geometry
3D geometry
sphere
analytic geometry
projective geometry
advanced fields
advanced fields unsolved

I cannot understand the sentence below:

"$ P^1$ can be thought of as $ C$(complex plane) together with a single point $ \infty$ (i.e. a copy of $ P^0$) and

thus can be identified with the Riemann sphere $ C \cup \lbrace \infty \rbrace$", where  $ P^n$ is complex projective space.

Who can explain it to me?

3ks

Which part don't you understand?  The Riemann sphere, or the identification of $ CP^1$ with the Riemann sphere?

The part i cannot understand is that why $ P^1$ can be thought of as $ C$ with $ \infty$ and the identification of  $ CP^1$ with the Riemann sphere.

For your first question, $ \mathbb{C}P^1$ can be split up into two parts:  if a point $ (x : y)$ has the property that $ x \neq 0$, it can be identified with $ \left( 1 : \frac {y}{x} \right)$.  Otherwise, it can be identified with the point $ \left( 0 : 1 \right)$.  The former is a copy of $ \mathbb{C}$; the latter is called the point at infinity.

The second question is a matter of [url=http://en.wikipedia.org/wiki/Stereographic_projection]stereographic projection[/url].  Imagine you place the south pole of the Riemann sphere directly on top of the origin in $ \mathbb{C}$.  Now, start from the north pole and draw rays that intersect both the sphere and $ \mathbb{C}$.  For every point on the Riemann sphere (except the north pole itself), there is a unique line through that point and through the north pole, and that line intersects $ \mathbb{C}$ in a unique point, and vice versa.  Thus we can identify $ \mathbb{C}$ with the Riemann sphere minus one point.

What is that point?  $ \infty$ (specifically, complex infinity)!  If we could project from the "edge" of the complex plane, the resulting line would be tangent to the north pole, and that's the missing point.

Thank you t0rajir0u! I think I can understand your explanation.
But a little problem still exists: for the explanation of the 1st Q, if $ x\equal{}0$, why can $ \mathbb{C}P^1$ be identified with
$ \left( 1,0 \right)$? :oops:

If a point can't be put into the form $ (1 : t)$, it has zero $ x$-coordinate and is of the form $ (0 : t)$.  Since $ (0 : 0)$ is not a valid point in projective space, $ t \neq 0$, so we can divide by $ t$ and identify any such point with $ (0 : 1)$.  In general,

$ P^n \simeq A^n \cup P^{n\minus{}1} \simeq A^n \cup A^{n\minus{}1} \cup ...$

where $ A^n$ is affine $ n$-dimensional space (over some field).


Place 2005 points on the circumference of a circle. Two points $P,Q$ are said to form a pair of neighbours if the chord $PQ$ subtends an angle of at most 10 degrees at the centre. Find the smallest number of pairs of neighbours.

The answer is $25 \cdot \binom{57}{2} + 10 \cdot \binom{58}{2}.$ This works by taking $35$ evenly-spaced "clumps" of $57, 57, 57, \cdots, 57, 58, 58, \cdots, 58$ points respectively, with $25$ of $57$ points and $10$ of $58$ points.

Let's show that this is optimal. Consider the graph where we connect two points if they are neighbors. Now, consider the complement of said graph, say $G.$ Clearly, $G$ has no $K_{36}$. Therefore, Turan implies that $G$ has the maximal amount of edges in our construction, so we're done.

$\square$


What is the smallest prime factor of 1821?

Start with 2 and try the numbers. 2 doesn't work but three does.


CANMOO!!!!! :blush:

Hey, who has Canmoo right now? Are they going to bring him to Seminar, or are they just going to bring him to IMO only?  :(

I do. I am not coming to Seminar though, but Canmoo will go to IMO for sure :D

Alex, how about we meet sometime next week and you hand the canmoo over to me so that I can bring it to the Seminar. (or if you don't trust me with canmoo, give to Jon)

I received the invitation to Seminar from my teacher only today, so hopefully I will still be allowed to come. In that case, I will come to seminar, and Canmoo too. :D

[quote="soledad42"]Guys we should bring Canmoo to Winter Camp (Both if possible). Who has them now?[/quote]

I have one.

errr....in that case, should I somehow alter my last post?

Wow, Seminar invites are out already?  Back in my day, they didn't come out till at the EARLIEST the year of!

no, the invites are no way out yet. The posts above are for last year...

[quote="sunchips"][quote="Mathgod"][quote="carpo"]SHHH

THERE IS, HAS BEEN, AND EVER WILL BE ONLY ONE CANMOO[/quote]

Ok, fine, I have the CANMOO. It's sitting on my bed.[/quote]

Yoh Danny, are you bringing him to Winter?

And LMAO you would sleep with the moose  :P[/quote]

I will bring CANMOO to the winter camp.


algebra
polynomial
modular arithmetic
number theory proposed
number theory

Let $ m,n \in N$.
Find all $ (m,n)$ that is corresponding to $ n^5\plus{}n^4\plus{}1\equal{}7^m$
Hint: $ n^5\plus{}n^4\plus{}1\equal{}(n^2\plus{}n\plus{}1)(n^3\minus{}n\plus{}1)$

"Find all solution in $ \mathbb{N}$ of $ n^5\plus{}n^4\plus{}1\equal{}p^k$ where $ p$ is a prime"

Let $ \omega$ be a third complex root of unit, so $ \omega^5\plus{}\omega^4\plus{}1\equal{}\sum_{i\equal{}0}^2{\omega^i}\equal{}0 \implies n^2\plus{}n\plus{}1 | n^5\plus{}n^4\plus{}1$.
By polinomial division we get the factorization $ n^5\plus{}n^4\plus{}1\equal{}(n^2\plus{}n\plus{}1)(n^3\minus{}n\plus{}1)$, both factors are $ >1$ if $ n>1$.
$ gcd(n^2\plus{}n\plus{}1,n^3\minus{}n\plus{}1)\equal{}(n\minus{}2,7)\equal{}7$ iff $ n \equiv 2 \pmod 7$, so it exists $ N \in \mathbb{N}$ such that $ n\equal{}7N\plus{}2$.
Let $ (a,b) \in \mathbb{N}^2$ s.t. $ n^2\plus{}n\plus{}1\equal{}7^a \text{ and } n^3\minus{}n\plus{}1\equal{}7^b$ then we get $ 7N^2\plus{}5N\plus{}1\equal{}7^{a\minus{}1}$ and $ 7^2N^3\plus{}7\cdot 6 N^2 \minus{}11N \plus{}1 \equal{} 7^{b\minus{}1}$, so we can say that $ 7$ divides both $ 5N\plus{}1$ and $ 11N\minus{}1$, contradiction if $ N>1$.

So the unique solutions are $ (n,p,k)\equal{}(1,3,1) \text { and } (2,7,2)$  :lol:


I was just looking at a question from Hong Kong TST. Prove:
\[\text{If: }a+b+c+d=1\text{ then }6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}\]

My thought was let:
\[f(a,b,c,d)=a+b+c+d-1\text{ and }g(a,b,c,d)=6(a^{3}+b^{3}+c^{3}+d^{3})-a^{2}-b^{2}-c^{2}-d^{2}-\frac{1}{8}\]
\[\Delta g=\lambda \Delta f\]. Therefore, at the critical point(s) of g:
\[18a^{2}-2a=18b^{2}-2b=18c^{2}-2c=18d^{2}-2d\]
Since the function \[18x^{2}-2x\] is strictly increasing above x=1/18, then either a=b=c=d or a,b,c,d<1/9. Clearly this second case violates the condition so a=b=c=d=1/4 is the only critical point of g with the given condition. I know how to show that it is not a maximum, but how do I show that it is not a saddle point?

Here is my solution
with the Chebyshev's or Power mean we know $a^3+b^3+c^3+d^3\ge\frac{1}{4} (a^2+b^2+c^2+d^2)(a+b+c+d)=\frac{(a^2+b^2+c^2+d^2)}{4} $ and $a^3+b^3+c^3+d^3\ge \frac{1}{16}(a+b+c+d)^3=1/16$ now we done


calculus
derivative
function
geometry
LaTeX
calculus computations

The function S(w, h) = (15.63w^0.425) * (h^.0725) gives the  approximate surface area (in sq. inches) of a person h inches tall who weighs w pounds.

1) Now at his full height of 5'9", he goes to the gym and bulks up.  At what rate is his body surface increasing when he reaches 140 pounds?

Use $LaTeX$ please, it makes things a lot easier to read.

[hide="Solution"]All right, from your problem, we see that at this point in time, $h$ is a constant equal to $69$ inches (converting). This immediately gives us a direction for our directional derivative, because we want the direction in which $h$ is not changing. This infers that we want $\frac{\partial S}{\partial w}$ evaluated at $w = 140$, since $\frac{\partial S}{\partial w}$ gives us the rate of change of $S$ if $h$ is held constant.

Using power rule, we find that $\frac{\partial S}{\partial w}= 6.64275w^{-.575}h^{.0725}$

Evaluating at $(w, h) = (140, 69)$ yields $.5268 \frac{in^{2}}{lb}$.

Rather strange units, yes, but they make sense if you think about it.[/hide]


limit
function
Functional Analysis
real analysis
real analysis unsolved

Consider banach space $l^\infty$, $C$ is subset of
$l^\infty$ of all $x \in l^\infty$ and $lim_{n\to \infty}x(n)$ exists.

(1) Show $C$ is closed subspace.

(2) Define linear functional on $C$ by $\varphi(x)=lim_{n\to \infty}x(n)$. Show $\parallel \varphi\parallel =1$.
(So $\varphi$ can be extended to $\Psi$ on  $l^\infty$ and $\parallel \Psi\parallel =1$.)

(3) Show for some $x \in l^\infty$, there is no $y \in l^{1}$ such that $\Psi(x)=\sum_{n=0}^{\infty}x(n)y(n)$

(I need help on hints for (1) and (3), thanks)

For (1): use the notation developed in (2) and write $\varphi(x)=\lim_{n\to\infty}x(n).$

Now suppose you have a sequence of sequences $x_{k}$ that is a Cauchy sequence in $\ell^{\infty}.$ That means that by making $k,m$ sufficiently large, we can make $\|x_{k}-x_{m}\|_{\infty}=\sup_{n}|x_{k}(n)-x_{m}(n)|<\epsilon.$ Since, $|\varphi(x_{k})-\varphi(x_{m})|\le\|x_{k}-x_{m}\|_{\infty},$ we have that $\{\varphi(x_{k})\}$ is a Cauchy sequence that converges to some number $L.$

Now define $y(n)=\lim_{k\to\infty}x_{k}(n).$ The one thing we have left to prove is that $\lim_{n\to\infty}y(n)=L,$ hence $y\in C.$ This proof isn't too hard, but it is where we need the uniform convergence. Study the proof that the uniform limit of continuous functions is continuous for inspiration, if you need to.

