probability
calculus
integration
real analysis
function
floor function
derivative

Let $y$ be an arbitrary positive number. Define $g(y)$ to be the expected number of independent uniform $[0,1]$ random variables needed for the the sum to first exceed $y.$

Compute $g(y)$ and discuss the asymptotic behavior of $g(y)$ as $y\to\infty.$

A bit of intuition about that asymptotic behavior: The expected value of each of our random variables is $\frac12.$ In order for the sum to exceed $y$, perhaps the sum of the expectations should be about $y$. In other words, don't be too surprised if $g(y)$ works out to something like $2y$ for large $y.$

Of course, I'm looking for something more detailed than that.

I think it's been proven in the previous thread that for $y$ between $0$ and $1$, $g(y)=e^y$. Another way to prove that:

The probability that the number of turns is $\le n$ is $1-\frac {y^n}{n!}$, because $\frac{y^n}{n!}$ is the volume of the small simplex in the corner of the $n$-hypercube, having mutual orthogonal sides of length $y$ (this is where we use $y\le 1$; if $y>1$, the shape in the corner is no longer a simplex) along the sides of the cube. The probability that the number of turns is $n$ is thus $\frac{y^{n-1}}{(n-1)!}-\frac{y^n}{n!}$, and from here it's really easy: we add up expressions of the form $n(\frac{y^{n-1}}{(n-1)!}-\frac{y^n}{n!})$ and use the series expansion of $e^y$.

I don't know what the right way to approach the general case is though.

Hint: try conditioning on the value of the first term in the sequence of random variables.

(Yes, $g(y)=e^y$ for $0<y\le1.$ If convenient, you may also let $g(y)=0$ for $y<0.)$

Another variation on this problem- what if we're adding independent exponential random variables of mean $k$? The answer is particularly nice in that case.

The asymptotic behavior is accessible for almost any positive continuous random variable, in exactly the expected form (x/mean)+(constant)+(goes to 0)

[quote="jmerry"]Another variation on this problem- what if we're adding independent exponential random variables of mean $k$? The answer is particularly nice in that case.[/quote]

You're right!  Nice observation.

In fact, let N be the number of turns.  Then it seems that N - 1 follows a Poisson distribution.  This reminds me of results concerning waiting times and Poisson processes, but I don't these well enough to say.

[quote="jhaussmann5"]Then it seems that N - 1 follows a Poisson distribution.[/quote]
That is correct. Would you please go ahead and state the exact formula for $g(y)$ you would derive from that observation?

[b]However[/b], that observation only applies to jmerry's variant which uses the exponential distribution. It does [i]not[/i] apply the original case which uses the uniform distribution, or to other cases using other distributions.

I'll go ahead an post the first part of my solution, although I will leave part of it for a few days to see if anyone else has any ideas.

We have a sequence of independent identically distributed nonnegative continuous random variables. The density of this distribution is $f(x)$, where $f$ is supported on $[0,\infty).$ Let $g(y)$ be the expected number of these random variables required for the sum to exceed $y.$ As a notational convenience, define $g(y)=0$ for $y<0.$

Condition on the first random variable. It has the value $x,$ which leaves the remaining target as $y-x.$ Thus the expected number of remaining terms will be $g(y-x),$ making the expected overall number of terns given that the first terms was $x$ equal to $1+g(y-x).$ (Note that this holds even if $x>y.$) But the overall expected value is the expectation of the conditional expectations. Hence we have the following integral equation for $g$:

$g(y)=1+\int_0^yf(x)g(y-x)dx.$

We recognize the integral as a convolution, and abbreviate this equation as $g=1+f*g.$

Since we have a linear convolution equation, a transform method looks attractive. With everything happening on $[0,\infty)$, I turn to the Laplace tranform.

Let $F(s)=\int_0^{\infty}f(x)e^{-sx}dx$ and $G(s)=\int_0^{\infty}g(y)e^{-sy}dy.$ Note that $F(s)$ is, except for a sign change, the "moment generating function" for the underlying random variable.

Apply the Laplace tranform to our integral equation to get

$G(s)=\frac1s+F(s)G(s)$

We rearrange this to $G(s)=\frac1{s(1-F(s))}.$

Try this for jmerry's example: the exponential distribution of mean $k$. Then $f(x)=\frac1ke^{-x/k}$ and $F(s)=\frac1{1+ks}.$ Then our equation leads to $G(s)=\frac{1+ks}{ks^2}=\frac1k\cdot\frac1{s^2}+\frac1s.$ From that we derive that $g(y)=\frac yk+1.$

Now back to the original case, in which the random variable was uniform $(0,1)$. In that case, $F(s)=\frac{1-e^{-s}}s.$ From this we derive $G(s)=\frac1{s-1+e^{-s}}.$

To get anywhere with this we expand this as a geometric series:

$G(s)=\frac1{s-1}-\frac{e^{-s}}{(s-1)^2}+ \frac{e^{-2s}}{(s-1)^3}+\cdots$

Let $H(y)$ be the Heaviside unit step function - that is, $H(y)=1$ for $y\ge0$ and $H(y)=0$ for $y<0.$ The inverse Laplace transform of the series above gives us that

$g(y)=\sum_{k=0}^{\infty}H(y-k)\frac{(-1)^k(y-k)^k}{k!}e^{y-k}.$

This can be restated as

$g(y)=\sum_{k=0}^{\lfloor y\rfloor}\frac{(-1)^k(y-k)^k}{k!}e^{y-k}.$

The function is smooth except at the integers. It is continuous at 1, has a continuous first derivative at 2, a continuous second derivative at 3, and so on.

Still to be addressed: the asymptotic behavior as $y\to\infty.$ I'll wait a few days to give others a chance to work on that.

The time has come to finish off this topic.

For the original case, the function $g(y)$ is asymptotically equal to $2y+\frac23.$ That is, $g(y)=2y+\frac23+q(y)$ where $\lim_{y\to\infty}q(y)=0.$

For the same problem in which the uniform (0,1) distribution is replaced by a "nice enough" continuous positive random variable with mean $\mu$ and variance $\sigma^2$, then the $g(y)$ for that case is asymptotically equal to $\frac{y}{\mu}+\frac12+\frac{\sigma^2}{2\mu^2}.$

I'm not sure what "nice enough" means. Bounded with square integrable density would be sufficient but is certainly stronger than it needs to be. But some condition is needed, because discrete distributions would follow a somewhat different (but related) pattern.

We will do this by Laplace transform methods. For the original case, from above, we see that $g(y)$ has Laplace transform $G(s)=\frac1{s-1+e^{-s}}.$ Expanding this as a Laurent series about 0, we find that $G(s)=\frac2{s^2}+\frac23\cdot\frac1s+Q(s)$ where $Q(s)$ has a removable singularity at 0 and behaves like $\frac1s$ at $\infty.$ The inverse transform of this is $g(y)=2y+\frac23+q(y).$ The inverse Laplace transform procedure gives us that

$q(y)=\frac1{2\pi}\int_{-\infty}^{\infty}Q(it)e^{ity}dt.$

Since $Q(it)\in L^2(\mathbb{R})$ we can say that $q(y)\in L^2.$ Furthermore, since $Q$ is smooth, $q$ behaves itself at infinity. From this, I argue that $\lim_{y\to\infty}q(y)=0.$

For the more general case, the random variable has density $f(x)$ with Laplace transform $F(s)=\int_0^{\infty}f(x)e^{-sx}dx = 1 - \mu s + \frac{\mu^2+\sigma^2}2s^2 + O(s^3).$ The function $g$ has Laplace transform

$G(s)=\frac1{s(1-F(s))}= \frac1{\mu s^2-\frac{\mu^2+\sigma^2}2s^3+\cdots} = $
$\frac1{\mu}\cdot\frac1{s^2}+\frac{\mu^2+\sigma^2}{2\mu^2}\cdot \frac1s +Q(s)$

where once again, $Q(s)$ has a removable singularity at zero. If we assume nice enough behavior for $Q$, the asymptotic result follows.

