The length of a rectangle is increased by $ 10\%$ and the width is decreased by $ 10\%$. What percent of the old area is the new area?

$ \textbf{(A)}\  90  \qquad
\textbf{(B)}\   99  \qquad
\textbf{(C)}\   100  \qquad
\textbf{(D)}\   101  \qquad
\textbf{(E)}\   110$

[hide="Click for solution"]
Let the original length be $ l$ and the original width be $ w$.  Then, the original area is $ lw$.

The new length is $ 1.1l$ and the new width is $ 0.9w$.  Thus, the new area is $ 0.99lw$, which is $ 99 \%$ of the old area; answer choice $ \boxed{\textbf{(B)}}$.
[/hide]

[quote=Louis_Vuitton][quote=IDKorean]I set up a quick example of this problem

original l=10
original width= 10

After scaling

l= 11
width=9

11x9=99

Therefore, its 99% of the old area[/quote]

i don't get it
how did you know the rectangle had original length of 10 and width of 10???? the problem did not tell you???[/quote]

for these types of problems since they don't give you the length and width of the recatngle you can assume any number
the only reason they choose 10 is because it is easy to work with perecentages

[hide]
I usually substitute arbitrary values as the length and width of this rectangle in order to simplify calculations knowing that the area will always decrease by a certain constant regardless of the length and width due to the fact that this problem states that the statement is axiomatic. Because the numbers will be increased and decreased by $10\%$, I will set $l = 10$ and $w = 10$.

$10\times1.1 = 11$ and $10\times0.9 = 9$. The area of the scaled rectangle would be $9\times11 = 99$. Because $\frac{99}{100} = 0.99$, the final answer is $99\%$, or $\boxed{B}$
[/hide]

[quote=Louis_Vuitton][quote=IDKorean]I set up a quick example of this problem

original l=10
original width= 10

After scaling

l= 11
width=9

11x9=99

Therefore, its 99% of the old area[/quote]

i don't get it
how did you know the rectangle had original length of 10 and width of 10???? the problem did not tell you???[/quote]

1. Please stop bumping old threads
2. We can just let the side length be 10 because for these problems it should work for any side length.

Let the side length of the square be x.

$(1-\frac{1}{10})x \times (1+\frac{1}{10})x=\frac{9}{10}\times \frac{11}{10}=\frac{99}{100}$

Therefor the new rectangle is 99% of the square.

Did I seriously waste my 200th post on this  :help:  :wallbash_red:  :weightlift:  :furious:  :censored:
 :bomb:


analytic geometry
graphing lines
slope
geometry unsolved
geometry

I'm in need of a specific formula involving distance and slope on a graph:

I have a point on a line. I know the value of the point and the slope of the line. My problem is that I need to find a point on that line that's a specific distance from the first point. What formula would I use to do that?

Any help is appreciated.

Take two points $A,\ B$ on a line with slope $m.$ Let $x_{A},\ x_{B}$ be the absiccae of the points $A,\ B$.Make a right triangle $ABH$ with legs $AH,\ BH.$ Since $\frac{BH}{AH}=\frac{1}{|m|}$, we have $AB=\sqrt{1+m^{2}}AH\Longleftrightarrow \boxed{AB=\sqrt{1+m^{2}}|x_{A}-x_{B}|}$.

Specifically:  Suppose the point is (3,5), the slope 1/2, and the distance 7 [there will be two such points!]

Equation of the line: y - 5 = (1/2)(x - 3)

Distance formula [Pythagoras, using (x,y) and (3,5)] :

\[(x-3)^{2}+(y-5)^{2}= 7^{2}\]

Basically 2 variables 2 equations.

Thanks, guys. I'm starting to understand this better now. I think you answered my question, but for further clarity--say I have the slope, [i]m[/i], (like 1/2 for instance) and I have the point x1 and y1, (like 3, 5, like you mentioned)--what equation do I use to determine x and what equation do I use to determine y where the x and the y are alone on one side of those equations?

Thanks again in advance!


function
advanced fields
advanced fields solved

Show that the function

\[
\mu(z) = \sum z^{2^n} = z + z^2 + \ldots
\]
is analytic in the open unit disc $\mathbf{B}^2 \setminus \mathbf{S}^1$, ie for all $|z| < 1$, and that it cannot be analyticaly continued to any larger open set.

Convergence in $|z| < 1$:

$\displaystyle{\sum |z^{2^n}| \leq \sum |z^n| = \frac 1{1-|z|}}$ for $|z| < 1$, so $\mu(z)$ converges absolutely in $|z| < 1$.

For the second, I'll demonstrate that the set of singularities of $\mu$ is dense on $\mathbf S^1$, which shows that no convergence circle of a power series for $\mu$ around some $z, |z| < 1$ can exceed $\mathbf S^1$.

Every $\zeta$ satisfying $\zeta^{2^N} = 1$ for some N is a singularity of $\mu$, for $\displaystyle{\mu(\zeta) = \sum_{n=0}^{\infty} \zeta^{2^n} = \sum_{n=0}^{N-1}\zeta^{2^n} + \sum_{n=N}^{\infty} \zeta^{2^n} = \sum_{n=0}^{N-1}\zeta^{2^n} + \sum_{n=0}^{\infty} (\zeta^{2^N})^{2^n} = \underbrace{\sum_{n=0}^{N-1}\zeta^{2^n}}_\text{convergent} + \underbrace{\sum_{n=0}^{\infty} (1)^{2^n}}_\text{divergent}}$, and so $\mu(\zeta)$ diverges. But the set of such $\zeta$s is dense in $\mathbf S^1$.


Given $ a, b, c > 0$. Prove that: $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc} \geq\ \frac {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}$

Wait for a the nice proofs  :)
@can_hang: [hide]Anh ko dc lam` bai` nay` dau nhe' :P[/hide]

I think we can prove this ineq by SOS (of course, it isn't my way and I haven't tried with this way) :maybe:

Of course, indeed, SOS is a possible way but rather ugly, this problem is an example. The works to prove $ Sb \plus{} Sc \geq\ 0$ is not nice any more :oops:
My proof is only Am-GM  :)

[quote="BG Yoda"]$ (a^2 \plus{} bc)(b \plus{} c) \equal{} a^2b \plus{} a^2c \plus{} b^2c \plus{} c^2b \equal{} c(a^2 \plus{} b^2) \plus{} b(a^2 \plus{} c^2)$ $ \ge 2\sqrt {bc(a^2 \plus{} b^2)(c^2 \plus{} a^2)}\dots$  :wink:[/quote]
Yes, that's it, my friend. You saw my way to creat this ineq  :oops:

Very nice,BG yoda.I think almost problem of nguoivn can prove by AM-GM,but it is very hard to saw his way.Thank you BG yoda and nguoivn.

[quote="BG Yoda"]$ (a^2 \plus{} bc)(b \plus{} c) \equal{} a^2b \plus{} a^2c \plus{} b^2c \plus{} c^2b \equal{} c(a^2 \plus{} b^2) \plus{} b(a^2 \plus{} c^2)$ $ \ge 2\sqrt {bc(a^2 \plus{} b^2)(c^2 \plus{} a^2)}\dots$  :wink:[/quote]

BRILLIANT solution, BG yoda... a nice problem with an even NICER solution!! :clap2:

Hope nguovin will give us more problems like this! :lol:


How can one prove $ \dim(W_1\plus{}W_2)\equal{}\dim(W_1)\plus{}\dim(W_2)\minus{}\dim(W_1\cap W_2)$, where $ W_1$ and $ W_2$ are finite-dimensional subspaces of a vector space?

Rough outline: Find a basis for $ W_1\cap W_2.$ This is an independent set in $ W_1$ and hence can be extended to a basis for $ W_1.$ Go back to the basis for the intersection and this time extend it to a basis for $ W_2.$ Show that we have a basis for $ W_1\plus{}W_2.$ Count the elements of the various bases.


If $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}3abc$, then
$ \frac{ab}{9ab\minus{}1}\plus{}\frac{bc}{9bc\minus{}1}\plus{}\frac{ca}{9ca\minus{}1}\leq\frac{3}{8}$

[quote="lipako"]If $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3abc$, then
$ \frac {ab}{9ab \minus{} 1} \plus{} \frac {bc}{9bc \minus{} 1} \plus{} \frac {ca}{9ca \minus{} 1}\leq\frac {3}{8}$[/quote]

let $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c}, x \plus{} y \plus{} z \equal{} 3$ , so it's equivalent to proof

$ \frac {1}{9 \minus{} xy} \plus{} \frac {1}{9 \minus{} yz} \plus{} \frac {1}{9 \minus{} zx} \minus{} \frac {3}{8}\geq 0,$ $ \frac {3}{8} \minus{} \frac {1}{9 \minus{} xy} \minus{} \frac {1}{9 \minus{} yz} \minus{} \frac {1}{9 \minus{} zx}$

$ \equal{} \frac {243 \minus{} 99xz \minus{} 99yz \minus{} 99xy \plus{} 19yz^{2}x \plus{} 19yx^{2}z \minus{} 19y^{2}xz \minus{} 3y^{2}x^{2}z^{2}}{8(9 \minus{} xy)(9 \minus{} yz)(9 \minus{} zx)}$

$ 9 \minus{} xy\geq 0,9 \minus{} yz\geq 0,9 \minus{} zx\geq 0,$  it's obvious .so it's equivalent to proof

$ 243 \minus{} 99xz \minus{} 99yz \minus{} 99xy \plus{} 19yz^{2}x \plus{} 19yx^{2}z \plus{} 19y^{2}xz \minus{} 3y^{2}x^{2}z^{2}\geq 0$

let $ x \plus{} y \plus{} z \equal{} p,xy \plus{} yz \plus{} zx \equal{} \frac {p^{2} \minus{} q^{2}}{3},xyz \equal{} r$,so $ 0 < q < p \equal{} 3, r < 1 .$  $ f(p,q,r) \equal{} 243 \plus{} 19pr \minus{} 3r^{2} \minus{} 33p^{2} \plus{} 33q^{2}$

$ f(3,q,r) \equal{} \minus{} 54 \plus{} 33q^{2} \minus{} 3r^{2} \plus{} 57r,$  if $ p\leq 2q, f(3,q,r)\geq f(3,q,0)\geq f(3,\frac {3}{2},0) \equal{} \frac {81}{4}\geq 0 .$

if $ p\geq 2q$ ,$ f(3,q,r)\geq f(3,q,\frac {(3 \minus{} 2q)(3 \plus{} q)^{2}}{27})$

$ \equal{} \frac {q^{2}(8748 \plus{} (81 \minus{} (2q)^{4}) \plus{} 18(27 \minus{} (2q)^{3}) \plus{} 81(9 \minus{} (2q)^{2} \plus{} 1836(3 \minus{} 2q)}{972}\geq 0.$

The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that:
\[ \frac{1}{6\minus{}ab}\plus{}\frac{1}{6\minus{}ac}\plus{}\frac{1}{6\minus{}bc}\leq\frac{3}{5}\]

[quote="arqady"]The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {1}{6 \minus{} ab} \plus{} \frac {1}{6 \minus{} ac} \plus{} \frac {1}{6 \minus{} bc}\leq\frac {3}{5}\]
[/quote]

$ \frac {3}{5} \minus{} \frac {1}{6 \minus{} ab} \minus{} \frac {1}{6 \minus{} bc} \minus{} \frac {1}{6 \minus{} ca}$

$ \equal{} \frac {108 \minus{} 48ca \minus{} 48bc \minus{} 48ab \plus{} 13bc^{2}a \plus{} 13a^{2}bc \plus{} 13ab^{2}c \minus{} 3a^{2}b^{2}c^{2}}{5(6 \minus{} ab)(6 \minus{} bc)(6 \minus{} ca)}$

$ 6 \minus{} ab\geq 0,6 \minus{} bc\geq 0,6 \minus{} ca\geq 0,$ it's obvious.so it's equivalent to proof

$ 108 \minus{} 48ca \minus{} 48bc \minus{} 48ab \plus{} 13bc^{2}a \plus{} 13a^{2}bc \plus{} 13ab^{2}c \minus{} 3a^{2}b^{2}c^{2}\geq 0.$

let$ a \plus{} b \plus{} c \equal{} p,ab \plus{} bc \plus{} ca \equal{} \frac {p^{2} \minus{} q^{2}}{3},abc \equal{} r$ so $ 0 < q < p \equal{} 3, r < 1.f(p,q,r) \equal{} 108 \minus{} 3r^{2} \plus{} 16q^{2} \minus{} 16p^{2} \plus{} 13pr,$

$ f(3,q,r) \equal{} \minus{} 36 \minus{} 3r^{2} \plus{} 16q^{2} \plus{} 39r,$   if $ p\leq 2q, f(3,q,r)\geq f(3,q,0)\geq (3,\frac {3}{2},0) \equal{} 0 .$

 if $ p\geq 2q ,$ $ f(3,q,r)\geq f(3,q,\frac {(3 \minus{} 2q)(3 \plus{} q)^{2}}{27})$

$ \equal{} \frac {q^{2}(81 \minus{} (2q)^{4}) \plus{} 18(27 \minus{} (2q)^{3}) \plus{} 81(9 \minus{} (2q)^{2}) \plus{} (1188(3 \minus{} 2q)}{972}.$

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114583


$2^n + 153= a^2$        where $n, a$  is positive integers

If n is odd analyze the equation mod 6 and you will get a contradiction.
Let n=2k the equation becomes 153=(a-(2^k))(a+(2^k)) obviously k<=7 so check all these cases(sorry, no time to type all of it)


Hi, I need help on this problem:

Find the sum:

$\displaystyle \frac 1 {\sqrt[3]1 + \sqrt[3]2 + \sqrt[3] 4} + \frac 1 {\sqrt[3] 4 + \sqrt[3] 6 + \sqrt[3] 9} + \frac 1 {\sqrt[3] 9 + \sqrt[3] {12} + \sqrt[3] {16}} $

Well I just notice that those 3 sequences of numbers are similar to x 2 +xy+y 2 ... So I multiplied by x-y so the denominator will become the form of $x^3-y^3$ but I got the same answer.