The quantifiers aren't quite right in the statement of (3). If we're allowed to let $y$ depend on $x,$ then we can always engineer a $y$ to make $\langle x,y\rangle$ equal anything we want it to. The correct statement of (3) should be this:

Show that there is no $y\in\ell^{1}$ such that for every $x\in\ell^{\infty},$ $\Psi(x)=\sum_{n=0}^{\infty}x(n)y(n).$

One comment on proving this: note that the extension promised in (2) is an explicit invocation of the Hahn-Banach Theorem, and Hahn-Banach depends on the Axiom of Choice. As long as we stay in $C,$ we know that $\Psi=\varphi.$ But once $x$ leaves $C,$ we really don't know what $\Psi(x)$ does. Therefore in proving (3), we should stay with examples in $C.$

Hint: consider the action of $\varphi$ on $e_{k}=(0,0,\dots,1,0,\dots)$ for each $k.$ Then consider the action of $\varphi$ on $(1,1,1,\dots).$


Let $ n\left(A\right)$ be the number of distinct real solutions of the equation $ x^{6} \minus{}2x^{4} \plus{}x^{2} \equal{}A$. When $ A$ takes every value on real numbers, the set of values of $ n\left(A\right)$ is

$\textbf{(A)}\  \left\{0,1,2,3,4,5,6\right\} \\ \textbf{(B)}\  \left\{0,2,4,6\right\} \\ \textbf{(C)}\  \left\{0,3,4,6\right\} \\ \textbf{(D)}\  \left\{0,2,3,4,6\right\} \\ \textbf{(E)}\  \left\{0,2,3,4\right\}$

Is it asking for the roots? If it is, then it should be the following:

Factor A:

$ x^6-2x^4+x^2=x^2(x^4-2x^2+1)
\&=x^2(x^2-1)^2=0$

$ x^2=0$         $ (x^2-1)^2=0$
$ x=0$             $ x^2-1=0$
                     $ x=\pm1$

So the set $ n(A) = {0,\pm1}$.

For some reason that's not one of the choices...

no, it is not. it asks for number of solutions. $ n(A)$ is the number of distinct solutions for $ A$. The problem asks for the set of $ n(A)$ for every $ A$.
For your special case, $ A \equal{} 0$. Thus $ n(0) \equal{} 3$. The set in choices must include $ 3$. We get this result.

Draw the curve $ y \equal{} x^{6}\minus{}2x^{4}\plus{}x^{2} \equal{} x^{2}(x^{2}\minus{}1)^2$.
Then draw the possible lines $ y \equal{} A$. You will easily get $ \left\{0,2,3,4,6\right\}$ for the set of $ n\left(A\right)$.

If you want exact results,
$ A < 0 \Rightarrow n\left(A\right) \equal{} 0$.

$ A \equal{} 0 \Rightarrow n\left(A\right) \equal{} 3$.

$ 0 < A < \frac{9}{64} \Rightarrow n\left(A\right) \equal{} 6$.

$ A \equal{} \frac{9}{64} \Rightarrow n\left(A\right) \equal{} 4$.

$ A > \frac{9}{64} \Rightarrow n\left(A\right) \equal{} 2$.

If you wonder where $ \frac{9}{64}$ is coming from, it is the value of function at $ x \equal{} \pm \frac12$

$ \boxed D$


abstract algebra
vector
invariant
group theory
superior algebra
superior algebra unsolved

Let G a finite abelian group of n element. Let
\[ x\equal{}a_{1}\cdot a_{2}\cdot a_{3}\cdot ......\cdot a_{n}\]

\[ a_{i}\in G, 1\leq i\leq n\]
if it exists two or more elements b so that
\[ b\in\ G\]

\[ b^{2}\equal{}e\]

\[ b\neq e\]
to prove that x=e

Is there something missing? x can be every element of G with the given information. And b has no connection to x...

"For every finite abelian group containing at least two elements of order 2, the product of all elements in the group equals the neutral element." owk

ugly proof: ;)

let $ G$ be a finite abelian group. if there is no element of order $ 2$, we observe that $ \sum_{x\in G}x \equal{}\sum_{x\in G}\minus{}x$, thus $ 2\sum_{x\in G}x \equal{} 0$ and we're done. now for the general case, we use the structure theorem of abelian groups. if $ G$ and $ H$ are groups s.t. $ \sum_{x\in G}x \equal{} 0$ and $ \sum_{x\in H}x \equal{} 0$, then also $ \sum_{(x,y)\in (G,H)}(x,y) \equal{}\sum_{x\in G}(|H|x,\sum_{y\in H}y) \equal{} |H|\sum_{x\in G}(x,0) \equal{} (0,0)$. thus we only have to look at $ \mathbb{Z}/2^{e_{1}}\oplus ...\oplus\mathbb{Z}/2^{e_{k}}$ where $ k\geq 2$ and $ e_{i}\geq 1$. this implies $ e_{1}\plus{}...\plus{}e_{k}\minus{}1\geq e_{i}$ for all $ i$.

first, we have $ \sum_{x\in\mathbb{Z}/2^{n}}x \equal{} 2^{n\minus{}1}(2^{n}\minus{}1)$. applying this many times as above, we get

$ \sum_{x\in\mathbb{Z}/2^{e_{1}}\oplus ...\oplus\mathbb{Z}/2^{e_{k}}}x \equal{} (2^{e_{1}\plus{}...\plus{}e_{k}\minus{}1}(2^{e_{1}}\minus{}1),...,2^{e_{1}\plus{}...\plus{}e_{k}\minus{}1}(2^{e_{k}}\minus{}1)) \equal{} 0$.

You do not need the structure theorem. Elements not having order $ \le2$ cancel in pairs $ x\plus{}(\minus{}x)\equal{}0$ (written additively), so $ \sum_{x\in G}x\equal{}\sum_{\substack{x\in G\\2x\equal{}0}}x$. Therefore, it suffices to consider groups annihilated by 2, i.e., $ \mathbb F_{2}$-vector spaces. It is clear that the sum of all elements is invariant under automorphisms, but it is also easy to see that in a vector space of dimension at least 2, there are no vectors invariant under all automorphisms except the zero vector. owk


Let $ f(x)\minus{}x$ and $ f(x)\minus{}x^3$ be two monotonic increase functions on R. Prove that the function $ f(x)\minus{}\frac{\sqrt3}2 x^2$ is also a monotonic increase function on R

$ g_1(x)\equal{}f(x)\minus{}x, g_2(x)\equal{}f(x)\minus{}x^3$

$ g_1'(x)\equal{}f'(x)\minus{}1\ge 0, g_2'(x)\equal{}f'(x)\minus{}3x^2\ge 0$

so $ f'(x)\ge \max(1, 3x^2)$

wts: $ \max(1,3x^2)\ge x\sqrt 3$

$ x\le \frac{1}{\sqrt 3} \to 1 \ge x\sqrt 3$

$ x \ge \frac{1}{\sqrt 3} \to 3x^2 \ge x\sqrt 3$

so $ f'(x)\minus{}x\sqrt 3 \ge 0$ thus $ f(x)\minus{}\frac{\sqrt 3}{2} x^2$ monotone increasing


trigonometry
inequalities
algebra
function
domain
inequalities unsolved

Prove that [tex] \forall x \in (0,1) [/tex] ,we have:
 [tex] \frac{sin{(\pi\cdot x)}}{\pi\cdotx} \geq \frac{1-x^2}{1+x^2} [/tex]

Please Edit it,because $\frac{sin\pi}{\pi}$ is a constant..

pro with same subject:
find p.q satisfying: for every 0 <= x <= 1 we have: / :sqrt: 1-x 2  -px-q/ <=( :sqrt: 2-1)/2
to hung: thi cu the nao roi

[quote="Myself"]The inequality isn't true for x=0.[/quote]But 0 is not in the domain of the function ;)

The inequality is not true. Certainly if $x=1/2$, $\frac{1}{\pi}<\frac35$ :?


If 3 to the 35th power is multiplied out, what is the units (ones) digit of the product?

wooo, totient FTW
[hide]
Euler's totient theorem says that odd numbers to the fourth power where the number isn't a multiple of 5 end with 1.
$3^{24}$ ends with 1.
So our answer is 1.[/hide]

[quote="math6genius"][hide]The pattern is...

$3^{1}=3$
$3^{2}=9$
$3^{3}=...7$
$3^{4}=...1$

Then it repeats again.  So...

$\frac{35}{4}$ has a remainder of 2.  Therefore the answer is $\boxed{7}$[/hide][/quote]

Um, if it had a remainder of 2, then it would end in 9...?

EDIT: after doing it myself, youre right, i guess that 2 was a typo.

[quote="Art of Owna"]uh..its 3^35, not 24..[/quote]
:rotfl: :rotfl:

answer is 7

if we observe the units digits:
$3^{0}= 1$, $3^{1}= 3$, $3^{2}= 9$, $3^{3}= 27$, $3^{4}= 81$

the pattern 1, 3, 9, 7 continues indefinatly
the 35th root is the 36th in the sequence since we start with 0 so it follows:
$36 \equiv 4 \mod_{4}$

Thus $3^{35}$ would have $7$ in the units digit

math6genius got it right.

The answer is 7.


Hi, everyone,

I see this forum has so far had few posts on one of the most interesting applications of linguistics: learning languages more correctly and efficiently. Let's discuss in this thread which languages you are studying and how to study them better. For starters, I'll post a link

http://learninfreedom.org/languagebooks.html

about useful resources for language learners. Other links of this kind are very welcome in this thread. Let's share tips on how to learn languages better--that saves more time for math.   :D

Here's one that I think might be usefull: [url=http://www.yourdictionary.com/]http://www.yourdictionary.com/[/url]. This site has, among many other utilities, online dictionaries and grammars of pratically all languages of the world. And of course, it also has a forum.

I seem to disagree with the language education research that's been going on these days. I generally take a few days to learn a lot of grammar before I learn very many words. But I also find it very difficult to find useful words to learn. Most modern language courses emphasize WAY too much on memorizing phrases and obscure words like "supermarket" and "scuba dive", while having not the faintest idea why they mean what they do. On the other hand, classical language classes do not even attempt to make students compose speech or writing. Solus tamen illo uti conor. Maybe someday, I'll come up with a good wordlist to try to learn for new languages.

Have you heard of Morris Swadesh's word list, which he used for glottochronology investigations?

I found it easy to learn Food Tongue because I had a dedicated teacher and we started simply. I learned me, you, yes, no, what, ... Also, Food Tongue I couldn't really memorize words but had to figure them out from several contexts. It helped that Food Tongue has no grammar though.

Hope this is useful. If anyone has suggestions for a Spanish curriculum like that, I'd greatly appreciate it :).

Here are good sites for learning these languages :

French: [url]http://www.ielanguages.com/french1.html[/url]
It has French I, II, III, Informal French and Slang, and French-Italian.

German: [url]http://www.ielanguages.com/German1.html[/url]
German I, II, III tutorials

Spanish: [url]http://www.ielanguages.com/spanish1.html[/url]
Spanish I, II, and III

Italian:[url]http://www.ielanguages.com/italian1.html[/url]
I, II, III and French and Italian (again)

There are others.. just roll your mouse over language tutorials to see a drop down menu with Romance, Germanic, Slavic, & Non I-E languages

I found the Spanish I tutorial helpful last year when starting Spanish   :lol:

I  think the best way to learn a functional language is by emulating the natural way. It is not going to give you a lot of language theory and "deep-meaning" but is functional and, as a part of daily life, a "deep-knowing" subject.

