algebra
polynomial
trigonometry
superior algebra
superior algebra unsolved

a) The order as before ;)
Basis: $1, \sqrt{2} , \sqrt{3} , \sqrt{6}$.

b) well... ummm... trivial¿

c) didn't we have that already¿¿¿

d) $\epsilon = \frac{-1 i \cdot \sqrt{3}}{2}$, so not nessary to adjunct anymore. Rest as before.

e) It gets a bit boring to post all the calculat-only-stuff, doesn't it¿

[quote="Cezar Lupu"]Prove that:

  a) \[ [\mathbb{Q}(\sqrt{2},\sqrt{3}): \mathbb{Q}]=4;  \]
      Find a basis in $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$.

  b) \[ \mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3}),  \]
       because $\sqrt{2}+\sqrt{3}$ is not of degree $2$ over $\mathbb{Q}$.

  c) Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$.

  d) \[ [\mathbb{Q}( i, \sqrt{3}, \epsilon): \mathbb{Q}]=4,  \]
       where $\epsilon=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}.$[/quote]

(a) $[Q(\sqrt{2},\sqrt{3}): Q] \leq [Q(\sqrt{2}): Q][Q(\sqrt{3}): Q] = 4$
Yet, we see that $[Q(\sqrt{2},\sqrt{3}): Q(\sqrt{2})] = [Q(\sqrt{3}): Q]$ because $\sqrt{3}$ is not in $Q(\sqrt{2})$. Thus the equality holds.
A possible basis is $\{1,\sqrt{2},\sqrt{3},\sqrt{6} \}$

(b) It suffices to show $Q(\sqrt{2},\sqrt{3}) \subset Q(\sqrt{2}+\sqrt{3}) = K$
$K$ contains $5+2\sqrt{6} \Rightarrow \sqrt{6} \in K$. This further implies $\sqrt{6}(\sqrt{2}+\sqrt{3}) = 2\sqrt{3}+3\sqrt{2} \in K$, which trivially follows $\sqrt{2}, \sqrt{3} \in K$.

(c)$x = \sqrt{2} + \sqrt{3} \\ x^2 = 5 + 2\sqrt{6} \\ x^4 - 10x^2 + 25 = 24 \\ x^4 - 10x^2 + 1 = 0$
The polynomial is minimal because it's degree 4.

(d) $Q(i,\sqrt{3},\epsilon) = Q(i,\sqrt{3})$. By similar argument in (a), $[Q(i,\sqrt{3}): Q] = 4$

Just to give c) http://www.mathlinks.ro/Forum/viewtopic.php?t=89607


Find all $ m \in R$ such that the equation $ \sqrt[4]{1\plus{}x}\plus{}\sqrt[4]{1\minus{}x} \equal{}m$ have solution

[quote="tenniskidperson3"]Let $ a\equal{}\sqrt[4]{1\plus{}x}$ and $ b\equal{}\sqrt[4]{1\minus{}x}$.  We have $ a\plus{}b\equal{}m$ and $ a^4\plus{}b^4\equal{}2$.  Let $ ab\equal{}n$.  We have $ m^4\minus{}4m^2n\plus{}2n^2\equal{}2$.  We factor out the $ m^2$ to get $ 2\equal{}m^2(m^2\minus{}4n)\plus{}2n^2\geq 2n^2$, so $ n\leq 1$.  Also $ n\geq 0$ because it is the product of two nonnegative factors.

Complete the square to get $ (m^2\minus{}2n)^2\minus{}2n^2\equal{}2$.  Now we add the $ 2n^2$ to the other side and manipulate:

$ 2\leq 2\plus{}2n^2\equal{}(m^2\minus{}2n)^2\equal{}2\plus{}2n^2\leq 4$

$ \sqrt{2}\leq m^2\minus{}2n\leq 2$

$ \sqrt{2}\leq\sqrt{2}\plus{}2n\leq m^2\leq 2\plus{}2n\leq 4$

$ \sqrt[4]{2}\leq m\leq 2$.

We can show that both of these are obtainable: if $ x\equal{}1$ then the lower bound is attained.  If $ x\equal{}0$ the upper bound is obtained.[/quote]
 Can let  x=cost  and t in [0;pi/2]


Q)

Five pirates and a monkey are shipwrecked on an island. The pirates have collected a pile of coconuts which they plan to divide equally among themselves the next morning. Not trusting the others, one pirate wakes up during the night and divides the coconuts into five equal parts with one left over, which he gives to the monkey. The pirate then hides his portion of the pile. During the night, each of the other pirates does exactly the same thing by dividing the pile he finds into five equal parts leaving one coconut for the monkey and hiding his portion. In the morning, the pirates gather and split the remaining pile of coconuts into five equal parts and again one is left over for the monkey. What is the smallest number of coconuts the pirates could have collected for their original pile?

 :D

Lets count the backwards, from the last split.
the morning quantity $ 1\plus{}5p_6\equal{}4p_5$
before the $ 5$ pirate $ 1\plus{}5p_5\equal{}4p_4$
before the $ 4$ pirate $ 1\plus{}5p_4\equal{}4p_3$
before the $ 3$ pirate $ 1\plus{}5p_3\equal{}4p_2$
before the $ 2$ pirate $ 1\plus{}5p_2\equal{}4p_1$, $ p_1\equal{}\frac{1}{4}\plus{}\frac{5}{4}p_2$
before the $ 1$ pirate $ 1\plus{}5p_1\equal{}n$

Replacing you get:

$ 1024n\equal{}11529\plus{}15625 p_6$


Yeah Myth, basically I would associate the transpose ${A^T}$ to $\mathsf{T}:\mathsf{P}_n(\mathsf{F})\rightarrow\mathsf{F}^{n+1}$ defined by $\mathsf{T}(f)=(f(c_0),f(c_1),\ldots,f(c_n))$; then $\mathsf{T}$ is an isomorphism by Lagrange polynomials, where $\mathsf{P}_n(\mathsf{F})$ is the set of polynomials with degree less than or equal to $n$ defined on the field $\mathsf{F}$ and $f$ is a Lagrange polynomial


Let $ a\ge0$. Find root of this system: $ \begin {cases}x \plus{} y \plus{} xy \equal{} a^2 \plus{} 2a \\
x^4 \plus{} y^4 \equal{} 2a^4 \end {cases}$

[quote="Alex_ITA"]$ x \equal{} y \equal{} a$?[/quote]

can you show the solution?

[hide="Solution"]
To find the roots of an equation, the equations must first be set to zero. Therefore, we must first solve the system of equations:

$ \begin{cases} a^2\plus{}2a\equal{}0\\ 2a^4\equal{}0\end{cases}$

Therefore, $ a\equal{}0$.

We must now solve the system of equations:

$ \begin{cases} x\plus{}y\plus{}xy\equal{}0\\ x^4\plus{}y^4\equal{}0\end{cases}$

By inspection, the only solution to this system is $ (x,y)\equal{}(0,0)$

Therefore, $ x\equal{}y\equal{}a\equal{}0$[/hide]

since there is symmetry in $ x^4\plus{}y^4\equal{}2a^4$ that means $ x\equal{}y\equal{}a$
because you can change $ x^4$ with $ y^4$ and it wont make a difference and thats the symmetry therefore $ x\equal{}y$.
and if $ x^4\equal{}y^4$ then  $ x^4\plus{}x^4\equal{}2x^4\equal{}2a^4$ again symmetry between $ 2x^4\equal{}2a^4$ therefore $ x\equal{}y\equal{}a$

@JZambrano201 why did you set it to 0 ? where did that come from ?although the solution is correct but i dont get how you equal it to 0

[quote="binomial_4eva"]since there is symmetry in $ x^4 \plus{} y^4 \equal{} 2a^4$ that means $ x \equal{} y \equal{} a$
because you can change $ x^4$ with $ y^4$ and it wont make a difference and thats the symmetry therefore $ x \equal{} y$.
and if $ x^4 \equal{} y^4$ then  $ x^4 \plus{} x^4 \equal{} 2x^4 \equal{} 2a^4$ again symmetry between $ 2x^4 \equal{} 2a^4$ therefore $ x \equal{} y \equal{} a$

@JZambrano201 why did you set it to 0 ? where did that come from ?although the solution is correct but i dont get how you equal it to 0[/quote]

Thank you very much. :P

[quote="azjps"][quote="JZambrano201"][hide="Solution"]
To find the roots of an equation, the equations must first be set to zero. Therefore, we must first solve the system of equations:

$ \begin{cases} a^2 + 2a = 0 \\
2a^4 = 0\end{cases}$

Therefore, $ a = 0$.

We must now solve the system of equations:

$ \begin{cases} x + y + xy = 0 \\
x^4 + y^4 = 0\end{cases}$

By inspection, the only solution to this system is $ (x,y) = (0,0)$

Therefore, $ x = y = a = 0$[/hide][/quote]

Root refers to the solutions of the system (not zeroes; note also that $ a$ is by convention a defined constant).

[quote="binomial_4eva"]since there is symmetry in $ x^4 + y^4 = 2a^4$ that means $ x = y = a$
because you can change $ x^4$ with $ y^4$ and it wont make a difference and thats the symmetry therefore $ x = y$.
and if $ x^4 = y^4$ then  $ x^4 + x^4 = 2x^4 = 2a^4$ again symmetry between $ 2x^4 = 2a^4$ therefore $ x = y = a$

@JZambrano201 why did you set it to 0 ? where did that come from ?although the solution is correct but i dont get how you equal it to 0[/quote]

Symmetry does not work like that; you can change $ x^4, y^4$, but that does not imply that $ x = y$ (notice also that you didn't use both equations).

[hide="Solution"]
For $ x,y \ge 0$, by Power Mean,
\[ \frac {|x| + |y|}{2} \le \sqrt [4]{\frac {x^4 + y^4}{2}} = a \Longrightarrow \boxed{|x| + |y| \le 2a}\qquad (1)
\]
with equality when $ |x| = |y| = a$. By AM-GM,
\begin{align*} \left(\frac {x + y}2 \right)^2 + 2 \left(\frac {x + y}2\right) & \ge x + y + xy = a^2 + 2a \\
\left(\frac {x + y}{2} + 1\right)^2 & \ge (a + 1)^2 \end{align*}
And since we assumed that $ x,y,a \ge 0$, and $ f(m) = (m + 1)^2$ is strictly increasing when $ m \ge 0$, we have $ \frac {x + y}{2} \ge a \Longrightarrow \boxed{x + y \ge 2a}\qquad (2i)$, again with equality when $ x = y = a$. Then $ (1)$ and $ (2i)$ yield $ x + y = 2a \Longleftrightarrow \boxed{\boxed{x = y = a}}$. A similar approach on the negatives (actually, the above works on $ x,y \ge - 1$, and then we consider $ x,y \le - 1$) gives $ \boxed{x + y \le - 2a - 4}\qquad (2ii)$, which shows no other solutions.
[/hide][/quote]
Thank you. :rotfl:


[quote]We are seeing an increased volume of traffic by some malware software. In order to protect our customers from damage from that malware, we are blocking your query. A few legitimate queries may get flagged, and for that we apologize. Please be assured that we are hard at work on this problem and hope to get it resolved even better as soon as possible.
If you are using phpBB, please check out the phpBB downloads site http://www.phpbb.com/downloads.php and make sure you are not vulnerable.
- MSN Search Team[/quote]
After I clicked on the goto page 28 of the necropost game I in the FF.

And then it wouldn't let me post.

I tried that and it worked fine.  I posted a message(then deleted it because it was spam)

Maybe you're using a bad web browser.  I use Chrome, what do you use?

If you look at the message, you would notice it is MSN  :wink:

I never heard of an MSN browser.  I was thinking the MSN was your primary search engine and they found some link error.

The search engine itself is Bing, but our mail and toolbar, etc. are MSN.


Quadrilateral ABCD is inscribed in a circle center O, and $AC \perp BD$.  Prove that $[AOCB]=[AOCD]$.

Thanks

[hide]WLOG assume $AOCD$ is convex. Let $M$ be the midpoint of $BD$. Then $OM\perp BD\implies OM\parallel AC$, which means $[AOCD]={1\over 2}AC\cdot MD={1\over 2}AC{BD\over 2}={1\over 2}[ABCD]$. The conclusion follows.[/hide]

I don't understand your proof, how do you calculate the area of $[AOCD]$, which is a quadrilateral? Thanks...  :blush:

Split it into two triangles - $AOC$ and $ACD$. If you take $AC$ as base, then you need O-altitude of the former and D-altitude of the latter, or to be precise, you just need their sum. It's been proven that $M$ and $O$ are equidistant from $AC$, hence the altitudes of those two triangles add up to $MD$.


Galois Theory
superior algebra
superior algebra unsolved

So while reviewing for my final I was trying to come up with an example of when the fixed field (notated in my book as) $E^{\text{Gal}(E/F)}\neq F$. This is one of the several ways you know that a field extension $F\subset E$ is a Galois extension. So any field extension that isn't Galois should work. So from previous example I know that the splitting field $E$ of $x^{4}-2x^{2}-2$ in $\mathbb{C}$ over $\mathbb{Q}$ isn't Galois since $E=\mathbb{Q}(\sqrt{1+\sqrt{3}}, \sqrt{1-\sqrt{3}})$ and $8=\big|\text{Gal}(E/\mathbb{Q})\big|\neq [E: \mathbb{Q}]=4$. It seems to me though that $E^{\text{Gal}(E/\mathbb{Q})}=\mathbb{Q}$, does anyone know what else is fixed so that this isn't equal?