The reality game show Survivor is played with 16 people divided into two tribes of 8. In the first episode, two people get homesick and quit. If every person has an equal chance of quitting, and the probability that one person quits is independent of the probability that any other person quits, what is the probability that both people who quit are from the same tribe?

The probability that they're from different tribes is $ \frac {8\cdot8}{\dbinom{16}2} \equal{} \frac8{15}$. Thus, the probability that they're from the same tribe is $ 1 \minus{} \frac8{15} \equal{} \boxed{\frac 7{15}}$.

Edit: Thanks, Yo--tinytim. :)

There are $ 16C2 = 120$ ways two of the 16 people to go home. There are $ 8C2 + 8C2 = 56$ ways for the two people to be in the same tribe. So the answer is actually ${ \frac {56}{120} = \frac{7}{15}}$.

In your solution, you assumed that the order the 2 people were sent home mattered so the numberator of your fraction should actually be $ 8\cdot{8}\cdot{2}$

Or you can just pick the first person who quits and then for the other person to be picked, there are 7 team members who can quit and 8 others from the other team, so 15 people together. Thus, 7/15.

[quote=Lorleolando]Solution 2 is awesome.[/quote]

so much more of a short cut than solution 1

[hide=solution]
Say we have $$\star \star \star \star \star \star \star \star | \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar.$$

WLOG, assume that one of the $\bigstar$s get homesick.
$$\star \star \star \star \star \star \star \star | \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \cancel{\bigstar}.$$

In order to meet the desired condition, we must pick one of the ones (that are boxed) on the $\bigstar$'s team.
$$\star \star \star \star \star \star \star \star | \boxed{\bigstar} \boxed{\bigstar} \boxed{\bigstar} \boxed{\bigstar} \boxed{\bigstar} \boxed{\bigstar} \boxed{\bigstar} ~ \cancel{\bigstar}$$

Out of our $15$ possible choices, $7$ (the boxed ones) meet our condition. Therefore we have $\boxed{\frac{7}{15}}.$
[/hide]


trigonometry
geometry
trapezoid
geometry unsolved

:? [b]Hi everyone!I'm a new member and I've composed a problem, but I don't know to solve it.Her's the problem:
Let's take the squares ABCD, DCFE and EFGH(A-D-E-H and B-C-F-G the order).Find m(<AIB), where {I}=AF intersectated whith BD.Could anyone help me?[/b]

[quote="Virgil Nicula"]You didn't use the third square ![/quote]
Yes, you are right.I didn't use the third square :blush: .

[color=darkblue][b]Generalization 1.[/b] Given are $n\ge 2$ squares $A_kB_kB_{k+1}A_{k+1}\ ,\ k\in \overline {0,n-1}$ .

For $1\le m<n$ denote the intersection $I\in A_0B_m\cap B_0A_n$ .

Prove that $\boxed {\ \tan \widehat {A_0IB_0}=\frac{m+n}{mn-1}\ }\ .$[/color]

[color=darkred][b]Generalization 2.[/b] Given is the trapezoid $ABCD$ for which $AD\parallel CD$ ,

$AB\perp AD$ , $AB=1$ , $BC=m$ , $AD=n$ , where $1\le m<n$ .

Denote the intersection $I$ of the diagonals $AC$ and $BD$ .

Prove that $\boxed {\ \tan\widehat {AIB}=\frac{m+n}{mn-1}\ }\ .$[/color]

[quote="Virgil Nicula"][color=darkblue][b]Generalization 1.[/b] Given are $n\ge 2$ squares $A_kB_kB_{k+1}A_{k+1}\ ,\ k\in \overline {0,n-1}$ .

For $1\le m<n$ denote the intersection $I\in A_0B_m\cap B_0A_n$ .

Prove that $\boxed {\ \tan \widehat {A_0IB_0}=\frac{m+n}{mn-1}\ }\ .$[/color]

[color=darkred][b]Generalization 2.[/b] Given is the trapezoid $ABCD$ for which $AD\parallel CD$ ,

$AB\perp AD$ , $AB=1$ , $BC=m$ , $AD=n$ , where $1\le m<n$ .

Denote the intersection $I$ of the diagonals $AC$ and $BD$ .

[b]Prove that $\boxed {\ \tan\widehat {AIB}=\frac{m+n}{mn-1}\ }\ .$[/color][/quote[/b]]

    [b] Mr Nicula, thank you for the demonstration.It's a very interesting demonstration.You are a very good professor.I wish I had had a good professor like you. :lol: [/b]

[quote="Virgil Nicula"][hide="Here are two proofs of the generalization 2."]
[color=darkblue][b][u]The first method ($7^{th}$ class).[/u][/b] Denote $\phi=m(\widehat {AIB})$.
$AC=\sqrt{1+m^2}\ ,\ BD=\sqrt{1+n^2}\ ,\ \triangle AID\sim\triangle CIB\Longrightarrow$
$1.\blacktriangleright\frac{IA}{n}=\frac{IC}{m}=\frac{AC}{m+n}\Longrightarrow IA=\frac{n}{m+n}\cdot \sqrt{1+m^2}\ .$
$2.\blacktriangleright\frac{IB}{m}=\frac{ID}{n}=\frac{BD}{m+n}\Longrightarrow IB=\frac{m}{m+n}\cdot \sqrt{1+n^2}\ .$
$\cos \phi =\frac{IA^2+IB^2-AB^2}{2\cdot IA\cdot IB}=\frac{mn-1}{\sqrt{(1+m^2)(1+n^2)}}$.
But $(1+m^2)(1+n^2)=(m+n)^2+(mn-1)^2$.
Therefore, $\sin \phi =\frac{m+n}{\sqrt{(1+m^2)(1+n^2)}}\Longrightarrow \boxed {\ \tan \phi=\frac{m+n}{mn-1}\ }\ .$[/color]

[color=darkred][b][u]The second method ($9^{th}$ class).[/u][/b] Denote $u=m(\widehat {ADB})$ and $v=m(\widehat {ACB})$.
$\tan u=\frac 1n\ ,\ \tan v=\frac 1m\ ,\ \phi =u+v\Longrightarrow \tan \phi=\frac{\tan u+\tan v}{1-\tan u\cdot \tan v}\Longrightarrow$
$\tan \phi =\frac{\frac 1n +\frac 1m}{1-\frac{1}{mn}}\Longrightarrow \boxed {\ \tan \phi=\frac{m+n}{mn-1}\ }\ .$[/color][/hide][/quote]
     [b]It is very good... [/b]:roll:


inequalities
function
calculus
calculus computations

Okay. So, I have to prove this:

lim(x,y)->0,0 _(xy^2)/(x^2+y^2+y^3)= 0
So, I divide the top and bottom by (y^2) and get:
lim(x,y)->0,0 _x/((x/y)^2+1+y)

So since (x/y)^2 must be positive, the limit evaluates to a positive number.
So, the whole thing becomes after you plug in: 0/(positive number)

That equals 0.
Is that a correct way to do it? Mostly the part of the 0/0 indeterminate form.

In my opinion, always use inequality like below
\[ 0 \leqslant \left| {\frac{{xy^2 }}
{{x^2  \plus{} y^2  \plus{} y^3 }}} \right| \leqslant \frac{{\left| x \right|y^2 }}
{{3\left| x \right|^{\frac{2}
{3}} \left| y \right|^{\frac{5}
{3}} }} \equal{} \frac{1}
{3}\left| {xy} \right|^{\frac{1}
{3}} .
\]

Err... How do you get the denominator for the greater than f(x,y) part?

We think of $ y^3$ as less significant than $ x^2 \plus{} y^2$ when $ y$ is small. But if we just drop it from the denominator, the function can change in either direction. We want to drop something nonnegative from the denominator. So we drop $ x^2$ and get
$ \frac {|x|y^2}{x^2 \plus{} y^2 \plus{} y^3}\le \frac {|x|y^2}{y^2 \plus{} y^3} \equal{} \frac {|x|}{1 \plus{} y}$... but only if $ y\ne 0$. However, this is not a problem, because $ \frac {|x|y^2}{x^2 \plus{} y^2 \plus{} y^3}\le\frac {|x|}{1 \plus{} y}$ even if $ y \equal{} 0$.

The possibility of $ y$ being zero is why I would avoid writing $ (x/y)^2$.


In an urn we have red and blue balls. A person has invented the next game: he extracts balls until he realises for the first time that the number of blue balls is equal to the number of red balls. After a such game he finds out that he has extracted 10 balls, and that there does not exist 3 consecutive balls of the same color. Prove that the fifth and the sixth balls have different collors.

Since the game lasts for 10 extractions, the first two balls have the same color, wlog suppose it's blue. Then the sequence is BBRB, given that no three consecutive balls have the same color. Now suppose the fifth and sixth have the same color. It must be red. Hence the sequence is BBRBRR and the game is over after six extractions. Contradiction.


I post this problem againt .
Prove that: $cos^{n}x.cosnx = \frac{1}{2^n}\sum\limits_{i=0}^{n}\large{n\choose i} cos2ix$, for all $n\in N*$ and $x\in R$

[hide="Here is a  proof using the complex numbers."]You can use the complex numbers. Denote $z=\cos x+i\sin x$ and define $A=\sum_{k=0}^n C^k_n\cos 2kx$, $B=\sum_{k=0}^nC^k_n\sin 2kx\Longrightarrow$ $A+iB=$ $\sum_{k=0}^n C^k_n z^{2k}=$ $\left(1+z^2\right)^n=$ $(1+\cos 2x+i\sin 2x)^n=$ $\left(2\cos^2x+2i\sin x\cos x\right)^n=$ $2^n\cos ^n x(\cos x+i\sin x)^n=$ $2^n\cos ^n x(\cos nx+i\sin nx)$ $\Longrightarrow$ $A=\boxed {\sum_{k=0}^n C^k_n\cos 2kx=2^n\cos ^n x\cos nx}$ $B=\boxed {\sum_{k=0}^n C^k_n\sin 2kx=2^n \cos^n x\sin nx}\ .$[/hide]


probability
expected value
probability and stats

With a binomial distribution each event has the same probability. I'm looking for a distribution where the probability for each event is different.

For example: There are 3 independent events and the probability of success for each event is different. What is the probability of at least 2 successes?

What kind of distribution would I use for that or how would I alter the binomial distribution formula?

Thanks in advance.

Let's say there are three non-identical trials.

The probability of three successes is simple: $ p_1p_2p_3.$

The probability of exactly two successes is a little more complicated. Here it is in three different algebraic forms:

$ p_1p_2(1 \minus{} p_3) \plus{} p_1(1 \minus{} p_2)p_3 \plus{} (1 \minus{} p_1)p_2p_3$

$ \equal{} p_1p_2 \plus{} p_1p_3 \plus{} p_2p_3 \minus{} 3p_1p_2p_3$

$ \equal{} p_1p_1p_3\left(\frac1{p_1} \plus{} \frac1{p_2} \plus{} \frac1{p_3} \minus{} 3\right)$

You can see that that would reduce to the binomial probability $ 3p^2(1 \minus{} p)$ if $ p_1 \equal{} p_2 \equal{} p_3 \equal{} p.$ You can probably also see how to get the probability of exactly one success.