The same result can also be obtained by considering the "excess" - that is, by examing the distribution of the amount by which the running sum exceeds $y$ when we reach the stopping point.


superior algebra
superior algebra unsolved

in a previous post, we have found all the automorphism of $S_n$, for $n \neq 6$.
Now, prove that any automorphism of $A_n$, $n \neq 6$, is the restriction of one automorphism of $S_n$ .

I think this is the right idea, but I didn't have the patience to write it:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphisms&t=43233[/url]


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 A
\/closed

Down at the bottom, there is a section that says "Who is Online"  and it shows hidden users.  What are hidden users?

there's a choice when you go to ur profile to not show ur online status.. people who select "no" for that choice are shown as hidden users.

is there a reason why you would want ot be hidden?

like if ur math coach signs on to yell at u for failing the AIME.. u might just want to hide.. (cough cough) lol

lol your math coach sounds like a strict person  our teacher is happy that two ppl from our school made it to aime ( we couldve had three but the other person missed #1 on amc 10 a)

yea.. our math coach is pretty strict... but then he does have a pretty big reputation to keep... *sigh*
we had a lot of AIME qualifiers... but USAMO ppl.. aie.. *shrugs*

i got a 5.. im a freshman.. first time taking it.. i thought i did pretty well for my own standards.. but of course my coach thinks anything less than a 10 is terrible... *sigh*


inequalities
quadratics
algebra
polynomial
inequalities proposed

$\bigstar$ [i][Hungkhtn][/i]
Let $a,b,c$ are positive numbers such that $a+b+c=3$. Prove that:
$\sqrt{a^{3}}b+\sqrt{b^{3}}c+\sqrt{c^{3}}a\le 3$
My brother (Toanhocmuonmau) have a very nice solution without expanding. At the moment, I am waiting for everybody.

This problem does not belong to me  :blush:

[quote="zaizai-hoang"]$ \bigstar$ [i][Hungkhtn][/i]
Let $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \sqrt {a^{3}}b \plus{} \sqrt {b^{3}}c \plus{} \sqrt {c^{3}}a\le 3$
[/quote]
Very ugly, but proof.
We'll prove that $ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2\leq3,$ where $ a,$ $ b$ and $ c$ are non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ 3u^2 \minus{} 2v^2 \equal{} t^2u^2,$ where $ t\geq0.$
Hence, $ 1\leq t\leq\sqrt3$ and $ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2\leq3\Leftrightarrow6 \minus{} \sum_{cyc}(a^3b^2 \plus{} a^3c^2)\geq\sum_{cyc}(a^3b^2 \minus{} a^3c^2)\Leftrightarrow$
$ \Leftrightarrow6\left(\sqrt {3u^2 \minus{} 2v^2}\right)^5 \minus{} \sum_{cyc}a^2b^2\sum_{cyc}a \plus{} abc\sum_{cyc}ab\geq(a \minus{} b)(a \minus{} c)(b \minus{} c)\sum_{cyc}ab\Leftrightarrow$
$ \Leftrightarrow2\left(\sqrt {3u^2 \minus{} 2v^2}\right)^5 \minus{} 9uv^4 \plus{} (6u^2 \plus{} v^2)w^3\geq(a \minus{} b)(a \minus{} c)(b \minus{} c)v^2\Leftrightarrow$
$ \Leftrightarrow8t^5u^3 \minus{} 9u^3(3 \minus{} t^2)^2 \plus{} 2(15 \minus{} t^2)w^3\geq2(3 \minus{} t^2)(a \minus{} b)(a \minus{} c)(b \minus{} c).$
$ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\geq3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3} \equal{}$
$ \equal{} u^3\left(\frac {3(3 \minus{} t^2)}{2} \minus{} 2 \minus{} 2\sqrt {\left(1 \minus{} \frac {3 \minus{} t^2}{2}\right)^3}\right) \equal{} \frac {u^3}{2}\left(5 \minus{} 3t^2 \minus{} \sqrt {2(t^2 \minus{} 1)^3}\right).$
Hence, $ 8t^5u^3 \minus{} 9u^3(3 \minus{} t^2)^2 \plus{} 2(15 \minus{} t^2)w^3\geq$
$ \geq(8t^5 \minus{} 9(3 \minus{} t^2)^2 \plus{} (15 \minus{} t^2)(5 \minus{} 3t^2) \minus{} (15 \minus{} t^2)\sqrt {2(t^2 \minus{} 1)^3})u^3\geq0$
because $ 8t^5 \minus{} 9(3 \minus{} t^2)^2 \plus{} (15 \minus{} t^2)(5 \minus{} 3t^2) \minus{} (15 \minus{} t^2)\sqrt {2(t^2 \minus{} 1)^3}\geq0\Leftrightarrow$
$ \Leftrightarrow8t^5 \minus{} 6t^4 \plus{} 4t^2 \minus{} 6\geq(15 \minus{} t^2)\sqrt {2(t^2 \minus{} 1)^3}\Leftrightarrow$
$ \Leftrightarrow2(4t^4 \plus{} t^3 \plus{} t^2 \plus{} 3t \plus{} 3)\geq(15 \minus{} t^2)\sqrt {2(t \minus{} 1)(t \plus{} 1)^3},$ which is true for $ t\geq1.$
Thus, it remains to prove that
$ \left(2\left(\sqrt {3u^2 \minus{} 2v^2}\right)^5 \minus{} 9uv^4 \plus{} (6u^2 \plus{} v^2)w^3\right)^2\geq(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2v^4.$
But $ \left(2\left(\sqrt {3u^2 \minus{} 2v^2}\right)^5 \minus{} 9uv^4 \plus{} (6u^2 \plus{} v^2)w^3\right)^2\geq(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2v^4\Leftrightarrow$
$ \Leftrightarrow\left(2\left(\sqrt {3u^2 \minus{} 2v^2}\right)^5 \minus{} 9uv^4 \plus{} (6u^2 \plus{} v^2)w^3\right)^2\geq$
$ \geq27v^4(3u^2v^4 \minus{} 4v^6 \plus{} (6uv^2 \minus{} 4u^3)w^3 \minus{} w^6)\Leftrightarrow$
$ \Leftrightarrow(9u^4 \plus{} 3u^2v^2 \plus{} 7v^4)w^6 \minus{} \left(45uv^2 \minus{} (6u^2 \plus{} v^2)\left(\sqrt {3u^2 \minus{} v^2}\right)^5\right)w^3 \plus{}$
$ \plus{} (3u^2 \minus{} 2v^2)^5 \minus{} 9uv^2\left(\sqrt {3u^2 \minus{} 2v^2}\right)^5 \plus{} 27v^{10}\geq0\Leftrightarrow$
$ \Leftrightarrow8(7t^4 \minus{} 48t^2 \plus{} 117)w^6 \minus{} 4(4t^7 \minus{} 45t^6 \minus{} 60t^5 \plus{} 405t^4 \minus{} 1215t^2 \plus{} 1215)w^3 \plus{}$
$ \plus{} 5t^{10} \minus{} 72t^9 \plus{} 405t^8 \plus{} 432t^7 \minus{} 2430t^6 \minus{} 648t^5 \plus{} 7290t^4 \minus{} 10935t^2 \plus{} 6561\geq0.$
If $ 4t^7 \minus{} 45t^6 \minus{} 60t^5 \plus{} 405t^4 \minus{} 1215t^2 \plus{} 1215\leq0$ then our inequality is proven.
But $ 4t^7 \minus{} 45t^6 \minus{} 60t^5 \plus{} 405t^4 \minus{} 1215t^2 \plus{} 1215 > 0$ and $ 1\leq t\leq\sqrt3$ gives $ 1\leq t < 1.227...$
Id est, it remains to prove that $ (4t^7 \minus{} 45t^6 \minus{} 60t^5 \plus{} 405t^4 \minus{} 1215t^2 \plus{} 1215)^2\leq2(7t^4 \minus{} 48t^2 \plus{} 117)\cdot$
$ \cdot(5t^{10} \minus{} 72t^9 \plus{} 405t^8 \plus{} 432t^7 \minus{} 2430t^6 \minus{} 648t^5 \plus{} 7290t^4 \minus{} 10935t^2 \plus{} 6561),$
which is equivalent to $ 27(t \minus{} 1)^2(2t^{12} \minus{} 20t^{11} \plus{} 93t^{10} \plus{} 366t^9 \minus{} 801t^8 \minus{} 2304t^7 \plus{} 1998t^6 \plus{} 6588t^5 \minus{}$
$ \minus{} 972t^4 \minus{} 8748t^3 \minus{} 2187t^2 \plus{} 4374t \plus{} 2187)\geq0,$ which is true for $ 1\leq t < 1.228.$

