x_1\plus{}a_2x_2\plus{}...\plus{}a_nx_n$ for some nonnegative integers $ x_1, x_2, ..., x_n$.

A nice problem. I can't solve it yet, but I can provide this perhaps helpful lemma.
We show that any sufficiently large positive integer can be written as a linear combination of the $ a's$ with non-negative coefficients (call this a "positive representation").

Since $ a_1, ..., a_n$ are coprime we can write

$ 1 \equal{} a_1y_1 \plus{} ... \plus{} a_ny_n \: \: (y_i \in \Bbb{Z})$

Let

$ k \equal{} a_1|y_1| \plus{} ... \plus{} a_n|y_n|$.

Suppose $ j > k^2$. Then we can write, by the division algorithm,

$ j \equal{} qk \plus{} r$

where $ q\geq k, 0\leq r < k$. This becomes

$ j \equal{} q(a_1|y_1| \plus{} ... \plus{} a_n|y_n|) \plus{} r(a_1y_1 \plus{} ... \plus{} a_ny_n)$
$ \equal{} a_1(q|y_1| \plus{} ry_1) \plus{} ... \plus{} a_n(q|y_n| \plus{} ry_n)$
and since $ q\geq k > r$, each of the coefficients is positive. Hence every integer $ j > k^2$ has a positive representation.

It's also easy to see that $ p$ must be the greatest unrepresentable number. Indeed, $ 0$ is representable hence $ p \minus{} 0 \equal{} p$ must not be representable. But $ \minus{} 1, \minus{} 2,...$ are not representable, hence $ p \plus{} 1, p \plus{} 2, ...$ must all be representable.
It's also easy to see that $ u$ and $ p \minus{} u$ cannot both be representable, or else their sum $ p$ would also be representable, contradicting the above.
The difficulty lies in showing that, for all $ 0 < u < p$, at least one of $ u$ or $ p \minus{} u$ is representable. For all other values of $ u$ the theorem is obvious.


calculus
derivative
function
calculus computations

Let $ f$ and $ k$ be differentiable functions such that $ f'(x)\equal{}k(x)$ and $ k'(x)\equal{}\minus{}f(x)$.  Let $ A(x)\equal{}\left [f(x) \right ]^2 \plus{} \left [k(x) \right ]^2$.  Find $ A'(x)$.

hello, we get by the chain rule
$ A'(x)\equal{}2f(x)f'(x)\plus{}2k(x)k'(x)\equal{}2f(x)k(x)\plus{}2k(x)(\minus{}f(x))\equal{}0$
Sonnhard.

Just by substituting that $ f'(x) \equal{} k(x)$ and $ k'(x) \equal{} \minus{} f(x)$ into the expression for the derivative of $ A(x)$ which is:

$ A'(x) \equal{} 2f(x)f'(x) \plus{} 2k(x)k'(x)$ you get that $ A'(x) \equal{} 2f(x)k(x) \minus{} 2k(x)f(x) \equal{} 0$.

An example of these functions could be $ f(x) \equal{} sinx$ and $ k(x) \equal{} cosx$

Ps sorry I hadn't seen Dr.Sonnhard's post


Fish?  Once I went fishing at a pond.  I caught just one fish.  Several months later . . . all the fish were gone.  For me anyway, non-linear differential equations answer many of the puzzling questions about our world in way nothing else can.  And if I can impart the most significant reason for studying them to students in here, it is that one. :)

Here's one in case some of you are into Biology:

I hold the entire foundation of Stephen J. Gould's Theory of "Punctuated Equilibrium" rests on the bedrock of non-linear dynamics.

[quote="shawtoos"]I hold the entire foundation of Stephen J. Gould's Theory of "Punctuated Equilibrium" rests on the bedrock of non-linear dynamics.[/quote]
I don't know- I see it as more of a statistical fact. The rate of change of the mean in an undirected random walk is inversely proportional to the square root of the population size, so small populations change rapidly and large populations remain stable. The "controversial" part is that natural selection ordinarily keeps species stable over long periods of time.

If the system is sufficiently complex, random models usually work well. Evolutionary biology is that complex.

Hello JMerry.  Here's the part I'm referring to:  "...evolution takes place in rapid bursts, separated by long periods in which little change occurs."

The rapid bursts part is due in my opinion to catastrophe:  a critical point in the dynamics of the system is reached which often leads to qualitatively different change. Take for example the cubic differential equation:

$ \frac{dx}{dt}\equal{}a\minus{}bx\minus{}x^3$

It exhibits the cusp-catastrophe.  The plot below is the equilibrium surface for that equation.  I imagine 550 or so million years ago,  all of life on earth represented by the yellow trajectory on the top fold.  Something, as yet unknown, pushed life towards the edge of the cusp (towards the catastrophe set).  It didn't just move onto the underside of the surface, but rather, fell off abruptly along the green trajectory (the catastrophe)  . . . disparate clans emerged shortly thereafter along the red trajectory.  I realize evolution is not this simple but the principle in my opinion does a nice job explaining the theory; same for the divergence of man from apes.

[img]http://img91.imageshack.us/img91/306/cubicdejv4.jpg[/img]

On a thread about whether paths were closed in phase space for a differential equation, I posted a simple Lotka-Volterra system. After that, the discussion got off track. I'm creating this topic as a place to move some of those posts.