If you want more info without having to work it out $\text{Gal}(E/\mathbb{Q})\cong D_{4}$. Thanks.

Edit: Actually I just realized that probably $\sqrt{3}$ is fixed. So $\mathbb{Q}(\sqrt{3})\subseteq E^{\text{Gal}(E/\mathbb{Q})}$ at least. Is this correct?

Splitting fields over perfect fields like $\mathbb{Q}$ are always Galois extensions ;) (and $[E: \mathbb{Q}]=8$)
But how do you define Galois groups if the bigger field is not a Galois extension¿ (there are multiple ways)

So the reason I couldn't find any easy examples was because I was considering only splitting fields over $\mathbb{Q}$.

Hmmm...so going a different route $\text{Gal}(\mathbb{Q}(2^{1/3})/\mathbb{Q})\cong \{e\}$, so this definitely isn't Galois because $\mathbb{Q}(2^{1/3}) \ncong \mathbb{Q}$. Then for the fixed field part of the example $E^{\text{Gal}(\mathbb{Q}(2^{1/3})/\mathbb{Q})}= \mathbb{Q}(2^{1/3})\neq \mathbb{Q}$ since the Galois group only contains the identity mapping.

Thanks, for some reason I wasn't really thinking along these lines.


A set of marbles can be divided in equal shares among 2, 3, 4, 5, or 6 children with no marbles left over.  What is the least number of marbles that the set could have?

[hide]LCM of the numbers is 60.[/hide]
I quit for now  ;)
all 35 topics started by you math92.


In this latex: [code]\documentclass[16 pt]{article}
\pagestyle{empty}
\begin{document}

What number is less than 35^2, greater than 34^2, divisible by 19, and a multiple of 9?

\vspace{28\baselineskip}

Alfreda hiked half her planned trip the first day; tiring, she only covered one-third the remaining distance the second day; rain slowed her to one-forth the remaining distance the third day, and the fourth day she hiked three miles, which was one-fifth of what she had left.  How many miles did she hike during the four days?

\end{document}[/code]
How do you increase the font size.  I tried changing the text size up top, but to no avail.

Try these ones:

\tiny tiny font
\scriptsize very small font
\footnotesize quite small font
\small small font
\normalsize normal font
\large large font
\Large larger font
\LARGE very large font
\huge huge
\Huge largest

Best,


Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(n+1) > f(f(n)).\]

The first step is to show that f(1) < f(2) < f(3) < ... . We do this by induction on n. We take $ S_n$ to be the statement that f(n) is the unique smallest element of { f(n), f(n+1), f(n+2), ... }.

For m > 1, f(m) > f(s) where s = f(m-1), so f(m) is not the smallest member of the set {f(1), f(2), f(3), ... }. But the set is bounded below by zero, so it must have a smallest member. Hence the unique smallest member is f(1). So $ S_1$ is true.

Suppose $ S_n$ is true. Take m > n+1. Then m-1 > n, so by $ S_n$, f(m-1) > f(n). But $ S_n$ also tells us that f(n) > f(n-1) > ... > f(1), so f(n) ≥ n - 1 + f(1) ≥ n. Hence f(m-1) ≥ n+1. So f(m-1) belongs to { n+1, n+2, n+3, .. }. But we are given that f(m) > f(f(m-1)), so f(m) is not the smallest element of { f(n+1), f(n+2), f(n+3), ... }. But there must be a smallest element, so f(n+1) must be the unique smallest member, which establishes $ S_{n \plus{} 1}$. So, $ S_n$ is true for all n.

So $ n \leq m$ implies $ f(n) \leq f(m)$. Suppose for some m, $ f(m) \geq m\plus{}1$, then $ f(f(m)) \geq f(m\plus{}1)$. Contradiction. Hence $ f(m) \leq m$ for all m. But since $ f(1) \geq1$ and f(m) > f(m-1) > ... > f(1), we also have f(m) ≥ m. Hence f(m) = m for all m

let $x_1$ such as that $f(x_1)$ be the minimum of $f(N) \implies f(x_1) > f(f(x_1-1))$ which is a contradiction if $x_1 \ne 1$

so $x_1=1$ and the function is injective for $f(1)$ which means if $f(a)=f(1) \iff a=1$. now take the 2nd smallest

value of $f(N)$ and assume it occurs for $x_2$ then $x_2>1\implies f(x_2)>f(f(x_2-1))$ which implies that

$f(f(x_2-1))=f(1)$ since $f$ is injective for $f(1)$ then

$f(x_2-1)=1 \le f(1)$ and we also know that $f(1)\le f(x_2-1)$ then $f(1)=1$ and $x_2=2$ and now we shall utilize

induction in order to prove the statement,

statement: $f$ is injective for $A=\{f(1),f(2),...,f(n)\}$ and it reaches its $i^{th}$ smallest value

with $i$ and $\forall i\in A -\{f(n)\}:f(i)=i$

we have already proved the it for $n=2$, now for the induction step, let $x_{n+1}$ be such as that

$f(x_{n+1})$ be that $(n+1)$th smallest values in $f(N)$.

so $f(x_{n+1}) > f(f(x_{n+1}-1)) \implies \exists k, 1\le k \le n:  f(f(x_{n+1}-1))=f(k)$

$\implies x_{n+1}=k+1$ and if $k \le n-1$ we can see the contradiction with induction hypothesis so

$k=n \implies x_{n+1}=n+1$

$f(x_{n+1}-1) =f(n) \ge n$ and $f(n+1) > f(f(n))$ so $f(f(n))=k$ for some $k$ in $\{1,2,...,n\}$ and if $k \ne n$

then it will be in contradiction with $f(n) \ge n$ so $f(f(n))=n \implies f(n)=n$


My "friend" asked me to help him on a problem (which is kinda odd, seeing that I rarely talk to this person). This site is awesome, btw.

Anyway,

1) Solve $xr-xm=z+y-i$, $zy-zi=x+r-m$, $a=r-m$, $b=y-i$, for $(x,z,a,b)$ so that ${x,z,a,b}$ are positive integers.

[quote="theawesomest23"]My "friend" asked me to help him on a problem (which is kinda odd, seeing that I rarely talk to this person). This site is awesome, btw.

Anyway,

1) Solve xr-xm=z+y-i, zy-zi=x+r-m, a=r-m, b=y-i, for (x,z,a,b) so that {x,z,a,b} are [b]positive integers[/b].[/quote]

[quote="Treething"]substituting in gets you the following equations:

xa = z + b
zb = x + a

Now, if z and b are both greater than or equal 2, then zb is greater than or equal to xa. Similarly, if x and a are both greater than or equal to 2, xa is greater than or equal to zb. This obviously doesn't work out unless xa = zb, meaning a, b, x, and z are all equal to 2.

(2, 2, 2, 2).[/quote]

What about x=1, a=5, z=2, b=3
or
x=2, a=3, z=1, b=5
or
x=3, a=2, z=5, b=1

etc.

I think I've seen this type of problem somewhere. Forgot if I ever solved it.

[hide]The equations are equivalent to: \[\left\{\begin{array}{l}ax = z+b \\ bz = x+a\end{array}\right.\]Sum them up to get $(a-1)x+(b-1)z = a+b$.
If $a=1$, then $x = z+b$ and $bz = x+1 = z+b+1$, so $(b-1)(z-1) = 2$, so $(b,z) \in\{(2,3),(3,2)\}$ (Because $b-1$ and $z-1$ are non-negative divisors of $2$.) You'll get $x=5$. So $(a,b,x,z) \in\{(1,2,5,3),(1,3,5,2)\}$ are the solutions when $a=1$.
If $b=1$, $x=1$, $z=1$ you'll get (because of the symmetry) the following solutions: $(a,b,x,z) \in\{(2,1,3,5),(3,1,2,5),(5,2,1,3),(5,3,1,2),(2,5,3,1),(3,5,2,1)\}$. So now suppose that $a,b,x,z \geq 2$.
We know that $(a-1)x \geq 2(a-1) \geq a$. But $b-(b-1)z = (a-1)x-a \geq 0$. Therefore, $b \geq (b-1)z$. But $(b-1)z \geq 2(b-1) \geq b$, so we should really have that $b = (b-1)z = bz-z = x+a-z$, so $ax = x+a = z+b = zb$. Rewrite those equations as $(a-1)(x-1) = 1$ and $(b-1)(z-1) = 1$ to see that $(a,b,x,y) = (2,2,2,2)$.
And we're done.[/hide]

hmm, didn't consider those cases, i figured testing where all four variables are 1 or 2 would work.


inequalities
function
calculus
derivative
algebra solved
algebra

Prove that if x,y,z>0 satify xy+yz+zx=1 then
 sum 1/(x^3+x)>=9sqrt(3)/4.

[i]unknown source[/i]

Put x = cotg A, y = cotg B, z = cotg C with A + B + C = 180.

Then 1/(x3 + x) = (cotg A . (cotg2 A + 1))-1 = tg A sin2 A = (sin3 A)/(cos A).

The result follows by applying Jensen on the convex function

f(x) = (sin3 x)/(cos x).

Am I right?

I hope that function is convex-I really don't want to find the second derivative. Now try to prove the inequality without any derivatives. That's the interesting part.

Actually the function is only convex on [0, pi/2).  So you can't apply Jensen's.

Hi all,

Valiowk was right. The second derivative of sin(x)^3/cos(x) is equal to

  6 sin(x) cos(x) + 4sin(x)^3/cos(x) + 2sin(x)^5/cos(x)^3

This inequality can be proven without calculus. I know that Harazi like the sum x^3 + x, so I find following solution:

By Cauchy-Schwarz
Sum (1/(x^3+x)).Sum((x^2+1)/x) >= (Sum 1/x)^2
=> Sum (1/(x^3+x)) >= (Sum 1/x)^2/Sum((x^2+1)/x) = 1/xyz(xyz(x+y+z)+1).

Now, it easy to show that xyz <= sqrt(3)/9, xyz(x+y+z) <= 1/3 and we are done.

Namdung

I indeed like sum x^3+x, but my approach was different:
   Consider E=sum 1/(x+x^3). Then xyzE=sum yz/(x^2+1)=sum yz/(x+y)(x+z)=1-2xyz/(x+y)(y+z)(z+x)>=3/4. Also, use the fact that xyz<=1/3sqrt(3).

My solution is like Harazi solution.

We can rewrite the inequality as

sum zy/(x+y)(x+z) \geq (9 \sqrt 3/4)xyz

but 1=(xy+yz+zx)^3 \geq 27(xy)(yz)(zx) so xyz \leq 1/3 \sqrt 3

so it sufficient to prove that

sum zy/(x+y)(x+z)  \geq 3/4

Supposing x \geq y \geq z we have x/(y+z) \geq y/(x+z) \geq z/(x+y)

we can apply Cheybechev

sum zy/(x+y)(x+z)  \geq 1/3[sum x/(y+z)]^2=(1/3)(9/4)=3/4

I hope it is correct.


geometry
perpendicular bisector
geometry proposed

Let $ ABC$ be a triangle with $ \angle ABC\equal{}\angle ACB\equal{}70^{\circ}$. Let point $ D$ on side $ BC$ such that $ AD$ is the altitude, point $ E$ on side $ AB$ such that $ \angle ACE\equal{}10^{\circ}$, and point $ F$ is the intersection of $ AD$ and $ CE$. Prove that $ CF\equal{}BC$.

The problem is obvious,because $ \triangle ABC$ is isosceles,hence $ AF$ is a perpendicular bisector,so $ CF\equal{}FB$,but $ \angle FCB\equal{}60$,thus $ CF\equal{}FB\equal{}CB$,we are done.

we get $\angle DFC=30,\angle DCF=60$,hence $2.CD=CF$,but $D$ is the midpoint of $BC$.hence done.

angle ABC= angle ACB=70°
SO, AB=AC, AD  is altitude from A onto BC
So, ∆ABD≈∆ACD
So, BD=CD
angle BAD= angle CAD=20°
angle ACE=10°
So, angle CFD=20°+10°=30°
So, in ∆FDC, sin (angle CFD)=CD/CF
sin 30°=CD/CF
So, CF= 2CD=BC
@ KRISHIJIVI


How do I modify my title? For example, my language teacher requires a heading like this:
                                                                  ThinkFlow
                                                                  Tuesday, December 16th
                                                                  Language, period 4
in the top-right corner. Naturally, I also don't want author information under my title.
Thanks,

ThinkFlow

You don't get an author if you put \author{} before \maketitle
But you don't say what document class you are using or whether you need a title page or not.
If you aren't using a separate title page, what is wrong with making your own title

\begin{flushright}
ThinkFlow \\
Tuesday, December 16th \\
Language, period 4
\end{flushright}

and not using \maketitle?

Sorry for not giving the class and more detail. I am using the article class, and I do not want a separate title page. My teacher insists that the title be in the format that I wrote, but left justified, in the tip-right corner. I was wondering if this was possible. Thanks.

\begin{flushright}
\parbox{4cm}{
ThinkFlow \\
Tuesday, December 16th \\
Language, period 4}
\end{flushright}

Change the 4cm to whatever works for your document.


The incircle of triangle $ ABC$ is tangent to $ BC,CA,AB$ at $ D,E,F$. Suppose the incircle intersects $ AD$ again at a point $ X$, and $ AX\equal{}XD$. $ XB,XC$ intersect the incircle again at points $ Y,Z$. Show that $ EY\equal{}FZ$.

