function
limit
complex analysis
complex analysis solved

find all functions $f:C\rightarrow C,$ entires such that exists $n\in N$ with the property $f(z^n)=(f(z))^n$

i'll assume [tex]n>1[/tex] for the rest, [tex]n=1[/tex] is trivial.

1st step: [tex]f(\mathbf S^1) \subseteq \mathbf S^1[/tex], if [tex]f[/tex] is not [tex]\equiv 0[/tex]

indirect proof: assume [tex]z \in \mathbf S^1, f(z) \notin \mathbf S^1[/tex]. then [tex]lim_{n \to \infty} f(z^n) = lim_{n \to \infty} f(z)^n = \infty \text{ or } 0[/tex], depending on whether [tex]|f(z)| > 1[/tex] or [tex]|f(z)| < 1[/tex]. Since [tex]\mathbf S^1[/tex] is compact, [tex]z^n[/tex] has a convergent subseries, let [tex]Z[/tex] be its limit. Then, if [tex]|f(z)| > 1[/tex], [tex]f(Z) = \infty[/tex], which is impossible, since [tex]f[/tex] is entire. If [tex]|f(z)| < 1[/tex], [tex]f(Z) = 0[/tex], and [tex]f(\sqrt[n^k]Z) = \sqrt[n^k]0 = 0[/tex] for all [tex]n^k[/tex]-th roots of [tex]Z[/tex], [tex]k \in \mathbb N[/tex]. but since the [tex]n^k[/tex]-th roots of [tex]Z[/tex] are dense in [tex]\mathbf S^1[/tex], [tex]f \equiv 0[/tex] on [tex]\mathbf S^1[/tex], which implies [tex]f \equiv 0[/tex] on [tex]\mathbb C[/tex].

2nd step: finding all functions from [tex]\mathbf S^1[/tex] to itself that satisfy the functional equation.

Let's write [tex]f: \exp(i\theta) \mapsto \exp(ig(\theta))[/tex], where [tex]g: \mathbb R/(2\pi i \mathbb Z) \to \mathbb R/(2\pi i \mathbb Z)[/tex]. Then [tex]g[/tex] is uniquely determined by [tex]f[/tex], and [tex]g(n\theta)[/tex] = [tex]ng(\theta)[/tex]. [tex]g[/tex] is real-analytic, so it must be linear: [tex]g(\theta) = m\theta + \frac{2\pi i k}{n-1}, k \in \mathbb Z[/tex]. [tex]m[/tex] must be a whole number for g to be well-defined. Thus [tex]f(z) = z^m \cdot \zeta[/tex], where [tex]\zeta[/tex] is a [tex](n-1)[/tex]th root of unity. This can only be continued to an entire function if [tex]m \geq 0[/tex].

Thus we finally have: [tex]f(z) = \zeta z^m, m \in \mathbb N_0, \zeta^{n-1} = 1[/tex] or [tex] f(z) = 0[/tex]

Ugh. I like abusing TeX.

[quote="luqash"]Ugh. I like abusing TeX.[/quote] :D that's why it's here for ;)
anyway, you can use the normal dollar tags now  like this
[code] $ \displaystyle \frac {x+y}{z^2} $ [/code] which will generate $   \displaystyle \frac {x+y}{z^2} $ and also double dollar tags.

Nice solution! My ideea was preety much the same:
It is easy to prove that f verifies the condition
\[f(z^{n^p})=(f(z))^{n^p},(\forall)p\in N,(1)\] If $f(z)\neq 0,$ for all z in C, then $f(z)=e^{g(z)},$ with g entire, nonconstant. The relation is
$n^pg(z)=g(z^{n^p}),$ for all z in C, wich it's equivalent with
$g(z)=\frac{g(z^{n^p})}{n^p},$ for all p in n. For |g(z)|<1, we have:
$g(z)=\lim\limits_{p\rightarrow\infty}\frac{g(z^{n^p})}{n^p}=0.$ So g=0.
If exists $z_0\neq 0$ such that f(z)=0, then every solution of the equation $z^{n^p}=z_0$ is a solution of the equation f(z)=0. Furthermore every such solution is contained in $B=B(0,|z_0|).$ If f have an infinite number of solutions in B, we have a limit point and f=0 (f is entire). So f must have a finite number of solution in B, which is not posible because of (1).
So the unique solution of f is 0 and $f(z)=z^sg(z),$ with g entire and $g(z)\neq 0$, for all z in C.Relation in hypotesis became
$g(z^n)=(g(z))^n,g(z)\neq 0.$ From the above we obtain that g is constant. then $f(z)=cz^s,$ with $c^n=c$ :cool:

[quote="caugust"]The relation is
$n^pg(z)=g(z^{n^p}),$ for all z in C.[/quote]

In the begin, I thought so too, but in the end, you have to consider $n^pg(z)=g(z^{n^p})+2\pi i k, k \in \mathbb Z$, where $k$ may vary with $p$. and that is where the problems start. in the end, you should not only have $g = 0$ as a solution, but also $g = \frac {2 \pi i k}{n-1}$. This solution comes from nirvana in your proof (in mine, it does as well), and this is where i am not convinced of both, and my statement [quote="luqash"] $g$ is real-analytic, so it must be linear: $g(\theta) = m\theta + \frac{2\pi i k}{n-1}$[/quote] is still [b]very[/b] suspect to me, though i feel there should be some kind of proof for it.

But perhaps there is none, and there are other zero-free solutions... gonna have to think about it again.

O.K.! Let,s try to fix it.
Indeed, we have the relation:
\[ n^pg(z)=g(z^{n^p})+2\pi ik_p,k_p\in N.(1) \]
Let's  write (1) for $z^s$ where $s\in N.$ We obtain
\[ n^pg(z^s)=g(z^{sn^p})+2\pi ik_p,k_p\in N.(2) \]for all $ s\in N. $ Consider |z|<1, and make $s\rightarrow\infty.$ We obtain
$ n^pg(0)=g(0)+2\pi ik_p, $ or $ g(0)(n^p-1)=2\pi ik_p, $ for all $p\in N.$ Then
\[ g(0)=\frac{2\pi ik_p}{n^p-1}=\frac{2\pi ik_1}{n-1}, \] so
\[ 2\pi ik_p=2\pi i(n^{p-1}+n^{p-2}+...+n+1)k_1. \] rewrite relation (1) as
\[ g(z)=\frac{g(z^{n^p})}{n^p}+\frac{2\pi i(n^{p-1}+n^{p-2}+...+n+1)k_1}{n^p}, \] and take $p\rightarrow\infty.$ for |g(z)|<1. We obtain that g=0.
I hope this is correct and complete. :D

If you take $p \to \infty$ in \[ g(z)=\frac{g(z^{n^p})}{n^p}+\frac{2\pi i(n^p-1)k_1}{n^p(n-1)}, \], you obtain $g(z) = \frac{2 \pi i k_1}{n-1}$, which actually is the correct result.


LaTeX
real analysis
real analysis unsolved

Please someone to help me for post this problem (atachement)  in latex!Thanx!

To present symbols in latex write your text between dollar signs (the symbol above the 4 on your keyboard):

[code]$write the mathematical symbology here$[/code]

For example, here's the first mathematical symbols on your attachment:

$ f_{n}\equal{}\left[(x^{2}\minus{}1)^{n}\right]^{(n)}$

Scroll the mouse over any latex to see how it was written. More information is in [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url], specially paragraph 4.


hello, rewriting your first inequality in the form
$ 0<\arcsin(\frac{1}{x})-\frac{20\pi}{63x}$
and defining
$ f(x)=\arcsin(\frac{1}{x})-\frac{20\pi}{63x}$
we have
$ \lim_{x \to 1+}f(x)=\frac{1}{126}\frac{63\pi^2-40}{\pi}$
we have $ f'(x)<0 \Leftrightarrow -\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}+\frac{20}{63\pi x^2}<0$
and additionally is $ f^{''}(x)>0$ for $ x>1$ our function is strictly convex, and has no zero points, so we have
$ 0<\arcsin(\frac{1}{x})-\frac{20\pi}{63x}$ or
$ \frac{63x}{20}>\frac{\pi}{\arcsin(\frac{1}{x}})$
for $ x>1$
Sonnhard.


abstract algebra
Ring Theory
group theory
superior algebra
superior algebra unsolved

Let $ M$ be a finitely generated module over a PID. Then $ M$ satisfies the ACC for submodules.

Ok, so let me ask the following question first:
Why is a submodule also finitely generated? In the case of $ \mathbb{Z}$-modules this reduces to whether a subgroup of a finitely generated abelian group is finitely generated. I've never thought about this before, and it seems like it should be obvious, but I find it surprisingly difficult to prove. If we take any subgroup $ K$ we can use the structure theorem and intersect $ K$ with the torsion part and the free part of the bigger group, but is it obvious that $ K$ is the direct sum of these?

Ohps, I realize how easy this is now, it doesn't even need the structure theorem. We can write $ M \equal{} D^n/K$ for some submodule $ K$ of $ D^n$. So if $ M_1$ is a submodule of $ M$ it can be identified with $ N/K$ where $ N$ is a submodule of $ D^n$. Since $ D$ is a PID, it follows that $ N$ is free of finite rank, and hence $ M_1$ is finitely generated.

The ACC now follows because if we take the union of a chain of submodules we get a new module that is finitely generated, and ACC follows as usual.

I have a new question: Suppose $ M_1$ and $ M_2$ are two free modules of finite rank (maybe modules over a PID? I don't know what is required). Suppose that $ M_1$ is a proper submodule of $ M_2$. Can I say that the rank of $ M_2$ is strictly greater than that of $ M_1$? EDIT: not true, $ 2\mathbb{Z} \subset \mathbb{Z}$


For each integer $ n\geq 4$, let $ a_n$ denote the base-$ n$ number $ 0.\overline{133}_n$. The product $ a_4a_5 \dotsm a_{99}$ can be expressed as $ \frac {m}{n!}$, where $ m$ and $ n$ are positive integers and $ n$ is as small as possible. What is the value of $ m$?

$ \textbf{(A)}\ 98 \qquad \textbf{(B)}\ 101 \qquad \textbf{(C)}\ 132\qquad \textbf{(D)}\ 798\qquad \textbf{(E)}\ 962$

good problem manipulating the telescopic series  :lol:

I got the expression for $a_{n}$ similar to how frt did it, but I had a different way of making the expression telescope:

[hide="Another Method"]Notice that

$n^{2}+3n+3=\frac{n^{3}+3n^{2}+3n}{n}=\frac{(n+1)^{3}-1}{n}$.

Then

$a_{n}=\frac{n^{2}+3n+3}{n^{3}-1}$
$a_{n}=\frac{\frac{(n+1)^{3}-1}{n}}{n^{3}-1}$
$a_{n}=\frac{(n+1)^{3}-1}{n(n^{3}-1)}$

And then everything cancels nicely.  :) [/hide]

In the problem, is $ 0.\overline{133}_n$ the same as $ \frac{133}{999}_n$?

[quote="314.math"]In the problem, is $ 0.\overline{133}_n$ the same as $ \frac {133}{999}_n$?[/quote]

Be careful: 999 is only $ 1000 \minus{} 1$ in base 10, so the more accurate answer would be $ \left(\frac{133}{[n\minus{}1][n\minus{}1][n\minus{}1]}\right)_n$.


function
integration
inequalities
real analysis
real analysis unsolved

Let $ f$ is a continuity funtion and $ f: [0;1] \rightarrow [-1;1]$. Prove that:

$ \int_{0}^{1}\sqrt{1-f^{2}(x)}dx \le \sqrt{1-\displaystyle \left [\int_{0}^{1}\sqrt{1-f(x)}dx  \right ]^2}$

[quote="Extremal"]it's looks like Jensen's inequality. what can you say about $ L(x) \equal{} \sqrt {1 \plus{} x^{2}}$ is it convex ? :roll:[/quote]
? :maybe:

I mean $ L(x) \equal{} \sqrt{1\minus{}x^{2}}$ and use Jensen for $ L$ as a convex function. but I don't see full solution , it's need time to check  :(

Let $ f(x) \equal{} x$.
\[ \int_0^1{\sqrt {1 \minus{} x^2}dx} \equal{} \frac {\pi}{4} > 0.7853...
\]

\[ \sqrt {1 \minus{} \left(\int_0^1{\sqrt {1 \minus{} x}dx}\right)^2} \equal{} \sqrt {1 \minus{} \left(\frac {2}{3}\right)^2} \equal{} \frac {\sqrt {5}}{3} < 0.745...
\]
Unless I made some obvious mistake, the inequality doesn't seem to hold.

Edit: On second thought, let $ f(x)\equal{}0$.

[quote="thanhnam2902"][size=150][color=darkblue]Can You  post solution?[/color][/size][/quote]

Hey [b]thanhnam2902[/b],

Did you read the previous post? If we choose f(x) = 0, the expression does not hold :(

So problem is wrong  :P

I think right inequality is

\[ \int_{0}^{1}\sqrt{1-f^{2}(x)}dx \le \sqrt{1-\displaystyle \left [\int_{0}^{1}f(x)dx \right ]^2}\]


function
trigonometry
calculus
calculus computations

How do you find natural parametrisation of KuKloeides. this ${f(t)=(rt-rsint,r-rcost})$. I know it's simpl, but I want know wich (standart or nonstandart) method you use. I know that it's not natural parametrisation
 ${f(t)=(tr-rsint,r-rcost})$ because $f'(t)=(r-rcost,rsint)$ and $|f'(t)|=\sqrt{r^{2}(2-2cost)}=r\sqrt{2(1-cost)}$ wich is not equal 1, therefore it's not natural parametrisation.

Let's look for a function $t(s)$ such that $\vec f(t(s))$ is an arclength parametrization. By the Chain Rule $\left|\frac{d}{ds}\vec f(t(s))\right|=\left|\vec f'(t(s))\right|\left|\frac{dt}{ds}\right|$. Equating this to $1$, obtain $\frac{dt}{ds}=\frac{1}{r\sqrt{2(1-\cos t)}}$. This is a separable equation. Simplifying ${\sqrt{2(1-\cos t)}}=2\sin(t/2)$, get $2r\sin(t/2)dt=ds$, hence $-4r\cos(t/2)=s$ is a solution.

