inequalities
logarithms
inequalities proposed

Let $ x\neq1,$ $ y\neq1$ and $ x\neq1$ such that $ xyz\equal{}1.$ Prove that:
\[ \frac{{x^2 }}{{\left( {x \minus{} 1} \right)^2 }} \plus{} \frac{{y^2 }}{{\left( {y \minus{} 1} \right)^2 }} \plus{} \frac{{z^2 }}{{\left( {z \minus{} 1} \right)^2 }} \ge 1\]
When the equality occurs?

I have  4 solutions, How did you prove the inequality, arqady?

The technique $ \ln xyz\equal{}1$ is not useful, is it?

[quote="arqady"]Let $ x\neq1,$ $ y\neq1$ and $ x\neq1$ such that $ xyz \equal{} 1.$ Prove that:
\[ \frac {{x^2 }}{{\left( {x \minus{} 1} \right)^2 }} \plus{} \frac {{y^2 }}{{\left( {y \minus{} 1} \right)^2 }} \plus{} \frac {{z^2 }}{{\left( {z \minus{} 1} \right)^2 }} \ge 1
\]
When the equality occurs?[/quote]

$ \Leftrightarrow \sum_{cyc} \frac{x^2}{(1\minus{}x)^2}\ge 1$

setting $ a\equal{}\frac{x}{1\minus{}x}, b\equal{}\frac{y}{1\minus{}y}$ and $ c\equal{}\frac{z}{1\minus{}z}$

we have $ abc\equal{}(a\plus{}1)(b\plus{}1)(c\plus{}1)\Leftrightarrow ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\plus{}1\equal{}0\Leftrightarrow p\plus{}q\plus{}1\equal{}0$

we need prove that $ p^2\minus{}2q\ge 1$

$ \Leftrightarrow p^2 \plus{}2(p\plus{}1)\ge 1$

$ \Leftrightarrow (p\plus{}1)^2\ge 0$ which is true :wink:

[quote="kunny"]Comment

$ \boxed{x \plus{} y \plus{} z \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1\Longrightarrow (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \equal{} 0}\ (xyz\neq 0)$

Here is my solution.

[color=blue]Solution 1[/color]

By $ xyz \equal{} 1$, we have $ \frac {x}{x \minus{} 1} \plus{} \frac {y}{y \minus{} 1} \plus{} \frac {z}{z \minus{} 1} \minus{} \left\{\frac {xy}{(x \minus{} 1)(y \minus{} 1)} \plus{} \frac {yz}{(y \minus{} 1)(z \minus{} 1)} \plus{} \frac {zx}{(z \minus{} 1)(x \minus{} 1)}\right\}$

$ \equal{} \frac {x(y \minus{} 1)(z \minus{} 1) \plus{} y(z \minus{} 1)(x \minus{} 1) \plus{} z(x \minus{} 1)(y \minus{} 1) \minus{} xy(z \minus{} 1) \minus{} yz(x \minus{} 1) \minus{} zx(y \minus{} 1)}{(x \minus{} 1)(y \minus{} 1)(z \minus{} 1)}$

$ \equal{} \frac {x(y \minus{} 1)(z \minus{} 1 \minus{} z) \plus{} y(z \minus{} 1)(x \minus{} 1 \minus{} x) \plus{} zx(y \minus{} 1 \minus{} y)}{(x \minus{} 1)(y \minus{} 1)(z \minus{} 1)}$

$ \equal{} \frac {x \plus{} y \plus{} z \minus{} (xy \plus{} yz \plus{} zx)}{(x \minus{} 1)(y \minus{} 1)(z \minus{} 1)}$

$ \equal{} \frac {x \plus{} y \plus{} z \minus{} (xy \plus{} yz \plus{} zx) \plus{} xyz \minus{} 1}{(x \minus{} 1)(y \minus{} 1)(z \minus{} 1)}$

$ \equal{} \frac {(x \minus{} 1)(y \minus{} 1)(z \minus{} 1)}{(x \minus{} 1)(y \minus{} 1)(z \minus{} 1)} \equal{} 1\ \because x\neq 1,\ y\neq 1,\ z\neq 1$.

$ \therefore \frac {x^2}{(x \minus{} 1)^2} \plus{} \frac {y^2}{(y \minus{} 1)^2} \plus{} \frac {z^2}{(z \minus{} 1)^2}$

$ \equal{} \left(\frac {x}{x \minus{} 1} \plus{} \frac {y}{y \minus{} 1} \plus{} \frac {z}{z \minus{} 1}\right)^2 \minus{} 2\left\{\frac {xy}{(x \minus{} 1)(y \minus{} 1)} \plus{} \frac {yz}{(y \minus{} 1)(z \minus{} 1)} \plus{} \frac {zx}{(z \minus{} 1)(x \minus{} 1)}\right\}$

$ \equal{} \left(\frac {x}{x \minus{} 1} \plus{} \frac {y}{y \minus{} 1} \plus{} \frac {z}{z \minus{} 1}\right)^2 \minus{} 2\left(\frac {x}{x \minus{} 1} \plus{} \frac {y}{y \minus{} 1} \plus{} \frac {z}{z \minus{} 1} \minus{} 1\right)$

$ \equal{} \left(\frac {x}{x \minus{} 1} \plus{} \frac {y}{y \minus{} 1} \plus{} \frac {z}{z \minus{} 1} \minus{} 1\right)^2 \plus{} 1\geq 1$.

The equality holds when $ \boxed{\frac {x}{x \minus{} 1} \plus{} \frac {y}{y \minus{} 1} \plus{} \frac {z}{z \minus{} 1} \equal{} 1 \Longleftrightarrow \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 3}$.

[color=blue]Solution 2[/color]

(a) Let $ x \plus{} y \plus{} z \equal{} a,\ xy \plus{} yz \plus{} zx \equal{} b,\ xyz \equal{} 1$, $ x,\ y,\ z$ are the roots of the cubic equation : $ t^3 \minus{} at^2 \plus{} bt \minus{} 1 \equal{} 0$.

If $ t \equal{} 1$ is the roots of the equation, then we have $ 1^3 \minus{} a\cdot 1^2 \plus{} b\cdot 1 \minus{} 1 \equal{} 0\Longleftrightarrow a \minus{} b \equal{} 0\ \therefore t\neq 1\Longleftrightarrow a \minus{} b\neq 0$.

Thus the cubic equation with the roots $ \alpha \equal{} \frac {x}{x \minus{} 1},\ \beta \equal{} \frac {y}{y \minus{} 1},\ \gamma \equal{} \frac {z}{z \minus{} 1}$ is $ \left(\frac {t}{t \minus{} 1}\right)^3 \minus{} a\left(\frac {t}{t \minus{} 1}\right) \plus{} b\cdot \frac {t}{t \minus{} 1} \minus{} 1 \equal{} 0$.

$ \Longleftrightarrow (a \minus{} b)t^3 \minus{} (a \minus{} 2b \plus{} 3)t^2 \minus{} (b \minus{} 3)t \minus{} 1 \equal{} 0\ \cdots [*]$. Let $ a \minus{} b \equal{} p\neq 0,\ b \minus{} 3 \equal{} q$, we can rerwite the equation as $ pt^3 \minus{} (p \minus{} q)t^2 \minus{} qt \minus{} 1 \equal{} 0$.

By Vieta's formula, we have $ \alpha \plus{} \beta \plus{} \gamma \equal{} \frac {p \minus{} q}{p} \equal{} 1 \minus{} \frac {q}{p},\ \alpha \beta \plus{} \beta \gamma \plus{} \gamma \alpha \equal{} \minus{} \frac {q}{p}$.

$ \therefore \frac {x^2}{(x \minus{} 1)^2} \plus{} \frac {y^2}{(y \minus{} 1)^2} \plus{} \frac {z^2}{(z \minus{} 1)^2} \equal{} \alpha ^2 \plus{} \beta ^2 \plus{} \gamma ^2$

$ \equal{} (\alpha \plus{} \beta \plus{} \gamma)^2 \minus{} 2(\alpha \beta \plus{} \beta \gamma \plus{} \gamma \alpha)$

$ \equal{} \left(1 \minus{} \frac {q}{p}\right)^2 \minus{} 2\left( \minus{} \frac {q}{p}\right)$

$ \equal{} \left(\frac {q}{p}\right)^2 \plus{} 1\geq 1$, the equality holds when $ \frac {q}{p} \equal{} 0\Longleftrightarrow q \equal{} 0\Longleftrightarrow b \equal{} 3$, which completes the proof.

I really have been enjoying to solve the problem.

[color=blue]Solution 3[/color]

Here is a natural solution in comparison with the posted solution before.

Since $ x,\ y,\ z$ aren't equal to 1, we can set $ x \equal{} a \plus{} 1,\ y \equal{} b \plus{} 1,\ z \equal{} c \plus{} 1\ (abc\neq 0)$.

$ \frac {x^2}{(x \minus{} 1)^2} \plus{} \frac {y^2}{(y \minus{} 1)^2} \plus{} \frac {z^2}{(z \minus{} 1)^2} \equal{} \frac {(a \plus{} 1)^2}{a^2} \plus{} \frac {(b \plus{} 1)^2}{b^2} \plus{} \frac {(c \plus{} 1)^2}{c^2}$

$ \equal{} 3 \plus{} 2\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$

$ \equal{} 3 \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{abc} \plus{} \frac {(ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(a \plus{} b \plus{} c)}{(abc)^2} \cdots [*]$.

Let $ a \plus{} b \plus{} c \equal{} p,\ ab \plus{} bc \plus{} ca \equal{} q,\ abc \equal{} r\neq 0$, we have $ xyz \equal{} 1\Longleftrightarrow p \plus{} q \plus{} r \equal{} 0$. since $ r\neq 0$,

we have $ [*] \equal{} 3 \plus{} \frac {2q}{r} \plus{} \frac {q^2 \minus{} 2rp}{r^2} \equal{} 3 \plus{} \frac {2q}{r} \plus{} \left(\frac {q}{r}\right)^2 \minus{} 2\frac {p}{r}$

$ \equal{} 3 \plus{} \frac {2q}{r} \plus{} \left(\frac {q}{r}\right)^2 \plus{} 2\frac {q \plus{} r}{r} \equal{} \left(\frac {q}{r}\right)^2 \plus{} 4\frac {q}{r} \plus{} 5$

$ \equal{} \left(\frac {q}{r} \plus{} 2\right)^2 \plus{} 1\geq 1$.

The equality holds when $ q \equal{} \minus{} 2r$ and $ p \plus{} q \plus{} r \equal{} 0\ (r > 0)\Longleftrightarrow p: q: r \equal{} 1: ( \minus{} 2): 1$. $ Q.E.D.$

[/quote]

[quote="kunny"]I have  4 solutions, How did you prove the inequality, arqady?
[/quote]
My proof the same of the bunhiacovski's proof.

This is my first solution:

Let $ a \equal{} \frac {x}{x \minus{} 1},\ b \equal{} \frac {y}{y \minus{} 1},\ c \equal{} \frac {z}{z \minus{} 1}\ (x\neq 1,\ y\neq 1,\ z\neq 1)$

$ \Longleftrightarrow x \equal{} \frac {a}{a \minus{} 1},\ y \equal{} \frac {b}{b \minus{} 1},\ z \equal{} \frac {c}{c \minus{} 1}$.

$ \therefore xyz \equal{} 1\Longleftrightarrow abc \equal{} (a \minus{} 1)(b \minus{} 1)(c \minus{} 1)$, yielding $ ab \plus{} bc \plus{} ca \equal{} a \plus{} b \plus{} c \minus{} 1$.

$ \therefore \frac {x^2}{(x \minus{} 1)^2} \plus{} \frac {y^2}{(y \minus{} 1)^2} \plus{} \frac {z^2}{(z \minus{} 1)^2} \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca)$

$ \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(a \plus{} b \plus{} c \minus{} 1) \equal{} (a \plus{} b \plus{} c \minus{} 1)^2 \plus{} 1\geq 1$. $ Q.E.D.$

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113999
I have proved this inequality 2 years ago!  :D

I have seen the solution. haven't you seen [url=http://www.mathlinks.ro/viewtopic.php?p=1191514#1191514]this forum?[/url]

But you have already solved 2 years ago. :lol:


Whenever I go to a different page on this site, I get a dialog box that says that an error in Flash Player is causing this script to be unresponsive.  The page doesn't come up till I hit abort.

You must be using the AoPS style right? Please give more details about your computer's configuration (OS, Browser, Flash Player version) to evaluate your problem.

[quote="topper"]I have windows XP, but I'm not sure what browser this is, as the ISP (juno) has a strange interface.  I have a dialup connection. and flashplayer 6.

[hide] nice icons!  :)  :(  :?  :D  :lol:  ;)  :blush:  :P [/hide][/quote]First of all, please do not make tests in this forum. Use the [url=http://www.mathlinks.ro/Forum/index.php?f=224]test forum[/url] for that.

Secondly, you must tell me what your browser / browser version is. Without that detail (which is the most important out of all the other details) I really can't figure out what's going on.

Furthermore, I suggest you upgrade your system to Macromedia Flash Player 7. This might solve the problem ...  :?

I agree with Valentin -> go to macromedia.com and install the most recent version of Flash.  Sounds like whatever you're using possibly has a messed-up plug-in for flash.  Reinstalling should hopefully clean it up.


Hi folks,

I receive a lot of Reply Notification in my mail on yahoo.