Unbelievable! Brilliant arqady.....
 :omighty:

Please post your Brother's proof zaizai toang...

This is an inequality in problem P 2.12 in Algebraic Inequalities (page 91).
This is my solution.

We have a proof by Rearrangement and AM-GM Inequality. Here...

[quote="Sunkern_sunflora"]Nice proofs guys.... :coolspeak:

But where is zaizai-toang's brother's proof??!! :maybe:[/quote]
His proof (also Canhang's one ) is very nice (and intelligent,of course)
He  used Vasc ineq,Cauchy Swart,and AMGM,he only needs a few lines to write it  :lol:

[quote="arqady"]Beautiful proof!  :lol:  My congratulations dduclam![/quote]
[quote="Sunkern_sunflora"]Nice proofs guys.... :coolspeak:
 [/quote]
Thank you ! :lol:

[quote="Allnames"] His proof (also Canhang's one ) is very nice (and intelligent,of course)
He  used Vasc ineq,Cauchy Swart,and AMGM,he only needs a few lines to write it  :lol:[/quote]

Allnames, Can_hang's proof is approximate to proof on Vasc's (by Cauchy-Schwarz, you can to see again in above Vasc's file, and see the following Can_hang's file). But I have a nother proof, more simply plenty more ! :)

[quote="dduclam"][quote="arqady"]Beautiful proof!  :lol:  My congratulations dduclam![/quote]
[quote="Sunkern_sunflora"]Nice proofs guys.... :coolspeak:
 [/quote]
Thank you ! :lol:

[quote="Allnames"] His proof (also Canhang's one ) is very nice (and intelligent,of course)
He  used Vasc ineq,Cauchy Swart,and AMGM,he only needs a few lines to write it  :lol:[/quote]

Allnames, Can_hang's proof is approximate to proof on Vasc's (by Cauchy-Schwarz, you can to see again in above Vasc's file, and see the following file). But I have a nother proof, more simply plenty more ! :)[/quote]
My friend, your proof is very nice and I like it. :)
But as I said, there still have 2 other proofs which one using the fact $ c(b\minus{}c)(b\minus{}a)(ab\plus{}bc\plus{}ca) \le 0$ and one using Cauchy Schwarz (different from the known Cauchy Schwarz one) :)

[quote="dduclam"][quote="can_hang2007"][quote="dduclam"]
Allnames, Can_hang's proof is approximate to proof on Vasc's (by Cauchy-Schwarz, you can to see again in above Vasc's file, and see the following file). But I have a nother proof, more simply plenty more ! :)[/quote]
My friend, your proof is very nice and I like it. :)
But as I said, there still have 2 other proofs which one using the fact $ c(b \minus{} c)(b \minus{} a)(ab \plus{} bc \plus{} ca) \le 0$ and one using Cauchy Schwarz (different from the known Cauchy Schwarz one) :)[/quote]

Really!?! I want to know your beautiful proof very much, my friend !  :)[/quote]
I will write the solution by the above fact here, for the solution with Cauchy Schwarz, there is some security reasons that I cannot post it here, sorry my friend.
Assume that $ a \ge b \ge c$, then we have
\[ c(b\minus{}c)(b\minus{}a)(ab\plus{}bc\plus{}ca) \le 0\]
Therefore
\[ a^3b^2\plus{}b^3c^2\plus{}c^3a^2 \le a^3b^2\plus{}b^3c^2\plus{}c^3a^2\minus{}c(b\minus{}c)(b\minus{}a)(ab\plus{}bc\plus{}ca) \equal{} b(c^2\plus{}a^2)(ab\plus{}bc\plus{}ca) \minus{} abc(a^2\plus{}b^2\plus{}c^2) \equal{} b(3\minus{}b^2)[b(c\plus{}a)\plus{}ca]\minus{}3abc\]
Setting $ x\equal{}c\plus{}a\Rightarrow c\equal{}\frac{x^2\plus{}b^2\minus{}3}{2}$, we need to prove
\[ b(3\minus{}b^2)[bx \plus{}\frac{x^2\plus{}b^2\minus{}3}{2}]\minus{}3b\cdot \frac{x^2\plus{}b^2\minus{}3}{2} \le 3\]
\[ \Leftrightarrow f(x)\equal{}b^3x^2\plus{}2(b^4\minus{}3b^2)x\plus{}b^5\minus{}3b^3\plus{}6 \ge 0\]
Examine $ f(x)$ as a quadratic polynomial with $ x$, then the highest coefficient of it is positive and
\[ \Delta^{\prime}_f \equal{}(b^4\minus{}3b^2)^2\minus{}b^3(b^5\minus{}3b^3\plus{}6)\equal{}\minus{}3b^3(b\plus{}2)(b\minus{}1)^2 \le 0\]
Hence $ f(x) \ge 0$ and we have done. :)

The following inequality is also true.
Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
\[a^3b^2+b^3c^2+c^3a^2\leq\frac{1}{3}\sum_{cyc}(2a^3b+2a^3c-a^2bc)\]


In a triangle, ABC, AB = AC, angle A = 40 degrees. Point  i within the triangle with angle OBC = angle OCA. The number of degrees in angle BOC is:

a. 110
b. 35
c. 140
d. 55

A cube is made by soldering twelve 3-inch lengths of wire poperly at the vertices of the cube. If a fly alights at one distance it would travel before coming to any vertex a second time, without retracing any distance, is:

a. 24
b. 12
c. 30
d. 18

[quote]Point i within the triangle with angle OBC = angle OCA[/quote]
Shouldn't it be point O within the triangle....