This needs some clean-up (absolute values and such).


If a set of markers is placed in rows of 4 each there are 2 left over; if the markers are placed in rows of 5 each, there are 3 markers left over. If they are in groups of 7 there are 5 left over.  How many markers  are in the set?

[hide]I guess I use a kind of modified Modulus, more or less logic. If there are three left over when put in groups of five, the number ends in either an 8 or a three. If it has two left over when put in groups of 4, then it has to be even, so the number has to end in 8. Look at the last clue. That shows the number minus five has to be a multiple of 7, so you need to look for a multiple of seven that ends in three. that obviously gives you 63, so your answer is 63+5 or 68 Wait.. my b EDIT: But 68 is a multiple of four, so the next one you get is 133 (63+70). Add five and get 138 for your answer.[/hide]

[quote="Rep123max"][hide="This is a solution using the Chinese Remainder Theorem"] In this case, it probably (I haven't spent time looking for another solution, but I'm sure there is one) isn't the fastest.

So basically, we want to find the common mod base when we have $x=2\pmod 4$, $x=3\pmod 5$ and $x=5\pmod 7$. We can express $x$ in several ways, namely $2+4r$, $3+5s$, and $5+7t$.

Let's start with the first two:

$2+4r=3+5s$
$4r=1+5s$
$4r\equiv 1\pmod 5$
$4r\equiv 16\pmod 5$
$r\equiv 4\pmod 5$
$r=4+5a$

$2+4(4+5a)=3+5s$
$2+16+20a=3+5s$
$18+20a=3+5s=x$
$x\equiv 18\pmod {20}$ or $x=18+20b$.

Now we combine this with the third:
$18+20b=5+7t$
$13+20b=7t$
$7t\equiv 13\pmod {20}$
$t\equiv 19\pmod {20}$
$t=19+20c$

$18+20b=5+7(19+20c)$
$18+20b=5+133+140c$
$x=18+20b=138+140c$
$x\equiv 138\pmod {140}$

So the smallest value of $x$ would be 138. Hockeyadi23 - 68 is divisible by 4. Better look over your logic.[/hide][/quote]
oops.. didn't read the question fully


a) Find a bijective function which takes $[0,1]$ in $(0,1)$.

b) Find a bijective function which takes $(a,b)$ in $[a,b)$.

c) Find a bijective function which takes $[a,b)$ in $(a,b]$.

d) Find a bijective function which takes $(a,b]$ in $R$.

Thank you very much.

just the first one, I'm sure you will guess the other ones :
[hide]
$f(0) = 1/2$
$f(1) = 1/3$
$f(1/n) = 1/(n+2), \forall n \in 2,3, ...$
$f(x) = x$ otherwise[/hide]


When are we allowed to discuss Round 1 solutions?

chess64: We should not discuss the problems until next wenesday. Teachers can give  out the exam on Tuesday the 25th, too.

Oh, okay. I don't this is considered discussing the problems: I got the highest score in my school! :) (if it is, then a mod can delete it)

Shoot. I messed up. That's okay, though.

How have you guys taken it already?  The official date is the 25th...  Did your school arrange to take it early?

joml88: Yeah, I guess.

dakyru: :blush: no, surprisingly (i made a very careless mistake on one of them). and even more surprisingly, nobody else even got the same score as me :huh:

Yeah, I'm really mad.  I misread #2 and got Sam confused with the person who had the most scares.  There goes my perfect score for the year that I was hoping for.

So...er...are we allowed to discuss now? The competition's over... on the east coast, the time is 5:40PM. On the west coast, the time is 2:38 PM. hmm...maybe some schools are still taking it. I think we should wait a few hours.

it's 7:18 on the east coast.. i doubt schools on the west coast are still taking it. let's discuss.

Ah crap! I didn't realize that either.. :(

wow, this stinks.

[quote="chess64"]#1, #3, #4, #5, #6!!!!!!

My incorrect solution for #2!! :( :( :( :(

[hide]
$t=2s, s=3r \text { or } r=s/3$.

The sum is less than or equal to 2005.

$t+s+r \le 2005\\ 2s+s+s/3 \le 2005\\ s \le \boxed {601}$

NOOOO.....

FRACTIONAL PEOPLE DON'T EXIST :wallbash_red:
[/hide]

(hey, since when did they limit the # of smilies you can have? :P)[/quote]

hey cool, me too.  :(

Mistakes are always something learned.  I consider my mistakes on the first competition worth many points saved in other competitions.  Thinking of it like this will make you happy.  ;)

5/6,

got number 6 wrong
I even had the right answer written in my notes
arghhhh!

[hide]$3.5\leq r\leq 4$[/hide] :mad:

[quote="lorddude"]
[hide]$3.5\leq r\leq 4$[/hide] :mad:[/quote]

huh?  That's not the answer to #6.

The right answer was $3.5 \text { or } 4$. For some reason, most people I know put either 3.5 or 4, but not both. (I think I'm the only one who got that right in my school :P) But it's kind of obvious that there is more than one solution, especially when they say "[b]all[/b] possible values..." :roll:

[quote="chess64"] (I think I'm the only one who got that right in my school :P) [/quote]

Ooh, you're special. I wish I was as special as you were. I think I'll go to work on being special now.

I don't like this contest much, but that's not their fault.


Let $ a_0,a_1,\ldots,a_{n\minus{}1}$ be integers. Show that $ \displaystyle\prod_{s\equal{}1}^{n\minus{}1}s!$ divides $ det\|a^{t}_k\|(0\leq k,t\leq n\minus{}1)$.

[quote="Ibrogimov"]Let $ a_0,a_1,\ldots,a_{n \minus{} 1}$ be integers. Show that $ \displaystyle\prod_{s \equal{} 1}^{n \minus{} 1}s!$ divides $ det\|a^{t}_k\|(0\leq k,t\leq n \minus{} 1)$.[/quote]

Since $ \prod_{s \equal{} 1}^{n \minus{} 1}s! \equal{} 1\cdot\prod_{s \equal{} 1}^{n \minus{} 1}s! \equal{} 0!\cdot\prod_{s \equal{} 1}^{n \minus{} 1}s! \equal{} \prod_{s \equal{} 0}^{n \minus{} 1}s! \equal{} \prod_{i \equal{} 1}^n\left(i \minus{} 1\right)!$ (here we substituted $ i$ for $ s \plus{} 1$) and $ \det\left(\left(a^t_k\right)_{0\leq k\leq n \minus{} 1,\ 0\leq t\leq n \minus{} 1}\right) \equal{} \prod_{0\leq j < i\leq n \minus{} 1}\left(a_i \minus{} a_j\right)$ (by the Vandermonde determinant formula), this rewrites as $ \prod_{i \equal{} 1}^n\left(i \minus{} 1\right)!\mid \prod_{0\leq j < i\leq n \minus{} 1}\left(a_i \minus{} a_j\right)$. This was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=56803 . (See also http://www.mathlinks.ro/viewtopic.php?t=4858 for an equivalent version.)

  darij


If $ \frac{x}{y}\equal{}\frac{34}{12}$, $ \frac{y}{z}\equal{}\frac{24}{102}$, and $ \frac{z}{w}\equal{}\frac{9}{12}$, what is $ \frac{w}{x}$?

If you think this is so ridiculously easy, please refrain from posting.

[quote="alanchou"]If $ \frac {x}{y} \equal{} \frac {34}{12}$, $ \frac {y}{z} \equal{} \frac {24}{102}$, and $ \frac {z}{w} \equal{} \frac {9}{12}$, what is $ \frac {w}{x}$?

If you think this is so ridiculously easy, please refrain from posting.[/quote]

I don't want to!  :(
[hide="The Solution"]
Well...
You convert the fractions so that in each fraction the variables equal to the other.
So...
$ \frac{x}{y}$ Would be converted to $ \frac{204}{72}$

$ \frac{y}{z}$ Would be converted to $ \frac{72}{306}$

$ \frac{z}{w}$ Would be converted to $ \frac{306}{408}$

So then,

$ \frac{w}{x}$ Would be equal to $ \frac{408}{204}$

Which reduces to $ 2$
 :lol:
[/hide]

many enters in the solution don't make it a big solution :)

Multiply all numbers and note that $ \frac{z}{w} \frac{x}{y} \frac{y}{z}\equal{}\frac{x}{w}$

Or using numbers $ \frac{34}{12} \frac{24}{102} \frac{9}{12}\equal{}\frac{1}{2}$

The number that you want is the inverse of this. so this number is $ 2$

Please use hide tags.

And I never mentioned size in the comment, I just mentioned if it's nice or not

Both answers are correct, btw


I'm not sure what a polygonal line is, polygonal means having many sides (I think), and a line does not have any sides, so I am not sure...

Polygonal Line: not a straight line, I guess..


Let $ a,b,c$ be any positive integers,prove that
${ 3^{a+b+c}a^{a}b^{b}c^{c}\ge {(a+b+c)}}^{a+b+c}$

Consider the following function

$ f(x)\equal{}x\ln(x)$ has the property as follows

* $ f''(x)\equal{} \frac{1}{x} >0$

so the above function is convex on the interval $ (0,\infty)$ and the rest is shown by Jensen inequality.

Can i assume that $ a\plus{}b\plus{}c\equal{}3$ :?:

yes, you can, my friend. That makes the inequality more interesting and solvable.

[hide="My Solution"]
Assuming $ a\plus{}b\plus{}c\equal{}3$ we are left to prove $ a^{a}b^{b}c^{c}\ge 1$
Taking log both the sides we get $ a\log_{abc}a\plus{}b\log_{abc}b\plus{}c\log_{abc}c\ge \log_{abc}abc$.
Applying the chebycheff's we get $ a\log_{abc}a\plus{}b\log_{abc}b\plus{}c\log_{abc}c \ge\frac{(a\plus{}b\plus{}c)(\log_{abc}a\plus{}\log_{abc}b\plus{}\log_{abc}c)}{3}$

$ (\log_{abc}a\plus{}\log_{abc}b\plus{}\log_{abc}c)\equal{}\log_{abc}abc$
Hence Proved[/hide]

I tried Chebychev at the very first time but it doesn't work indeed.

Let's say after using Chebychev, what is your next step? It turns out that the inequality has inversed its sign since

$ \frac{a\plus{}b\plus{}c}{3} \ge \sqrt[3]{abc}$

[quote="great math"]I tried Chebychev at the very first time but it doesn't work indeed.

Let's say after using Chebychev, what is your next step? It turns out that the inequality has inversed its sign since

$ \frac {a \plus{} b \plus{} c}{3} \ge \sqrt [3]{abc}$[/quote]

Why to apply A.M-G.M to $ a\plus{}b\plus{}c$ just substitute $ a\plus{}b\plus{}c\equal{}3$ and 3 cancel out and then we are left with log terms,use the properties of log and u r done!!

So what do you do with $ log_{abc}abc$? because the right hand side is 0.

How it is zero?? we have to prove this $ a^{a}b^{b}c^{c}\ge 1$
The RHS is 1 and $ \log_{abc}abc\equal{}1$

No,I mean that $ log_{abc}1$ is different from $ log_{abc}abc$

[hide="Another Solution "]
The given inequality can be written as ${ a^{\frac{a}{a+b+c}}b^{\frac{b}{a+b+c}}c^{\frac{c}{a+b+c}}}\ge\frac{a+b+c}{3}$
The A.M of the inequality is $ \frac{\frac{a}{a}+\frac{b}{b}+\frac{c}{c}}{a+b+c}=\frac{3}{a+b+c}$
Then G.M is $ \sqrt[a+b+c]{\frac{1}{a^{a}b^{b}c^{c}}}$
Apply $ A.M \ge G.M$
[/hide]


I'm having difficulty on this one Engel NT problem; its like:

Find the last three digits of $ 7^{9999}$.

So I understand until $ 7x\equal{}1\mod 1000$. Now he's like find inverses with $ 7x\plus{}1000y\equal{}1$. I don't get this method at all.

But then he's like for this problem its easiest to do like $ 1001\equal{}7*11*13$. 1001 = 1 mod 1000. So 143*7=1 mod 1000 so x is 143. I don't get why this works either.

help?

[quote="#H34N1"]I'm having difficulty on this one Engel NT problem; its like:

Find the last three digits of $ 7^{9999}$.

So I understand until $ 7x \equal{} 1\mod 1000$. Now he's like find inverses with $ 7x \plus{} 1000y \equal{} 1$. I don't get this method at all.

But then he's like for this problem its easiest to do like $ 1001 \equal{} 7*11*13$. 1001 = 1 mod 1000. So 143*7=1 mod 1000 so x is 143. I don't get why this works either.

help?[/quote]

[hide]If we find such x,y, we can take both sides mod 1000 to get that 7x is congruent to 1 mod 1000[/hide]

How would one find inverses like that though?

Euclidean Algorithm.

Also, a quick search gives: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=271260822&t=207630[/url]

$ 7a\plus{}1000b \equal{} 1$
$ 6b\equiv 1 \bmod 7$
$ 6b\plus{}7c\equal{}1$
$ c\equiv 1 \bmod 6$
$ c\equal{}1, b\equal{}\minus{}1, a\equal{}1001/7\equal{}143.$


Does she conduct any math research? Will she be the first female to win a Fields Medal?

Eugenia Malinnikova
IMO 1989 41
IMO 1990 42
IMO 1991 42

Some of her research papers can be found here:
http://www.math.ntnu.no/~eugenia/
I think she was the best female IMO contestant ever.


function
geometry
geometric transformation
calculus

Hey...I'm having some trouble with Algebra II; if anyone has taken it can you help me with this problem? Thanks so much.

Given f(x) = { x^2 if x<1
{ 4x if x >= 1
[The two functions above are all in one { } ...sorry I couldn't figure out how to type that.] ;]

Write the rule for each function:
a) g(x), a vertical compression of f(x) by a factor of 1/4
b) a(x), a horizontal stretch by a factor of 2
c) h(x), a vertical translation 3 units down

A vertical stretch of a function $ f(x)$ by $ k$ units is $ kf(x)$. If $ k < 0$, it also flips vertically.
A vertical translation of a function $ f(x)$ by $ k$ units is $ f(x)\plus{}k$.
A horizontal stretch of a function $ f(x)$ by $ k$ units is $ f(x/k)$. If $ k < 0$, it also flips horizontally.
A horizontal shift of a function $ f(x)$ by $ k$ units is $ f(x\minus{}k)$.