I'm not sure I can see any less cumbersome algebraic representations - I think you're pretty much stuck expressing things this way.

If I do this for $ p_1 \equal{} 0.3, p_2 \equal{} 0.5, p_3 \equal{} 0.7$ we get the following probabilities:

$ P(3\text{ successes}) \equal{} 0.105$
$ P(2\text{ successes}) \equal{} 0.395$
$ P(1\text{ success}) \equal{} 0.395$
$ P(\text{ successes}) \equal{} 0.105$

With an expected value of $ 1.5.$ Note that the expected value is simply $ \sum p_i,$ which in this case is $ .3 \plus{} .5 \plus{} .7 \equal{} 1.5.$

If you made all three $ p$'s equal to $ .5$ you would get a binomial distribution with mean $ 1.5.$ The particular probabilities of that would be $ .125, .375., .375.$ and $ .125.$

The variance of the binomial is $ np(1 \minus{} p).$ The variance for this distribution would be $ \sum p_i(1 \minus{} p_i).$ In this particular instance, the variance in the binomial $ p \equal{} .5$ case would be $ 3(.5)(.5) \equal{} 0.75$ while the variance for the varied case with $ p_i \equal{} .3,.5,.7$ respectively is $ (.3)(.7) \plus{} (.5)(.5) \plus{} (.7)(.3) \equal{} 0.67.$

The overall picture is this: we are talking about a random variable that can be given as the sum of several independent random variables. The individual pieces are well-understood, being Bernoulli trials, but the since the pieces differ from each other some of the formulas, particularly for the probabilities, don't simplify all that much. The mean and variance are easy, since they're simply the sums of the means and variances of the individual pieces, respectively.

Thank you very much for the reply.

Well, that solution is perfect for 3 trials, but I was hoping there was some generic way to find the probability for n non-identical trials and exactly k successes. For identical trials with probability P, it is simply (as you mentioned) C(n, k) * P^k * (1-P)^(n-k). I was hoping there was some way to alter that formula using P1, P2, ... , Pn (where Pn is probability of nth trial being successful) instead of P.

Is there any way to do this or is it too difficult to generalize?

EDIT: Can P just be replaced with P1/k * P2/k * ... * Pn/k in the binomial expansion? I think that might work?

It's a pretty nasty mess in general. It's a struggle to find a reasonable notation to represent it.

Suppose we have $ n$ trials, with probability of success $ p_i$ for $ 1\le i\le n.$ Let $ Y$ be the number of successes.

I'm going to give a formula for $ P(Y\equal{}k).$ This formula is going to involve a sum over all subsets of $ \{1,2,\dots,n\}$ with exactly $ k$ elements. Let's adopt the notation $ \Sigma_k$ for the set of all such subsets - that is, the set of all subsets of $ \{1,2,\dots,n\}$ with exactly $ k$ elements.

With all of that said, this is the formula:

$ P(Y\equal{}k)\equal{}\sum_{S\in\Sigma_k}\left(\prod_{i\in S}p_i\right)\left(\prod_{i\notin S}(1\minus{}p_i)\right)$

This is a sum with $ \binom{n}{k}$ terms, each term of which is the product of $ n$ factors. The reason the binomial distribution can be expressed in nicer terms is that if all the $ p_i$ are equal, then all of the terms are equal and the sum can be replaced by $ \binom{n}{k}$ times one sample term. I would welcome another representation of this that is either more algebraically elegant or more computationally tractable, but I don't see it myself.

The expectation and the variance are much simpler to represent. As I said above,

$ E(Y)\equal{}\sum_{i\equal{}1}^np_i$ and $ \text{Var}(Y)\equal{}\sum_{i\equal{}1}^np_i(1\minus{}p_i).$

For large $ n$ (and for most $ p_i$ not too close to either zero or one), we should be able to use that to estimate cumulative probabilities for $ Y$ by assuming that its distribution is approximately normal.

That is a pretty ugly formula, but I see it's derived directly from a probability tree and you're manually finding all the combinations that have at least k successes. Actually, this is exactly how I calculate these type of probabilities now, I just never realized that you could actually write that out in a formula. I also hoped that there was a shortcut, but I guess not.

Well, I have a better understanding now, thanks for your help.

notice also that if all the probabilities are quite small, you can approximate what you are looking for by a Poisson random variable:
http://en.wikipedia.org/wiki/Le_Cam%27s_theorem

Yes, when I added the qualifier about the $ p_i$ not being close to zero or one for the normal approximation, I was thinking if they were all close to zero or one, we'd use a Poisson approximation. (For $ p_i$ near $ 1,$ flip the definitions of "success" and "failure" so as to trade $ p_i$ for $ 1\minus{}p_i.$)

Yeah, that makes sense, I never thought of that (though not applicable for the data I encounter).

Thank you both for your help. :)


calculus
derivative
algebra
system of equations
algebra unsolved

Solve the system of equations:
$ x\plus{}\sqrt{x^2\minus{}2x\plus{}2}\equal{}3^{y\minus{}1}\plus{}1$ and $ y\plus{}\sqrt{y^2\minus{}2y\plus{}2}\equal{}3^{x\minus{}1}\plus{}1$

let $ u\equal{}x\minus{}1, v\equal{}y\minus{}1$. then the equations become

$ u\plus{}\sqrt{u^2\plus{}1}\equal{}3^v$
$ v\plus{}\sqrt{v^2\plus{}1}\equal{}3^u$

note that  $ 3^x$ and $ x\plus{}\sqrt{x^2\plus{}1}$ are both strictly increasing on $ \mathbb{R}$, so we must have $ u\equal{}v$. so we must have $ u\plus{}\sqrt{u^2\plus{}1}\equal{}3^u$. clearly $ u\equal{}0$ is a solution. now, the derivative of $ h(u)\equal{}3^u\minus{}u\minus{}\sqrt{u^2\plus{}1}$ is $ 3^u\ln3\minus{}1\minus{}\frac{u}{\sqrt{u^2\plus{}1}}$, which is positive for all $ u$, which means $ h(u)$ is strictly increasing and thus $ u\equal{}0$ is the only solution. so $ (x,y)\equal{}(1,1)$ is the only solution.


search
number theory unsolved
number theory

For all positive integer $n>1$, let $p(n)$ be the greatest prime divisor of $n$.
Prove that there are infinitely integers $n$ with
$p(n)<p(n+1)<p(n+2)$

[quote="Rust"]It posted before.[/quote]
course I searched before(bulgaria, p(n)<p(n+1), etc.) but I coudn't find it.
link please

dont whine mate, im thinking about it atm :)

http://www.mathlinks.ro/Forum/viewtopic.php?t=2005
try more effective search key words

how can you up this post?
this post was under

It is easy problem.
It is obviosly $p(m^{k})=p(m)$ and there are infinetely many m, suth that $p(m-1)<p(m)$ (for example m is prime).
If $(p(m+1)>p(m)$, then work$n=m-1$, else$p(m^{2}-1)=max(p(m-1),p(m+1))<p(m)=p(m^{2})$. If $p(m^{2}+1)>p(m)$, then work$n=m^{2}-1$, e t.c. If$\forall k\in N \ p(m^{2^{k}}+1)<p(m)$we have contradition with$(f_{k},f_{l})=1 \ or \ 2$, were $f_{k}=m^{2^{k}}+1$.  :lol:


[quote="Bomboyage"]12. There are purple and green marbles in a bag in an unknown ratio, but when 15 green marbles are removed, the ratio becomes 2 purple marbles for each green marble. Then, when 45 purple marbles are additionally removed, the ratio becomes 5 green marbles for each purple marble. How many green marbles were in the bag before any were removed?[/quote]

[hide] so lets say that # of purple is x and # of green is y.

so y/x-15=2/1

solve for y
Cross Multiply:

y=2x-30

and y-45/x-15=1/5

so subsitute it into the equation

2x-30-45/x-15=1/5

2x-75/x-15=1/5

10x-375=x-15

10x=x+390
9x=390

wait, it seemed right. What did I do wrong?

anyways, I'll do it with guess and check.

so 60 p 45 g
60p/30g
15p/30g
doesn't work, try a smaller number

50p/40g
50p/25g
5p/25g

so 40 green marbles.

By doing it this way, it took me like 30 seconds ;)
[/hide]

on problems like these, solutions may seem really long, but its very quick if you know what to do

[quote="Bomboyage"][quote="Ihatepie"]

10x-375=x-15

10x=x+390
9x=390

wait, it seemed right. What did I do wrong?

[/quote]
It's rather 360.

Okay, I got it. Thank you for your help.[/quote]
 :blush: I hate it when I do the entire problem right but mess up from a simple arithmetic error.

BTW where did you get the 1999 state sprint?


calculus
derivative
calculus computations

Given $ F(x,y) \equal{} f(u, v)$, where $ u \equal{} x^2 \minus{} y^2$ and $ v \equal{} 2xy$ and suppose that $ \frac {\partial^2 f}{\partial u\partial v} \equal{} \frac {\partial^2 f}{\partial v \partial u}$.
Find $ \frac {\partial F}{\partial x}$ and $ \frac {\partial^2 F}{\partial x^2}$ in terms of the first or second order partial derivatives of $ f$ about $ u$ or $ v$.

[hide="My answer"]
$ \frac {\partial F}{\partial x} \equal{} 2x\frac {\partial f}{\partial u} \plus{} 2y\frac {\partial f}{\partial v},$

$ \frac {\partial^2 F}{\partial x^2} \equal{} 2\frac {\partial f}{\partial u} \plus{} 4x^2\frac {\partial^2 f}{\partial u^2} \plus{} 8xy\frac {\partial^2 f}{\partial u\partial v} \plus{} 4y^2\frac {\partial^2 f}{\partial v^2}.$

Please point out my mistakes if there is any.
[/hide]

Well on your second line, it should be $ \frac{\partial^{2}f}{\partial u^{2}}$.


If $z_1$ and $z_2$ are the roots of the equation $\alpha z^2 + 2 \beta z + \gamma = 0$ , then prove that

$| \alpha | ( |z_1| + |z_2| ) = | \beta + \sqrt{\alpha \gamma} | + | \beta - \sqrt{ \alpha \gamma} |$

[quote="Singular"]I think it is wrong.  For example, consider x^2 - 3x + 2.  It has roots 1,2.

So then 3  must equal  $| -3 + \sqrt{2} | + | -3 - \sqrt{2} |$ which isn't true.[/quote]

$\beta=\frac32$
so $3=| -\frac32 +\sqrt2 |+|-\frac32-\sqrt2|$


find n is positive integer n \geq 0 so that if x \geq y \geq z \geq 0 then
   \sum  x^n*y/z \geq \sum   x^n     :?  :D


I need help to know the translation of
[quote]AD AMOREM SUPERNORUM TRAHE DESIDERIUM[/quote] In English? I saw it on the doorstep to a church.

I dont know what SUPERNORUM means.

TO LOVE SUPERNORUM TO DRAG WISH???

That is exactly that I was thinking... I think i spelled it wrong but i don't know...

I think it might be SUPER NOVUM or Above New or something like that But still

TO LOVE ABOVE NEW DRAG(imperative) WISH

is wierd

AD AMOREM should be To Love
DISIDERIUM deffinately means wish and being in accusative it would be a direct object.

how about... bring your wish to the supernal love

edit--actually more likely it is to the love of the supernals... but eh I don't actually know what the word means so w/e

I found an online latin translator and it looks like supernum means "above, superior"

[quote][b] probability1.01 :[/b]edit--actually more likely it is to the love of the supernals... but eh I don't actually know what the word means so w/e[/quote]
I agree with you. I think that the traslation should be: bring your wish to the love of the superior things. Where the "superior things" stands for some values, or something that goes beyond the senses or something like that.

I would guess that in this case, the superior thing would be God, since this in the context of a church.


I need to find the solution to:

$\mathbf{x}'=\begin{bmatrix}4 & -2 \\ 8 & -4\end{bmatrix}\mathbf{x}$

I have found the eigenvalues, which is $\lambda=0$ with a multiplicity of two.  Because of this, I assumed that the solution would be in the form of:

$\mathbf{x}=\mathbf{m}t+\mathbf{n}$

When I plug this into the equation $\mathbf{x}'=A\mathbf{x}$, I come up with the following two equations:

$A\mathbf{m}=0$

$\mathbf{m}=\mathbf{n}$

Finding the eigenvectors the usual way I come up with:

$\mathbf{m}=\mathbf{n}=\begin{bmatrix}1 \\ 2\end{bmatrix}$

From here, I return to $\mathbf{x}=\mathbf{m}t+\mathbf{n}$ and substitute the vectors I found, coming up with:

$\mathbf{x}=t\begin{bmatrix}1 \\ 2\end{bmatrix}+\begin{bmatrix}1 \\ 2\end{bmatrix}$

I am unsure of how to procede from here, however.  Any ideas?