Okay, I'm having a little trouble with this so please help. This is Problem #8 on Workout 7 of 06-07.

I have two one-quart bottles; the first is filled with water, the second bottle is empty. I pour half of the contents of the first bottle into the second, then one-third of the contents of the second bottle into the first, then one-quarter of the contents of the first bottle into the second, and then one-fifth of the contents of the second into the first, and so-on. How much water is in the first bottle after the 2007th pour? Express your answer as a common fraction.

Notice that every time you pour something out of the first bottle you have 1/2 remaining. So the answer is 1/2.


probability
number theory
prime factorization
puzzles

A standard die with sides labeled 1 through 6 is rolled three times. Find the probability that the product of the three numbers rolled is divisible by 8. Express your answer as a rational number in lowest terms.

[hide]You can verify ur solution here http://www.projecteureka.org/problem/question/283 [/hide]

We see that there has to be at least three 2's in the prime factorization of the product because the prime factorization of 8 is $ 2^3$. We can do casework to find the answer, but we can also do complementary counting. I am not good at counting and I am almost sure that I will make a mistake, so I won't post a solution.


Let $ x, y \in R$ and $ x^2\plus{}y^2\equal{}1$. Find minimum of $ A\equal{}x\sqrt{1\plus{}y}\plus{}y\sqrt{1\plus{}x}$.

It's very easy.
Clearly we only consider in the cases $ \minus{} 1\leq x\leq 0, \minus{} 1\leq y \leq 0$.
We have $ \minus{} x \minus{} y \equal{} |x| \plus{} |y|\geq x^2 \plus{} y^2 \equal{} 1\Rightarrow x \plus{} y\leq \minus{}1$
Therefore $ |A|\leq \sqrt {(x^2 \plus{} y^2)(2 \plus{} x \plus{} y)}\leq\sqrt {2 \minus{} 1} \equal{} 1$
Hence A get minimum value equal -1 if x=-1,y=0

[quote="thanhnam2902"]Let $ x, y \in R$ and $ x^2 \plus{} y^2 \equal{} 1$. Find minimum of $ A \equal{} x\sqrt {1 \plus{} y} \plus{} y\sqrt {1 \plus{} x}$.[/quote]
hint :
 $ x^2 \plus{} y^2 \equal{} 1$ thus  $ \minus{} \sqrt {2} \leq x \plus{} y \leq \sqrt {2}$
 let $ x \plus{} y \equal{} u$  then  $ xy \equal{} \frac {u^2 \minus{} 1}{2}$
 $ 2A^2 \equal{} 2(xy(x \plus{} y) \plus{} x \plus{} y \plus{} 2xy\sqrt {1 \plus{} x \plus{} y \plus{} xy}) \equal{} u^3 \minus{} u \plus{} \sqrt {2}(u^2 \minus{} 1)|u \plus{} 1|$
if $ u \leq \minus{} 1$  then $ 2A^2 \equal{} u^3 \plus{} u \plus{} \sqrt {2}(1 \minus{} u^2)(1 \plus{} u)$
if$ u \geq \minus{} 1$ then $ 2A^2 \equal{} u^3 \plus{} u \plus{} \sqrt {2}(u^2 \minus{} 1)(1 \plus{} u)$

[quote="hien"]It's very easy.
Clearly we only consider in the cases $ \minus{} 1\leq x\leq 0, \minus{} 1\leq y \leq 0$.
We have $ \minus{} x \minus{} y \equal{} |x| \plus{} |y|\geq x^2 \plus{} y^2 \equal{} 1\Rightarrow x \plus{} y\leq \minus{} 1$
Therefore $ |A|\leq \sqrt {(x^2 \plus{} y^2)(2 \plus{} x \plus{} y)}\leq\sqrt {2 \minus{} 1} \equal{} 1$
Hence A get minimum value equal -1 if x=-1,y=0[/quote]

[size=150][color=darkblue]Hi, Hien! You are wrong.[/color][/size][size=150][color=darkblue]This is my solution:[/color][/size]

[color=darkblue]$ \bullet$ If $ xy\ge 0 \Longrightarrow minA\equal{}\minus{}1$, when $ x\equal{}0, y\equal{}\minus{}1$ OR $ x\equal{}\minus{}1,y\equal{}0$.

$ \bullet$ If $ xy<0$, let $ t\equal{}x\plus{}y \Longrightarrow$ $ \begin {cases} xy\equal{} \frac{t^2\minus{}1}{2}<0\\\minus{}1<t<1\end {cases}$

But $ x^2\plus{}y^2\equal{}1 \Longrightarrow 2A^2\equal{}f(t)\equal{}(1\plus{}\sqrt{2})t^3\plus{}\sqrt{2}.t^2\minus{}(1\plus{}\sqrt{2})t\plus{}2\minus{}\sqrt{2}$

$ f'(t)\equal{}0 \Longleftrightarrow t_1\equal{}\minus{}\frac{1\plus{}\sqrt{2}}{3} , t_2\equal{}\sqrt{2}\minus{}1$