It is better in my opinion to understand and know the reasoning behind each of these, rather than just memorizing the formulas. Playing around with transformations of linear functions may help with this, as it's true that you can get from any line to any other line with just these transformations.

Well, I understand the rules for each different transformation, but the part I don't understand is how the parts x<1 and x>=1 change with each transformation.

From my interpretation, you're talking about a [b]Piecewise[/b] defined function.
Simply put, this is a function defined with more than one formula (each formula applies to a specific interval). In your case the first formula applies when x is strictly less than 1, and the second applies when x is greater than or equal to one.

So the function (being defined piecewise) undergoes a few transformations.

This means that BOTH of the "formulas" need to undergo the transformation.

So for part one, you're compressing both formulas by a factor of 1/4.
This means that we can piecewise define it like this:
$ g(x) ={\frac{x^{2}}{4}\forall x < 1\wedge x\forall x\ge 1 }$

A bonus question (not really related to the transformation problem):

Is the function f(x) continuous at $ x = 1$?
If you know some calculus you can do it with a limit definition.

Um...Thanks for the help...but I did not understand any of that equation you wrote...plus I know absolutely no calculus.
Sorry...could you please explain it in terms of algebra?
 :surrender:

[quote="DiscreetFourierTransform"]
$ g(x) ={\frac{x^{2}}{4}\forall x < 1\wedge x\forall x\ge 1 }$[/quote]
Holy squirrels, man, do you really have to confuse the Algebra II student?

Anyway, what this says is that $ g(x) =\frac{x^{2}}{4}\text{, if }x < 1$ and $ g(x) = x\text{, if }x\geq1$ (I, too, do not know the piecewise notation  :blush: ). This just multiplies $ f(x)$ by $ \frac{1}{4}$, so it could be said $ g(x) =\frac{f(x)}{4}$.

As for $ h(x)$, $ h(x) = x^{2}-3\text{, if }x < 1$ and $ h(x) = 4x-3\text{, if }x\geq1$. That is to say $ h(x) = f(x)-3$.

For $ a(x)$, $ a(x) =(\frac{x}{2})^{2}\text{, if }x < 1$ and $ a(x) = 2x\text{, if }x\geq1$. Or $ a(x) = f(\frac{x}{2})$.

[quote="DiscreetFourierTransform"]A bonus question (not really related to the transformation problem):

Is the function f(x) continuous at $ x = 1$?
If you know some calculus you can do it with a limit definition.[/quote]
You don't need Calc. for this question... Just plug $ 1$ in for $ x$ in both and if both values agree (and if one of the signs is $ \geq$ or $ \leq$) then it is continuous. If the tangent values agree is a different story.

[quote="starcellist"]Oh, okay, I get it now... Thanks so much!  :D[/quote]
You're welcome!  :lol:


AMC
AIME
geometry
rhombus
trigonometry
trig identities
Law of Sines

Find the area of rhombus $ABCD$ given that the radii of the circles cirumscribed around triangles $ABD$ and $ACD$ are $12.5$ and $25$, respectively.

use area / semiperimeter  = radius.  you get 3 variables and 3 equations, which can be solved in an ugly way.

cool problem.  i found a simple solution :P

[hide]draw in AC and BD, and let $a$ be angle ABD, $b$ be angle ACD, and $x$ be the side of the rhombus.

by the law of sines, $\frac {x}{\sin b}=25$ and $\frac{x}{\sin a}=50$
$x=25\sin b=50\sin a \Rightarrow \sin b=2\sin a=2\cos b$
$\sin^2 b + \cos^2 b=1 \Rightarrow (2\cos b)^2 + \cos^2 b = 5\cos^2 b =1$
$\cos b = \frac{\sqrt{5}}{5}=\sin a$ and $\sin b = \frac{2\sqrt{5}}{5}$
plugging back in, $x=10\sqrt{5}$

area of rhombus is $x^2\sin 2a=x^22 \sin a \cos a =(10\sqrt{5})^2(2(\frac {\sqrt{5}}{5})(\frac {2\sqrt{5}}{5}))=\boxed{400}$[/hide]

Here's one involving Law of Cosines - it's rather brute force, but it's easy to work with.

[hide]Let the intersection be point $E$, and let the center of the bigger circle be $O$ and the smaller circle be $O'$. Let $\angle AOD = \theta$.

Let the side of the rhombus be $c$, $BE = x$, and $O'E = y$.

$AD^2 = AO^2 + DO^2 + 2(AO)(DO)\cos{\theta}$ (1)

We know $\bigtriangleup ABD \sim \bigtriangleup ODA$. So, $\frac{c}{2x} = \frac{25}{c} \Rightarrow c^2 = 50x$.

Subsituting into (1),
$50x = 50(25) - 50(25)\cos{\theta} \Rightarrow x = 25 - 25\cos{\theta}$ (2)

We also know that $\angle AOD \cong \angle BAD \cong \angle EO'B$. So, $\cos{\theta} = \frac{y}{12.5}$.

Substitute into (2) to get $x = 25 - 2y$.

We know $x^2 + y^2 = \displaystyle\left(\frac{25}{2}\right)^2$.

Substitue and solve to get $y=7.5$, so $AE = 20$ and so $x = 10$.

Our area is $2(AE)(x) = 400$.[/hide]

P.S. Peter, I see you're registered.

I don't remember having to use any trig or $\frac{abc}{4R}$ when I did this problem that year....

This one's strikingly similar to solution #2, though it never bothers to find the length of the sides.

Let $BD=d_1$, $AC=d_2$, $\angle BAD=\alpha$, $\angle ABC= \beta$.
Note that $\alpha=180-\beta$.
LOS on side $BD$ and angle $\angle BAD$ of $\triangle ABD$ to get $\frac{d_2}{\sin(\alpha)}=2\times 12.5$.
Similarly, we get $\frac{d_1}{\sin(\beta)}=2\times 25$ by LOS on $\triangle ABC$.
Note that $\sin(\beta)=\sin(\alpha)$.
Thus, we have $d_1=2d_2$ by comparing our two equations.
Note that $\cos(\beta)=-\cos(\alpha)$
Do Law of Cosines on $\triangle ABD$ to get $s^2+s^2-2s^2\cos(\alpha)=d^2_2$  where s is the length of a side of the rhombus.
On $\triangle  ABC$ to get $s^2+s^2+2s^2\cos(\alpha)=d_1^2=4d^2_2$
Solve these two equations to get $\cos(\alpha)=\frac{3}{5}$.
Law of Sines to find $d_1$ and $d_2$, multiply them, divide by 2, and we get 400.  yay!

[quote="behemoth571"] [color=darkred]Find the area of rhombus $ ABCD$ given that the circumradii $ r$ , $ R$ of the triangles $ ABD$ and $ ACD$ respectively.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ m(\widehat {ABC})\equal{}2\phi$ . Thus, using the [b]Sinus' theorem[/b] obtain $ \left\|\begin{array}{c}
AC\equal{}2R\sin 2\phi\\\
BD\equal{}2r\sin 2\phi\\\
AB\equal{}2r\sin\phi\end{array}\right\|$ . Therefore,

$ 4\cdot AB^2\equal{}AC^2\plus{}BD^2$ $ \implies$ $ 4r^2\sin^2\phi\equal{}(R^2\plus{}r^2)\sin^22\phi$ $ \implies$ $ \left\|\begin{array}{c}
\cos\phi\equal{}\frac {r}{\sqrt {R^2\plus{}r^2}}\\\\
\sin \phi\equal{}\frac {R}{\sqrt {R^2\plus{}r^2}}\end{array}\right\|$ . In conclusion,

$ \sin 2\phi \equal{}2\sin\phi\cos\phi \equal{}\frac {2Rr}{R^2\plus{}r^2}$ and $ [ABCD]\equal{}\frac 12\cdot AC\cdot BD\equal{}2Rr\sin^22\phi$ $ \implies$ $ \boxed {\ [ABCD]\equal{}\frac {8R^3r^3}{\left(R^2\plus{}r^2\right)^2}\ }$ .

[b]Remark.[/b] If $ r\equal{}\frac R2$ , then $ [ABCD]\equal{}\left(\frac {4R}{5}\right)^2$ . In this case, for $ R\equal{}25$ obtain $ [ABCD]\equal{}400$ .[/color]


Find the exponent of 2 from the factorization of number ${(2n)! \over n!}$, $n \geq 1$.


counting
distinguishability
geometry
3D geometry
octahedron
rotation
geometric transformation

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
[asy]import three;
import math;
size(180);
defaultpen(linewidth(.8pt));
currentprojection=orthographic(2,0.2,1);

triple A=(0,0,1);
triple B=(sqrt(2)/2,sqrt(2)/2,0);
triple C=(sqrt(2)/2,-sqrt(2)/2,0);
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
triple F=(0,0,-1);
draw(A--B--E--cycle);
draw(A--C--D--cycle);
draw(F--C--B--cycle);
draw(F--D--E--cycle,dotted+linewidth(0.7));[/asy]$ \textbf{(A)}\ 210 \qquad \textbf{(B)}\ 560 \qquad \textbf{(C)}\ 840 \qquad \textbf{(D)}\ 1260 \qquad \textbf{(E)}\ 1680$

Grr! I hate these types of problems... :mad:

That diagram took me forever to reproduce, so appreciate this one. :)

EDIT: Turns out I got it wrong anyway.  The solution is on the wiki, and I don't want to claim it as my own.

[hide="Nice diagram, worthawholebean"]
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
[asy]unitsize(1.5cm);
defaultpen(0.5);
import three;
import math;
currentprojection=orthographic(2,0.2,1);
triple A=(0,0,1);
triple B=(sqrt(2)/2,sqrt(2)/2,0);
triple C=(sqrt(2)/2,-sqrt(2)/2,0);
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
triple F=(0,0,-1);
draw(A--B--E--cycle);
draw(A--C--D--cycle);
draw(F--C--B--cycle);
draw(F--D--E--cycle,dotted+linewidth(0.7));
fill(A--B--C--cycle,rgb(1,.6,.6));[/asy]
There are $ 7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face.
[asy]size(8cm);
defaultpen(0.5);
import three;
import math;
currentprojection=orthographic(2,0.2,1);
triple A=(0,0,1);
triple B=(sqrt(2)/2,sqrt(2)/2,0);
triple C=(sqrt(2)/2,-sqrt(2)/2,0);
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
triple F=(0,0,-1);
picture p = new picture, r = new picture, s = new picture;
draw(p,A--B--E--cycle);
draw(p,A--C--D--cycle);
draw(p,F--C--B--cycle);
draw(p,F--D--E--cycle,dotted+linewidth(0.7));
fill(p,A--B--C--cycle,rgb(1,.6,.6));
fill(p,A--B--E--cycle,rgb(1,1,.6));
add(scale(2.2)*p);
draw(r,A--B--E--cycle);
draw(r,A--C--D--cycle);
draw(r,F--C--B--cycle);
draw(r,F--D--E--cycle,dotted+linewidth(0.7));
fill(r,A--B--C--cycle,rgb(1,.6,.6));
fill(r,A--C--D--cycle,rgb(1,1,.6));
add(scale(2.2)*shift(2*right)*r);
draw(s,A--B--E--cycle);
draw(s,A--C--D--cycle);
draw(s,F--C--B--cycle);
draw(s,F--D--E--cycle,dotted+linewidth(0.7));
fill(s,A--B--C--cycle,rgb(1,.6,.6));
fill(s,B--C--F--cycle,rgb(1,1,.6));
add(scale(2.2)*shift(4*right)*s);[/asy]
Thus, the answer is $ 7!/3 = 1680 \Rightarrow \mathrm{(E)}$.
[/hide]

If you know that a cube/octahedron has $ 24$ symmetries, from a study of group theory or something, then this problem is a matter of dividing $ 8!$ by $ 24$. :|

http://en.wikipedia.org/wiki/Platonic_solid#Symmetry

I don't know how many people this applies to, but a lot of people on my team are great at visualizing cubes and tetrahedrons, and terrible with anything else.

Remember, we can map the faces of an octahedron to the vertexes of a cube, and the vertexes of a cube to the faces of an octahedron. Such pairs of polyhedrons are called [url=http://en.wikipedia.org/wiki/Dual_polyhedron]dual polyhedrons[/url].

[quote="MellowMelon"]If you know that a cube/octahedron has $ 24$ symmetries, from a study of group theory or something[/quote]

I don't think you need group theory - just choose one of the 8 faces to go on top, in one of 3 orientations, so there are 24 symmetries.  :wink:

if you're like me and you don't know anything about symmetry or group theory or visualizing
you could just do 8c4 (8 choose 4) which is 70
that would be for the top half

and the bottom half would just be 4!=24 ways

Therefore it would be 70*24 which is [b]1680[/b].

Your computation only gets the right answer by a freak coincidence -- choosing four colors for the top and an order for four colors on the bottom does not describe a coloring of the octahedron up to symmetry.

Alright, I know this is a revival of sorts, but I have come up with an insight that none of the above posters found.

Imagine a point at the center of each of the octahedron's faces. These points, when connected together, form a cube. Thus, the [b]number of rotations of the octahedron is the same as the number of rotations as a cube.[/b]
To find the number of rotations of a cube, simply note that There are 6 faces to place on the bottom as a base, and 4 ways to turn the cube for each of these configurations, for a total of $4 \cdot 6 = 24$ rotations. Next, there are 8! ways to color the faces without rotations, so the answer is $\dfrac {8!}{24} = \boxed {1680}$, so the answer is E.

That's essentially the same as what MellowMellon wrote ;)

We can also do this with Burnside's Lemma. First, we make a reduction that is easier to think about. Because the cube and octahedron are dual, this problem is equivalent to counting number of ways we can assign each vertex of a cube a unique color so that the cubes are distinguishable under rotation. There are 24 symmetries because there are 8 options for any vertex to go, and then 3 ways to orientate the cube, and clearly only the identity leaves any configuration fixed because all of the colors are distinct, so the answer is just 8! divided by 24 which is 1680.