[hide]First draw the triangle. It is an isosceles,
$ \angle BAC=40^{o}\quad \Longrightarrow \angle ABC=\angle ACB=70^{o}$

If $ \angle ABO=x$, $ \angle OBC=z$, $ \angle OCB=y$,
$ x+z=70^{o}$, $ y+z=70^{o}$ $ \Longrightarrow x=y$

Inside the triangle, to find $ \angle BOC$,
$ 180^{o}-70^{o}$
$ =\boxed{110}$

$ \boxed{A}$[/hide]

The second question is not very clear. Could Re-edit it?

I think he means "what is the maximum number of edges the fly could traverse before coming to the some vertex again?"


Ας συζητήσουμε στο topic αυτό τα αυριανά θέματα του "ΕΥΚΛΕΙΔΗ".

[color=red][size=150][b]ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΕ ΟΛΟΥΣ!!![/b][/size][/color]

Αλέξανδρος

[color=red][b]Βάζω τα θέματα 2008 για να δουν όσους δεν εμπλέκονται άμεσα. [/b][/color]

Έχω λύσει αρκετά από Α΄Λυκείου και Γ Λυκείου.Δεν πρόλαβα (11:31) να κοιτάξω της Β'. Καλά θεματάκια. Έλειπε ίως μια αποδεικτική γεωμετρία στη Β αλλά εντάξει  ! Θα τα αξιολογήσετε μόνοι σας. Δεν χρειάζονταν δυσκολότερα , με εξαίρεση ίσως ένα θέμα .

[color=red][b] Καλά αποτελέσματα και Καλή συνέχεια !![/b][/color]

egw sthn b lukeiou egrapsa, kai nomizw pws ta themata htan arketa pio eukola apo auta pou eihan valei persi sthn bita...Sto 1 apla ypswsa sto tetragwno k meta tis praxeis klp, sto deutero eipa pws s kathe agwna tha dinodai 4 vathmoi k mu vghke 4*(n(n-2))/2=364 ara n=14, sto triro ta ekana teleia tetragwna k meta eipa epeidh sto deutero melos evgaine 1 oti prepei to kathe ena na einai mikrotero apo ena k ekana kapoies praxeis k vrhka m=3, k sto 4 to i vrhka me pythagoreio auta pu dn hxera k vghke, kai meta to xrhsimopoihsa k evgala to ii.....Swsta einai??

Εγώ έλυσα 2-3 θέματα..  :ninja:

Το 4ο δε το 'δα καθόλου , μόνο το διάβασα.. Δεν είχα υπομονή να το παιδέψω , είδα ότι μπορεί να χρειαζόταν και Ceva που δεν τον θυμόμουν και το άφησα :P

Γενικά επειδή δεν είχα προετοιμαστεί καθόλου δεν είχα σκοπό να πάω, αλλά με ξύπνησαν που με ξύπνησαν .. :P

Καλά αποτελέσματα σε όλους  :)

[quote="KAZA"] sto triro ta ekana teleia tetragwna k meta eipa epeidh sto deutero melos evgaine 1 oti prepei to kathe ena na einai mikrotero apo ena k ekana kapoies praxeis k vrhka m=3[/quote]

Γιαυτό δεν είμαι σίγουρος αλλά νομίζω ότι θέλει Cauchy-Swartz ...(δηλάδή σίγουρα έτσι βγαίνει απλά δεν ξέρω για τον τρόπο σου)

Για την Γ Λυκείου τα θέματα είναι ωραία (το πρώτο μόνο ήθελε πράξεις) και τα άλλα απαιτούσαν βασική άλγεβρα
(δηλ παραγοντοποιήσεις και ταυτότητες)
Το τελευταίο θέμα ήταν πολύ ωραίο . Να το σκεπτικό της λύσης μου
$ 2x^2y^2\plus{}2y^2z^2\plus{}2z^2x^2\minus{}x^4\minus{}y^4\minus{}z^4\equal{}(x\plus{}y\plus{}z)(x\plus{}y\minus{}z)(y\plus{}z\minus{}x)(z\plus{}x\minus{}y)$
kai $ x\plus{}y\geq z$ , $ y\plus{}z\geq x$ $ z\plus{}x\geq y$ .

 :)

exei kaneis tis lyseis tis 2as lykeioy??
elysa ta 2 prwta kai to 1o erwtima apo to 4o,ta alla aplws ta palepsa

odws lathos einai t 3 pu vrhka...gt ama einai 3 einai ola megista...vlakeia ekana pali

Για Β Λυκείου:
Πρώτο θέμα: Πανεύκολο σχολικού επιπέδου. Όλοι θα έπρεπε να το γράψουν. :D
Δεύτερο θέμα: Επίσης εύκολο ήθελε μόνο να δεις ότι οι αγώνες ήταν 91 και ίσοι με ν ανά 2.
 :lol: Τρίτο θέμα: Με βάση α^2+β^2+γ^2>=(α+β+γ)^2/3 βγαίνει σε δύο γραμμές ότι το m είναι 6-ρίζα3 :wink:
Τέταρτο θέμα: Πρώτο ερώτημα γελοίο αλλά μετά δυσκόλευε. :(  Εγώ πήρα να το κάνω με πράξεις, Cauchy Scwarz και τριώνυμο αλλά έκανα λάθος στην Cauchy Scwarz (αν είναι δυνατόν :mad: ) και μου βγήκε στο 1/4 αντί στο 1/3. Η λύση ήταν με το συμμετρικό του Δ... :!:

ΕΥΧΟΜΑΙ ΣΕ ΟΛΟΥΣ ΚΑΛΑ ΑΠΟΤΕΛΕΣΜΑΤΑ... :wink:  :first:

Με συγχωρείται για το spam. Κατ' αρχάς αναρτήθηκαν οι λύσεις στο site για όποιν θέλει και κατά δεύτερον μπορεί κανείς να κρίνει πώς θα κυμανθούν οι βάσεις και τα βραβεία αναλογικά με πέρσι? Αληθεύει ότι πέρσι το πρώτο βραβείο ήταν και με 3 θέματα???

Και εγώ 4 εγραψα.Protonie για την  4 ακριβώς η ιδέα ειναι με συμετρικο του Δ

Mια σύντομη λύση των μιγαδικών της Γ΄Λυκείου
Θέτουμε το κλάσμα λ πραγματικό και χιαστά ΄βρίσκουμε ευκολα εξίσωση της μορφής az^4+bz^2+a=0 και οι συζυγείς ρίζες ως προς z^2 αν πολ/σθούν δίνουν (χωρίς πράξεις απο Vieta) a/a=1 άρα ...το ζητούμενο

Να η δική μου λύση για το 4. Κάπου στο τέλος έκανα ένα αριθμητικό και πήρε φαλάγγι και το 3ο ερώτημα. Τι μπορεί να πάθει κανείς. Αν έχω κάποιο λάθος πείτε μου.

Έστω $ AM \equal{} x$

To ζητούμενο άθροισμα:

$ sqrt(x^2 \plus{} a^2) \plus{} sqrt[(2a \minus{} x)^2 \plus{} 4a^2]$

Αρκεί υψώνοντας στο τετράγωνο να ελαχιστοποιήσουμε το:

$ 2x^2 \plus{} 9a^2 \minus{} 4ax \plus{} 2sqrt{(x^2 \plus{} a^2)[(2a \minus{} x)^2 \plus{} 4a^2)]}$

Όμως το ριζικό από Cauchy Scwarz ελαχιστοποιείται όταν
$ (x/2a \minus{} x) \equal{} a/2a$, δηλαδή $ x \equal{} 2a/3$, το ζητούμενο. Μετά απλά επαληθεύουμε αν φεύγουν τα $ x$ από την παράσταση που όντως φεύγουν. :wink:

Ωραία η λύση στο θέμα των μιγαδικών xp . Ωραία είναι και η λύση της ΕΜΕ σε αυτό το θέμα .