Thank you very much :).

Let's back up several steps.  What are the eigenvalues? :)

Reinforcing blahblahblah's objection: the sum of the eigenvalues of a matrix is the trace of that matrix. The trace of this matrix is $-2.$

Another check: a matrix has zero as an eigenvalue if and only if it is singular. Does that matrix look singular to you?

Fix that. There is no eigenvalue of multiplicity 2. There's also no real eigenvalue.

I apologize, I copied the problem down incorrectly.  I have fixed the original post, where the (1,1) entry should be a 4 instead of a 2.  The rest of my work should coincide with the edited post.  Any ideas now?  Sorry about that, and I appreciate any help :).

I have studied this case in a basic course about dynamical systems

your assumption about the $t \vec{m}+\vec{n}$ is true, and I also found $\vec{m}=0$ but then I did not find $\vec{m}=\vec{n}$ I found $A\vec{n}=\vec{m}$

Oh, so it may just be an error in my algebra.  I am going to re-try this, and post what I come up with later.  Thank you for the help.

Every matrix could be write in the following product of matrices:

A=(C^-1)*J*C, where J is a Jordan matrix.

So x´=(C^-1)*J*Cx -> Cx´=J(Cx). Make u=Cx -> u´=Cx´-> u´=Ju and now is more easy to solve this problem. It´s useful when matrix A isn´t a diagonalizing matrix.

Note: A Jordan matrix has a real number fixed onto the diagonal, 1 at the sup-diagonal and zero in other elements.

I didn't have an initial condition, but upon re-doing the problem I come up with:

$\mathbf{x}=c_{1}\begin{bmatrix}1 \\ 2\end{bmatrix}+c_{2}\left(t\begin{bmatrix}1 \\ 2\end{bmatrix}-\begin{bmatrix}0 \\ \frac{1}{2}\end{bmatrix}\right)$

If we have the initial condition $\mathbf{x}(0)=\begin{bmatrix}a\\b\end{bmatrix},$

then we can solve for $c_1$ and $c_2$ to find $c_1=a,\,c_2=4a-2b.$ Putting that into the last calculation, we get

$\mathbf{x}(t)=a\begin{bmatrix}1\\2\end{bmatrix}+(4a-2b)\left(t\begin{bmatrix} 1\\2 \end{bmatrix}-\begin{bmatrix}0\\ \frac12\end{bmatrix}\right) = a\begin{bmatrix} 1+4t\\8t \end{bmatrix}+b\begin{bmatrix} -2t\\1-4t\end{bmatrix}.$

This agrees exactly with what I found on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65733]amcavoy's other thread.[/url]


calculus
integration
trigonometry
function
calculus computations

Divide the numerator and denominator of the integrand by $x$ and let $t=\frac{1}{x^{2}}-1,$ we have that your integral is $-\frac{1}{2}\int \frac{1}{t}dt$

A more "mainstream" approach, as a typical American calculus textbook would teach it: Make the substitution $x=\sin\theta$ so that $dx = \cos\theta\,d\theta$ and $\sqrt{1-x^{2}}=\cos\theta.$ The integral becomes

$\int\frac{\cos\theta}{\sin^{2}\theta\cos\theta}\,d\theta= \int\csc^{2}\theta\,d\theta=-\cot\theta+C$

At this point, drawing a triangle to help us, we see that if $\sin\theta=x,$ then $\cot\theta=-\frac{\sqrt{1-x^{2}}}x.$

This gives us an answer of $-\frac{\sqrt{1-x^{2}}}x+C.$

- and the potential for this answer to be incomplete or misleading is quite high. The integrand is defined as a real-valued function on two intervals disconnected from each other: $(-1,0),$ and $(0,1).$ The "$+C$" doesn't do a particularly good job of saying that such a constant would only be valid on one of those intervals at a time. In order to converge, a definite integral cannot include $x=0$ inside or on the boundary of the interval of integration.


limit
real analysis
real analysis solved

Compute the limit: lim(n-->oo) [n^(1/n)]^a;
where a=1+1/2+...+1/n.

This one doesn't seem to be too difficult  :)

H(n) =  \sum 1/i
H(n) equivalent to ln(n)

Taking log we get something like ln(n)/n  \lim 0.

So lim(n-->oo) [n^(1/n)]^a = 1.

I am toooo slow! :( That's the correct limit Belenos! :)
I am going to write a detailed proof :) I already had it written, but Belenos was faster! :)
lim(n-->oo) n^(1/n)= e^(lim ln/n)=e^0=1
because ln/n=0 this is easy just apply Cesaro-Stolz and you get the result! :)

now, back to our limit, it can be written like so:

lim (1+1/2+..+1/n)*ln(n)/n=L

apply Cesaro Stolz to it we get that:
lim [(1+1/2+..+1/n+1/(n+1))*ln(n+1)-(1+1/2+..+1/n)*ln(n)]/(n+1-n)=
=lim[ (1+..+1/n)ln(n+1)-(1+..+1/n)ln(n)+ln(n+1)/(n+1)]=
=lim[ (1+...+1/n)*ln((n+1)/n)]=lim [n(1+..+1/n) ln(1+1/n)*1/n]=
=lim (1+..+1/n)/n * ln(1+1/n)^n=(Cesaro-Stolz again)=
=e*lim [(1+..+1/(n+1)-1-..-1/n]/(n+1-n)=e*lim 1/(n+1)=0
so we get that L=0 and out limit is 1.

cheers! :D :D


geometry
perimeter
incenter
geometry unsolved

A triangle $ABC$ is divided into two parts of the same area and perimeter by a line $d$.Prove that the incenter of triangle $ABC$ is on the line $d$

WLOG $d$ intersec $AC$ and $BC$ at $M,N$ respectivly. Let bisector of $\angle ACB$ mets $d$ in $I$. Denote by $x$ distance for $I$ to sides $AC,BC$ and by $y$ distance from $I$ to $AB$. Then $CN+CM=AB+AM+BN$ and $x(CN+CM)=x(AM+BN)+yAB$, which implies
$xAB=yAB$ i.e. $I$ is incenter.


Hey, just wondering -- what kind of music does everyone listen to and what are your favorite songs right now?

Right now I'm listening to a band called [url=http://unbunny.com]Unbunny[/url]. They play sort of indie, low-fi, catchy music.

Well, right now I'm listening to [url=http://www.strungout.com]Strung Out[/url]. It's really hard to classify them, but I'll have to say that they're metal, hardcore, and maybe punk.

For some reason, no one seems to like my taste in music.  :(

[quote="Tunio"]I like classical music! Right now i'm listening to Simple Plan though, cause I think they're also cool :P[/quote]
I love it too...But now, I am listening to some old Chinese songs...

I'm listening to my computer's no soundcard no mixerdevice speaker.  Otherwise I'd be listening to NPR.

[quote="Iversonfan2005"]I'm listening to 2001 by Dr. Dre. The song is "The Next Episode"[/quote]

Mmm oldschool :D That song is awesome.

I'm listening to the latest Greenday CD. It's soooooo good!

I'm totally addicted to the track Vadim Zhukov - Exit.  Vadim is such a talented artist... the amazing quality of his simple, beautiful productions is astounding.  He's a very talented composer musically as well.  Robert Nickson's remix is quite well-produced as well.

I'm listening to Elyot's node93 music. It's okay compared to rap

[quote="Tunio"]The Star Wars Episode III soundtrack is simply wonderful! John Williams is godly :roll:[/quote]

Oh my gosh wow I just bought this cd. Holy crap! It's amazing! Wow it's sooooooo good! :10:

I was just listening to Tin Man, by America. Right now I'm listening to the noise of the Cuisinart ice cream maker!

Currently listening to a playlist over and over again.

By the Way - Red Hot Chili Peppers
In the End - Linkin Park (Listened to this a billion times after failure at state, the song is so true)
Mr. Brightside -- The Killers (fav song)
Adam's Song -- Blink 182
Speed of Sound -- Coldplay
Lose Yourself -- Eminem

Hmm... nothing from the same band, heh.

In the End and Breaking The Habit by Linkin Park

I agree wuth Fanatic.
These songs are so true.

woah,fanatic listed almost the exact playlist i have...other than those

Coheed and cambria-Welcome home,therasmus-inthe shadows,eminem-without me

[quote="José"]Nestor en Bloque (live)[/quote]

If anyone wants true argentinian music, the best one, please PM me

Right now I am listening Anastacia but it is not my favorite; I prefer Loreena McKennit and the soundtrack of EVITA

did anybody try listening to the site i posted?

Basically anything by matchbox twenty is good.  :D (Except 1 or 2 songs)

Right now I'm listening to their song Black and White People.

[quote="h_s_potter2002"]My Hips Don't Lie, by Shakira[/quote]

i really love its clip. :wink:

I think it's time to lock this thread so that we focus more on performing and creating music on this forum.


Rita makes a list of numbers in which each number after the first is 3 more than twice the number that precedes it.  If the third number is 19, what is the first number on the list?

MY WORK:

x = first number

2x + 3 = second number

19 = third number

I got x = 8 as the first number but the book tells me that the answer is x = 2.5

How can this be?

Please, explain.

[hide]The If the first number is a, the second number is 2a + 3. The third number is 2(2a+3) + 3 = 4a + 9 = 19 --> 4a = 10 --> a = 2.5[/hide]

[quote="sharkman"]I got confused when the question stated that the third number is 19. I assumed that 19 should follow the second number which is 2a + 3 on the list.[/quote]
Use recursion, our reccurrence relation is
$a_{n}=2a_{n-1}+3$
The characteristic polynomial is
$x^{2}=2x+3$
The roots of that is $x^{2}-2x-3=0$
$x=1 \pm 2$
x=3, -1
So our formula is $3^{k}\cdot \alpha+3^{k}\cdot \beta$ where $\alpha$ and $\beta$ are invariants.
You can finish the problem from here :)


probability
limit
real analysis
real analysis unsolved

Find the "probability" for 2 integers to be relatively prime(more exactly the limit of the probability of choosing 2 such integers from the set $[1..n]$when $n$ goes to $\infty)$

I hope it wasnt already posted here.

My 2 cents:

$p(n) = \frac{\phi(n)(\phi(n)-1)}{n(n-1)}$

where $\phi(n)$ is Euler's totient function.

$\lim_{n \to \infty} p(n) = ?$

Mike

the right answer is $6/\pi^2$. But solution which i knew is wery long.... And it's more to algebra and numbers theory then to probability...


find if this function is increasing or decreasing
$ f(x)\equal{}ln(x^2\plus{}1)\minus{}e^{\minus{}x}\plus{}1$

[hide=" sure, whatever"]We take the differential of the equation:

$ f'(x) \equal{} \frac {2x}{x^2 \plus{} 1} \plus{} e^{ \minus{} x}$

For positive real numbers $ x$ this value is positive, so it is increasing. However, when $ x$ is negative, with the exception of the small values of $ x$ less than zero, the given equation decreases.
[/hide]

Yes, for values greater than 0, this is trivial. However, for values less than 0, it's not.

The function is in fact increasing everywhere. This is equivalent to proving that
\[ e^{\minus{}x} > \minus{}\frac {2x}{x^2 \plus{} 1}
\]

when $ x \in \mathbb R^\minus{}$ (negative real numbers)

or equivalently,
\[ e^x > \frac {2x}{x^2 \plus{} 1}
\]

when $ x \in \mathbb R^\plus{}$

which is true because $ \frac {2x}{x^2 \plus{} 1}$ will always be less than or equal 1, while $ e^x$ will always be greater than 1.


function geometry perimeter algorithm search

There has been a lot of partition talk before, and I just have to bring it up again. People tried to find a pattern, but from what I read, no one has found a formula regarding the number of partitions a number has. An organized list might help,  but now a 5th to 6th grade problem wants me to find the number of partitions of 2000. Now how would a list help?

Hope you all know what partitions are, just look on the 2001 state sprint.

This is obviously something I ought to know, but what are "partitions"?

So they're ways to add different integers to get the number where order doesn't matter?