There are $8!$ ways to color an octahedron if different orientations are distinct. However, this is not the case; for each octohedron, we can orient it in $24$ ways (we have $6$ ways to chose the bottom vertex and from there, $4$ ways to rotate it). Hence our answer is simply $\frac{8!}{24}=\boxed{1680}$.

@above where did you get $18$ from...

Also the reason why we divide by 6 is because for any configuration we make, we can rotate it 60,120,180,240,300,360 degrees to make equally valid configurations. So each configuration is counted 6 times...

Oh that makes sense.

I got 18 as you could rotate it 3 ways not including itself (90,180,270), and there are 6 vertices $3*6$. Thanks!

Ok. The important thing to remember is that you should "include itself" (i.e. you can rotate it 4 ways)

[b]Solution:[/b] Notice that there are three "axes" the octahedron can rotate in. We can ignore two of the rotations by labeling the top center triangle face as $A$. Then notice that there are $7!$ possibilities. However, every $3$ of them will be identical as one can also rotate the octahedron about the axis where $A$ faces the same direction but all the other sides spin around it. Hence our answer is $7!/3 = \boxed{E}$.

[quote=smartguy888]I am kinda stuck on why we divide by $24$, not $18$. Like for example, we need to know how many ways can we place 6 keys in a keychain. There are 5 ways to rotate it, 120,240,360...600 degrees. If we do 720 it will just look the same right? So why is the answer $6!/6$, not $6!/5$?[/quote]

why you dum


To a party Asites some men and women. If two men are greeted with a handshake. But if a man lies with a woman or two women are, then you are greeted with a kiss on the cheek. We know that everyone who attended the party greeted each other (two by two) and the number of men exceeded the number of four women. If the number of kisses on the cheek exceeded by 75 the number of handshakes, how many people attended the party?.

Greetings  :D

No matter the proficiency one has in the English language, I think one should strive to make the statement of her problem at least intelligible.

So assume at a party there are $ m$ men and $ w$ women, the men outnumbering the women by $ 4$, i.e. $ m \equal{} w \plus{} 4$. Two men greet each other by a handshake; two women, or a man and a woman, greet each other by a kiss on the cheek. As everyone greets everyone else, there are $ 75$ more kisses than handshakes, i.e. $ \genfrac {(} {)} {0pt} {} {m\plus{}w} {2} \minus{} \genfrac {(} {)} {0pt} {} {m} {2} \equal{} \genfrac {(} {)} {0pt} {} {m} {2} \plus{} 75$. This last equation yields $ w^2 \plus{} (2m \minus{} 1)w \minus{}m^2 \plus{} m \minus{} 150 \equal{} 0$, of discriminant $ \Delta \equal{} (2m\minus{}1)^2 \plus{}4(m^2 \minus{}m \plus{}150) \equal{} 2(2m\minus{}1)^2 \plus{} 599 \equal{} x^2$, leading to some Pell equation.
But since we are given that $ m \equal{} w \plus{} 4$, it all boils down to a simple substitution, with unique solution $ w\equal{}9$ and $ m\equal{}13$.


quadratics
number theory unsolved
number theory

If I have a prime number $ p$ of the form $ p\equal{}4k\plus{}3$, how can I show that $ a$ is a primitive root modulo $ p$ if, and only if, $ \minus{}a$ has order $ (p\minus{}1)/2$ in $ \left(\mathbb{Z}/p\mathbb{Z} \right)^{\times}$?

[hide="hint"]If $ g$ is a primitive root, then $ \minus{}1 \equiv g^{\frac{p\minus{}1}{2}}$[/hide]

[quote="K81o7"][hide="hint"]If $ g$ is a primitive root, then $ \minus{} 1 \equiv g^{\frac {p \minus{} 1}{2}}$[/hide][/quote]

I see, thank you.  And so would this follow from the fact that in $ \left( \mathbb{Z}/p\mathbb{Z} \right)^{\times}$, if $ a$ is a primitive root modulo $ p$, then $ \bar{1} \equal{} \bar{a}^{p \minus{} 1} \equal{} \bar{a}^{(p \minus{} 1)/2} \cdot \bar{a}^{(p \minus{} 1)/2}$, which can only be the case if $ \bar{a}^{(p \minus{} 1)/2} \equal{} \overline{ \minus{} 1}$?  The result would then follow, right, since I can simply multiply both sides of $ \bar{a}^{(p \minus{} 1)/2} \equal{} \overline{ \minus{} 1}$ by $ \overline{ \minus{} 1}$.  That gives the $ \Rightarrow$ implication.  It would seem that the converse can be proven similarly (though I haven't given serious thought to it yet), but I'm confused since it appears that nothing in the argument used the fact that $ p$ is of a specific form, namely $ p \equal{} 4k \plus{} 3$.

What's wrong with this?  I'm assuming I have to use the fact that $ p\equal{}4k\plus{}3$.

Use the fact that $ \minus{}1$ is not a quadratic residue for primes of the form $ 4k \plus{} 3$


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry unsolved

Here is a problem of my friend :
 Let a triangle $ ABC$ in plane which have the orthocenter $ H$ and the circumcenter $ I$.The perpendicular bisector of $ AH$ cut $ AC,AB$ at $ E,D$ correspondence.Prove that : $ \hat{AOD}\equal{}\hat{AOE}$
 [hide] I am so sorry about my poor English ! [/hide]

△ADH∽△AOC  =>  △ADO∽△AHC  =>  ∠AOD=∠ACH
Analogously, ∠AOE=∠ABH
=>∠AOD=∠AOE

very nice proof  :lol:

this is a different proof (i think).. the first picture is a "graphic" demostration that the perpendicular bisector of $ AH$ passes through the reflections of $ O$ on sides $ AB, AC$... the second one "completes" the proof...


Draw an angle $\angle ABC$ of $30^\circ$, and put point $M$ on ray $\stackrel{\longrightarrow}{BC}$ 1 unit away from the vertex $B$; put point $X$ on  $\stackrel{\longrightarrow}{BA}$ 2 units away from the vertex. Then, continue plotting points on the angle by summing the distance from the last plotted point to the vertex, switching over to the opposite ray, and placing the point at a distance from the last point on that ray. Continue placing points, always going further and further away from the vertex, until you get points $X,Y,Z$ on $\stackrel{\longrightarrow}{BA}$ and points $M,N,P$ on $\stackrel{\longrightarrow}{BC}$.

Draw lines $PX$, $NY$ and $MZ$. Find the area of the triangle enclosed by the three lines.

If possible, generalize for starting distances $a$ and $c$, with $a \ge c$.

it is trivial to find the areas of the little quadrilaterals formed by two of the cevians, and you subtract those areas from BYN


function
quadratics
calculus
derivative
special factorizations

[i]1) 518, 2) 607 (these numbers are only for my own use, plz just skip it)[/i]

I have a few problems i don't know how to solve:

[b]1) Two functions (f and g) are as follows:
f(x)=ax+b
g(x)=2ax-2b
Choose a and b, so that the two functions intersect in (1;1).[/b]

I believe i have to exchange f(x) and g(x) with 1 and exchange x with 1 since that's the intersections, but it isn't enough to do as far as i can see when doing so.

[b]2) Find the unknown numbers for a and c in the following function:
f(x)=(ax^2)-6x+c[/b]

The brackets are only for showing that the -6x+c is connected with the ^2. Hope you can help me.

In advance, thx for any help on any of the two problems.

[quote="Sporally"] The brackets are only for showing that the -6x+c is connected with the ^2. [/quote] I suppose you mean that the brackets are for showing that the ^2 is connected with the ax?
[hide="Anyway, for the first one..."] You let $(x,y)$ be $(1,1)$ and this gives you two equations:
$a+b=1$ and $2a-sb=1$. Solving, you get $a=\frac{3}{4}$ and $b=\frac{1}{4}$.[/hide]
For the second one, I don't have a clue... it could be anything :huh:  Or is nr. 2 somehow connected to nr. 1?

Yes that was what i meant - ^2 is connected to ax. The two functions do not have a connection.

:huh: Are you sure the second question doesn't involve any another conditions? e.g. f is always positive etc. ?

Oh damn, sorry i forgot to tell you. I don't know the english word for it, but the 'top spot' of the function is (2; -8). Sorry i forgot to tell you - hope this helped you to help me :)

[quote="Sporally"]Oh -, sorry i forgot to tell you. I don't know the english word for it, but the 'top spot' of the function is (2; -8). Sorry i forgot to tell you - hope this helped you to help me :)[/quote]
I assume you mean $f(x)=(ax)^2-6x+c$ . Completing the square you get

$f(x)=a^2(x-\frac{3}{a^2})^2+c-\frac{9}{a^2}$

Ok , I assume again the " top spot " is the maximum . Which mean $\frac{3}{a^2}=2$ and $c-\frac{9}{a^2}=8$ and $a<0$ because the function is a convex quadratic function . Solve it to get $a=-\frac{\sqrt{6}}{2}$ and $c=14$

[quote="Sporally"]And i knew i should never had done anything with the brackets - all that i wanted to show with the brackets was that the -6x+c wasn't a part of the squared 2-thing. Let's try that second one from the beginning:

f(x)=ax^2-6x+c   turning point (2;-8)

I think 'turning point' is a better expression. In x=2 and y=-8 the graph is turning. Hope this helped.[/quote]

ok then still same . Completing the square

$f(x)=a(x-\frac{3}{a})^2+c-\frac{9}{a}$ . So $\frac{3}{a}=2$ and $c-\frac{9}{a}=-8$ . Solve and get $a=\frac{3}{2}$ , $c=-2$ ;)

[quote="Sporally"]I kind of a rookie. Could you show it step by step and tell what you do?[/quote]

So , do you know how to completing a square ?

like $f(x)=ax^2+bx+c$ . Completing the square will give u

$f(x)=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$  . The turning point would occur when $x+\frac{b}{2a}=0$ which gives you the maximum/minimum = $c-\frac{b^2}{4a}$ . (We also call this point $(-\frac{b}{2a},c-\frac{b^2}{4a})$ as turning point) .

So just compare them and solve it .

Or another way is by using differentiation to find $f'(x)=2ax-6$ and the turning point is when $f'(x)=0$ which is $2ax-6=0$ so $x=3a$ . But the turning point is $(2,-8)$ .Hence $3a=2$ . And substitute back to get your $c$ and such .

Can you prove the step from:

f(x)=ax^2+bx+c

to:

f(x)=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}

Let's just get back to this one another day, it's simply not clear for me how you do each step. Could you or anyone else do the first task i wrote in the first thread then? Maybe i will learn something from this one i can use on the second task.


More curiosity than anything else but can anyone tell me whether in Romania the medium of teaching is English or Romanian. Could Valentin or someone from  Romania reply? thanks.

Normally, classes are taught in Romanian. There are also some schools, private schools, in which English is the main language . Very few go to private schools, and very few of those are good at math ( except thos from The International Turkish Highschool in Bucharest, ICHB).
Anyway, there are some classes whose profile include an extensive program of English, but  Romanian is still the basic language :P.


For positive reals $ a_1,a_2,...,a_n,b_1,b_2,...,b_n$ $ (n\geq 3)$ with the $ b_i$ pairwise distinct, denote $ S\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n,T\equal{}b_1b_2...b_n$.

(a) Let $ f(x)\equal{}(x\minus{}b_1)(x\minus{}b_2)...(x\minus{}b_n)\sum_{j\equal{}1}^n\frac{a_j}{x\minus{}b_j}$. Find the number of distinct real zeroes of the polynomial $ f(x)$.

(b) Prove that $ \frac{1}{n\minus{}1}\sum_{j\equal{}1}^n(1\minus{}\frac{a_j}{S})b_j>(\frac{T}{S}\sum_{j\equal{}1}^n\frac{a_j}{b_j})^\frac{1}{n\minus{}1}$

[quote="easternlatincup"]For positive reals $ a_1,a_2,...,a_n,b_1,b_2,...,b_n$ $ (n\geq 3)$ with the $ b_i$ pairwise distinct, denote $ S \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n,T \equal{} b_1b_2...b_n$.

(a) Let $ f(x) \equal{} (x \minus{} b_1)(x \minus{} b_2)...(x \minus{} b_n)\sum_{j \equal{} 1}^n\frac {a_j}{x \minus{} b_j}$. Find the number of distinct real zeroes of the polynomial $ f(x)$.

(b) Prove that $ \frac {1}{n \minus{} 1}\sum_{j \equal{} 1}^n(1 \minus{} \frac {a_j}{S})b_j > (\frac {T}{S}\sum_{j \equal{} 1}^n\frac {a_j}{b_j})^\frac {1}{n \minus{} 1}$[/quote]
a) Obviosly $ deg(f(x))\equal{}n\minus{}1$ and $ f(b_j)\equal{}a_j\prod_{i\not \equal{}j}(b_j\minus{}b_i)$, therefore if $ 0<b_1<b_2<...<b_n$, then $ f(b_{n\minus{}2k})>0, f(b_{n\minus{}1\minus{}2k}<0$. It give n-1 real roots $ 0<b_1<x_1<b_2<...<x_{n\minus{}1}<b_n$.
b) Consider $ f(x)\equal{}Sx^{n\minus{}1}\minus{}\sum_j b_j(S\minus{}a_j) x^{n\minus{}2} \plus{}(\minus{}1)^{n\minus{}1}T\sum_j\frac{a_j}{b_j}$.
Therefore \[ \sum_{i\equal{}1}^{n\minus{}1} x_i\equal{}\sum_{j\equal{}1}^nb_j(1\minus{}\frac{a_j}{S}), \prod_{i\equal{}1}^{n\minus{}1}x_i\equal{}\frac{T}{S}\sum_{j\equal{}1}^n\frac{a_j}{b_j}.\]
Your condition equavalent to $ \frac{1}{n\minus{}1}\sum_i x_i >(\prod_i x_i )^{1/(n\minus{}1)}$, it is true because $ x_i$ are distinct positive numbers.


Here are the 2 math problems I need help solving

1. A basketball team plays exactly 5 players in a game. The team made an average of 0.625 of each of the shots taken in a particular game.  Edgar notices that the results published in the school paper forgot to mention how many shots he made during the game.  Alex made 5 out 11 shots, Brian made 8 out of 9, Chris made 9 out of 15, and Derrick made 12 out of 22 shots. He also knows he took more shots than Brian but fewer than Derrick.  How many baskets did Edgar make in the game.  (The answer is 11)  I need help solving the problem.