"Let's share tips on how to learn languages better--that saves more time
for math."

Read math books in your target language!  Then you can do both at once! :lol:

Seriously!  I learn better when I read things I'm actually interested in reading.  I hate slogging through a lesson sentence only to discover it means "The horses eat grass" or something lame like that!  I have quite a collection of foreign language books and magazines I've amassed from trips to Europe, but you could probably find websites about what you're interested in.  You could also try finding a pen pal-- you write them in X language, they write you in English, and you correct each other's mistakes.[/quote]


From 6 am to 7 am, Sam drove 40 miles an hour.

From 7 am to 8 am, Sam drove $ 60 \minus{} 40 \equal{} 20$ miles an hour.

From 8 am t0 9 am, Sam drove $ 100 \minus{} 60 \equal{} 40$ miles an hour.

From 9 am to 10 am, Sam drove $ 120 \minus{} 100 \equal{} 20$ miles an hour.

From 10 am to 11 am, Same drove $ 160 \minus{} 120 \equal{} 40$ miles an hour.

$ \frac{40 \plus{} 20 \plus{} 40 \plus{} 20 \plus{} 40}{5} \equal{} \boxed{32}$

I had this problem where Bob was driving 60 mph, but drove the same distance at 40 mph. Then the average was 48 mph. In this problem, is it safe to just say he drove 160 miles for 5 hours, so his average is 32 mph?

Yes; I was just showing in unnecessary detail.

However, you would do it dragon's way in FTW because you're going for speed.  160/5 is much faster than what you did.


calculus integration geometry limit trigonometry

We will define the sequence of geometric figures ${f_{n}}$. The figure $f_{n}$ is inscribed in the figure $f_{n-1}$. $f_{2n-1}$ are circles for all positive integral $n$, and $f_{1}$ is the unit circle. $f_{2n}$ is a regular $\text{n+2 gon}$ for all positive integral $n$. If $a_{n}$ is the area of the figure $f_{n}$, find $\lim_{n\to\infty}a_{n}$.

I hope the answer isn't $0$, that would be boring.  :huh: :lol:

Sorry if this isn't the right forum.

[hide="Mid-result"]The limit is $a=\pi\left(\prod_{n=3}^{\infty}\cos{\pi\over n}\right)^{2}$, and I still have to calculate the product. Then I will post the full solution. (Comp calculation shows that $a\neq 0$.)

Of course, if someone has the product already, please post it.[/hide]

[hide="Appendix"]According to [url=http://mathworld.wolfram.com/PolygonCircumscribing.html]this[/url], there's no closed formula for the desired limit.[/hide]


qs 10
napthalene with an aldehyde group at 1 position

no wrong  :D   :P   :wink:

Atleast that's what my book says.
Anyway write ur mechanism and i'll see

Wat do you mean by ur last post?  :maybe:

[quote="anitsahu"]sorry i had also taken heat into account
its coming cinnamaldehyde's higher carbon member
in the chain part
alpha acidic hydrogen of acetaldehyde is abstracted and it attacks at the carbonyl carbon
it attacks again after connected my sigma bond when another acidic hydrogen is abstracted
this gives the ans according to me
if heat is given u will get my previous ans for sure on account of 1,5 sigmatropic reaction[/quote]

Do you mean to say an aldol will take place???  :maybe:

And by basic medium i mean HO- and not a bulky base   :D

yup very much an aldol
the product will change significantly if heat or light is given

an aldol is wrong (according to my book)  :D  :D
That's why i gave this question  :D

yes damn sure bcos its my own question with idea from a book  :D  :D  :D

i think i got it
its micheal addition
i m really bogus
because in case of alpha beta unsaturated compounds michael addition takes place


calculus
integration
trigonometry
function
calculus computations

Calculate:
$ \int e^x\frac{2\plus{}\sin x}{3\plus{}\cos x} dx$

It cannot be described by elementary functions.

J.Y.Choi
How do you know it?? :blush:

I mean:
I have already solve it USing the subsitution:
$ \sin x\equal{}\frac{e^x\minus{}e^{\minus{}x}}{2i}$
$ \cos x\equal{}\frac{e^x\plus{}e^{\minus{}x}}{2}$

and is there any method to say turely that whether a function can be integrate or not without using program?

$ \sin x \equal{} \frac{e^{ix} \minus{} e^{\minus{}ix}}{2i}$

$ \cos x \equal{} \frac{e^{ix} \plus{} e^{\minus{}ix}}{2}$.

Your expression is incorrect.

I tried and tried, but I couldn't compute your integral. Finally, I used Wolfram Mathematica program.

http://integrals.wolfram.com/index.jsp
http://www.wolframalpha.com/

I don't know Mathematica can be wrong... I mean, I don't know human brain can solve or not what Mathematica don't know how to find.

[quote="atomicwedgie"]$ \sin x \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$

$ \cos x \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$.

Your expression is incorrect.[/quote]

I'm sorry for making such a stupid mistake.- -

hello, Maple 13 gives the following expression
$ \minus{}i{{\rm e}^{x}}\plus{}\int \!{\frac { \left( {\frac {4}{13}}\plus{}{\frac {6}{13}}
\,i \right) {{\rm e}^{x}} \left( 13\,{{\rm e}^{ix}}\plus{}3\plus{}2\,i \right) }{6
\,{{\rm e}^{ix}}\plus{} \left( {{\rm e}^{ix}} \right) ^{2}\plus{}1}}{dx}$
Sonnhard.


i received this by e-mail. Do you think its genuine or some random crap? [also, plz try it. its worth the effort.] Anyway, i think its interesting:

Start:
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>  I know! Calculations are hard work, but it's nearly over..
>Come on, one more...
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>NOW! THINK ABOUT A TOOL AND A COLOR!
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>  You just thought about a red hammer, didn't you?
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>  If this is not your answer, you are among 2% of people who have a different,
if not abnormal, mind.  98% of the folks would answer a red hammer while doing
this exercise.

I had a green hammer :(

perhaps I did it too fast :(

That's very neat.  I said aloud "Red Screwdriver", but the image of a red hammer flashed through my mind.

Blue Screwdriver - That's what I thought of.

hmm... i came up with a blue saw.. *shrugs* lol

[quote="BHorseMath"]I don;t get it.[/quote]

There are a lot of tools in the world. A hammer doesn't come naturally to me.  :D

hmm... I thought about green, cause that's my favorite color, and Ti-89... cause well, that's my favorite tool.

Oh well.

I think this test was meant for normal people.. :D

Yeah, I got red hammer.  Dammit, I'm normal.

Lol Magnara - it's ok to be normal. I think people who took this test thought about the tool and the color too much. Don't spend more than one second thinking about both - first image. I think that's how it works, but I don't understand the concept behind this test.

Perhaps it's meant to show that the red hammer is the most used tool in America by 98 percent??? :lol:  :lol:  :lol:

orange hammer.  Now for me, that raises the question: what about colorblind people who can't see red?  Would brown or grey mean they also are part of the 98% who are 'normal'?  I'm not colorblind, but orange is pretty close to red and for someone who doesn't see green it would pretty much look the same as a red one.

I got a red wrench... I suspect that (whoever made that e-mail) just made up the statistic about 98% immediately thinking of red hammers.  Then again, as most of us are math nerds, perhaps we are all part of the abnormal 2% :D.

I see colors perfectly, I didn't think even half a second and still I took a green hammer. I just saw a flash of a hammer on a flashy green background for some reason...

Something similar

You will be asked a series of simple math questions. Try to do the math in your head as fast as you can.

It may help to say the answers aloud quietly.

Follow these instructions one at a time and as QUICKLY as you can!

What is:

2+2?

4+4?

8+8?

16+16?

Quick! Pick a number between 12 and 5.

Got it?

Now scroll down...

The number you picked was 7, right?

Isn't that weird???

[hide]Wow....I picked 7! Must be because 12 - 5 = 7...[/hide]

The above is in spoiler because anyone who goes directly to the most recent post (currently this one) would have the surprise of beta's test ruined. Try beta's test, and then see my response.

I got 6 too. Maybe i spent too long thinking of a number...(i think i took about a second or two)

You are supposed to say 7 because 12+5 = 7mod10.

yeah, i had a red hammer :D, i got no clue why but i couldn't think of anything else.
just saw the other test, yeah i took 7. i seem to be one of most normal guys here ;)

Peter

[quote="yif man12"]I picked 14. :?  Not kidding, either. I guess that must mean I picked a number too quickly.[/quote]

I picked 13, 5-12-13 triangle came to me too fast

The 98% statistic about the red hammer has been used by similar "tests", but i don't know about beta's, if there is a percent

98% statistics aren't really accurate since many of those tests are taken on groups like 200 people... don't know about this particular one, but many tests are.

[quote="beta"][quote="joml88"]Since when are 13 and 14 between 12 and 5???[/quote]

We, of course, didn't read the question carefully[/quote]

Yeah, I got 13 as well.  5-12-13 came to mind, and I couldn't think of another number for a couple of seconds.

Yes but you are EXTREMELY abnormal too :D

(j/k) ;)

Nah really when I had to figure I had 13... not between 12 and 5 so alternative... 7

so don't worry you're just a bad reader :P

for the red hammer thing, if you asked a normal dude to pick a tool and a color, (chances are)they will say red hammer. over any other combination(tested it at school b4)--
-edit- yes u ur right whitehorse

[quote="Scrambled"]for the red hammer thing, if you asked a normal dude to pick a tool and a hammer, (chances are)they will say red hammer. over any other combination(tested it at school b4)[/quote]

 :shock: ??? You mean tool and color right?

why would a "normal" person think about red hammer? I have never seen a red hammer in real life(I have seen criminals kill people with a hammer on TV and blood...)!

[quote="devious_"][quote="BHorseMath"]I got black TI 83 plus.  Thats the part I don't get.[/quote]

I got that too. In the second test I got 7.[/quote]

I think I got black TI because I was wishing I had a calculator doing that math.

I thought of an orange hammer.  Then I thought of a yellow hammer and then a yellow screwdriver.

A Hammer is what most people think as a tool, and I guess that mean red is what most people thinks as a color.

And I got 7 on beta's one too.  Weird.


trigonometry
integration
function
calculus
calculus computations

I can work out the aveage distance b/w 2 points on a circle

If we take $x_0 = (1,0)$ then
$d(x_0,p(\theta))=\sqrt{(1-\cos\theta)^2+\sin^2\theta}$

The aveage distance is then
$\int_{-\pi}^{\pi} \sqrt{(1-\cos\theta)^2+\sin^2\theta}\frac{d\theta}{2 \pi}=\frac4\pi$

My question is, is there a simple expression for the aveage distance b/w 2 points on a square?

[b]Edit[/b]: Sorry, this may be in the wrong forum

You would probably have to make a piece-wise function for the sqaure.  I would tilt the square so that its verticies are on the axes, in order to eliminate the vertical lines.