@protonios : Πέρσυ στην Αθήνα πράγματι το πρώτο βραβείο ήταν με 3 θέματα γιατί το θέμα 3 δεν το έλυσε κανείς από εκεί . Σε όλη την Ελλάδα εξάλλου δύο άτομα το κάναν ο Ράπανος και ο Βαφείδης ...  :wink:

Δε ξέρω πώς το λύνει  η ΕΜΕ (no time to take a look) πάντως θυμάμαι έγραψα τον αριθμητή 2(παρανομαστή) + κάτι ψιλά και η άσκηση απλοποιήθηκε κατά πολύ.

[quote="bilot4"][quote="silouan"]
Το τελευταίο θέμα ήταν πολύ ωραίο . Να το σκεπτικό της λύσης μου
$ 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2 \minus{} x^4 \minus{} y^4 \minus{} z^4 \equal{} (x \plus{} y \plus{} z)(x \plus{} y \minus{} z)(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)$
kai $ x \plus{} y\geq z$ , $ y \plus{} z\geq x$ $ z \plus{} x\geq y$ .

 :)[/quote]

Κατ' αρχάς το όλο θέμα είναι να ξέρει κάποιος αυτήν την ταυτότητα (νομίζω λέγεται De Moivre) ή να την αποδείξει.
Αυτό που ήθελα να σημειώσω, όμως, ιδιαίτερα είναι ότι το
$ x \plus{} y\geq z$ ,για παράδειγμα, προκύπτει πολύ απλά απ' την ισότητα :

$ \frac {x \plus{} y}{z} \equal{} \frac {SB}{SB'}\geq 1$ από υπόθεση... γιατί το site της ΕΜΕ έχει αρκετά πιο μπερδεμένη λύση.[/quote]

ακριβώς...Η ανισότητα προκύπτει ακριβώς όπως αναφέρεις και μετά είναι απλή εφαρμογή της ταυτότητας de Moivre.


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geometry
number theory unsolved
number theory

I have this snake i cant kill:

Let $n>4$ be an integer, and $(a_i,b_i)$, $i=1,2,3...,n$, different pairs of integers such that for all $i$:

$|a_ib_{i+1}-a_{i+1}b_i|=1$ and $(a_{n+1},b_{n+1})=(a_1,b_1)$.

prove there exist $i$ and $j$, with $1<|i-j|<n-1$ and $|a_jb_i-a_ib_j|=1$

Kill the evil snake! I think the poor snake has already been killed by Pierre. I am sure he posted a solution for this problem.

I tried to search for it, and i didnt find it, does someone knows where is it?

My Alzheimer is still alive...At first sight, this is a geometrical problem : Consider the points $A_i(x_i,y_i)$. Thus $[OA_iA_{i+1}] = \frac 1 2$, where $[.]$ denotes area, and Pick's formula is probably around here...

Pierre.


The calculuator is most likely experiencing something known as Roundoff Error.
This occurs when the accuracy of the number in question must be so high (think like 30-40 decimals) that the calculator can no longer manipulate it effectively. This is because the maximum bit size of most calculators puts a physical limit on the size of the numbers that can be manipulate.

If you use a more modern language, like C++ or Java, then I think that there's support in those languages for as arbitrary precision (It just uses more memory...)

As for the specific question, the calculator happens to be evaluating two different things.
One of them is the 100th root of $i$, and one is the 4th root of $i$.

This is done with natural logarithms, but it is a taxing process for the calculator to go through and evaluate it.

No, that's not the problem at all.  Your calculator is experiencing something called, "You are asking it to give you a single numerical answer to a question with multiple values, so it gets confused."  (There's also an order-of-operations issue: it's not generally true in the complex numbers that $(x^{a})^{b}= (x^{b})^{a}$.)

Your calculator has correctly (approximately) calulated one of the 100th roots and one of the 4th roots of $i$; it just happens that the given 100th root and given 4th root are different.

To add to JBL and  to make it clear,

Consider something like $({2^{4})^{(1/2)}}$ as applied to a dumb calculator whose dumbness I can control.
(And see two methods as  (Method 1: $(16)^{(1/2)}$)   vs (Method 2 :  or $2^{(4/2)}= 2^{2}$)

[b][i]one[/i] [/b]of the square-root of  $2^{4}$ is $-4$,
but  if you "calculate" $2^{2}$ you are going to get only $+4$ .

IOW there are 100 possible values of $i^{(1/100)}$    one of them is what you have listed first, Another (and there are 98 still more to go) one is the one you listed i as a the second number..

In fact, if you call  the first number $x$, then $x , x^{5}, x^{9}, \dots x^{397}$ are valid roots, the second number is nothing but $x^{25}$.

to beat a dead horse:

let $\epsilon = e^{\frac{i\pi}{200}}$

then $\epsilon^{100}= i$

but we also have $\epsilon^{500}=i$

and $\epsilon^{900}= i$

and etc.  this follows because $e^{i\theta}$ moves around the unit disk in the complex plane as $\theta$ increases and makes a full circle as $\theta$ grows to $\theta+2\pi$

what calculator is that?

[quote]what calculator is that?[/quote]
Guess any calculator worth its $NaCl$ would do that.  Maple, and even my stone-age TI (and an old H65 too)   understands  complex numbers and would do these calculation..:)  (With my bamboo slide rule,  or   Micosoft XP calculator, I have to set "grad" and take cos  and sin of 1 grad to get those numbers :)


AMC USA(J)MO USAMO trigonometry inequalities induction geometry

Well, we might as well put up solutions we got that we thought were nice.  The only one I've done so far is number 1, so here it is:

Let AB = a, BC = b, CD = c, and DA = d.  We know that [tex]\displaystyle{a + c = b + d}[/tex], since ABCD is circumscribed about a circle.  Also, constructing diagonal BD and applying the law of cosines, we have
(1) [tex]\displaystyle{a^2+d^2-2ad\cos(\angle A)=b^2+c^2-2bc\cos(\angle C)}[/tex].
Now, we know [tex]\displaystyle{60^o \leq A, C \leq 120^o}[/tex], which means that [tex]\displaystyle{-1/2 \leq \cos A, \cos C \leq 1/2}[/tex].  The maximum value in the left side of (1) is greater than or equal to the smallest possible value for the right hand side:
(2) [tex]\displaystyle{a^2+c^2+ac \geq b^2+c^2-bc}[/tex]
Since [tex]\displaystyle{a + c = b + d}[/tex],
[tex]\displaystyle{a - d = b - c}[/tex](3),
and [tex]\displaystyle{a^2 + d^2 -2ad = b^2 + c^2 - 2bc}[/tex] (4), and by the trivial inequality,
[tex]\displaystyle{b^2+c^2-2bc \geq 0}[/tex] (5).

Adding (2) to twice (4) and adding twice (5), we should be left with the inequation
[tex]\displaystyle{3a^2 + 3ad + 3d^2 \geq b^2 + bc + c^2}[/tex].