$ \Longrightarrow maxf(t)\equal{}f(t_1)\equal{}\frac{76\minus{}12\sqrt{2}}{27}$

$ \Longrightarrow |A| \le \sqrt{\frac{38\minus{}6\sqrt{2}}{27}}$

$ \Longrightarrow minA \equal{}\minus{}\sqrt{\frac{38\minus{}6\sqrt{2}}{27}}<\minus{}1$.[/color]

[color=darkblue]I think my problem can done by other way. Who can give me other way?[/color]

[quote="Flakky"]
So $ \min(x\sqrt {y + 1} + y\sqrt {x + 1}) = - \sqrt {2 + \sqrt {2}}$
equality $ x = y = - \frac {\sqrt {2}}{2}$
[/quote]

Sorry I have tried to input $ x=y=-\frac{\sqrt{2}}{2}$ into $ x\sqrt{1+y}+y\sqrt{1+x}$, and it seems to me that the output will be $ -\sqrt{2-\sqrt{2}}$, not $ -\sqrt{2+\sqrt{2}}$.

Sergey Markelov
markelov@mccme.ru

[quote="Flakky"]Okay here is my proof:
if we find some $ x_0$ and $ y_0$ that $ x_0\sqrt {y_0 + 1} + y_0\sqrt {x_0 + 1} = min(x\sqrt {y + 1} + y\sqrt {x + 1})$ $ x_0$ and $ y_0$ must negative otherwise contradiction. So it's sufficiently to find $ A = max(x\sqrt {y + 1} + y\sqrt {x + 1})$, $ x > 0$, $ y > 0$ (cases x=0 or y=0 are trivial)
By Cauchy–Schwarz inequality we have:
$ (x\sqrt {y + 1} + y\sqrt {x + 1})^2\le(x^2 + y^2)(y + 1 + x + 1) \Longleftrightarrow x\sqrt {y + 1} + y\sqrt {x + 1} \le \sqrt {x + y + 2}$
But $ \frac {(x + y)^2}{2}\le x^2 + y^2 = 1 \Longrightarrow x + y\le \sqrt {2}$
Hence $ x\sqrt {y + 1} + y\sqrt {x + 1} \le \sqrt {2 + \sqrt {2}}$ $ \Longrightarrow$ $ A = \sqrt {2 + \sqrt {2}}$
So $ min(x\sqrt {y + 1} + y\sqrt {x + 1}) = - \sqrt {2 + \sqrt {2}}$
equality $ x = y = - \frac {\sqrt {2}}{2}$
$ - \sqrt {2 + \sqrt {2}} < - \sqrt {\frac {38 - 6\sqrt {2}}{27}}$[/quote]
[color=darkblue]Thank you very much. I will see again my solution. I think you are true, because: $ - \sqrt {2 + \sqrt {2}} < - \sqrt {\frac {38 - 6\sqrt {2}}{27}}$[/color]

[quote="Flakky"]Okay here is my proof:
if we find some $ x_0$ and $ y_0$ that $ x_0\sqrt {y_0 + 1} + y_0\sqrt {x_0 + 1} = min(x\sqrt {y + 1} + y\sqrt {x + 1})$ $ x_0$ and $ y_0$ must negative otherwise contradiction. So it's sufficiently to find $ A = max(x\sqrt {y + 1} + y\sqrt {x + 1})$, $ x > 0$, $ y > 0$ (cases x=0 or y=0 are trivial)
By Cauchy–Schwarz inequality we have:
$ (x\sqrt {y + 1} + y\sqrt {x + 1})^2\le(x^2 + y^2)(y + 1 + x + 1) \Longleftrightarrow x\sqrt {y + 1} + y\sqrt {x + 1} \le \sqrt {x + y + 2}$
But $ \frac {(x + y)^2}{2}\le x^2 + y^2 = 1 \Longrightarrow x + y\le \sqrt {2}$
Hence $ x\sqrt {y + 1} + y\sqrt {x + 1} \le \sqrt {2 + \sqrt {2}}$ $ \Longrightarrow$ $ A = \sqrt {2 + \sqrt {2}}$
So $ min(x\sqrt {y + 1} + y\sqrt {x + 1}) = - \sqrt {2 + \sqrt {2}}$
equality $ x = y = - \frac {\sqrt {2}}{2}$
$ - \sqrt {2 + \sqrt {2}} < - \sqrt {\frac {38 - 6\sqrt {2}}{27}}$[/quote]
 u're wrong.
 i think that if $ A^2 \geq a$  and $ A < 0$ then $ A \geq - \sqrt {a}$

But $ \frac {(x + y)^2}{2}\le x^2 + y^2 = 1 \Longrightarrow x + y\le \sqrt {2}$ then  occur equal  if and only if : $ x=y =\sqrt{2}/2$ and  obvious $ x,y > 0$

[quote="kunny"]Let $ x + y = s,\ xy = t$, $ x^2 + y^2 = 1\Longleftrightarrow s^2 - 2t = 1$ and $ s^2 - 4t\geq 0\Longleftrightarrow t = \frac {s^2 - 1}{2}$ and $ - \sqrt {2}\leq s\leq \sqrt {2}$.