[quote="Phoenix257"]Thomas Greenthumb wants to expand his nursery operations by adding more greenhouses.  HE needs to calculate the surface area above ground and volume for heating and cooling purposes for the following greenhouse that is shaped like half a cylinder laid on its side.  What is the ratio of surface area to volume?  (The ends of the greenhouse are semicircles with radius BD)  The pic below is what it looks like unfolded (A net diagram)[/quote]

First sentence was unrelated but made the whole problem's wording confusing ...

[hide]So
- radius = 30ft
- length = 80ft

The surface area is equal to
$80 \cdot 2 \cdot 30 \pi+30^{2}\pi = 5700 \pi$.

The volume is
$\frac{1}{2}\cdot 80 \cdot 900 \pi = 36000 \pi$.

So the ratio is $\frac{57}{360}=\boxed{\frac{19}{120}}$.[/hide]


group theory
abstract algebra
geometry
geometric transformation
superior algebra
superior algebra unsolved

[quote="vess"]Let $G$ be a finite group with $|G|>n$. Show that $G$ is isomorphic to a transitive subgroup of $S_n$ if and only if $G$ has a subgroup $H$ of index $n$ such that neither $H$ nor any proper subgroup of $H$ is normal in $G$.[/quote]

What is the group acting you have in mind ?
Because for translations, all groups are transitive :-D
- indeed $y=g\cdot x$ with $g=y\cdot x^{-1}$.

Fabrice

[quote="grobber"]I think it's p[retty clear what it means here :?. A subgroup of permutations of $S_n$ is transitive if it's action on a set of $n$ elements is transitive.[/quote]

Thanks for the hint .
This was not clear at all for me.
Now I can try to do the problem.

Fabrice

It actually turned to be a pretty simple problem; I solved it a little while after I posted it, so I know a solution. But I won't post it yet --- I wouldn't like to spoil the fun.  :D

Pretty simple indeed, but we haven't touched it for like a year :).

First assume that $H\le G$ with the mentioned properties ($[G: H]=n$, and no non-trivial subgroup of $H$ is normal in $G$). For each $g\in G$, consider the permutation $\alpha_g$ of the set $S$ of left cosets of $H$ defined by $\alpha_g(xH)=gxH,\ \forall xH\in S$. Clearly, the map $\alpha$, sending each $g\in G$ to $\alpha_g$ is a homomorphism from $G$ into $S_n$, and $\alpha(G)$ is transitive. Let's find the kernel of this homomorphism: $g\in\ker\alpha\iff gxH=xH,\ \forall x\in G\iff g\in xHx^{-1},\ \forall x\in G$, which means that $g\in\bigcap_{x\in G}xHx^{-1}$. However, $\bigcap_{x\in G}xHx^{-1}$ is a subgroup of $H$ which is normal in $G$, so it must be trivial, and this means that $\ker\alpha$ is trivial. In conclusion, $\alpha$ is an isomorphism of $G$ onto a transitive subgroup of $S_n$.

Conversely, let $G$ be a transitive subgroup of $S_n$, which acts on the $n$-element set $X=\{x_i\}_{i=1}^n$. Let $H$ be the subgroup of $G$ consisting of those permutations in $G$ which fix $x_1$. It's clear that $H$ has index $n$ in $G$ (any set of $n$ elements of $G$, one sending $x_1$ to every $x_i$, is a system of representatives for the cosets of $H$). If there are non-trivial subgroups of $H$ which are normal in $G$, then the normal interior $H_G$ of $H$ in $G$, which is $\bigcap_{u\in G}uHu^{-1}$ and is the largest subgroup of $H$ normal in $G$, must also be non-trivial. Let $g$ be a non-trivial element of $H_G$. We get $guH=uH,\ \forall u\in G$, which in this context means that $g$ fixes all the $x_i$'s. We thus have a contradiction, since no non-trivial permutation of $\{x_i\}$ can fix all the $x_i$'s.


[quote="Jimmy"]Does $q | n$ mean that $q$ divides $n$?[/quote]Yes, but I just looked at the Latex guide and I think \mid is the proper way to make the | symbol. $q\mid{n}\Leftrightarrow{q}\text{ divides }n$. Also $q\nmid{n}\Leftrightarrow{q}\text{ does not divide }n$


Consider the problem, prove that (a^3)is congruent mod 6.

I realized that this means (a^3)-a must be divisible by 6.  So a^3 - a factors into a(a+1)(a-1).  So by induction it is impractical to show that the equation works for a=1, so it is shown for 2, and assumed that it works for a.  Then, I tried to prove that a(a+1)(a+2).  Is it acceptable to say that since it is assumed that a(a+1)(a-1) is divisble by 6, for all integers a, then since a+1 must be an integer, then when a+1 is substituted for a the equality should still hold?

Would that be an accpetable proof.  If not please list other methods of doing this problem.  If so let me know also.  Thank you very much.

[quote="mna851"]Consider the problem, prove that (a^3)is congruent mod 6.

I realized that this means (a^3)-a must be divisible by 6.  So a^3 - a factors into a(a+1)(a-1).  So by induction it is impractical to show that the equation works for a=1[/quote]

Its not impractical for a = 1 - its perfectly correct. $1^3 - 1$ is divisible by 6. You don't need to start with a=2 (it of course, depends what the question is - do you have to prove it for all integers? Positive integers? Positive integers greater than 1?)

[quote="mna851"]
so it is shown for 2, and assumed that it works for a.  Then, I tried to prove that a(a+1)(a+2).  Is it acceptable to say that since it is assumed that a(a+1)(a-1) is divisble by 6, for all integers a, then since a+1 must be an integer, then when a+1 is substituted for a the equality should still hold?
 [/quote]

With induction you don't assume that $a(a+1)(a-1)$ is divisible by 6 for all integers a. You assume its divisible by 6 for $a=k$ for some integer k and try to prove its divisible by 6 for $a=k+1$. So no, you can't just assume that it works for k+1, thats what you have to prove.

[quote="mna851"]
Would that be an accpetable proof. If not please list other methods of doing this problem. If so let me know also. Thank you very much.
 [/quote]

If you follow the above prove through you should be able to solve it by induction. You can also solve it by thinking about what a product of three consecutive integers must be a multiple of.

Yes, but If you keep the case where a=1, then you can prove that a^3 - a is congruent to all numbers because you will get a zero in the numerator always.

I'm sorry, can you show me how to prove this one by induction, I'm new to it.

I won't solve it for you, but I'll lay out the main steps of induction:

Proving that $a^3-a$ is divisible by 6: (I'll do it for positive integers, though its true for all integers too)

Base case: $a=1: a^3 - a = 1-1 = 0$, which is divisible by 6. Thus the statement is true for a = 1.

Assume that $a^3-a$ is divisible by 6 for $a=k$. Then $k^3 - k$ is divisible by 6. We want to prove that $(k+1)^3 - (k+1)$ is divisible by 6. Try expanding that and see what happens. Remember you are ASSUMING that $k^3 - k$ is divisible by 6.

Do that, and you've proven it for all positive integers.

When you expand that out you get k^3 + 3k^2 +2k +1 - k -1
So you can rearrange and get (k^3 - k)/6 + (3k^2 + 2k)/6.  So by the assumption the term on the left is divisble by 6, but how can the term on the right be proven.  This is where I get stuck when I induct.

You expanded wrong:
$(k+1)^3=k^3+3k^2+3k+1$
From there, all you need to do is show $2|k^2+k$. Easy enough if you factor.

[hide="No induction necessary"]$a^3-a=a(a+1)(a-1)$
That's a product of three consecutive integers. One of them must be a multiple of 3 and at least one of them must be a multiple of 2. Then their product is a multiple of 6.[/hide]
Please try to figure it out before looking.

Ok, thank you all, I get it very well now.  I was able to do the problem the way bictor did it.  But just curious, would a proof like that be acceptable on a test like the USAMO or any other proof oriented olympiad.

[quote]But just curious, would a proof like that be acceptable on a test like the USAMO or any other proof oriented olympiad.[/quote]

Of course! The proofreaders will accept logic like... of any three consecutive integers, one will be divisible by three.

Another approach is to note that $\frac{a^3 - a}{6} = {{a+1} \choose 3}$, which is an integer.


[b]given a triangle ABC, if point P divides segment AB into lengths of p(AB), and (1-p)(AB), point Q divides segment BC in a similar manner, and point R divides AC in a similar manner, prove that the area of triangle PQR= (the area of triangle ABC)*(pqr+(1-p)(1-q)(1-r)).[/b]
Not too hard, but is pretty useful.

[hide="Solution"]
$\frac{\text{Area}(\triangle APR)}{\text{Area}(\triangle ABC)}= p(1-r)$
$\frac{\text{Area}(\triangle BPQ)}{\text{Area}(\triangle ABC)}= q(1-p)$
$\frac{\text{Area}(\triangle CQR)}{\text{Area}(\triangle ABC)}= r(1-q)$

Hence
$\text{Area}(\triangle PQR)$
$=\triangle ABC-(\triangle APR+\triangle BPQ+\triangle CQR)$
$=\triangle ABC\left(1-\left(\frac{\triangle APR}{\triangle ABC}+\frac{\triangle BPQ}{\triangle ABC}+\frac{\triangle CQR}{\triangle ABC}\right)\right)$
$=\triangle ABC (1-(p(1-r)+q(1-p)+r(1-q)))$
$=\triangle ABC (1-p-q-r+pq+pr+qr)$
$=\triangle ABC (pqr+(1-p)(1-q)(1-r))$[/hide]


Hello everyone!

Welcome to isabella2296's Super Tourney, A Tourney Like No Other!

Each Round contains 8 questions, and for each one you must get a certain amount of questions right to move on! There will be an indefinite number of rounds, until there are 2 people left. They will have to Countdown each other on the FTW to determine the winner.

So sign up today! Sign-ups will close July 25th, Friday at midnight, and the first questions will be posted on Saturday!

SIGN-UP LIST:

1. AIME15
2. Urc
3. BlackSwordsman
4. pinkpiglet
5. Neji0313
6. crazy4math
7. vallon22
8. moogra
9. ernie
10. Bachukas
11. shaun ong
12. xcenario

I will join!

[size=50]Can you post answers and solutions to your 10 round tournament?[/size]

Uh, is this really classroom math? Cause I'm just wondering what level problems it will have

well, i would guess it would be classroom math level  :D

[quote="isabella2296"]Similar to MathCounts level, but not National questions or anything. Easier State problems and chapter.[/quote]

Then I'll be in XD

So any panda avatars...

Anyway, if they're so easy, you're going to see A LOT of perfect scores.

[quote="AIME15"]
Anyway, if they're so easy, you're going to see A LOT of perfect scores.[/quote]

I meant some of the easier state problems, no uber-hard ones that are impossible, or the easier nats. It's not going to be quite so EASY...

none of the problems is impossible...
will we have to submit answers, or solutions?

why would we WANT to submit solutions?

sorry for the spam :( :(

idk
but then again, it said that the last two will cd for the title, so that probably had some part in it

Same with AIME15's Mock Mathcounts!! I pmed levans, hopefully I'll get a reply!!!

wait, AIME15's mock mathcounts was moved?
when?
and where?

my mock mathcounts is still on mathcounts...

hm...
idk what isabella was talking about then
o well

@vallon and aime:

I meant that AIME also uses a countdown at the end.