Now, multiplying on both sides by (3) in absolute value, we get
 [tex]\displaystyle{3|a^3-d^3| \geq |b^3-c^3|}[/tex], and the given result follows in full by symmetry.

Soln to #4:

Bob can win.

Divide the 6x6 grid into three pairs of rows.

We want to force that the order of terms along the bottom row of each row pair is shifted circularly two the the right from the order of terms on the top row.

If Alice places a number in the top row, Bob places a number in the bottom row shifted two columns to the right (taken circularly mod 6), and if Alice places a number in the bottom row, Bob places a number in the top row shifted two columns to the left. Bob can always maintain ordering in a particular row, since there are infinitely many rationals between any two distinct rationals.

Therefore, Bob can force the black square in the bottom row of a row pair to be shifted two to the right of a black square in the top row of a row pair. No line can connect these two squares, so Bob wins.

hey guys...i got #1 and #4 fully... #2 partly...#3 less partly.   However, I was very stupid in that i didn't read that we were supposed to write the equality conditions for #1 :-(  how many points do you think i'll lose?

Thank you!

And yes, Bob wins :-p

[quote]How do I tex a  sign?[/quote]

\geq or \leq

Oh by the way, did anyone get

0 < k < 1/2   ,   2 < k

for #3?  I don't wanna write my proof up, because I'm not sure if it's right.

And does anyone have any clue as to how to get #6?  I might have been able to take a look at it, but because of my struggles on question 5, I only had about 40 minutes for all of question 6.  And in that 40 minutes all I wrote was a big question mark in my answer sheet.

[quote="GreenTwilight"]Oh by the way, did anyone get

0 < k < 1/2   ,   2 < k

for #3?  I don't wanna write my proof up, because I'm not sure if it's right.

[/quote]

I'm sorry, but it isn't right. The answer is all positive reals except k=1. I'm not exactly sure how to do it, it involves a fractaling technique, which for k=1 becomes infinite. I have been convinced of the correctness of the answer, however.

compucomp, I'm not saying I don't believe you, but can you direct me to a proof of that?  It would be interesting to see why the answer is what it is.

[quote="GreenTwilight"]This post may be long but...

#2 (Maybe a bit unclear.  I'm gonna shorthand it because people on this site are smart)

The trick is to prove that every set must have both the numbers 1 and 0, and that with those numbers you can get any integer.  Here's why:

(Condition C)

0 + 1 = 1, which are all in set S, so
0 - 1 = -1.

(Lowest # you have) + (1) are all in the set, so (Lowest # you have) - (1) = (the next lowest integer)  Continue for all negative integers.
Substitute with Highest # you have and  negative 1 for all positive integers.

Okay, now there are two possibilities for your original set of numbers.  Either it is just the # 1 or at least two relatively prime numbers.  If it's the #1, then with condition B you can say that (1 - 1) is in the set, so 0 is in the set.  (This can also show you that 0 is in EVERY set S, just take a number and subtract it from itself in condition B)

If you have two relatively prime numbers, then you know the positive difference is in set S according to condition B.

(I wrote up a longer, clear proof for the rest, but you guys can just use your imagination to figure out what I'm talking about)

The rest follows under condition C

You have 3 numbers in set S, one being the difference of the other 2.  The sum of the two lowest numbers so far adds up to the other, so the positive difference must be in set S.  Now you have a lower set of 3 numbers, still following the condition that one is the difference of the other 2.  You can keep finding lower and lower positive integers with this pattern until you get to 1.  (If you don't get to 1, then you started with 2 numbers that were not relatively prime to each other.  Just trust me on that one)

Alright there's #2 (I'm pretty sure).  If anything's unclear, say something and I'll try to clear it up[/quote]

WHY did everyone make this mistake?

It says the greatest common factor of (a_1, a_2, a_3, ... , a_n) is 1.

This does NOT mean that there are i and j such that a_i is relatively prime to a_j.

Take for example the set (2*3, 3*5, 2*5). There are no two relatively prime numbers in this set, yet the greatest common factor of the entire set is 1.

!@#$!@\#$!@#$!@\#$!@#$
AAAHHHH NOOOOOOO
Great.  USAMO fooled me on 2 questions.
I fell for the trap in question 2 and question 3 so far.  You know what's even more aggravating?  I know EXACTLY how to fix my error in that problem.
!@\#$%@#)*&%!@#$

CRAP. I did it GreenTwilight's way (but a little less complete because I was running out of time). I hope I still get [i]some [/i]partial credit for it....

I did #2 the same way as well. Apparently the only one I got right was #4 :(

Pah. Those evil people who wrote #2. Those evil, evil people.

I fell into the same trap (though with a slightly different approach, so my proof works up to the last page.) However, the way I phrased my induction to n from 2 offers up a pretty simple way to fix the error, so I think I'll still get partial credit.

Hey smart people...
I'm still waiting for solutions to question 3 and question 6

I'd like to know those answers too.

Number six scared me so much that all I put for my answer was "The desired result is intuitive." And no, I don't expect any credit for it, thanks for asking.

I don't know anyone that got 3 or 6 correct.

I was going to write an anecdote for #6, just for fun, but didn't feel like it because I was just too tired to care at that point.

How do we know that the answer is k!=1 if we don't know anyone who got it right?

I don't  know anyone that got it right firsthand, but know someone that knows someone that got it right and he told me the answer was k!=1.

Is it possible that he just got it wrong?

Here's my partial number 3, even though I didn't get anywhere near this far in the test, and now I do it in a matter of minutes :-(.  My lesson I guess is not to attack the impossible problems but rather to do the easier ones :-(.  Anyway:

Ok I'm still working on a diagram, so the first equation you'll just have to believe me or try working it out (it shouldn't be too hard if you understand what I mean).

[hide]If we partition into two right concave hexagons (hard to explain, but I think most of you know what I mean), with the sides coinciding with the 1 by k rectangle being of lengths 1, x, and x+y where y is non-negative.  With some ratios (the two hexagons are similar) we have k = 1/(y+x) + x.  Now if x is close to k, we can increase y a lot to make 1/(y+x) small.  Not 100% rigorous, but it's the basic idea.  Grrr... I can't believe I didn't do any of this during the test...[/hide]

Ok I realize this is hard to understand, so I'll try to post a diagram if I have time.

Hmm...

Whenever you get that diagram up, I'd be happy to take a look.  I actually have no idea what you're talking about, but you seem to be right.

Actually, I just realized that I was missing a few limitations to eliminate congruity, but basically, I think you should use ratios to find k, and prove that for all k save some there are x and y or whatever variables you want to use.

Yeah, I solved number 1 today
Because of the tangents
AB+CD=AD+BC
AB-AD=BC-CD
Apply law of cosines
AD^2+AB^2-2(AB)(AD)cosx=BC^2+CD^2-2(BC)(CD)cosy=BD^2
where x=angle A and y=angle C
Rewrite it
(AD-AB)^2+2(AB)(AD)(1-cosx)=(BC-CD)^2+2(BC)(CD)(1-cosy)
(AB)(AD)(1-cosx)=(BC)(CD)(1-cosy)

Note that 2(BC-CD)^2 :ge: 0
and 3(BC)(CD)(3-(1-cosy)/(1-cosx)) :ge: 0 because max of
(1-cosy)/(1-cosx) is 3 (when y=120 and x=60)
3(BC)(CD)(3-(1-cosy)/(1-cosx)) =9(BC)(CD)-3(AB)(AD)
2(BC-CD)^2=3(BC-CD)^2-(AB-AD)^2
add them up, we have
3(BC-CD)^2-(AB-AD)^2+9(BC)(CD)-3(AB)(AD) :ge: 0
3(BC-CD)^2+9(BC)(CD) :ge: 3(AB)(AD)+(AB-AD)^2
3(BC^2+CD^2+(BC)(CD)) :ge:(AB^2+AD^2+(AB)(AD))
multiply AD-AB on RHS, and BC-CD on LHS
we have
3((BC)^3-(CD)^3) :ge: (AB)^3-(CD)^3

i think there's a typo, because i don't quite follow the algebra.