Now let $ X = x\sqrt {1 + y},\ Y = y\sqrt {1 + x}$, $ X^2 + Y^2 = 1 + st\Longleftrightarrow X^2 + y^2 = 1 + s\cdot \frac {s^2 - 1}{2} = \frac {1}{2}s^3 - \frac {1}{2}s + 1\ ( - \sqrt {2}\leq s\leq \sqrt {2})$.

$ f'(s) = \frac {3}{2}s^2 - \frac {1}{2})$, $ f( - \sqrt {2})\leq f(s)\leq f(\sqrt {2})\Longleftrightarrow 1 - \frac {1}{\sqrt {2}}\leq X^2 + Y^2\leq 1 + \frac {1}{\sqrt {2}}$, yielding $ - \sqrt {2 + \sqrt {2}}\leq A = X + Y\leq \sqrt {2 + \sqrt {2}}$.

The desired minimum value is $ - \sqrt {2 + \sqrt {2}}$ when $ (X,\ Y) = ( - \frac {\sqrt {2 + \sqrt {2}}}{2}, - \frac {\sqrt {2 + \sqrt {2}}}{2} )$.[/quote]

Thank Kunny.


[quote="G-UNIT"]What forbade us from drawing a line perpindicular to two parallel ones?[/quote]
I'm not sure I understand your question. If it's what I think it means, then it's not correct because in a square ABCD AB is perpendicular to both AD and BC, which are two parallel lines.  :wink:

I think he meant this: Without this particular statement, you cannot draw a line perpendicular to two parallel lines. What is this statement?

using euclid's first 3 postulates, you can easliy construct a line that is perpendicular to one of the lines, but this does not prove that the line is also perpendicular to the second line, i think you either need the parallel postulate and the fact that right angles are equal (the next 2 postulates of euclid)

of couse i am not sure about this solution, it is sort of an open ended question

Although I've never paid much attention to the matter, I recently came across a passage stating that it was impossible to prove two alternate interior angles formed when a transversal cuts across two parallel lines is equal.  Of course if we could draw a line perpindicular to each of the two parallel lines we'd form a trapezoid - knowing two of the angles are 90 the remaining same-side interior angles would have to add up to 180 implying that the alternate interior angles are equal.  Consequently, the extra line must be illegal and being heavy critic of the government am I interested in knowing why.

That's the problem with the definition of parallel lines. Euclid had trouble defining clearly what parallel lines are. In fact, there exist parallel lines that can intersect but that is in another system. You definitely need Euclid's 5th Postulate (Parallel Postulate) to show that.


calculus
integration
trigonometry
function
calculus computations

$ \int_0^{\frac {\pi }{2}} {\frac {{\sqrt {\cos x} }}{{\sqrt {\sin x} \plus{} \sqrt {\cos x} }}dx}$

hello, we have
$ \frac{\sqrt{\cos(x)}}{\sqrt{\sin(x)}\plus{}\sqrt{\cos(x)}}\equal{}\frac{1}{\sqrt{\tan(x)}\plus{}1}$, setting $ z\equal{}\sqrt{\tan(x)}$ we get $ x\equal{}\arctan(z^2)$ and our integral will be
$ 2\cdot \int\frac{z}{(1\plus{}z)(1\plus{}z^4)}\,dz$.
Sonnhard.

$ I\equal{}\int_0^{\frac {\pi }{2}} {\frac {{\sqrt {\cos x} }}{{\sqrt {\sin x} \plus{} \sqrt {\cos x} }}dx} \equal{} \int_0^{\frac {\pi }{2}} {\frac {{\sqrt {\sin x} }}{{\sqrt {\sin x} \plus{} \sqrt {\cos x} }}dx} \\ \\
\therefore 2I \equal{} \int_0^{\frac {\pi }{2}} dx \equal{} \frac{\pi}{2} \\ \\
\therefore \boxed {I \equal{} \frac{\pi}{4}}$

hello, ok, your result is correct.
Sonnhard.

Interesting question:
Is $ \int_{}^{} {\frac {{\sqrt {\cos x} }}{{\sqrt {\sin x} \plus{} \sqrt {\cos x} }}dx}$ can be represented in elementary functions?

I don't think that's an interesting question. With no further investigation, I'd rate that as "highly unlikely." I don't see the motivation for going beyond that.

this technique must be quite well know:

$ \int^\frac {\pi}{2}_0 \frac {\sqrt {\cos x}}{\sqrt {\cos x}+\sqrt {\sin x}}dx$ = $ \int^\frac {\pi}{2}_0 \frac {\sqrt {\cos (\frac {\pi}{2}-x)}}{\sqrt {\cos x}+\sqrt {\sin x}}dx$

= $ \int^\frac {\pi}{2}_0 \frac {\sqrt {\sin x}}{\sqrt {\cos x}+\sqrt {\sin x}}dx$

then we denote each integral by I. then adding the two integrals give

$ I + I = 2I =  \int^\frac {\pi}{2}_0 \frac {\sqrt {\sin x}+\sqrt {\cos x}}{\sqrt {\cos x}+\sqrt {\sin x}}dx = \int^\frac {\pi}{2}_0 1 dx
= \frac {\pi}{2}$

so,

$ I = \frac {\pi}{4}$ :clap2:


geometry
trigonometry
number theory
relatively prime

[quote="Hustla25"]Triangle ABC has an in-radius of 5 and a circum-radius of 16. If 2 cosB =
cosA + cosC, then the area of triangle ABC can be expressed as (a(b^1/2))/c,
where a, b, and c are positive integers such that a and c are relatively
prime and b is not divisible by the square of any prime. Compute a + b + c.[/quote]