SIGN-UP LIST:

1. AIME15
2. Urc
3. BlackSwordsman
4. pinkpiglet
5. Neji0313
6. crazy4math
7. vallon22
8. moogra
9. ernie
10. Bachukas
11. shaun ong
12. xcenario
13. shentang

idk
but it doesn't really matter how many pplz sign up
at the end, it has to be down to 2 pplz, soo...

Yes but I want a lot of people!

THIS THREAD DOES NOT BELONG HERE!

You could make a signup thread on the mathcounts forum.

that's a good idea
just kill this one and restart it on MC forum

That's exactly what I'm going to do...I'll post the link soon.

Here's the new link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1196569#1196569

And yes, I pmed levans AND D3m0n SHad0w....no response.  :(

Woah...I'm confused. So this ISNT a FTW game?


There is a bag of 50 white marbles and 50 black marbles. I choose two at a time with no replacements. Every time I get two of the same colour, I put back in a white marble. Each time its different colours, I put back in a black one. What will be the last marble remaining?

Is this probability or will there definitely be a certain color left at the end?

[hide]Each time you take out two white marbles, you still have the same number of black marbles remaining.  Each time you take out two black marbles, you have two less black marbles remaining.  Each time you take out one marble of each color, you put the black marble back in, so you still have the same number of black marbles remaining.

Black marbles can therefore only be removed two at a time, so you will always have an even number of black marbles in the bag.  You cannot ever have one black marble in the bag, so the last marble remaining must be white.[/hide]


The sides of a right triangle are 20, x, and hypotenuse y.
The sides of another are 30, x-5, and hypotenuse y+5.
Find x.

EDIT: Typo sry everyone. :oops:

[hide="Well..."]
$400+x^{2}=y^{2}$
$900+(x-5)^{2}=(y-5)^{2}$

Expanding the second equation, we have

$x^{2}-10x+925=y^{2}-10y+25$

We subtract the first equation to get

$-10x+525=-10y+25$
$-2x+105=-2y+5$
$2x-2y=100$
$x-y=50$
$x=50+y$

For $y$ to be the hypotenuse, $y>x$, and clearly the above equation cannot be true. Thus, there are no solutions.[/hide]

[quote="Totally Zealous"]The sides of a right triangle are 20, x, and hypotenuse y.
The sides of another are 30, x+5, and hypotenuse y-5.
Find x.[/quote]

If $x$ and $y$ are restricted to positive reals, as they usually are, then there is no solution.

[hide="If they are not..."]
then the solution is $x=-21$, $y=-29$.
Solve for $y$ in the first, then substitute it into the second. Square, rearrange, square again and you're done.[/hide]

From the given triangles,
\[20^{2}+x^{2}= y^{2}\]
and
\[30^{2}+(x+5)^{2}= (y-5)^{2}\]
.
\[y = \sqrt{400-x^{2}}\]
$900+x^{2}+10x+25 = (\sqrt{400-x^{2}}-5)^{2}\\ \&= 400-x^{2}-10\sqrt{400-x^{2}}+25\\ x^{2}+5x+250 = 5\sqrt{400-x^{2}}\\ x^{4}+10x^{3}+525x^{2}+2500x+62500 = 10000-25x^{2}\\ x^{4}+10x^{3}+550x^{2}+2500x+52500 = 0$
X must be greater than 5, and must be a root to this equation.  If x is rational, it must be a factor of $52500$.  Have fun!

Three people replied while I was typing this up?! :huh:  :o  :oops_sign:

Edit: apparently my way of solving this problem would be unnecessarily tedious to work out.

I believe the original problem should read $x-5, y+5$.

[hide="Then..."] The $20$ should immediately make you think $20, 21, 29$, which is a very popular Pythagorean triple to use on contests.  Indeed,

$30^{2}+(21-5)^{2}= (29+5)^{2}$

(This is a multiple of $8, 15, 17$, which is another little-known triple that people like to sneak in on contests.)

So $x = 21, y = 29$. [/hide]

[quote="sapphyre571"][hide="solution"]
we know $x^{2}+400=y^{2}$
so $y^{2}-x^{2}=400$
We also know $(x+5)^{2}+900=(y-5)^{2}$
so $x^{2}+10x+925=y^{2}-10y+25$
so $y^{2}-x^{2}=10x+10y+900$
this means $10x+10y=-500$
which has no positive solutions.
[/hide][/quote]
I assume that your last equation is a typo, because $10x+10y =-500$ obviously has real solutions, specifically the line $y =-x-50$.

Edit: the master (tOrajirOu) has rendered this point moot.

There's no need to learn Pythagorean triples by heart or to make statistics on what is "popular on the contests" (that strategy kind of degrades the point of mathematics by promoting grinding over creative thinking).

$400+x^{2}=y^{2}$
$900+(x-5)^{2}=(y+5)^{2}$

Expanding the second equation:

$900+x^{2}-10x+25=y^{2}+10y+25$

$900+x^{2}-y^{2}=10x+10y$

From the first we have $x^{2}-y^{2}=-400$:

$500=10x+10y$

$x+y=50\qquad(*)$

Again from the first:

$y^{2}-x^{2}=400$

By factoring:

$(y-x)(y+x)=400$

Taking $(*)$ into account:

$y-x=8$

Now we have

$y+x=50$
$y-x=8$

Subtracting the second from the first we get $2x=42\iff x=21$

Adding them together we get $2y=58\iff y=29$

Sorry, I have a propensity to grab for the fastest solution to problems, and often the fastest solution to problems of this type start by assuming the answers are integer  :)  Point taken, though.

roadnottaken, your substitution, which seems to be taken from similar approaches when dealing with linear systems, is needlessly difficult when dealing with problems of this type.  sapphyre's and Farenhajt's cancellations of the $x^{2}, y^{2}$ terms, on the other hand, produced a system of linear equations rather than a quartic.

[quote="roadnottaken"]From the given triangles,
$20^{2}+x^{2}= y^{2}$
and
$30^{2}+(x+5)^{2}= (y-5)^{2}$
$y = \sqrt{400 \ \boxed{-}\ x^{2}}$
$\vdots$[/quote]

The work for this solution isn't that tedious if you don't accidentally flip a sign, like you did here. It should be $\boxed{+}$, not $-$.

And that is what I get for trying to solve an equation at 9:30 at night. :oops:


Prove that for any positive real numbers $a,b$ and $c$,
\[\sum_{cyc}\frac{a^{2}+bc}{a(b+c)}-\sum_{cyc}\frac{a(b+c)}{a^{2}+bc}\ge \frac{a+b+c}{\sqrt[3]{abc}}-3 \]


geometry
rectangle
graph theory
combinatorics proposed
combinatorics

a $n*(n-2)$ board is filled with numbers $1,2,3...,n$ in such a way that no number appears twice in the same row or column. prove one can add two more rows (to obtain an $n*n$ board) filled with numbers $1,2...,n$ in such a way that property mentioned above is preserved.

The last two columns must consist of $1,2,\ldots,n$ each. We also know what combination must lie on each row of length $2$ in the $n\times 2$ array that we must add. The only freedom we have is reversing the order of the two numbers in each row.

Consider the graph having $1,2,\ldots,n$ as vertices and having $i,j$ connected iff in some row of length $2$ we have $i$ and $j$. Each vertex has degree at most $2$ in this graph, so its connected components are some cycles and some simple paths. It's easy to see that these connected components are "independent" (in the sense that we can deal with each one of them separately), and that both a cycle and a simple path can be placed in the two columns, thus giving the result we want.

[quote="seshadri"]in fact more is true:for any $r<n$,if we have an $r$ x $n$ rectangle with this property,one can extend it to an $(r+1)$ x $n$ rectangle with this property still retained(i.e. any latin rectangle can be completed into a latin square).[/quote]

I wanted to post this a few days ago, but I found it while looking through older problems :).

Consider the $n$ sets consisting of the numbers that are mising from each column of the matrix. We now form a bipartite graph having these $n$ sets as one part of the bipartition, and the $n$ numebrs as the other part. Two vertices (one corresponding to a set, and one to a number) are connected iff the set contains the number. This graph must be $n-r$-regular, because on the one hand every set contains $n-r$ numbers, and on the other hand every number appears once on every row, so $r$ times in total, which means that it must appear $n-r$ more times among the numbers which are missing. Since every regular bipartite graph has a perfect matching, for every set we can find a number to add to the column corresponding to the set such that these $n$ numbers are all distinct. This way, we form the $r+1$'th row of the latin rectangle.


Let M be a compact manifold without boundary.
Is there an Immersion f:M-->R^m?
(m=dimM)

My 'educated' guess would be No.
I think an immersion in this case implies a diffeomorphism,
which is of course impossible (since R^m is not compact).
But I can't really prove this...

you just need two facts about $ f(M)$:
- it's closed (why?);
- it's open (why?).
and this concludes (again, why?).

[hide="answers"]i don't think they're needed, at this point. but still, here we go with some more detailed "hints" (or with the solution itself, if you prefer)...
- $ M$ is compact, $ f$ is continuous and $ \mathbb{R}^m$ is T2;
- as you said, in this case immersion implies [i]local[/i] diffeomorphism, which in turn implies $ f$ open.
and finally, $ \mathbb{R}^m$ is connected.[/hide]

f smooth, M compact  ==> f(M) compact
M compact ==> M is a finite union of open sets Vi
==> f(Vi) open ==> f(M)=U{f(Vi)} open

f(M) open and compact in R^m ==> contradiction.

correct?

well, you don't need compactness to show that $ f(M)$ is open.
and the contradiction.. well, yes: $ f(M)$ is open, closed (because a compact subset of an hausdorff space is closed), and nonempty. so by connectedness of $ \mathbb{R}^n$ it's the whole space, which is not compact.

i'm just being really pernickety here, just to make sure you got things straight...

In fact, I can't use compactness, because it doesn't ensure there is an open cover at all.
But the no-boundary condition does...
So M is some union of open sets Vi, And then f(M)=Uf(Vi) is a union of open sets.
Q.E.D.

Appreciate your help!

Or better yet:
No boundary  <==>  M open  (?)

[quote="Gilad.R"]In fact, I can't use compactness, because it doesn't ensure there is an open cover at all.
But the no-boundary condition does...[/quote]
uhm.. let's make things clear: every topological space has an open cover, every infinite (as a set) hausdorff space has plenties of them, and compactness has nothing to do with it.
what you need is that there's an open cover [i]with subset homeomorphic to[/i] $ \mathbb{R}^n$ (with $ n\equal{}\dim M$), and this, together with the immersion condition, gives you that the [i]image[/i] is open (you need some further attention, since you might need to consider a refinement of the cover you've chosen, but this is the idea).

you need to be a bit careful about what's open and what's not, or where it's open and where not.. here our intuition tends to fail, at times.


modular arithmetic
number theory unsolved
number theory

If d = gcd(x, y) then d(x + y) is a perfect square... or something like that, don't remember exactly, but it is a classical exercise, and I am sure that you can find it on the forum.