2.  A ball is dropped from a height of 32 ft. and rebounds 4/5 of its previous height on each bounce.  How far will the ball have traveled (in total) by the time it hits the ground for the 9th time? Round to the nearest foot.  The answer is 245 but im not sure how to get that one.

Please help and reply if you know how to help me solve the, Thanks :)

Here are the 2 math problems I need help solving

1. A basketball team plays exactly 5 players in a game. The team made an average of 0.625 of each of the shots taken in a particular game. Edgar notices that the results published in the school paper forgot to mention how many shots he made during the game. Alex made 5 out 11 shots, Brian made 8 out of 9, Chris made 9 out of 15, and Derrick made 12 out of 22 shots. He also knows he took more shots than Brian but fewer than Derrick. How many baskets did Edgar make in the game. (The answer is 11) I need help solving the problem.

2. A ball is dropped from a height of 32 ft. and rebounds 4/5 of its previous height on each bounce. How far will the ball have traveled (in total) by the time it hits the ground for the 9th time? Round to the nearest foot. The answer is 245 but im not sure how to get that one.

Please help and reply if you know how to help me solve the, Thanks
Let me try the second question
It's a power sequence with n= 9 ( strating from the 1 st hit the ground)
The total distance was =(2 x S9) + 32
S9 is the sum of power sequence
= a(1-(4/5)^9/(1-4/5)
where a is the fisrt height after bouncing = 32 x 4/5
am I rite?

$ n \equal{} 9$ should be $ n \equal{} 8$, giving $ 245.05050327...$

For the first problem, if Edgar made $ e$ out of $ f$ shots then we have

$ \frac{5\plus{}8\plus{}9\plus{}12\plus{}e}{11\plus{}9\plus{}15\plus{}22\plus{}f}\equal{}\frac{5}{8}$

$ 272\plus{}8e\equal{}285\plus{}5f$

$ 8e\equal{}5f\plus{}13$

But $ 9<f<22$

Therefore $ f\equal{}15$ and $ e\equal{}11$.

[hide="1."] We know that the team made $ .625\equal{}\frac{5}{8}$ of their shots. Adding up everybody's shot totals as a fraction, we have $ \frac{5\plus{}8\plus{}9\plus{}12\plus{}x}{11\plus{}9\plus{}15\plus{}22\plus{}y}\equal{}\frac{34\plus{}x}{57\plus{}y}$, where $ x$ and $ y$ are the totals of Edgar's made shots and total shots attempted. Since the above fraction is equal to $ \frac{5}{8}$, the denominator must be a multiple of $ 8$. We are given that $ 9<y<22$, so we look for the value of $ y$ which will give us a multiple of $ 8$. We see that $ 57\plus{}15\equal{}72$, so $ y\equal{}15$. We can now solve for $ x$ as follows:

$ \frac{34\plus{}x}{72}\equal{}\frac{5}{8} \implies 34\plus{}x\equal{}45 \implies x\equal{}\boxed{11}$.[/hide]

I guess you beat me to it, gaussintraining, at 7.06pm.


Ring Theory
linear algebra
matrix
algebra
binomial theorem
superior algebra
superior algebra unsolved

An element a in a commutative ring is said to be nilpotent if a^n=0 for soem positive interger n ( which may depend on a). Prove that the set of all nilpotent elements in a commutative ring R is an ideal of R.

Please help....

What can't you do ? About all the axioms can be immediately checked ...

i am confused on what a nilpotent is... is it possible to say that the set of nilpotent elements in a communtative ring R is a subring of R... because i would still not understand how to prove

ar is in I and ra is in I...

maybe just list hypothetical elements in the case would help me visualize the problem.

thanks

ok let $a^n=0$,  what can u tell abt $(ra)^n$? since ur ring is commutative $(ra)^n=(r^n)(a^n)$, so ra lies in I. similarly ar. this should do it

Well, that nilpotent elements are closed under addition, is not so immediate.

When $a^n = b^m = 0$, then after the binomial theorem
\[ (a+b)^{m+n} = \sum\limits_{k=0}^{m+n} \begin{pmatrix} m+n \\ k \end{pmatrix} a^k b^{m+n-k} = 0  \]
since $0 \leq k < n \Rightarrow m+n - k > m \Rightarrow b^{m+n-k} = 0$ and $n \leq k < m+n \Rightarrow a^k = 0$.


trigonometry
ratio
geometry
angle bisector
trig identities
Law of Sines
geometry proposed

find all triangle $\triangle ABC$ such that the median $AM$,the bisector $BD$ and the altitude $CH$ are
concurent.

[quote="marko avila"]i remember this problem. the answer is only the equilateral triangles.[/quote]

Wrong. There exist infinitely many triangles ABC such that the median from the vertex A, the angle bisector from the vertex B and the altitude from the vertex C are concurrent. These are all triangles ABC satisfying $\frac{\sin C}{\sin B}=\frac{\cos A}{\cos B}$. One random example for such a triangle is the triangle with the sidelengths a = 12, $b=\sqrt{145}+3$, $c=\sqrt{145}+7$.

Does the problem ask for a (parametric) Euclidean construction of all such triangles ABC ? Well, I don't have a really nice purely geometric construction, but here is an algebraic way to proceed: Rewrite the equation $\frac{\sin C}{\sin B}=\frac{\cos A}{\cos B}$ in the form $\frac{\cos B}{\sin B}=\frac{\cos A}{\sin C}$. But by the sum of angles in triangle ABC, we have A = 180° - (B + C), and thus cos A = - cos (B + C) = - (cos B cos C - sin B sin C) = sin B sin C - cos B cos C. Thus,

$\frac{\cos A}{\sin C}=\frac{\sin B\sin C-\cos B\cos C}{\sin C}=\sin B-\cos B\frac{\cos C}{\sin C}=\sin B-\cos B\cot C$.

Thus, the equation $\frac{\cos B}{\sin B}=\frac{\cos A}{\sin C}$ becomes $\frac{\cos B}{\sin B}=\sin B-\cos B\cot C$. From this equation, if the angle B is given, we can easily find the angle C. Hence, we know two angles of triangle ABC; thus, we can construct the triangle (up to similitude, of course). Hence, we are able to construct the triangle ABC for every possible value of the angle B.

  Darij

DARIJ:  so stupid¡¡¡ very sorry¡¡ it just is so similar to a british tst problem¡¡ but now that i see the british one i see i am wrong.
anyway nice solution¡¡¡

it was a problem of Poland Math Olympiad of 1961.
My answer is:$tgB=\frac{\sin{C}}{\cos{A}}$

[quote="shobber"]My answer is:$tgC=\frac{\sin{C}}{\cos{B}}$[/quote]

That's wrong. The correct condition for the median from the vertex A, the angle bisector from the vertex B and the altitude from the vertex C to be concurrent is $\frac{\sin C}{\sin B}=\frac{\cos A}{\cos B}$.

[i]Proof.[/i] Let the median from the vertex A meet the side BC of triangle ABC at a point M; let the angle bisector from the vertex B meet the side CA at a point D; let the altitude from the vertex C meet the side AB at a point H. Then, by the Ceva theorem, the median from the vertex A, the angle bisector from the vertex B and the altitude from the vertex C are concurrent if and only if $\frac{BM}{MC}\cdot\frac{CD}{DA}\cdot\frac{AH}{HB}=1$. But the point M is the point where the median from the vertex A of triangle ABC meets the side BC; in other words, the point M is the midpoint of this side BC, and thus $\frac{BM}{MC}=1$. Also, since the angle bisector of an angle in a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{CD}{DA}=\frac{a}{c}$. By the law of sines, $\frac{a}{c}=\frac{\sin A}{\sin C}$, so that $\frac{CD}{DA}=\frac{\sin A}{\sin C}$. Finally, since the point H is the foot of the altitude from the vertex C of triangle ABC, the triangles CHA and CHB are right-angled, and thus $AH=CH\cdot\cot\measuredangle CAH=CH\cdot\cot A$ and $HB=CH\cdot\cot\measuredangle CBH=CH\cdot\cot B$, so that $\frac{AH}{HB}=\frac{\cot A}{\cot B}=\frac{\cos A/\sin A}{\cos B/\sin B}$.

Hence, we have the following chain of equivalent assertions:

{The median from the vertex A, the angle bisector from the vertex B and the altitude from the vertex C are concurrent}
$\Longleftrightarrow$ {$\frac{BM}{MC}\cdot\frac{CD}{DA}\cdot\frac{AH}{HB}=1$}
$\Longleftrightarrow$ {$1\cdot\frac{\sin A}{\sin C}\cdot\frac{\cos A/\sin A}{\cos B/\sin B}=1$}
$\Longleftrightarrow$ {$\frac{\sin B}{\sin C}\cdot\frac{\cos A}{\cos B}=1$}
$\Longleftrightarrow$ {$\frac{\cos A}{\cos B}=\frac{\sin C}{\sin B}$}
$\Longleftrightarrow$ {$\frac{\sin C}{\sin B}=\frac{\cos A}{\cos B}$}.

And the proof is complete.

  Darij


Triangle $ ABC$ is isosceles with angle $ B$ congruent to angle $ C$. The measure of angle $ C$ is four times the measure of angle $ A$. What is the number of degrees in the measure of angle $ B$?

Let x be the number of degrees in angle A. Thus $ x\plus{}4x\plus{}4x\equal{}180\Rightarrow x\equal{}20$, and the measure of angle $ B\equal{}\boxed{80}$.


a,b,c are rationals , prove that  : $ \frac{1}{(a\minus{}b)^2} \plus{} \frac{1}{(a\minus{}c)^2 } \plus{} \frac{1}{(b\minus{}c)^2}$ is a square of rational number

nice Dr Sonnhard Graubner , but  $ \frac{(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)}{(a\minus{}b)(b\minus{}c)(a\minus{}c)}$  is rational ? why ??

hello, because the quotient of rational numbers is rational.
Sonnhard.

thanks Dr Sonnhard Graubner  :10:  so you can resolve my system ( other post )

[quote="linda2005"]a,b,c are rationals , prove that  : $ \frac {1}{(a - b)^2} + \frac {1}{(a - c)^2 } + \frac {1}{(b - c)^2}$ is a square of rational number[/quote]
Let $ A = \frac {1}{(a - b)} + \frac {1}{(a - c)^2 } + \frac {1}{(b - c)^2}$.
Then we have,
$ A = (\sum\frac {1}{(a - b)})^2 - 2(\sum\frac {1}{(a - b)(b - c)})$
$ = (\sum\frac {1}{(a - b)})^2 - 2\frac {a - b + b - c + c - a}{\prod(a - b)}$
$ = (\sum\frac {1}{(a - b)})^2$,
Which is RATIONAL. :)
QED :blush: .


Gauss
geometry
3D geometry
sphere
calculus
integration
ratio

A point charge $ q$ is placed at the centre of a hollow sphere of radius $ R$. Find flux through the shaded region.

[hide]I found the area using the common technique (by integration - the donkey method), but I want to know if there's another method, a faster one.[/hide]

What is the "donkey method"?  I would do a standard Gauss' law setup, find the electric field due to the point charge, find the area of the flux region and proceed as normal.

[quote="JRav"]find the area of the flux region [/quote]
that is the question.. :D
and i too wud lik to kno the donkey method :rotfl:

[quote="JRav"]
What is the "donkey method"?

[/quote]

[quote="Rituraj"]
And I too wud lik to know the donkey method

[/quote]

I call all methods used as last resort as donkey-methods: the tough ones you have to plough through with great effort like a donkey.:blush:

$ A\equal{}\int\limits^{\frac{R}{2}}_0 2\pi\sqrt{R^2\minus{}x^2}dx \\ \\ \\
\therefore A \equal{} \pi R^2$

My professor did this using the solid angle stuff. But I understood only $ 50$% of it.

Perhaps this should have been in Calculus Computations...
Does anyone know an easier way out. I am very lazy when it comes to integration. :oops:

actually solid angle is the easiest (very easy infact) way to do it....
but i'll hav to think for it ,so wait for some time (actually i hav stopped thinking currently for my PMT preparations :rotfl: )

and the method u call "Donkey method" is actually called as "[b]BRUTE FORCE method[/b]" :)

[quote="rituraj007"]
And the method u call "Donkey method" is actually called as "[b]BRUTE FORCE method[/b]" :)[/quote]

What is in a name?
-Albert Einstein  :wink:

 :rotfl:

"identity"

--Infinity squared
 :rotfl:  :rotfl:

Actually there is a mathematical result (I don't know how to prove it, I should try to find a way w/o calculus, should one exist.) That equates the ratio of the surface area of such a constant-height rotation to the entire surface area to the ratio of the height of the rotation to the diameter of the sphere. In this case, R/2 is 1/4 of the diameter, so the surface area of our shaded region should be $ \frac{4 \pi R^2}{4}\equal{}\pi R^2$.

I will look at the derivation of that result later. Now I am going to sleep.

I have one stupid way but I want you to check it and tell me if it is right or wrong (most probably wrong).

Let's divide the region into $ n$ cylinders of height $ h$ such that $ n\to\infty$ and $ h\to0$. Now, $ \frac{R}{2n} \equal{} h$.

Now the area is given by $ 2\pi Rh\times n$. But $ nh \equal{} \frac{R}{2}$, hence $ 2\pi Rhn \equal{} \not{2}\pi\frac{R^2}{\not{2}} \equal{} \pi R^2$

Tada!


Putnam
calculus
integration
college contests

Putnam 1984/A5) Let $R$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x+y+z\leq 1$. Let $w=1-x-y-z$. Express the value of the triple integral

\[\iiint_{R}xy^{9}z^{8}w^{4}\ dx\ dy\ dz\]

in the form $a!b!c!d!/n!$ where $a,b,c,d$ and $n$ are positive integers.