It's the same problem as the one proposed at Ibero America 1995 with the diference that there they used "$ AX\equal{}XD$ if and only if $ EY\equal{}FZ$".

We have that $ DE$ is the polar of the point $ C$ with respect to the incircle of the triangle $ ABC$.Denote $ \{T\}\equal{}DE \cap CX$, therefore we have that the division $ (XZTC)$ is harmonic .Hence the pencil $ E(XZTC)$ is harmonic.Intersecting the line $ AD$  with the harmonic pencil we obtain that the division $ (ADXZ')$ is also harmonic , where $ \{Z'\}\equal{}AD\cap EZ$.I wil use the original statement "$ AX\equal{}XD$ if and only if $ EY\equal{}FZ$".
Therefore $ AX\equal{}XD$ if and only if the division $ (ADX\infty)$ is harmonic , but we proved that $ (ADXZ')$ is also harmonic, so it is equivalent to $ (ADX\infty)\equal{}(ADXZ')$ if and only if $ AD\cap EZ\equal{}\infty$ meaning $ AD\parallel EZ$.
Similar we obtain that $ AX\equal{}XD$ if and only if $ AD\parallel FY$.
So $ AX\equal{}XD$ if and only if $ EZ\parallel FY$ wich is equivalent to $ EY\equal{}FZ$ (because the trapezium $ EFYZ$ is isoscele)

Q.E.D

Right. That's a nice solution.  :)

What about not using polars and projective geometry?

I think I must have misunderstood something.
Andrei, the converse of this problem doesn't seem to be true; if $ ABC$ is equilatrel, $ AX\equal{}XD$ doesn't hold but $ EY\equal{}FZ$ does, by symmetry. Where have I gone wrong?

You are wright epitomy01!From  the statement $ AX\equal{}XD$ we conclude that $ EY\equal{}FZ$, but if the traingle $ ABC$ is isoscel with $ AB\equal{}AC$ we have that $ EY\equal{}FZ$ but not necessarily $ AX\equal{}XD$.So it is essential to take $ AB$ not equal to $ AC$ if we want to have $ AX\equal{}XD \Leftrightarrow EY\equal{}FZ$.
Thank you for your remark :D


How to prove the limit

$ \pi = [\lim_{n\to \infty}\frac {2{}^2{}^n\cdot(n!)^2}{(2n)!\cdot\sqrt {n}}]^2$

Thanks!

It's a direct consequence of Stirling's approximation $ n! \sim n^n e^{\minus{}n}\sqrt {2\pi n}$


let ABCD be a quadrilateral with an incircle .then the incircles of the triangles ABC and CDA are tangent.


LaTeX
superior algebra
superior algebra unsolved

Let $E$ be a field such that $\mathbb{Q} \subset E \subset \mathbb{C}$ and $[E : \mathbb{Q}] = 2$.

Show that there exists $d \in \mathbb{Z}$ such that $E = \mathbb{Q}(\sqrt{d})$.

A nice and fundamental problem.

let a be an element in $E \backslash \mathbb{Q}$
then  1 and a form a basis of E over $\mathbb{Q}$ (as the degree is given to be two)
now $a^2$ is in E too, so  it has to be of form $a^2=p a+q$ with p and q rational
so a is of form $\frac{\sqrt{p^2-4q}+p}{2}$ where $p^2-4q$ is no rational square

now it is clear that $\mathbb{Q}(a)$ is part of $E$
however, $\mathbb{Q}(a)=\mathbb{Q}(1,a)$ and therefore is $E$

now it is easy to check that $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{p^2-4q})$

now all there is left to prove is that every $\mathbb{Q}(\sqrt{a})$ with a rational nonsquare, is of form
$\mathbb{Q}(\sqrt{d})$ with d integer

this is easy if you note that $\sqrt{\frac{p}{q}}=\frac{ \sqrt{p q}}{q}$

actually you can even make the theorem stronger by demanding that d is squarefree

[quote="fredbel6"]let a be an element in $E\Q$[/quote]

This is a detail, but you surely mean $E \backslash \mathbb{Q}$, right? :)

(LaTeX hint: use \mathbb{Q} to get $\mathbb{Q}$, the same works for $\mathbb{R}$, $\mathbb{C}$, ...)

I actually wrote that but my LaTeX use was bad.

Thanks for the tips Arne

There is a generalisation however : if $[ E : K ]$= 2 and the characteristic of $K$ is not two, then
$E=K(a)$ with $a^2$ in $K$


articles blogs ratio percent geometry perimeter MATHCOUNTS

Obviously, this is not the forum for me since you folks are all pretty high level math experts.
I'm an adult, 29 years old, who never placed much emphasis on formal education. I need to start working on improving my math skills. I've purchased some childrens math workbooks and as best I can tell, my math skills are on about a 5th or 6th grade level.

I desperately want to improve this.

Does anyone know of any forums that deal with remedial adult math skills?
Can anyone suggest any workbooks, programs or plans that I can use to help bring my skill level up?
Can anyone suggest any strategies for someone like me to eventually be capable of college level math?

Thanks for any help

Stupid in Florida

Heh. First: don't call yourself stupid. I don't know the reason that you don't know much math beyond middle school level, but I wouldn't be surprised if it had more to do with the atrocious math "education" that you were subjected to when you went through school.

Second: this forum is pretty unconventional in terms of math education, but it's one of the best math forums online. This community is oriented around sharing knowledge and talking about math. There are very bright students who will help you with things (they may even be trying to learn the same things you are; who cares if they're way younger?), coaches and parents of students who do math competitions at all levels (middle school through college), and math teachers and professors. No matter what you're trying to do, there will be both peers and mentors here for you.

Third: Why should learning math be different according to how old you are when you do it? Kumon packets are good for rote practice; I actually learned how to do all math up through basic algebra by doing Kumon packets on my own. If you keep to the timing and accuracy requirements that they give you, you'll probably improve a fair amount at your basic math skills. It's important to have good fundamentals (arithmetic and manipulation of numbers) before you start trying to understand abstract maths concepts. My dad always talks about how hard it is to teach algebra to the 9th graders at his school because the middle school classes fail to teach them decimals and fractions and stuff well enough.

I'm no expert on math education, so I'm not going to say anything more. Good luck! Don't treat learning math as a painful thing that you must put yourself through in order to become a more successful person. You can actually enjoy it and still get the same benefits too!

One thing you mustn't do is to go for workbooks.
Check this out (this was posted in the forum by somebody else:
http://www.maa.org/devlin/devlin_03_08.html

What you need is a mathematician friend, with enough passion for the subject to bring it alive to you (and make sure he reads that article too before he starts on you  :) )

We at the AoPS are always ready to help whenever you're stuck somewhere...

Try taking online classes that appeal to you. Or go to your local bookstore and buy some books on the math topics your trying to learn and ask a friend to help you out in the learning. Remember your never to old to start learning new things (lol thats so cheesy, probably shouldn't say that though   :|) Good luck!  :D

First thing: back away from the workbooks and learn your concepts. Workbooks are good but getting a method down, but don't subject yourself to a million and one problems. Who needs a repeat of "formal education", anyways?

We're all here for you--and we've all got different ways to teach things. Chances are, you'll find a way that works for you.

You should also make sure to go to link as posted above--the one that leads to Lockhart's Lament. It's fantastic. (It's what I've been thinking ever since I entered pre-school...:P)

As other people have said, thinking you can't do it will never do the job. Think positively, even post a million times in your blog that "I can do it" will help you progress. I'd focus on one small goal, like doing one problem without referring to the textbook or using manipulatives. AoPS books can help you do the job. Even getting a patient private tutor will work, even if it costs money. I wouldn't recommend it, though, and still try to learn from self. If you do that, you can achieve almost anything by yourself. (Unless you need two people to do it).

Can anyone suggest beginners literature?

It seems that everything I'm finding is either too basic (1+1 = 2) or too advanced (using symbols that I don't even know what they are). Sadly, all of my friends aren't qualified to help me with math. I love them as people, but I fully acknowledge that they aren't the smartest people around and I don't want to live my life like them, thus a part of my desire to get better at math.  : :)

[quote="WantToGetBetter"]It seems that everything I'm finding is either too basic (1+1 = 2) or too advanced (using symbols that I don't even know what they are). Sadly, all of my friends aren't qualified to help me with math. I love them as people, but I fully acknowledge that they aren't the smartest people around and I don't want to live my life like them, thus a part of my desire to get better at math.  : :)[/quote]

I hope I am not offending you in any way, but I am an elementary student and considered a genius in my class, but I seem just like you. Looking at everyone else I don't want to end up like them. I know the basics of what a standardized fifth grader or very very standard sixth grader should know. With everything shifting around in my life, I got to wait until it settles down (hopefully Spring Break will do the trick). Just try going to the library and renewing those math textbooks for 7th or 8th graders.

[quote="WantToGetBetter"]It seems that everything I'm finding is either too basic (1+1 = 2) or too advanced (using symbols that I don't even know what they are).[/quote]
We could probably direct you to something if you would help us figure out where you are. So, tell us which things in this list you are comfortable with and which ones cause you trouble:

1) Addition and subtraction of large numbers
2) Multiplication of large numbers
3) Long division
4) Manipulation of fractions (arithmetic with fractions)
5) Decimal representations of fractions
6) Ratios and percents
7) Word problems -- understanding how to use the methods from 1-6 to solve real-world math problems.
8) Basics of geometry -- the basic shapes and how to find their areas, perimeters, etc.
9) Using variables (having problems with letters in them to represent unknown numbers) and solving linear equations (like $ 5x \plus{} 7 \equal{} 22$)

If you can already do all of this, you probably want to look for a good intro algebra or geometry book (if you're willing to spend money, check the bookstore on this site!) or some guides online. Someone who's involved in MATHCOUNTS/AMC8 coaching might know good resources for that stuff.

[quote="Xevarion"]Someone who's involved in MATHCOUNTS/AMC8 coaching might know good resources for that stuff.[/quote]

Well, I'm not even an assistant coach for MathCounts yet, bu next year I will be.

Check out all the Introductory books here: http://www.artofproblemsolving.com/Books/AoPS_B_Texts_Intro.php

McGraw Hill taught me long multiplication and division pretty well, and MathCounts helped me get through simpler multiplications a lot faster. http://www.teachchildren.com/specmatcol7.html   I can't find 8th grade, though.

And search around this forum, you might get a lot more resources.

I've always thought that thinking visually helped me nail arithmetic and basic algebra skills. I imagine numbers as existing on the number line or as "bars" corresponding to their size... that's kind of vague, but I'd be happy to try to explain in more detail what's worked for me if you contact me by PM (or email or AIM? I forget whether I made those available) with more specifics on where you stand currently.

[quote="Xevarion"]

1) Addition and subtraction of large numbers
2) Multiplication of large numbers
3) Long division
4) Manipulation of fractions (arithmetic with fractions)
5) Decimal representations of fractions
6) Ratios and percents
7) Word problems -- understanding how to use the methods from 1-6 to solve real-world math problems.
8) Basics of geometry -- the basic shapes and how to find their areas, perimeters, etc.
9) Using variables (having problems with letters in them to represent unknown numbers) and solving linear equations (like $ 5x \plus{} 7 \equal{} 22$)

[/quote]

I'm pretty comfortable up to #2, I drop off very quickly at #3. Oddly enough, I'm pretty good at word problems and I totally understand the concepts behind Algebra, it's just that my fundamentals are so sorely lacking that I constantly "hit a wall" whenever I try to solve a problem.  :maybe:

I suggest that you try to find a good explanation of how to do long division and how to manipulate fractions. Then practice. A LOT. Especially with fractions. Understanding what fractions represent and how they work is really fundamental to arithmetic and very important for everything that comes later.

As for a reference -- I'm afraid I can't provide a good one. I'm not familiar with maths education resources except at college level. I am sure someone else here will be able to help, though.

It sounds like you were never taught the reasons behind why long division works. This might be a good intuitive way to understand it:

Let's say I want to divide 1845 by 15, or in other words to find the number of 15s that I can add to get 1845. This is the same as finding the number of times I can subtract 15 from 1845 until I reach zero.

At this point, draw two columns, the left one to tally how many times you have subtracted 15, and the right one to keep track of what number you're on. For example, you start with no marks in the left column and 1935 in the right column. After you subtract 15 once, you would add a tally to the left column and replace 1845 with 1830 in the right column (but don't do this yet). After doing this a bunch of times, the right column would be zero, and there will be a number of tallies in the left column that corresponds to how many subtractions you made.

Now, subtracting 15s one by one would be a very tedious process, so we might try to do things more efficiently. Imagine what would happen if we subtracted 15 one hundred times in a row. Since 15*100 = 1500, that would be the same as subtracting 1500. So instead of doing this a hundred times, you can just make a symbol in the left column that means "100 tallies" and subtract 1500 from the right column. Similarly, you can make a (different) symbol for "10 tallies" that corresponds to subtracting 150.

Then the process for the number 1845 might go something like this:

Left column: 0 tallies, Right column: 1845
Left column: 100 tallies, Right column: 345 (after doing 100 subtractions at once)
Left column: 110 tallies, Right column: 195 (after doing 10 subtractions at once)
Left column: 120 tallies, Right column: 45 (after doing another 10 subtractions at once)
Left column: 121 tallies, Right column: 30 (after doing one subtraction)
Left column: 122 tallies, Right column: 15 (after doing one subtraction)
Left column: 123 tallies, Right column: 0 (after doing one subtraction)

As you can see, at each step you try to do as many subtractions at once as possible. Since you had to subtract 15 a total of 123 times to get from 1845 to 0, you know that 1845 divided by 15 equals 123. Once you understand this process, you can read about how to do long division, which does essentially the same thing but is more efficient and less intuitive.