[quote="filip"]Given natural numbers $x,y,z$ such that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$. Prove that $x+y$ is divisible by perfect square $>1$.[/quote]

It's a simple problem, the equation is equivalent to $(x+y)z=xy$ so, if $t|(x+y)$ with  $t$ a prime, then $t|x$ or $t|y$ but then $t|y$ or $t|x$ because $t|(x+y)$, now by definition of the problem can assume that $(x,y,z)=1$ so $t\not |z$ and hence $t^2|(x+y)$ with $t>1$.

Best Regards.
RX the Crazy Math

This question is trival since all positive integers can be expressed in the form of $\alpha^2\beta$ where $\alpha, \beta$ are positive integers, and $\beta$ is not a perfect square. But if we add the condition that $gcd(x, y, z) = 1$, then the question becomes proving that $x+y$ is a perfect square, which would not make the question so trival.

Now, if we have this additional condition:
$\frac{1}{x} + \frac{1}{y} = \frac{1}{z} \Leftrightarrow yz+zx=xy$ By considering congruences modulo $x$, $y$ and $z$ separately, we get $yz \equiv 0 \pmod{x}$, $zx \equiv 0 \pmod{y}$ and $xy \equiv 0 \pmod{z}$ respectively. Thus we have $x \mid yz, y \mid zx, z \mid xy$

Let $x=ab$ such that $a \mid y$ and $b \mid z$, $a,b$ are positive integers, then it follows that we can let $y = ka$ and $z = lb$ for some positive integers $k, l$. Since $gcd(x, y, z) = 1$, it follows that $gcd(a, l) = gcd(b, k) = 1$. Now, we have the following:
\[y \mid zx \Rightarrow ka \mid lab^2 \Rightarrow k \mid lb^2 \Rightarrow k\mid l\]
\[z \mid xy \Rightarrow lb \mid ka^2b \Rightarrow l \mid ka^2 \Rightarrow l \mid k\]
Thus $k=l$ as both $k, l$ are positive integers.
It follows that $xyz = k^2a^2b^2$. Now from $yz + zx = z(x+y) = xy$, we have $x+y = \frac{xyz}{z^2}$ which is a perfect square. And since $x, y$ are positive integers, $x+y > 1$ and we are done.

[quote="dblues"]:blush: Didn't notice that RobertuX has already posted the soln...[/quote]

Don't worry, I think is really nice give different solutions to same problem to understand better how to solve problems, and how the people thinks.   ;)  ;)

Best regards.
RX the  Crazy Math


How do you convert $ (R)\minus{}2\minus{}heptanol$ into $ (R)\minus{}2\minus{}heptanethiol$?

I think $ \ce{SOCl_2}$ in prescence of pyridine to give (S)-2-chloroheptane and then treat it with NaSH which follows a $ SN_2$ pathway and so you get (R)-2-heptanthiol.

One possible strategy that can be used is the following: two SN2 reactions, since in each there is inversion of configuration. Note that several well known methods, like the Mitsunobu reaction, Grignard reagents, and use of $ \ce{SOCl2}$ alone, cannot be used on this case (why?). Here's my proposal:

[hide="Answer"]1. $ (R) \minus{} \ce{C5H11 \minus{} CHOH \minus{} CH3} \plus{} TsCl \longrightarrow (R) \minus{} \ce{C5H11 \minus{} CHOTs \minus{} CH3}$

2. $ (R) \minus{} \ce{C5H11 \minus{} CHOTs \minus{} CH3} \plus{} \ce{I \minus{} } \longrightarrow (S) \minus{} \ce{C5H11 \minus{} CHI \minus{} CH3}$

3. $ (S) \minus{} \ce{C5H11 \minus{} CHI \minus{} CH3} \plus{} \ce{HS \minus{} } \longrightarrow (R) \minus{} \ce{C5H11 \minus{} CHSH \minus{} CH3}$[/hide]

Use of $ \ce{SOCl_2}$ alone follows an $ \ce{SN_i}$ mechanism which is intramolecular and takes place without inversion and so that's why Pridine is used to convert alcohol to alkyl chloride with inversion

Yes, that's correct. How about the Grignard?

I thought grignard take away the active hydrogen of alcohol and so not used. :maybe:
Or is ur question was treatment of grignard to alkyl halides? :huh:  I don't get u carcul.

Of course, I was referring to treating an alkyl halide with Mg to get a Grignard reagent. Then the thiol could be made from this organometallic.


Let  $ x \equal{} 1,y \equal{} 2,z \equal{} 0,$ then

        LHS-RHS=$ \minus{}\frac{19}{32}!$

But we have
\[ \sum_{cyc}\frac {1}{a^{2} \plus{} bc}\ge\frac {3}{4}(ab \plus{} bc \plus{} ca)\left(\frac {1}{a^{3}b \plus{} b^{3}c \plus{} c^{3}a} \plus{} \frac {1}{ab^{3} \plus{} bc^{3} \plus{} ca^{3}}\right)
\]

[quote="fjwxcsl"]Let  $ x \equal{} 1,y \equal{} 2,z \equal{} 0,$ then

        LHS-RHS=$ \minus{} \frac {19}{32}!$

But we have
\[ \sum_{cyc}\frac {1}{a^{2} \plus{} bc}\ge\frac {3}{4}(ab \plus{} bc \plus{} ca)\left(\frac {1}{a^{3}b \plus{} b^{3}c \plus{} c^{3}a} \plus{} \frac {1}{ab^{3} \plus{} bc^{3} \plus{} ca^{3}}\right)
\]
[/quote]

Sorry, $ a,b,c$ be sidelengths of a triangle :( I edited my post. Thank you fjwxcsl.

[quote="dduclam"]Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\frac3{4}(a^2 \plus{} b^2 \plus{} c^2)\left(\frac1{a^3b \plus{} b^3c \plus{} c^3a} \plus{} \frac1{ab^3 \plus{} bc^3 \plus{} ca^3}\right)
\]
[/quote]
Very nice inequality, Dduclam. My solution proved that
$ \sum \frac{1}{a^2\plus{}bc} \ge \frac{12(a^2\plus{}b^2\plus{}c^2)}{3\sum ab(a^2\plus{}b^2)\plus{}2\sum a^2b^2} \ge \frac3{4}(a^2 \plus{} b^2 \plus{} c^2)\left(\frac1{a^3b \plus{} b^3c \plus{} c^3a} \plus{} \frac1{ab^3 \plus{} bc^3 \plus{} ca^3}\right).$
But I think it is not nice. I think you have a better proof, may I see it, my friend? :)

[quote="fjwxcsl"]Let  $ x \equal{} 1,y \equal{} 2,z \equal{} 0,$ then

        LHS-RHS=$ \minus{} \frac {19}{32}!$

But we have
\[ \sum_{cyc}\frac {1}{a^{2} \plus{} bc}\ge\frac {3}{4}(ab \plus{} bc \plus{} ca)\left(\frac {1}{a^{3}b \plus{} b^{3}c \plus{} c^{3}a} \plus{} \frac {1}{ab^{3} \plus{} bc^{3} \plus{} ca^{3}}\right)
\]
[/quote]

Hello, I think you must edit x, y, z into a, b, c.


Mamy prostokąt, który możemy podzielić na małe prostokąty o conajmniej jednym boku całkowitym. Udowodnić, że wyjściowy prostokąt też musi mieć conajmniej jeden bok całkowity.

Bardzo fajny problem... Znam jego rozwiązanie, więc może tylko podpowiem, że bardzo ładny jest sposób korzystający z gry HEX...


This is the NEW thread for the game Elect the Executioner!!!

Signups here!!!

PS: please someone update the list of the people who had signed up.

/in

i am checking who has already signed up now

Edit:

CURRENT SIGNUPS:

chenhsi
lotsofmath
pacman2812
iLord

[size=9]/in[/size]

The message is too small. Please make the message longer before submitting.

It says [hide][size=25]/in[/size][/hide]

Still can't see?
[hide][size=42]/in[/size][/hide]

Now, [hide][size=200]/in[/size][/hide]

[quote="dannyhamtx"][code][size=1]/in[/size]

The message is too small. Please make the message longer before submitting.

It says [hide][size=3]/in[/size][/hide]

Still can't see?
[hide][size=5]/in[/size][/hide]

Now, [hide][size=24]/in[/size][/hide][/code][/quote]

Yawn...wait danny and budi, won't AMP hamper your access after a while?

I'm bored.
I will start classes again in about one or two weeks :(

CURRENT SIGNUPS:

chenhsi
lotsofmath
pacman2812
iLord
budi713
dannyhamtx

How many people do we want to play?

Q: How many players?
A: Any odd amount will do. I'll take 9 to begin with.

Q: How do we play?
A: Each round, each player has 3 "voting points". He may give those points to other players, by posting give 1 point to <name>. You can't give points to yourself, and you can't sit on your points. You MUST give them to someone else within an alloted time-limit, or you're warned, and then modkilled. Once all points have been distributed, the player with the most points becomes the Executioner. He chooses to eliminate someone from the game, by posting execute <name>. Then, a new round will begin, with 1 less player.

Q: What is the proper way to give points?
A: Post in Bold, like this. I give __ points to ________.

Q: What's the length of a round?
A: The point-giving stage lasts a maximum of 10 days. (After that stage, if you have any points left, you're warned. Three days after you're warned, you will be modkilled). The execution deliberations last a maximum of 5 days (if the Executioner takes longer than that to decide, he's warned. Three days after he's warned, he will be modkilled).

Q: What happens when there are only 2 players left? How can an Executioner be chosen?
A: The 7 players previously eliminated from the game become the jury. They vote for the winner.

Q: What happens at the end of the round, if two players are tied for points?
A: The other players in the game hold a vote to make one of the two an Executioner.

Q: What if this vote is tied too?
A: The player most recently executed casts the deciding vote.

Q: What if more than two players are tied for points?
A: The round will continue until one of those players decides to give a point to another player (thus making him the Executioner).

Q: What if I will be gone for a while, for vacation/buisness trip/finals?
A: You can PM me your actions in advance, for as long as you will be gone.

Q: What's the winner's prize?
A: To mod the next game, of course!

Q: So, basically it's a game of politics?
A: Well, duh.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205339[/url] for an example game.

lets have 9 players again this time

CURRENT SIGNUPS:

chenhsi
lotsofmath
pacman2812
iLord
budi713
dannyhamtx
myyellowducky82

/in

The message is too small. Please make the message longer before submitting.

All right, let the games begin!

[u]Number of Points Left[/u]

chenhsi - 3
lotsofmath - 3
pacman2812 - 3
iLord - 3
budi713 - 3
dannyhamtx - 3
myyellowducky82 - 3
BOGTRO - 3
AIMEman - 3

[u]Number of Points Given[/u]
None Yet!

(Oh yeah, you guys can make point counts as well, I am not officially modding this.)