[hide="hint"]

$ 2\cos{B}\equal{}\minus{}\cos{B}$
$ B\equal{}90$

[/hide]

[quote="azjps"][quote="CircleSquared"][quote="Hustla25"]Triangle ABC has an in-radius of 5 and a circum-radius of 16. If 2 cosB =
cosA + cosC, then the area of triangle ABC can be expressed as (a(b^1/2))/c,
where a, b, and c are positive integers such that a and c are relatively
prime and b is not divisible by the square of any prime. Compute a + b + c.[/quote]

[hide="hint"]

$ 2\cos{B} \equal{} \minus{} \cos{B}$
$ B \equal{} 90$

[/hide][/quote]

Err.. $ \cos A \plus{} \cos B \plus{} \cos C \neq 0$.[/quote]

Ehh hold on there.  Something weird happened with my parentheses.


geometry algebra function domain inequalities integration calculus

[quote="Hong Quy"]find the area of the figure, which occupies the plane domain $ XOY$, defined by the inequality:
$ \frac {\pi}{2} \leq \arctan \frac {y}{x \minus{} 2} \minus{} \arctan \frac { y}{x \plus{} 2} \leq \frac { 5 \pi}{6}$[/quote]

Simplify and you get:

$ \left\{ \begin{gathered}
   \minus{} 2\sqrt 3  \minus{} \sqrt {16 \minus{} x^2}  \leqslant y \leqslant  \minus{} \sqrt {4 \minus{} x^2} , \hfill \\
   \minus{} 2 \leqslant x \leqslant 2. \hfill \\
\end{gathered}  \right.$

So

$ A \equal{} \int\limits_{\minus{}2}^2 \Bigl(\minus{}\sqrt {4\minus{}x^2}\plus{}\sqrt{16\minus{}x^2}  \plus{} 2\sqrt 3\Bigl)dx \equal{} \ldots \equal{} 12\sqrt 3 \plus{} \frac{2\pi }{3}.$

I believe $ A\equal{}2\left(\frac{\pi}{3}\plus{}4\sqrt{3}\right)$.

[quote="kunny"]I believe $ A \equal{} 2\left(\frac {\pi}{3} \plus{} 4\sqrt {3}\right)$.[/quote]

 :)

See carefully this picture

[img]http://s002.radikal.ru/i199/1001/21/699d943bc2ee.png[/img]

I'm very sorry, my calculation was wrong.

I forgott to add the area a half of equitriangle with side length of $ 4$.

$ A\equal{}2\int_0^2 (\sqrt{16\minus{}x^2}\plus{}2\sqrt{3}\minus{}\sqrt{4\minus{}x^2})\ dx$

$ \equal{}2\left(\frac 12 \cdot 4^2\cdot \frac{\pi}{6}\plus{}\frac 12\cdot \frac{\sqrt{3}}{4}\cdot 4^2\plus{}2\sqrt{3}\cdot 2\minus{}\frac{\pi}{4}\cdot 2^2\right)$

$ \equal{}\text{Your answer}$ :blush:

Thank you for attaching the graph.

kunny


Sper ca cineva sa imi raspunda la acest mesaj! :)

Cum se poate trimite o problema la Gazeta Matematica?

Multumesc anticipat!
Lagrangia.

Foarte simplu: se pune problema intr-un plic (se scrie pe ea problema propusa pentru clasa cutare) evident cu tot cu solutie, si se expediaza pe adresa gazetei matematice (vezi pe coperta gazetei exact care e).


USAMTS
geometry
3D geometry
tetrahedron
inequalities
ARML
analytic geometry

I don't know if this topic is necessary, but how'd you guys all do on the USAMTS?  I just looked at the solutions, and I was thrilled to find that I got the answer to uber-brute force number 1 right, although with a really inelegant solution.  Then there was number three, which I also solved inelegantly.  For some reason, I thought this year's first round was much easier, but maybe that's cuz I smartified a bit. :\

I think, in general, the problems are easier for me (with the exception of #1, which I failed the counting and did not double-check). Also, my solutions are comparatively shorter and cleaner than last year's (again, with the exception of #1, which took 4 pages of diagrams and explanations...to end up with a wrong answer) In the next round, I should work on these problems a bit earlier...

Wrong on #1, uglified on the |ax^2+bx+c| problem, otherwise I think I wrote decent stuff.

all right, but my lack of diagrams on #1 makes me worry

Yeah, I got #1, but my explanations of the pairings were vague and hard to understand.

As for the ax^2+bx+c thingy, I got the right answer and showed that I could get that value, but I made some unsettling unrigorous statements. :(

I like my proof for number five though.  It was pretty...pretty?

I think the first problem is easier with [[Burnside's Lemma]].  (Maybe the [url=http://en.wikipedia.org/wiki/Burnside's_lemma]wikipedia entry[/url] is more accessable if you haven't done much group theory.)

For problem 4, I would probably inscribe the tetrahedron in a cube; that tends to make things easier.