Thanks. That was fast.

i will look for a pattern:

partitions of 1:
1

partitions of 2:
1+1
2

partitions of 3:
1+1+1
1+2
3

partitions of 4:
1+1+1+1
1+1+2
2+2
1+3
4

partitions of 5:
1+1+1+1+1
1+1+1+2
1+1+3
1+2+2
1+4
2+3
5

one of them is 1+n-1 where n is the number, so that adds the number of partitions of n-1. Then there's 2+n-2 ... down to where the first number is n, but the second number is either a) greater than the first number or b) zero. so we have to subtract where n-x<n/2 where x is the first number. also, we need to subract f(x) where x != 1 and then add 1 for when the partition is the number itself.

n-f(n)
1-1
2-2
3-3
4-5
5-7

defining f(1)=1 and f(2)=2. we get f(3)=f(2)+f(1) and f(4)=f(3)+f(2)+f(1)-f(1) and f(5)=f(4)+f(3)-f(2)+f(2)+f(1)-f(2)-f(1)+1

Thus, by the pattern, we'd have f(6)=f(5)+ f(4)-f(2)+f(3)-f(3)+f(2)+f(1)-f(2)-f(1)+1 which gives f(6)=7+5-2+3-3+2+1-2-1+1= 11

checking:
1+1+1+1+1+1
1+1+1+1+2
1+1+1+3
1+1+4
1+1+2+2
1+5
1+2+3
2+4
2+2+2
3+3
6

which is 11.

Thus, the pattern is as follows:

$f(n)=2 + \sum_{i=1}^{\frac n2} f(n-i)-f(i)$

I will check it but if anyone else also wants to feel free.

What partitions really are, are recursive functions.
Partitions of n
1+{partitions of n-1}
2+{partitions of n-2} and so on and delete the overlapping ones

look at my post :P

anyway, for the partitions of 7, my function says there are 19, lets check.

1+1+1+1+1+1+1
1+1+1+1+1+2
1+1+1+1+3
1+1+1+2+2
1+1+1+4
1+1+2+3
1+2+2+2
1+1+5
1+3+3
1+2+4
1+6
2+5
2+2+3
3+4
7

which is only 15, so my formula's off somewhere.

My thought is that instead of subtracting f(n-x) we need to subtract f(x-1) so i'll try that.

My new formula says 16 so either i'm missing one or my formula's still off.

Mathematics is known to have mathematical problems that can be understood easily but the solution may well be very difficult, or even unsolved sometimes like Goldbach's Conjecture. The topic of partition theory is still growing with many areas of research. Finding the number of partitions of 2000 is of course [u]not[/u] within the grasp of any middle schooler, although maybe some high school student might attempt to apply Hardy-Ramanujan's formula. There is such formula but it is well beyond the scope of most forums here, perhaps I can post it in College Playground. All in all, I wanted to say that this problem is hard and whoever asked it did not know the difficulty of it, it's like asking a high school student to prove Fermat's Last Theorem.

[hide="BTW click here if you want to see the first term of Hardy-Ramanujan formula to satisfy your curiosity"]
\[p(n)\approx\frac1{2\pi\sqrt2}\frac{d}{dn}\left(\frac{e^{\frac{2\pi}{\sqrt6}\sqrt{n-\frac1{24}}}}{\sqrt{n-\frac1{24}}}\right)\][/hide]

Check this out:
[url]http://www.gcschool.org/pages/program/abacus/6_2_00.html[/url]

Also this
[url]http://mathforum.org/library/drmath/view/56244.html[/url]

[quote="236factorial"]Check this out:
[url]http://www.gcschool.org/pages/program/abacus/6_2_00.html[/url][/quote]
Compare the difficulty between B.169 and B.171. Obviously the problem was not stated the way it should be for the intended level (I feel sorry for the 5th and 6th graders who tried to solve it in year 2000 when that challenge was administered). I wonder what they were thinking when they graded the answers and saw the large number in the answers, or the problem that no one answered...   :D  :rotfl:

I do agree with you. In fact, I'm now working on the 7th and 8th grade set, and there's a problem involving partitions much easier than this one. Strange, isn't it?

I don't expect any 5th or 6th grader to be worrying about partition formula, especially when they are still in elementary school learning about perimeter and long division!!!
Would this be appropriate for high school level?

[quote="236factorial"]I do agree with you.[/quote]
This must surely be the worst ABACUS problem, in terms of difficulty.  The answer is a 40- or 50-digit number!  Good grief!

So, can I please request that you avoid posting these overly difficult "ABACUS International" problems to the Middle School Forums?

You are a very good problem solver, 236!.  So, if you can't solve the problems, and if you don't see any reasonable approach that's at a Middle School level, maybe these should be posted to a "High School" Forum in the future.

it's a 46 digit number :)

of course, I didn't actually work it. I just found it on a website :P

I agree with everyone in that this problem was completely misplaced in dificulty. Perhaps if it were only 20....even then it would be hard.

[quote="nat mc"]all I can say is that partitions are nasty[/quote]

and fat... but MathWorld has lots and lots of information about Partitions that I can't understand at all... http://mathworld.wolfram.com/PartitionFunctionP.html

I wonder what the graph of partitions would look like.  It would probably be almost exponential

[quote="mpoppins"]the problem states
How many different ways can you get 2000 as the sum of positive whole numbers? (You may use the same number more than once in the same sum.)
they are looking for the # of different ways (not partitions) so 3 has 4 ways (order matters!)
3, 1+2, 2+1, 1+1+1  . this is then not very hard

answer[hide]2^1999[/hide] :)[/quote]

That doesn't work.  ;)

I don't see why 2+1 and 1+2 are different ways to sum up to 3.  But it does make sense, your answer does.  I guess we all made a mistake thinking it was partitiions

[quote="mpoppins"]the problem states
How many different ways can you get 2000 as the sum of positive whole numbers? (You may use the same number more than once in the same sum.)[/quote]

Which problem states?

Abacus feb 2000 grade 5-6

http://www.gcschool.org/pages/program/abacus/6_2_00.html

1 has 1 way ie 1
2 has 2 ways: 1+1, 2
3 has 4 ways: 1+1+1, 2+1, 1+2 , 3
4 has 8 ways: 1+1+1+1, 2+1+1, 1+2+1, 1+1+2, 2+2, 3+1, 1+3, 4
and you should be able to convince yourself that n has 2^(n-1) ways.

as you can see there are many more "ways" than partitions which are much harder!

for an example of middle school kids solving a hard partition problem without resorting to generating functions see

http://www.themathcircle.org/focusarticle.htm :)

Ok, now I agree its right since it is for middle schooolers

Since $n=1+1+1+1 ... \text {(n times)}$ then we can place some dividers.

For example if $n=3$ then we have $n=1+1+1$

$1 \ | \ 1 \ 1$ represents $1+2$ and $| \ 1 \ 1 \ 1$ represents $0+3$. There are two places where a divider can separate a 0 from the rest of the group and yield the same permutations/combinations so we must count only one-half once we calculate how many ways the divider can be placed because obviously $0+n=n+0=n$.

We then have $\binom n0 + \binom n1 + \binom n2 + \cdots + \binom nn = 2^n$

We must divide by two to not double count so $\frac {2^n}{2} = 2^{n-1}$

Thanks to mpoppins for finding the original wording of the Abacus problem.  Yes, this *is* a reasonable Middle School problem.

Here's an interesting anecdote about 100-digit partition numbers, for those who care.
[hide]About 10 years ago, a company (RSA), used to have a "partition factoring challenge".  They published some partition numbers starting at about p(8000) or p(9000).  [The published numbers were in the range of about 100 digits to about 150 digits.]

The challenge was to factor these relatively large (100+ digit) numbers, which were for all intents and purposes "random".  The idea was to spur research into better and faster factoring algorithms.  (And also to help build confidence in the RSA encryption algorithm, which hinges on the difficulty of factoring 150- to 300-digit numbers.)  Cash prizes were awarded.

The company wanted integers in the range of 100 to 150 digits, but they also needed a way to choose those integers, so that people could not say the numbers they picked for the challenge were somehow "rigged" to be too easy or too hard to factor.  Partition numbers were just what they wanted -- seemingly random and neither too plentiful nor too scarce in any digit-range.

I'm sure if you do a Google search, you can still find the prime factorizations of many partition numbers.[/hide]

[quote="rcv"]You are a very good problem solver, 236!.  So, if you can't solve the problems, and if you don't see any reasonable approach that's at a Middle School level, maybe these should be posted to a "High School" Forum in the future.[/quote]

Why thank you  :P  But we all know that's not true.  ;)


Let $x,y,$ be real numbers such that $(x+\sqrt{1+y^2})(y+\sqrt{1+x^2})=1$, prove that $(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1$.

Hint: prove that the hypothesis implies that $x+y=0$.

When y is fixed for only one x:  $(x+\sqrt{1+y^2})(y+\sqrt{1+x^2})=1$. It means is x=-y. If x=-y, then
$(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1$

[hide]

Fix , such that it is a constant. The equation is then first degree in , and as such has one unique solution. Thus, each  has a unique  such that .

Let  and , such that . These restrictions allow the two values to take on all real numbers, as desired. Pluggin this in, we get

Changing it all to  and , clearing denominators and expanding, we get:

Bringing the first term on the RHS over to the left, we get

on the left hand side. On the right, bring over the  term to the left, and then replace the  terms with  terms, to get

Factor both sides of the equations to get

Observe that the right side has a root when . If , then , so  causes both sides to be zero, and it's a root of the equation (the only root, as we first proved). Observe that if , then . Going back to original equation and plugging our trigonometric expression back in, we get

as desired.[/hide]

On a real Olympiad, would the 'Observe that  is a root of the equation' part deduct alot of points from my paper? I couldn't figure out how to rigorously show that  is a root.


trigonometry
vector
inequalities
linear algebra
matrix
ratio
geometry

I would consider the cross product far more mysterious. The dot product is everywhere; it's one side in the Cauchy-Schwarz inequality, it's an entry in a matrix product, and it's central to all sorts of things. Taking a pointwise product and adding the terms is one of the most natural operations we can do with a pair of lists of numbers, especially when the sum of the squares matters.
The geometric interpretation says that this ratio $ \frac{A\cdot B}{|A\parallel{}B|}$, which is between $ \minus{}1$ and $ 1$ by C-S, is the cosine of the angle between the two vectors. This is the fundamental tool we use to translate between algebra and geometry; it's very hard to measure angles in higher dimensions without this tool. We can even use it to impose Euclidean geometry on spaces that didn't come with it; $ f\cdot g\equal{}\int_a^b f(x)g(x)\,dx$ is a perfectly good dot (inner) product on the space of "nice" functions on $ [a,b]$.
(Note: There is a slightly different version for vectors in $ \mathbb{C}^n$ or other complex vector spaces)

The cross product, in contrast, suffers from being tied to a small number of dimensions. In two dimensions, the cross-product is a $ 1$-dimensional object, and it can be interpreted as the area operator. In three dimensions, it's a $ 3$-dimensional object, and we can put it in the same space as the original vectors- but we shouldn't really, because that depends on an arbitrary choice of orientation. It's still the area operator, but area has a direction in three dimensions, usually indicated by the normal to the surface. In four dimensions, the cross product is a six-dimensional object, and we just can't treat it like any other vector.

All I have to say is wow. My physics teacher presented the cross and dot product but went into very little detail in it. We could pretty much say he wrote it on the board for us to memorize and use on our homework assignment, which is proof you can apply formulas more than understanding what the heck you're doing. I suppose we're not supposed to understand it in much detail since it's elementary physics (college physics level 1, not sure what that is in the U.S.) but it still bothers me that formulas this far into my education can still remain so obscure.

I know that in my physics class, I was never presented the algebraic aspect to the dot product. (That $ (a,b) \cdot (c,d) \equal{} ac \plus{} bd$)
We were only told that $ u \cdot v \equal{} |u\parallel{}v|\cos\theta$
My physics teacher never bothered with the coordinate aspect since it had limited application in the context of what we were working on.

Once you see that the dot product can be expressed both ways, it becomes a link between the algebraic coordinates and the geometric angles. It becomes even more useful in higher dimensions.

The cross product's application (especially in the context of physics) is to find a third vector that is perpendicular to the two that are given. This interpretation is only relevant in three dimensions. It's used in things like torque since torque's direction is defined to be perpendicular to the plane of rotation. ($ \tau \equal{} r \times F$)

Putting the cross product in the same space as the vectors it came from is just a convenient abstraction- note that it never makes sense to add something like torque or angular momentum to anything that wasn't defined using a cross product. There is a meaning that fits with $ u\cdot (v\times w)$; that "vector triple product" is the (oriented) volume operator in $ \mathbb{R}^3$. The orthogonality comes from the fact that three linearly dependent vectors give zero volume.