Long division is easier to understand with polynomials in my opinion. Besides, OP just wants basics so I think it's ok to just learn it as a mechanical process.

By the way, in this context (middle school type mathematics), if you wanted to divide 1346 by 15 would you still begin by removing 1500 and accepting getting a negative number in the next step, or would you remove a smaller multiple of 15?

Sometimes it helps to just surf the web for math advice or even look at online textbooks...[url]http://www.classzone.com/cz/index.htm[/url] this is a really good site for beguinners and advanced people alike...Just PM me if you need a password and username... I might be able to help..small chance though..

It seems like Kumon might help. I've grade some papers and it's repetitive tedious work but people actually learn things (to some extent). While grading the papers, the people really seem to learn it after a while.

A read the book "Math Smart Jr." a while back when I was learning. I though it was pretty good, it tried to explain concepts with stories about some fictional kids and what they were doing, whether it was making lemonade or pizza - the stories might be kinda kiddy-like, but it helped me to understand the basics. I don't remember if it had exercises in it, you'll probably want a workbook to practice with, but you might want to look into it. Or the version for older people might be more suitable, Math Smart, without out the stories.


Find all natural numbers $ \ge10$, such that if we remove the second digit, the resulting number divides the original number.

Find all natural numbers $ \ge10$, such that if we remove the third digit, the resulting number divides the original number.

I'll just give you an idea for the first part - essentially, the second part seems to be similar, but I haven't really looked at it closely, so I might be wrong. Seems like it, though.

Let $ t$ be that number. Then, $ t \equal{} 100a \plus{} 10b \plus{} c$, $ a\geq 0$, $ 0\leq b,c\leq 9$. When we remove the second digit, we get $ 10a \plus{} c$. So, $ 10a \plus{} c|100a \plus{} 10b \plus{} c$ ($ 100a \plus{} 10b \plus{} c \equal{} k(10a \plus{} c)$). Since $ b$ is small, if $ a > 0$ you can clearly limit $ k$ to, in the worst case, $ 19$. Also, you can limit it on the other site, to, say, $ 6$. I'm sure much better limitations are possible. So, either $ a \equal{} 0$ and then you have a two-digit number (which you can handle in any number of ways) or $ k$ is limited, which you can handle, if not anything else, case-by-case (some of them will probably go away in an instant due to divisibility).

Here, I think that's the general idea. Of course, probably a better and much shorter solution exists.


Prove that if you choose 51 numbers from 1-100, there will always be two such that the absolute value of their difference is 50.

[hide]We can pair up 1-51, 2-52 etc. That is 50 pairs, by pigeon hole we get at least one pair of numbers into our 51. The pair is 50 apart.[/hide]

I put yours in spoiler to give others a chance, if you don't mind.


Let $ a_1,a_2,...$ be an infinite sequence of positive real numbers such that $ a_n < a_{n \plus{} 1} \plus{} a_{n^2}$ for all positive integers $ n$.

Show that the sequence $ a_1,a_1 \plus{} a_2,a_1 \plus{} a_2 \plus{} a_3,...$ is unbounded.

??? But if the $ a_i$'s are positive integers then, for all $ n$, we have $ a_1\plus{} \cdots \plus{}a_n \geq n$ hence result.... :wink:

Pierre.


geometry
rectangle
Asymptote
inequalities
similar triangles

Two congruent rectangles are positioned in a right triangle of legs $ a$ and $ b$ as shown. Both rectangles have a vertex on the hypotenuse and a long side on a leg. If $ a$ and $ b$ vary and the short side of each rectangle is $ kb$ for real numbers $ k$, compute the largest possible value of $ k$.

How do you use asymptote in the forum? I haven't tried to use it here before...

[quote="aufha"]Two congruent rectangles are positioned in a right triangle of legs $ a$ and $ b$ as shown. Both rectangles have a vertex on the hypotenuse and a long side on a leg. [b]If $ a$ and $ b$ vary[/b] and the short side of each rectangle is $ kb$ for real numbers $ k$, compute the largest possible value of $ k$.

How do you use asymptote in the forum? I haven't tried to use it here before...[/quote]

Vary...inversely?  Directly?  Actually change values in the given problem? :D

They actually just say vary as in you can change a and b but there is always a largest possible value of k. Or rather, a and b are variables, not constants.

This is a maximization problem I think... The solution mentions the inequality $ \frac{a}{b}\plus{}\frac{b}{a} \ge 2$ where a and b are positive.

[hide]
Let the rectangle have dimensions $ m \times n$, $ m \le n$. Then by similar triangles, we have $ b \equal{} n \plus{} m \plus{} \frac {mn}{n \minus{} m}$, and so (using some manipulations) $ \frac {1}{k} \equal{} \frac {b}{m} \equal{} 1 \plus{} \frac {n}{m} \plus{} \frac {n}{n \minus{} m} \equal{} 1 \plus{} \frac {n \minus{} m}{m} \plus{} 1 \plus{} \frac {m}{n \minus{} m} \plus{} 1$ $ \equal{} 3 \plus{} \frac {m}{n \minus{} m} \plus{} \frac {n \minus{} m}{m}$ $ \ge 3 \plus{} 2\sqrt {\frac {m}{n \minus{} m} \cdot \frac {n \minus{} m}{m}} \equal{} 5$ where the inequality follows from [[AM-GM]]. Thus $ k \le \boxed{\frac {1}{5}}$, with equality when $ m \equal{} n \minus{} m \Longrightarrow n \equal{} 2m$.

[asy]real m = 2, n = 4, b=n+m+(m*n)/(n-m); pointpen = black; pathpen = black + linewidth(0.7);
D((0,0)--(0,m)--(n,m)--(n,0)--cycle); D((0,m)--(0,n+m)--(m,n+m)--(m,m)--cycle); D((0,m)--(0,b)--(n,m));
MP("m",(0,m/2),W); MP("n",(0,(2*m+n)/2),W); MP("n-m",(3m/2,m)); MP("m\left(\frac{n}{n-m}\right)",(0,(b+m+n)/2),W);[/asy]
where the Asymptote code is:
[code][asy]
real m = 2, n = 4, b=n+m+(m*n)/(n-m); pointpen = black; pathpen = black + linewidth(0.7);
D((0,0)--(0,m)--(n,m)--(n,0)--cycle); D((0,m)--(0,n+m)--(m,n+m)--(m,m)--cycle); D((0,m)--(0,b)--(n,m));
MP("m",(0,m/2),W); MP("n",(0,(2*m+n)/2),W); MP("n-m",(3m/2,m)); MP("m\left(\frac{n}{n-m}\right)",(0,(b+m+n)/2),W);
[/asy][/code]

[/hide]

Wait.. so what was your main strategy and motivation for your steps? Was there a reason to believe that AM GM or something like that would have shown up?


The first is not true. :P See $x=y=1,z=0$ :P

The second is also wrong, see $x=y=1,z=3$.

[quote="mathboy_cntHDVN"][quote]The first is not true. :P   See   $x=y=1,z=0$ :P [/quote]
oh , $x,y,z>0$  :maybe:[/quote]
You can let $x=y=1, z \rightarrow 0$ :P.

[quote="toanhocmuonmau"]You can let $x=y=1, z \rightarrow 0$ :P.[/quote]
Careful here. We have $z>0$ so $z\rightarrow 0+$.

[quote="gemath"]That's right, but the following is true
$(x+y+z)^{6}\ge 27(x^{2}+y^{2}+z^{2})(xy+yz+zx)^{2}$ for all $x,y,z>0$  :D and I find a nice solution from somebody![/quote]
Let $a=(x+y+z)$ and $b=x^{2}+y^{2}+z^{2}$ and $c=xy+yz+zx$ we now that $a,b,c>0$ 'cause $x,y,z>0$ and we now that $a^{2}=b+2c$.
Then we want to prove that $a^{6}\ge 27bc^{2}$ which is equivalent to
$(b+2c)^{3}\ge 27bc^{2}$<=> $b^{3}+6b^{2}c+12bc^{2}+8c^{3}\ge 27bc^{2}$ using $MA\ge MG$ we now that $b^{3}+2c^{3}\ge 3bc^{2}$ and  $6b^{2}c+6c^{3}\ge 12bc^{2}$
 q.e.d. :clap2:


Would you consider the internet to be intelligent?

do I consider the internet to be a sentient being in its own right or do I consider it capable of supporting intelligent thought ?

Well I always thought the internet was intelligent, then I found out about places like 4chan (be very warned, I probably shouldn't even be mentioning it) so now I'm not sure.

The Internet is a collection of metals and electrons zipping through those metals. It is incapable of creating, destroying or modifying anything on its own. Humans have created all that is on the Internet.
Robots and automated programs do not count as sentient, or even intelligent creatures. They were created by humans too, and their functions were explicitly dictated by those humans.

Did you even bother to read through the site given?  Keep an open mind.  Obviously the internet is not intellegent as of now.  Did you ever think of what [b]The Matrix [/b]was about?

The Matrix was not created by a form of worldwide communication. It was created by the [url=http://en.wikipedia.org/wiki/Technological_singularity]Singularity[/url], and one of the resulting paths.
I think that "computers" and "the Internet" are different concepts. I am completely sure the Singularity will occur, but I believe that the Internet, as grand as it is, is simply a form of communication. If it were to become "intelligent", then it would no longer be the Internet. The messenger did not create the message. Note that if a computer were to post information on the Internet by itself, it would be the computer, not the Internet, that was intelligent.
And yes, I did read the link. The characters described are robots, which are computers, not the Internet.

[quote="evilhamster"]I am completely sure the Singularity will occur[/quote]
I don't understand this viewpoint (read about it on the Wikipedia article). How can an intelligent entity increase their own intelligence? If they could, they would have to know what characterizes the state of further intelligence, which in turn implies they are already intelligent to that level? Sorry it just became a bit complicated. :P

Humans have increased their own intelligence. We now understand the laws of physics and chemistry, as well as electricity, computerization, social and political systems, supply and demand chains, architecture, and civilization.
And, of course, math  :P .

At the same time, our brain has evolved very little from 50,000 years ago, before we understood these in full detail. You are right that when we understand the next level of intelligence, then we are intelligent to that level. However, in human history progress has often been scorned. Examples: Heliocentric model of the universe, plate tectonics, the Earth's unspecialness in the universe, multiple universes, the wave theory of light, relativity, dark matter, global warming, etc., etc. These represented a higher level of thinking when they were conceived, but they were not accepted for some time.


topology
advanced fields
advanced fields theorems

Hello

This one seems obvious, but it isn't? Let $X$ be a simple connected space : this means that it is pathconnected and that in every point, any path beginning and ending there is homotopic with just the path that stays in that point

now let $p_1$ be a path beginning in $a$ and stopping in $b$
and let $p_2$ also be a path beginning in $a$ and stopping in $b$

show that they are homotopic

Seems extremely obvious, there are 'no holes', but on the other hand, the exact definition only says something about paths with same first and last point, so how can I get a homotopy between my two paths $p_1$ and $p_2$?

Belated hint: $p_{1}$ is homotopic to $p_{1}p_{2}^{-1}p_{2}$.

:ewpu: It's a lot simpler than one would think.
Thanks  :roll:


geometry
topology
email
Princeton
college
complex analysis

I didn't know exactly which thread to put this topic under, but can anybody with much more experience than I have give me some good examples of Fuchsian Groups to study?  I am a senior in high school who is reading up on complex analysis, mobius transformations, PSL(2,R) and hyperbolic geometry stuff for a paper.  Unfortunately, I don't know much topology at all, so it is hard for me to proceed through some of the books I have at any reasonable pace.  One example is our favorite PSL(2,Z).

I am not even sure if this question makes sense, but I am just sort of lost right now.

Thanks for any help!

I have no idea, but the Internet is your friend. Wikipedia has what seems to me to be a decent article about them:

http://en.wikipedia.org/wiki/Fuchsian_group

There's also an article in the Annals of Math, "Convergence Groups are Fuchsian Groups" by David Gabai. If you want to read it and you don't have access to JSTOR, I'll email it to you. (You'll have access through Princeton soon enough; I don't see anything wrong with using my Princeton access to JSTOR to get stuff for you :P).


Let X1,X2,.....,Xn denote a random sample from the uniform distribution on the interval (K,K+1). Let r1=Xbar -1/2, and r2=Xn -n/(n+1)

Show that r1 and r2 are unbaised estimators of K.

I assume that $ \bar X\equal{}\frac 1n\sum_jX_j$ and $ X_n\equal{}\max_j{X_j}$. In both cases all you need is to show that the expectation or $ r_{1,2}$ equals $ K$.


limit
calculus
integration
ratio
calculus computations

For the following, which values of p determine a convergent series?

(a) $\Sigma_{n=1}^{\infty} \frac{1}{n^p}$ (I do not know how to make the Sigma larger and put the infinity on top and n=1 on bottom....)

I know that this is our basic p-series and that it is convergent when $p > 1$ and divergent when $p \leq 1$

I think the way to prove this is to see that

If $p < 0$, then $\lim_{n \rightarrow \infty} \frac{1}{n^p} = \infty$

If $p = 0$, then $\lim_{n \rightarrow \infty} \frac{1}{n^p} = 1$

In both cases, the limit doesn't equal zero so the series' sum cannot converge.