I'm talking about JBL's solution for #1 at the beginning.

It seems that everyone did #1 using law of cosines. Can somebody tell me if this is correct?

Let AB = x+y, BC = y+z, CD = z+t, DA = t+x
Let I be the center of the circle, and let IE = the perpendicular to CD.

 :ang: DIE = 1/2  :ang: ADC, so 30  :le:   :ang: IDE  :le:  60.

Thus 1/ :rt3:   :le:  t/r  :le:   :rt3: , since t/r = cot(IDE).

Similarly, 1/ :rt3:   :le:  x/r, y/r, z/r  :le:   :rt3: .

Adding gives 2/ :rt3:   :le:  (x+y)/r  :le:  2 :rt3: , or 2r/ :rt3:   :le:  AB  :le:  2r :rt3: .

Similarly, 2r/ :rt3:   :le:  BC, CD, DA  :le:  2r :rt3: .

So the max value of 1/3 (AB :^2: +(AB)(AD)+AD :^2: ) = 12r :^2:  = the min value of BC :^2: +(BC)(CD)+CD :^2: .

Multiplying by AB-AD = BC-CD gives the left inequality, and the right side holds by symmetry.


For $ a,b,c \geq 0$
$ xy^3 \plus{} yz^3 \plus{} zx^3 \geq x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2$

take $ x\equal{}10$, $ y\equal{}1$, $ z\equal{}0$ :lol:


geometry
circumcircle
IMO Shortlist
power of a point
radical axis
perpendicular bisector

Can anyone post solutions to Round 2?

And what are the sources of the problems?

I thought Q1 was really lovely once I had decided to draw a table with the elements of the union of all our sets along the top and the 'i' values along the left, writing 'j' in a cell (a_k,i) if a_k is in A_(i,j).

I'll attach full solutions to Q1 and Q3 in PDF format tonight. My Q3 method was to spot that AA' etc pass through the external similitude centre of incircle and circumcircle, and then areal co-ordinate bash the rest.

[quote="Carlez Tevos"]And what are the sources of the problems?[/quote]
Problem 1 is a generalization of [url=http://www.mathlinks.ro/viewtopic.php?p=874985#874985]IMO Shortlist 2006 problem C3[/url], apparently due to Tusza (1989).

Problem 2 is IMO Shortlist 2007 problem N7.

Problem 3 was proposed by me and yetti. It is a particular case of some stuff we study in a future paper in Forum Geometricorum. Our solution is based on the inversion $ \Psi$ with respect to the incircle of triangle $ ABC$. I'm really curious if someone found a neat synthetic proof without inversion!

Tom, if you want to check your equations, see Paul Yiu's sketch from [url=http://tech.groups.yahoo.com/group/Hyacinthos/message/15607]Hyacinthos message 15607[/url].

[quote="pohoatza"][quote="Carlez Tevos"]And what are the sources of the problems?[/quote]
Problem 1 is a generalization of [url=http://www.mathlinks.ro/viewtopic.php?p=874985#874985]IMO Shortlist 2006 problem C3[/url], apparently due to Tusza (1989).

Problem 2 is IMO Shortlist 2007 problem N7.

Problem 3 was proposed by me and yetti. It is a particular case of some stuff we study in a future paper in Forum Geometricorum. Our solution is based on the inversion $ \Psi$ with respect to the incircle of triangle $ ABC$. I'm really curious if someone found a neat synthetic proof without inversion![/quote]
Where can i find these problems?

[quote="pohoatza"]What do you mean? You didn't participate to Round 2?

In any case, I think Valentin will attach all this year's problems to the Mathlinks Contest subforum soon.[/quote]
No,i didn't...Why doesn't somebody else post these problems?

[quote="pohoatza"]Problem 2 is IMO Shortlist 2007 problem N7.[/quote]
2007 Shortlisted problems should'n be posted before IMO 2008, should it?

Ok, attached are my solutions for Q1 and Q3. I think my Q1 solution is quite nice, although if it were handwritten it would probably be much clearer (I could have drawn out some tables - I think it's best to visualise it as an 'Excel' spreadsheet where rows can be added to the middle of the table). My Q3 solution isn't particularly nice, basically doing one quite nice thing (using some homotheties to 'locate' A',B',C') and then just bashing with areals. My equation for the common radical axis is the same (up to multiplying by $ \minus{}a^2b^2c^2$ as that Paul gets, which seems promising :)).

Albanian Eagle, what is your beautiful solution?

I solved Q1 by induction..,
I don't like Q2.. This is so easy if you could observe that $ v_{p}(ap^{k} \plus{} b) \equal{} v_{p}(ap^{k}) \plus{} v_{p}(b)$ ($ b < p^{k}$)..
First I struggle to solve it with $ v_{p}(ap^{k} \plus{} b) \equal{} a \frac {p^{k} \minus{} 1}{p \minus{} 1} \plus{} v_{p}(a) \plus{} v_{p}(b)$. But suddenly I found above fact, and then.....

I found many beautiful facts about Q3, but I couldn't got it....
When is the posting limit? I always forgot and when I want to see a problem to try more on that, I find "Round 0".

Pohoatza, do you have a reference to Tusza's paper? Is it also proved there that the constant is sharp?

Dear Cosmin could you post your solution to the problem 3 of round 2? Thank you .

Hm....
I don't know original labels of points, so I renamed some points.
$ M$ is a midpoint of arc $ BC$, and $ N$ is the tangent point of the circle in the problem.
Then obviously, when $ X$ denotes $ MN \cap BC$, $ A,D,N,X$ lie on a circle. ($ \angle ANX \equal{} \angle ADX$ by easy angle chasing)

We can show $ XI \perp AI$ by some ways. The fastest method is applying a Pascal Theorem to $ BCLNMA$ ($ T$ is a midpoint of arc $ AB$)
But if $ P,Q,R$ is tangent point of incircle to $ BC, CA, AB$, We can show that $ NAI \sim NIP$.
($ \angle NAI \equal{} \angle NIP$ is easy. we can show $ \angle ANI \equal{} \angle INP$ by using the fact that $ AN$ pass through the isogonal conjugate of Nagel Point.)
These triangles are also similar to $ MIM'$ ($ M'$ is midpoint of $ BC$). Using this, we can show that $ AI\perp XI$.
Inversion with a circle whose center is $ A$ also works.

Now invert all thing with respect to incircle. Then $ I$ is the orthocenter of $ A'B'C'$, $ A', D'$ is symmetric wrt $ B'C'$.
Because $ \angle AIX \equal{} 90$, $ I, X'$ is symmetric wrt a perpendicular bisector of $ B'C'$.
A circle passing through $ A,D,N$ become a circumcircle of $ A'X'D'$. If $ A''$ is antipodal of $ A'$, $ A',X',D',A''$ lie on a circle.
The circumcenter of $ A'B'C'$ is one of radical center of these three circles. So remain is to show that the center of circles lie on a line.
But it is easy to show that it is the intersection point of $ B'C'$ and the perpendicular bisector of $ A'A''$.
So these points are all lie on a orthotransversal of circumcenter of $ A'B'C'$. it is easy to show.
....


algebra
polynomial
function
topology
Ring Theory
superior algebra
superior algebra solved

Let $(K,v)$ be a complete valuated field. Let $R$ be the valuation ring of $v$. Let $f$ be a monic irreducible polynomial in $K[X]$. Prove that $f\in R[X]$ if, and only if, $f(0)\in R$.