I think a reasonably good motivation for introducing the scalar product when working with concrete vectors in euclidean space is that we need operators that can be used to deal with angles. Addition transforms angles in a very complicated way, and for higher dimensions we don't have something nice like complex number multiplication

Somewhat related: jmerry I noticed that you included that the dot product as an entry in the matrix product was one of its many mysteries, and I was wondering if you could go into some more detail as to the connection? I always regarded it as a coincidence of how we defined multiplication to agree with composition of linear transformations? I never thought to think anymore of it as whether or not we establish an inner product for our vector space is largely irrelevant to it being in our definition of multiplication.

After studying some linear algebra this topic came back to mind.

You see, it really bothers me when an abstract definition is given then later some intuitive properties emerge. When you do that, the definition seems to have been formulated out of nowhere and thus any interesting properties it holds seem to be mere coincidences rather than the consequences of a sensible construct.

Here's an example of what I mean:

For scalar products, my book defines it as $ |u\parallel{}v|\cos(\theta)$ then derives the component meaning of it as $ u_{1}v_{1} \plus{} u_{2}v_{2} \plus{} \cdots$. But then I have to wonder "what does $ |u\parallel{}v|\cos(\theta)$ actually mean? Why not define it as $ |u\parallel{}v|\tan^{2}(\theta)$? That would seem equally as arbitrary to me." However, if we started with the component meaning, then we're simply multiplying the products of the components and adding the results, which seems like a very reasonable definition to work from.

Now imagine trying to give an intuitive definition for a cross product. At the moment I'm having trouble with it. In 2-space, I guess you could start with defining the cross product as a vector whose magnitude is the area of a paralallogram determined by $ u$ and $ v$ then set its direction perpendicular to the xy plane.

Suffice to say, I'm not very content with $ u \times v \equal{} \Bigg( \det\begin{bmatrix} u_{2} & u_{3} \\
v_{2} & v_{3} \end{bmatrix}$ , $ \minus{} \det\begin{bmatrix} u_{1} & u_{3} \\
v_{1} & v_{3} \end{bmatrix}$, $ \det\begin{bmatrix} u_{1} & u_{2} \\
v_{1} & v_{2} \end{bmatrix} \Bigg)$ as a starting definition for 3 space. Again, I'm thinking "why are we taking a 2x3 matrix, deleting a column, taking the determinant of the result and setting that as a co-ordinate of a resultant vector? How in any way does this relate to the concept of multiplication?"

I guess what I'm asking then is, what would be the most intuitive definition of a cross product to start from?

[quote="Prime factorization"]This has been bothering me a lot as of late. I sort of understand that $ \vec{A} \circ \vec{B} \equal{} |\vec{A}\parallel{}\vec{B}|\cos(\theta)$ as being "the projection of vector A on vector B multiplied by the magnitude of vector B" and vice versa (pretty sure it's equivalent to the projection of B on A times the magnitude of A), but I haven't the faintest clue what the application of this fact is. Okay, I can see the use of transforming a vector into a number of sorts, but why is it done this way? The purpose of this and, to a lesser extent, the cross product, confound me immensely.[/quote]

It might be helpful for you to think about how work is defined for a force acting on an object that moves in a straight line instead of the projection of one vector in the direction of another vector. It's easy to see that $ Fcos(\theta)$ is the component of the force in the direction of motion so that the work done is $ Fcos(\theta)$ x displacement. I have a computer graphics book that explains the dot product in the clearest way for me.

Computer graphics book's explanation.

[url=http://img515.imageshack.us/i/vectordotproduct001.jpg/][img]http://img515.imageshack.us/img515/7807/vectordotproduct001.th.jpg[/img][/url]
[url=http://img39.imageshack.us/i/vectordotproduct002.jpg/][img]http://img39.imageshack.us/img39/6200/vectordotproduct002.th.jpg[/img][/url]

Calculus 3 textbook's explanation of dot product:

[url=http://img34.imageshack.us/i/vectordotproduct003.jpg/][img]http://img34.imageshack.us/img34/4425/vectordotproduct003.th.jpg[/img][/url]
[url=http://img115.imageshack.us/i/vectordotproduct004.jpg/][img]http://img115.imageshack.us/img115/4373/vectordotproduct004.th.jpg[/img][/url]

[quote="Prime factorization"]After studying some linear algebra this topic came back to mind.

You see, it really bothers me when an abstract definition is given then later some intuitive properties emerge. When you do that, the definition seems to have been formulated out of nowhere and thus any interesting properties it holds seem to be mere coincidences rather than the consequences of a sensible construct.

Here's an example of what I mean:

For scalar products, my book defines it as $ |u\parallel{}v|\cos(\theta)$ then derives the component meaning of it as $ u_{1}v_{1} + u_{2}v_{2} + \cdots$. But then I have to wonder "what does $ |u\parallel{}v|\cos(\theta)$ actually mean? Why not define it as $ |u\parallel{}v|\tan^{2}(\theta)$? [/quote]

Two cases arise naturally in physics. For work, you want to know the component of a force in the direction of motion and that would be $ |F|\cos(\theta)$ and for torque you want to know the component of a force perpendicular to a lever arm and that would be $ |F|\sin(\theta)$. It can then be shown algebraically that $ |F|\cos(\theta)|d|$ = $ F_{1}d_{1} + F_{2}d_{2} + F_{3}d_{3}$ and that $ |F|\sin(\theta)|r|$ = ${ |r_{2}F_{3} - r_{3}F_{2}, r_{3}F_{1} - r_{1}F_{3}, r_{1}F_{2} - r_{2}F_{1}}|$.

I've only passed an introductory course in linear algebra and they tend to leave out all the theory so the only information that I have for the motivation for defining the vector cross product as $ v\times w$ = ${ (v_{2}w_{3} - v_{3}w_{2}, v_{3}w_{1} - v_{1}w_{3}, v_{1}w_{2} - v_{2}w_{1}})$ is this:

[url=http://img530.imageshack.us/i/vectorcrossproduct001.jpg/][img]http://img530.imageshack.us/img530/2993/vectorcrossproduct001.th.jpg[/img][/url]
[url=http://img188.imageshack.us/i/vectorcrossproduct002.jpg/][img]http://img188.imageshack.us/img188/4497/vectorcrossproduct002.th.jpg[/img][/url]

[url=http://en.wikipedia.org/wiki/Exterior_algebra]Exterior algebra.[/url]

With wedge products, alternating multilinear forms, and all of that.


[quote="Timothy"]A tree is now 12 feet tall and grows 18 inches a year.  In how many years will the tree be 36 feet tall?[/quote]
[hide]
The tree grows $1\frac{1}{2}$ feet a year. It needs to grow $36-12=24$ feet.

$\frac{24}{\frac{3}{2}}=24*\frac{2}{3}=16$ years.[/hide]

[quote="Timothy"]A tree is now 12 feet tall and grows 18 inches a year.  In how many years will the tree be 36 feet tall?[/quote]

[hide]24 by 3/2= 48/3=16 years[/hide]

[quote="risinglegend"]i think it takes 16 yrs. :?:[/quote] please hide your answers and dont just give an answer - actually explain why.

[hide]i think the ans. is 16 years because :
36-12=24ft left to grow.
1.5ft=18inches
36/1.5=16[/hide]

 :)


conics
parabola
rotation
quadratics
function
analytic geometry
geometry

[color=cyan]This question is an invention of a friend of mine.  We couldn't figure out any solution to it, other than to know how it was generated in the first place.  Also, I'm a bit worried I have the numbers wrong.  But that shouldn't matter.  Anyhow, without further ado:

Jan has a parabola with equation y = ax^2.   Jan rotates his parabola about the origin, and finds that the rotated parabola passes through the points (20, 40) and (7, 24).  Find a.

okay, I know the second pair isn't right.  I'll edit as soon as I get the correct pair.[/color]

I think you could probably draw x^2 + y^2 = 2000 and x^2 + y^2 = 625 and see where they intersect the parabola. You get two quadratics in y whose roots are a function of a. These should give you the coordinates in terms of a of the points that map to (20, 40) and (7, 24). Find the Euclidean distance between them and set it equal to sqrt(13^2+16^2) = 5sqrt(17).

Come to think of it, it's highly likely that this "solution" involves solving quintic equations.

I'm thinking more along the lines of rotating the [i]axis[/i] along with the parabola. I think it would be cool if we could write your points (whatever they may be) in polar form, and then just tack on a (+theta) to their angle to signify the variable rotation, and then solve, but I don't know the polar equation for a parabola, so we'll just have to rotate the points in Cartesian coordinates using ugly sin/cos etc. Hopefully it should simplify.

Hmm...just trying to find the polar equation of y=ax^2; that converts to (if I remember right) r sin :theta: = a (r cos :theta:):^2:, or
sin :theta: = ar cos:^2: :theta:. Hmmmm.... sorta weird with the cos ^2 but only the sin (to the first power), but it's still probably simpler than rotating points in cartesian coordinates.

Ok, school has started and I don't have much time for this, so I will not solve this, but I think I know how it would be done... I will leave it for someone else to do.

y = ax^2 + Bxy where b=0

is also the equation of the parabola. Find a new B value to that rotates the graph until the points will fit.... then solve for A... Don't forget point (0,0).

Also, if you want to try this is polar form, which is don't think is doable but I think much harder, you will need to use Kepler's Law.

a base parabola opening upward is r=1/(1-sin(:theta:)) but this has a focus at the origin, and will not give you an exact curve... not only this, polar parabola's are graphed using the eccentricity and directrix, neither of which show up in the rectangular equation. Another problem is that rotating polar graphs is not in any way shape or form easy. These two reasons lead me to believe the problem is not solved using polar coordinates, but rather by simply rotating using xy term.

True, true.... you are probably right, but I just want to make one comment: I did not suggest rotating the parabola, I wanted to rotate the points in the opposite direction (this should be quite easy, just add a constant to their angle). This is why I thought of it, but if you say graphing polar parabolas is hard, then I guess we should stick with rectangular.

I'm sorry for misunderstanding... rotating the points may be easier, even though I am not sure how it would be done. Rectangular is my best suggestion.

[color=cyan]Rotating is not that difficult -- MysticTerminator showed exactly how to do it.  The problem is that this does not give a nice result that you can solve when you plug in the values given.

The directrix and focus are contained in the equation, spoon -- given the equation of a parabola, it defines a unique focus and directrix.  In this case, it's even easy to find out that they are the point (0, a/4) and y = -a/4.

Teratomato, I don't see why the intersection of x^2 + y^2 = 2000 and x^2 + y^2 = 625 with y = ax^2 should be the point that rotate to (7, 24) and (20, 40).  Although I agree that it will not be easy to solve, if you do do it that way.

Spoon, are you sure that changing the value of B results in a rotation of the parabola?  It doesn't change the shape?[/color]

[color=cyan]Okay, the original problem had the points (20, 40) and (24, -7).  And they were supposed to be on opposite legs of the parabola.  But is was created with an arithmetic mistake.  But that does mean that it's still solvable, just not a nice answer.[/color]

[quote="JBL"]Okay, the original problem had the points (20, 40) and (24, -7).  And they were supposed to be on opposite legs of the parabola.  But is was created with an arithmetic mistake.  But that does mean that it's still solvable, just not a nice answer.[/quote]

Okay, the points after rotation (I am rotating the [b]points[/b], not the parabola) are (20 cos :theta: - 40 sin :theta:, 20 sin :theta:+40 cos :theta:) and (24 cos :theta:+7 sin :theta:, 24 sin :theta:-7 cos:theta:). So, now we have two variables (a, :theta:) and two equations. This shouldn't be [i]that[/i] difficult.

Plugging my points into the equation y=ax:^2:, we get 20 sin :theta:+40 cos :theta: = a (400+1200 sin :^2: :theta: - 800 sin 2:theta:) and 24 sin :theta:-7 cos:theta:= a(49+527cos:^2::theta:+168sin2:theta:)

Hmmm....now what? Would multiplying the first equation by 7 and adding it to the second equation times 40 help? Hmmm....

[color=cyan]Yes, you get two variables and two unknowns.  That doesn't really make it solvable.  Particularly when you have several different trig functions of one of the variables.  That's how far we got, too :).  By the way, there really isn't a big difference between rotating the points and the parabola itself -- you should end up with the same thing either way.[/color]

[quote="JBL"]Teratomato, I don't see why the intersection of x^2 + y^2 = 2000 and x^2 + y^2 = 625 with y = ax^2 should be the point that rotate to (7, 24) and (20, 40).  Although I agree that it will not be easy to solve, if you do do it that way.[/quote]

Rotation preserves distance, and you're rotating about the origin. x^2+y^2=2000 is the locus of points at (20,40)'s distance from the origin. Since (20,40) is on your rotated parabola, the point that rotates into it is obviously on y=ax^2. So it is on an intersection of the two curves. Also, you are now saying that the two points are on different legs, which makes the problem more or less completely solvable this way - assuming the algebra does not become too intense.