Then we can apply the integral test because $\int_1^\infty \frac{1}{x^p} ,\ dx$ converges if $p > 1$ and diverges if $p \leq 1$

So that is that, right?

Part (b) asks for the values that make the following series convergent

$\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^p}$

I know that there is a theorem that states
If $lim_{n \rightarrow \infty} | a_n | = 0$, then $lim_{n \rightarrow \infty} a_n = 0$.

I do not know how to connect this information to answer the question.

Another question I came across which I was unsure about was:

Assume a is not equal to 0.
What's the difference between $\Sigma_{n=0}^{N} ar^n$ and $\Sigma_{n=1}^{N} ar^{n-1}$? Is there a difference?

As far as I can tell, there is not, but I am not sure. I don't see why it would be n-1 if it were approaching infinity....

I am trying hard to understand series, but I feel that they are very tricky and there is a lot to look out for. I am not trying to memorize what converges and what doesn't, I'd rather understand why.....Any help is appreciated

Notation: write \sum_{n=1}^{\infty} \frac 1{n^p} to get $\sum_{n=1}^{\infty} \frac 1{n^p}$

$\sum_{n=1}^{\infty} \frac 1{n^p}$ : Your reasoning is sound. The integral test works well. (So does the condensation test, but you only need one argument.)

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$:

Let's start with two questions:

Do the terms tend to zero?

As you mentioned, $|a_n|\to0$ iff $a_n\to0.$ Hence, you've already answered this question with your work on the previous problem. For $p\le0,$ the terms do not tend to zero and the series diverges. For $p>0,$ the terms do tend to zero, so we must ask more questions.

Does the series converge absolutely?

Again, you have already answered: if $p>1,$ the series converges absolutely; if $p\le 1,$ it does not.

These two questions have now settled the issue except for the range $0<p\le 1.$ In those cases, the series does not converge absolutely, although the terms do tend to zero. Could the series converge anyway? (That is, converge conditionally.)

The usual array of tests taught in a Calculus II course include a number of comparison tests: comparison, limit comparion, the integral test, the root test, the ratio test. (The "condensation test" I mentioned above isn't in the usual calculus books, but is another member of this list, another comparison.) All comparison tests are test for absolute convergence; they involve the size of the terms. Yes, the ratio test and the root test are comparison tests - comparisons to geometric series.

So, do you know any test that is [i]not[/i]  comparison test? In the usual calculus textbooks, there is only one: the alternating series test. (In more advanced works, you may see references to "Dirichlet's test" or to what you can determine by "summation by parts." Those should be viewed as generalizations of the alternating series test.)

Assume $0<p\le 1.$  Does the series satisfy the conditions of the alternating series test? Let's see: terms tend to zero - yes. Terms decrease in size - yes. Terms alternate in sign - yes. So the test applies and the series converges. The convergence is conditional because we have already denied absolute convergence.

The alternating series test is also valid for $p>1,$ it's just that we didn't need it there.

[quote]What's the difference between \Sigma_{n=0}^{N} ar^n and \Sigma_{n=1}^{N} ar^{n-1}?[/quote]

There's one term worth of difference.

Make the change of variable $m=n-1,$ meaning $n=m+1$ in the first sum. Since $n$ goes from $0$ to $N,$ then $m$ will go from $1$ to $N+1.$

$\sum_{n=0}^Nar^n=\sum_{m=1}^{N+1}ar^{m-1}.$

Now term those $m$'s back into $n$'s; this isn't the old $n,$ it's a new $n.$

$=\sum_{n=1}^{N+1}ar^{n-1}=\sum_{n=1}^Nar^{n-1}+ar^N.$

And that's the extra term. If we made these infinite sums (with $|r|<1$ so they converge), they'd be exactly the same:

$\sum_{n=0}^{\infty}ar^{n}=\sum_{n=1}^{\infty}ar^{n-1}$


function
induction
calculus
derivative
integration
algebra unsolved
algebra

find all function$f : R\to R$  such that :
$f(xy+f(y))=yf(x) +f(y)$ forall $x,y$ in real numbers

well i have gone through many problems of functional equations and seen that most of the problems (approx. 60%) can be solved by finding $f'(x)$ in two different ways and equating the expressions....
other problems involve ubique methods such as induction etc...
generally the answer is $f(x) = x$ only {which seems to be the case here also.}
This one seems to be easier one(find the derivative)...

It's an easy one....
	$f(xy + f(y)) = yf(x) + f(y)$

Taking y as a constant value and Differentiating above equation w.r.t $x$ we get

	$f'(xy + f(y)) (y + 0) = yf'(x) + 0$
(as f(y) is a constant, it's derivative is zero)
	$f'(x) = f'(xy + f(y))$    ( y cancels)
This is the supreme relation we wanted..

now, put y = 0 (a constant) we get

	$f'(x) = f'(f(0)) = a$ $constant = k$$(say)$

	i.e $f'(x) = k$
Integrating both sides w.r.t x

	$f(x) = kx + c$ (where c is constant of integration  c= f(0))
Now substituting in main equation,

	$f(xy + f(y)) = yf(x) + f(y)$
	$k(xy + ky + c) + c = y(kx + c) + ky + c$
	$kxy + k^2y + kc + c = kxy + y(c + k) + c$
As this is an identity in x,y comparing coefficients of y and constant terms,we get
	$kc + c = c$
	$k^2 = c + k$
First equation gives $c=0$ or $k = 0$
	$c=0$ yields $k = 0$ or $k =1$ from second equation
	$k=0$ yields $c = 0$ from second equation
Hence there are only two functions satisfying the given equation,
(1) $f(x) = 0$ for all $x \in R$
(2) $f(x) = x$ for all $x \in R$
Q.E.D

the problem didn't say $f$ was differentiable.

We have $\ f(xy+f(y))=yf(x)+f(y)$  $\ (*)$
Then $\ f(yx+f(x))=xf(y)+f(x)$. If $\ f(x)=f(y)\neq0$ then $\ yf(x)+f(y)=xf(y)+f(x) \Rightarrow f(x)(x-y)=0 \Rightarrow x=y$. So $\ f$ is injective on $\ \mathbb{R^*}$.
We have $\ f(f(0))=f(0)$.
Assume $\ f(0)\neq0$ then $\ f(0)=0$ (because f is injective on $\ \mathbb{R^*}$), which is clearly false.
Then $\ f(0)=0$. From $\ (*)$ we have $\ f(f(y))=f(y)$. If $\ f(y)\neq0$ then $\ f(y)=y$ because of the injectivity.
So $\ f(x)\in\{0,x\}$.

Assume $\ f(1)=1$ and $\ f(y_0)=0$ for $\ y_0\neq0$. By plugging $\ x=1$ in $\ (*)$ we get $\ f(y_0)=y_0f(1)=y_0$. This is false, so $\ f=1_R$.

Assume $\ f(1)=0$ and $\ f(y_0)\neq0$ for $\ y_0\neq0$. Then $\ f(y_0)=y_0$. By plugging $\ x=1$ in $\ (*)$ we get $\ f(y_0+f(y_0))=f(y_0)\Rightarrow f(2y_0)=y_0\in\{0,2y_0\}$ which is false because $\ y_0\neq0$. So $\ f=0_R$.


After thousands of years of endless wandering, two leaders (which are going to be eternal, by the way) decide to settle down and form towns. They were the Egyptians and the Sumerians.

It is now up to you, the player to decide the fate of these nations... and many more to come.

Signups:

2 players needed right now, more empires will be added as the game goes along (trust me, this isn't impromptu.)

Yellow: Egypt
Blue: Sumer

[quote="james4l"]Pick: Egypt or Sumer.

Turn		Ancient Age Start
	1	EGYPT
	1	SUMER
	8	INDIA
	9	CHINA
	18	HITTITES
	22	BABYLON
	28	ETRUSCANS

A summary of the first seven countries.[/quote]
what do the numbers mean?
i choose etruscans

Lol, somehwhat copy of my game.

Babylon.

Lol is correct.

But it's historical! And educational!

Larger number = come in later in the game.

Some countries will have to put down rebellions (or go with the flow of the new regimen), like the Romans revolting against the Etruscans about 10 turns later*

*representative of non-historicalness

Macedon and Greece will be separate entities, but Sparta will not be apart from Athens.

I wanna be Hittites so I can hit people.

1 EGYPT - pythag011
1 SUMER - nt811581147
8 INDIA - AIME15
9 CHINA - BOGTRO
18 HITTITES - 1=2
22 BABYLON - Dojo
28 ETRUSCANS - funtwo

[b]4000 BC - Turn 1[/b] It is dawn of humankind. pythag011 decides to settle at the foot of the Nile. pythag011, the citizens of Egypt have vested absolute power to you.

Meanwhile nt811581147 decides to settle in the banks of Tigris and Euphrates. nt811581147, the citizens of Sumer have also vested absolute power to you.

Player Country Sci Tax Government
pythag011	Egypt			1 (1)	25 (1)	Early Settlement
nt811581147	Sumer			1 (1)	25 (1)	Early Settlement

Pick a technology to research:
[Cost]
[6] S1: Mathematics - You can now enlist siege weapons in your army. Build: Library, GREAT LIBRARY
[7] C1: City State - Unlocks City State Government. Build: Temple
[5] M1: Conscription - Allows conscription of troops.
[8] I1: Bronze Working - Metal can be used. Build: THE PYRAMIDS
[6] E1: Barter - Trade with other civilizations is possible. Build: Bazaar, THE COLOSSUS (req. dock)
[color=red][9] N1: Map Making - Allows production of Naval transport [Requires: Conscription][/color]

Science: Each tech makes science 9% cheaper.
Civic: City limit +1 each time, Territory limit +15 each time (except last tech, allows 100%)
Military: Unit limit [10/30/60/100/150/200/300/500/1000/inf]
Industry: Each tech makes construction 5% cheaper and faster.
Economic: Each tech increases taxes (based on population + area)
Seafaring: +1/+1 Attack, Defense to all ships. Every 3 techs = +1 ship movement

aw dang i made a mistake. not only do i come in last, but i get a rebellion -.-

Um, yeah, that is a problem... don't worry, many other countries will eventually face the same problems...

For example: Greece will have to go against the Macedons
China, India, Japan, Khmer will have to go against the Mongols
Mesopotamia (Persia, Hittites, Sumer, Babylon) will be crowded
Egypt will have to face the Carthage AND Kushite threats
Mesoamerica (Maya, Aztecs, Inca) will be semi-crowded AND late-arriving
Rome will have to face breakup of Byzantines, Holy Rome, Spain, Britain, Teutons, Russia, etc.

So you have Civ 4...

I'm thinking about it. Probably not under normal circumstances*

*excluding: long periods of anarchy, high economic depression, extremely low population, etc.

AIME15: Research Conscription first.

[hide="TECHNOLOGY ADVISOR"]
TECHS
Science: Each tech makes science 9% cheaper.
1: Mathematics - You can now enlist siege weapons in your army. Build: Library, GREAT LIBRARY
2: Timekeeping - +25% all resources
3: Astronomy - Build: Observatory
4: Scientific Method - Techs research 25% faster
5: Physics - Build: University
6: Biology - Disease is no longer occurent.
7: Atomic Theory - Build: Research Facility
8: Genetic Manipulation - Build: CURE FOR CANCER, LONGEVITY
9: Quantum Theory - +100% all resources
10: Artificial Intelligence - 2000% speed at creating troops

Civic: City limit +1 each time, Territory limit +10 each time (except last tech, allows 100%)
1: City State - Unlocks City State Government. Build: Temple
2: Empire - Unlocks Monarchy, Totalarianism, Republic
3: Imperialism - Can sign NAPs, alliances, MPPs. Unlocks Theocracy, Feudalism, Serfdom Build: Courthouse, Provincial Capitol
4: Divine Right - Build: Cathedral, Basilica, SISTINE CHAPEL
5: Nationalism - Unlocks Democracy, Parliamentary Monarchy. Build: Embassy
6: Realpolitik - Unlocks Fascism, Communism, Socialism, Capitalism Build: Spy HQ
7: International Law - Build: UNITED NATIONS
8: Extremist Policies - Rebellion Chance decreases by 90%, Build: Police Station, Interpol
9: Interplanetary Diplomacy - Build: Interplanetary Embassy
10: World Government - Rebellion Chance decreases to 0%

Military: Unit limit [10/30/60/100/150/200/300/500/1000/inf]
1: Conscription - Allows conscription of troops.
2: Professional Army - Allows advanced troops. Build: TERRACOTTA ARMY, SUN TZU'S ART OF WAR
3: Imbued Metalwork - Attack +2, Defense +2 Build: Tower, Barracks
4: Firearms - Attack +4
5: Metallurgy - Defense +6, Siege units do double damage. Build: Military Complex (Upgraded Barracks)
6: Motorized Transport - Allows production of Tanks. All Military Troops gain one movement. Build: CLAUSEWITZ'S ON WAR
7: Flight - Allows production of airplanes, missiles, and flak. Build: Airport, AA Turret
8: Unconventional Warfare - Defense +8. Allows production of Nukes (with Atomic Theory) and Helicopters. Increases range for missiles, bombers, and fighters by 1.
9: Laser Dischargers - Attack +10. Allows production of attack spaceships, stealth bombers, and ICBM. Build: Starport
10: Integrated Defense - Build: NUCLEAR SHIELD, SDI Defense

Industry: Each tech makes construction 5% cheaper and faster.
1: Bronze Working - Metal can be used. Build: THE PYRAMIDS
2: Iron Working - Attack +2, Build: Mine, Forge
3: Labor Theory - +1 Happiness. Build: Farm, Fishing Nets
4: Printing Press - Can trade technologies
5: Machinery - Buildings construct 50% faster
6: Industrialism - Coal, rubber can be used. Build: Factory
7: Combustion - Oil, aluminum can be used. Build: Coal Plant, Hydro Plant, Solar Plant
8: Synthetic Fibers - Uranium can be used. +1 Happiness. Ivory is obsolete. Build: Nuclear Plant, Manufacturing Plant
9: Synthetic Fossil Fuels - Coal, Oil obsolete. Build: Heavy Water Plant
10: Nuclear Protonization - Aluminum, Uranium, Rubber are homologous.