I give [b]budi713 3 points[/b]

Number of Points Left

chenhsi - 3
lotsofmath - 3
pacman2812 - 3
iLord - 3
budi713 - 3
dannyhamtx - 0
myyellowducky82 - 3
BOGTRO - 3
AIMEman - 3

Number of Points Given
budi713-3

[b]I give budi713 3 points[/b], because he is on my team for the tug-of-war.

[u]Number of Points Left[/u]

chenhsi - 0
lotsofmath - 3
pacman2812 - 3
iLord - 3
budi713 - 3
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 3

[u]Number of Points Given[/u]

budi713: 6
iLord: 1

No, you have 3 points per round, and you can give them all at once if you want.

uuhh... 2 to iLord because all must have a number of the form 3x

Number of Points Left

chenhsi - 0
lotsofmath - 3
pacman2812 - 1
iLord - 3
budi713 - 3
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 3

Number of Points Given

budi713: 6
iLord: 3

3 to iLord

Number of Points Left

chenhsi - 0
lotsofmath - 3
pacman2812 - 1
iLord - 3
budi713 - 0
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 3

Number of Points Given

budi713: 6
iLord: 6

[b]3 to budi713[/b], because I owe it to him from survivor.

chenhsi - 0
lotsofmath - 3
pacman2812 - 1
iLord - 3
budi713 - 0
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 0

Number of Points Given

budi713: 9
iLord: 6

2 to iLord, 1 to budi

chenhsi - 0
lotsofmath - 0
pacman2812 - 1
iLord - 3
budi713 - 0
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 0

Number of Points Given

budi713: 10
iLord: 8

[b]3 to Lotsofmath[/b]

chenhsi - 0
lotsofmath - 0
pacman2812 - 1
iLord - 0
budi713 - 0
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 0

Number of Points Given

lotsofmath: 3
budi713: 10
iLord: 8

my last point is for iLord.

1 to iLord :wacko: I like this smiley!! :D

chenhsi - 0
lotsofmath - 0
pacman2812 - 0
iLord - 0
budi713 - 0
dannyhamtx - 0
myyellowducky82 - 2
BOGTRO - 3
AIMEman - 0

Number of Points Given

lotsofmath: 3
budi713: 10
iLord: 9

Now, it's between iLord and budi. Yellowducky, BOGTRO, please define it quicly!! :)

Ppl plz hurry up.......

I am getting bored on this thread now sadly

:wacko: :wacko: :wacko:

P.S.: I am not spamming! This emoticon is too cool!

bump

you guys, this thread is basically dead now

this ended in a failure (pretty much) unless someone posts again

We could just say that iLord wins the round.

No, because theroically they are modkilled so budi wins.


[quote="MCNational!"]The chickens and pigs in Farmer McCoy's barn have a total of 36 heads and 98 legs. How many pigs are in the barn?[/quote]

Suppose we had all pigs. There would be $ 36 \times 4 \equal{} 144$ legs. However, there are only 98 legs. That means we need $ (144 \minus{} 98) \div 2 \equal{} 23$ chickens, and there are then 36 - 23 = $ \boxed{13}$ pigs.

This isn't a terribly different solution but I just used algebra instead, in fact, Isabella's way is probably faster.
x=number of chickens,
$ 2x \plus{} 4(36 \minus{} x) \equal{} 98$ so you get $ 2x \plus{} 144 \minus{} 4x \equal{} 98$ then just simplify to get $ \minus{} 2x \equal{} \minus{} 46$
So the number of chickens you end up with is $ 23$ and since there are $ 36$ heads, you get $ 36 \minus{} 23 \equal{} 13$ pigs.


[quote="wudud"]xyzklm are nonnegative real numbers Show that

(1-x)(1-y)+(1-z)(1-k)+(1-l)(1-m)>=0[/quote]

what do you mean? It's obviously false isn't it?! Or do you miss some conditions?


I received an announcement of a Math Jam scheduled for yesterday (November 3) about a couple of upcoming AoPS classes.  I don't think I see the transcript for that math jam.

Fixed.  The transcript is available now; thanks for pointing it out.


geometry
circumcircle
parallelogram
geometry proposed

Let $ H$ be the orthocenter, $ O$ the circumcenter, and $ R$ the circumradius of an acute-angled triangle $ ABC$. Consider the circles $ k_a,k_b,k_c,k_h,k$, all with radius $ R$, centered at $ A,B,C,H,M,$ respectively. Circles $ k_a$ and $ k_b$ meet at $ M$ and $ F$; $ k_a$ and $ k_c$ meet at $ M$ and $ E$; and $ k_b$ and $ k_c$ meet at $ M$ and $ D$.
$ (a)$ Prove that the points $ D,E,F$ lie on the circle $ k_h$.
$ (b)$ Prove that the set of the points inside $ k_h$ that are inside exactly one of the circles $ k_a,k_b,k_c$ has the area twice the area of $ \triangle ABC$.

actually, M=O
AH//MD and AH=MD  AHDM is a paralleogram

S(EAFBDC)=2S(ABC)


Given trangle $ABC$ with circumcircle $\omega$, let $P$ be an arbitrary point on $\omega$, and let $A'$, $B'$, and $C'$ be the orthogonal projections of $P$ onto $BC$, $AC$, and $AB$ respectively, prove that $A', B',$ and $C'$ are colinear.

The line passing through $A'$, $B'$, and $C'$ is the Simson line. :wink:

[hide]first i want to subtituid $F,E,D$INSTEAD OF $A',B',C'$
$\angle APC+\angle B=180,\angle B+\angle DPF=180$SO$\angle DPF=\angle APC$AND $\angle DPA=\angle FPC$BECAUSE$\angle DPA=\angle DEA , \angle FPC=\angle FEC$THEN $\angle DEA=\angle FEC$THEREFORE $D,E,F$ARE COLINEAR![/hide]


group theory
abstract algebra
modular arithmetic
superior algebra
superior algebra unsolved

[quote="michael.jackson_1958"]show every groups with the order $ %Error. "display" is a bad command.
\ 2p$ ($ %Error. "display" is a bad command.
\ p$ is prime)is homomorphism with one of these groups:$ %Error. "display" is a bad command.
\mathbb{Z}_{2p}$or $ \ D_p$.[/quote]
Every group of order $ 2p$ is isomorphic to either $ \mathbb{Z}_{2p}$ or $ D_p$ (if $ p>2$ - you forgot to mention that  :wink: ).
We know that there exists an element $ a$ of order $ p$ and an element of order $ 2$. Form the subgroup $ \left< a\right>$ which is an index 2 subgroup so it is a normal subgroup. This means the conjugate $ bab^{-1} = bab$ is contained in this group. Thus, $ bab = a^k$ . This means $ b(bab)b = ba^kb=a^{k^2}$, thus $ a=a^{k^2}$. This tells us that $ k\equiv 1 \pmod p$ so $ k=1$ or $ k=-1$. Now depending whether it is $ \pm 1$ you get either $ \mathbb{Z}_{2p}$ or $ D_p$.

The statement is still true if $ p\equal{}2$: the Klein four group is the same as $ D_2$.

We can go further and show that every group of order $ pq$, $ p > q$ both prime, is a semi-direct product of cyclic groups of order $ p$ and $ q$. Clearly there are group of order both $ p$ and $ q$ by Cauchy's theorem. Consider the action of $ G$ on the right cosets of a group $ H$ of order $ p$ by right-multiplication. This gives a homomorphism of $ G$ into $ S_q$, and since $ p$ does not divide $ q!$, the kernel of this homomorphism must contain a group of order $ p$. However, the kernel is not the entire group because the kernel must be contained in $ H$ (since we need $ Ha \equal{} H$ for any $ a$ in the kernel). It follows that the kernel is, in fact, $ H$, so that $ H$ is a normal subgroup. Since there is a group of order $ q$ $ H$ has a complement, so $ G$ is in fact a semi-direct product.


Let $ a,b,c$ be positive numbers such that $ abc\equal{}1$.Prove that:
$ \frac{a}{ab^2\plus{}2b^6}\plus{}\frac{b}{bc^2\plus{}2c^6}\plus{}\frac{c}{ca^2\plus{}2a^6}\geq\ 1$

no one interested in this inequality??? :(  :(

I think it's very easy so nobody want to solve it  :wink:
Let $ a\equal{}\frac{x}{y}$,... and using Cauchy Schwarts, we have:
$ x^{12}\plus{}y^{12}\plus{}z^{12} \geq\ xyz \sum\ x^6y^3$
(obviously trues by Am-Gm)

[quote="nguoivn"]I think it's very easy so nobody want to solve it  :wink:
Let $ a \equal{} \frac {x}{y}$,... and using Cauchy Schwarts, we have:
$ x^{12} \plus{} y^{12} \plus{} z^{12} \geq\ xyz \sum\ x^6y^3$
(obviously trues by Am-Gm)[/quote]
if you put $ a\equal{}\frac{x}{y}$....this ine become:
$ \sum{\frac{z^6x}{xy^2z^4\plus{}2y^7}}\geq\ 1$;
if you use Cauchy-Schwarz,this ine is false :wink:  :oops:  :roll:

[quote="nguoivn"]Oh, sorry, my friend. I'm so hurry, and I saw it to:
Prove that: $ \sum\ \frac {1}{ab^2 \plus{} 2b^6} \geq\ 1$
That's interesting because this ineq is also trues (my proof above is for it)  :D
Now, with your ineq, let $ a \equal{} \frac {y}{x}$,...
It's becomes: $ \sum\ \frac {x^9}{x^7y^2 \plus{} 2zx^2y^6} \geq\ 1$
Using Cauchy Schwarts, we have:
$ \sum\ x^{12} \geq\ \sum\ x^{10}y^2$
So, we only need to prove: $ \sum\ x^6y^6 \geq\ xyz \sum\ x^4y^5$
Let $ a \equal{} yz$;... it's becomes: $ a^6 \plus{} b^6 \plus{} c^6 \geq\ a^5b \plus{} b^5c \plus{} c^5a$ (obviously trues)
We have q.e.d  :)[/quote]
now,your solution is true :wink:


ratio
trigonometry
analytic geometry
geometry
similar triangles

Two congruent circles, $ \omega_1$ and $ \omega_2$, are tangent to each other, and they are both tangent to line $ \over arrow{AB}$ with $ \omega_1$ tangent to line $ AB$ at $ C$ and $ \omega_2$ tangent to line $ AB$ at $ D$. Point $ E$ lies on line $ AB$ such that $ C$ is between $ E$ and $ D$. Point $ F$, distinct from point $ C$, lies on $ \omega_1$ such that line $ EF$ is tangent to $ \omega_1$. Point $ G$ lies on $ \omega_2$ and point $ H$ lies on line $ EF$. Given that $ ED=250$ and that the radius of $ \omega_1$ is $ 75$, compute the minimum possible length of $ \overline{GH}$.