[color=cyan]Hmm. . . I somehow had supposed that x^2 + y^2 = 2000 was a parabola . . . I only see you getting quartics, not quintics . . . might be doable.  Give it a try.[/color]

yes, but in said quartics, you never have an x or x:^3: term, do you? So that simplifies to equations of quadratics.

Scratch what I just said. Simply replace the x:^2: in your circular equations with y/a. No terms to the fourth power now.

[color=cyan]But you need to convert everything to a . . . so you can solve for x in terms of a twice, and set them equal (I think), but then you end up with nastiness in terms of a.  Unless you get very lucky.  I'll try this later tonight.  Dinner now :).[/color]

yes the value of B is a rotational value, it is more complicated than you might think, and unfortunately I lent the only book I had that dealt with rotation to a friend, so I can't look it up myself.

I am bumping this because it is a rather interesting unaswered question. Does anybody want to tackle it?

[hide="nicer posts"]
[quote="MysticTerminator"]Okay, the points after rotation (I am rotating the points, not the parabola) are $(20 \cos \theta-40 \sin \theta , 20 \sin \theta+40 \cos \theta )$ and $(24 \cos \theta+7 \sin \theta , 24 \sin \theta-7 \cos \theta )$. So, now we have two variables $(a, \theta )$ and two equations. This shouldn't be that difficult.

Plugging my points into the equation $y=ax^{2}$ , we get $20 \sin \theta+40 \cos \theta = a (400+1200 \sin^{2}\theta-800 \sin 2 \theta )$ and $24 \sin \theta-7 \cos \theta = a(49+527 \cos^{2}\theta+168 \sin 2 \theta )$

Hmmm....now what? Would multiplying the first equation by 7 and adding it to the second equation times 40 help? Hmmm....[/quote]
[/hide]


ratio
geometry
rectangle
geometry proposed

Let $ ABCD$ be a rectangle and $ P$ be a point on the line $ AC$ but not belong to segment $ AC$ satisfies: $ \widehat{PBC}\equal{}\widehat{DPB}$. Calculate the ratio $ \frac{PC}{PB}$.

[b]HINT.[/b] [hide]$ \frac{PC}{PB}\equal{}\frac{1}{\sqrt{2}}$[/hide]


[quote="The Zuton Force"]Was just wondering if there any competative rowers on this forum, and what people think of rowing in general[/quote]

what's rowing?

You mean like crew?
I went to the first practice for my university's team and liked it alot.
Then I quit because the exercise program was so intense. That might have just been at my school, but rowing seemed really fun.

I did rowing last summer with the local club.  Fun, but hard work.  I never actually competed, though... There was never a place for me in a boat. :(

I'm kind of grounded here in Wyoming, it is probably always too windy or the water we have is frozen over, but my family is doing a sabatical in Melbourne, Australia this fall and I'm looking into rowing crew while I'm there to get in shape for my ski season...  Don't know all that much about it except it looks like fun and is hard work and great conditioning.

[quote="José"]Can anyone explain me what is rowing exactly?[/quote]
You and possibly others get into a boat and start paddling. Whoever crosses the finish line first wins.

[quote="Bictor717"]You and possibly others get into a boat and start paddling. Whoever crosses the finish line first wins.[/quote]
Subdividing that answer to make it more precise: If you're rowing, you're facing backwards (facing opposite the direction the boat is going). The oar or oars in your hands are attached to the side of the boat at a fulcrum. The rower pushes his or her body toward the front of the boat (and the specialized racing boats have sliding seats) causing the blade end of the oar to move in the opposite direction, propelling the boat. They make racing rowboats (sculls) for 1, 2, 4, or 8 rowers. The 8's and some of the 2's and 4's carry an extra person, called the coxswain, who steers and gives instructions but does not row.

If you're facing forward, and holding a paddle that is not attached to the side of the boat, then you're in a canoe or kayak (or, on a very large scale, a dragon boat). Canoes and kayaks are also raced.

Racing sculls are narrow, low-sided boats meant to be used on calm water in light winds. To have a club or team that rows, you need not merely water, but sheltered water, in which it's possible to hold a race 1000 or 2000 meters long. Some variations of canoe and kayak racing also use calm water, but there are also whitewater events.

is rowing really a hard thing to do? I know some friends in crew right now and every day they say its painful...

Well I joined my school's crew team this year. It's a lot of work and requires commitment, but it's a very enjoyable experience. Our school rows mainly 8-sweeps (8 rowers, each with one oar on one side of the boat, and alternating sides, plus one coxwain). I plan on doing crew in college and buying a single-person scull boat to pass time in the future.


Someone asked for some links so here are a few:

[url=http://www.jgiesen.de/Quiz/QuizEnglish.html]Physics Quizes[/url]
[url=http://web.mit.edu/6.002/www/current/]Problems & Solutions[/url]
[url=http://www.ece.utexas.edu/~friedric/032704.htm]more resources[/url]
[url=http://www.tkk.fi/Misc/Electronics/wire_mains.html]for electricians[/url]
[url=http://staff.southwest.tn.edu/kfoster/links_4.htm]links to tutorials[/url]
[url=http://murray.newcastle.edu.au/users/staff/gunilla/elec130/resources.html]html & powerpoint files[/url]
[url=http://www.ibiblio.org/kuphaldt/electricCircuits/]electric circuit stuff[/url]
[url=http://www.proaxis.com/~iguanalabs/maintut.htm]electronics tutorials[/url]

Rogue physicist: Thank you sooooooooooooooooo much!  :D
Moderator: Put it into a sticky!


ABC triangle. Prove that:

la*lb+lb*lc+lc*la \leq p* \sqrt (3r 2 +12Rr)

What do you mean with la, lb, lc ?
And p = (a + b + c)/2, I suppose ?

la is the bisector corespondig to angle BAC

Yes :)  p=(a+b+c)/2 and la is the internal bisector of angle A.

Cheers :)

This problem was given for the selection tests in the Republic of Moldavia in 2003.

Thanks for the info! I didn't know that.

Have a nice day! :)

Here is a solution .. I think it's good!

la*lb+lb*lc+lc*la <= p* sqrt (3r^2 +12Rr) and you raise to the power of 2 and apply Cauchy- Buniakowski
(la*lb+lb*lc+lc*la)^2<=(la^2+lb^2+lc^2)^2<=p^2(3r^2 +12Rr) and we have to decide whether the last inequality is correct.
so we prove now that
(la^2+lb^2+lc^2)^2<=p^2(3r^2 +12Rr)

using [b]Skopet's[/b] inequality la^2+lb^2+lc^2<=p^2 we prove that:
(la^2+lb^2+lc^2)^2<=p^4<=p^2(3r^2 +12Rr) is also an intermidiate inequality
all we have is to prove that p^2<=3r^2+12Rr which is obious true as p^2=12Rr-2r^2... so the sequence of intermediate inequalities is true  and so the inequality is proved!!!

cheers! Merry X-mas!! :D :D

just a quick unharmfull notice: usually in standard international (English that is :) ) notation wa is the length of the interior angle bisector.

Second, Lagrangia your solution is not good. Just see that you have non-homogenous inequalities.

Further more you can see that p2 \geq 3r2 + 12 Rr , because ab+bc+ca = p2 + r2 + 4Rr => 4p2 \geq 3p2 + 3r2 + 12Rr that is (a+b+c)2 \geq 3(ab+bc+ca) which is true, so your finish is also not valid ...

there was a typing error instead of (la+lb+lc)^2 , everywhere it should have been (la^2+lb^2+lc^2)^2 (i have reedited my post) and the ending is good
p^2=2r(6R-r)<= 3r^2 + 12 Rr which is true isn't it?! :?

i still don't understand the mistake i am making.. :(

the mistake is simple, I have already explained it above.
your equality p2 = 12Rr-4r2 is not valid, as easily shown by the case in which r=1, R=2, a=b=c=2\sqrt 3 => p = 3\sqrt 3 => 27 = 24 - 4!

[quote="Maverick"]ABC triangle. Prove that:

\[l_a l_b+l_b l_c+l_c l_a \leq p \sqrt {3r^2 +12Rr}.\][/quote]

Stewart theorem gives $\displaystyle l_a^2 = bc[1-(\frac{a}{b+c})^2] = \frac{bc}{(b+c)^2} [(b+c)^2-a^2] = \frac{4bc}{(b+c)^2} p (p-a) \leq p (p-a)$.

So $(l_a l_b+l_b l_c+l_c l_a )^2 \leq 3 (l_a^2 l_b^2+l_b^2 l_c^2+l_c^2 l_a^2) $

$\leq 3p^2[(p-a)(p-b) + (p-b)(p-c) + (p-c)(p-a)] = 3p^2[(ab+bc+ca) - p^2]$.

On the other hand $\displaystyle r^2 + 4Rr = (\frac{S}{p})^2 + \frac{abc}{S}\cdot \frac{S}{p} = \frac{(p-a)(p-b)(p-c)+abc}{p}  = (ab+bc+ca) - p^2$.

Therefore $(l_a l_b+l_b l_c+l_c l_a )^2 \leq 3p^2(r^2 + 4Rr) = p^2 (3r^2 + 12Rr )$.

Q.E.D.


[quote="tkhtn"]Let $a,b,c\ge 0$ .
Prove that: $\sum \frac{a}{\sqrt{a+b}}\le \frac{5}{4}\sqrt{a+b+c}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90923


After I download the install file for the Go2PDF program and try to run it, I get the error "This program requires Windows version 5.0 or later." Does anyone know what Windows version 5.0 is supposed to mean, or why I am getting this error? (I have Windows 98.)

Windows version 5 is Windows 2000 or later, so Windows 98 isn't included. See http://www.phm.lu/Documentation/Windows/Versions.asp for details of versions

Hmm... That would explain it. Is there anything else I can download that would be compatible for my comp? I really only need the Go2PDF to include images in my TeX.

If you can make .eps graphic files, I don't think you need Go2PDF, though I never use extra graphics in LaTeX, so I don't know much about it.  You should be able to make .eps files on an old computer if you have the right software.

edit:  Also, there is a package called epstopdf, which looks like it converts eps to pdf.

I'm not sure you can include .eps files in PDF docs (but you should be fine if you are producing .ps or .dvi's).  It's not too easy to make .eps docs in PC's unfortunately.

I'd recommend googling around for other PDF solutions, or investing in Windows XP Professional, which has the one feature that (to me) finally makes Windows usable - namely, a 'turn back the clock' feature that allows you to reset system settings to some point in the past (without losing any of your files) whenever Windows decides to break.

[quote="rrusczyk"]Windows XP Professional, which has the one feature that (to me) finally makes Windows usable - namely, a 'turn back the clock' feature that allows you to reset system settings to some point in the past (without losing any of your files) whenever Windows decides to break.[/quote]
XP home has this feature as well. Unfortunately, it only works when Windows breaks a little bit. When it breaks a lot, you still need to reinstall. Someday, I'll convert to Linux, and my computer will work. But not for now.

zfzf - try [url=http://www.cutepdf.com/Products/CutePDF/writer.asp]CutePDF[/url]

n1337 - is that on a PC or a Mac.  PC's are not as .eps friendly as Macs.

[quote="rrusczyk"]
n1337 - is that on a PC or a Mac.  PC's are not as .eps friendly as Macs.[/quote]

I was able to do it on my old Windows XP.  Now I have a Mac, but I don't have any graphics software on it yet.


Hi

Could anybody post problems where the [b]fermat point[/b] of a triangle is [b]useful[/b]???

What about the [b]hessian[/b] points of a triangle??

Thanks a lot

Dear Mathlinkers,
see  for example

http://perso.orange.fr/jl.ayme   vol.2 La fascinante figure de Cundy

where the Fermat's points and the hesse's points are considered.
Sincerely
Jean-Louis


Consider the matrices
\[A=\left[\begin{matrix}{1}& 2\\ 0& 1\\{}\end{matrix}\right]\qquad B=\left[\begin{matrix}{1}& 0\\ 2& 1\\{}\end{matrix}\right]\]

Prove that, for $n > 1$, there aren't integers $a_{1},a_{2},a_{3},\ldots,a_{n}$ and $b_{1},b_{2},\ldots,b_{n-1},b_{n}$ with $a_{2},a_{3},\ldots,a_{n}$ and $b_{1},b_{2},\ldots,b_{n-1}$ nonzero such that
\[A^{a_{1}}\cdot B^{b_{1}}\cdot A^{a_{2}}\cdot B^{b_{2}}\cdot\ldots\cdot A^{a_{n}}\cdot B^{b_{n}}= I,\]
where $I$ is the identity matrix of order $2$.