Economic: Each tech increases taxes (based on population + area)
1: Barter - Trade with other civilizations is possible. Build: Bazaar, THE COLOSSUS (req. dock)
2: Currency - Caravan available. Build: Marketplace
3: Mercantilism - Caravans move 2x fast
4: Banking - Luxuries generate +1 income. Build: Bank
5: Economics - incur 1% interest (capped at 100). Build: SMITH'S TRADING POST
6: The Corporation - Unlocks Capitalism
7: Stock Exchange - Build: Stock Exchange, WALL STREET
8: Commercialization - incur 2% interest (capped at 250).
9: Planetary Trade - Can trade between planets.
10: Global Prosperity - incur 5% interest (capped at 1000).

1: Map Making - Allows production of Naval Transport
2: Navy - Allows production of Attack Ship, Demolition Ship. Build: Dock, Harbor, Drydocks, Lighthouse, GREAT LIGHTHOUSE
3: Navigation - SA+1, SD+1, SM+1
4: Magnetism - SA+1, SD+1, SM+1 Build: MAGELLAN'S VOYAGE
5: Ironclads - Allows production of Ironclad
6: Engines - Converts AS => Destroyer, SA+3, SD+1, SM+2.
7: Amphibious Warfare - Allows production of Marines, Submarines, Battleships, upgrades Destroyers => AEGIS Destroyer, Aircraft Carrier
8: Naval Supremacy - SA+5, SD+3, SM+1. Submarines can carry nukes. [/hide]

[hide="FOREIGN ADVISOR"]
[hide="Diplomacy"]Egypt - Sumer - India = Tripartite Truce (NON-BINDING)[/hide]

[hide="Signups"][1] Ancient Age Start
1 EGYPT - pythag011
1 SUMER - nt811581147
8 INDIA - AIME15
9 CHINA - BOGTRO
18 HITTITES - 1=2
22 BABYLON - Dojo
28 ETRUSCANS - funtwo
[31]	Classical Age Start
31	GREECE - lotsofmath
33	MACEDONS
34	CARTHAGE - sup3rcrasher
35	MINAEA
37	ETRUSCANS=?>ROME
38	HITTITES=?>PERSIA
38	KUSH
41	BANTU
43	KUSH=?>AKSUM
47	JAPAN - dragon96
55	MAYA
56	ROME=sp>BYZANTINE
56	ROME=sp>WEST ROME
58	FRANKS
[61]	Medieval Age Start
61	CELTS
61	BRITONS
61	GAUL
64	MINAEA=?>ARABIA
64	SRIVIJAYA
67	SPAIN
70	VIKINGS
71	KHMER
72	BULGARIA
73	FRANKS=?>HOLY ROME
73	KIEVAN RUS
73	ARABIA=?>SELJUKS
77	PORTUGAL
[81]	Renaissance Start
81	MALI
81	INCA
81	MONGOLS
86	AZTECS
87	SELJUKS=?>OTTOMANS
[96]	Enlightenment S.
96	GAUL=?>FRANCE
96	IROQUOIS
109	VIKINGS=?>SWEDEN
116	(ENGLAND-COLONY)=?>USA
[126]	Industrial Start
128	AZTECS/(SPAIN-COLONY)=?>MEXICO
129	INCA=?>ARGENTINA
130	(PORTUGAL-COLONY)=?>BRAZIL
131	SWEDEN=?>SCANDINAVIA
131 ZULU
133	(ENGLAND-COLONY)=?>CANADA
140	HOLY ROME=?>GERMANY
[146]	Modern Start
146	(ENGLAND-COLONY)=?>AUSTRALIA
155	ZULU=?>SOUTH AFRICA
157	SRIVIJAYA=?>INDONESIA[/hide]

=?> Revolt
[/hide]

[b]3100 BC - Turn 18[/b]
New "technology!" I will now automate tech-pointing out so long as you have posted something substantial!*
*No exitus. However, reduced technology progress to ceiling(0.7*tech)
1=2, Hittites formed. (Starting: E1 tech, City Size 2). Pick Tech.
2 turns.
On deck: Dojo, Babylon (Babylon), Turn 22.

[hide="EGYPT (pythag011)"][5/6, +2] (assumed) City State
[3/20, +3] Thebes* will grow in 6 turns (to size 5)
Thebes: Palace
[0/0] You can build workers. You may build 0 more units.
[5/10] You may not expand your territory.
[+2] Treasury: 48 Gold.[/hide]

[hide="SUMER (nt811581147)"][2/14, +1] Professional Army
[12/15, +2] Ur* will grow in 2 turns (to size 4)
Ur: Palace
[2/25] You may conscript units from cities. Cost: Each population point equals 10 conscripts.
[4/10] You may expand your territory. (1 square)
[+3] Treasury: 54 Gold.[/hide]

[hide="INDIA (AIME15)"][4/8, +2] Conscription
[5/20, +2] Delhi* will grow in 8 turns (to size 5)
Delhi: Palace
[0/0] You cannot build units.
[3/10] You may expand your territory. (1 squares)
[+1] Treasury: 35 Gold.[/hide]

[hide="CHINA (BARBARIAN)"] (Does not use stuff like tech and cities. Has military)[/hide]

[hide="HITTITES (1=2)"][0/0 ,+2] tech...
[0/10, +2] Hattusas* will grow in 5 turns (to size 3)
Delhi: Palace
[0/0] You cannot build units.
[2/10] You may expand your territory. (2 squares)
[+1] Treasury: 50 Gold.[/hide]

Yellow - Egypt
Blue - Sumer
Gray - India
Orange - China
Turquoise - Hittites

[hide="Report (Turn 16)"]MOST POPULOUS COUNTRY:
India (pop: 305,000)
Egypt (pop: 285,000)
Sumer (pop: 265,000)

MOST MILITARIZED COUNTRY:
Sumer (army size: 14)
India (army size: 5)
Egypt (army size: 4)

MOST ADVANCED COUNTRY:
Egypt (techs: 3)
Sumer (techs: 2)
India (techs: 1)[/hide]

EGYPT: S1, M1, I1
SUMER: M1, N1
INDIA: I1
HITTITES: E1

[quote="james4l"][quote="pythag011"]Research Conscription.

I need a bigger army..[/quote]

Conscription is already researched...

City state: more cities, better government, more money, etc.[/quote]

How do I get a bigger army?

[quote="james4l"]... you just tell me how many troops to conscript.

My apologies, mz94. Will put up in next turn-report.[/quote]

Conscript 10 troops.

whats the best square for me to conquer, james?

[quote="sup3rcrash3r"]James, won't the people that started first have an advantage?

Or are later nations formed with some sort of tech?

Because every new nation in history didn't start all over.[/quote]

If you read the profile for the Hittites, they started off with Barter and a larger city.

[quote="BOGTRO"]i drop out. no point to this game[/quote]

You don't need to say that. You're already dead in this game...


Here's a problem I was given in pre-calc today:

If Chicken McNuggets were made in boxes of 6, 9, and 20, what is the greatest number of nuggets that can't be purchased in whole boxes?

[/hide]

since 6=3*2 and 9=3*3, you can make every multiples of 3 with 6 and 9 except 3.(2*3-2-3=1)

theres also 20, so you can make every number in forms of 20x+3y where y is not 1.

however, 20*3+3*1=63 which is a multiple of 3. 63 can be made with 6, 9 and 20, and 63+20k can be made with one more 20 before that.

so the largest number that cannot be made with 6, 9 and 20 is 20*2+3*1=43

answer : 43??

[quote="Lazarus"]see the above guy's post - this question is basically equivalent to sayin like "if chicken mcnuggets came in boxes of 20s and 3s, what's the largest amount you can't buy?"

and $ 20\cdot3 \minus{} 20 \minus{} 3 \equal{} \boxed{37}$.[/quote]

try and make 43 with 6, 9 and 20.  :D

I agree. The answer is 43. (hehe...that rhymed)  :lol:
Lazarus, you're forgetting that you can't make numbers of the form $ 20x \plus{} 3$ when $ x \equal{} 0$, $ x \equiv 1 mod(3)$, or $ x \equiv 2 mod(3)$.

oh man dang my bad lol

i did not think about that. like, at all

This is in fact an [i]extremely[/i] [url=http://en.wikipedia.org/wiki/Coin_problem#McNugget_numbers]well-known problem[/url] (possibly because of its relevance to real life?).


In rectangle $ ABCD$, $ AB\equal{}5$ and $ BC\equal{}3$. Points $ F$ and $ G$ are on $ \overline{CD}$ so that $ DF\equal{}1$ and $ GC\equal{}2$. Lines $ AF$ and $ BG$ intersect at $ E$. Find the area of $ \triangle{AEB}$.
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

pair A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3);
pair E=extension(A,F,B,G);
draw(A--B--C--D--A--E--B);

label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$E$",E,N);
label("$F$",F,SE);
label("$G$",G,SW);
label("$B$",B,SE);

label("1",midpoint(D--F),N);
label("2",midpoint(G--C),N);
label("3",midpoint(B--C),E);
label("3",midpoint(A--D),W);
label("5",midpoint(A--B),S);[/asy]$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ \frac{21}{2} \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ \frac{25}{2} \qquad
\textbf{(E)}\ 15$

Obviously $ \triangle AEB \sim \triangle FEG$. Thus we have $ \frac {h}{2} \equal{} \frac {h \plus{} 3}{5} \rightarrow h \equal{} 2$.
$ \frac {5\cdot 5}{2} \equal{} \frac {25}{2} \rightarrow \boxed {D}$

[quote="crazypianist1116"]Obviously $ \triangle AEB \sim \triangle FEG$. Thus we have $ \frac {h}{2} \equal{} \frac {h \plus{} 3}{5} \rightarrow h \equal{} 2$.
$ \frac {5\cdot 5}{2} \equal{} \frac {25}{2} \rightarrow \boxed {B}$[/quote]

You do mean $ \boxed{D}$, right?

[hide="Alternatively,..."]

$ [ABFG]=[ABCD]-[ADF]-[BCG]=15-\frac{3}{2}-3=\frac{21}{2}$. Since $ \triangle{AEB}\sim\triangle{FEG}$, $ \frac{\frac{21}{2}+[EFG]}{[EFG]}=\left(\frac{5}{2}\right)^2=\frac{25}{4}$. Solving, $ [EFG]=2$, so the area of $ \triangle{AEB}$ is $ \frac{21}{2}+2=\frac{25}{2}\Longrightarrow\mathrm{ (D) \ }$

[/hide]

[asy]unitsize(6mm);
    defaultpen(linewidth(.8pt)+fontsize(8pt));

    pair A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3);
    pair E=extension(A,F,B,G);
    draw(A--B--C--D--A--E--B);

    label("$A$",A,SW);
    label("$B$",B,SE);
    label("$C$",C,NE);
    label("$D$",D,NW);
    label("$E$",E,N);
    label("$F$",F,SE);
    label("$G$",G,SW);
    label("$B$",B,SE);

    label("1",midpoint(D--F),N);
    label("2",midpoint(G--C),N);
    label("3",midpoint(B--C),E);
    label("3",midpoint(A--D),W);
    label("5",midpoint(A--B),S);[/asy]

$ FG\equal{}5\minus{}2\minus{}1\equal{}2$.  Hence, the ratio of sides in triangle $ EFG$ and triangle $ EAB$ is $ \frac{2}{5}$.  Thus, letting the altitude of triangle $ EFG$ be $ \kappa$, we have that $ \frac{\kappa}{\kappa \plus{}3}$=$ \frac{2}{5}$.  Hence, $ {\kappa}\equal{}2$, and the altitude of $ \Delta AEB$ is 5.  Thus, [$ \Delta AEB$]=$ \frac{25}{2}$.  $ \boxed D$

Sorry for the bad bump, but here’s a cool solution with infinite summations:

Since $\Delta{ABE}\sim{\Delta{FGE}}$ then $[AFGB]\sim{[FXYG]}$, where $X$ and $Y$ are ponts on $EF$ and $EG$ respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of $\frac{FG}{AB}=\frac{2}{5}$:
\begin{align*}
[AEB] &= [AFGB]+[FXYZ]+\dots\\
[AEB] &= [AFGB]+\frac{2}{5}[AFGB]+\left(\frac{2}{5}\right)^2[AFGB]+\dots
\end{align*}
We have to find the ratio of the areas when the sides have shrunk by length $\frac{2}{5}l$
[asy]
unitsize(0.6 cm);

pair A, B, C, D, E, F, G;

A = (0,0);
B = (5,0);
C = (5,3);
D = (0,3);
F = (1,3);
G = (3,3);
E = extension(A,F,B,G);

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

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$E$", E, N);
label("$F$", F, SE);
label("$G$", G, SW);
label("$2/5$", (D + F)/2, N);
label("$4/5$", (G + C)/2, N);
label("$6/5$", (B + C)/2, dir(0));
label("$6/5$", (A + D)/2, W);
label("$2$", (A + B)/2, S);
[/asy]
Let $[AFGB]'$ be the area of the shape whose length is $\frac{2}{5}l$
$$[AFGB]'=[ADCB]-[ADF]-[BCG]$$
Substituting values gives us
$$[AFGB]'=\frac{12}{5}-\frac{6}{25} - \frac{12}{25}\implies [AFGB]'=\frac{42}{25}$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get
$$\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}$$By applying an infinite summation
$$[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{\left(\frac{4}{25}\right)^n}$$
we have that
$$S=\frac{a_1}{1-r}\implies \boxed{[AEB]=\frac{25}{2}}.$$

Coordinate bash along with shoelace theorem worked for me.


quadratics
calculus
integration
modular arithmetic

Find integral solutions to $x^{2}+3 = 2y^{2}$.