[hide="A solution"]\[\iiint_{R}xy^{9}z^{8}w^{4}\ dx dy dz = 4\iiint_{R}\int_{0}^{1-x-y-z}xy^{9}z^{8}w^{3}\ dw dx dy dz = 4\iiiint_{Q}xy^{9}z^{8}w^{3}\ dw dx dy dz\]

where $Q=\left\{ (x,y,z,w)\in\mathbb{R}^{4}|\ x,y,z,w\geq 0, x+y+z+w\leq 1\right\}$, which is a Dirichlet integral giving

\[4\iiiint_{Q}x^{1}y^{9}z^{8}w^{3}\ dw dx dy dz = 4\cdot\frac{1!9!8!3!}{(2+10+9+4)!}= \frac{1!9!8!4!}{25!}\][/hide]

Recall that a Dirichlet Integral is this: if $V^{n}=\left\{ (x_{1},x_{2},\ldots ,x_{n})\in\mathbb{R}^{n}|\ x_{1},x_{2},\ldots ,x_{n}\geq 0, \sum_{k=1}^{n}\left( \frac{x_{k}}{\alpha_{k}}\right)^{\beta_{k}}\leq 1\right\}$, then

\[\iint\cdots\int_{V^{n}}x_{1}^{\gamma_{1}-1}x_{2}^{\gamma_{2}-1}\cdots x_{n}^{\gamma_{n}-1}\ dx_{1}dx_{2}\cdots dx_{n}= \frac{\prod_{k=1}^{n}\left[\frac{\alpha_{k}^{\gamma_{k}}}{\beta_{k}}\Gamma \left( \frac{\gamma_{k}}{\beta_{k}}\right)\right]}{\Gamma \left( 1+\sum_{k=1}^{n}\frac{\gamma_{k}}{\beta_{k}}\right)}\]

The answer above uses the properties of the Dirichlet distribution to find the solution, this hints that we can interpret the integral in a probabilistic manner.

Consider the segment $[0,1]$ and choose $22$ random numbers from it. Then partition the interval into the four contiguous segments of random length respectively.

Notice that the integral $$I = \frac{1}{1/3!} \iiint_{R} \frac{22!}{1!9!8!4!} x^{1}y^{9}z^{8}(1-x-y-z)^{4}\ dx\ dy\ dz $$ computes the average probability of having chosen $1$ element from the first segment, $9$ from the second, $8$ from the third and $4$ from the last in any order.

Now, let's calculate the probability in another way. Label the numbers chosen, from smallest to biggest, as $p_1, p_2,\ldots, p_{22}$, note that we want $$0 \le p_1 \mid p_2, \ldots , p_{10} \mid p_{11} , \ldots , p_{18} \mid p_{19}, \ldots, p_{22} \le 1$$Since this $23$ numbers are uniformly chosen at random we can observe that the probability of the bars being on those places (which is the same as the probability we are trying to compute) is $$I = \frac{3!}{23 \cdot 24 \cdot 25}$$ so we conclude that $$\iiint_{R} x^{1}y^{9}z^{8}(1-x-y-z)^{4}\ dx\ dy\ dz = \frac{1!9!8!4!}{25!}$$


Show, that $ f(x)\equal{}\frac{x}{\sqrt{1\minus{}\left\|x\right\|^2}}$ is continuous on $ \mathrm{Int}( \mathrm{B}^n)$.

This is just an application of the arithmetic theorems and the continuity of the square root and norm functions. As long as the quantity inside the square root is nonnegative and we're not dividing by zero, the combination of continuous functions is continuous.

Now: can you show that is it one-to-one, that it maps the open unit ball onto all of $ \mathbb{R}^n,$ and that it is, in fact, a homeomorphism between the ball and $ \mathbb{R}^n?$

I was trying to show the continuity using just the definition with open sets, $ f^{ - 1}(U)$ open in $ \mathrm{B}^n$ for any open $ U\in\mathbb{R}^n$.

But it goes for Euclidian topology, so it suffices to use the standard $ \epsilon - \delta$ definition of continuity.
[b]
Continuity:[/b]

So for $ x = (x_1,\ldots,x_n)$ and $ y = (y_1,\ldots,y_n)$:

\begin{eqnarray*} d(f(x),f(y)) & \leq & d(f(x_1,x_2,\ldots,x_n),f(y_1,x_2,\ldots,x_n)) + d(f(y_1,x_2,\ldots,x_n),f(y_1,y_2,x_3,\ldots,y_n)) + \\
& + & \cdots + d(f(y_1,\ldots,y_{n - 1},x_n),f(y_1,y_2,\cdots,y_n)) \end{eqnarray*}

Now the new functions can be observed as the functions of one variable:

$ \begin{eqnarray*} f(x) & = & g(x)h(x) = (x_1,\ldots,x_i,\ldots,x_n)\frac {1}{\sqrt {(1 - x^2_1 - \ldots - x^2_{i - 1} - x^2_{i + 1} - \ldots - x^2_{n}) - x_i^2}} = \\
& = & (x_1,\ldots,x_i,\ldots,x_n)\frac {1}{\sqrt {c - x_i^2}} = g(x_i)h(x_i) = f(x_i)$

Both $ g(x_i)$ and $ h(x_i)$ are continuous, therefore $ f(x_i)$ is continuous. So for each $ \epsilon > 0$ there exists $ \delta > 0$:

$ d(f(x),f(y)) < \frac {\epsilon}{n} + \ldots + \frac {\epsilon}{n} < \epsilon$

whenever $ \left\|x - y\right\| < \delta$.

[b]Bijection:[/b]

We take an arbitrary unit vector $ e\in \mathbb{R}^{n}$ and $ t\in ( - 1,1)$:

$ f(te) = \frac {te}{\sqrt {1 - t^2\left\|e\right\|^2}} = \frac {t}{\sqrt {1 - t^2}}e = g(t)e = h_{e}(t)$

$ g(t)$ is a bijection of $ ( - 1,1)\rightarrow \mathbb{R}$, since it is continuous, strictly increasing and $ \lim_{t \to \pm1} = \pm \infty$.

So $ h_{e}(t)$ is a bijection of $ ( - 1,1)e\rightarrow \mathbb{R}e$.

Finally $ f(x)$ is a bijection of $ ( - 1,1)\mathrm{S}^0\rightarrow \mathbb{R}\mathrm{S}^0$.

[b]Homeomorphism:[/b]

Inverse of $ f(x)$:

$ \begin{eqnarray*} f(x) = f(te) = g(t)e \Rightarrow f^{ - 1}(x) & = & f^{ - 1}(te) = g^{ - 1}(t)e = \frac {t}{\sqrt {1 + t^{2}}}e = \\
& = & \frac {te}{\sqrt {1 + \left\|te\right\|^{2}}} = \frac {x}{\sqrt {1 + \left\|x\right\|^{2}}} \end{eqnarrray*}$

Now $ f^{ - 1}(x)$ is continuous, therefore $ f(x)$ is open and hence open continuous bijection-homeorphism between $ \mathrm{B}^n$ and $ \mathbb{R}^n$.


The consecutive angles of a trapezoid form an arithmetic sequence. If the smallest angle is $75^\circ$, then the largest angle is

$\text{(A)} \ 95^\circ \qquad \text{(B)} \ 100^\circ \qquad \text{(C)} \ 105^\circ \qquad \text{(D)} \ 110^\circ \qquad \text{(E)} \ 115^\circ$

[hide]If the common difference is $n$, $75+(75+n)+(75+2n)+(75+3n)=360$.

Solving gives $n=10$.

Thus the largest angle is $105^\circ \implies \boxed{C}$[/hide]

[hide]Put the angles in terms of 75 and x, where x is the difference between each increasing number: 75+75+x+75+2x+75+3x=360. Simplify and get x=10, so the largest angle is 105, answer C.[/hide]

[hide]If they form and arithmetic sequence, the first plus the last would be the average.  The average is 360/2=180

75+x=180, x=105 so thats the last angle.  (c)[/hide]


The possitive real numbers a,b,c ,x,y,z are such that
\[ \left\{ \begin{array}{l}
 cy \plus{} bz \equal{} a \\
 az \plus{} cx \equal{} b \\
 bx \plus{} ay \equal{} c \\
 \end{array} \right.
\]
 Fint the smaillest possible value of the epression
\[ P \equal{} \frac{{x^2 }}{{1 \plus{} x}} \plus{} \frac{{y^2 }}{{1 \plus{} y}} \plus{} \frac{{z^2 }}{{1 \plus{} z}}
\]

( for mathematics and youth Magazine VietNam)


Show that for $ n > 3$:

$ \frac{n^5 \minus{} 1}{n \minus{} 1}\neq k^2$ for any integer $ k$

i.e. it is not a perfect square.

Proving nonexistence results, as opposed to existence results, is pretty hard to do by induction.  There isn't a particularly good way to go from "this equation has no solutions" to "therefore, this equation must not have solutions."  One way to do this is to prove the contrapositive, but then it becomes [b]infinite descent[/b] instead.

No, there's a direct proof here.

[quote="MNrule"]Show that for $ n > 3$:

$ \frac {n^5 \minus{} 1}{n \minus{} 1}\neq k^2$ for any integer $ k$

i.e. it is not a perfect square.[/quote]

Direct proof:

For $ n$ even, $ (n^2 \plus{} \frac {n}{2})^2 < \frac {n^5 \minus{} 1}{n \minus{} 1} < (n^2 \plus{} \frac {n}{2} \plus{} 1)^2$

For $ n$ odd, $ (n^2 \plus{} \frac {n \minus{} 1}{2})^2 < \frac {n^5 \minus{} 1}{n \minus{} 1} < (n^2 \plus{} \frac {n \plus{} 1}{2})^2$. Done.

[quote="MNrule"][quote="stephencheng"][quote="MNrule"]Show that for $ n > 3$:

$ \frac {n^5 \minus{} 1}{n \minus{} 1}\neq k^2$ for any integer $ k$

i.e. it is not a perfect square.[/quote]

Direct proof:

For $ n$ even, $ (n^2 \plus{} \frac {n}{2})^2 < k^2 < (n^2 \plus{} \frac {n}{2} \plus{} 1)^2$

For $ n$ odd, $ (n^2 \plus{} \frac {n \minus{} 1}{2})^2 < k^2 < (n^2 \plus{} \frac {n \plus{} 1}{2})^2$. Done.[/quote]

I couldn't understand the proof at all! And how does it prove our case. Can anyone explain?[/quote]

It shows that value of $ \frac {n^5 \minus{} 1}{n \minus{} 1}$ is between two consecutive squares no matter $ n$ is even or odd , so is impossible to be a square. Sorry for the poor presentation.

The more interesting question is, how did you come up with those inequalities? What thought process led you to them?

Experience. A commonly used technique to show that a number isn't a perfect square is to bound it between two consecutive perfect squares.


Can anyone find four distinct integers $ a, b, c, d$ such that:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 45^{50}$

Observing this in $ \mod4$, we find out that 3 of the squares have to be multiples of 4 and the remaining one has to be 1 more than a multiple of 4. I wouldn't know how to proceed from here, just that you have

$ 4w\plus{}4x\plus{}4y\plus{}(4z\plus{}1) \equal{} 45^{50}$ ...

Use [url=http://mathworld.wolfram.com/EulerFour-SquareIdentity.html]this[/url] over and over again.  Alternately, give $ a, b, c, d$ a large common factor to reduce the problem as much as possible.  (Actually, that second idea is much better.  Are you aware that every positive integer can be written as a sum of four squares?)

What do you mean give them "a large common factor?" I cannot assign them numbers because $ 45^{50}$ is such a large number (obviously I can't use a calculator). How would you proceed with this?

[hide="What t0r meant, and solution with that method"]Let $ a \equal{} 45^{24} a_1$, etc. to reduce the problem to
$ a_1^2 \plus{} b_1^2 \plus{} c_1^2 \plus{} d_1^2 \equal{} 45^2$.

The hardest part from here is not reducing it so much that we can't get distinct integers anymore. Here's a way that does work: $ a_1 \equal{} 5 a_2$, etc. giving
$ a_2^2 \plus{} b_2^2 \plus{} c_2^2 \plus{} d_2^2 \equal{} 81$
Now $ (a_2,b_2,c_2,d_2) \equal{} (0,1,4,8)$ works, so the original solution would be $ (a,b,c,d) \equal{} (0, 45^{24} \cdot 5, 45^{24} \cdot 20, 45^{24} \cdot 40)$.[/hide]

I know I'm being dumb right now, but I still don't understand. Any possible chance you can explain it in simpler terms?

[quote="MellowMelon"] so the original solution would be $ (a,b,c,d) \equal{} (0, 5\cdot 45^{24} , 20\cdot 45^{24} , 40 \cdot45^{24} )$.[/quote]

This style (in the last set of parentheses) can read as more natural
(as a suggestion.)


geometry
geometric transformation
reflection
circumcircle

[quote="the future"]Prove that a point H inside triangle ABC is the orthocenter if and only if angle(A) + angle(BHC) =180 degrees and angle(B)+angle(AHC)=180 degrees[/quote]
[hide="Solution"]
Reflect $ H$ across $ AC$ and $ BC$ to form $ X$ and $ Y$ respectively.  First we will show that if $ H$ is the orthocenter, then the condition holds.

Let the foot of the perpindicular from $ C$ to $ AB$ be $ F$, that of $ B$ to $ AC$ be $ E$, and let $ HX$ meet $ CB$ at $ D$.  Now, we see that $ A$, $ H$, and $ D$ are collinear, and as are $ E$, $ H$, and $ B$ and $ F$, $ H$, and $ C$.  Now because $ X$ is $ H$ reflected over  $ BC$, it follows that $ \triangle BDH\cong \triangle BDX$, which implies that $ \angle CBE \equal{} \angle HBD \equal{} \angle DBX \equal{} \angle CBX$.  Since $ \angle ADB \equal{} 90 \equal{} \angle AEB$, it follows that $ AEDB$ is cyclic, so $ \angle XBC \equal{} \angle EBC \equal{} \angle DAC \equal{} \angle XAC$.  Hence, $ X$ lies on the circumcircle of $ \triangle ABC$ as does $ Y$.  This means that $ \angle BAC \plus{} \angle BXC \equal{} 180\implies \angle BHC \plus{} \angle BAC \equal{} 180$ because of the reflection.  Similarly, $ \angle AHC \plus{} \angle ABC \equal{} 180$.