---------------------------------------------------------------------
you no longer wish to watch this forum you can click "Stop watching
this forum" found at the bottom of the page.
--------------------------------------------------------------------

But I don't find at the bottom of the page

Could you help me ?

thank

It appears for me - its at the very bottom, just above the "Powered by PHPBB" part, below where it explains what the symbols next to posts are.

You must go into the forum that you are receiveing email notices and you will  find the stop button there :)

[quote="Moubinool"][quote="TripleM"]...just above the "Powered by PHPBB" ...[/quote]

I did not find    :?[/quote]Have you looked in the right subforum?! You need to be in the same subforum as your notifications come.

There is no "stop watching" above php, in the subforum

Valentin I don't want to receive "repply notification" anymore

here one of repply notification
http://www.mathlinks.ro/Forum/viewtopic.php?p=138183#138183

[quote="Moubinool"]There is no "stop watching" above php, in the subforum.

Valentin I don't want to receive "repply notification" anymore[/quote]Dear God, I'll use my super-duper power to I'll login as you and remove it myself :D Just tell me from which forum you receive reply notifications :)


Να βρεθεί η μέγιστη τιμή της παράστασης
$A=xy+x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}-\sqrt{1-x^{2}}\cdot \sqrt{1-y^{2}}$

Νίκο

ετοιμάζομαι για έξοδο και η απάντηση μου λίγο βιαστική .
Προφανώς για θετικά χ , ψ η παράσταση Α παίρνει την μέγιστη τιμή της .
Θέτουμε  χ = συνα και ψ = συνβ .
Οπότε
            Α = συνασυνβ + συναημβ + συνβημα - ημαημβ =
               = συν(α + β) + ημ(α+β) .
Και επειδή
                 2(χ^2 + ψ^2) >= (χ + ψ)^2
άρα η μέγιστη είναι η ρίζα του 2 .

Έτσι την έκανες και εσύ ?

Νίκος Ρ

Αλλιώς  $\sin x+\cos x = \sqrt{2}\cdot \sin \left( x+\frac{\pi}{4}\right)$ .

Ναι καλύτερα έτσι , Ramanujan .


1. Prove there no function f defined for all real numbers and taking real numbers as values such that for every real number r:
(i) f(r^2) - (f(r)^2)) >= 1/4
(ii) the equation f(x)=r has at most one solution for every r.

2. In a certain village there are m houses around a circular pond, where m>1 is odd. In each house there are at least n people. Show that in the village there is either a female with at least n female neighbours or a male with at least n male neighbours.

3. The real numbers ... a(-2), a(-1), a0,a1,a2 ... satisfy
(i) a1=1
(ii) a(m+n) - mn = a(m) + a(n) + 1 for all integers m,n.
Find a closed formula for a(n).

4. Let ABC be a triangle. Let A',B' and C' lie on BC, CA and AB (extended if necessary) respectively. Let lA,lB and lC be lines in the plane ABC through A',B' and C' perpendicular to BC, CA and AB respectively. Prove:
lA,lB and lC are concurrent iff:
A'C^2 + B'A^2 + C'B^2 = A'B^2 + B'C^2 + C'A^2

(3)
assume $n>1$ first.

$f(n+1)=f(n)+f(1)+n+1=f(n)+n+2$
thus we have
$f(n)=f(n-1)+n+1$
$f(n-1)=f(n-2)+n$
$f(n-2)=f(n-3)+n-1$
...
$f(2)=f(1)+3$
sum up the $n$ equalities above, we get
$f(n) = f(1) + 3+4+\cdots+n+(n+1)=1+\frac{(n+4)(n-1)}{2}$.

and it is easy to verify that this works for negative $n$.

Problem 2 :

Consider any house. Assume its inhabitants are not all males or all females.
Assume that the neighbouring houses have :
Neighbour 1 : [tex]$M_1$[/tex] males  [tex]$F_1$[/tex] females
Neighbour 2 : [tex]$M_2$[/tex] males  [tex]$F_2$[/tex] females

Now, since each house has at least n people, so
           [tex]$ M_1 + F_1  \geq n $[/tex]  and,  [tex]$ M_2 + F_2  \geq n $[/tex]
           [tex]$ \Rightarrow M_1 + F_1 + M_2 + F_2 \geq 2n $[/tex] .

Let us try to construct the village satisfying the requirements, such that no male has n neighbouring males, and no female has n neighbouring females. So by requirement,
           [tex]$ M_1 + M_2  < n $[/tex]  and,  [tex]$ F_1 + F_2  < n $[/tex]

so that, in our chosen house a male doesn't have n neighbouring males and a female doesn't have n neighbouring females.

Adding last two inequalities, we have
           [tex]$ M_1 + F_1 + M_2 + F_2 < 2n $[/tex], contradicting our first inequality.

So, the chosen house has either only males or only females.

Now, if the neighbouring houses have a mixture of males and females, then as the chosen house has n of same sex, a person of the same sex cannot live in the neighbouring houses.

So, we see we have to build the village such that, starting from one house consisting of people of same sex, the neighbouring houses should have people all of the opposite sex, and we go on adding houses similarly. So, total number of houses is
One chosen house (of males, say)  + 2 neighbours (of females) + Their 2 neighbours (of males) + ...... = m

As the number of houses m (> 1) is odd, so finally the sum will end in 2, i.e., we have two neighbouring houses having all people of same sex. So, any one of these people will have n neighbours of his/her own sex. So, we can't build such a village.

Thus, in every such village, there exists one female with at least n female neighbours or there exists one male with at least n male neighbours.

*****************************************************************
Sorry for the long explanation....  :(

*****************************************************************
Sambit

An apology to all :

Sorry, I am new to this forum, and tried my hand at posting my message using TeX for the first time, and ended up problems. I thought I will preview first, but somehow, I submitted the message and the whole post looks messy..

 :(
Sambit

hey, you don't need dollar sign between [code][tex] [/tex][/code]...
just use [code][tex] [/tex][/code] or the dollar signs.

For #1:

[hide]

Take x = 0

f(0) 2 - f(0) + 1/4 <= 0

1/4 * (2f(0) - 1) 2 <= 0

This implies f(0) = 1

Take x = 1...etc...implies f(1)=1 contradiction.

[/hide]

Problem 2 :

Consider any house. Assume its inhabitants are not all males or all females.
Assume that the neighbouring houses have :
Neighbour 1 : [tex]M_1[/tex] males and [tex]F_1[/tex] females
Neighbour 2 : [tex]M_2[/tex] males and [tex]F_2[/tex] females

Now, since each house has at least n people, so
     [tex]M_1 + F_1 \geq n[/tex] and, [tex]M_2 + F_2 \geq n[/tex]
     [tex]\Rightarrow  M_1 + F_1 + M_2 + F_2 \geq 2n [\tex]

Let us try to construct the village satisfying the requirements, such that no male has n neighbouring males, and no female has n neighbouring females. So by requirement,
     [tex]M_1 + M_2 < n[\tex] and, [tex]F_1 + F_2 < n[\tex]

so that, in our chosen house a male doesn't have n neighbouring males and a female doesn't have n neighbouring females.

Adding last two inequalities, we have
    [tex]M_1 + M_2 + F_1 + F_2 < 2n [\tex], contradicting our first inequality.

So, the chosen house has either only males or only females.

Now, if the neighbouring houses have a mixture of males and females, then as the chosen house has n of same sex, a person of the same sex cannot live in the neighbouring houses.

So, we see we have to build the village such that, starting from one house consisting of people of same sex, the neighbouring houses should have people all of the opposite sex, and we go on adding houses similarly. So, total number of houses is
One chosen house (of males, say) + 2 neighbours (of females) + Their 2 neighbours (of males) + ...... = m

As the number of houses m (> 1) is odd, so finally the sum will end in 2, i.e., we have two neighbouring houses having all people of same sex. So, any one of these people will have n neighbours of his/her own sex. So, we can't build such a village.

Thus, in every such village, there exists one female with at least n female neighbours or there exists one male with at least n male neighbours.

*****************************************************************
Sorry for the long explanation.... [b]Thanks to "liyi"[/b] for the TeX support... I finally figured out how to post in TeX

*****************************************************************
Sambit


trigonometry
function
symmetry
analytic geometry
articles
ratio
geometry

Is there any way to calculate sines and cosines of angles without having to memorize them for each angle (I mean is there any function $ f(x)$ which can be used to calculate sin, cos, or tan of $ x$? I have a rather limited understanding of trigonometry.

No. it isn't any [b]function[/b].
But there is a serie related to taylor .
by the serie, we can estimate sin with any precision.
It is
\[ \sin x \equal{} x \minus{} \frac {{x^3 }}{{3!}} \plus{} \frac {{x^5 }}{{5!}} \minus{} \frac {{x^7 }}{{7!}} \plus{} \cdots \plus{} ( \minus{} 1)^n \frac {{x^{2n \plus{} 1} }}{{(2n \plus{} 1)!}} \plus{} \cdots
\]

You can find more informations about this serie in college play ground.

Thank you for the post, but how do we evaluate that. Under, ordinary circumstances, I would evaluate that sigma-equation but I feel brain-dead.  :(  Just out of curiousity, are there any other formulas for this (like calculating tan. obviously through this we can calculate cos. through this; my trigonometry knowledge is EXTREMELY limited).

hello, maybe this will help you
http://mathworld.wolfram.com/TaylorSeries.html
Sonnhard.

Hmm, usually you only have to know the common values on the unit circle and half those angles. Like for example, you usually don't need sin 40 (If you did, you'd have to use a Taylor series). For those values on the unit circle, don't memorize them, just picture where the angle intersects the circle. You know angles 30 and 60 meet the circle at $ (\frac {\sqrt {3}}{2}, \frac {1}{2})$ and $ (\frac {1}{2}, \frac {\sqrt {3}}{2})$, so this isn't too hard to remember by symmetry, and 45 meets it at $ (\frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2})$. All the other angles you can just find the sin and cos using symmetry, picture the circle in your head. I can see that the point at 150 degrees has the same y-coordinate as 30, for example, so I know sin 150=$ \frac {1}{2}$.

You could just use the sum and difference theorems for sine and cosine. With these, you should be able to find the most common angles.

I've seen many people say that $ \sin{30^{\circ}} \equal{} \frac {1}{2}$ and $ \sin{60^{\circ}} \equal{} \frac {\sqrt {3}}{2}$ without justification. I feel like giving you justification, so I made a proof that consists entirely of pure geometry.
I'm sorry that I cannot give you any diagrams, but I shall try to be as clear as possible.

[hide="Click me!"]We have triangle ABC with $ \angle A \equal{} 30^{\circ}$, $ \angle B \equal{} 60^{\circ}$, and $ \angle C \equal{} 90^{\circ}$. We draw point $ D$ such that $ BD$ bisects $ \angle CBA$. We now have $ \angle CBD \equal{} 30^{\circ}$ and $ \angle DBA \equal{} 30^{\circ}$. Since $ \angle BDA \equal{} \angle DAB$, triangle $ DAB$ is isosceles with $ BD \equal{} DA$, and $ \angle BDA \equal{} 120^{\circ}$. $ \angle CDA \equal{} 180^{\circ} \equal{} \angle CDB \plus{} \angle BDA \equal{} \angle CDB \plus{} 120^{\circ}$, so $ \angle CDB \equal{} 180^{\circ} \minus{} 120^{\circ} \equal{} 60^{\circ}$. Note we have corresponding angles:

$ \angle BDC \equal{} \angle ABC$
$ \angle DCB \equal{} \angle BCA$
$ \angle CBD \equal{} \angle CAB$

Therefore $ \triangle BDC\sim \triangle ABC$.

Let $ CD \equal{} x$ and $ BD \equal{} y$. From the [[Pythagorean Theorem]], $ BC \equal{} \sqrt {y^2 \minus{} x^2}$. Since $ BD \equal{} AD$, we have $ AD \equal{} y$. Since $ \triangle BDC\sim \triangle ABC$, we have $ \frac {BC}{AC} \equal{} \frac {DC}{BC}$, which is the same as $ \frac {\sqrt {y^2 \minus{} x^2}}{x \plus{} y} \equal{} \frac {x}{\sqrt {y^2 \minus{} x^2}}$. Cross-multiplying, we have $ y^2 \minus{} x^2 \equal{} x^2 \plus{} xy$, or $ 2x^2 \plus{} xy \minus{} y^2 \equal{} 0$. This can be factored into $ (x \plus{} y)(2x \minus{} y) \equal{} 0$. Since $ x$ and $ y$ are positive, $ x \plus{} y\neq 0$, so $ 2x \minus{} y$ must equal 0. Therefore $ y \equal{} 2x$. Plugging in $ 2x$ for $ y$ in $ BC \equal{} \sqrt {y^2 \minus{} x^2}$ gives $ BC \equal{} \sqrt{3x^2} \equal{} x\sqrt {3}$.

In triangle $ BCD$, $ \sin{30^{\circ}}$ is defined as $ \frac {CD}{BD} \equal{} \frac {x}{y}$, which is $ \frac {x}{2x} \equal{} \frac {1}{2}$. $ \sin{60^{\circ}} \equal{} \frac {BC}{BD} \equal{} \frac {x\sqrt {3}}{2x} \equal{} \frac {\sqrt {3}}{2}$.[/hide]

In a 45-45-90 triangle, the legs are equal since the triangle is isosceles, and you can use the pythagorean theorem to see that the hypotenuse of the triangle is $ \sqrt {2}$ times the leg, and $ \sin{45^{\circ}} \equal{} \cos{45^{\circ}} \equal{} \frac {\sqrt {2}}{2}$.

People are also mentioning the sum/difference formulas for the sine and cosine, but I don't know how to prove them, so I'd like to see a proof.