Problem 3 isn't too bad; it's just kind of unusual.

[hide]You can get a value of 300 if you set $ (a,b,c) = (200,0,-100)$.  Now, suppose you can get more than 300.  Setting $ x=\pm 1,0$, we have
\begin{align*}
\lvert a+b+c \rvert &\le 100, \\
\lvert a-b+c \rvert &\le 100, \\
\lvert c \rvert &\le 100.
\end{align*}
Now, suppose that $ a,b,c$ is a counterexample; since we may replace $ (a,b,c)$ with $ (-a,-b,-c)$, and we may replace $ b$ with $ -b$, we may assume without loss of generality that $ a,b \ge 0$.

Since $ \lvert c \rvert \le 100$, it follows that $ \lvert a + b \rvert  = \lvert a \rvert + \lvert b \rvert > 200$.  Therefore by the triangle inequality,
\[ \lvert a+ b+c \rvert \ge \lvert a+b \rvert - \lvert c \rvert > 100, \]
a contradiction.  Therefore no counterexample exists.  $ \blacksquare$

Questions :  What if we let $ a,b,c$ be complex?  What if we let $ x$ be complex?[/hide]

i'm not from georgia, but i think i had perfect solutions except for #1 which i got 23 instead of 22.

[quote="Boy Soprano II"]I think the first problem is easier with [[Burnside's Lemma]].  [/quote]

Yeah, I looked around in the USAMTS forum and discovered the existence of Burnside's the day the solutions were posted.

As for #4, yeah, I should have learned my lesson about tetrahedrons when you showed me inscribing a tetrahedron in a cube during the 2nd ARML practice. :(  Instead, I resorted to analytic geometry-bashing, which actually worked after two pages of very unpleasant algebra. :maybe:

Can anyone give a bijective proof for #1 on the first round. :D

[quote="azn6021023"]How do you inscribe a tetrahedron in a cube?  Where are its vertices in relation to the cube?[/quote]
Let the cube's vertices be $ (\pm 1, \pm 1, \pm 1)$.  Of those, pick the coordinates with an even number of $ \minus{}1$s.  These are the vertices of a regular tetrahedron.  (Can you prove this?)  This is a very nice relation that sort of relies on the fact that 3-space doesn't have very many dimensions.

This tends to make some things a lot easier, including coordinate bashing.

Oh...I thought that all of the vertices lay on the cube.   :blush:
Well, you can show that if the points made a tetrahedron, then their side lengths are congruent, which is true.
And through vectors, you can show that the angles between the are congruent?

[quote] This is a very nice relation that sort of relies on the fact that 3-space doesn't have very many dimensions. [/quote]
I'm glad that you haven't lost your sardonic sense of humor, Miles.

Man, if I had used such simplified coordinates for the tetrahedron's vertices, coordinate bashing would be a proverbial piece of cake.

And it would have shortened my proof to like -1 pages. =|

[quote="Boy Soprano II"]
Let the cube's vertices be $ (\pm 1, \pm 1, \pm 1)$.  Of those, pick the coordinates with an even number of $ \minus{} 1$s.  These are the vertices of a regular tetrahedron.  (Can you prove this?)  This is a very nice relation that sort of relies on the fact that 3-space doesn't have very many dimensions.

This tends to make some things a lot easier, including coordinate bashing.[/quote]

You could also use 4 space coordinates with the vertices being the permutations of $ (1,0,0,0)$. This makes generalizing coordinates of a regular n-tetrahedron (n vertices with each vertex connected to each other vertex) easy.

[quote="azn6021023"]Oh...I thought that all of the vertices lay on the cube.   :blush: [/quote]
They do---the tetrahedron's vertices are also vertices of the cube. :?:

[quote]Well, you can show that if the points made a tetrahedron, then their side lengths are congruent, which is true.
And through vectors, you can show that the angles between the are congruent?[/quote]
How about properties of isosceles triangles?

[quote][quote] This is a very nice relation that sort of relies on the fact that 3-space doesn't have very many dimensions. [/quote]
I'm glad that you haven't lost your sardonic sense of humor, Miles.[/quote]
I wasn't trying to be humorous or sardonic.  Constructions like this wouldn't really work in general $ n$-space.  In 3-space there's just enough room to fit a regular tetrahedron inside a cube.  There isn't enough room in 2-space and there's too much room in 4-space.  I think it has something to do with the structure of $ S_3$, which is nicely compact and yet in ways non-degenerate.

By the way, "$ n$-simplex" is a better name than $ n$-tetrahedron, since tetra-hedron means "four points" or something like that.

At the same time, it has to do with the fact that in 3-space, simplices (generalized tetrahedra) have a number of vertices that is a power of 2, since the number of vertices of a hypercube is always a power of 2 and for a symmetric inscription (which I think is necessary?) the number of vertices in the simplex must divide the number of vertices in the hypercube.  So maybe it happens in $ 2^n\minus{}1$-space.

Note also that it is a little difficult to connect these two arguments on the surface.  Perhaps this is part of something deeper in finite group theory.