Now, if the condition holds, then we see that $ \angle BXC \plus{} \angle BAC \equal{} 180$ and $ \angle ABC \plus{} \angle AYC \equal{} 180$, so $ X$ and $ Y$ lie on the circumcircle of $ \triangle ABC$.  Now, since $ H$ is the reflection of $ X$ over $ BC$ and $ X$ lies on the minor arc $ BC$, it follows that $ H$ must lie on the minor arc $ BC$ reflected over $ BC$.  Similarly, $ H$ lies on the minor arc $ AC$ reflected over $ AC$, and thus $ H$ is the intersection between these two arcs.  Yet, these two arcs already intersect at a point, namely $ A$.  Hence, they can only intersect at one more point.  Before, we have shown that the orthocenter satisfies the same conditions as $ H$, which means that the second point which the two arcs intersect at is the orthocenter, so $ H$ is the orthocenter.  [/hide]


Evaluate dy/dt for the function at the point.
  xy + x = 12 ; dx/dt = -3 , x = 2 , y = 5

$ xy \plus{} x \equal{} 12$
differentiate by $ x$,,we find:
$ y \plus{} x\frac {dy}{dx} \plus{} 1 \equal{} 0$
$ \frac {dy}{dx} \equal{} \minus{} \frac {(1 \plus{} y)}{x}$

$ \frac {dx}{dt} \equal{} \minus{} 3 \rightarrow dx \equal{} \minus{} 3dt$

$ \frac {dy}{ \minus{} 3dt} \equal{} \minus{} \frac {(1 \plus{} y)}{x} \rightarrow \frac {dy}{dt} \equal{} \frac {3 \plus{} 3y}{x}$

insert the value of the point we get::
$ \frac {dy}{dx} \equal{} 9$

sorry if I'm wrong

a simpler way would be differentiating with respect to $ t$ directly in the first step :)


number theory
relatively prime
number theory open

1. Let a and b be relatively prime positive integers with a>b then jacobi's symbol (a/b) can be evaluated using n(([log_2]b)) bit operations.

2.Let [S(j)]^n denote the position of wire in nth section spliced to jth wire of the first section then [S(j)]^n = 1+(j-1).S^(n-1) (mod m) (show)

Dear friend, where are you gonna use these question...(i only wondered :) )

Solution2. Sol: i) For n=2 by the rules of the splicing system and [S(j)]^2 = 1+(j-1).S (mod m) so it's true
...    ii) Assume [S(j)]^n = 1+(j-1).S^(n-1) (mod m)
...   iii) Since we have that the wire in position [S(j)]^n spliced to the wire in position
.... ... [S(j)]^(n+1) = 1+([S(j)]^n -1)S (mod m)
...                          = 1+(j-1)S^n       (mod m)

....

Solution1: Using(a/b)= (-1)^[s_1(R_1)^2-1)/8+...+s_(n-1).(R_(n-1))^2-1)/8+(R_1 - 1)/2+...+(R_(n-1) - 1)/2], where  R_j,s_j are integers for a>b.
Perform a sequence: n(x) divisions such that x=log of b to the base 2. The number of divisions does not exceed the number of divisions needed to find (a,b) using the euclidean algorithm. Thus by Lema's Theorem, use n(x) divisions are needed. Each division can be done using n(x^2) bit operations. So R_j and s_j can be found using n(x) bit operations.
n(x^3) bit operations are required to find the integers R_j,s_j from a and b. The exponent of -1 in the expression for (a/b) by the first theorem will be evaluated using the last three bits in the binarry expznsions of R_j and last bit in the binary expansions of s_j.
Therefore using n(x) additional bit operation to find (a/b). Since
n(x^3)+n(x)=n(x^2)


The bases of the trapezoid are $ 6$ and $ 19$ , and its diagonals are $ 15$ and $ 20$ . Calculate the perimeter of the trapezoid.

I keep finding $ 57$ as an answer but the answer key shows $ 52.98$ . Can anybody please help me with this problem?

[hide="Cheap way"]
Let $ ABCD$ be the trapezoid with $ AB = 6,CD = 19,BD = 15,$ and $ AC = 20$. Drop the perpendiculars to $ AX$ and $ BY$. Let $ AX = BY = h$, $ BC = y$, $ AD = z$, and $ CY = x$. We have the system of 4 equations:
\begin{align} x^2 + h^2 & = y^2 \\
(13 - x)^2 + h^2 & = z^2 \\
(19 - x)^2 + h^2 & = 225 \\
(6 + x)^2 + h^2 & = 400 \end{align}

We can solve for $ x$ and $ h$ subtracting $ (3)$ from $ (4)$ to get $ 2x- 13= 7\implies 2x= 20\implies x = 10$. Here we can see by the 3-4-5 right triangles that $ h = 12$, so $ y = \sqrt {10^2 + 12^2} = \sqrt {244} = 2\sqrt {122}$ and $ z = \sqrt {3^2 + 12^2} = \sqrt {153} = 3\sqrt {17}$.

So the perimeter is $ 25 + 2\sqrt {122} + 3\sqrt {17}$, which is about $ 52.98981622866629043772367503944 \approx 52.99$.[/hide]

First draw the two heights from the top vertices to the lower base.
The middle portion is 6 and the two others we can call $ a$ and $ 13 \minus{} a$.
We can call the height $ h$.

So...

$ (6 \plus{} a)^2 \plus{} h^2 \equal{} 15^2$
               and
$ (19 \minus{} a)^2 \plus{} h^2 \equal{} 20^2$

Subtract the two equations when only variables are on one side and numbers on the other and you get...

$ 50a \equal{} 150$ therefore $ a \equal{} 3$

Plug $ 3$ into one of the equations and you get...

$ 81 \plus{} h^2 \equal{} 225$
$ h^2 \equal{} 144$ therefore $ h \equal{} 12$

$ (13 \minus{} 3)^2 \plus{} 12^2 \equal{} 244$
             and
$ 3^2 \plus{} 12^2 \equal{} 153$

The side lengths are $ 3\sqrt{17}$, $ 2\sqrt{61}$, 6, and 19.

Perimeter=52.9898

Let the trapezoid be ABCD, with AB=6, CD=19, and X the intersection of the diagonals. AXB and CXD are similar, so $ AX\equal{}\frac{6}{6\plus{}19} \cdot 20\equal{}\frac{24}{5}$, $ BX\equal{}\frac{6}{6\plus{}19} \cdot 15\equal{} \frac{18}{5}$, similarly $ CX\equal{}\frac{76}{5}$ and $ DX\equal{}\frac{57}{5}$. But we find by Pythag that AXB is right. So all four triangles with X as a vertex are right, we can use Pythag to find $ AD\equal{}\sqrt{153}$, $ BC\equal{}\sqrt{244}$, so the answer is 52.98.

can you show the proof of your 52.99. any one also me myself have the answer of 52.9898 exactly by the way which geotrig goes from.

Uh, 52.9898 rounds to 52.99
We want an answer rounded to the hundredths place.


recurrence relation
polynomial
algebra
roots
IMO
chebyshev polynomial
imo 1976

Let $P_{1}(x)=x^{2}-2$ and $P_{j}(x)=P_{1}(P_{j-1}(x))$ for j$=2,\ldots$ Prove that for any positive integer n the roots of the equation $P_{n}(x)=x$ are all real and distinct.

[u]Solution.[/u] Clearly, $ P_n(x)>P_{n\minus{}1}(x)\ \forall n>1, x\in\mathbb{R}$.  If $ x\notin [\minus{}2,2]$, $ P_n(x)>x$.  Therefore, all real solutons of $ P_n(x)\equal{}x$ are in $ [\minus{}2,2]$.  Notice that $ P_n(x)\equal{}x$ has at most $ 2^n$ real roots.  Let $ x\equal{}2\cos t$,  Then it follows that $ P_n(x)\equal{}2\cos 2^nt$.  It follows that $ P_n(x)\equal{}x\iff \cos 2^nt\equal{}\cos t\iff 2^nt\equal{}2\pi k\plus{}t\iff t\equal{}\frac{2\pi k}{2^n\minus{}1},$ where $ k\equal{}0,1,\hdots, 2^n\minus{}1$.  And the conclusion follows. $ \Box$

[u]Motivation.[/u] Consider the Chebyshev function $ T_n(x)$ and $ 2T_n\left(\frac{x}{2}\right)$.

My solution:
We have that a square plus 2=x. Subtracting 2 and square rooting we get another expression $=\pm \sqrt{2+x}$ If we keep on doing this we will get 2^n solutions. Now substitute x on the LHS for in the fractions an infinite number of times. To prove that all of these are real, we need to show that 2-the expression>0, but the expression is trivially <2

Can anyone write the solution in more elaborative way the above solution is pretty good but I am unable to track down the whole thing.
Would be helpful for me.
Thanks in advance.

$\textbf{\textcolor{red}{Claim:-}}$ all real roots of $P_{n}(x)=x$ lies in $[-2,2]$\\

$\textbf{\textcolor{blue}{Proof:-}}$ clearly through induction we can show that $p_{n}(x)>x$ $\forall$ $x>2$  and for $x<-2$ we have $P_{n}(x)>2$ $\forall$ $n$ hence we have al roots of $P_{n}(x)=x$ lies in [$-2,2]$ as claim follows $\square$\\

now we substitute $x=2\cos{\theta}$ , we have $P_{1}(x)=2\cos{2\theta}$ from again induction we have $P_{n}(x)=2\cos{2^{n}\theta}$\\

so we need to solve $\cos{2^{n}\theta}=\cos{\theta}\implies 2^{n}\theta= 2\pi m \pm \theta \implies \theta=\frac{2\pi m}{2^{n}-1}$ and $\theta=\frac{2\pi m}{2^{n}+1}$ where $m \in \mathbb{Z^{*}}$ hence we have all $2^{n}$ roots of $P_{n}(x)$  real and distinct $\blacksquare$


geometry
3D geometry
tetrahedron
parallelogram

Show that if the four faces of a tetrahedron have the same area, then all four faces are congruent.

Any takers?  :( I really want a solution or hint for this one... or at least something saying it isn't true.

[hide]Let $ AB\equal{}c, AC\equal{}b, BC\equal{}a, AD\equal{}ax, BD\equal{}by, CD\equal{}cz$ Then you can bash through herons for the four triangles to find four solutions $ (x,y,z)\equal{}(1,1,1)$ and permutations of $ (\frac{\sqrt{2b^{2}\plus{}2c^{2}\minus{}a^{2}}}{a},1,1)$. In the latter case, $ AD\equal{}\sqrt{2b^{2}\plus{}2c^{2}\minus{}a^{2}}$.

It's clear that if we fix the lengths of all sides except $ AD$, $ AD$ would reach its maximum when $ ABDC$ is a degenerate tetrahedron (parallelogram) and that maximum would be $ AD\equal{}\sqrt{2b^{2}\plus{}2c^{2}\minus{}a^{2}}$. Since the tetrahedron isn't degenerate, $ AD<\sqrt{2b^{2}\plus{}2c^{2}\minus{}a^{2}}$ and we must have $ (x,y,z)\equal{}(1,1,1)$.  :D [/hide]


ratio
logarithms
geometric sequence
geometric series

x^x^x^x^x... = 1/2 solve for x can you try to solve it with log?

It's just an infinite geometric sequence. Since the first term is the same as the common ratio:

$\frac {r}{1-r} = \frac 12$.

Solve and you get $\frac 13$.

I don't know how you applied geometric sequence.

My solution.
[hide]

$\displaystyle x^{x^{x^{...}}}=\frac12 \Rightarrow x^\frac{1}{2}=\frac12 \Rightarrow \sqrt{x}=\frac12 \Rightarrow x=\frac14.$[/hide]

You can apply geometric sequence because it is an infinate sequence, therefore it's okay to use that formula.

yea i solved it like hiitsme and it is 1/4 but i want to know how to solve it using logs

I agree with tarquin since it is an infinite geometric sequence you can use the formula $\frac {r}{1-r}$ = $\frac {1}{2}$ it comes out r=$\frac {1}{3}$

How is it an infinite geometric series?
$1+\frac12+\frac14+\frac18+\cdots$ is an infinite geometric series.
The problem given has only one term, which involves exponentiating infinitely many times. The problem does not ask for a sum of a geometric sequence. Please make sure you have your facts straight.

I think he made some kind of mistake and somehow added the powers.


Prove that: $ \sqrt{1\plus{}3\plus{}5\plus{}...\plus{}2009} \in \mathbb{Q}$.

[hide="short and sweet"]The sum of consecutive odd numbers from 1 to n is n^2, which can be proven by induction.

So $ \sqrt{1\plus{}3\plus{}5\plus{}...\plus{}2009}\equal{}\sqrt{2009^2}\equal{}2009$,

which is obviously rational.[/hide]

Almost correct, you have the idea, but

$ 1\plus{}3\plus{}\cdots \plus{}2009\not\equal{} 2009^2$

The formula is $ 1\plus{}3\plus{}\cdots \plus{}(2n\minus{}1)\equal{}n^2$, therefore $ 1\plus{}3\plus{}\cdots\plus{}2009\equal{}1005^2\implies \sqrt{1\plus{}3\plus{}5\plus{}...\plus{}2009}\equal{}1005$


geometry
3D geometry
sphere
function
analytic geometry

With the square root function, that's to find the radius, and then cincodemayo5590 just used $\frac{4}{3}\pi^{3}$ to get $\frac{32\pi}{3}$ for the volume. You can't really explain it further...

OH I SEE

So basically you just subtract and add the three coordinates to find the radius. Then, you use the formula for the volume of a sphere. Gotcha  :lol:

[quote="Go Around the Tree"]The center of a sphere is located at the point (-1,3,2). A point on the surface of the sphere is located at (-1+rt2, 2, 3). Find the volume of the sphere.[/quote]

[hide]

I use the distance formula $\sqrt{(-1+\sqrt{2}+1)^{2}+(2-3)^{2}+(3-2)^{2}}$.

This equals 2, which is the radius of the sphere.

Plug it in to the equation, $\frac{4}{3}\cdot \pi \cdot 2^{3}$.

[/hide]

I think the three-coordinate thing brought me off track when I took the test . I see how simple it really is now.


Summation
number theory
binomial coefficients
Divisibility
modular arithmetic
IMO
IMO 1974

Prove that for any n natural, the number \[ \sum \limits_{k=0}^{n} \binom{2n+1}{2k+1} 2^{3k}  \]
cannot be divided by $5$.