This depends on how you defined the functions $ \cos$ and $ \sin$ in the first place.

If you start off with $ e^{i x} \equal{} \cos x \plus{} i \sin x$ a lot of results follow easily from rules of arithmetic regarding exponents. For example, one has:

$ e^{i a}e^{i b} \equal{} (\cos a \plus{} i \sin a)(\cos b \plus{} i \sin b)$
$ \equal{} (\cos a \cos b \minus{} \sin a \sin b) \plus{} i (\sin a \cos b \plus{} \sin b \cos a)$

However, we have $ e^{i a} e^{i b} \equal{} e^{i (a \plus{} b)}$, so also

$ e^{i a}e^{i b} \equal{} \cos(a \plus{} b) \plus{} i \sin(a \plus{} b)$

From this we derive that

$ \cos(a \plus{} b) \equal{} \cos a \cos b \minus{} \sin a \sin b$
$ \sin(a \plus{} b) \equal{} \sin a \cos b \plus{} \sin b \cos a$

See also [url=http://en.wikipedia.org/wiki/List_of_trigonometric_identities]this wiki article[/url] and the accompanied [url=http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities]proofs page[/url].

No, I just wanted to know if there was any method to quickly evaluate the sigma equation. The 30-60-90 triangle thing was helpful, but I think it a lot easier simply to say that we construct a median through an equilateral triangle and get 2 30-60-90s (but overall this is a specific case).

The sigma equation is in my opinion unwieldy and the geometric interpretations should be used for most if not all problems.

Also I had no idea that the 30-60-90 ratios could be proved so easily. :(

[quote="1=2"]I've seen many people say that $ \sin{30^{\circ}} \equal{} \frac {1}{2}$ and $ \sin{60^{\circ}} \equal{} \frac {\sqrt {3}}{2}$ without justification. [/quote]

Let $ ABC$ be a triangle with $ A,B,C\equal{}30,60,90$, respectively. Reflect $ B$ over $ C$ to $ B'$. Easily, $ ABB'$ is equilateral and $ BC$ is half of $ AB$. The other part follows by the pythagorean theorem.

As I said, 30-60-90s are a specific case. I am inexperienced with sigma equations, so is there any general method for solving the sigma quickly, or do you actually (very unlikely but) brute force it?

I really doubt that this is High School Basics level guys.


number theory
greatest common divisor
relatively prime

Prove that p! and (p-1)! -1 are relatively prime iff p is an odd prime.
I am posting my soln but I am sure there would be a more elegant simple solution.

[hide]Suppose they are not coprime.
Let GCD of p! and (p-1)! -1 is d.
d divides p! and d divides (p-1)!-1..........
There exist integer k such that
kd=p! or p(p-1)!=kd....(1)
Also (p-1)! -1=k'd (say)    So (p-1)!=k'd+1......(2)
Putting the value of (p-1)! in (1).......
We get p(k'd+1)=kd or d(k-pk')=p....
d divides p or d=1...
If possible let p=d....
d divides (p-1)!-1  or  p divides (p-1)!-1...
But by Wilson's theorem,.... we have p divides (p-1)!+1....
===>  p divides 2.......
But p is an odd prime  and hence contradiction........

Thus our assumption is wrong and hence they are relatively prime......[/hide][/code]

Arvind, Had this question been true, you would have to prove the only if part too, which I believe is not true.
Anyway the idea of your proof isn't long, it just looks that way 'coz you wrote it that way with assumptions and stuff.

$ p!$ and $ (p \minus{} 1)! \minus{} 1$ are relatively prime for all integers $ p > 1$
[b]Proof:[/b]
For composite $ p > 3$, All the prime factors of $ p!$ are less that $ p$, and all those prime factors divide $ (p \minus{} 1)!$ hence none of those prime factors divide $ (p \minus{} 1)! \minus{} 1$. Hence $ p!$ and $ (p \minus{} 1)! \minus{} 1$ has no no common prime factors. So They are rel. prime.
For odd primes $ p$, the only factor of $ p!$ that doesn't divide $ (p \minus{} 1)!$ is $ p$ itself, in that case by Wilson's theorem, $ (p \minus{} 1)! \plus{} 1$ is divisible by $ p$ implying $ (p \minus{} 1)! \minus{} 1$ is not divisible by $ p$. Again there is no common prime factor of $ p!$ and $ (p \minus{} 1)! \minus{} 1$
So $ p!$ and $ (p \minus{} 1)! \minus{} 1$ are relatively prime in all cases where $ p > 1$. QED.


How do I change the meaning of the \char command? I'm working a lot with finite fields, so I often need to write $ \text{char }\mathbb{K}$ and similar. I would like to write "\char\mathbb{K}" as command, but if I add to my header [code]\DeclareMathOperator{\characteristic}{char}
\renewcommand{\char}{\characteristic}[/code]then I get an incredible amount of errors... so how should I do this instead?

You could use something like [code]\DeclareMathOperator{\chark}{char\,\mathbb{K}}[/code] then
[code]$\chark$[/code]

Currently I already have \c for that, but I keep writing \char accidently, so I want to make it \char. :)
Should I conclude from your reply that it is impossible?

\char is a TeX primitive* so you are strongly advised not to redefine it. Yes [code]\renewcommand{\char}{\text{char}\,\ensuremath{\mathbb{K}}}[/code] works but then it could break something else. Somewhere (eg a package) may need the original meaning of \char which could cause problems.
You could use \Char instead.

* To see what \char does follow it by a number between 0 and 255 eg \char37. You get the character at position 37 in the current font.


In triangle $ ABC$ with angles $ \alpha, \beta, \theta$, suppose $ \alpha \equal{} 2 \beta$. Demonstrate that $ A^2\equal{}B(B\plus{}C)$.

Note: $ \angle BAC \equal{} \alpha, \angle ABC \equal{} \beta, \angle BCA \equal{} \theta$.

hello, we get
$ \gamma\equal{}\pi\minus{}3\beta$
and after the sine theorem we get
$ a\equal{}\frac{b\sin(2\beta)}{\sin(\beta)}$
$ c\equal{}\frac{b\sin(3\beta)}{\sin(\beta)}$
and your theorem is equivalent to
$ \sin^2(2\beta)\minus{}\sin^2(\beta)\equal{}(1\plus{}2\cos(2\beta))\sin^2(\beta)$
$ \sin^2(3\beta)\equal{}(1\plus{}2\cos(2\beta))^2\sin^2(\beta)$
But it is not the same?
Sonnhard.

[quote="Agr_94_Math"]This problem can also be done by bisecting angle A  in ot $ \beta$ and $ \beta$ and it becomes very obvious after that.[/quote]

Good, that was the solution I was waiting for.

@Dr Sonnhard Graubner, are you sure it is not true?

Wolfram alpha, shows this as "True". If you want to test it, enter "sin^2(3x) = (1+2*cos(2x))^2*sin^2(x)" into wolfram alpha.

[quote="tonypr"]Wolfram alpha, shows this as "True". If you want to test it, enter "sin^2(3x) = (1+2*cos(2x))^2*sin^2(x)" into wolfram alpha.[/quote]I'm going to post this trig identity in Intermediate forum :P (the non-square version). Also, be a bit more specific about what $ a,b,c$ are... I'm assuming those are the lengths of the sides of the triangle opposite the respective vertices. The upper cases refer to points, so it would be nice if the equality is written as $ a^2\equal{}b(b\plus{}c)$.


give your mathcounts profile for those of you who have taken it.

my profile:

 :( , and  :mad: , and  :stretcher:

6th grade: 12th chapter (i think), 30th states (i know)
7th grade: 8th chapter, 64th states :stretcher:
8th grade: 2nd (lost in tiebreaker to Justin Ahmann) chapter, 13 states  :( (really hoped to do better)

6th grade: 12 chapter 47th state
7th:  5th chapter 8th state :mad:

6th grade:18th at chapter :blush:  didn't make it to state

7th grade: 4th at chapter, and 34th at state :(

6th grade: forget what I got on school...that's all
7th grade: 40/2nd Chapter, 6th State
8th grade: Wait till next year!

umm actually i think you were 46th last year. i was 47th since you had more sprint than me.

8th: 3rd Chapter, 17th State

..

yeah, 1 year.. it sucks.

[quote="undefined117"][quote="mwpl11"]ha  :lol: . it's okay. not doing it is probably better than shaming urself and dragging down ur team by getting 64th. now i know what it feels like to be like a certain few ppl on our team...[/quote]

O_O do you mean to imply something?[/quote]

no way! if he was here, that totally would've been directed at michael, lol. i don't care if u tell him, cuz i think me, u, and karen have rubbed it in pretty deep already!

ZOMG MATHCRAZED = LYNDON JI [4th place] LOLLOLOLOLL ZOMG ZOMG LOL LOL EXPOSXED!!!! XD

/lyndonspeak in real life.

Anyhow,

6th grade
Chapter: 9th
State: 43rd

7th grade:
Chapter: 1st [44]
State: 5th [26/12... 2nd highest sprint score TT]

8th grade:
Chapter: 1st [28/16]
State: 1st [btw mathcrazed = lyndon ji]

[quote="mathcrazed"]lol... uhh... 6th: 32nd at chapter (3rd countdown)  :blush: , and 42nd at state (which was the best 6th grader -_- sadly enough) :o
7th: chapter... lost tiebreaker (for 2nd to last place)...
state:  :rotfl:[/quote]

were u in sixth grade last year?
my friend got 14th as a sixth grader last year.

[quote="mathcrazed"]i think you might mean 2 years ago or maybe 3, because there was a girl who got 14th as a sixth grader, shawn do you remember her?[/quote]

Christine Hong-she got 18th (as a 6th grader)-she went to my school, along with all those other kids who go to the Sunday class..yea she was pretty good, she beat David, Yingxue, Roufan, and even Paul Lee (who made USAMO) at state...I didnt know about mathcoutns back then so I got 8th in school, 23rd chapter, didnt make state..

O yea..tell your dad I coundt go to Sunday class today.

Wait..how do you know you were the top 6th grader?  What about yuqing and jeff hou?  and what about those 7th graders THIS YEAR that were 6th graders last year that are going to nationals

[quote="Fanatic"]He probably had a list. I figured out I was 2nd or 3rd sixth grader in 05, at 41st place.

Christine scared me with her way higher ranking, but this year she got 7th in North Carolina with what I presume is a lower score than what I got at state.[/quote]

No I was just wondering because I thought Yuqing got like a 27 (or 29 or so) and I think he tied Lyndon or something..

Well, 7th isnt [i]that[/i] bad, but I wouldve expected her to do better, as I thought she would make nationals or something like that after she got 18th at state.

[quote="indianamath"]huh? i heard she got 10th in NC.

yeah i tied with lyndon with a 27 but he had more sprint right.[/quote]

wtf how do all you people know her.  I know her because she lived in my neighborhood and my mom and her mom still talk..

[quote="indianamath"]she's an AoPser. lol her username is asianhottie8293[/quote]

No what I meant is how do you even know her (I mean someone can have an acocunt but you still have to know that person to talk to her).  I mean its not like she was oh hi, i used to go to clay middle school randomly.  I know fanatic and mathcrazed both went to her classes if I remember correctly, but you were in 5th grade when she was in 6th.

Dr. shen used to talk about her. and in one of the AoPS classes, she was asking if anyone there was jeffrey and said that she was from NC. so, i asked her if she was christine hong.

This was a while ago (back with the Benjamins)

6th: Chapter: 13th,   State: 21st, team 1st

7th: Chapter: 5th,    State: 8th, team 1st

8th: Chapter: 1st,   State: 5th, team 5th (Benjamins moved for mathcounts)

In 8th grade I got a 28/12 but missed being on nationals, but when I took the national test, I would've gotten 7th place, and I don't get nervous.  The guy that beat me at state got 54th place, and Indiana lost on a tiebreaker and got second. Just think if he hadn't beaten me.

Whoa..a 28/12-40, with a tiebraker of 28 sprint and you didnt make nationals...wow I gotta feel bad for you because the only thing worse than failing at state is getting 5th (and you proalby wouldve made nationals anywhere else)

Wow, that must have been a REALLY hard year not to make nats with a 40...makes me feel like I have to improve A LOT if I want to stand a chance...Did the 4th place guy get more than 40 or was it a tiebreaker?

Yeah I remember that.

26/12 didn't cut it in 2006 (but 27/12 or 26/14 would've)... problems I missed were 7 and 8 on the target, with a last second switch from right to wrong on the latter.

 5th place is pretty good company, IMO... Phil, me, Jeffery XD.

6th: 2nd chapter, 2nd state by tiebreaker, ~218th nationals
7th: 1st chapter, 1st state, ~98 nationals
8th: 1st chapter, 1st state, unknown nationals (will find out in a few weeks)

Apparently, I don't come from a very good state.

what are we talking about... -_- ... i remember learning about christine hong from dr. chi... he was like... she got 18th in 6th grade! and i was like... -_-... i suck :D and we were all happy happy... guess what yuqing... i missed 4 of the target questions last year XD... and ... our team *coughcoughsuxcoughcoughcough* .


Has anyone noticed all the new games are basically spam? We have what, 4 Tug of Wars? We have a counting game, a countdown game, all of these are just Spam-a-thons =\ Your thoughts?

Yeah I agree especialy because everyone posts at the same time and then don't really care about the rules of it but some topics are still good.

I think that the countdown is spam spam and spam... Just to increase number of posts or whatever...
The same holds for the tugs of war, but it would at least be ok when there would be just one, not more.

But others (like questions&answers) are ok.

its not the spam and fun factory its the [i]spam and more spam factory[/i]

what's wrong with this section of the forum? who cares about post numbers? i post here to have fun.

you know, i really don't see the point in getting your posts up. there is no benefit from having the most number of posts here.

[quote="jhollenbeck"]Yes, I doubt that people here truly care about their post numbers (except the ones who complain that others do).   I am just here to have fun.  Because the tug of wars only include so many people, we need more than one.  I agree that the countdown is kindof pointless though.[/quote]

Ok 2, fine, 3, err.. maybe, 4 is getting way too overboard considering 2 of them allow everybody to play and 1 of them is currently accepting signups and the other is playing. You don't see 3 mafia games do you?

I do care about post numbers somewhat. But my favourite is still The Never Ending Story!

By the way, AoPS as I said is an Arena of Professional Spamming!

I agree with jhollenbeck, there are too many Tug-Of-Wars.  They really need to be limited.  I also think the mega countdown is entirely spam.

I HEREBY MAKE A PATENT THAT AFTER COUNTDOWN1 ENDS, I WILL START COUNTDOWN2 WHERE IT IS TIMED, SO IT IS NOT JUST WRITING POINTLESSLY NUMBERS.

There should be a separate game factory where posts do not get counted.

Yo, the most popular topics in aops are all spam. we have tug of wars( many of them), mega countdown, and tonnes of pretty retarded topics like the or game. I encourage people to create more interesting and fun topics, like "You are an Aopser if...", which i think is the best topic in this forum. The best topics are some of the ones in the Canadian Community forum.Those things are creative, interesting, hilarious, ..........................

[quote="jhollenbeck"]Now that there are 5 Tug-of-Wars...I say too much!  They are getting out of hand.[/quote]

how is it hurting you? don't go to those topics if you don't want to.

Wow, that's really sad that people care about their post count as much as you do ;), you do realize that the point of Art of Problem Solving is for well PROBLEM SOLVING, although games and stuff are ok, but just doing this simply for the sake of getting post count up is just plainly pathetic. Not only does it ruin your integrity since you're crying over keeping your posts but it also ruins the integrity of this forum. I'm not saying that Games and Fun Factory is bad, but come on, do something constructive with your time rather than just posting nonsense all day long ;)

[quote="bubka"]Ok let's get this straight then:
Art of Problem Solving begins and ends in Games and Fun Factory.

Do something constructive rather than post nonsense? [i]Indeed?[/i] Much as you know about nonsense! Take an example: Lewis Carrol was one of the greatest mathematicians. How did [i]he [/i]use his time? Yes, writing [i]nonsense[/i]. How much nonsense have you read I ask? Keep out of things you have no idea about. Nonsense does not penetrate skulls of all thicknesses, true, but keeping mum is the least you can do! Putting some emoticons looks smart, but it does not make you smart.[/quote]

Well, your argument is flawed. Art of Problem Solving does not "begin and end" in Games and Fun Factory since Games and Fun Factory was established long after the website was established and if that were the case, the website would be named "The Art of Random Musings." Lewis Carrol should be spelled Lewis Carroll and his "nonsense" was quite different from your constant nagging over not getting rid of the G&FF posts, and is quite contradictory to your origninal argument and it seems you care more about postcount rather than the G&FF forum, since AoPS begins and ends in G&FF right? And lastly, emoticons are not there to make you look smart, their purpose is to express an emotion you can't with words. (or for the efficiency).

Anyway, I think this topic should be locked since it seems that it has done nothing but provide heated arguments.

Before Games and Fun Factory AoPS was but a name. It was born the day G&FF was born.

I have read most of Lewis Carrol[b]l[/b] nonsense and I am perfectly well informed about his works, thank you. Though my nonsense is actually more Wodehouse and DNA oriented, but Lewis Carroll is my hero.

And it is not post count I am fighting for but justice. I can never stand segegation.

:o  :|  :huh:  :rotfl:
(sorry I couldn't resist ;) )

You should be awarded the "most-SPAM-and-spam-award"  :D

I think the personal attacks should be kept out of this. Okay, here's the deal. Games and Fun Factory is not in and of itself bad. It is a less serious area where newer members might feel more comfortable posting. However, when people get fixated on increasing post count, they often post meaningless things. Zetax makes a good point; a topic just takes up space if no one but the people who post there to increase post count would read the stuff. I imagine it's difficult to moderate for innappropiate things if there are so many space fillers.

Look, AoPS/Mathlinks is a MATH forum above ALL else.

Additinally, as Valentin already stated, this is NOT a democracy.  This forum is the private property of the administrators.  What they say goes.  They are very reasonable and listen to the opinion of the users.  However, when it comes to issues about how the board should be used, they have complete jurisdiction.

Personally, I think that old topics in GaFF should be deleted.  No one's seriously going to read through millions of posts of spam in the future.

Daniel, you're doing a great job moderating this forum.  Keep up the great work and don't give in to any pressure.

Btw, even if Arnav were added as another moderator, I'm not really sure how that would affect very much.  It seems that he would just leave everything alone.  But miths could still just go and delete stuff like usual and mods can't undo what other mods delete.

I would exclude mafia threads from that then, a lot of people go through to read the old mafia threads especially the ones participating in a current stalling mafia game.

Congratulations ZetaX on reaching your 2000 posts. You must be feeling great now.

And as for topics, there are many topics in other forums which nobody has bothered to reply to. The only certain way to lose interest in maths is to get serious about it.

I never support deleting old threads. They contain many brilliant gems of wit that are mush more worth treasuring than old maths problems.

I know that there are a lot unanswerde topics, but I doubt that this is because they are to serious; more exactly, some things like the followings can be reasons for not answering:
- to easy (yes that's indeed a reason!)
- to hard
- very bad writen (especially no $\LaTeX$)
- bad headlines/sources (ever tried to find some old problem again¿)
- ugly problems ;)
- to much spam...

But well, back to topic (or not):
I'm not really interested in the 2000 posts (well they look nice  :D ), and would surely fall a good part down when all posts are un-counted here. But it doesn't bother me.
And the old threads... well... some ideas:
The rule that someone should not relive old topics should have exclusions, especially for polls, and additionally a poll-collecting-thread should be made. That would stop the reoccuring of all these polls (often even the same people answer).
Similar things for other stuff that could repeat.
But for the really unintersting stuff (there is a lot!), it should be simply deleted. And as said above, the mods decide what to delete ;)

Like $the$ $or$ $game$, $answer-a-question-ask-a-question$, $the$ $inactive$ $topics$, and simply locked topics.

What's the point in deleting them? Don't you see you are removing a part of history?

Also if you delete the inactive topics no one here will have any posts as all there old ones will be deleted

This topic will be closed now since it has gotten (surprise surprise) just about as spam-filled as any of the other ones. We're not voting on what to allow and what not to allow. We're not taking opinions, suggestions, or requests. Nor are we doing this by majority rule. Spam will be deleted under the current conditions and that's final.

:-) Yes and Joseph points out something I think you guys missed a while back: Arnav won't fix anything.

And in case you guys missed my point, we're not having another debate either. This is a notification, not a negotiation.


Asymptote
geometry
geometric transformation
email

Hello,

A beginning of translation of the package [url=http://piprim.tuxfamily.org/asymptote/travaux/index.html]geometry_dev[/url] for Asymptote is available [url=http://svnweb.tuxfamily.org/filedetails.php?repname=piprim/asymptote&path=%2Fbranches%2Fgeometry_en.pdf&rev=0&sc=0]here[/url].
Any help is welcome for rereading and correcting it.

Thanks in advance.

Also, in what stage of development is the package? (alpha, beta, rc, final)? I would like to give it a spin on the site asy :)

Let RC because the package is still little used so little tested.

All right, so I will add this to the custom packages on this site (so codes using it will work, just put the import geometry_dev command).

[asy]import geometry_dev;
size(7cm);

triangle t=triangleabc(5,3,4);
draw(t);
label(t);
draw(incircle(t));

triangle it=intouch(t);
dot(it);
label( "$A^{\prime}$", "$B^{\prime}$", "$B^{\prime}$", it);[/asy]

Also please let me know by email whenever you release new versions of graph_pi and/or geometry_dev so that I can update them on the site.

The versions used on the site can be viewable at http://www.mathlinks.ro/latexrender/asy/geometry_dev.asy and http://www.mathlinks.ro/latexrender/asy/graph_pi.asy

Hello,
The complete English translation of the documentation geometry_dev package is now available [url=http://piprim.tuxfamily.org/asymptote/geometry_en.pdf]HERE[/url].
As the authors do not have a good English language, proofreading and corrections are welcome.

Note: geometry_dev.asy is available [url=http://piprim.tuxfamily.org/asymptote/travaux/index.html]here[/url].

Hello,

A new version of [b]geomtry_dev[/b] is available [url=http://piprim.tuxfamily.org/asymptote/travaux/modules/geometry_dev.asy]HERE[/url]
This version is needed for the new version of Asymptote (1.44) and for SVN version.
[b]This version of geometry_dev is not compatible with Asymptote 1.43[/b]


[quote="joejia"]A stack of 100 nickels has a height of 6.25 inches. What is the value, in dollars, of an 8-foot stack of nickels?[/quote]

[hide]  100nickels = 6.25 inches

          x nickels    = 96 inches

96 / 6.25 = 15.36; 15.36 * 100= x = 1536 nickles= [b]$76.80[/b]

[/hide]

[hide]
$\frac{6.25}{100}=0.0625$                                           Each nickel is 0.0625 in. high

$\frac{8\cdot12}{0.0625}=1536$                                         There are 1536 nickels in the stack

$1536\cdot0.05=76.8$

The pile is worth $76.8[/hide]

[quote="joejia"]A stack of 100 nickels has a height of 6.25 inches. What is the value, in dollars, of an 8-foot stack of nickels?[/quote]

[hide]$\text{A nickel has a height of (in inches)}$

$\frac{6.25}{100}=0.0625$

$\frac{0.0625}{12}$
$\text{feet}$

$8\div\frac{0.0625}{12}=1536$

$1536*0.05=76.80$[/hide]

[hide]8 feet is 96 inches. 96/6.25 = 15.36, so there are 1536 nickels. 1536*5 = 7680. So the stack is worth $76.80[/hide]


calculus
integration
calculus computations

Let $ I_k \equal{} \int_{ \minus{} 1}^1 {n\choose k}(1 \plus{} x)^{n \minus{} k}(1 \minus{} x)^k\, dx$. Integrate by parts to see that if $ k>0$ then $ I_k \equal{} I_{k\minus{}1}$. Hence $ I_k\equal{}I_0 \equal{} \frac{2^{n\plus{}1}}{n\plus{}1}$.


On a number line, what is the coordinate of the point that is between point $ A$ at $ \frac{3}{x}$ and point $ B$ at $ \frac{7}{2x}$ and is one-third the distance from $ A$ to $ B$? Express your answer as a common fraction in terms of $ x$.

A is also equivavlent to 6/2x, so 1/3 of 1/2x is 1/6x. 6/2x+1/6x=19/6x


For two orthologic trinagles $ A_1A_2A_3$ and $ B_1B_2B_3$, consider $ O_1$ to be the $ A_1A_2A_3$ orthology centre{point of concurrence of the the perprendiculars from $ A_1A_2A_3$ and $ O_2$ to be $ B_1B_2B_3$ orthology centre.{{point of concurrence of  the perprendiculars from $ B_1B_2B_3$}.
I read that $ O_1$ and $ O_2$ are isogonal conjugates with respect ot each other. But isogonal conjugates for which triangle? Like we have isogonal conjuagte of a point with respect to  a particular trinagle only rite? So can someone please tell which trinagle with respect to whihc the the two orthology centres are isogonal conjuagtes of each other?

Thanks in advance.
Aravind.

Dear mathlinkers,
I haven't see something like this in the geometric litterature...
Sincerely
Jean-Louis

[quote="Agr_94_Math"]Here is the link Mr. Jayme:
[url]http://mathworld.wolfram.com/OrthologyCenter.html[/url]
I dont think Wolfram Mathworld can be wrong.[/quote]
[quote="Wolfram Mathworld"]Orthology Center

Given a pair of orthologic triangles, the point where the perpendiculars from the vertices of the first to the sides of the second concur and the point where the perpendiculars from the vertices of the second to the sides of the first concur are the orthology centers of these triangles. [color=red]The points are isogonal conjugates of each other[/color].

SEE ALSO: Orthologic Triangles

This entry contributed by Rudolf Fritsch.[/quote]

I think the red words have not a meaning, because two points may be isogonal conjugates of each other, only with respect to an angle ( the line segments connecting these points with the vertex of the corresponded angle ), or with respect to a triangle.

Kostas Vittas.


integration
induction
real analysis
real analysis unsolved

It's enough to prove the case $ n=2$, the rest follows by induction. Also, we can assume $ r=1$; the general case follows by considering $ |f_{i}|^{r}$. So we want to prove that $ \int |fg|\le \|f\|_{p}\|g\|_{p'}$... you should be able to find a proof in any introductory analysis textbook.

the case $ n=2$ and $ r=1$ I can do myself. I'll try induction


geometry
rectangle
combinatorics proposed
combinatorics

Here is a fine problem:

Let $f(n)$ denote the maximum number of rectangles into which a square can be divided in such a way that every line which is parallel to on of the sides of the square intersects the interiors of at most $n$ of the rectangles. Show that
\[
3 \cdot 2^{n-1} - 2 \leq f(n) \leq n^n.
\]

For the lower bound:

For $n=2$, dissect the square in four equal squares. Call them $1,2,3,4$, according to which quadrant they represent. Now, for $n=3$, dissect the smaller squares $1,3$. For $n=4$, dissect the quadrants $1,3$ of each of the quadrants $1,3$ of the original square, and so on.

And an opposite problem:

Let $g(m,n)$ denote the minimum number of rectangles into which a square can be divided in such a way that every line which is parallel to the horizonal side intersects at least $m$ rectangles, and and every line which is parallel to the vertical side intersects at lest $n$ rectangles. Find $g(m,n)$.

For the upper bound.

Suppose the contrary. Consider the horizonal side and at most $n$ rectangles which stay on it. They divide the whole segment into $n$ segments. It follows that at least one of this segments has interior points with at least $n^{n-1}+1$ projections of our rectangles ($CD$ on the figure below). Now bound our consideration from the whole square to the rectangle $GHFE$. It is clear that $GHFE$ satisfies the same condition as the initial square and it has $n^{n-1}$ rectangles inside itself. In particulary, side $GH$ contains at most $n$ rectangle's sides. So we will find segment $IJ$ which intersect with at least $n^{n-2}$ projections. And so on. Thus we will perform $n+1$ steps. And now it is clear how to find vertical line which intersects at least $n+1$ rectangle.

Hope it is correct.

Is it possible to construct some example from this solution?


$ a,b$ are positive integers. If for each $ n$, we have $ rad(a^n\minus{}1)\equal{}rad(b^n\minus{}1)$, then show that $ a\equal{}b$.

$ \text{rad}(N)$ is the product of the prime divisors of $ N$.

the assertion is obviously equivalent to $ \forall p\mbox{ primes }: ord_p(a)\equal{}ord_p(b)\Longleftrightarrow a\equal{}b$ where

$ ord_p(z)\equal{}\left\{\begin{array}{ll}\min\{n\in\mathbb{N}|z^n\equiv 1\bmod{p}\} & \mbox{if }p\not|z\\
\infty & \mbox{otherwise}\end{array}\right.$


$ 3^(3 \minus{} x) \equal{} 2^(2x \plus{} 1)$
Find x

Sorry, if might look like 3(3-x)=2(2x+1), but it's actually 3^(3-x)=2^(2x+1)

Take  the $ \ln$ on both sides.. This gives you:

$ (3\minus{}x)\ln3 \equal{}(2x\plus{}1)\ln 2$

$ 3\ln 3\minus{}\ln2\equal{}x(\ln3\plus{}2\ln2)$

$ \ln\left(\frac{3^3}{2}\right)\equal{}x\ln(12)$

$ x\equal{}\log_{12}\left(\frac{27}{2}\right) \approx 1.0474$

LaTeX note:

when using long inputs in places like powers and index places, use {} to specify what you want. So 3^(3-x) would be 3^{3-x}


Euler
function
number theory
totient function
number theory unsolved

[quote="alexilic"]Is the equation
\[ 2 \cdot \varphi (n) \equal{} n \plus{} 1\]
solvable? I noticed first solutions involve prime Fermat numbers,
\[ n \equal{} F_k \minus{} 2 \equal{} 2^{2^k} \minus{} 1\]
for $ k \equal{} 0, 1, 2, 3, 4, 5$.[/quote]

Hummm, what is $ \varphi(n)$ in your problem ?. If it is the Euler totient function, then $ \varphi(n) < n$ $ \forall n > 1$ and so $ \varphi(n) \equal{} n \plus{} 1$ has no solution in $ \mathbb N$

Pco, you forgot about the factor $ 2$ in the front.

BTW, solutions are given in Sloane's [url=http://www.research.att.com/~njas/sequences/A050474]A050474[/url].

If you have time, take a look at this paper: http://www.ams.org/bull/1932-38-10/S0002-9904-1932-05521-5/S0002-9904-1932-05521-5.pdf . Can you use something out of it?

[quote="hsiljak"]Pco, you forgot about the factor $ 2$ in the front.[/quote]

 :blush:  Ooooops, sorry. Thought it was a kind of number from a list of equation. (I should have looked also at the subject).


Every Spanish citizen is assigned an ID that consists of eight digits (0–9) and one letter. The letter is not assigned randomly: only 23 letters are used, which correspond to the 23 possible remainders on dividing the eight-digit number by 23. The letter acts as a "check symbol": if somebody makes a mistake when entering the number, chances are that the resulting ID will not be valid (the letter will be wrong) and the error will be detected right away.

a) Show that if one of the eight digits is entered incorrectly (e.g., 12349678 in place of 12345678) or if two digits are mistakenly swapped (e.g., 12845673 in place of 12345678), the letter will always be wrong.

b) 23 is called the "modulus" of the check symbol. The Spanish alphabet has 27 letters, so the check symbol could have been based on any modulus up to 27. Show that 23 is the largest possible modulus less than or equal to 27 for which both of the conditions in (a) hold. Find another such modulus, and show that it works as well.

meh not exactly my favorite problem but I believe this is right.

[hide="a"] basically the idea is let the two swapped digits be $ d_{1},d_{2}$, let their place values be $ 10^{a_{1}},10^{a_{2}}$. Then the difference between the two ID numbers is $ (d_{1}\minus{}d_{2})(10^{a_{1}}\minus{}10^{a_{2}})$ or something close to that. It's not hard to see that neither of the products can be 0 mod 23 (just check the residues of the 10^i's for 0-7).

Then if we enter a digit incorrectly, we see that $ (d_{1}\minus{}d_{2})10^{a}$ can't be 0 mod 23 either.[/hide]

[hide="b"] it's not too difficult to find counterexamples for 24, 25, 26, and 27, just look for two $ i$'s between 0 and 7 such that $ 10^{a_{1}}\equal{}10^{a_{2}}$ as before.

Also, the proof that 19 works is similar to that in part a.[/hide]

[quote]It's not hard to see that neither of the products can be 0 mod 23 (just check the residues of the 10^i's for 0-7). [/quote]

You can narrow down your search space greatly with Fermat's little theorem, which states that, if a^d-1=0 (mod p), then p-1=0 (mod d)

17 also works.

Was it necessary to write 10 to a power of an unknown variable? If I remember correctly I did get this proof almost correct except I simply wrote 10 without a power...

Yep, you have to represent place value in there.


If $ x$ is from $ [0,\frac{\pi}{2}]$, then $ \sin x\geq\frac{x}{x\plus{}1}$.

It is well known that $ \tan x\geq x$ in the given interval.
So it is sufficient to show that $ \sin x\geq \frac{\tan x}{\tan x \plus{}1}\equal{}\frac{\sin x}{\sin x\plus{}\cos x}$
$ \iff \sin x\plus{}\cos x\geq 1\iff \cos ^2x\geq (1\minus{}\sin x)^2$
$ \iff 1\minus{}\sin ^2x\geq (1\minus{}\sin x)^2\iff \sin x(\sin x\minus{}1)\leq 0$ which is obvious.


Από το βιβλίο του τίτλου βάζω την παρακάτω άσκηση.Δεν πρόλαβα καν να την κοιτάξω , αλλά με τις ασκήσεις στον εγγεγραμμένο κύκλο έχω μια προτίμηση .Ελπίζω να μη μας νικήσει !

[b] ΑΣΚΗΣΗ[/b]

 Ο εγγεγραμμένος κύκλος ενός τριγώνου $ ABC$ εφάπτεται της $ BC$ στο $ E$ και έχει κέντρο $ I$.Αν $ AK$   είναι ύψος  του τριγώνου $ ABC$ και $ D$ είναι το μέσο της $ BC$ , να αποδείξετε ότι
  [b]α)[/b] η εθεία  $ DI$ τέμνει το $ AK$ σε σημείο $ Z$ ώστε  $ AZ\equal{} \rho$
  [b] β)[/b] H ευθεία $ DI$ διχοτομεί το τμήμα  $ AE$.

Μπάμπης

Κατ΄αρχάς έστω $ S$ το σημείο τομής $ ID,AE$ και $ P$ το συμμετρικό του $ E$, ως προς το $ D$ (κι άρα σημείο επαφής του παραγγεγραμμένου κύκλου με τh $ BC$).

Θα αποδείξω πρώτα το (β) ερώτημα και μετά από αυτό το (α) θα είναι τετριμμένο λόγω της ισότητας των τριγώνων $ AZS, SIE$.

Εστω $ N$ το αντιδιαμετρικό του $ E$ (στον εγγεγραμμένο κύκλο).Τότε επειδή ως γνωστόν υπάρχει μια ομοιθεσία με κέντρο $ A$ που πηγαίνει τον εγγραγραμμένο κύκλο στον παραγγεγραμμένο το $ N$ θα πηγαίνει στο $ P$ κι άρα $ A,N,P$ συνευθειακά. Αρα στο τρίγωνο $ EPN$ θα ισχύει $ DI//NP$ αφού $ D,I$ μέσα των $ EP, EN$. Αρα τώρα απ΄το τρίγωνο $ AEP$  $ DI//AN\iff DS//AP \iff AS\equal{}SE$ κι άρα το ζητούμενο. :)

Οι ασκήσεις που προτείνετε κ. Στεργίου είναι πάντα υπέροχες!


Hi,

My son is in middle school and doing Algebra I and my daughter is in
high school and doing Algebra II. I was thinking of enrolling them in same classes here, but here is my dilemma.

My daughter comes home with lot of work and she is already struggling for time, is it a good idea to register her for Algebra II again here?

My son on the other had got your book and started working on it and he is thru' first 55+ pages on his own and should be be joined in Algebra I or is there a different course that you would recommend?

Thank

I think you should have them take the pre-tests and post-tests for the courses and see how well they do.  It can help you to make you decision.

I think starting with Intro C & P can be a good choice.

Your son would probably be a good fit for the Algebra 1 course if he's already worked a bit in the book -- working ahead in the book is one of the biggest keys to success in our courses.  If he has already completed an algebra 1 class in school, then probably the best course would be to let him continue with our book, and have him enroll in the Intro Counting course.

As for your daughter, if she's already strapped for time, then I wouldn't exacerbate that problem unless she herself is very interested in doing more math, and is willing to give up some other activities to pursue it.  Time is a very scarce resource for high school students today, and far too many of them are pushed into far too many activities, at the expense of activities they would choose themselves and thereby benefit from much more.  If she does want to take a class with us, then the Intro Counting class would also probably be very good for her.  It would expose her to important areas of mathematics that she has probably not encountered much in school.

Wouldn't the later parts of Intro to Algebra help? And part of Alg 3? I'm using Alg 3 for the Algebra part of Precalc


algebra
polynomial
inequalities
number theory proposed
number theory

Assume $a_0,a_1\dots,a_n$ are digits and $(\overline{a_na_{n-1}\dots a_0})_{10}$ is prime number. Prove that the polynomial $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0$ is irreducible in $\mathbb Z[x]$

This result is named Cohn's theorem or Cohn's irreducibility criterion.

The reference above gives a solution only when $a_{n}\geq 2$, so here we put the general solution, following the same guidelines.

Let $z\in\mathbb C$ be such that $|z-10|\leq 1$, then $|z|\geq 10-|z-10|\geq 9$ so that \[|a_{n-2}z^{n-2}+\dots+a_{1}z+a_{0}|\leq 9\frac{|z|^{n-1}-1}{|z|-1}\leq\frac{9}{8}(|z|^{n-1}-1)<\frac{9}{8}|z|^{n-1}\leq|z|^{n}\] We also have ${\rm Re}(1+\frac{a_{n-1}}{a_{n}}\frac{1}{z})=1+\frac{a_{n-1}}{a_{n}}\frac{{\rm Re}(z)}{|z|^{2}}\geq 1$ because ${\rm Re}(z)=10+{\rm Re}(z-10)\geq 10-|z-10|\geq 9$, so that \[|a_{n}z^{n}+a_{n-1}z^{n-1}|= a_{n}|z|^{n}\left|1+\frac{a_{n-1}}{a_{n}}\frac{1}{z}\right| \geq a_{n}|z|^{n}{\rm Re}(1+\frac{a_{n-1}}{a_{n}}\frac{1}{z}) \geq a_{n}|z|^{n}\] Putting together both inequalities above yields \[|p(z)| \geq |a_{n}z^{n}+a_{n-1}z^{n-1}|-|a_{n-2}z^{n-2}+\dots+a_{1}z+a_{0}| >a_{n}|z|^{n}-|z|^{n}\geq 0\] Conclusion: If $\alpha_{1}$, $\alpha_{2}$,..., $\alpha_{n}$ are the complex zeros of $p(x)$, then $|\alpha_{1}-10|>1$, $|\alpha_{2}-10|>1$,...,$|\alpha_{n}-10|>1$.
Given any non-trivial factorization $p(x)=f(x)g(x)$ in $\mathbb C[x]$, $f(x)$ and $g(x)$ having integer leading coefficients, so that $f(x)=A(x-\alpha_{i_{1}})\dots(x-\alpha_{i_{s}})$, $g(x)=B(x-\alpha_{i_{s+1}})\dots(x-\alpha_{i_{n}})$ with $i_{1},\dots,i_{n}$ a permutation of $1,\dots,n$ and $A,B\in\mathbb Z$, we have \[|f(10)|=|A\parallel 10-\alpha_{i_{1}}|\dots |10-\alpha_{i_{s}}|>1\]  \[|g(10)|=|B\parallel 10-\alpha_{i_{s+1}}|\dots |10-\alpha_{i_{n}}|>1\] but being $p(10)=f(10)g(10)$ a prime number, either $f(10)\not\in\mathbb Z$ or $g(10)\not\in\mathbb Z$, so that $f(x)\not\in\mathbb Z[x]$ or $g(x)\not\in\mathbb Z[x]$.


[size=150]1. the union of closed interval [1/n, 2 - 1/n] = open interval (0,2).

2. the intersection of open interval (-1/n, 1 + 1/n] = closed interval [0,1].[/size]

[color=green]Locked for double posting.[/color]

1. $ I_n\equal{}[\frac{1}{n}, 2\minus{}\frac{1}{n}]$
if $ x\in (0,1)$
then $ \exists n$ s.t. $ nx \ge 1$ so $ x\ge \frac{1}{n}$ so $ x\in [\frac{1}{n}, 1) \subset I_n$ for at least one n
if $ x \in (1,2)$ then  $ 2\minus{}x \in (0,1)$
and $ \exists n$ s. t $ n(2\minus{}x)\ge 1$ so $ x\le 2\minus{}\frac{1}{2}$ so $ x\in (1, 2\minus{}\frac{1}{n}]\subset I_n$ for at least one n
if $ x\equal{}1$, then $ I_1\equal{}\{1\}$ so x is in $ I_1$
thus $ (0,2) \subset \bigcup_{n\equal{}1}^\infty I_n$

if $ x\le 0$, then $ x \le \frac{1}{n}$ for all n, so $ x\notin \bigcup_{n\equal{}1}^\infty  I_n$
if $ x \ge 2$ then $ x \ge 2\minus{}\frac{1}{n}$ for all n, so $ x\notin \bigcup_{n\equal{}1}^\infty  I_n$

thus $ \bigcup_{n\equal{}1}^\infty I_n \equal{} (0,2)$ exactly

2. $ J_n \equal{} (\minus{}\frac{1}{n}, 1\plus{}\frac{1}{n}]$
if $ x\in [0,1]$ then $ \frac{\minus{}1}{n} < x < 1\plus{}\frac{1}{n}$ so $ x\in J_n$ for all n, so $ [0,1] \subset \bigcap_{n\equal{}1}^\infty J_n$
if $ x<0$ then $ \exists n$ s.t. $ n(\minus{}x) >1$, so $ x<\frac{\minus{}1}{n}$ and $ x\notin J_n$ for at least one n
if $ x>1$ then $ \exists n$ s.t. $ n(x\minus{}1)>1$, so $ x> 1\minus{}\frac{1}{n}$ and $ x\notin J_n$ for at least one n

thus we have $ \bigcap_{n\equal{}1}^\infty J_n \equal{} [0,1]$ and we are done


conics
parabola
calculus
derivative
analytic geometry
graphing lines
slope

The line $ x\equal{}c$ where $ x>0$ intersects the cubic $ y\equal{}3x^{3}\plus{}3x^{2}\minus{}9$ at point P and the parabola $ y\equal{}4x^{2}\plus{}4x\plus{}5$ at poin Q.

a) If a line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q, find the value of $ c$ where $ c>0$.

b) Also, find the equations of the two tangent lines described above.

I'm not going to work everything out here right now, but I will show my ideology.

a) I find the derivative of both equations and I set them equal to each other. I solve for x and if there are more than one answer, I take the positive one.

b) With the answer from above, I find the slope at that point and make an equation using the slope and the coordinates of the point.

Thank you in advance.

hello, where is your work? i can't find any equation.
Sonnhard.

Like I said, I wasn't going to show every step. I was just making a rough outline. I will do the actual computations and such later.  :)


The amount of $ \$72$ is to be divided among Mr. Abbondanzio, Ms. Barta and Ms. Conders in the ratio $ 6: 2: 1$, respectively.  How much should Ms. Barta receive?

We divide $ \$72$ by $ 6\plus{}2\plus{}1$ to get $ 8$. Ms. Barta received $ 8\cdot2\equal{}\boxed{16}$.

[quote="$ LaTeX$"]We divide $ \$72$ by $ 6 \plus{} 2 \plus{} 1$ to get $ 8$. Ms. Barta received $ 8\cdot2 \equal{} \boxed{16}$.[/quote]
but 6+2+1does not equal 8 but it is equal to 9
ans should be 18
since
it is $ 6: 2: 1$
$ 6/9$+$ 2/9$+$ 1/9$=$ 81$
$ 2/9$ of $ 81$=$ 18$
Mrs. Barta should get $ 18$ dollars

[dollar]LaTeX[dollar] did not say that $ 6\plus{}2\plus{}1$ equals $ 8$, he said that $ \frac{72}{6\plus{}2\plus{}1}\equal{}8$.  So actually, the correct answer is $ \boxed{16}$

72÷(1+2+6)=8
8*2=16
akishaj is going to be here now!!!!!     :[  i suck!!!!!     crying%$:&

The answer is $72 \cdot \frac{2}{9}=16$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

a friend asked me this question, and i got nicely surprised by its solution...
as the title says, you can't use sylow theorems.
given a group of order 108, and a subgroup of order 27, show that there's a normal subgroup of order 9 or 27.

if you want, you can give a generalization (which is pretty obvious, after you've solved it).

Consider the operation on the left cosets of the given subgroup. This yields an homomorphism $G\to S_{4}$. The kernel of this homomorphism is a subgroup of the given one and has at least 9 elements (look at the orders) and hence it has 9 or 27 elements.


I am a high school student. I am an American citizen living in India. I will have to apply for the universities by this december. Which of them, in your opinion, have good mathematics departments? My SAT score is 2060( math 720 reading 680 writing 660). My GPA is too bad. Do I stand a chance for any into the universities? I know that I stand no chance for the top colleges like MIT or Caltech. Could you suggest where i should apply? I don't have any qualifications except the national olympiad (didn't do well at the selection camp for the IMO) and a generalization of a theorem of Posa that I could attach to my application. Do I stand a chance for any of the Univesity of California branches?( I mean UCLA, UC San Diego, Uc Irvine etc)
I would be thankful if anyone would reply to this post.

Do you have an opportunity to retake the SAT test?

Your SAT scores should be good enough for admission to most regional schools.  I don't think any North Carolina state school would reject you based on your SAT scores.  Most California state schools wouldn't either.

Grades might be a bigger issue for you with respect to college admission.  Does your high school transcript indicate that you are unwilling to work on academics?  Generally, high school grades are viewed as an indicator of willingness to work hard.  If you are attending high school in India, it might be difficult for many admissions officers at regional schools to make this interpretation by looking at your transcripts.  You ought to provide some information regarding the competitiveness of the students at your high school.  If you are in the bottom half of a class at a highly competitive high school then poor grades may not be viewed as negatively as they would otherwise for purposes of college admission.

Even with poor grades, I suspect that most of the regional schools in NC would admit you.  The average SAT (reading + math) scores for entering freshmen at UNC-Charlotte and similar regional schools in NC are typically under 1200.  California schools might have slightly higher scores than NC schools, but not enough to make a difference for you.

My school (a Cal State school, CSULB) would admit you coming from a U.S. high school with that SAT if your GPA were about 2.7 or higher. As it is, you'd get handled as an international applicant, where the standards are a little less formula-driven.

Thanks for replying to my post. I am taking the SAT Reasoning test for the second time this May. Is recommendations an important factor for admissions? I mean, are they as important as the GPA or the SAT?

The value of recommendations varies widely by college. For many state schools, including the UC's, they don't matter at all.

I should have included this in the earlier post. Is the statement of purpose important? I mean, is it worth to spend considerable time writing it?

Yes, you should spend at least as much time writing your essays and/or "statements of purpose" as is necessary to convince an admissions officer that you know what you'd like to accomplish in college and that the institution has the programs to help you meet your goals.

Poorly written essays are potential reasons to reject applicants, especially those with poor grades or other marginal qualities to the application.  This might especially be true for you.  As mentioned above, you might be treated as an international applicant and not admitted through the typical formula-driven process used at many regional state schools.

The only good news is that they will treat me as a domestic student.( I sent e-mails to some universities to ask about that).
Could you give any suggestions as to which universities I should apply to? I will need plenty of financial aid. Hence I must apply to places will good financial programmes. Does anyone know about the University of Southern California? Does it have good financial aid programmes?

Apply to CSU Long Beach and some UC's like the ones you'd want to go to (I don't know, you might not want to go to Merced or something, so apply wherever you think you'd be happy). I think tuition at UC's for a non-CA resident will not be worth going there. There are some scholarships, but they are mostly insufficient...

I don't know how bad your gpa is (i sense some modesty in this - i bet it's not as bad as you make it out to be), but you can also apply to some ivies if you want (i think it's safe to say, however, that if your gpa is lower than 3.0, you'd probably be wasting the money spent for the application fees at the ivies). Harvard, Princeton, and Yale will treat you same as the domestic students and will give lots of financial aid.

I think it'd be hard to get any money at USC, especially because I doubt they know much about math competitions, so they'd judge you mostly on grades and SAT's. USC costs about $\$$50,000 a year. I sincerely regret applying there - it was one of the more stupid things I did...

The best advice: Work really hard in school right now and raise your grades. It's good to show improvement.

I retook the SAT as tokenadult advised. The new score is 2250 (math 800, critical reading 760, writing 690). Do I now stand better chances of being accepted by good universities? Which colleges should I apply to? I probably could get my 11th grade gpa changed. But not the 9th or 10th grade gpa.


Find all integers $ (m,n)$ such that \[ 3^{m}\equal{}2mn^{3}\plus{}1.\]

Here what I did , maybe it is wrong since at the night my mind stops and forget to start again...
Write the equation as $ 3^m\minus{}1\equal{}2mn^3$ (1)

We know that if $ 2^a\parallel{}\frac{x^2\minus{}y^2}{2}$ and $ 2^b\parallel{}n$ then $ 2^{b\plus{}a}\parallel{}x^n\minus{}y^n$

Let $ 2^a\parallel{}m$ and $ 2^k\parallel{}n$ We have that $ 2^3\parallel{}\frac{3^2\minus{}1}{2}$ and $ 2^a\parallel{}m$ so $ 2^{a\plus{}3}\parallel{}3^m\minus{}1$

so the greatest power of 2 that divides the  RHS of (1) is $ a\plus{}1\plus{}3k$ and this must be equal with $ a\plus{}3$ which is absurb
except $ m\equal{}n\equal{}1$ and we are done

We rewrite equation as $ 3^{m}\minus{}1\equal{}2mn^{3}$. Let us write $ m\equal{}2^{a}b$, where $ b$ is odd, and $ n\equal{}2^{c}d$, where $ d$ is odd. Hence the RHS of rewritten equation is divisible by $ 3c\plus{}a\plus{}1$. Now if I have understood Hensel`s lemma correctly, then the power of $ 2$ dividing $ 3^{m}\minus{}1$ is $ a\plus{}2$, if $ a>0$. So, if $ a>0$ then $ a\plus{}2\equal{}3c\plus{}a\plus{}1 \rightarrow 3c\equal{}1$, which is a contradiction. Hence $ a\equal{}0$.

Now we write $ m\equal{}p_{1}p_{2}...p_{s}$, where $ p_{i}$ are odd primes greater than 3, since LHS is not divisible by 3, if greater than 0. All primes $ p_{i}$ divide the RHS of the equation. Hence we must check if they also divide the LHS. Now we choose the smallest from $ p_{i}$-s (not necessarily unique). We write $ 3^{b}\equal{}(3^{p_{1}})^{p_{2}...p_{s}} \equiv 3^{p_{2}...p_{s}} (mod p_{1})$ (Fermat`s theorem).

If we want that $ p_{1}$ divides LHS, then $ 3^{p_{2}...p_{s}}\minus{}1 \equiv 0 (mod p_{1}) \rightarrow 3^{p_{2}...p_{s}} \equiv 1 (mod p_{1})$. Now since from Fermat`s theorem we know, that $ 3^{p_{1}\minus{}1}\equal{}3^{(ord_{p_{1}} (3) \cdot k_{1})} \equiv 1 (mod p_{1})$, and from our problem that $ 3^{p_{2}...p_{s}}\equal{}3^{(ord_{p_{1}} (3) \cdot k_{2})} \equiv 1 (mod p_{1}) \rightarrow  gcd(p_{1}\minus{}1, p_{2}...p_{s} ) \equal{} ord_{p_{1}}(3) > 1$. However, since all $ p_{2}, p_{3},..., p_{s}$ are prime and greater than $ p_{1} \rightarrow gcd(p_{1}\minus{}1, p_{2}...p_{s})\equal{}1$. Contradiction.

Hence no prime can divide $ m$, which is possible only if $ m\equal{}1$. Hence $ (1,1)$ is the only solution to the given equation.


$f(x) = \sqrt(2*(1+\sqrt(1-x^2)))$
$g(x) = 2x^2+7x+2$
$h(x)=(2-\sqrt(2))x+2$

$f(0) = g(0)$
$f$ is decreasing for $x \geq 0$ whereas $g$ is increasing, so there are no solution for $x > 0$

$f$ is concave (convex down in english ?) so $f(x) \geq h(x) > g(x)$ for $x < 0$  so there are no solutions for $x < 0$

hey, do you have any solution ? Remember that
                Sqrt(2(1+sqrt(1-x^2)))=sqrt(1-x)+sqrt(1+x)


Please explain how to solve this...

A company wishes to provide ID cards for its employees.  Each identification number will begin with a letter of the alphabet followed by two digits.  How many different ID numbers can be made?

I know the answer is 2600, but I don't how to get that number.

THANKS!  I really need this ASAP.

There are $26$ different letters that could be used, $10$ different numbers for the first digit, and $10$ again for the second digit.  Since each space is independent of the other, you can just multiply them:

$26\times 10\times 10=2600$

Wow- that was too easy.  THanks so much!


Okay i learned that having a smaller level of significance corresponds to a higher z-score limit.

But z-score measures how spread out data is, so shouldn't a higher accuracy actually correspond to lower z-score?

That doesn't make sense. Before you do something in statistics, you need to know precisely what question you're asking. These are two unrelated questions, which cannot be answered by the same data.

Elaborating on that:

"How spread out the data is" is a property of a distribution as a whole. A Z-score is a property of a single observation.

Take an example: suppose your distribution has the value $ 0$ repeated $ 99$ times and the value $ 100$ which occurs once. The mean of this distribution is $ 1,$ while the standard deviation is approximately $ 10.$ That - the standard deviation - is a measure of how spread out the distribution is.

Z-scores are essentially a rescaling of the units of measurement involved, shifting the axis so that the mean is at zero and measuring everything in units of standard deviation.

Each point at $ 0$ scales to a Z-score of approximately $ \minus{} .1$. These points all lie "very close" to the mean. On the other hand, the point at $ 100$ has a Z-score of approximately $ \plus{} 10.$ It lies "very far" from the mean, and would be considered an outlier.

Everything I've just said is probability, not statistics. I'm not asking to draw any inferences.


limit
calculus
integration
inequalities
real analysis
real analysis unsolved

Prove that if $(a_{n})_{n\geq 1}$ is a sequence of positive numbers with $\sum_{n=1}^{\infty}a_{n}<\infty$, then for all $p\in (0,1)$
\[\lim_{n\to\infty}n^{1-1/p}({a_{1}^{p}+a_{2}^{p}+\ldots+a_{n}^{p})=0. }\]

${\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\alpha_{i}\right)=L}$
If $0<p<1$
${\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\alpha_{i}^{p}\right)\leq L}$
${\alpha_{i}^{p}\leq \alpha_{i}}$
So the last limit converges
${\lim_{n\to \infty }\, \left(n^{1-\frac{1}{p}}\sum_{i=1}^{n}\alpha_{i}^{p}\right)=\left(\lim_{n\to \infty }\, n^{1-\frac{1}{p}}\right) \left(\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\alpha_{i}^{p}\right)\right)}$
But
${\lim_{n\to \infty }\, n^{1-\frac{1}{p}}=0}$

the right limit converges to a value lesse than L.

is missing something..?!

humpf

only holds for $\alpha >1$

hohoho. how stupid i am huhu.. but this proof is wrong.

[quote="Yustas"]It looks more natural if we write integrals instead of sums(however, it makes no difference):
\[f(x)>0, \ \int_{0}^{\infty}f(x)dx<\infty \rightarrow \ \lim_{n\to\infty}n^{1-1/p}\int_{0}^{n}f^{p}(x)dx=0. \]
To prove this it is enough to apply Holder's inequality: $\int_{0}^{n}f^{p}(x)dx\leq \left(\int_{0}^{\infty}f(x)dx\right)^{p}n^{1-p}$. Now note that $p+1/p>2 \ \forall \ p\in(0,1)$.[/quote]
Congratulations! your Idea is Excellent, But I can't understand : How do you change sum by integral?

[quote="Yustas"]
It looks more natural if we write integrals instead of sums(however, it makes no difference):
[/quote]
am I right ? because
$\sum \leq \int$?

You can apply Holder's inequality to the sum directly. Or, if we let $f(x)=a_{i}, \ x\in(i-1,i]$, it is obvious that $\int_{0}^{\infty}f(x)dx=\sum a_{i}$.

Yes I agree with you I post sulution but it's will be "your" sulution without integral:
By Holders inequality We can prove next inequality:
$(\sum_{i=1}^{n}b_{i}^{q})n^{q-1}\geq(\sum_{i=1}^{n}b_{i})^{q}$
where $q>1$
after substitution :
$b_{i}=a_{i}^{\frac{1}{q}}$ and $\frac{1}{q}=p$ where $p$ will be $0<p<1$
we will be take :
$(\sum_{i=1}^{n}a_{i})n^{1-p}\geq \sum_{i=1}^{n}a_{i}^{p}$
and it's will be true if we take limit. after is simpl.


Given a sequence $a_{1},a_{2},...,a_{9}$ of one-digit numbers with the property that if for some $n$ we have $a_{n}= 1$, then $a_{n+1}$ ≠ $2$; and if for some $n$ we have $a_{n}= 3$, then $a_{n+1}$ ≠ $4$. Prove that there exist two numbers $\large k,l \in{1,2,3,...,98}$ such that  $a_{k}= a_{l}$ and $a_{k+1}= a_{l+1}$.

Sorry about my poor English!

Solution is similar to http://www.mathlinks.ro/Forum/viewtopic.php?t=135582

Construct a de Bruijn graph and show that there is no arc-disjoint path of length 97.


Hello. I found this interesting thing on a chain mail. It actually works. Try it! After you try it, prove it. I know you don't prove things in Mathcounts or MOEMS, but this is pretty simple.
-------------------------
Don't tell me your age;
The Hershey Man will know!
YOUR AGE BY CHOCOLATE MATH
This is pretty neat.
1. First of all, pick the number of times a week that you would like to have chocolate
2. Multiply this number by 2
3. Add 5
4. Multiply it by 50
5. If you have already had your birthday this year add 1759 ... If you haven't, add 1758.
6. Now subtract the four digit year that you were born.
You should have a three digit number:
The first digit of this was your original number (i.e., how many times you want to have chocolate each week).
The next two numbers are YOUR AGE!
-------------------------
Try it and prove it.  :D  :D  :D

BTW: Hide your proofs so you don't spoil it.

[hide="hint 1"]
Let $ x$ be the number from step $ 1$.  Go through the steps with algebra.
[/hide]
[hide="hint 2"]
The exact method will become invalid after this year ends.
[/hide]

Yes it will. Nice observation. Find out what you can tweak in these steps to make it work for the next year.  :D

For the next year you'd have to add 1760 or 1759 instea, I believe..

[hide]
x is the # of chocolate you eat a week and y is the year you were born on:

x
2x
2x+5
100x+250
100x+2008 (maybe +1)
100x+2008-birthday (maybe +1)

So it works
[/hide]

[quote="Utter Loser"]I've seen one of these, but for the one I saw, you couldn't have chocolate/whatever it was more than 10 times a week. I wonder what the difference is, because I don't see it....... :huh:[/quote]

Lets say you want to have chocolate 1583217510289375892137508912375890123750982137509812374893217508932175081327 days of the week. Do this algebra. Will you have a three digit number?

The problem works if you say that the last 2 digits will be your age.

This doesn't work if your age is 3 digits. :o

Not many people live to 100 years old. I guess that was made for children, simply because they will try to find out why it works.


[quote="samus"]I'm self-studying, and I really don't understand:
1. What is the equation of the plain which goes through point (11,7,-2) and contains the line (x-10)/5 = (y-3)/7 = (z-1)/-2.

2. Find the area of the triangle whose vertices are given:
X(8,-4,2);Y(-6,10,18);Z(8,0,-6).[/quote]

Neither of these are actually multivariable calculus questions, but they are straightfoward.

A direction vector for l in q1 is <5,7,-2>.  We then find another point in the desired plane, <10,3,1>.  Hence <1,4,-3> is a direction vector which is parallel to the plane.

We take the cross product of these two direction vectors to find the normal vector to the plane, and it's straightforward from there.

q2:

Find vectors XY,XZ, take the norm of the cross product, divide by two, done.


[quote="bkunchanapalli"]p,q,r are given real numbers.Consider the polynomials (q-r)(x-q)(x-r)+(r-p)(x-r)(x-p)

+(p-q)(x-p)(x-q)+k for all real k.Then

A)there exists a k such that the degree of the polynomial is 2
B)there exists a k such that the degree of the polynomial is 1
C)there exists no k such that the degree of the polynomial is 0
D)there exists no k such that the degree of the polynomial is undefined
E)none of these[/quote]

doubt:what is k here

[quote="qqwref"]Our polynomial is:
$(q-r)(x-q)(x-r)+(r-p)(x-r)(x-p)+(p-q)(x-p)(x-q)+k$

Multiplying and simplifying, it is equivalent to:
$q^2 r - q r^2 + r^2 p - r p^2 + p^2 q - p q^2 + k$

There is no x whatsoever and the degree must be either 0 (a constant polynomial) or undefined (the polynomial is 0), depending on whether the value of k makes that polynomial equal to 0 or not.

So A is false, B is false, C is false, D is false. The answer must be E.[/quote]

thanks


algebra solved
algebra
functional equation

It's been posted here:

[url]http://www.mathlinks.ro/viewtopic.php?t=5344[/url]

There's no solution yet, but I just wanted to mention that it was already on the forum. However, I'm afradid that I can't remember my proof  :blush:  :blush: I'll work on it and post it as soon as I get it (I should say [b]if[/b] I get it :)). I hope I don't find out it was wrong to begin with..

I think I've got a very important part: showing that $f$ is decreasing, surjective and continuous. The solution can't be too far. I'm really embarrassed about posting a problem in the "Proposed Problems" forum and the forgetting the solution, but I'm only human, and the problem has been there for quite some time..

$f(f(x)+\frac ax)=xf(a+1)\ (*)$, so if we fix $a$ we find $f$ to be surjective. Let's assume we can find $x>y\in (0,\ \infty)$ s.t. $f(x)>f(y)$. We take $a=\frac{f(x)-f(y)}{\frac 1y-\frac 1x}$ in $(*)$, getting $f(x)+\frac ax=f(y)+\frac ay\Rightarrow f(f(x)+\frac ax)=f(f(y)+\frac ay)\Rightarrow xf(a+1)=yf(a+1)\Rightarrow x=y$, which is a contradiction, so $f$ is decreasing (not necessarily strictly decreasing). From here and from the surjectivity we find $f$ to be continuous as well.

Here's something extra: if we make $a\rightarrow 0$ in $(*)$, we get $f(f(x))=xf(1)$, so $f$ is actually injective, so it's bijective and strictly decreasing. Then $f(f(1))=f(1)$, so it has a fixed point, and call that point $x_0$. Then $f(f(x_0))=x_0=x_0f(1)\Rightarrow f(1)=1\Rightarrow f(f(x))=x$.

We're really close now, but I have to go so I'll have to let someone else finish it off.

Thank you very much grobber.
I realized that my solution WAS correct .(I wanted to send it at Proposed questions but I sent it in unsolved problem!!!!!)
you see:
f(f(x)+f(y))=xf(1+xf(y))=yf(1+yf(x))
And :
f(x+y)=f(f(f(x))+y)=f(x)f(1+yf(x))=f(y)f(1+xf(y))
So xf(x) is constant.
So xf(x)=1f(1)=1 an f(x)=1/x. :P  :P  :P

How did you derive the fact that $xf(x)$ is constant from those relations? (sorry if I'm missing the obvious).

Of, I've got it. Like I said, sorry for asking that :). Here's what I did:

Put $y=1$ and get $f(1+f(x))=xf(1+x)$. In this relation we put $f(x)$ instead of $x$ and we get $f(1+f(f(x)))=f(1+x)=f(x)f(1+f(x))=xf(x)f(1+x)$. If we simplify this with $f(1+x)$ we find $xf(x)=1$.


Solve

$ \begin{cases}|\,|x\plus{}y|\minus{}|2y|\,|\equal{}2\\ |x\plus{}y|\plus{}|2y|\equal{}6\end{cases}$

[quote="Farenhajt"]Solve

$ \begin{cases}|\,|x \plus{} y| \minus{} |2y|\,| \equal{} 2 \\
|x \plus{} y| \plus{} |2y| \equal{} 6\end{cases}$[/quote]
[hide] Call $ |x \plus{} y| \equal{} a$ and $ |2y| \equal{} b$.  So, $ |a \minus{} b| \equal{} 2$ and $ a \plus{} b \equal{} 6$.

If a-b>0, then $ a\minus{}b\equal{}2, a\plus{}b\equal{}6$.  we add our equations together, we obtain that $ 2a \equal{} 8 \rightarrow a \equal{} 4$.  We can thus find b=2.  Now, we have that $ |x \plus{} y| \equal{} 4$ and $ |2y| \equal{} 2$.  So, $ y \equal{} \pm 1$.  If $ y \equal{} 1$, then $ |x \plus{} 1| \equal{} 4$, so $ x \equal{} \minus{} 5, 3$.  Also, if $ y \equal{} \minus{} 1$, then $ |x \minus{} 1| \equal{} 4$, so $ x \equal{} \minus{} 3, 5$.

If a-b<0, then we have that $ b\minus{}a\equal{}2, a\plus{}b\equal{}6$.  We again add our equations together and divide by two to get that $ b\equal{}4$ and $ a\equal{}2$.  So, $ y\equal{}\pm 2$.  Thus, either $ |x\plus{}2|\equal{}2$, so $ x\equal{}0, \minus{}4$ or $ |x\minus{}2|\equal{}2$, so $ x\equal{}0, 4$.

So, $ (x,y)\equal{}\boxed{(0,2), (\minus{}4,2), (0,\minus{}2), (4,\minus{}2), (\minus{}3, \minus{}1), (5,\minus{}1), (\minus{}5, 1), (3,1)}$.
  [/hide]
EDIT: Sorry, I did... I will edit it.

[hide="substitution"]
$ |6\minus{}2|2y\parallel{}\equal{}2$
So $ 2|2y|\equal{}4$ or $ 2|2y|\equal{}8$
So $ y\equal{}\minus{}1,1,\minus{}2,2$
Substitute back into equation 1 to get the ordered pairs, check for extraneous roots, etc
[/hide]

[quote="Farenhajt"]Solve

$ \begin{cases}|\,|x \plus{} y| \minus{} |2y|\,| \equal{} 2 \\
|x \plus{} y| \plus{} |2y| \equal{} 6\end{cases}$[/quote]

[hide="elimination"]
if $ |x \plus{} y| \minus{} |2y| > 0$

add both equations to get
$ |x \plus{} y|\equal{}4$
$ |2y|\equal{}2$
$ |y|\equal{}1$

if $ |x \plus{} y| \minus{} |2y| < 0$

add both equations to get
$ |2y|\equal{}4$
$ |y|\equal{}2$
$ |x \plus{} y|\equal{}2$

find ordered pairs
$ |y|\equal{}1$ & $ |x \plus{} y|\equal{}4$

find ordered pairs
$ |y|\equal{}2$ & $ |x \plus{} y|\equal{}2$

[/hide]


inequalities
trigonometry
inequalities proposed

Let a,b,c be stricly positive real numbers such that :
           $(a+b+c)^{2}+\frac{(a+b)(b+c)(c+a)}{4}=\frac{1}{2}\sum_{cyc}(a+b)(b+c)+1$
Prove the ineguality: $\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\geq\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}$.

At first sight, this is inequality should have a computational solution based on the substitutions: $a+b=2\cos{C}, b+c=2\cos{A}, c+a=2\cos{B}$, where $A,B,C$ are angles of a triangle. (because the condition could be written as $\sum{(a+b)^{2}}+\prod{(a+b)}= 4$) .

Actually the substitution $cosC=\frac{a+b}{2}, cosA=\frac{b+c}{2}, cosB=\frac{c+a}{2}$ is the key to the problem , because the rest is based on the Lema :
In any acute triangle $ABC$ , for any real numbers $x,y,z$ we have the ineguality:
\[x^{2}+y^{2}+z^{2}\geq 2xy cosA+2yz cosB+2zx cosC\]
Then, by puting $x=\sqrt{\frac{b}{ca}}, y=\sqrt{\frac{c}{ab}}, z=\sqrt{\frac{a}{bc}}$ , and with the help of our substitution, the ineguality is solved :)
So a computational solution isn't quite necesary, but you could try it by substituing :
$a=cosB+cosC-cosA , b=cosC+cosA-cosB , c=cosA+cosB-cosC$(Tell me if it works  :) )

[quote="pohoatza"]substitutions: $a+b=2\cos{C}, b+c=2\cos{A}, c+a=2\cos{B}$, where $A,B,C$ are angles of a triangle[/quote]

[quote="Vrajitoarea Andrei"]Actually the substitution $cosC=\frac{a+b}{2}, cosA=\frac{b+c}{2}, cosB=\frac{c+a}{2}$ is the key to the problem.[/quote]

Didn't I say the same thing ? :)

Sorry, i tried to say that the substitution is the hole problem...and that no other calculations that involve trigometry are necesary.When i said "actually" i ment to say that you have already solved the problem  :D . Sorry if I ofended you or anything

No problem, nothing about that :), nice inequality. :wink:


symmetry
advanced fields
advanced fields unsolved

Could someone explain some of the details of the following proof to me?

For given $ n,d>0$, let $ M_0, \ldots, M_N$ be all the monomials of degree $ d$ in the $ n\plus{}1$ variables $ x_0, \ldots, x_n$, where $ N\equal{}{n\plus{}d \choose n}\minus{}1$.  Define a mapping $ \rho_d: \mathbb{P}^n \rightarrow \mathbb{P}^N$ by sending the point $ P\equal{}(a_0, \ldots, a_n)$ to the point $ \rho_d(P)\equal{}(M_0(a), \ldots, M_N(a))$ obtained by substituting the $ a_i$ in the monomials $ M_j$.  This is called the $ d$-uple embedding of $ \mathbb{P}^n$ in $ \mathbb{P}^N$.

The claim is that $ \mathbb{P}^n \cong \mbox{im } \rho_d$.

I found the following proof:

The image $ Y: \equal{}\rho_d(\mathbb{P}^n)$ is contained in the open affine cover $ V_i\equal{}V_I$, where $ I$ ranges over the pure powers, $ I\equal{}(0, \ldots, 0,d,0, \ldots, 0)$ (that is, $ X^I\equal{}X_i^d$).  The inverse image is then the open affine $ U_i$, $ X_i \neq 0$.  Dehomogenising, we certainly then get a map $ U_i \xrightarrow{f_i} V_i$, whose entries are polynomials.  We now try to write down the inverse map.  We first do this in the case of $ f_0$, so that we want to define a map $ g_0: V_0 \rightarrow U_0$ a morphism, which will turn out to be the inverse of $ f_0$.  In this case, we send $ [Z_I] \mapsto [Z_{(d,0, \ldots, 0)}: Z_{(d\minus{}1, 1, 0, \ldots, 0)}: Z_{(d\minus{}1,0,1,0, \ldots, 0)}: \cdots : Z_{(d\minus{}1, 0, 0, \ldots, 1)}]$, and this is a morphism.  Moreover, it is not hard to see that it is the inverse of $ f_0$.  By symmetry this gives us maps $ g_i$, inverses of $ f_i$, and since the inverse map $ g\equal{}f^{\minus{}1}$ is unique, provided it exists, compatibility on overlaps is guaranteed.

I have a few questions.  First, what is meant by 'pure powers', and thus, what do the sets $ V_i\equal{}V_I$ look like?  What do they consist of?  Why is the inverse image $ U_i$ and what is the map $ f_i$?  And what is meant by $ [Z_I] \mapsto [Z_{(d,0, \ldots, 0)}: Z_{(d\minus{}1, 1, 0, \ldots, 0)}: Z_{(d\minus{}1,0,1,0, \ldots, 0)}: \cdots : Z_{(d\minus{}1, 0, 0, \ldots, 1)}]$?  I guess I'm just confused about the notation used here, but also by the questions I've asked here.  Thanks in advance for your response.

$ I$ denotes the sequence of exponents of a monomial. if all the $ a_i^d$ vanish, then also the $ a_i$ vanish: thus the image of $ \rho_d$ is contained in the union of the sets $ \{X_I \neq 0\}$, where $ I$ runs through the pure powers. if $ X^I\equal{}X_i^d$, the only condition for $ a \in \mathbb{P}^n$ to be in the preimage of $ \{X_I \neq 0\}$ is that $ a_i^d \neq 0$, i.e. $ a_i \neq 0$. thus the preimage is the standard open affine $ U_i \equal{} \{X_i \neq 0\}$. $ [Z_I]$ denotes just a typical element in $ \mathbb{P}^N$, i,e. a sequence of scalars $ Z_I$ for every monomial $ I$.


1.
$ \sum_{i\equal{}3}^{n}( i ( \sum_{j\equal{}i\plus{}1}^{n} 3 ) )$

Inner summation:

$ 3(n\minus{}j)$
$ 3( (n\minus{} (i\plus{}1)) \plus{} 1)$
$ 3( n\minus{}i\minus{}1\plus{}1 )$
$ 3( n\minus{}i )$

Outer sumation becomes:

$ \sum_{i\equal{}3}^{n} ( i ( 3(n\minus{}i) ) )$

Final solution:

$ 3 [ (\sum_{i\equal{}3}^{n} in \minus{} \sum_{i\equal{}3}^{n} i^{2} ) ]$  *with a subtraction you can separate...right?

$ 3 [ ( \frac{n(n\plus{}1)(n)}{2} \minus{} 3n ) \minus{} (\frac{n(n\plus{}1)(2n\plus{}1)}{6} \minus{} 5) ]$  *all that's left is to factor and simplify...right?

2.
$ \sum_{i\equal{}0}^{n\minus{}1} \sum_{j\equal{}0}^{4i} (2n\minus{}i)$

Inner summation:

$ (4i\plus{}1)(2n) \minus{} (4i\plus{}1)(i)$
$ (8in \plus{} 2n) \minus{} (4i^2 \plus{} i)$

Outer summation becomes:

$ \sum_{i\equal{}0}^{n\minus{}1} [ ( 8in \plus{} 2n ) \minus{} (4i^2 \minus{} i ) ]$
$ n \sum_{i\equal{}0}^{n\minus{}1} (8i \plus{} 2) \minus{} \sum_{i\equal{}0}^{n\minus{}1} (4i^2 \plus{} i )$ *break them apart because of the subtraction, than factor out n from the left one
$ [ n( 8(n)(n\minus{}1) \plus{} 2n ) ] \minus{} [ (4( \frac{ (n\minus{}1(n))  (2(n\minus{}1) \plus{} 1) }{6} ) ] \plus{}  [ \frac{(n\minus{}1)n}{2} ) ]$ *expand and simplify...right?

Damn, and here I thought I understand summations...Thanks!  I will analyze it further and find out where I went wrong.  Does that mean my #2 is correct?

[quote="frankiecostello"]-, and here I thought I understand summations...Thanks!  I will analyze it further and find out where I went wrong.  Does that mean my #2 is correct?[/quote]

I don't know what you're worrying about -- your solution to (1) is correct and gives the same answer as Yuriy's.


$\triangle{ABC}$ is an isoscele triangle with $\angle{C}=90^o$. $D$ is the midpoint of $BC$, $E$ is on $AB$ such that $AE: EB=2: 1$.

Prove $CE \perp AD$.

let M be the midpoint of AB and H=ADxCM. Then AH:HD=2:1, and as AE:EB=2:1 from Thales we get HE is parallel to DB. This means HE is perpendicular to AC. As CM is perp to AB it follows then H is the orthocenter of AEC, so AH is perp to CE

[quote="Farenhajt"]If we put the coordinate system such that $C$ is the origin and the coordinates of $A$ and $B$ are $A(a,0), B(0,a)$, then we have $D(0,{a\over 2}), E(\frac{a}{3},\frac{2a}{3})$. Hence $\overrightarrow{CE}=[\frac{a}{3},\frac{2a}{3}]$ and $\overrightarrow{AD}=[-a,{a\over 2}]$. Then $\overrightarrow{CE}\cdot\overrightarrow{AD}=-a\cdot\frac{a}{3}+\frac{a}{2}\cdot\frac{2a}{3}=0$, which means $CE\perp AD$[/quote]

Oh Farenhajt, you're very good at math.

[quote="Hokkage"]Doesn't Farenhajt's solution only prove it for an isoceles right triangle, as the sides (in his proof) have the exact same length? I think it loses generality (I could be wrong though).[/quote]

the problem says isosceles

Confer the bolded part in the initial post:

[quote="shobber"]$\triangle{ABC}$ is an [b]isoscele triangle[/b] with $\angle{C}=90^o$[/quote]

[b]The first method.[/b] Put w.l.o.g. $CA=CB=1,\ AB=\sqrt 2$. Thus, $AE=\frac{2\sqrt 2}{3}$, $BD=CD=\frac 12$, $BE=\frac{\sqrt 2}{3}$ and $DE^2=BE^2+BD^2-2\cdot BE\cdot BD\cdot \cos 45^{\circ}$, i.e. $DE^2=\frac 14-\frac 19$. Therefore,
$CE\perp AD\Longleftrightarrow CA^2-CD^2=EA^2-ED^2\Longleftrightarrow 1-\frac 14=\frac 89-\left(\frac 14 -\frac 19\right)$ a.s.o.

[b]Remark.[/b] Define the middlepoint $M$ of the side $[AB]$, and the intersection

 $L\in AD\cap CE$. The points $A$ and $E$ are conjugate harmonical w.r.t. the points

$M$ and $B$ because $\frac{AM}{AB}=\frac{EM}{EB}=\frac 12$. Therefore $CE\perp AD$ $\Longleftrightarrow$

$LA\perp LE$ $\Longleftrightarrow$ $\boxed {\widehat {MLE}\equiv \widehat {BLE}}\ .$

Haha, whoops. Didn't read the problem closely enough. In that case, nice solution Farenhajt.

[b]The second method.[/b] Define the point $F\in CE$, $FB\parallel AC$. Then
$\triangle ACE\sim \triangle BFE$ $\Longrightarrow$ $\frac 12=$ $\frac{EB}{EA}=$ $\frac{FB}{AC}=$ $\frac{FB}{2\cdot CD}$ $\Longrightarrow$
$FB=CD$ $(\ AC=BC\ )$ $\Longrightarrow$ $\triangle ACD\sim\triangle CBF$ $\Longrightarrow$ $CE\perp AD$.

[b]Remark.[/b] Let $ACBS$ be a square and the midpoints $D,F$ of the sides $[CB],[BS]$
respectively. Then $AD\perp CF$, i.e. $AD\perp CE$ and the point $E\in AB\cap CF$
has the property $AE=2\cdot EB$. Probably (in my opinion), the author "cleaned" this
problem (by elimination of the segments $BF$ and $EF$) and obtain the initial problem.

[b]Generalization.[/b] Given are the isosceles and right triangle $ABC$
$(CA=CB,\ CA\perp CB)$ and the points $E\in (AB)$, $D\in (BC)$.
Then $AD\perp CE$ $\Longleftrightarrow \frac{EA}{EB}=\frac{CB}{CD}\ .$
[hide="For Hokkage."]Hokkage, you didn't read the my last remark. What is your opinion ? Generally, a problem isn't abandoned quickly, although you solved it. You try and another lines of action, i.e. you find another methods.[/hide]