So Santhosh, your post doesn't make any sense.  What is a permutation of $ (1,0,0,0)$?  Also, I'm pretty sure you can't embed a simplex in a hypercube in 4-space.  On the other hand, you have pointed out a convenient way of embedding an $ n$-dimensional tetrahedron in $ (n\plus{}1)$-space (specifically, in the subset of coordinates the sum of whose vertices is 1), and I imagine that this is useful in a general setting.

[quote]
They do---the tetrahedron's vertices are also vertices of the cube.[/quote]

Oh, I see it now.  I forgot that technically $ 0$ is also an even number.

Scores are up.  55455=24=$ PHAIL^{PHAIL}$...

But gj to Santhosh and Andrei for getting 25s (as expected).

[quote="Boy Soprano II"]So Santhosh, your post doesn't make any sense.  What is a permutation of $ (1,0,0,0)$?  Also, I'm pretty sure you can't embed a simplex in a hypercube in 4-space.  On the other hand, you have pointed out a convenient way of embedding an $ n$-dimensional tetrahedron in $ (n \plus{} 1)$-space (specifically, in the subset of coordinates the sum of whose vertices is 1), and I imagine that this is useful in a general setting.[/quote]

Sorry, I was too lazy to list out $ (1,0,0,0)$, $ (0,1,0,0)$, $ (0,0,1,0)$, and $ (0,0,0,1)$.
I wasn't trying to embed a 4-simplex in a hypercube.
Yes, putting an $ n$-simplex (yay, I now know the term for it) in $ (n\plus{}1)$ dimensional space can make things easier.
I solved #5 of the North Fulton Invitational Power Round this year (Proving that PA^2+PB^2+PC^2+PD^2 was constant for all P on the circumsphere of ABCD) using that set of coordinates.

[quote="Xantos C. Guin"]
I solved #5 of the North Fulton Invitational Power Round this year (Proving that PA^2+PB^2+PC^2+PD^2 was constant for all P on the circumsphere of ABCD) using that set of coordinates.[/quote]

I wrote that problem with a feeling that someone was going to use coordinates in his/her proof. :lol:

Todo correcto para mí.

My solutions to problems 1, 2, and 4 were pretty solid, though I had to resort to citing Catalan's Conjecture on number 4 for the reason why 8 and 9 are the only consecutive powers of numbers.  But someone did write a proof for it, and the rest of the solution was sound, so I think I got full credit for it.  As for number 3, I wrote a total bs solution (like I did for round 1's number 3, which I got 4 points on :)) that gave me the right answer.  Finally, for number 5, I used a java program that ran for 2 and a half days, and it eventually gave me the correct answer (and I proved that it did, as well).

So I'm guessing I got an ugly score for number 3 but decent scores for the rest.  In the words of Billy, "that happifies me."

arrgh

I realized my solution to #4 was wrong 2 hours before the deadline, when I was about to send my solutions. I came up with some stuff that reduced to needing that fact about 8 and 9, but...

Otherwise the solutions seem to be good.


This diagram is the //gram in the complex number plane. How do I prove that the 2 triangles are congruent?

[url=http://img110.imageshack.us/my.php?image=55165357gc0.png][img]http://img110.imageshack.us/img110/3938/55165357gc0.th.png[/img][/url]

The triangles do not have to be congruent.
They are congruent iff
[hide="HINT"]
The $ \vec{PR}$ is a translated $ \vec{OL}$ and $ \vec{RX}$ is a tranlated $ \vec{LN}$.
[/hide]

Can you describe the intended picture, cuz the picture you have is a little crude. It doesn't even appear like those 2 triangles are even close to [i]similar[/i].


[quote="coffeym"]Prove that

$p\sin{x}+q\cos{x}\equiv\sqrt{p^{2}+q^{2}}\sin{\left({x+\tan^{-1}{\frac{q}{p}}}\right)}$

where $p,q\not=0$[/quote]

[hide]consider a right triangle with legs $p,q$ and hypotenuse $c$ and angle $x$ opposite side $p$ and angle $y$ opposite side $q$

$p\sin{x}+q\cos{x}\equiv\sqrt{p^{2}+q^{2}}\sin{\left({x+\tan^{-1}{\frac{q}{p}}}\right)}$
$\Leftrightarrow\frac{p^{2}}{c}+\frac{q^{2}}{c}\equiv c\times \sin{(x+y)}$
$\Leftrightarrow\frac{c^{2}}{c}\equiv c\times \sin{90^\circ}$
$\Leftrightarrow c\equiv c$[/hide]

Strictly speaking, it is not a great idea to work at two sides of an equation and find they equate.  Perhaps try harmonic form.  :lol:

[quote="coffeym"]Prove that

$p\sin{x}+q\cos{x}\equiv\sqrt{p^{2}+q^{2}}\sin{\left({x+\tan^{-1}{\frac{q}{p}}}\right)}$

where $p,q\not=0$[/quote]
[hide] \begin{eqnarray*}\sqrt{p^{2}+q^{2}}\sin\left(x+\arctan\frac{q}{p}\right) &=& \sqrt{p^{2}+q^{2}}\left(\sin x\cos\arctan\frac{q}{p}+\sin\arctan\frac{q}{p}\cos x\right) \\ &=& \sqrt{p^{2}+q^{2}}\left(\frac{p\sin x}{\sqrt{p^{2}+q^{2}}}+\frac{q\cos x}{\sqrt{p^{2}+q^{2}}}\right) \\ &=& \boxed{p\sin x+q\cos x}\end{eqnarray*} [/hide]


Can a triangular number be expressed as the sum of one triangular number and one non-triangular number?