Everything that follows takes place in $\mathbb F_5(\sqrt 2)$, i.e. the field we get by adjoining a root of $x^2-2=0$ to $\mathbb F_5$, the field with $5$ elements.

We have $\sum_{k=0}^n\binom{2n+1}{2k+1}2^{3k}=\sum_{k=0}^n\binom{2n+1}{2n-2k}3^k=\sum_{k=0}^n\binom{2n+1}{2(n-k)}2^{-k}$. Now, this is zero iff it's zero when we multiply it by $2^n$, so we may as well prove that $\sum_{k=0}^n\binom{2n+1}{2(n-k)}\sqrt 2^{2(n-k)}\ne 0$. The LHS is $\alpha$ from $(1+\sqrt 2)^{2n+1}=\alpha+\beta\sqrt 2,\ \alpha,\beta\in\mathbb F_5$. We have $(1-\sqrt 2)^{2n+1}=\alpha-\beta\sqrt 2$, so by multiplying them we get $-1=\alpha^2-2\beta^2$. If we were to have $\alpha=0$, then we would get $1=2\beta^2,\ \beta\in\mathbb F_5$, and this is impossible, since it would make $3=2^{-1}$ a square $\beta^2$ in $\mathbb F_5$ (i.e. $3$ would be a quadratic residue modulo $5$, and it's not).

how do you make the pass from $2^{3k}$ to $3^k$ and then to $2^{-k}$?

He works $\mod 5$ I think... (or better to say: in it's quadratic extension $\mathbb{F}_{25}$)

he says he works in $\mathbb F_5 (\sqrt{2})$. how is that the same thing?

$\mathbb{F}_5$ is just a synonym for the field of integers $\mod 5$.
There is no solution $a^2 \equiv 2 \mod 5$, so adjoin $\sqrt 2$,  which means just consider all numbers of the form $a+b\sqrt 2 \mod 5$ with $a,b$ residue classes $\mod 5$ (all together written as $\mathbb{F}_5[\sqrt 2]$).
There are $25$ of them and they build up a field, also called $\mathbb{F}_{25}$.

Thanks. I finally understood the solution.

i have a simpler (though a little longer) solution:
re-write your sum:
\[ a_n \equal{} \sum_{k \equal{} 0}^{n} \binom{2n \plus{} 1}{2k \plus{} 1} 2^{3k} \equal{} \frac {(1 \plus{} 2\sqrt2)^{2n \plus{} 1} \plus{} (1 \minus{} 2\sqrt2)^{2n \plus{} 1}}{2}
\]
now, what i've wrote is a sequence, generated by recurrence: $ a_{n \plus{} 1} \equal{} 18a_n \minus{} 49a_{n \minus{} 1}$, whose first terms are $ a_0 \equal{} 1, a_1 \equal{} 11$.
it's an easy check to show that no terms can be $ 0 \mod 5$, since sequence has a short period...

[quote="ma_go"]
now, what i've wrote is a sequence, generated by recurrence: $ a_{n \plus{} 1} \equal{} 18a_n \plus{} 49a_{n \minus{} 1}$
[/quote]

That should be $ a_{n\plus{}1}\equal{}18a_n\minus{}49a_{n\minus{}1}$ and it will work indeed :P
Pretty nice solution anyways.


calculus
integration
algebra
polynomial
function
real analysis
real analysis unsolved

[quote="borislav_mirchev"]Find:
$ \int\frac{x^{4}\plus{}4x^{2}\minus{}1}{2x^{6}\minus{}3x^{4}\plus{}4x^{2}\plus{}1}dx$[/quote]
Mathcad can  give me the answer in terms of $ \arctan(\sqrt{2}x),\,\arctan(x^{3}/\sqrt{2}\minus{}3\sqrt{2}x/4)$, and another messy $ \arctan$ term.

What's the point?

The idea here is not to use MatCad. If it is useful for you - it is the source of the problem. (If you don't speak Russian I may translate.)

http://www.umsolver.com/cgi-bin/concur.pl?l=rus&s=math&n=2

It is the answer:
(-1/2)*sqrt(2)*arctg((-5/4)*sqrt(2)*x^3+(1/2)*sqrt(2)*x^5+(5/4)*sqrt(2)*x)+(1/2)*sqrt(2)*arctg((-3/4)*sqrt(2)*x+(1/2)*sqrt(2)*x^3)+(1/2)*sqrt(2)*arctg(sqrt(2)*x)

[quote="borislav_mirchev"]The idea here is not to use MathCad. [/quote]Hmm ... that doesn't sound like an [b]idea[/b] to me.    :huh:

They said on this site - there should be "short" solution whatever that means. I think there should be some tricks with the polynomials or some good substitution. What do you think? Any ideas for solution? In some problems when you know the answer it is a good starting point to find the solution. May integration by parts be used?

Is this integral so difficult, or it is boring?

[quote="borislav_mirchev"]If you don't speak Russian I may translate
 [/quote]
What about you,borislav mirchev,do you speak russian?(I know russian,so if you know it,it will be much easier for me to explain you solution of V.1996).

Yes, I speak russian. I'm from Bulgaria. There weren't enough English teachers so I studied 8 years Russian in school vs. 3 years English. I'm not good in grammar but I may without problems watch russian TV - Russian is similar to Bulgarian.

$ \left(arctg\left(\frac {P}{Q}\right)\right)^{'} \equal{} \frac {1}{1 \plus{} (\frac {P}{Q})^{2}}.\left(\frac {P}{Q}\right)^{'} \equal{} \frac {1}{1 \plus{} (\frac {P}{Q})^{2}}.\frac {P^{'}.Q \minus{} PQ^{'}}{Q^{2}} \equal{} \frac {P^{'}Q \minus{} P.Q^{'}}{P^{2} \plus{} Q^{2}}$

We'll try to find $ P, Q,$ such that: $ (*) P^{2} \plus{} Q^{2} \equal{} 2x^{6} \minus{} 3x^{4} \plus{} 4x^{2} \plus{} 1$.
We'll try to find $ P$ in the form: $ \sqrt {2}(x^{3} \plus{} ax);$ $ Q \equal{} bx^{2} \plus{} c$ where $ a, b, c$ are not defined coefficients.
From $ (*)$ follows that the following equations are right: $ 4a \plus{} b^{2} \equal{} \minus{} 3, 2a \plus{} 2bc \equal{} 4, c^{2} \equal{} 1$.
After solving them as a system we choose one good solution: $ a \equal{} \minus{} 1, b \equal{} 1, c \equal{} 1$.
So our functions are: $ P \equal{} \sqrt {2}(x^{3} \minus{} x), Q \equal{} x^{2} \plus{} 1$.
$ P^{'}Q \minus{} P.Q^{'} \equal{} \sqrt {2}(x^4 \plus{} 4x^2 \plus{} 1)$.
It means that $ arctg\left(\frac {P}{Q}\right)^{'} \equal{} \frac {1}{\sqrt {2}}.\frac {P^{'}Q \minus{} P.Q^{'}}{P^{2} \plus{} Q^{2}} \equal{} \frac {x^{4} \plus{} 4x^{2} \minus{} 1}{2x^{6} \minus{} 3x^{4} \plus{} 4x^{2} \plus{} 1}$
So the answer is: $ \int \frac {x^{4} \plus{} 4x^{2} \minus{} 1}{2x^{6} \minus{} 3x^{4} \plus{} 4x^{2} \plus{} 1}\,dx \equal{} \frac {1}{\sqrt {2}}arctg\left(\frac {\sqrt {2}(x^{3} \minus{} x)}{x^{2} \plus{} 1}\right) \plus{} C$.

It is the solution that is assumed the competitor should gave.

We can explain why the asked antiderative is in the form "k.arctan(?)":
$ 2x^3-3x^2+4x+1$ has a real negative root and 2 conjugate complex roots, then $ 2x^6-3x^4+4x^2+1=2(x^2+u^2)(x^2-vx+w)(x^2+vx+w)$ where $ u,v,w$ are real and $ v^2-4w<0$.
$ P=\dfrac{Q}{R}=\dfrac{x^4+4x^2-1}{2x^6-3x^4+4x^2+1}$ is an even function, therefore $ P=\dfrac{a}{x^2+u^2}+\dfrac{bx+c}{x^2-vx+w}+\dfrac{-bx+c}{x^2+vx+w}$ where $ a,b,c$ are real.
The antiderative doesn't contain any "log" iff $ b=0$.
$ \dfrac{bx+c}{x^2-vx+w}=\dfrac{\alpha}{x-\rho}+\dfrac{\bar{\alpha}}{x-\bar{\rho}}$ and the condition is ${ \alpha+\bar{\alpha}}=0$ i.e. $ \alpha^2<0$. $ \alpha=\dfrac{Q(\rho)}{R'(\rho)}$ where $ R(\rho)=0$. With these 2 relations Maple  finds (*): $ \alpha^2+1/8=0$ (You can calculate with the hand). Then $ b=0$ and an $ P$-antiderative is in the form: $ \dfrac{a}{u}arctan(?)+\dfrac{2c}{\sqrt{4w-v^2}}(arctan(?)+arctan(?))$. A sum or a difference of "arctan" is an "arctan" then it's sufficient to satisfy this condition: $ \dfrac{a^2}{u^2}=\dfrac{4c^2}{4w-v^2}$.
LHS: $ \dfrac{a}{x^2+u^2}=\dfrac{ia/(2u)}{x+iu}-\dfrac{ia/(2u)}{x-iu}$; (*) implies that $ \dfrac{-a^2}{4u^2}+1/8=0$ or $ \dfrac{a^2}{u^2}=1/2$.
RHS: $ c=-\alpha\bar{\rho}-\bar{\alpha}\rho= (with (*)) \dfrac{i}{2\sqrt{2}}(\rho-\bar{\rho})=\dfrac{i}{2\sqrt{2}}i\sqrt{4w-v^2}$ and the RHS is also $ 1/2$.     QED
Conclusion: the key is the (*) relation.

Very good idea.
I asked the moderators of the site where this integral is from for a solution. I wanted to know the solution because this integral had looked very good. I think that if some man have very good experience with integrals and differentials he will guess the solution. Integrals are very interesting subject because you know answer of your problem and then to start with finding good arguments for solutions. I believe that there may be another approach for solving this integral. You may try to solve integrals 2. and 3. from the following link and to find good arguments for them.
http://www.mathlinks.ro/Forum/viewtopic.php?t=168846


Well I will be applying to graduate school in a few years.  Since I would like to become a quant I am wondering if it would be better for me to apply to masters in mathematical finance programs or an applied math PhD program.  Also since I don't currently go to a school which has a prestigious math department I was wondering what kinds of advice you guys could give me about getting into a good school.

If you want to be a quant, why not just start working right away? I don't have the data available to do a cost-benefit analysis, but I'm willing to bet that the gain of starting to work 4-5 years earlier significantly outweighs the gain from the higher starting salary that PhDs would get. (You can still start at a pretty good salary with just a bachelor's.)

Also, why are you deciding so early? If you're doing math as your major, all three options (your two, plus the one I just suggested) should remain feasible all the way until the time to do the applications actually comes up. (I mean, right now its the spring of my junior year, and I still could do any of those three.)

[quote="maokid7"]I did consider trying to start work right after completing my undergraduate degree.  However, since I'm not attending a "big name" school I'm not too sure how successful my attempt would be.[/quote]
It can't hurt to try. If you don't get any decent internships for the summer after junior year then start applying to grad school. If you do get an internship junior summer then you'll probably be able to work for the same place after you graduate.


Two ladders, both 6 meters in length, are leaned up against
opposite vertical walls in a 3-meter-wide corridor, as shown. How
far above the ground do the two ladders cross? Express your
answer as a decimal to the nearest tenth.    http://i43.tinypic.com/24320k6.jpg

[hide="Click for solution"]
[img]http://i43.tinypic.com/24320k6.jpg[/img]
The ? is just the altitude of a equilateral triangle with side length $ 3$, or $ \boxed{\frac{3\sqrt{3}}{2}}$.
[/hide]

[quote="motiroti"]Two ladders, both 6 meters in length, are leaned up against
opposite vertical walls in a 3-meter-wide corridor, as shown. How
far above the ground do the two ladders cross?  [u][i][b]Express your
answer as a decimal to the nearest tenth.[/b] [/i][/u]  http://i43.tinypic.com/24320k6.jpg[/quote]

Read the question  :P

$ 2.6$.


this is one of tiplers problems i don't know why whatever i do doesn't give the right answer...

A Helium balloon is used to lift a load of $110$  $N$. The weight of the balloon's skin is $50$  $N$ and the volume of the balloon when fully inflated is $32$ $m^{3}$ . The temperature of the air is $0^{ o}C$ and the atmospheric pressure is $1$  $atm$.The balloon is inflated with sufficient Helium gas so that the net upward force on the balloon and its load is $30$  $N$. Neglect changes of temperature with altitude.

[b]Find the number of moles of Helium gas in the balloon.[/b]

and find the max altitude that the balloon can go.


Find the circumference of a circle inscribed in an isosceles trapezoid with bases 8 and 18 inches long.

A.12$ \pi$
B.7$ \pi\sqrt{2}$
C.10$ \pi$
D.13$ \pi$

[quote="Lulze"]Hmm, correct me if I'm mistaken, but it doesn't seem like that's enough information to solve the problem, because even if we know the base lengths, we don't have any way to figure out the lengths of the other sides or the altitudes, so this trapezoid could be of any size.[/quote]But a circle cannot be inscribed in ALL trapezoids...

[hide]Draw a diagram. You will see that you can use the two tangent theorem on all four vertices of the trapezoid, which gives you that legs measure $ \frac{8}{2}\plus{}\frac{18}{2}\equal{}4\plus{}9\equal{}13$. Using the Pythagorean Theorem, we get that the height of the trapezoid is $ \sqrt{13^2\minus{}5^2}\equal{}12$. Since the height of the trapezoid is also a diameter of the circle, the circumference is $ 12\pi \Rightarrow \boxed{A}$.[/hide]


Given a positive integer n, let p(n) be the product of the non-zero digits of n. (if n has only one digit, then p(n) is equal to that digit.) Let

S = p(1) + p(2) + p(3) + ... + p(999).

What is the largest prime factor of S?