[hide="Hint 1"] There are none.  :D [/hide]

[hide="Hint 2"] Try $\bmod 8$. [/hide]

[quote="SplashD"][hide]$x^{2}+3\equiv 3,4,7 \pmod 8$
$2y^{2}\equiv 0, 2 \pmod 8$

Therefore, there are no integral solutions.[/hide]
I was wondering, why mod 8?[/quote]

I think its because:
since the equation is quadratic, we consider them in mod $2^{n}$. So first we consider mod 4 so that the squares are equal to 1 or 0.
Then the LHS=0,3 mod 4 and RHS=0,2. So this tells us that there might be a solution.Then things get a bit more complicated, we get lazy and we try mod 8 which gives us the answer :D


How does: $ 2^{4(2^x)}$ lead to $ 16^x \equal{} 4(2^x)$. This is from AoPS:1 Chapter 1 Problems, #10.

The original question is: "Find $ x$ if $ 2^{16^x}\equal{}16^{2^x}$."

get both in terms of 2^z and then set equal. That leads to the desired equation


Let ABC be a triangle such that the cevians AD,BE,CF concurs.Let M,N and P be points on the sides FE, FD,and ED respectively.Show that the lines CP(extended), AM and BN concurs if,and only if,the lines EN, DM and FP concurs.

Observing that this is a pure incidence theorem might give us some advantage. So without loss of generality we can apply some projection to the plane of the figure so that $AB$ becomes the line at infinity, and we get the following equivalent problem (we preserve the original notation):

Let $CDXE$ be a parallelogram. We take points $M$ and $N$ on the parallel lines to $CX$ through $D$ and $E$, respectively. The lines $DM$ and $EN$ intersect at $O$. The parallel line to $CE$ through $M$ and the parallel line to $CD$ through $N$ meet at $O'$. Let $P$ be a point on $DE$. Show that $P$, $C$ and $O'$ are collinear if and only if $OP$ is parallel to $CX$.

Let $CD=a$, $CE=b$, $DE=c$, $CX=d$, $EM=m$, $DN=n$. Suppose first that $OP$ is parallel to $CX$, and denote by $Y$ and $Y'$ the intersections of $EX$ with $CP$ and $CO'$, respectively. By Thales' theorem we have $EY/a=EP/PD=m/n$, so $EY=am/n$. On the other hand, trace the intersection of $DN$ and $CE$, and let $Z$ be the intersection of $CE$ and $NO'$. If $CY=p$, again by Thales' theorem we have $p/(b-p)=n/(d-n)$, or $p=bn/d$. Now trace the intersection of $EM$ and $CD$, and let $W$ be the intersection of $MO'$ and $CD$. Letting $CW=q$ it's clear that $q/(a-q)=m/(c-m)$, or $q=am/c$. Observe that $O'Z=CW$; therefore $O'Z=am/c$. Finally, $EY'/OY'=b/p$, so $EY'=(d/bn)(b/p)(am/d)=am/n$. Hence $EY=EY'$, that is, $Y=Y'$, and we are done.

For the converse, observe that $EP/PD=EY/CD=(am/n)/a=m/n$. But clearly we have $MO/OD=m/n$; consequently, $EP/PD=MO/OD$. This implies that $OP$ is parallel to $EM$, and the result follows :)


let $p>3$ be prime number and $n<2p$. Prove that polynom $x^{2p} + px^n - 1$ is ireducibile over $Z[X]$.

[quote="gopherhole112"]eisensteins criterion? don't remember it, but its exactly what you want.[/quote]
Eisenstein needs something divisible by $p$ as constant term.

But it was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=43023 (without solution).


integration
real analysis
real analysis unsolved

hello, i think this website is very helpful for you
http://mathworld.wolfram.com/SineIntegral.html
Sonnhard.


complex analysis
function
calculus
derivative
complex numbers
complex analysis unsolved

Here is a nice problem related to holomorphic self maps of the unit disk.

Lemma: If $\psi$ is a holomorphic self-map of the unit disk such that

$\frac{1-|z|^{2}}{1-|\psi(z)|^{2}}\leq \frac{1}{1-|\psi(0)|^{2}}$ for all $z\in D$,

then either $\psi(0)=0$ or $\psi'(0)=0$.

I believe this lemma should hold for every holomorphic function defined in a neighborhood of $0$ and bounded by $1$ in absolute value.

Assume to the contrary, that $\psi(z) = a+bz+O(z^{2})$, where $a$ and $b$ are nonzero complex numbers. Let $z_{t}= (a/b) \cdot t$. Then $\psi(z_{t}) = (1+t) a+O(t^{2})$. By the condition,
\[\frac{1-|\psi(z_{t})|^{2}}{1-|z_{t}|^{2}}\geq 1-|\psi(0)|^{2}= 1-|a|^{2}.\]
But the expression in the left hand side equals
\[1-(1+t)^{2}|a|^{2}+O(t^{2}) = 1-|a|^{2}-2 |a|^{2}\cdot t+O(t^{2}) < 1-|a|^{2}\]
(for sufficiently small $t$). We get a contradiction.

nice solution.
Here is another one:

assume that $\psi(0)\neq 0$. Then let $C=1-|\psi(0)|^{2}$, then the condition of the problem is equivalent to:

$f(z)=|\psi(z)|^{2}+C(1-|z|^{2})\leq 1$ for all $z \in D$, with equality when $z=0$.
Pass to the real and imaginary parts by writing $z=x+iy$ and $\psi=u+iv$.

Viewing $f$ as a function of two variables we get

$f(x,y)=u^{2}(x,y)+v^{2}(x,y)+C(1-(x^{2}+y^{2}))$,

we know that the origin is one of its critical points. Evaluating the partial derivatives with respect to x and y at the origin , we get that

$u_{x}(0,0)u(0,0)+v_{x}(0,0)v(0,0)=0$
$u_{y}(0,0)u(0,0)+v_{y}(0,0)v(0,0)=0$

Using cauchy -Riemann equations we get that

$u_{x}(0,0)u(0,0)+v_{x}(0,0)v(0,0)=0$

$-v_{x}(0,0)u(0,0)+u_{x}(0,0)v(0,0)=0$

Since $\psi(0)\neq 0$ we know that $(u(0,0), v(0,0))\neq (0,0)$ so the determinant above must be zero, implying that $u_{x}(0,0)=v_{x}(0,0)=0$. It follows that $\psi'(0)=0$.


Please follow this link to the forum . It's regarding a prestigious Indian Scholarship of IISc, Bangalore and IIT,Bomaby for high schoolers and undergrads.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=323536#323536]Forum[/url]

[url=http://www.iisc.ernet.in/kvpy]Website[/url]

Some people had asked about previous year papers. Well, as far as I know, they have never (and never will) release past papers. It's an IIT style paper made by IIT Profs. It really tests your understanding! That's all I can tell you - just the part of what I thought! Are there any other KVPY Fellows on this forum???

I have the matter they taught us in the summer camp at IISc, Bangalore. However, it's too advanced for XI and XII and I doubt it to be of any use in the exam! Just use the basics and you should get the answer!!! (Note:- I can post this if you want!)

Nevertheless could you atleast post the Math section of it.
 It would be useful perhaps for IMO.
   Ashwath

yes
pleaese do tell about the maths section atleast.


How many integers $ m \neq 0$ satisfy the inequality $ \frac{1}{|m|}\geq \frac{1}{8}$?

hello, your inequality is for $ m\ne 0$ equivalent to
$ \minus{}8\le  m \le 8$ and so we have $ 16$ integer solutions.
Sonnhard.


Given any sequence of n letters, each chosen from {A, B, C, D, E, F, G, H, I, J} show that we can find a bijection {A, ... J} <-> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} such that the n-digit number corresponding to the sequence does not have a leading zero and is divisible by 9.

is this problem from the Australian Mathematical Olympiad (AMO) or was it shortlisted from the AMO 2002?

if it was on the 2002 paper, then go here:

[url]http://www.kalva.demon.co.uk/austral.html[/url]


I think there must be a mistake in your problem, because it looks like a nonsense. Maybe you should express yourself more precisely.

[quote="mychrom"]let be two circles $C_{1}$ (interior to$C_{2}$) and $C_{2}$. Two tangents at $C_{1}$ intersect the $C_{2}$ in $A,B$ (first tangent) and in $C,D$ ( the second one ). Prove that  $AB=CD$.[/quote]

i think this is only true when C_1 and C_2 have the same center


probability
combinatorics unsolved
combinatorics

A footballer shoots 4 times to score a goal! We know that the probability for him to score a goal is 0,3. What is the probability to miss the target the first three times, but score the last time? What is the probability to score twice?

I have a few questions:

By 0,3 do you mean $\frac{3}{10}$?

Are you asking for the probability, given that he scores a goal, or the probability at all?

In some countries, 0.3 is written like 0,3.


logarithms
calculus
integration
Euler
real analysis
real analysis solved

Evaluate:
$\sum_{i=1}^{\infty}{\frac{1}{2^n n^2}}$
I think it easy for the experienced mathlinkers.

Yes you are right.I knew that it would be easy for you.Did you use integrating method or something else?

You can find the expression of $\sum\frac {x^n}{n^2}$ from $\sum x^n$ through integration, right? :? I didn't compute the value here, because I was too lazy to compute the integrals :).

Edit: I hadn't seen the posts above :).

Yes you're right grobber.I managed to do the integrating(though it is really boring)because i needed to solve another [b]important[/b] problem. :)

$f(x)=\sum_{n\geq 1}\frac{x^n}{n^2}$

Here a way derivate $f$ and integrate to get $f(x)=\frac{\pi^2}{6}-\int_{1}^{x}\frac{\ln(1-t)}{t}dt$ for  $0 \leq x < 1$

as moubinool write if  $ f(x)=\sum_{n\geq 1}\frac{x^n}{n^2} $
then the derivate of $f(x)$ is $-{\frac {\ln  \left( 1-x \right) }{x}}$
Now the derivate of  $ f(x)+f(1-x)+ln(x)ln(1-x) $ is 0 .
Take the limit $x \to 0$ and use  $f(1)=\frac{\pi^2}{6}$.
We deduce Euler formula
$f(x)+f(1-x)=\frac{\pi^2}{6}-ln(x)ln(1-x)$
To conclude put x=1/2...


Prove that
$ \sum_{cyc}\sqrt{\frac{a}{b\plus{}c}} \ge 2$.
for all nonnegative reals $ a,b,c$.

I was able to prove it myself via Holder's by taking this sequence two times and the third one $ \sum a^2(b\plus{}c)$. In the end it reduces to Schur's with equality iff two are equal and third is 0.
I want to see a better solution to this problem, so keep posting solutions as soon as you find them.

$ \sqrt{xy}\le \frac{x\plus{}y}{2} \Longrightarrow \sqrt{\frac{b\plus{}c}{a}} \equal{} \sqrt{\frac{b\plus{}c}{a}\bullet 1} \le \frac{1}{2}(\frac{b\plus{}c}{a} \plus{} 1) \equal{} \frac{a\plus{}b\plus{}c}{2a} \Longleftrightarrow \sqrt{\frac{a}{b\plus{}c}} \ge \frac{2a}{a\plus{}b\plus{}c} \Longrightarrow \sqrt{\frac{a}{b\plus{}c}} \plus{}  \sqrt{\frac{b}{c\plus{}a}}  \plus{}\sqrt{\frac{c}{a\plus{}b}}  \ge \frac{2a}{a\plus{}b\plus{}c} \plus{} \frac{2b}{a\plus{}b\plus{}c} \plus{} \frac{2c}{a\plus{}b\plus{}c} \equal{} ... \equal{} 2$  :wink:


MATHCOUNTS
percent
geometry
3D geometry
perimeter
trapezoid
probability

What? No MOEMS Marathon?? That can be fixed.

The rules here are similar to the ones in all other marathons:

1. Answer a problem.
2. Post another problem.
3. That's it!

[size=200]Problem the First[/size]

A firefighter stood on the middle rung of a ladder, went up 3 rungs, was forced down 5 rungs, and then went up 7 rungs to extinguish the fire. Then the firefighter climbed the remaining 6 rungs to the top of the ladder. How many rungs are there on the entire ladder.
Source: 2000 MOEMS Division E

Just for curiosity, I like to ask small question on this part.

[quote].....2. Post another problem. [/quote]

Do you mean problem from MOEMS? Just be sure. :)

I guess it would have to be at a MOEMS (4-8 grade) level, not necessarily from MOEMS. Most MATHCOUNTS Chapter and possibly State problems would work.