[hide="sure, why not?"]1+2=3, 1+5=6, 1+9=10, etc  :P
Of course, I'll see how to prove that.... [/hide]

Oops.. lol other than 1 are there any?  I didn't find any yet..

[hide]Say there's triangular $t$ who is equal to $1+2+3+\cdots+a$. If you add $a+1$ to $t$, you'll get another triangular number, $s$.

This is assuming that $a+1$ isn't triangular and by observation, we see that here are an infinite number of those.  :) [/hide]


function
induction
search
algebra unsolved
algebra

[quote="Math pro"]$ f: (2,4) \minus{} > (2,4),g: (2,4) \minus{} > (2,4)$ such that:
i.$ f(g(x)) \equal{} g(f(x)) \equal{} x$
ii.$ f(x).g(x) \equal{} x^2$
prove that:: $ f(3) \equal{} g(3)$[/quote]
I dont think you didnt solve it,because here is our homework and it is in many books about function equation
More clearly,you can prove $ f(a^nx)\equal{}a^{n\plus{}1}a$ where $ f(x_0)\equal{}ax_0$by induction
since $ f(x)\in (2,4)$ we have $ f(x)\equal{}x)$and similarly $ g(x)\equal{}x$
[hide="to mathpro"]A spam type[/hide]

[quote="Allnames"][quote="Math pro"]$ f: (2,4) \minus{} > (2,4),g: (2,4) \minus{} > (2,4)$ such that:
i.$ f(g(x)) \equal{} g(f(x)) \equal{} x$
ii.$ f(x).g(x) \equal{} x^2$
prove that:: $ f(3) \equal{} g(3)$[/quote]
I dont think you didnt solve it,because here is our homework and it is in many books about function equation
More clearly,you can prove $ f(a^nx) \equal{} a^{n \plus{} 1}a$ where $ f(x_0) \equal{} ax_0$by induction
since $ f(x)\in (2,4)$ we have $ f(x) \equal{} x)$and similarly $ g(x) \equal{} x$
[hide="to mathpro"]A spam type[/hide][/quote]
I know that,all names but the problem I want to mention is not only this problem!
the problem is much harder!:
find $ f: R\minus{}>R$ such that:
$ f(\frac{x^2}{f(x)})\equal{}x$
I think you should think before you post!
[hide]Almost posts of you are spam![/hide]

[quote="Math pro"][quote="Allnames"][quote="Math pro"]$ f: (2,4) \minus{} > (2,4),g: (2,4) \minus{} > (2,4)$ such that:
i.$ f(g(x)) \equal{} g(f(x)) \equal{} x$
ii.$ f(x).g(x) \equal{} x^2$
prove that:: $ f(3) \equal{} g(3)$[/quote]
I dont think you didnt solve it,because here is our homework and it is in many books about function equation
More clearly,you can prove $ f(a^nx) \equal{} a^{n \plus{} 1}a$ where $ f(x_0) \equal{} ax_0$by induction
since $ f(x)\in (2,4)$ we have $ f(x) \equal{} x)$and similarly $ g(x) \equal{} x$
[hide="to mathpro"]A spam type[/hide][/quote]
I know that,all names but the problem I want to mention is not only this problem!
the problem is much harder!:
find $ f: R \minus{} > R$ such that:
$ f(\frac {x^2}{f(x)}) \equal{} x$
I think you should think before you post!
[hide]Almost posts of you are spam![/hide][/quote]
so you should write it in your post,we cant understand what do you want(  I am very stupid,so your smart ideas make me not understand)
With your new problem,i am sure we can kill it because the condition is quite clearly
[hide="to mathpro"]I think they are not spam http://www.mathlinks.ro/search.php?search_author=Math+pro
with you,all post with some sentences are not spam ,thank[/hide]

to All names: [hide]I think you are pro at spaming http://www.mathlinks.ro/Forum/search.php?search_author=Allnames
and I'm very bad at spaming   :rotfl: [/hide]
Any one can solve my problem?It 's from a book but the solution is very terrible! :( Any one can help me?

since $ f(g(x)) \equal{} g(f(x))$  so  $ g(x) \equal{} f^{ \minus{} 1}(x)$

so we have :

$ f(3).g(3) \equal{} 9$  and $ f(g(3)) \equal{} g(f(3)) \equal{} 3$

so : $ f(\frac {9}{f(3)}) \equal{} g(\frac {9}{g(3)}) \equal{} 3$

so $ \frac {9}{f(3)} \equal{} \frac {9}{g(3)} \rightarrow f(3) \equal{} g(3)$

any mistakes ??  :ninja:

[quote="mehdi cherif"]
so : $ f(\frac {9}{f(3)}) \equal{} g(\frac {9}{g(3)}) \equal{} 3$

so $ \frac {9}{f(3)} \equal{} \frac {9}{g(3)} \rightarrow f(3) \equal{} g(3)$

[/quote]
why??
Use my hint
Your problem is wrong ,mathpro,please,add any conditon,example,$ f$ is decreasing function