[quote="FightTheTide"]Two congruent circles, $ \omega_1$ and $ \omega_2$, are tangent to each other, and they are both tangent to line $ \over arrow{AB}$ with $ \omega_1$ tangent to line $ AB$ at $ C$ and $ \omega_2$ tangent to line $ AB$ at $ D$. Point $ E$ lies on line $ AB$ such that $ C$ is between $ E$ and $ D$. Point $ F$, distinct from point $ C$, lies on $ \omega_1$ such that line $ EF$ is tangent to $ \omega_1$. Point $ G$ lies on $ \omega_2$ and point $ H$ lies on line $ EF$. Given that $ ED = 250$ and that the radius of $ \omega_1$ is $ 75$, compute the minimum possible length of $ \overline{GH}$.[/quote]

[hide="approach"]
Draw the segment from w2 perpendicular to EF.  The point where that segment intersects w2 is the closest point to EF.
Let's call the point I, and the point where w2I intersects EF, J.

Also use similar triangles a lot.  Try to find a ratio, and see if you can draw new triangles in the right places so there are congruencies and easy ratios.

I couldn't. :(

[/hide]

[quote="FightTheTide"]Two congruent circles, $ \omega_1$ and $ \omega_2$, are tangent to each other, and they are both tangent to line $ \over arrow{AB}$ with $ \omega_1$ tangent to line $ AB$ at $ C$ and $ \omega_2$ tangent to line $ AB$ at $ D$. Point $ E$ lies on line $ AB$ such that $ C$ is between $ E$ and $ D$. Point $ F$, distinct from point $ C$, lies on $ \omega_1$ such that line $ EF$ is tangent to $ \omega_1$. Point $ G$ lies on $ \omega_2$ and point $ H$ lies on line $ EF$. Given that $ ED = 250$ and that the radius of $ \omega_1$ is $ 75$, compute the minimum possible length of $ \overline{GH}$.[/quote]
Diagram:
[asy]size(300);
int r=75;
pair O=(-r,0), P=(r,0), C=(-r,r), D=(r,r), E=(-r*8/3,r), F=tangent(E,O,r), H=2*F-E, G=IP(CR(P,r),H--(P.x+r,P.y),1);
D(CR(O,r)); D(CR(P,r)); D(F--O);
MP("75",midpoint(F--O),NNW);
draw(L(E,D,1/4,5/12),blue,Arrows);
draw(L(E,H,1/4,1/4),blue,Arrows);
draw(L(H,G,1/2,6/5),blue,Arrows);
MP("\omega_1",D(O),SE); MP("\omega_2",D(P),SE);
MP("C",D(C),N); MP("D",D(D),N); MP("E",D(E),NE); MP("F",D(F),SW); MP("H",D(H),WNW); MP("G",D(G),S);[/asy]

G is any point on the circle $ w_2$ and $ GH$ is not necessarily a tangent. This is a pretty nice problem. The way to minimize the distance from a circle to a line is to draw a line through the center of the circle, to the line. Lets see now, $ CD=75+75=150$. So $ EC=100$. That means that \[{ \tan\angle FEC=\frac{2\cdot\frac{3}{4}}1-\frac{3}{4}\cdot\frac{3}{4}}=\frac{24}{7}\]

Then if we affix $ E$ as the origin, $ DE$ as the positive $ x$ axis, and $ HE$ is $ y=\frac{24}{7}\cdot x$, we can use coordinates. The center of $ w_2$ is $ (250,75)$. Then we find the distance from that point to the line and  subtract the radius: \[ \frac{ |24x_0-7y_0+0|} {\sqrt{24^2+(-7)^2}}-r_\omega=\frac{|24*250-7*75|}{25}-75=144\]

We can finish it without coordinates, as well. The distance from $ \omega_2$ to $ EF$ is perpendicular to $ EF$.

The component of $ F\omega_1$ perpendicular to $ EF$ is $ r\equal{}75$, and the component of $ \omega_1 \omega 2$ perpendicular to $ EF$ is $ 2r \sin \angle DEF \equal{} \frac{48}{25} \cdot 75 \equal{} 144$.

Then, the distance from $ \omega _2$ to $ EF$ is $ 75\plus{}144$. We subtract $ 75$ for the radius, and get $ 144$

I realized that the coordinate approach made the problem a lot simpler than I had previously intended. Dang.

[hide="My lazy latexless solution"]
Let x be the measure of angle CEF.
realizing that CE=100, AC=75 and CA bisects CEF, we calculate that sin(x)=24/25, so tan(x)=24/7
Extend BD to meet EF at I. DI=(24/7)(250)=6000/7.
We then realize that minimalizing GH means that line GH passes through B, and that BH is the altitude to EI in triangle EBI. EI=250(25/7).
The area of triangle EBI we can already can calculate: it's (1/2)(250)(6000/7-75)
Also, letting y be our desired distance, we have Area (EBI) = (1/2)(250)(25/7)(y+75)

Setting the two area equations equal to each other, we get a lot of canceling and an eventual answer of y=144[/hide]


As most you [i]may[/i] have heard Pluto is no longer a planet.  The long debated over whether the it should be a planet ended last summer, and the new classifications for a planet and dwarf planets were made.

I say that Pluto should have the status as a planet, but science changes with the times, so the more specific classifications were eventually going to be thought up.
Even so, regardless of the word describing it, "Pluto is Still Pluto".
Crudely quoting Shakespeare, "A rose by any other name would smell as sweet".

Pluto should be dwarfed, it's way too small and never really fit all the standards.

Also, they have found dwarf-planets like Pluto.

Yeah...I found it kind of sad that Pluto was kicked out, but it didn't really fit in with all of the other planets (kind of a conformist message, isn't it? :P ). Plus, all of these other solar system objects--like Ceres and Pluto's largest moon, Charon--would have to be included, and I thought that would be a bit ridiculous after a while.

I did like the "Word of the Year" last year, though: pluto- to devalue or demote someone or something. :lol:

There would just be too many planets if Pluto's traits were the definition of a planet... so, I would have to go with it shouldn't be a planet :(

Hi! I have also read that if Pluto is regarded as a "planet", then so should Charon, because the center of gravity of the system Pluto-Charon lies outside of Pluto. So Charon actually isn't a satellite of Pluto, because they form a system of two bodies rotating around an external center of gravity.

Is this correct? :)

:wink: The people I have heard use that arguement said that since the center of gravity is pretty much between the two planets then you can not say one is a planet and the other is revolving around it.

Yeah, I think Pluto and Charon are considered a binary dwarf planet system.  So all Neptune-year round, it's the eighth planet from the Sun.  No more confusion as to which is the furthest.

Although many people don't agree with the decision of plutoing Pluto (:P) due to its size, composition, and placement, it can't be denied that it doesn't fit the mold of the other outer planets.

Although to be fair, this may be an urban myth, but the IAU apparently demoted Pluto in a session in which only 1/3 of its members were present.

I think Pluto is rightfully dwarfed because of its size. It is not as big as the other planets in the solar system. Also, Pluto is missing some of the things a planet has. A planet must follow a clear orbit, meaning it can not, under any circumstances, run into another object. However, Pluto constantly runs into the Quiper Belt. Besides, pluto is still a planet, but dwafazized.  =)


Joel collects comic books as a hobby. At a comic book convention he bought a Superman comic book and then sold it immediately for 20% more than the amount he paid for it. Next, he bought a Spiderman comic book and sold it immediately for 20% less than its purchase price. He received the same amount for the Superman and Spiderman comic. Assuming that each of the four sales was transacted in modern US currency, which of the following could represent the total Joel paid for the two comic books?
a) 30.30
b) 30.45
c) 30.60
d) 30.75
e) 30.90

I'm really confused with this question...

Let a superman comic book cost $ S_{p}$ dollars, and a Spiderman comic book $ S_{i}$ dollars.  If he sold the Superman comic book for $ 20%
$ more than he bought it, its price now is $ \frac {6S_{p}}{5}$.  If he sold the Spiderman comic book for $ 20%
$ less than he bought it, then he sold it for $ \frac C{4S_{i}}{4}$.  Since he received the same for both of these, their prices are equal.  Hence, you get that $ S_{p} = \frac {2S_i}{3}$.  The total amount he paid, therefore, is $ \frac {5S_{i}}{3}$.  So, for each sum, you can then find the value of $ S_{i}$ and the amount he sold the Spiderman book for.  Notice how the amount has to be divisible by $ 5$ (why).  Once you get the value for $ S_{i}$, THIS ALSO has to be divisible by $ 5$.  So, we are looking for the the answer that is divisible by $ 5^{2} = 25$. Our answer is therefore $ \boxed{D}$.


Prove that a graph is planar if and only if it does not contain a subdivision of $K_5$ or $K_{3,3}$.

A subdivision of a graph $G$ is the graph obtained by from $G$ by replacing some edges with paths.

Well, I don't think it is quite fair to propose such a problem. It is a really famous result but its proof is long, in particular it is boring to write it here... ;) .

I think you should have posted the result in the theorem section.

Pierre.

ok pierre so you should move it.
here is a short proof: http://www.cs.princeton.edu/~ymakaryc/papers/kuratowski.pdf


linear algebra
matrix
superior algebra
superior algebra solved

Here is what I have done( i hope it is good) .. but I haven't continued the idea because it is only good for n=2k+1;

because we can use the formula:
det(X+Y)+det(X-Y)=2(det(X)+det(Y))

observe that we write the relation like so:
det(A+B)+det(A-C(n,1)B)+det(A+C(n,2)B)+...+det(A+(-1)^nC(n,n)B)=0
and with:
det(A+(-1)^nC(n,n)B)+...+det(A+C(n,2)B)+det(A+B)=0
and add them both
and det(B)=1

det(B^2+1)=0 => det(B+iI)det(B-iI)=0=> det(B+iI)=0 or det(B-iI)=0

If det(B+iI)=0 , i is a root of P(x)= det(B+xI)=detB+ax+x^2 with real
coefficients since B is real P(x)=x^2+1 yes Lagrangia detB=1

It is well known that if AB=BA and ( A,B from M_2(R)) with det(A^2+B^2)=0 then det(A)=det(B)

then det(B^2+I2)=i*(-i)=1

isn't it? :?

yeah I see how to prove it
A,B commute then det(A^2+B^2)=0 => det(A+iB)det(A-iB)=0=>
det(A+iB)=0 or det(A-iB)=0 with q(x)=det(A+xB)=detA+kx+detBx^2 we have \pm i root of q(x) => k=0
q(i)=0 => detA-detB=0
Is it like this you prove it ?

This problem is really cute!!! Time to give a solution!
After quite some time spent on this problem here is what I have:

lets define P(x)=det(A+xB); then P(X)=ax^2+bx+c where c=det(A) and a=det(B).

we rewrite the problem like so:

sum(k=0,n) f((-1)C(n,k))=0 and using P(x) we get that det(B)*sum(k=0,n) (C(n,k))^2+det(A)(n+1)+a*sum(k=0,n) (-1)^k C(n,k)=0 so  (n+1)det(A)=-det(B)C(2n,n)=0
I proved that det(B)=1 above so det(A)=-1/(n+1)*(C(2n,n))

where C(n,k)= binomial coeficient
cheers!! :D :D

I think the sol is good! :D