AB=[5,2,2,1] and BA=[1,2,2,5] (row by row) ; thus these matrices are positive (>0); A=I+U and B=I+V where the elements of U,V are >=0.The studied expression D contains a product AB or a product BA; a left or right product of a positive matrix C by A or  B increases (>=) each element of C.
Conclusion: D is a positive matrix.
Moreover:  det(D)=1; let d=(a1+..an+b1+..bn)-2; then
 trace(D)>=6+2*d (easy); thus if d->infinity then the maximal
eigenvalue of D is trace(D)+O(1/d).

Excuse me luop_blanc can u plz tell me what is $D$ in your proof?

thx.

Hmm, sorry but $AB$ neither $BA$ must appear in the product. Take, for example, $D = AB^{-3}A^{4}$. Unless you want $ABB^{-4}A^{4}$, but I don't think your argument works because multiplying by the inverse does not make each element increase.

Maybe you didn't notice that the exponents might be negative.

Indeed I assumed (and Alekk also) that the exponents are positive.
Mea maxima culpa!

I don't think he assumed they are positive. Anyway, here is a proof. Take two matrices $A=(1,a,0,1)$, $B=(1,0,b,1)$ (first 2 elements are the first line). For integers $n_{i}, m_{i}$ define the matrices $M_{2i+1}=M_{2i}A^{n_{i}}$ and $M_{2i+2}=M_{2i+1}B^{m_{i}}$ with $M_{0}=I_{2}$. Now, let $f(i)$ be the coefficient in position $(1,1)$ of $M_{i}$ if $i$ is even and that of $(1,2)$ otherwise. You can easily see that $f(2n+1)=an_{n}f(2n)+f(2n-1)$ and $f(2n+2)=bf(2n+1)m_{n}+f(2n)$, which allows an immediate proof by induction that if $a,b$ are at least $2$, then $|f(n)|\geq n+1$.

I'm probably being stupid, but how do we prove the induction?  How do we know that $f(2n)$ and $f(2n-1)$ don't cancel out?

You are right, we need to impose that $m_{i}n_{i}$ is not zero. Anyway, we can easily see that it works now, because $|f(n)|$ is actually convex.

Okay, you are using convexity.

I had a different approach.  Given a point $z$ in $\mathbb{R}^{2}$, consider its four neighbors $Az$, $A^{-1}z$, $Bz$, and $B^{-1}z$.  For (almost) all $z$, three of those neighbors will be larger in norm than $z$, and one will be smaller.  (Use any reasonable norm.)  There are some degenerate points $z$ in which there are ties for norm, but we can ignore them.  (One way is to assume that the coordinates of $z$ are rationally independent.)

So we have a 4-regular graph on $\mathbb{R}^{2}$.  By the "3 neighbors larger" observation, the graph can't have any cycles.  That means that no nontrivial product of the matrices can be the identity.

Hi Ravi B, I don’t understand your proof .
 You choose the length of Oz as a potential in each point z of the graph; during each step the potential may decrease (probability 1/4) or increase (probability 3/4).  In a part of the plane a right product by A (for example) decreases the z potential; in another part this product increases the z potential. You (implicitly) say that we can't find a non trivial sequence  (on the graph) which contains a point of same potential than z (the starting point); why?

Let me say it another way.  Suppose that our matrix product is $C_{m}C_{m-1}\ldots C_{1}$.  Let $z_{0}$ be any point in the plane such that $C_{1}z_{0}$ is larger than $z_{0}$.  Consider the sequence of points $z_{1}= C_{1}z_{0}$, $z_{2}= C_{2}z_{1}$, and so on.

By the definition of $z_{0}$, we know that $z_{1}$ is larger than $z_{0}$.  Said backwards, $z_{0}$ is the one neighbor of $z_{1}$ that is smaller than $z_{1}$.  Thus $z_{2}$ must be larger than $z_{1}$.  By continuing, we see that our points are increasing in size.  Thus our matrix product can't be the identity.

A pretty proof from Nicolau Saldanha, Brazil:

We can use Hyperbolic Geometry to solve this problem, or even complex numbers...

Consider the semiplane $y>0$. Let the regions A+,A-,B+,B-,X on it, such that

A+: $y \geq 1$
B+:semicircle with diameter endpoints $(0,0)$ and $(1,0)$
A-: symmetric to A+ in relation to line $y=0$
B-: symmetric to B+ in relation to line $y=0$
X: complement of all above

It is easy to see that

$A(X U A+U B+U B-) = A+$
$B(X U A+U A-U B+) = B+$
$A^{-1}(X U A-U B+U B-) = A-$
$B^{-1}(X U A+U A-U B-) = B-$

And so $z = (A^{a_{1}}B^{b_{1}}... A^{a_{n}}B^{b_{n}})(i)$
is in A+ if $a_{1}> 0$,
is in A-  if $a_{1}< 0$,
is in B+ if $a_{1}= 0, b_{1}> 0$
is in B-  if $a_{1}= 0, b_{1}< 0$.


parameterization
LaTeX
complex analysis
function
complex analysis unsolved

Hi all

I have this homework problem I am stuck on. I know it has something to do with schwarz-christoffel formula but I can't figure out what to do. If anyone could give me a hint on how to get started, I would be greatly appreciative! (Please don't solve the problem for me however)

What is the image of the open unit disc under a map of the form f(z) = 1/z * (z-a_1)^b_1 * (z-a_2)^b_2 * ... * (z-a_n)^b_n

Where |a_i| = 1, 0<b_i<1 and the sum of the b_i's is 2.

Thanks for your help!

You can fairly easily figure out the images of the points $ a_i$. Now, what can you say about the image of each arc between two $ a_i$? (hint: count the arguments of the points in the image, don't pay too much attention to their absolute values). This will give you a complete qualitative picture. To find all the parameters in the resulting shape, you'll need to solve a few extremal problems, which, as far as I can tell, are unsolvable analytically so I do not think you are really expected to go that far.

By the way, since you are almost using the $ \text{\LaTeX}$ syntax anyway, why not to spend half an hour learning the proper way to write the formulae in your posts? :)

Thanks a ton Fedja. I see that the image of the arc between two zeros looks like a truncated rays emanating from zero b/c the argument is constant within each arc (Although not between points on different arcs). So the boundary looks sort of like a big asterisks * with many legs =D although obviously the rays need not match up with such symmetry.

I am guessing then that the image of the open disk is the complex plane minus the boundary image?? I'm not sure how you can deduce this though, I don't see why the boundary values (other then 0) couldn't be taken inside the disk  :huh: since the map isn't even holomorphic. Oh well, thanks very much for the help! As for \Latex I am currently in my 4th year of undergrad, last semester, taking 6 courses and TAing another, so time is a little short, but I definitely want to try to learn it before I head to grad school so that will be one of my summer projects.

The map is holomorphic if you consider it as a map on the Riemann sphere. After noticing that, you can just use the appropriate modification of the "principle of boundary correspondence" (or whatever the right name in English is), which says that if you have an analytic in the disk function that is continuous up to the boundary and the image of the boundary is a Jordan curve travelled once counterclockwise, then the function is one-to-one and the image of the disk is the inside of this curve (this is nothing but the argument principle in disguise). I leave it to you to properly modify it for your case (after all, you wanted just a hint, right? ;))


I have received an eror when I was trying to edit one of my posts that I made days ago.the eror was this:
[b]Sorry, but you cannot edit this post anymore. Ask a moderator to do it if it is really necessary.[/b]
I just wanna know whats the reason,and why does this eror occurs on this post,is it because its not my last post?
what does make it different from others?

P.S:I have looked all around ML to find the right place to post this porblem,and I think that this is the right place,so sorry if it isnt... :oops:

You cannot edit posts that are older than 36 hours. If that much time has passed it is time to post a new one if the information must be changed (or, if really imperative, you can ask a moderator to edit it - but only for good reasons!).

ow so I must post a new one I think...  :P
thanks...


a circle is inscribed in a right triangle with two legs both 12 units.find the area of the circle. sorry. no pics

[hide="doodling helps"]
Call the radius of the circle r.  Then draw perpendiculars from the center to all three sides.  These intersect the tangent points by the definition of an inscribed circle.  Then you will see a square, with 2 of the sides as radii.  Draw the diagonal with one end on the triangle's vertex and one end on the center of the circle.  That is part of the median of the triangle, and the other part is a radius of the circle.  Total length of the median - $ r(\sqrt{2}\plus{}1)$  The median is half the length of the hypotenuse, which is $ \sqrt{2}$ the length of a side, which is 12.  $ r(\sqrt{2}\plus{}1)\equal{}6\sqrt{2}$

[/hide]
Ah, too lazy to do the rest.  Also too lazy to explain it slowly enough that people can understand it.

[hide]The area of the triangle is $ 72$, but also $ rs$. The semiperimeter is $ 12\plus{}6\sqrt{2}$. The radius is $ 12\minus{}6\sqrt{2}$, and you can find out the area.[/hide]

That doesn't seem right, does it...

there's a simpler equation for the inradius:
$ \dfrac{a\plus{}b\minus{}c}{2}$ where a and b are the legs of the right triangle and c is the hypotenuse. i think that's right.


calculus
integration
real analysis
real analysis unsolved

From post 12 (generalization) of flip2004
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70062&sid=ae57ba510eb3fe6331e1be88bd7cc592

[quote="flip2004"]

 Consider $h\in C[0,1] , h([0,1]) \subseteq [0,1]$ and $K_n\in C[0,1] , \; K_n\ge 0$ on $[0,1]$.   If
i)  $\lim\limits_{n\to \infty}\int\limits_{0}^{1} K_n(t) =1$ ,
ii) there exists   $z_0 , z_0\in [0,1]$ ,  such that   $\lim\limits_{n\to \infty}\int\limits_{0}^{1} K_n(t)|h(t)|^j =z_0^j$   for  $j\in \{1,2\},$  then
$\[ \lim\limits_{n\to \infty}\int\limits_{0}^{1} K_n(t)f\left(h(t)\right) =f(z_0) \; \;,\; \; \forall f\in C[0,1] .$[/quote]

For any $\epsilon >0$ there exist A>0, for any x,y in [0;1] we have
$|f(x)-f(y)|\leq \epsilon + A(x-y)^2$

$|\int_{0}^{1}K_n(t)(f(h(t))-f(z_0))dt|\leq \epsilon \int_{0}^{1}K_n(t)dt+A(\int_{0}^{1}K_n(t)h(t)^2dt-2z_0\int_{0}^{1}K_n(t)h(t)dt+z_0^2\int_{0}^{1}K_n(t)dt)$

to finish tends n to +oo

[quote="Moubinool"] ... for any x,y in [0;1] we have  $|f(x)-f(y)|\leq \epsilon + A(x-y)^2$ ...[/quote]
Indeed this is the trick ! Nice solution.


geometry
trigonometry
CEMC
analytic geometry
calculus

I'm very very sorry that i inputed it wrongly.

Triangle ABC is any one of the set of triangles having base BC equal to a and height from A to BC equal to h, with h ＜ sqrt(3)/2a. P is a point inside the triangle such that the value of ∠PAB = ∠PBA =∠PCB. Show that the measure of ∠PAB, ∠PBA and ∠PCB is the same for every triangle in the set.

thank you

Try dropping an altitude from A and seeing what happens.  I did this with trigonometry but there may also be a synthetic solution.

one of the COMC (Open) questions!!  anyways yes i think a synthetic solution also exists.  you can get the official solution (i think) from the cemc or cms website

What do you mean by a "synthetic solution"? A solution using only pure Euclidean geometry?

Roughly, an analytic solution would be one that doesn't use algebraic geometry (coordinate systems and the like) or trigonometry or calculus or anything that isn't a direct result of Euclid's axioms.


Three quarters are tossed, and a tail appears on at least one of them. What is the probability that at least one head appears? Express your answer as a common fraction.

There are two things you must count, 1 head shows and 2 heads show up.

There are $2^3=8$ ways to flip the coins, but since at least one tail appears, the possibility $HHH$ cannot be included. Thus, there are only $7$ possibilites.
Out of the $7$, only $6$ of them have at least one head, $TTT$ doesn't satisfy the conditions.
Thus, the probability is:
$\boxed{\frac{6}{7}}$.

That's a much easier way (I should have known that though  :blush: )   :lol: