Να δειχθεί ότι
α)  δ$ (x^2-a^2)$=$ \frac{1}{2|a|}\[ [$δ$ (x-a)$+δ$ (x+a)\] ]$
β)  δ'(x)=$ \lim_{x\to{0}}{\frac{1}{a}}$$ [$δ$ (x+2a)$-δ$ (x-a)]$

όπου δ(x) είναι το συνρτησιακό Dirac δηλαδή έχει τις εξής ιδιότητες
δ(χ)=0 , όταν χ διαφορετικό του 0
 δ(0)=${ \displaystyle\infty}$ όταν χ=0
και $ \displaystyle\int^{+\infty}_{-\infty}$ δ$ (x)\,dx = 1$

[quote="spinos"]όπου δ(x) είναι το συνρτησιακό Dirac δηλαδή έχει τις εξής ιδιότητες
δ(χ)=0 , όταν χ διαφορετικό του 0
 δ(0)=${ \displaystyle\infty}$ όταν χ=0
και $ \displaystyle\int^{ + \infty}_{ - \infty}$ δ$ (x)\,dx = 1$[/quote]

Ελπίζω να καταλαβαίνεις ότι αυτός ο ορισμός είναι, στην καλύτερη περίπτωση, λίγο... προβληματικός ;)
(Among others, μία τέτοια συνάρτηση θα είχε αναγκαστικά ολ/μα 0 ;) )

Σε κάθε περίπτωση:

A) $ \int^\infty_{ - \infty} \delta(x^2 - a^2) f(x) dx = \int_{ - \infty}^0 + \int_0^\infty$
Όμως, η $ x^2 - a^2$ είναι 1-1 στο $ ( - \infty,0]$, οπότε μπορούμε να κάνουμε την αλλαγή μεταβλητής $ u = x^2 - a^2$... ομοίως για το άλλο μισό και προκύπτει το ζητούμενο.

Γενικά, αν $ g(x)$ συνάρτηση με απομονωμένα μηδενικά και παραγωγίσιμη σε αυτά με μη μηδενική και συνεχή παράγωγο εκεί, τότε $ \delta(g(x)) = \sum_{\xi: g(\xi) = 0} \frac {1}{|g'(\xi)|} \delta(x - \xi)$.

B) Υποθέτω χρησιμοποιούμε τον ορισμό: $ \langle\delta'|f\rangle = - f'(0)$, σωστά? Σε αυτήν την περίπτωση, το όριο που έχεις γράψει δεν είναι σωστό, αυτό που ισχύει είναι:
$ \delta'(x) = \lim_{a\to 0} \frac {1}{2a}\left(\delta(x + a) - \delta(x - a)\right)$.
Για να δειχθεί αυτό, άφησε την RHS να δράσει σε μία παραγωγίσιμη συνάρτηση $ f$, οπότε παίρνεις το όριο $ \lim_{a\to 0} \frac {f( - a) - f(a)}{2a}$ και... τέλος :lol:

Cheerio,

Durandal 1707


parameterization
trigonometry
calculus
calculus computations

If $x(t) = \sec^{2}(t)$ and $y(t) = 2+\tan(t)$, how would you eliminate the parameters and put it into rectangular form?


Let $R=gS-4$. When $S=8$, $R=16$. When $S=10$, $R$ is equal to:

$\textbf{(A)}\ 11 \qquad
\textbf{(B)}\ 14 \qquad
\textbf{(C)}\ 20\qquad
\textbf{(D)}\ 21 \qquad
\textbf{(E)}\ \text{None of these}$

[hide="Solution"]

Consider two equations:

$ 8g \minus{} 4 \equal{} 16$
$ 10g \minus{} 4 \equal{} ?$

From the first equation, we get $ 8g \equal{} 20$. Multiplying both sides by 10/8, we get $ 10g \equal{} 25$. Thus, R in the later equation is $ 25\minus{}4 \equal{} 21$. (D) [/hide]

Sorry for bumping a 15 year old thread, 8g-4=16 so 8g=20 we need to find 10g-4 which is 25-4=21 so we have (d)

This thread was created before most of us were born  :D

[hide=sol]$16=8g-4 \implies g = 5/2.$ $R = 10g - 4 = 21,$ so $\boxed{(D)}.$[/hide]

@above no? Most aops users are high schoolers


limit
calculus
derivative
calculus computations

Prove that:
$ \displaystyle\lim_{h\to 0}\ a^{kh} \minus{} 1 \equal{} k(\displaystyle\lim_{h\to 0}\ a^h \minus{} 1)$

I am looking for a proof that requires the least amount of "machinery" as possible.  In other words, try to avoid the use of theorems about derivatives such as the chain rule.  In particular, I am searching for a proof that does not rely on the fact that the derivative of a^x is ln(a)*a^x .

As you've written it, the two sides are equal because both are equal to 0 (well, assuming $ a > 0$).  Presumably what you really want is to show that
\[ \lim_{h \to 0} \frac {a^{kh} \minus{} 1}{a^h \minus{} 1} \equal{} k.\]
Then I suggest a substitution like $ a^h \equal{} \ldots$.  How far can you get with that?

Using your JBL's hint, I substituted $ 1 \plus{} e$ for $ a^h$ (as e approaches zero) (I'm not sure if this is a valid substitution, although it appears to at least give the correct answer).  The problem then reduces down to evaluating the derivative of $ x^k$ at $ x\equal{}1$, which of course gives the desired answer k.  Out of curiosity, is there a way to do this problem without relying on the fact that the derivative of $ x^k$ is $ kx^{k\minus{}1}$ for arbitrary k?

This [i]is[/i] that derivative, so the real question is how we can efficiently derive it in the first place.

For integer or even rational values of $ k$, we can factor: $ \frac{x^n\minus{}1}{x^m\minus{}1}\equal{}\frac{(x\minus{}1)(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}\cdots\plus{}1)}{(x\minus{}1)(x^{m\minus{}1}\plus{}x^{m\minus{}2}\plus{}\cdots\plus{}1)}\to \frac{n}{m}$ as $ x\to 1$.
For irrational values, we pretty much have to define it with the exponential. That brings the derivative of the exponential into the picture, but we need merely that it is differentiable and not the exact value.
$ \lim_{h\to 0}\frac{f(kh)\minus{}f(0)}{f(h)\minus{}f(0)}\equal{}\lim_{h\to 0}k\cdot\frac{f(kh)\minus{}f(0)}{kh}\cdot\frac{h}{f(h)\minus{}f(0)}\equal{}k\cdot f'(0)\cdot \frac1{f'(0)}\equal{}k$, as long as $ f'(0)$ exists and is nonzero.

jmerry, how are you getting that $ \lim_{h\to 0}\frac{f(kh)\minus{}f(0)}{kh} \equal{} f'(0)$ ?

Because $ \lim_{h\to 0}$ is the same as $ \lim_{kh\to 0}$.


I'm lost:

[b] (1) Four men and a monkey spend the day gathering coconuts on a tropical island.  After they have all gone to sleep at night, one of the men awakens and, not trusting the others, decides to take his share.  He divides the coconuts into four equal piles, except for one remaining coconut, which he gives to the monkey.    He then hides his share, puts the other piles together, and goes back to sleep.  Each of the other men awakens during the night and does likewise, and every time there is one coconut left over for the monkey.  In the morning all the men awake, divide what's left of the coconuts in to four, and again there is one left over that is given to the monkey.[/b]

This is my work so far:

start out w/C coconuts gathered

man1:  C = 4x + 1     =>  x = (C-1)/4  where x=the number of coconuts he took, and 3x = the number left in the pile

man2:  3x = 4q + 1    => q=(3x-1)/4=(3C-7)/16 where q=the number of coconuts he took, and 3q = the number left in the pile

man3: 3q = 4n + 1  => n=(3q-1)/4 = (9C-37)/64

man4: 3n = 4r + 1 => r=(3n-1)/4 = (27C-175)/256  so r=what's in the pile now

now they wake up, and divide the pile 1 more time, w/one left over for the monkey

all men: 3r = 4o + 1  => o=(3r-1)/4 = (81C-781)/1024, so there's 4 piles left of o coconuts in what's left in the original pile

so you need the minimum number that makes (81C-781)/1024 an integer, i dunno it might be helpful that 81=3^4, 781=11*71, and 1024=2^10, but I don't know how to do it unless you mindlessly plug in numbers.  maybe it's a mod problem, but I have not learned mods yet so there has to be another way to do it

[b] (2) An integer n is perfect if the sum of it's divisors (including 1 and itself) is 2n.  Show that if 2^p-1 is a prime number, then $n=2^{p-1}*(2^p-1)$ is perfect.  [/b]

No idea how to do this...if someone could help me out I would really appreciate it....

a) For first the answer is  -3  (Negative 3 )  If you give  1 to the monkey then -4 remains and you can always divide that in 4 equarl parts and if you hide your portion you  are where you started.. so this can happen  as many times as you want ...

Don;t like negative numbers?    Just add 4^5 to it that will also be an answer.

So   one answer is (4^5-3=1021).  (There are other answers byjust adding 4^5).

b) Just add all its factors.... straight-forward enough.
HTH

Let x+1 represent the number of cococnuts that they start with.
multiple 3/4 for each successive person since they will leave 3/4 of the coconuts, add one since each leaves an extra coconut
3x/4+1 will be the # of ccnts before the 1st guy
so x*(3/4)^n+1 will be the # of cocunts after that n people have gone
when 4 people have gone
x(3^4)/(4^4)+1
=81x/256+1
that is then divided by the four of them
=81x/1024+1
3^4/2^10 can not be reduced
therefore the lowest value of x that will yield an interger solution will be x=1024
total=x+1
=1024+1
=1025
therefore they started with 1025 coconuts

[quote]therefore the lowest value of x that will yield an interger solution will be x=1024
total=x+1
=1024+1
=1025
therefore they started with 1025 coconut[/quote] Check again .. 1025  does not work even the second  iteration.. regards.

1025
#1 256 taken, leaves 769 (256*3+1)
#2 192 taken leaves 577	(192*3+1)
#3 144 taken leaves 433	(144*3+1)
#4 108 taken leaves 325	(108*3+1)
325/4=81.25
each take 81 and the monkey gets one
I can't see my mistake

[quote="Misha123"]To (2):

I just give you a little hint: the divisors of your number are: 1,2,4,...,2^(p-1) and all of these numbers multiplied with (2^p-1). Now you just sum up all these numbers and get what you wanted.

Misha[/quote]

It seems like it would work, but when I do it I get the wrong answer:

2n = 2^p(2^p-1)

$\displaystyle\sum_{n=0}^{p-1}{2^n}+(2^p-1)=2^{2p}-2$

It's close but not the same, but I looked over my work a few times.....

[quote="Chigr"]1025
#1 256 taken, leaves 769 (256*[b]3[/b]+1)
#2 192 taken leaves 577	(192*3+1)
#3 144 taken leaves 433	(144*3+1)
#4 108 taken leaves 325	(108*3+1)
325/4=81.25
each take 81 and the monkey gets one
I can't see my mistake[/quote]  It is to be divided in 4 equal part.
After throwing 1 to the monkey and hiding  256 the pile now contains  contains 768  and that can, now be  evenly divided in 4 equal part;  leaving  no coconut for the monkey  for the second time aound.   Monkey may not be  happy:)  and   an unhappy monkey may make the  solution unsatiosfactory too. :)  right?

(Try with 1021 and you see why that works.  (1021-1)(3/4) = 765 and (765-1)(3/4)=573  and (573-1)*3/4 = 429 ... etc..)

Gyan, it's exactly what I meant with my hint.

Misha


If all positive integer a, b, c satisfying the following inequality can be a three sides of a triangle, find all k.

k(ab+bc+ca) > 5(a^2 + b^2 + c^2)

If $ k\le 10$, then for any $ a,b,c$ $ k(ab\plus{}bc\plus{}ca)\le 5(a^2\plus{}b^2\plus{}c^2$. Therefore $ k\le 10$ satysfyed your condition.
If $ k>10$, then $ D\equal{}(k\minus{}10)(k\plus{}10)(b^2\plus{}c^2)\plus{}2k(k\minus{}10)bc>0$. Therefore in interval $ a\in (\frac{k(b\plus{}c)\minus{}\sqrt D}{10},\frac{k(b\plus{}c)\plus{}\sqrt D}{10})$ exist $ a_0\equal{}\frac{k(b\plus{}c)}{10}>b\plus{}c$, suth that (a_0,b,c) give not triangle.
Therefore only $ k\in (\minus{}\infty ,10]$ is answer.


Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that:  $ 2\sqrt{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\plus{}5\sqrt{abc} \leq\ 7$
It's stronger than can_hang's old result  :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:  $ 2\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} 5\sqrt {abc} \leq\ 7$
It's stronger than can_hang's old result  :)[/quote]

$ p \equal{} a \plus{} b \plus{} c \equal{} 3$

$ ab \plus{} bc \plus{} ca \equal{} \frac {9 \minus{} q^2}{3} \\
( q \geq 0 )$

We have : $ r \equal{} abc \leq \frac {(q \minus{} 3)^{2}(2q \plus{} 3)}{27}$

$ LHS \minus{} 7 \equal{} f(r) \equal{} 2\sqrt {q^2 \plus{} r} \plus{} 5\sqrt {r} \minus{} 7$

We have : $ f^{'}(r) \equal{} {\frac {1}{\sqrt {{q}^{2} \plus{} r}}} \plus{} \frac {5}{2}\,{\frac {1}{\sqrt {r}}} > 0$

$ f(r) \leq f(\frac {(q \minus{} 3)^{2}(2q \plus{} 3)}{27})$

$ \equal{} \frac {2}{9}\,\sqrt {54\,{q}^{2} \plus{} 6\,{q}^{3} \plus{} 81} \plus{} \frac {5}{9}\,\sqrt {3}\sqrt {\left( q \minus{} 3 \right) ^{2} \left( 2\,q \plus{} 3 \right) } \minus{} 7 \equal{} g(q)$

We have :

$ g^{'}(q) \equal{} \frac {1}{9}\,{\frac {108\,q \plus{} 18\,{q}^{2}}{\sqrt {54\,{q}^{2} \plus{} 6\,{q}^{3} \plus{} 81}}} \plus{} { \frac {5}{18}}\,{\frac {\sqrt {3} \left( 2\, \left( q \minus{} 3 \right) \left( 2\,q \plus{} 3 \right) \plus{} 2\, \left( q \minus{} 3 \right) ^{2} \right) }{\sqrt { \left( q \minus{} 3 \right) ^{2} \left( 2\,q \plus{} 3 \right) }}} \leq 0$

$ g(q) \leq g(0) \equal{} 0$

And we have :

$ LHS \minus{} 7 \equal{} f(r) \leq g(q) \leq 0$

Done ! :)

[b]MQT yêu Quách Thị Hồng Dung[/b]

[quote="nguoivn"][quote="Materazzi"]
$ g^{'}(q) \equal{} \frac {1}{9}\,{\frac {108\,q \plus{} 18\,{q}^{2}}{\sqrt {54\,{q}^{2} \plus{} 6\,{q}^{3} \plus{} 81}}} \plus{} {\frac {5}{18}}\,{\frac {\sqrt {3} \left( 2\, \left( q \minus{} 3 \right) \left( 2\,q \plus{} 3 \right) \plus{} 2\, \left( q \minus{} 3 \right) ^{2} \right) }{\sqrt {\left( q \minus{} 3 \right) ^{2} \left( 2\,q \plus{} 3 \right) }}} \leq 0$
[/quote]
How did you prove it?  :maybe:[/quote]

Because $ 3 \geq q \geq 0$

And it is equivalent to :

$ (q\minus{}3)^{2}(126q^3\plus{}1026q^2\minus{}1296q\plus{}729) \geq 0$

Right !

And how about with the similar result:
$ \sqrt [3]{\frac {49}{9}(2\sum\ a^2 \plus{} 5\sum\ ab)(a \plus{} b \plus{} c)} \geq\ 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} 5 \sqrt [3]{abc}$
From it, we also obtain can_hang's result: $ \frac {7}{3}(a \plus{} b \plus{} c) \geq\ 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} 5 \sqrt [3]{abc}$  :)
@Materazzi: [hide]
[quote="Materazzi"]
And it is equivalent to :
$ (q \minus{} 3)^{2}(126q^3 \plus{} 1026q^2 \minus{} 1296q \plus{} 729) \geq 0$
[/quote] Em co the? viet ro~ hon doan. bien' doi? cua em dc ko, anh van~ chua nhin` ra n~ doan. danh' gia' do'. Thank kiu em  :) [/hide]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:  $ 2\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} 5\sqrt {abc} \leq\ 7$
[/quote]
[quote="nguoivn"]
$ \sqrt [3]{\frac {49}{9}(2\sum\ a^2 \plus{} 5\sum\ ab)(a \plus{} b \plus{} c)} \geq\ 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} 5 \sqrt [3]{abc}$
From it, we also obtain can_hang's result: $ \frac {7}{3}(a \plus{} b \plus{} c) \geq\ 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} 5 \sqrt [3]{abc}$  :)
[/quote]
Both of them can be solved by Am-Gm and Cauchy Schwarts  :)


geometry
perimeter
geometric transformation
reflection
power of a point
radical axis
geometry proposed

Let $ (I_B),(I_C)$ be the escribed circles of triangle $ ABC$
circle$ (I_B')$ is the reflection of $ (I_B)$ wrt the midpoint of $ AC$,$ (I_C')$ likewise
Prove that the radical axis of $ (I_B'),(I_C')$ bisects the perimeter of $ ABC$


$ x,y,z\neq 0$ ,altfel nu exista $ S$

Fie $ f: (0,1)\to \mathbb R$ ,$ f(x)=\frac{2x}{3x^{2}+1}$ ,$ f$ concava.
Din inegalitatea lui Jensen $ \Rightarrow$
$ \sum f(x)\leq 3\cdot f\left(\frac{x+y+z}{3}\right)=\frac{3}{2}\leq \frac{9}{8}+\frac{S}{24}$
Ultima inegalitate fiind echivalenta cu $ S\geq 9$ ,evident.

A doua inegalitate este echivalenta cu:
$ \sum \frac{1}{3x+\frac{1}{x}}\geq \frac{9}{3+S}$
Din CBS $ \Rightarrow$ $ \sum \frac{1}{3x+\frac{1}{x}}\geq \frac{9}{3\sum x+\sum \frac{1}{x}}\geq \frac{9}{3+S}$ $ \Leftrightarrow$
$ S\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3+\sum \frac{x}{y}+\sum \frac{x}{z}$
Notand $ \frac{x}{y}=a$ ,$ \frac{y}{z}=b$ si $ \frac{z}{x}=c$ ultima inegalitate devine:
$ \sum a^{2}+\sum \frac{1}{a}\geq 3+\sum a$ $ \Leftrightarrow$ $ \sum \left(a^{2}+\frac{1}{a}-a-1\right)\geq 0$ $ \Leftrightarrow$ $ \sum (a-1)^{2}(a+1)\geq 0$


The shape of the sign outside Bob's Burger Barn is a regular octagon. How many degrees are in the measure of an interior angle of this sign?


[quote="perfect_radio"][quote="metalnem"]Can somebody tell me formulation of Inflection Point Theorem (if it's not some kind of secret)? I can't find it on this site.[/quote]

i strongly doubt you'll find out what is IPT during your lifetime  :D[/quote]
definitely agree.  :D

IPT is Myth's variant of R(L)CF Theorem.
http://www.mathlinks.ro/Forum/topic-7942-140.html

Thanks! I know what is R(L)CF, but I didn't know that IPT is Myth's variant of R(L)CF.


trigonometry
LaTeX
function
direct proportion

[quote="Logg"]I know these are easy problems and I can do them on my calculator but I was wondering how am I able to do these without a calc or looking at a unit circle? Thanks.

Give the measure of the angle in radians and degrees.
(Inverse=Sin^(-1)

InverseSin(0.5)

InverseSin(-sqroot(2)/2)

InverseTan(-5)[/quote]

You kind of half to memorize them. For this reason, you should know by heart the sines and consines of common angles such as $ 30^{\circ}$, $ 45^{\circ}$, $ 60^{\circ}$, and $ 90^{\circ}$ and variations

Then to get the other angle values you can use sum and difference formulae for each the cosine, sine, and tangent.

ya but isn't there like a way to actually calculate them using algebra? Like for the angle u do Angle=S/R where S is arc length and R is radius

If you can't be bothered using latex then you should type arcsin, arctan etc. I have learned a table of exact values and I find that's the easiest way.

The range of the arcsin function is usually taken to be $ [-\frac{\pi}{2},\,\frac{\pi}{2}]$ and the same goes for the arctan function. So the angles should be within this range. I used to work in degrees and convert to radians but I now work entirely in radians. I find that the best way is to just memorise a table of values for sin, cos and tan for $ 0$, $ \frac{\pi}{6}$, $ \frac{\pi}{4}$, $ \frac{\pi}{3}$ and $ {\frac{\pi}{2}}$ but some people use triangles. For instance take the right angled, isoceles triangle with sides $ 1$, $ 1$ and $ \sqrt{2}$. The two other angles are $ 45^{o}$ or $ \frac{\pi}{4}$ and since the the triangle is right angled you can easily work out sin, cos and tan of $ \frac{\pi}{4}$. Another triangle which is useful can be constructed from an equilateral triangle with sides two. Cut the triangle in half along an altitude. Now you have a right angled triangle with other internal angles $ 30^{o}$ and $ 60^{o}$ or in radians $ \frac{\pi}{6}$ and $ \frac{\pi}{3}$. The hypotenuse is $ 2$, and the other sides are $ 1$ and $ \sqrt{3}$. Again sin, cos and tan of these angles can be easily calculated from this triangle.

So from this second triangle:

$ \sin{30^{o}}=\frac{1}{2}= 0.5$

Hence $ \arcsin{0.5}= 30^{o}$  or $ \frac{\pi}{6}$ I can convert between radians and degrees automatically. But if you have difficulty you can just use a direct proportion relationship: the angle in degrees over $ 360$ is equal to the angle in radians over $ 2\pi$. You can just rearrange to get an angle in degrees or radians.

For the others I am going to use the fact that the arcsin and arctan are odd functions. This means that $ \arcsin(-x) =-\arcsin(x)$ and the same is true for arctan.

${ \arcsin{(-\frac{\sqrt{2}}{2}})}=\arcsin{(-\frac{1}{\sqrt{2}})}=-\arcsin{\frac{1}{\sqrt{2}}}$

Using the first of the triangles described above: $ \sin{45^{o}}=\frac{1}{\sqrt{2}}$

Hence $ 45^{o}=\arcsin{\frac{1}{\sqrt{2}}}$

Hence ${ \arcsin{(-\frac{\sqrt{2}}{2}})}=-45^{o}$ or $ -\frac{\pi}{4}$

I would leave $ \arctan{(-5)}$ in that form as it not standard.

I have given loads of working here but none of it is required. After practice you will be able to instantaneously calculate these.

thanks for your work that really helps alot, I just need to memorize the unit circle again...I did it last year but I haven't had math since last spring so I forget :/


geometry
power of a point
radical axis
geometry proposed

Let $ABC$ be an acute-angled triangle inscribed in a circle $k$. It is given that the tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and $R$ be the second intersection point of the circle $k$ with the line $BM$. The line $PR$ meets again the circle $k$ at point $S$ different from $R$.

Prove that the lines $AP$ and $CS$ are parallel.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58603]www.mathlinks.ro/Forum/viewtopic.php?t=58603[/url]

The power of $M$ w.r.t. $k$ is $MA^2$.

Draw the circle $w$ passing through $P,B,R$

The point $M$ is on the radical axis of $w$ and $k$, so the power of $M$ w.r.t. the circle $w$ is $MA^2=MP^2$.

This means that the segment $MP$ is tangent to $w$.

So, the chord $PR$ is shown under equal angles from $P$ and $B$ : $\angle RPA = \angle PBR$

But $\angle PBR = \angle CBR = \angle CSR$ since $C,S,R,B$ are concyclic

So we find that $\angle RPA = \angle CSR$, hence $CS \parallel AP$

$MP^2=MA^2=MR\cdot MB\Longrightarrow MP^2=MR\cdot MB\Longrightarrow MPR\sim MBP\ (s.a.s.)$

$\Longrightarrow \widehat {PSC}\equiv \widehat {RSC}\equiv \widehat {RBC}\equiv \widehat {MBP}\equiv \widehat {MPR}\equiv \widehat {APS}$

$\Longrightarrow \widehat {PSC}\equiv \widehat {APS}\Longrightarrow AP\parallel CS.$

[b]Remark.[/b] I proposed this problem for J.B.M.O. 2005.

Consider an inversion of pole $P$ which invariates circle $k$.
Consequently, $SC$ will transform into the circle circumscribed to $\triangle{BRP}$.
Since $MA^2=MR\cdot{MB}=MP^2$ we get that the line $AP$
is tangent to the aforementioned circle and we are done!


How do you put $\Box$ at the end of a line? I mean, right intended like in most books/courses? Does it require a separate line? Or is there just something \rightintend or \rightbox or so?

On a side note, why does \box (with small b) give $\box$ ? :?

That was what I was looking for :) thanks!


let m,n be positive integers .
prove that $m^3 +n^3 +4$ is not perfect cube

The set of cubic residues $\mod{9}$ is $\{0, 1, 8\}$. Examining all posibilities of $m, n \mod{9}$ gives us contradiction.


Let $a,b,c$ be real numbers such that $a+b+c=0,a^2+b^2+c^2=1,a^3+b^3+c^3=0 \ (a\leqq b\leqq c)$.
Find the values of $a,b,c$.

Don't know if you need help, or just asking for fun ..
Root of the equation [hide] $x^3 - \frac {x}{2}  = 0  $ [/hide]are  [hide]$ -(1/\sqrt 2), 0 , + (1/\sqrt 2 ) $[/hide]

From the 1st equation,

$c = -(a + b)$
$c^2 = a^2 + 2ab + b^2$

Substituting $c^2$ to the 2nd equation,

$2a^2 + 2b^2 + 2ab = 1$
$a^2 + b^2 + ab = \frac 1 2$

and comparing with the 2nd equation

$c^2 = ab + \frac 1 2$

$c^3 = c \cdot c^2 = -(a + b)\left(ab + \frac 1 2\right)$

Factoring the 3rd equation:

$a^3 + b^3 + c^3 = (a + b)(a^2 - ab + b^2) - (a + b)\left(ab + \frac 1 2\right) =$

$= (a + b)\left(a^2 - 2ab + b^2 -  \frac 1 2\right) = (a + b)\left[(a - b)^2 -  \frac 1 2\right] = 0$

Either $a + b = 0$ or $(a - b)^2 = \frac 1 2$. Assume $a + b = 0$. From the 1st equation, $c = 0$ and from the 2nd equation,

$a^2 + b^2 = 2a^2 = 1$

$a = \pm \frac{\sqrt 2}{2}$, $b = \mp \frac{\sqrt 2}{2}$, $c = 0$

Since either $b < c < a$ or $a < c < b$, this solution does not satisfy the condition $a \le b \le c$ and has to be rejected. It would be possible  to solve assuming the 2nd facror is 0, i.e., $(a - b)^2 = \frac 1 2$. But from cyclic exchange of the variables $a, b, c$, it is clear that the solutions will be either

$a = 0$, $b = \pm \frac{\sqrt 2}{2}$, $c = \mp \frac{\sqrt 2}{2}$

or

$b = 0$,  $a = \pm \frac{\sqrt 2}{2}$, $c = \mp \frac{\sqrt 2}{2}$

The only solution satisfying the condition $a \le b \le c$ is $b = 0$ and $a < c$, i.e., $a = -\frac {\sqrt 2}{2}$, $c = +\frac {\sqrt 2}{2}$.

Thank you for your replies,everyone. :)

Here is my solutioin.

$ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=-\frac{1}{2}$. Let $abc=k$, since $a,b,c$ are roots of the following equation

\[x^3-0\cdot x^2-\frac{1}{2}x-k=0\Longleftrightarrow x^3=\frac{1}{2}x+k\].
Therefore we have the following equations.
\[a^3=\frac{1}{2}a+k\]
\[b^3=\frac{1}{2}b+k\]
\[c^3=\frac{1}{2}c+k\]
Adding up these equation,we get  $a^3+b^3+c^3=\frac{1}{2}(a+b+c)+3k\Longleftrightarrow 0=\frac{1}{2}\cdot 0+3k\Longleftrightarrow k=0$.

Hence our cubic equation is $x^3-\frac{1}{2}x=0$.Solving this,we obtain $x=\pm \frac{1}{\sqrt{2}},0$.

By $a\leqq b\leqq c$,the desired answer is $a=-\frac{1}{\sqrt{2}},b=0,c=\frac{1}{\sqrt{2}}$.


geometry
parallelogram
similar triangles
geometry proposed

[quote="james digol"]Let $ E$ and $ F$ be points on the sides $ AB$ and $ AD$ of a convex quadrilateral $ ABCD$ such that $ EF$ is parallel to the diagonal $ BD$. Let the segments $ CE$ and $ CF$ intersect $ BD$ at points $ G$ and $ H$, respectively. Prove that if the quadrilateral $ AGCH$ is a parallelogram, then so is $ ABCD$.   :lol:[/quote]

[i]Solution:[/i]

$ \bullet$ Let's suppose that the quadrilateral $ ABCD$ is a parallelogram Thus it follows that $ (EG) \parallel{} (AH)$ so in $ \triangle ABH$ we have
$ \frac {BG}{GH} \equal{} \frac {BE}{AE}$ and in $ \triangle ADG$ we have $ \frac {DF}{FA} \equal{} \frac {DH}{HG}$.On the other hand we have $ (EF) \parallel{} (BD)$ Hence in $ \triangle ABD$ we have $ \frac {DF}{FA} \equal{} \frac {BE}{AE}$ It follows that

$ \frac {DH}{HG} \equal{} \frac {BG}{GH} \Rightarrow DH \equal{} GB$

and since $ ABCD$ is a parallelogram Then $ AH \equal{} GC$ and $ \angle AHD \equal{} \angle BGC$ and Thus $ \triangle AHD \equal{} \triangle BGC$ which means that $ AD \equal{} BC$ and similarly we prove that $ \triangle AGB \equal{} \triangle DHC$ Thus $ AB \equal{} DC$ and the result follows  :)

Here's my solution...Hope it's correct..

Let $ EF$ intersect $ AG$ and $ AH$ at points $ I$ and $ J$, respectively. Note that $ EGHJ$ and $ FIGH$ are parallelograms. It follows that $ \bigtriangleup FHJ\cong \bigtriangleup IGE$, which implies that $ FJ \equal{} EI$. Using three pairs of similar triangles, $ FJ \equal{} EI$ implies that $ BG \equal{} DH$.

Let $ M$ be the midpoint of $ AC$. Since $ AGCH$ is parallelogram, $ M$ is the midpoint of $ GH$. Finally, with $ BG \equal{} DH$, we have $ BM \equal{} BG \plus{} GM \equal{} DH \plus{} HM \equal{} DM$. This means that the diagonals $ AC$ and $ BD$ bisect each other, and so $ ABCD$ is a parallelogram.   $ \square$


limit function geometry rectangle circumcircle inradius Euler

And Iranian National Mathematical Olympiad 2005 has begun here are the first day's problems

[b]First Day : Algebra Exam[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316311#316311]Problem 1[/url]
1. Suppose $a,b,c\in \mathbb R^{+}$. Prove that :
\[(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}\geq (a+b+c)(\frac1a+\frac1b+\frac1c) \]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316312#316312]Problem 2[/url]
2. Suppose $\{x_{n}\}$ is a decreasing sequence that $\lim_{n \rightarrow\infty}x_{n}=0$. Prove that $\sum(-1)^{n}x_{n}$ is convergent
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316313#316313]Problem 3[/url]
3. Find all $\alpha>0$ and $\beta>0$ that for each $(x_{1},\dots,x_{n})$ and $(y_{1},\dots,y_{n})\in\mathbb{R^{+}}^{n}$ that:
\[(\sum x_{i}^\alpha)(\sum y_{i}^\beta)\geq\sum x_{i}y_{i}\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316314#316314]Problem 4[/url]
4. Suppose $P,Q\in \mathbb R[x]$ that $deg\ P=deg\ Q$ and $PQ'-QP'$ has no real root. Prove that for each $\lambda \in \mathbb R$ number of real roots of $P$ and $\lambda P+(1-\lambda)Q$ are equal.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316315#316315]Problem 5[/url]
Suppose $a,b,c \in \mathbb R^{+}$and
\[\frac1{a^{2}+1}+\frac1{b^{2}+1}+\frac1{c^{2}+1}=2 \]
Prove that $ab+ac+bc\leq \frac32$
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316316#316316]Problem 6[/url]
5. Suppose $A\in \mathbb R^{m}$ is closed and non-empty . $f: A\to A$ is a lipchitz function with constant less than 1. (ie there exist $c<1$ that $|f(x)-f(y)|<|x-y|,\ \forall x,y \in A)$. Prove that $f$ has a unique point like $x$ that $f(x)=x$

[b]Second Day : Geometry Exam[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316324#316324]Problem 1[/url]
1. From each vertex of triangle $ABC$ we draw 3 arbitary parrallell lines, and from each vertex we draw a perpendicular to these lines. There are 3 rectangles that one of their diagnals is triangle's side. We draw their other diagnals and call them $\ell_{1}$, $\ell_{2}$ and $\ell_{3}$.

a) Prove that $\ell_{1}$, $\ell_{2}$ and $\ell_{3}$ are concurrent at a point $P$.

b) Find the locus of $P$ as we move the 3 arbitary lines.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316326#316326]Problem 2[/url]
2. Suppose $O$ is circumcenter of triangle $ABC$. Suppose $\frac{S(OAB)+S(OAC)}2=S(OBC)$. Prove that ditsance of $O$ (circumcenter) from radial axis of  circumcircle and 9-point circle is
\[\frac{a^{2}}{\sqrt{9R^{2}-(a^{2}+b^{2}+c^{2})}}\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316327#316327]Problem 3[/url]
3. Prove that in acute-angled traingle ABC if $r$ is inradius and $R$ is radius of circumcircle then:
\[a^{2}+b^{2}+c^{2}\geq 4(R+r)^{2}\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316328#316328]Problem 4[/url]
4. Suppose in triangle $ABC$ incircle touches the side $BC$ at $P$ and $\angle APB=\alpha$. Prove that :
\[\frac1{p-b}+\frac1{p-c}=\frac2{rtg\alpha}\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316325#316325]Problem 5[/url]
5. Suppose $H$ and $O$ are orthocenter and circumcenter of triangle $ABC$. $\omega$ is circumcirle of $ABC$. $AO$ intersects with $\omega$ at $A_{1}$. $A_{1}H$ intersects with $\omega$ at $A'$ and $A''$ is the intersection point of $\omega$ and $AH$. We define points $B',\ B'',\ C'$ and $C''$ similiarly. Prove that $A'A'',B'B''$ and $C'C''$ are concurrent in a point on the Euler line of triangle $ABC$.

[b]Second Day : Number Theory Exam[/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=317873#317873]Problem 1[/url]
1. Find all $n,p,q\in \mathbb N$ that:
\[2^{n}+n^{2}=3^{p}7^{q}\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=317874#317874]Problem 2[/url]
2. $a\in\mathbb N$ and $m=a^{2}+a+1$. Find the teh number of $0\leq x\leq m$ that:
\[x^{3}\equiv1(\mbox{mod}\ m) \]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=317877#317877]Problem 3[/url]
3. $p(x)$ is an irreducible polynomial in $\mathbb Q[x]$ that $\mbox{deg}\ p$ is odd. $q(x),r(x)$ are polynomials with rational coefficients that $p(x)|q(x)^{2}+q(x).r(x)+r(x)^{2}$. Prove that
\[p(x)^{2}|q(x)^{2}+q(x).r(x)+r(x)^{2}\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=317875#317875]Problem 4[/url]
4. $k$ is an integer. We define the sequence $\{a_{n}\}_{n=0}^{\infty}$ like this:
\[a_{0}=0,\ \ \ a_{1}=1,\ \ \ a_{n}=2ka_{n-1}-(k^{2}+1)a_{n-2}\ \ (n \geq 2) \]
$p$ is a prime number that $p\equiv 3(\mbox{mod}\ 4)$
a) Prove that $a_{n+p^{2}-1}\equiv a_{n}(\mbox{mod}\ p)$
b) Prove that $a_{n+p^{3}-p}\equiv a_{n}(\mbox{mod}\ p^{2})$
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=317878#317878]Problem 5[/url]
5. $a,b,c\in \mathbb N$ that $a,b\neq c$. Prove that there are infinitely many prime numbers like $P$ that there exist $n\in\mathbb N$ that $p|a^{n}+b^{n}-c^{n}$

I'll send the other days and the PDF Version soon

Actually these are not geometry problem  ;)  ;)  ;) .
these are algebra.  :(  ;)

Thanks for the PDF version Omid.
As fasr as i know ,there was a graf exam too(Instead of Combinatorics) and a exam about porabibility (or something i am not sure),May we have those problems too?

And here is the Final exam

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=320032#320032]Problem 1[/url]
We call the set $A\in \mathbb R^n$ CN if and only if every continuous $f:A\mapsto A$ ther exist : $x\in A$ that $f(x)=x$
a)Example: We know that $A=\{x\in\mathbb R^n| |x|\leq1\}$ is CN
b) circle is not CN.
Which one of thes sets are CN?
1)A=$\{x\in\mathbb R^3| |x|=1\}$

2)The cross $\{(x,y)\in\mathbb R^2|xy=0,\ |x|+|y|\leq1\}$

3)Graph of $f:[0,1]:\to \mathbb R$:
\[f(x)=sin\frac 1x\ \mbox{if}\ x\neq0,\ f(0)=0\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=320036#320036]Problem 2[/url]
$n$ vectors are on the plane. We can move each vector forward and backeard on the line that the vector is on it. If there are 2 vectors that their endpoints concide we can omit them and replace them with their sum (If their sum is nonzero). Suppose with these operations with 2 different method we reach to a vector. Prove that these vectors are on a common line

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=320035#p320035]Problem 3[/url]
$f(n)$ is the least number that there exist a $f(n)-$mino that contains every $n-$mino.
Prove that $10000\leq f(1384)\leq960000$.
Find some bound for $f(n)$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50581]Problem 4[/url]
a) Year 1872 Texas
3 gold miners found a peice of gold. They have a coin that with possibility of $\frac 12$ it will come each side, and they want to give the piece of gold to one of themselves depending on how the coin will come. Design a fair method (It means that each of the 3 miners will win the piece of gold with possibility of $\frac 13$) for the miners.

b) Year 2005, faculty of Mathematics, Sharif university of Technolgy
Suppose $0<\alpha<1$ and we want to find a way for people name $A$ and $B$ that the possibity of winning of $A$ is $\alpha$. Is it possible to find this way?

c) Year 2005 Ahvaz, Takhti Stadium
Two soccer teams have a contest. And we want to choose each player's side with the coin, But we don't know that our coin is fair or not. Find a way to find that coin is fair or not?

d) Year 2005,summer
In the National mathematical Oympiad in Iran. Each student has a coin and must find a way that the possibility of coin being TAIL is $\alpha$ or no. Find a way for the student.

And here the problems of the second day

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=332272#332272]Problem 1[/url]
An airplane wants to go from a point on the equator, and at each moment it will go to the northeast with speed $v$. Suppose the radius of earth is $R$.
a) Will the airplane reach to the north pole? If yes how long it will take to reach the north pole?
b) Will the airplne rotate finitely many times around the north pole? If yes how many times?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=332273#332273]Problem 2[/url]
We define a relation between subsets of $\mathbb R ^n$. $A \sim B\Longleftrightarrow$ we can partition $A,B$ in sets $A_1,\dots,A_n$ and $B_1,\dots,B_n$(i.e $\displaystyle A=\bigcup_{i=1} ^n A_i,\ B=\bigcup_{i=1} ^n B_i, A_i\cap A_j=\emptyset,\ B_i\cap B_j=\emptyset$) and $A_i\simeq B_i$.
Say the the following sets have the relation $\sim$ or not ?

a) Natural numbers and composite numbers.
b) Rational numbers and rational numbers with finite digits in base 10.
c) $\{x\in\mathbb Q|x<\sqrt 2\}$ and $\{x\in\mathbb Q|x<\sqrt 3\}$
d) $A=\{(x,y)\in\mathbb R^2|x^2+y^2<1\}$ and $A\setminus \{(0,0)\}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=332271#332271]Problem 3[/url]
For each $m\in \mathbb N$ we define $rad\ (m)=\prod p_i$ that $m=\prod p_i^{\alpha_i}.$

[b]abc conjecture[/b]
Suppose $\epsilon >0$ is an arbitary number, then there exist $K$ depinding on $\epsilon$ that for each 3 numbers $a,b,c\in\mathbb Z$ that $gcd (a,b)=1$ and $a+b=c$ then: \[ max\{|a|,|b|,|c|\}\leq K(rad\ (abc))^{1+\epsilon} \]
Now prove each of the foollowing statements with $abc$ conjectures :
a) Fermat's last theorem for $n>N$ that $N$ is a natural number
b)We call $n=\prod p_i^{\alpha_i}$ strong if and only $\alpha_i\geq2$
c) Prove that there are finitely many $n$ that $n,\ n+1,\ n+2$ are strong.
d) Prove that there are finitely many rational numbers like $\frac pq$ that: \[ \Big| \sqrt[3]{2}-\frac pq \Big|<\frac{2^ {1384}}{q^3}  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=332274#332274]Problem 4[/url]
Suppose we have some proteins that each protein is a sequence of 7 "AMINO-ACIDS" $A,\ B,\ C,\ H,\ F,\ N$. For example $AFHNNNHAFFC$ is a protein. There are some steps that in each step an amino-acid will change to another one. For example with the step $NA\rightarrow N$ the protein $BANANA$ will cahnge to $BANNA$("in Persian means workman"). We have a set of allolowed steps that each protein can change with these steps. For example with the
set of steps:
$1)\ AA\rightarrow A\\ 2)\ AB\rightarrow BA\\ 3)\ A\rightarrow$
Protein $ABBAABA$ will change like this:
$ABB\underline{AA}BA\\ \underline{AB}BABA\\ B\underline{AB}ABA\\ BB\underline{AA}BA\\ BB\underline{AB}A\\ BBB\underline{AA}\\ BBB\underline{A}\\ BBB$
You see after finite steps this protein will finish it steps.
Set of allowed steps that for them there exist a protein that may have infinitely many steps is dangerous. Which of the following allowed sets are dangerous.
a)$1)N)\rightarrow OONN$
b)
$1)HHCC\rightarrow HCCH$
$2)CC\rightarrow CH$
c) Design a set of allowed steps that change $\underbrace{AA\dots A}_{n}\rightarrow\underbrace{BB\dots B}_{2^n}$
d) Design a set of allowed steps that change $\underbrace{A\dots A}_n\underbrace{B\dots B}_m\rightarrow\underbrace{CC\dots C}_{mn}$
You see from $c$ and $d$ that we acn calculate the functions $F(n)=2^n$ and $G(M,N)=mn$ with these steps. Find some other calculatable functions with these steps

could you post the pdf of the above problems with their solution in pdf?
Thanks

hello.
I would like to know who are the problem designers of final exams?
I supose they could act better.
specially for geometry.
(bebakhshid ye khorde tond raftama!!!)
 thanks. :roll:  ;)

[quote="Omid Hatami"]c) Prove that there are finitely many $n$ that $n,\ n+1,\ n+2$ are strong.[/quote]

What does "strong" mean?

It is :
[quote="Omid Hatami"]
We call $n=\prod p_i^{\alpha_i}$ strong if and only if $\alpha_i\geq2$
[/quote]


The original inequality states if $x,y,z,p,q,r$ are non-negative real numbers such that:
$p+q+r=1$ and $x^p y^q r^z = 1$ then:
$ \frac{p^2 x^2}{qy+rz} + \frac{q^2 y^2}{rz+px} + \frac{r^2 z^2}{px+qy} \geq \frac{1}{2} $
My question is to find all the possible values for:
$ \sqrt{px} + \sqrt{qy} + \sqrt{rz} $
in the same conditions.


Initialy the sets {$ 1,2,3,...,n$} and {$ n \plus{} 1,n \plus{} 2,...,2n$} satisfies the required result.We can obtain any two sets by swaping elements pairwise two at a time from the two sets finitely many times.So it is sufficient to show that each swaping preserves the identity.
There exists some $ m$ for which $ a_1 < b_1,a_2 < b_2,...,a_m < b_m,a_{m \plus{} 1} > b_{m \plus{} 1},...,a_n > b_n$.
So $ \sum_{1}^{n} |a_i \minus{} b_i|(\rightarrow I) \equal{} \minus{} \sum_{0 < i\leq m} a_i \plus{} \sum_{0 < i\leq m} b_i \plus{} \sum_{m < i\leq n} a_i \minus{} \sum_{m < i\leq n} b_i(\rightarrow II)$
If $ (x \minus{} m)(y \minus{} m) > 0$ then swaping $ a_x$ and $ a_y$ in $ (II)$ will preserve the identity(because they have the same sign in $ (II)$).Then we can swap $ a_y$ and $ b_y$ in $ (I)$ which will not spoil the identity also.
If $ (x \minus{} m)(y \minus{} m) < 0$ then $ a_x$ and $ b_y$ have the same sign in $ (II)$. So in this case too identity is preserved.

The previous solution looks too messy.
Here is a better one.
http://www.mathlinks.ro/Forum/viewtopic.php?p=1152045#1152045

I am roaming in  Aops and Find this pretty old problem , Being an initial learner and a new user in Aops I want you friends and brothers to check my solution. Thanks . Is it good? :maybe:

Well there is a little mistake, I forget to type the word 'number' in the place after n. That is, $b_i$ is greater than exactly n, ('numbers'), I forget to type that word 'number' in that bracket part in my solution. Oops.
One may find that language error in 5th and 6th line, rest of the Solution is ok from my side


geometry
geometric transformation
reflection
inequalities
preparation for IMO

Hey guys! Don't know where to put it, so I'll just place it here :)

I wanna win gold in the IMO in 2010, and I think that some of you guys have done the same thing before? What is have you done and practiced to win medals?

As by now, I am pretty far away from gold. I would not even get a bronze if I was in this year, so I don't have a lot of knoweledge.

I have the book "Problem Solving Strageties" by Arthur Engel, but I haven't solved all the problems nor read the whole book.

So... Anyone wanting to help me with writing a plan?!:)

It does not work like that (make a plan ...). You should look weakily or monthly where you at, and then make plan for short times. You will also never finish because there always harder problems ;), but that's beautiful about it.

I have about two years. So I WILL acutally finish:P

Anyways. I understand your answer, and I agree with you to a certain degree. BUT ... I know for sure that some techniques will improve my problem solving skills. And I want to know these!
What we do in the danish team, is to practice, learning new techniques, and then ... practice some more.:)

So..! What techniques are needed in order to win IMO? :ninja:

the best way to learn problem solving is solving problems... My advise is try hard, do as much problems as you can and some theory might be useful. But you have to sit in the chair and focus on the problems...that's the best way I think  :)

Well... You are not really helping me...:(

Anyone with tips other than "practice more" will be apprieciated.

Do you want a list of theorems, techniques ?  :maybe:

There's no secret technique, or a mysterious plan that will make you a gold medalist. Look at IMO 2007 - only 39 gold medals (which means, that out of the world's 500 brightest young minds, only 39 were able to solve 4+ questions). Not even a single contestant solved all the questions.
As the other users told you, the only way to improve is to solve problems. Mathlinks is the ultimate place for just that...
I also recommend reading some subject-focused problem-solving books - such as Titu Andreescu's books. Mathematical Reflections is a great magazine (though there are too many inequalities articles in my opinion...), too.
It's more than possible to improve your skills in two years, and a silver medal is not an imaginary goal. but gold medal is extremely challenging. So, stop asking about plans, and start solving  :lol: and don't forget that it is just a contest. Do it for the fun

[url]http://reflections.awesomemath.org/[/url]
[url]http://www.math.ust.hk/excalibur/[/url]

[quote="bambaman"]There's no secret technique, or a mysterious plan that will make you a gold medalist. Look at IMO 2007 - only 39 gold medals (which means, that out of the world's 500 brightest young minds, only 39 were able to solve 4+ questions). Not even a single contestant solved all the questions.
As the other users told you, the only way to improve is to solve problems. Mathlinks is the ultimate place for just that...
I also recommend reading some subject-focused problem-solving books - such as Titu Andreescu's books. Mathematical Reflections is a great magazine (though there are too many inequalities articles in my opinion...), too.
It's more than possible to improve your skills in two years, and a silver medal is not an imaginary goal. but gold medal is extremely challenging. So, stop asking about plans, and start solving  :lol: and don't forget that it is just a contest.[/quote]
Thank you very much:) It is stuff like this I am asking for. (Well.. Not the thing about the 39 gold medals.. That completely blew my confidience:P:P:P:P)
Which if his books will you recommend the most? I really don't have the money to buy 'em all...
(Btw. Do you really think I would ask such a question if I was'nt solving already?:))
[quote="bambaman"]Do it for the fun[/quote]
Off course:) Life should be lived that way..!!

Amazigh: Do you have one?:) I know for an example that the IMO website states that inversion is not required. Yet, have i heard from danish participants that it is usefull sometimes.

Keep the good advise comming.

Hmmm... list of theorems and techniques? Read the [url=http://www.imomath.com/othercomp/Bfa/bfacts.pdf]second chapter of IMO Compendium[/url] to get some kind of full picture (I say some kind, as sometimes you can solve 2 or more problems on IMO without knowing most of these things, and sometimes you can be stuck with the problem which can be handled very easily with some "well-known" theorem).

And some kind of syllabus of IMO states:

[quote="olympiads.win.tue.nl/imo/"]Topics covered (see Unwritten Syllabus and Tutorials at Arkadii Slinko's Math Olympiad Website, currently unavailable):

    * Number Theory, including
          o Fundamental Theorems on Arithmetic
          o Linear and quadratic Diophantine equations, including Pell's equation
          o Arithmetic of residues modulo n, Fermat's and Euler's theorems
    * Algebra, including
          o Fundamental Theorems on Algebra, e.g. inequalities, factorization of a polynomial into a product of irreducible polynomials
          o Symmetric polynomials of several variables, Vieta's theorem
    * Combinatorics, including
          o Graph theory
    * Geometry, including
          o Properties of the orthocentre, Euler's line, nine-point-circle, Simson line, Ptolemy's inequality, Ceva and Menelaus etc.

Excluded topics:

    * Calculus (!)
    * Complex numbers (though present in the past)
    * Inversion in geometry
    * Solid geometry (though present in the past; may return)
[/quote]

But you should know that excluded topics aren't forbidden to use. It just means you won't get a problem which can't be used without knowing them.

You asked about good books... Here's what I can recommend:
-Problem Solving Strategies (Engel)
-Andreescu's books

Hey. I have actually seen that side :blush:  Thanks a lot for the book recommendations (:

I saw a link on this side to a book named "Winning Solutions (Problem Books in Mathematics)". I will read and solve that now:) I also have Engels book, so i should have something to do the next months :D

Suggestions for smarter studying are still welcomed.

What do you want to hear? Yes, we use some psychedelic matters... but they will realy open the doors of your perception and there will be no way back! Math is a very dangerous for human mind science, so think twice before standing on the road to IMO.

Realize that almost NOBODy will ever have the opportunity to get a gold in IMO.  How good are you, anyway? amc scores?

[url]http://www.imo-official.org/year_individual_r.aspx?year=2010[/url]
Well done, Mathias! Congratulations! :)

Haha, here's an example of where a bit of ambition and hard work can get you.

I feel blessed to have seen this thread!
Mathias,you are great :omighty:
@below ,he/she has not been using aops since nov 7,2015

[quote=Pukulu12345]Mathias well done...
Pls give some motivation..[/quote]

He's not active on AoPS anymore..

Same is my case like mathias just enough knowledge to qualify the regional level only.
I want some assistance and help and a list of handy theorems and books..
Please help.

[quote=MellowMelon][quote="AIME15"]amc scores?[/quote]
 :huh: Not everyone lives in the US.[/quote]
I'm really glad I'm not in this US-system which is too complicated and discouraging. You want to be an IMO medalist, but you may fail because of bad response in ABCD test. Umph...

Can someone please pin/ make it an announcement for atleast a minimal amount of time!?! It was so inspiring and motivating reading this [hide=thread] that too when I was feeling down myself, hopes losing if you say it..... [/hide].
Thank you for bumping this :)

What is the name of Mathias? I mean everyone is saying well done. Is he the one who got perfect score?

So, he got only HM?
http://www.imo-official.org/participant_r.aspx?id=19303
OOps there are three mathias LOL.
Here is the link of the person we are talking about:
http://www.imo-official.org/participant_r.aspx?id=18549
With one silver and one gold medal.


Solve the equation  in

$(log_ba)^x+(log_cb)^x+(log_ac)^x=-x^2+6(x-1)$

The logarithms should have been assumed positive (something like $a,b,c>1$ or $a,b,c\in (0,1)$).

[hide="Hint"]Use AM-GM.[/hide]

[quote="Raluca"]The logarithms should have been assumed positive

[/quote]

Yes.

Arne, why are you doing it?

  	 Log equation
Nice ? 	silouan 	3 	52 	Today, at 11:16 am
  	   	 [Moved]  Easy equation     	 arqady
 New posts 		View newest post Nice and easy question
A friend 	dandrei 	2 	59 	Today, at 10:05 am
Raluca View latest post
 New posts 		View newest post Zeta(3)=apery's constant

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68986
Why?

$\left( {\log _b a} \right)^x + \left( {\log _c b} \right)^x + \left( {\log _a c} \right)^x \ge 3 \ge - x^2 + 6\left( {x - 1} \right),\forall x \in R \Rightarrow$

$\Rightarrow \left[ {\left( {\log _b a} \right)^x + \left( {\log _c b} \right)^x + \left( {\log _a c} \right)^x = - x^2 + 6\left( {x - 1} \right) \Leftrightarrow x = - 3} \right] \Leftrightarrow$

$\Leftrightarrow \left[ {a = b = c} \right]$


This problem was confusing me, so any help would be greatly appreciated!

There is a theorem that assures us that if L: V->W is a linear transformation, then L(av_1 + bv_2) = aL(v_1)+bL(v_2), for all v1, v2 in V and all a, b in R. Prove that the converse of this statement is true by considering two cases: first a=b=1 and then b=0.

Suppose that $L(av_{1}+bv_{2}) = aL(v_{1})+bL(v_{2})$ for all $v_{1}, v_{2}$ in $V$ and all $a, b$ in $\mathbb R$. Plugging $a=b=1$ into the above identity, we get
(1) $L(v_{1}+v_{2}) = L(v_{1})+L(v_{2})$.
If instead we leave $a$ arbitrary and set $b=0$, the result is
(2) $L(av_{1}) = aL(v_{1})$.

Properties (1) and (2) constitute the definition of a linear transformation.


Point P is on $ \overline{AB}$. Point A has coordinates (3,11) and point B has coordinates (18,1). The ratio of $ AP : PB \equal{} 2 : 3$. What is the sum of the coordinates of point P?

$ P: \left(\frac {3 \cdot 3 \plus{} 2 \cdot 18}{2 \plus{} 3}, \frac {3 \cdot 11 \plus{} 2 \cdot 1}{2 \plus{} 3}\right) \equal{} \left(9, 7\right)$.
Thus, $ 9 \plus{} 7 \equal{} \boxed{16}$.

If point $P(x,y)$ lies on line segment $\overline{AB}$ (between points $A(x_1, y_1)$ and $B(x_2, y_2)$) and satisfies $AP:PB=m:n$, then we say that $P$ divides $\overline{AB}$ internally in the ratio $m:n$. The point of division has the coordinates
\[P=\left( \frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n} \right).\]


modular arithmetic
number theory
relatively prime
Putnam
college contests

Define a sequence ${a_i}$ by $a_1 = 3$ and $a_{i+1} = 3^{a_i}$ for $i \geq 1$.  Which integers between 00 and 99 inclusive occur as the last 2 digits in the decimal expansion of infinitely many $a_i$?

I think the last two digits of $3^n$ repeat with period 20. So that suggests we look at the power of 3 $\pmod{20}$...

(that's just my immediate thought when I see it. I don't know if this leads to a solution but I hope it helps :) )

That's the first step, not the last step. To know $a_n$ mod 100, we need to know $a_{n-1}$ mod 20 (or 40, if you'd rather just use Euler's theorem). To know $a_{n-1}$ mod 20, we need to know $a_{n-2}$ mod $\phi(20)=8$ (actually, 4 is good enough). To know $a_{n-2}$ mod 8, we need ...

With only two digits, there aren't very many steps in the chain before we get something that's obvious. In this case, the last two digits of $a_n$ have the same value for all $n\ge3$.

Suppose $a,b,k$ are integers with $a,b$ relatively prime, and let $a_{n+1}=a^{a_{n}}$, $a_1=a$. The last $k$ digits of $a_n$ are eventually constant.
In base 10, it's usually fairly easy to calculate these stable values.


I really want to join the AMC competitions at my own school, but that is unfortunately, not possible. I also do no want to ride a car for more than 20 minutes across the city to another school that gives the test. Is there any way to get the AMC competitions into the school even without the principal's consent?

I suggest you try thte ideas in Frequently Asked Question #4, at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


The Central European Olympiad in Informatics is next week in Hungary. If anyone would be interested in competing remotely, they have a simultaneous internet contest. http://team1.inf.elte.hu/ceoicontest/

You'll have to wake up at 2:00 a.m. EDT though, since the internet contest is simultaneous.

Or, 8:00 if you live in Europe (except for Britain and some other strange countries). ;)


Let:$k$is positive integer numberand $a,b,c>0$ satisfying :$a+b+c=3$
Prove that:
 $\large \sum \sqrt{a+k^{2}-1}\geq k(ab+bc+ca)$


I have a problem :)
I started digging  in the endless stuff we have here at my house,and I found a letter dated 98 :| UNOPENED!!! so I opened it,and it was in pakistani.if someone can help me translate I'd be obliged... the adress it should have been taken to is nearby so I will take it soon. thx
[img]http://66.98.192.31/~jeriko/baber%20mehrad.png[/img]

Someone help him please!!!I am not gud at translation otherwise I might have done it for u.

shouldn't you ask a parent?  that stuff just might be private.

i can help translating from urdu(pakistan national language) to engish if u want.

i can read urdu but not in this riting , im not a gud reader but i can read good n clean ritings  :D

hi my dear,

this letter is alittle unreadale. ;)

It is unreadable, parts of it aren't even urdu, i think.


If one quart of paint is exactly enough for two coats of paint
on a 9-foot by 10-foot wall, how many quarts of paint are
needed to apply one coat of paint to a 10-foot by 12-foot wall?
Express your answer as a common fraction.

The area of the first wall is 90ft squared.
The area of the second is 120ft squared.
120/90=4/3, the second is 4/3 the area of the second.
But u only want 1 coat, so divide by two to get 2/3.


How i sketch the complex region determined by $ |z\minus{}3|\plus{}|z\plus{}3|\leq 10$ ?

The geometric interpretation of this inequality is that the sum of the distances from z to -3+0i and 3+0i must be less than or equal to 10.  Therefore, the set of all such points is the interior and boundary of an ellipse with foci at (-3,0) and (3,0), major axis of length 10, and minor axis of length 8.

[hide]Just think about what the equation means, its |z-3| is the distance from z to 3 and |z+3| is the distance from z to -3. So this is just the set of all points for which the sum of those distances is less than 10,  well if you have the sum of distances from two points constant that is the lucus definition of an ellipse.  So from there its not hard to find that this is the interior of an ellipse centered at the origen with major axis 10 minor axis 8.[/hide]

Very nice [b]nateharman1234[/b] and [b]atomicwedgie[/b]! Maybe one can prove this in algebraic form... Much Thanks

[hide="Algebraically (a pain)"] Let $ z \equal{} x \plus{} yi$.  Then

$ |z \minus{} 3| \plus{} |z \plus{} 3| \equal{} \sqrt{(x \minus{} 3)^2 \plus{} y^2} \plus{} \sqrt{(x \plus{} 3)^2 \plus{} y^2} \equal{} 10 \implies$
$ (x \minus{} 3)^2 \plus{} y^2 \plus{} (x \plus{} 3)^2 \plus{} y^2 \plus{} 2 \sqrt{ ((x^2 \plus{} 9 \plus{} y^2)^2 \minus{} (6x)^2 } \equal{} 100 \Leftrightarrow$
$ \sqrt{x^4 \plus{} y^4 \plus{} 81 \plus{} 18x^2 \plus{} 18y^2 \plus{} 2x^2 y^2 \minus{} 36x^2} \equal{} 41 \minus{} x^2 \minus{} y^2 \implies$
$ x^4 \plus{} y^4 \plus{} 2x^2 y^2 \minus{} 18x^2 \plus{} 18y^2 \plus{} 81 \equal{} x^4 \plus{} y^4 \plus{} 2x^2 y^2 \minus{} 82x^2 \minus{} 82y^2 \plus{} 1681 \Leftrightarrow$
$ 64x^2 \plus{} 100y^2 \equal{} 1600 \Leftrightarrow$
$ \boxed{ \frac{x^2}{25} \plus{} \frac{y^2}{16} \equal{} 1 }$ [/hide]

Really good [b]t0rajir0u[/b]! i agree with you, it's a pain. It is nice to see the two sides...Much Thanks!!


geometry trigonometry calculus integration

The standard book used in our university proves the Kirchoff's law of radiation in an elaborate manner...I understand the proof.But I think using the same principles,this can be proved in a much better way:

Consider an enclosure is at temperature T and the enclosure is filled with temperature radiation emitted by the wall.

Let a body A has been placed inside an enclosure along with a blackbody B of IDENTICAL geometry.The enclosure have wall opaque to all wavelengths and is insulated thermally from the surroundings.In equilibrium state when A and B attain the enclosure temperature T,the total energy absorbed by them will be equal to the total energy emitted by them.

Let emissive power of absorptive power be given by e_λ and a_λ.
If the body A is of abnormal geometry different parts of the body may have different a_λ and e_λ...Suppose an amount of ∑∆U(λ) of radiation falls on the body A in a given time ∆t.As A and B are of identical geometrical shape the radiation falling on the blackbody B is also ∑∆U(λ) in the same time ∆t.

Then for the blackbody we have ∑∆U(λ)=k*∑(e_bλ) where k is a constant and ∑(e_bλ) is the radiation from the blackbody

For the body A,we have ∑a_λ*∆U(λ)=k*∑e_λ where ∑e_λ is the radiation from the body A.

The same proportionality constant has been used because the two bodies are of identical geometrical shapes and radiation emitted in the same time ∆t has been considered.

Hence from the above relations,

[∑a_λ*∆U(λ)/∑∆U(λ)]=[∑e_λ/∑(e_bλ)]

Since this equality should hold for each portion of the spectrum,we must have
a_λ=(e_λ/e_bλ) Which proves Kirchoff's law.

Please give me feedback.Any incorrect area or somewhere the logic should be more solid...

since it's a law we cannot proof that dude.

like maxwell equation, we cannot proof that. for pratical purpose it is rigth, but we cannot proof.

OK.........
I am worried mainly about exam purpose...
I did not like at all the sines and cosines and Integrations of the problem in that book...


Monica and Chandler plan to meet at Central Perk between 11 AM and 12 PM. However, they never decided when exactly to meet! Both will arrive at a random time between 11 AM and 12 PM, and will stay for exactly 15 minutes. What is the probability that they meet? It counts as meeting if one walks in the door as the other is leaving.

----------------

I posted it as a new topic because more people could see it. Could someone explain this to me please? Thanks.

[hide="easy"]7/16[/hide]

[color=blue]Kyyuanmathcount:  Important!!! LAST WARNING:  Post such as these w/ just an answer and no solution (especially w/ the word "easy" on it) will be deleted in the future![/color]

[quote="Fanatic"]Could you please post a solution?

TIA[/quote]
[hide="hint"]Geometric probability.[/hide]

Basically draw a square, one side for Monica and one for Chandler, and shade in the times in which Monica can come in respect to Chandler and find the area.

[hide]to make it easier, I will divide all times by 15.  first, draw a square of length 4, divided into 1 minute segments.  if you shade the area in which the two coordinates of the square differ by 1 minute or less, it makes a stripe from the top right to bottom left connecting the 1-minute marks on the bottom and left sides with the 3-minute marks on the right and top sides, respectively.  the total area of the two triangles not shaded is 9/2+9/2=9, so the probability that they do not meet is 9/16.  so, the probability that they do is 1-9/16=7/16[/hide]

Attached is a picture for those unclear.  The shaded region is where they arrive within 15 minutes of each other.


[quote="Silverfalcon"]A walnut-salesman knows that 20% of the nuts are empty. He has found a test for picking these out. This discards 20% of the nuts. However, when cracking the nuts that were discarded, one fourth of them were not empty after all. What proportion of the nuts that passed the test are then empty?
A. 4%
B. 6 and 1/4 %
C. 8%
D. 16%
E. None of these[/quote]
[hide="Solution"]
Let there be $ 100$ walnuts.  Thus, $ 20$ of them were taken out.  However, $ 20$ percent of these, or $ 4$, are good walnuts.  Therefore, there must be $ 4$ empty walnuts that passed the test.  So, $ \frac {4}{100} \equal{} 4$ percent passed the test.  Hence, the answer is $ \boxed{A}$. [/hide]

[hide]i get B

Solution:

If we say that the total number of nuts at the beginning is $ x$ then the number of nuts after the test is $ .8 \times x$ and the number of empty shells that passed the test is $ .05 \times x$.  $ \frac{.05}{.8}$ we get $ \boxed{.0625 \equal{} B}$[/hide]

and THE QUATTO MASTER 6000, the problem says that one fourth of the nuts that failed the test were good nuts not one fifth

[hide]Let there be 100 walnuts.  20 of these are empty.  So he takes out 20, but only 15 of these are actually empty. so 5/80 are still empty, which is 6.25%[/hide]

This problem has rather confusing wording. So, I'll post my own solution as well.

[hide="Solution"]

For convenience, let there be 100 walnuts. Now, here is what the first sentence of the problem says: the salesman KNOWS that 20 walnuts are empty and 80 walnuts are full. Here is what the second sentence says: when the test is run, it picks out 20% that are empty BUT it is obviously not correct all the time as we find out.

So, through the test, we pick out 20 walnuts in which we find 1/4 or 5 to be not empty. This implied that the other 5 empty walnuts must be in the 80 walnuts that passed the test. So, the answer is 5/80 or B. [/hide]

I think this is the answer:

[hide]1/4 of the 20% discarded, thus 5% of the total, were not empty, thus the missing 1/4 of all the empty nuts, and 5% of all of them, were mixed in with the 80% not discarded. So 5/80, or 1/16, of the ones that weren't discarded, were empty. Thus [b]6.25%[/b] of them were empty. I suppose the 16% answer was thrown in to fool you deliberately, since 6.25% is the same as 1/[i]16[/i].[/hide]

 :juggle:
-mbx24


Let $a_{1},a_{2},a_{3},a_{4},a_{5}$ be any distinct positive integers. Show that there exists at least one subset of three of these integers whose sum is divisible by 3.

So yes, there is a casework solution that is fairly simple and quick (whoever is up for it feel free to post it!) but personally I feel that there is probably a more elegant solution through pigeon-hole, my problem is I can't seem to figure out how to set it up. Any help towards this specific method of solution would be much appreciated!

This result is a special case of the Erdos-Ginzburg-Ziv theorem.  I gave an outline of one proof in the following thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10996[/url].


Once, I was thinking about systems of congruence equations: x :equiv: a1 (mod m1), x :equiv: a2 (mod m2), x  :equiv: a3 (mod m3), etc.

Then I wondered what these systems would be like in several variables:

3x + 5y :equiv: 7 (mod 11),
4x - y :equiv: 5 (mod 9), etc.

Of course, it would make me happiest to generalize this to m equations, n variables, and p moduli, but I have found even solving specific examples to be rather wearisome.

Thoughts? Conjectures? Proofs?

One way to approach your problem is to rewrite the system of congruences as:

3x + 5y = 11a + 7,
4x - y = 9b + 5, etc.

Then solve them by straight forward methods of linear algebra. You'll get x and y in terms of a and b. Using the condition that x and y are integers will give you a system of congruences for a and b relative to a common base. This system can be solved by elementary row operations (i.e. Gaussian elimination) modulo that base to get conditions on a and b. Then substitute those conditions back into your equations for x and y.

For example, take the congruence you mentioned above.

Solving gives x = (11a + 45b + 32)/23 and y = (44a - 27b + 13)/23.
Since x and y have to be integers, 11a + 45b :equiv: -32 (mod 23) and 44a - 27b :equiv: -13 (mod 23).

Solving these by elementary operations modulo 23, gives a :equiv: 18 (mod 23) and b :equiv: 0 (mod 23), or equivalently a = 23s + 18 and b = 23t.

Substituting these conditions on a and b back into the equations for x and y gives:
x = 11s + 45t + 10 and y = 44s - 27t + 35, for any integers s and t.


How can I make \Delta smaller than now?

[code]
\documentclass[twoside,A4paper]{book}

\def\grpedeset{\hbox{\Gamma\kern-5truept\raise3.3truept\hbox{\fontsize{1pt}{11pt}\selectfont \Delta}}}

\begin{document}
     \grpedeset
\end{document}
[/code]

Of course \Gamma and \Delta are in math mode.

Part which make me problem is:
[code]\fontsize{1pt}{11pt}\selectfont \Delta[/code]

I've tried all solutions. "tiny", "\scriptscriptstyle", but I need \Delta to be smaller.
Thanks.

The extsizes packages allows you font sizes as small as 8pt. Available from [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/extsizes.html?preferredCTAN=ctan.unsw.edu.au&action=/tools/cataloguesearch&catstring=extsizes]CTAN[/url] or the package manager in MiKTeX.

Thanks, that was almost the exact same question I had.


A friend of mine needs help and i have so little time, so i couldn't solve this:

$ (5p-52)x^{2}+(4-p)x+4=0$
$ (2q+1)x^{2}-5qx+20=0$

Find $ q$ and $ p$.

URGENT.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi
How can I show that G_H/G_N(H) is a Galois extension when I know that
H is a subgroup of Gal(G/E) and G/E is a Galois extension and N(H) is the Normalizer of
H in G.

Thank you for any help

Can any of the moderators help here?
There is a similar question on the forum with no answer from about 2 weeks ago.

Thanks

If $ L|K$ is some Galois extension, then we have for two intermediate fields $ A \subset B$:
$ B|A$ is a Galois extension iff $ \mathrm{Gal}(L|B)$ is normal in $ \mathrm{Gal}(L|A)$.

Proof:
We may assume $ A\equal{}K$ as this corresponds to restrict our looks onto $ \mathrm{Gal}(L|A) \subset \mathrm{Gal}(L|K)$ (and neither being normal nor being Galois is affected by that).
Thus we reduced it to:
If $ L|K$ is some Galois extension and $ B$ some intermediate field:
$ B|K$ is a Galois extension iff $ H\equal{}\mathrm{Gal}(L|B)$ is normal in $ G\equal{}\mathrm{Gal}(L|K)$.

But this is one of the statements of the main theorem of Galois theory. Nontheless a proof:
If $ B|K$ is a Galois extension, then every $ \sigma \in G$ maps $ B$ to $ B$ itself (as it is normal). Thus we get a restriction map $ \mathrm{Res}: G \to \mathrm{Gal}(B|K)$ with $ \sigma \mapsto \sigma_{|B}$, having $ H$ as kernel. Especially, as every kernel is normal, this subgroup is normal.
Conversely , if $ H$ is normal in $ G$, let $ \sigma$ be some $ K$-automorphism of $ B$ into some fixed algebraic closure. Then $ \sigma B \subset \sigma L \equal{}L$, thus we see $ \sigma$ as element of $ G$. Now the fix-group of $ \sigma B$ is obviously $ \sigma H \sigma^{\minus{}1}\equal{}H$ (as $ H$ is normal). As their fix-groups are the same, we get $ B\equal{}\sigma B$. Thus $ B|K$ is a normal and thus a Galois extension.


trigonometry
geometry
area of a triangle

Prove that if $ A, B, C$ are the angles of a triangle, then $ \tan A \plus{} \tan B \plus{} \tan C \equal{} \tan A \tan B \tan C$.  What is the geometric interpretation of this result?

I'm not sure, but you clearly can't use a right triangle because $ \tan90^\circ$ is undefined.
I guess you'd have to use an identity that extends trig to non-right triangles (the Laws of Sines, Cosines, and Tangents)...

[hide="hint"]
$ A\plus{}B \equal{} \pi \minus{} C$

[hide="solutions"]
$ \tan(C) \equal{} \tan(\pi \minus{} (A\plus{}B)) \equal{} \dfrac{\tan(\pi) \minus{} \tan(A\plus{}B)}{1\plus{}\tan(\pi)\tan(A\plus{}B)} \equal{} \minus{}\tan(A\plus{}B)$

$ \Rightarrow \minus{}\tan(C)\equal{} \tan(A\plus{}B) \equal{} \dfrac{\tan(A) \plus{} \tan(B)}{1\minus{}\tan(A)\tan(B)}$

$ \Rightarrow \tan(A)\tan(B)\tan(C) \equal{} \tan(A)\plus{}\tan(B)\plus{}\tan(C)$

QED

[/hide][/hide]

Umm, this is an algebraic/trigonometric proof, but I'm looking for a geometric one... any ideas  :?:

tan(a+b+c)= [tan(a) + tan(b) + tan(c) - tan(a)tan(b)tan(c)] /[1 -tan(a)tan(b) - tan(a)tan(c) - tan(b)tan(c)]
can easily be derived
tan(a+b+c)=0
because a+b+c= pie
so the denominator cancels out
and
tan(a) + tan(b) + tan(c) = tan(a)tan(b)tan(c)

From $ A \equal{} \frac {1}{2}ab\sin(C)$ and the cosine law we find $ \tan \equal{}\frac{4A}{a^2 \plus{} b^2 \minus{} c^2}$

Substituting this into the equation above and multiplying through $ (a^2 \plus{} b^2 \minus{} c^2)(a^2 \minus{} b^2 \plus{} c^2)( \minus{} a^2 \plus{} b^2 \plus{} c^2)$ yields

$ 4A\sum_{cyc} (a^2 \plus{} b^2 \minus{} c^2)(a^2 \minus{} b^2 \plus{} c^2) \equal{} 64A^3$

Expanding and factorising gives

$ (a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2 \minus{} c^2)(a^2 \minus{} b^2 \plus{} c^2)( \minus{} a^2 \plus{} b^2 \plus{} c^2) \equal{} 16A^2$

This is Herons Formula  :) So we conclude that equality of the sum and product of $ \tan$ relates the area of a triangle to the lengs of its sides.


calculus
derivative
function
percent
real analysis
real analysis theorems

Pardon a perhaps naive question from someone who is teaching himself calculus.

I understand that the derivative of a function can be used to estimate things like marginal cost, marginal production etc.

Yet with spreadsheets, or even function-enabled calculators, available to most people, I wonder -- why not just use the original function with new values, guaranteeing a fully correct answer, rather than using the derivative, which only gives an approximately correct answer?

Let me give an example from my textbook. (I won't show all the work as I'm not asking how to solve it).

Suppose z, the total weekly production of widgets, is a function of the number of skilled and unskilled workers (x and y respectively) as follows:

z = f(x,y) = 12,000 + 5,000y + 10x^2y - 10x^3.

Assume x = 40 and y = 60.

I understand (or think I do; as ever, please correct me if I'm wrong) that if you want to estimate the additional number of widgets produced when one more skilled worker is added to the work force, you can determine the partial derivative of x with respect to z, or

12,000 + 20xy - 30x^2

and plug in (40,60), which gives you the approximation of 12,000.

I also understand that if you want more than an estimate, you can "nail" the answer down by simply solving for

z = f(41,60) = 12,000 + 5,000y + 10x^2y - 10x^3
= 11,390.

The approximation provided by the partial derivative is about 5% too high.

In some cases, a 5% error is no problem. But what if the widgets in question are big-ticket items, meaning that lots of money is at stake? Five percent of a very big number is material.

So why, when Excel etc makes it so easy to calculate the same function while plugging in different values for the variables, would one even want to bother with the derivative method?

Because you don't always know that the function continues to hold in the future. Suppose, for example, that you're trying to model some real-world situation, like how much your employees produce in a day. You can take the data for 20 days and do a curve-fitting regression on it, and then you'll have some function. You can use this function to predict what'll happen in the future, but you don't actually know for sure that the workers will follow your function. It is just an approximation based on past behavior.


algebra
polynomial
calculus
integration
algebra unsolved

If anyone could help:

If P(x) is a monic polynomial with integral coefficients and k is any integer, must there exist an integer m for which there are at least k distinct prime divisors of P(m)?

Thanks.

If all prime divisors $P(m),m\in Z$ are $p_{1},p_{2},...,p_{n}$ then we have only $O((ln x)^{n})$ distinct means |P(m)| less than x. It give contradition for polinom. Therefore we have infinetely many prime divisors P(m).
Let $p_{1}|P(m_{1}),p_{2}|P(m_{2}),...,p_{k}|P(m_{k})$ By China theorem exist m: \[m_{0}=m_{i}(mod p_{i}),R=\prod_{i=1}^{k}p_{i}\] . If $m=m_{0}+rR,r\in Z$
we have $p_{i}|P(m)$ for all i. Therefore (if $P(m)\not =0$) P(m) has at least k distinct prime divisors.


where do you take the AIME? are there special testing sites or do you just do it at a school

In almost all cases, the AIME is given by the same school or other group that gave the AMC test. In any exceptions, the AMC organization needs to know so they can ship the test to the right place.


that was my problem too, but you can consider n>1


integration
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This is a multiple choice question from a past BC calc exam, that only 20% of the people answered correctly.  My friend in BC calc, was having problems with it, so if you're taking calc give it a shot.

If $\int_1^4 f(x) \ dx = 6$, evaluate $\int_1^4 f(5-x) \ dx$.   (I don't remember the choices).

Soultion:
[hide]Let $u=5-x$, then $du=-dx$, thus the integral can be rewritten as $-\int_4^1 f(u) \ du = -(-6) = 6$.[/hide]

absolutely right, and that is the proper way to see the answer really is correct for any such function f(x); on the other had,  if it's a multiple choice problem (so you know there must be a unique answer) one way to recognize the right answer is just to use a specific simple function for which the integral from 1 to 4 really is 6 [like, for example, the constant function f(x) = 2]  and try the new integral with that.

I have a slightly less specific version of the same idea that Ted Alper just posted. You know that the change of variables just flips the picture around, so the integral should be the same - a reflection doesn't change the area. But perhaps you're not sure whether the answer should be $6$ or $-6.$ (And I'm 99.9% sure that $-6$ was one of the possible multiple choice answers - whoever wrote that question didn't just fall off a turnip truck.) Let's see: the simplest way to get an integral that is postive would be for $f$ to be a postive function. So we assume that $f(x)>0$ for all $x$ that matter. (Note that this is really very similar to Ted's assumption that it was a specific function.) We're integrating $f(5-x)$ over some interval - but if $f$ is positive for all values that matter, then $f(5-x)>0$ where we're integrating it. Thus the integral has to be positive, and $-6$ is not a possible answer.

The answer that Jimmy posted takes a Calculus I/II attitude towards changes of variables, in which we write the limits of integration in order, and we keep track of that order. If we do that and use a decreasing function for a change of variables (which is what happened here), then two things happen: (1) you get limits of integration in the "wrong" order, and (2) there's a negative sign in the change from $dx$ to $du.$ If we work things correctly - as Jimmy did - these sign changes balance out. However, on a human scale, it does introduce the possibility of a sign error.

One could also take the Calculus III or "Jacobian" attitude towards the change of variables. Just take the absolute value of the derivative, so that in this case $\left|\frac{du}{dx}\right|=1$ and $du=dx$ and note that the function $x\mapsto5-x$ sends the set $[1,4]$ to $[1,4].$ The results are the same.

Personally, and this is a bit extreme, but I like the approach to integration by substitution that is taken by Darboux sums.  If we can find a partition that's a good approximation to $\int^4_1 f$, then $g=f(5-x)$ takes on the same values as $f$ and hence $\int^4_1 g$ has the same value.  While this is a bit too heuristic to be of much analytical value, it's not hard to firm it up.

Also, a more braindead approach would be to visualize it as an area under the graph of f. Then the second integral is the area of the mirror image of the first (reflected about x=5/2), since the line of reflection is vertical, the sign stays the same.


Prove or disprove:

There exists a bijective map $ a : \mathbb{N} \to \mathbb{Q}$ such that
if we let $ a_{n} \equal{} a(n)$ then $ \sum_{i \equal{} 1}^{\infty} (a_{n} \minus{} a_{n \plus{} 1})^{2}$ converges

[hide]
Not sure if I understood it correctly, but:

let $ a(n) \equal{} a_n \equal{} \frac{1}{n}$.

Now $ \sum_{n\equal{}1}^{\infty} (a_n \minus{} a_{n\plus{}1})^2 \equal{} \sum_{n\equal{}1}^{\infty} (\frac{1}{n}\minus{}\frac{1}{n\plus{}1})^2 \equal{} \sum_{n\equal{}1}^{\infty} \frac{1}{n^2(n\plus{}1)^2}$. That should converge.

[/hide]

I think it's not work because map $ a_{n}\minus{}>\frac{1}{n}$ is bijection from $ N$ to some part of $ Q$ numbers like $ \frac{1}{n}$ bijection inlcude surection therefore your  map is not bijection between $ N$ and $ Q$


A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outisde diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

$\text{(A) }171\qquad \text{(B) }173\qquad \text{(C) }182\qquad \text{(D) }188\qquad \text{(E) }210$


Hey all,

Is anyone aware of a theorem to handle the following question:

Suppose we are given a set of n positive real numbers, S. Is it possible to construct a polygon of n sides, where side i has length S[i]?

I don't want a solution which consists of just a single line (I'm not sure if that would even be a valid definition of a polygon).

I imagine there must be some necessary law as to the relationship of the magnitudes of the sides of a polygon, i.e. for triangles we have the cosine law.

Any ideas?
Thanks,
Sev

This is a double post. I answered it in the linear algebra section.


$\triangle ABC$ is a right-angle triangle such that $\angle ACB=90^{\circ}$. Let $D$ be the midpoint of $CB$ and $F,E$ are points on $AC$ such that $CF=FE=EA$. $G$ is the feet of the perpendicular drawn from $C$ to $AB$. $H$ is the circumcentre of $\triangle AEG$.

Prove that $\triangle ABC$ is similar $\triangle HDF$.

[hide="Hint"] If you are very desperate, try coordinate geometry.[/hide]

I am hoping for a nice Euclidean Method.

I solved it recently [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=510855][b]here[/b][/url].

Best regards,
sunken rock


function
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Functional Analysis
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real analysis unsolved

The family of all those functions from the Banach space
C[0, 1], which are nowhere differentiable on [0, 1], is the complement of a
first category set.

I messed this up on an exam once.

I think if we let $ D_\varepsilon$ be the set of all $ f$ such that for all $ x \in [0, 1 \minus{} \varepsilon]$ there is $ 0 < \delta \leq \varepsilon$ such that $ \frac {|f(x \plus{} \delta) \minus{} f(x)|}{|\delta|} > \frac {1}{\varepsilon}$, then the nowhere differentiable functions contains the intersection of these. Each $ D_\varepsilon$ is open (easy to see) and dense (approximate with smooth functions and then mess its derivative up by adding something similar to an iterate in the construction of the weierstrass function).

It seems to check out in my head right now, but given my posting history the smart money is on that it's incorrect. But it's something like that.

edit: Changed it ever so slightly to deal with Merryfield's last comment. Why is this problem so hard to do correctly?

I find interesting the following family of arguments: we have sets $ A\subseteq B,$ where the inclusion is obvious. But is $ A \equal{} B?$ More to the point, how could you prove that $ A\ne B?$

There are two broad ways of doing this. One is by explicitly exhibiting an element of $ B\setminus A.$ The other is by showing that $ B$ is "larger" than $ A$ in some important sense. I can think of three different senses of "larger" that can be used in such arguments:

1) cardinality
2) measure (or probability, which is the same thing)
3) Baire category.

Example: $ A$ is the (real) algebraic numbers and $ B$ is the real numbers. Are they the same? While it is possible to answer this by exhibiting a transcendental number, say $ \sum_{n \equal{} 1}^{\infty}10^{ \minus{} n!}$, it can also be answered (after Cantor) by cardinality: the algebraic numbers are countable, the real number uncountable.

Example: $ A$ is the set of linear functions from $ \mathbb{R}$ to $ \mathbb{R},$ and $ B$ is the set of additive functions (solutions of $ f(x \plus{} y) \equal{} f(x) \plus{} f(y).$ We really can't explictly exhibit one of monsters in $ B\setminus A$ but we can say that the sets have different cardinalities.

Then there's this one: $ B \equal{} C[0,1]$ and $ A$ is the set of continuous functions that are differentiable at at least one point. We can, of course, settle this by explicit construction, such as a Weierstrass-style lacunary trigonometric series. But can we also settle it by size?

This time, cardinality is no good as a size indicator: $ A$ and $ B$ have the same cardinality. But both of the other size indicators work. On the measure/probability side, we can talk about Brownian motion (in effect, a probability measure applied to the continuous functions) and argue that a Brownian path is almost surely nowhere differentiable. That's the harder size argument - more subject to learn, more details in the proof than what we see here.

Or there's using Baire category as the size indicator - as is asked for here. (I haven't read Kalle's proof carefully enough to decide whether he has it or not - but I think there's something to be said for dealing in one-sided upper and lower derivates.)


find all functional equation: $ R\rightarrow R$ such that:
$ f(x\minus{}y)\plus{}f(xy)\equal{}f(x)f(y)\plus{}f(x)\minus{}f(y)$

[color=green]$ \star x\equal{}y\equal{}0\longrightarrow f(0)^2\equal{}2f(0)$

$ \star y\equal{}0 \longrightarrow 2f(0)\equal{}f(0).f(x)$

- Case 1 $ f(0)\equal{}2$ then $ f(x)\equal{}2\;\forall x\in\mathbb{R}$

-Case 2. $ f(0)\equal{}0$

$ \star x\equal{}0\longrightarrow f(\minus{}y)\equal{}\minus{}f(y)$

$ \star x\equal{}y,y\equal{}x\longrightarrow f(y\minus{}x)\plus{}f(xy)\equal{}f(\minus{}x).f(\minus{}y)\plus{}f(y)\minus{}f(x)$

Then we have $ f(x)\equal{}x\;\forall x\in\mathbb{R}$

$ f(xy)\equal{}f(x).f(y)$

and $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ too

Then we have $ f(x)\equal{}x\;\forall x\in\mathbb{R}$

A bit harder if you replace $ f(x\minus{}y)$ bye $ f(x\plus{}y)$ and $ f(x)\minus{}f(y)$ by $ f(x)\plus{}f(y)$[/color]


Besides, we also have a nice result: $ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2 \geq\ (a^2\plus{}2bc)(b^2\plus{}2ca)(c^2\plus{}2ab)$
 :)

[quote="nguoivn"]Besides, we also have a nice result: $ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2 \geq\ (a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)$
 :)[/quote]

It's equivalent to $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\ge0$, obvious.  :wink:

[quote="dduclam"][quote="nguoivn"]Besides, we also have a nice result: $ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2 \geq\ (a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)$
 :)[/quote]

It's equivalent to $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge0$, obvious.  :wink:[/quote]
Yes, you are right. It's easy but nice ^^.
And please try to solve the first problem, it's one of the problems which I like best  :)
[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $ \sqrt [3]{a^2 \plus{} 2bc} \plus{} \sqrt [3]{b^2 \plus{} 2ca} \plus{} \sqrt [3]{c^2 \plus{} 2ab} \geq\ \frac {1}{\sqrt [6]{3}}(ab \plus{} bc \plus{} ca)\sqrt {a^2 \plus{} b^2 \plus{} c^2}$
 :)[/quote]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $ \sqrt [3]{a^2 \plus{} 2bc} \plus{} \sqrt [3]{b^2 \plus{} 2ca} \plus{} \sqrt [3]{c^2 \plus{} 2ab} \geq\ \frac {1}{\sqrt [6]{3}}(ab \plus{} bc \plus{} ca)\sqrt {a^2 \plus{} b^2 \plus{} c^2}$
 :)[/quote]
Besides, we have a similar result (with the same condition)
$ 5(\sqrt {a^2 \plus{} 2bc} \plus{} \sqrt {b^2 \plus{} 2ca} \plus{} \sqrt {c^2 \plus{} 2ab}) \geq\ 4(ab \plus{} bc \plus{} ca)\sqrt {a^2 \plus{} b^2 \plus{} c^2}\plus{}3\sqrt {3}$
PS: Both of them can be solved by Am-Gm :)


let  $ a,b,c >0$ such that $ abc \equal{} 32$

prove : $ \frac{a}{bc} \plus{} \frac{4}{c} \plus{} \frac{4b}{ac} \plus{} \frac{2c}{ab} \geq 3$

[quote="*210*"]let  $ a,b,c > 0$ such that $ abc \equal{} 32$

prove : $ \frac {a}{bc} \plus{} \frac {4}{c} \plus{} \frac {4b}{ac} \plus{} \frac {2c}{ab} \geq 3$[/quote]
$ \Leftrightarrow (a\plus{}2b)^2\plus{}2c^2\ge 96$
but $ (a\plus{}2b)^2\plus{}2c^2\ge 8ab\plus{}2c^2\equal{}4ab\plus{}4ab\plus{}2c^2\ge 96$  AM-GM  :wink: done

thanks  bunhiacovski !
it is exactly what I made.
you can make it without AM-GM ??

[quote="*210*"]let  $ a,b,c > 0$ such that $ abc \equal{} 32$

prove : $ \frac {a}{bc} \plus{} \frac {4}{c} \plus{} \frac {4b}{ac} \plus{} \frac {2c}{ab} \geq 3$[/quote]

$ \frac{a}{bc}\plus{}\frac{4}{c}\plus{}\frac{4b}{ac}\plus{}\frac{2c}{ab}\equal{}\frac{a}{bc}\plus{}\frac{2}{c}\plus{}\frac{2}{c}\plus{}\frac{4b}{ac}\plus{}\frac{c}{ab}\plus{}\frac{c}{ab}\geq6\sqrt[6]{\frac{16abc^{2}}{a^{3}b^{3}c^{4}}}\equal{}6\sqrt[6]{\frac{16}{(abc)^{2}}}\equal{}6\sqrt[6]{\frac{16}{32^{2}}}\equal{}3$

$ a\equal{}c\equal{}4;b\equal{}2$


Let $ A$ be nxn matrix.Show that if $ A^2 \equal{} I$ then $ rang(A \plus{} I) \plus{} rang(A \minus{} I) \equal{} n$, where $ I$ is unit matrix.

I won't solve it here, since you should be able to solve it yourself with a simple hint :

Let $ V$ be a vectorspace over $ \mathbb R$. If $ f: V\to V$ is linear and that $ f^2\equal{}id$ prove that $ V\equal{}im(id\plus{}f)\oplus im(id\minus{}f)$.


Let $x_1,x_2,...,x_n(n\geq 2)$ be positive real numbers such that $x_1^2+...+x_n^2=1$. Prove that\[\frac{2\sqrt{n}-1}{5\sqrt{n}-1}\leq \sum_{i=1}^n\frac{2+x_i}{5+x_i}\leq\frac{2\sqrt{n}+1}{5\sqrt{n}+1}\]

This is an incorrect inequality. I just check the most right ineq.
Take $x_1=\cdots=x_4=1/2$ for $n=4$. Then
\[
4\times (5/11)=\sum_{i=1}^4 \frac{2+x_i}{5+x_i}>5/11=\frac{2\sqrt{n}+1}{5\sqrt{n}+1}
\]

I think the inequality should be
\[
\sum_{i=1}^n \frac{2+x_i}{5+x_i}\leq n\left(\frac{2\sqrt{n}+1}{5\sqrt{n}+1}\right)
\]
or
\[
\sum_{i=1}^n \frac{2+x_i}{5+x_i}\leq \sum_{i=1}^n \frac{2\sqrt{n}+1}{5\sqrt{n}+1}
\]

In this case, I have a solution for the right inequality:).

BTW do we know each other ??? :D

I'm sorry, my fault. In case of your inequality, could you post your solution?

- Are you also from Indonesia? Where're you studying now? I'm Indonesian, but currently studying in Singapore -

$\sum \frac{2+x_i}{5+x_i}=n-\sum \frac{3}{5+x_i}$ and

\[
\sum \frac{3}{5+x_i}\geq 3\frac{\left(\sum 1\right)^2}{\sum 5 +\sum x_i}\geq \frac{3n^2}{5n+\sqrt{n}}
\]
THerefore
\[
\sum \frac{2+x_i}{5+x_i}\leq n-\frac{3n^2}{5n+\sqrt{n}}=n\left(1-\frac{3\sqrt{n}}{5\sqrt{n}+1}\right)=n\frac{2\sqrt{n}+1}{5\sqrt{n}+1}.
\]

Oh ya digitalboy,

saya juga dari indonesia, sekarang sedang belajar di US.

salam kenal