[quote="NoSoupForYou"]
A firefighter stood on the middle rung of a ladder, went up 3 rungs, was forced down 5 rungs, and then went up 7 rungs to extinguish the fire. Then the firefighter climbed the remaining 6 rungs to the top of the ladder. How many rungs are there on the entire ladder.
Source: 2000 MOEMS Division E[/quote]

11 rungs above the middle ($+3-5+7+6=11$), so 11 rungs under the middle, plus the middle is $11 \cdot 2 + 1=23 \text { rungs.}$

[size=150][b]Problem 2[/b][/size]

Ari, Barry, Cara, Dara, and Erin have a total of 85 dollars. Ari and Barry have a total of 40 dollars. Cara and Dara have a total of 30 dollars. Erin and Ari have a total of 20 dollars. How many dollars does Barry have?

Source: Sample MOEMS Division M

We have:

 $A+B+C+D+E=85$,
 $A+B=40$,
 $C+D=30$,
 $E+A=20$.

Now subtract $A+B=40$ from $A+B+C+D+E=85$
and get $C+D+E=45$. Then subtract $C+D=30$ from $C+D+E=45$
and get $E=15$.
That means that $A=5$ and $B=35$.

[size=200]Problem 3[/size]

In lowest terms, what is the product of this multiplication?
$\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot\frac{9}{11}\cdot\frac{11}{13}\cdot\frac{13}{15}\cdot\frac{15}{17}\cdot\frac{17}{19}\cdot\frac{19}{21}\cdot\frac{21}{23}\cdot\frac{23}{25}\cdot\frac{25}{27}$

Source: MOEMS Division M

All terms cancel except 2 - $\frac{3}{27}=\frac{1}{9}$. I don't feel like finding another problem, someone else do it.

Never fear!! The Soup Nazi is here!!

[size=200]Problem 4[/size]

What positive integer is 56 less than its square?

8
8^2 = 64
64-8=
56
I tried 1,2,3.....,8, and it worked

I don't have a problem.

[size=200][b]Problem 5[/b][/size]

Of all the mathletes at Wantagh Middle School, 80% own
computers and 40% are in band. However, 10% of all the
mathletes neither own computers nor are in band. What
percent of all the mathletes both own computers and are in
band?

[hide="da answer"]$.80-x + .40-x + x= .9
1.2-x=.9
x=.3$
Thus 30% percent of them are in both.[/hide]

well, then there is no solution obviously, because nothing else would work

Meh ok I'll tell you the answer then, it was a bit too hard.
$ x \equal{} \minus{} 3$
Ok
[size=200]New Problem 11[/size]
If $ f(x) \equal{} x^2 \plus{} x$ and $ g(x) \equal{} [10(x_6)]_{10} \plus{} x$
What is the value of $ g[f(120)]$
I like annoying people who don't have good calculators.

Peijin, this is moems, not a place where you can annoy people, ok?

And by the way, a multiplication of a number already expressed in base 6 and a number in base 10 is already in base 10?  $ x_{6}$ does not mean "convert $ x$ to base 6".  I think...

And the only way you'll get that solution $ x\equal{}\minus{}3$ on the old problem 11 is by solving a quartic, which isn't even Mathcounts level.

Looks like this marathon is above MOEMS level... I'll start with an easier problem.
Lisa plays her saxophone for 3 hours.  During that time, Bart goes out and skates on his skateboard and rides 5.4 miles.  At that same rate, how long will it take Bart to skate 11.7 miles?  Express your answer as hours to the nearest tenth.
____[u]Hours[/u]

My solution

[hide]Answer: $ \boxed{6.5 hours}$

Solution:

Ok i used proportions and becuase $ R \equal{} \frac {Distance}{Time}$ (i derived this from the formaula $ D \equal{} RT$ just by dividing both sides by T) and because the rate is the same for both trips we can set the fractions equal to each other.  Therefore $ \frac {5.4}{3} \equal{} \frac {11.7}{x}$ cross multiplying yeilds $ (5.4)(x) \equal{} (3)(11.7)$ dividing by 5.4 gives us $ x \equal{} \frac {(3)(11.7)}{5.4}$ evaluating the right side leaves us with $ \boxed{x \equal{} 6.5}$.[/hide]

Next problem (NOTE = HYPOTHETICAL -- COULDNT HAPPEN):  A plane that took off from Boston is traveling at a constant speed of 1000 mph (zoom zoom) after five hours it was directly over chicago and after 6 hours of total flight is was directly over San Francisco.  Boston is 4000 miles from Chicago.  How far is Chicago from San Francisco.  Round your answer to the nearest whole number.

Not counting wind resistance, engine overheating/failure, etc...
The answer would be 800.
But using your given rate of 1000mph it would be 1000miles.
Therefore, either the wind resistance slows the plane down by 200mph, or your problem was screwed.
[size=200]Problem 13[/size]
Two birds are flying towards each other.
Bird A flies at a rate of 10mph and Bird B flies at a rate of 12mph. The two birds start at the same point. Due to a bad wing, Bird B flies on a line with slope of -1, while Bird A flies straight on a line with a slope of 0. After 90 minutes, what is the area of the triangle formed by the point where the birds start, the point where Bird A is, and the point where bird B is. Assume that Bird B will not crash onto the ground.

You can't actually fly down...........and you didn't account for gravity.
You shouldn't post these problems....not really MOEMS level.

um since he says that the slope of the path is -1 and since it is in MOEMS you dont have to account for gravity.

Hao... Stop annoying people again.
 :(
You can do that during practice ok?

[quote="ZhangPeijin"]
[size=200]Problem 13[/size]
[u]Two birds are flying towards each other[/u].
Bird A flies at a rate of 10mph and Bird B flies at a rate of 12mph. [u]The two birds start at the same point[/u]. Due to a bad wing, Bird B flies on a line with slope of -1, while Bird A flies straight on a line with a slope of 0. After 90 minutes, what is the area of the triangle formed by the point where the birds start, the point where Bird A is, and the point where bird B is. Assume that Bird B will not crash onto the ground.[/quote]

Two birds are flying towards each other starting at the same point?  Ouch.

and the problem is impossible to solve the way it's written

yes i just realized that i assumed that the triangle created was acute therefore the bird is flying towards the ground.  But the bird could also be flying up on a slope of negative one yeilding i beleive a different area.

[u]Two birds are flying in the same direction[/u].
Bird A flies at a rate of 10mph and Bird B flies at a rate of 12mph. [u]The two birds start at the same point[/u]. Due to a bad wing, Bird B flies on a line with slope of -1, while Bird A flies straight on a line with a slope of 0. After 90 minutes, what is the area of the triangle formed by the point where the birds start, the point where Bird A is, and the point where bird B is. Assume that Bird B will not crash onto the ground and that neither gravity or wind will effect the birds' flying.

Ok... this topic is so completely not-MOEMS level recently, it's time for a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188792[/url]new marathon.[/url] Keep it MOEMS level, not algebra/graphing/Mathcounts level. :)

[Locked]


Consider an equilaiteral triangle $ ABC$ and let $ D$ a point on $ BC$. In the extension of $ AC$ consider a point $ E$ such that $ CE\equal{}BD$. Prove that $ DA\equal{}DE$.

Let $ F$ be a point on $ AC$ such that $ DC//AB$
Since then, $ DF\equal{}FC\equal{}CD ,\ AF\equal{}CE,\ \widehat{ACD}\equal{}\widehat{DCE}\equal{}120^\circ$
which implies $ \Delta ADF\equal{}\Delta EDC$ and so that $ DA\equal{}DE$ as desired

Another solution, less elegant is: if $M, P$ are projections of $B, D$ onto $AC$, then obviously $M$ is midpoint of $AC$, and $MP=\frac{BD}{2}=\frac{CE}{2}\implies AP=\frac{AE}{2}$, i.e. $D$ lies on the perpendicular bisector of $AE$, done.

Best regards,
sunken rock


Prove that there exists a positive integer $ n$ such that for every $ x\ge0$ the inequality $ (x\minus{}1)(x^{2005}\minus{}2005x^{n\plus{}1}\plus{}2005x^n\minus{}1)\ge0$ holds.

Your solution started in the right way and then went wrong... remember that if you take the root of a power, you have do divide the exponent, not substract from it.

The answer should be $ n \equal{} 1002$.


Why are demand curves usually plotted with quantity demanded on the "x-axis" and price on the "y-axis"?

It seems more natural that of the quantity demanded should depend on the price, and hence should be plotted as the dependent variable.

To my knowledge, this is just by historical convention, and there is no particular reason it is done this way instead of the other way.

Indeed, S and D graphs are usually set out that way, a long time practice.

a reason that ive heard deals with the fact that historically, in the short run a given quantity of goods acts more as the independent variable as it would dictate how price would fluctuate to reach an equilibrium. this could be due to the properties of the goods as most were agricultural.


[quote="sam-n"]assume that we put n point in n dimensional euclidean space randomly(suppose in bound rigion),consider their delony partition of space(assign to each point the rigion that is nearer to this point ),what is distribution of volume of one rigion when n tends to infinity?[/quote]If you have $n$ points in general position in $\mathbb{R}^n$, wouldn't all Delaunay regions be unbounded?

Yes,it is.but I assume points are in specific bounded rigion ,since randomly choosing points should be meaningful(with uniform distribution).


geometry
circumcircle
exterior angle
geometry solved

$ABC$ is a triangle and $X, Y, Z$ are centers of excircles $ABC$.
Proove that circumcircle of $ABC$ intersects midpoints of sides triangle $XYZ$

could anybody help me with this problem?

[quote="mateivld"]A,B,C are feet of the altitudes of triangle XYZ so you have the nine point circle. :lol:[/quote]
$ZY$ bisects the exterior angle of $A$, so $2\angle{BAZ}+2\angle{XAB}=180^o$. Hence: $\angle{XAZ}=90^o$. Similarly, we get that $A$, $B$, $C$ are the feet of perpendiculars from $X$, $Y$, $Z$ of $\triangle{XYZ}$. Thus the circumcircle of $\triangle{ABC}$ is the nine-point circle of $\triangle{XYZ}$, thus this circle passes through the midpoints of the sides of $\triangle{XYZ}$.


Hello everyone,

    The question is:

1) Prove that the equation (x - a)(x - b) = h^2 has always solutions, no matter what the numbers a, b, and h are. (Hint: Rearrange and calculate the discriminant Δ)

----

[b]My Work :[/b]

h^2 = (x - a)(x- b)

      = x^2 - bx - ax + ab

D = b^2 - 4ac

   = (-b)^2 - 4(1)(-a)

   = b + 4a

If this is right, how would b + 4a prove that this equation always has solutions? I know D must < 0 or > 0 to have real solutions.

I have it:
First, multiply everything out.  $x^{2}-(a+b)\cdot x+ab=h^{2}$
Move $h^{2}$...   $x^{2}-(a+b)\cdot x+(ab-h^{2})=0$
NOW we find the discriminant. $(a+b)^{2}-4(1)(ab-h^{2})$
Multiply out and rearrange. $a^{2}-2ab+b^{2}+4h^{2}$
$(a-b)^{2}+4h^{2}$
Since squares are always positive, the above expression is always positive.  Thus there are always real solutions.

Note: You can't just take discriminant anytime.  One side of the equation has to be 0, and the degree of the other side has to be 2.

[quote="CircleSquared"]
NOW we find the discriminant. $b^{2}-4ac=a^{2}+2ab+b^{2}-4ab+4h^{2}$
[/quote]

Thank you for your reply. I understand everything up to that equation.

For $-4ac$, how did you get ab as a and $4h^{2}$ as c?

I just can figure out how b^2 = a^2 [b]+[/b]2ab + b^2

I did the calculations many times and each time it comes out to : b^2 = a^2 [b]-[/b]2ab + b^2

Please explain your method more clearly.

Potential

Sorry, I've gone back and made it clearer (I hope)
For the equation, [b]in the discriminant[/b], a should be $1$
b should be $-(a-b)$
c should be $ab-h^{2}$.

The discriminant a, b, and c are different from the original quadratic's a and b.

Thank you very much to all those who helped, especially CircleSquared and Potential.

[quote="catalan"]Prove that the equation (x - a)(x - b) = h^2 has always solutions, no matter what the numbers a, b, and h are. [/quote]

This is easily done without any calculation of the determinant.  Simply note that for $x < min(a, b), x > max(a, b)$ the product $(x-a)(x-b)$ takes on all positive values.

Alternately, at $x = max(a, b)$ we have $(x-a)(x-b) = 0$, whereas at $x = h+|a|+|b|$ we have $(x-a)(x-b) \ge (h+|a|)(h+|b|) \ge h^{2}$, and since polynomials are continuous the function $(x-a)(x-b)$ must take on the value $h^{2}$ somewhere in between.


function
trigonometry
calculus
derivative
real analysis
real analysis unsolved

Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be the function given by $ f(x) \equal{} x^2\sin{\frac {1}{x}} \plus{} x$ if $ x$ is not equal to $ 0$;
and $ f(0) \equal{} 0$

Prove that there is no positive number $ c$ such that the function $ f: (\minus{} c,c)\rightarrow\mathbb{R}$ is increasing.

$ f'(x) \equal{} 1 \minus{} \cos(1/x) \plus{} 2x\sin(1/x)$ if $ x\ne0.$ For $ x_k \equal{} (\pi/2 \plus{} k\pi)^{ \minus{} 1}$ the derivative is alternatively positive and negative. Its continuity does the rest.

Once was enough:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316470
Don't double-post.