===========================================================================
Definitions:
In this case, a valuated field is a field $K$ together with a function $v: K\to\mathbb R\cup\{\infty\}$, called a valuation, such that:
\[ v(0)=\infty,  \] \[ v(K^*)\subseteq\mathbb R,  \] \[ v(xy)=v(x)+v(y),  \] \[ v(x+y)\geq\min\{v(x),v(y)\}.  \]

If $(K,v)$ is a valuated field, for every $1>\alpha>0$, one can associate a norm $\rho: K\to\mathbb R_+\cup\{\infty\}$ by $\rho(x)=\alpha^{v(x)}$ for every $x\in K$. These norms are all equivalent and they give a well defined metric space structure on $K$. If $K$ is complete with respect to this structure, we say that $(K,v)$ is a complete valuated field.

A more general definition can be given for a valuation by replacing $\mathbb R$ by an arbitrary ordered group.

Since it's complete, it's henselian (note that your $v$ is by definition not archimedian).
For those we have the following lemma, clearly implying the statement above:

$v(f) = \min ( v(a_0) , v(a_n) )$ with $f(x) = a_n x^n + ... + a_1 x + a_0$ an irreducible polynomial and $v(f) : = \min ( v(a_i) )$.

Proof:
WLOG $v(f)=0$ after multiplication.
Assume $\min ( v(a_0) , v(a_n) )>0$. Then there is some $m \neq 0,n$ with $v(a_m)=0$, so lets take the smallest such $m$. But then $f(x) \equiv x^m \cdot (a_n x^{n-m} + ... + a_m) \mod \mathfrak{p}$ with $\mathfrak{p} : = \{x \in K | v(x)>0 \}$ being the prime ideal of the valuated field.
But because of being henselian, this gives a factorisation $f(x)=g(x)h(x)$ with $g(x) \equiv x^m , h(x) \equiv a_n x^{n-m} + ... + a_m \mod \mathfrak{p}$, contradicting irreducibility.

Another proof:

Let $\widetilde K$ be the splitting field of the polynomial $f$, and let $\tilde v$ be the unique extension of $v$ to $\widetilde K$ ($(\widetilde K,\tilde v)$ is, in this case, also a complete valuation field). An element $\alpha\in\widetilde K$ is a $\tilde v$-integer iff the norm $N(\alpha)=N_{\widetilde K/K}(\alpha)$ is a $v$-integer. The norm of a root of $f$ is $f(0)$, hence all roots of $f$ are $\tilde v$-integers. This means that all coefficients of $f$ are $\tilde v$-integers (being precisely the elementary symmetric functions in the roots of $f$), and since they are elements of $K$, they are $v$-integers (exactly what we wanted to prove).


With no stops, it takes 3 hours for a motorist to make a trip driving at an average speed of 65 mph. When the motorist stops for gas and a sandwich on the same trip, he averages 60 mph. How many minutes long is the stop?

If it took him 3 hours to make that trip at 65 mph, then that means the trip was 195 miles long. But when he stopped, his speed became 60 mph, so the trip took him $ 195 \div 60 \equal{} 3.25$ hours, which is 15 minutes longer than usual.

I asked this question, but the Review said the answer was 180. I don't see how that's possible...

math154, read the "read this thread before posting" sticky first.

Maybe he was trying to be nice and support izzy because she may have had doubts as to the correctitude of her answer. Have you ever thought of that? Math154 is a kind and noble soul whom we should all try to emulate.


I've recently seen some posts recently deleted in the Round Table forum because they were ads, and I am wondering what links are allowed in signatures.

Would a link like this be considered an ad?
[quote][url=https://addons.mozilla.org/themes/moreinfo.php?id=784]Simpler theme for Firefox[/url][/quote]

Sorry if this is a silly question.

No that is not what we consider an ad. You can use it as a signature.


[quote="spix"]the solutions to $t^3-15t^2+71t-15=0$ so $(x,y,z)=(3,5,7)$ and permutations[/quote]

Even I got 3,5,7 but what the heck are permutations?That's what the book says as well but what do u mean by that?


Find all positive integers $ n$ such that $ n^{2008} \plus{} n \plus{} 1$ is a prime number.

It can be generalized.
$ n^k \plus{} n \plus{} 1$
Here's a hint for the solution:
[hide]
It is easily factorized by adding and substracting numbers $ n^{k\minus{}1}, n^{k\minus{}2},...,n^2$ and then show that one of the factors equals $ 1$ for $ n \equal{} 1$.
For example, $ k \equal{} 5$
$ n^5 \plus{} n \plus{} 1 \equal{} (n^2 \plus{} n \plus{} 1)(n^3 \minus{} n^2 \plus{} 1)$
$ n^3 \minus{} n^2 \plus{} 1 \equal{} 1$, for $ n \equal{} 1$
[/hide]
Tell me if you want the complete solution.

easier:
think cubic roots of unity  :wink: ...

@basted: how would you procede in your generalization with $ n^3\plus{}n\plus{}1$ or $ n^4\plus{}n\plus{}1$?


algebra
polynomial
group theory
superior algebra
superior algebra unsolved

Here is an easy but interesting result.
Let d and i be given integers. Find all k such that there exists an extension F over Q with irr. polynomial (x^k-a) that has a subfield with group  C(d) X C(d) X… X C(d) (i times) but none with  C(2) X   C(2) X... X C(2) (i+1 times). Here C(d) is the cyclic group of cardinality d.

[quote="Abel"]an extension F over Q with irr. polynomial (x^k-a)[/quote]

this is not precise. you mean some finite extension which has a primitive element whose minimal polynomial is $ x^{k}-a$ for some rational $ a$, right? so for short, some extension which is isomorphic to $ \mathbb{Q}[x]/(x^{k}-a)$ (this implicitely requires the irreducibility of the polynomial).

You are right. But before stating that the polynomial was irreducible, I had not said that it is the minimal polynomial. I meant the field obtained by adjoining all the zeros of x^k-a to Q. But that doesn't matter for this particular result as we have x^k-a irreducible.

A few things I should have mentioned. k is odd. Secondly d is not a prime power.
Actually, I don't have a solution (not even an incorrect one) for the case when d is a prime power.


number theory
Diophantine equation
number theory unsolved

It's so easy to write down diophantine equations and so difficult to solve them. I cannot understand why you post such a problem, knowing very well that even special cases are tremendously difficult and probably some of them unsolved. Even in the case where this problem really interests you, I don't understand why you don't read some of the latest articles in this field (Mignotte, Bugeaud, Bennet and so on). Such problems will never be solved on the forum, so why posting them?  :maybe:

This isn't my problem, of my friend . He sent to me as question and I think it have solution. Maybe we cann't slove it, but what you know on this problem?  :)

Here Pacific J. Math. 207 (2002), 61-75. you will find an article of Shorey and Bugeaud discussing this problem. Probably it will show its complexity (the authors are well known for their contributions in this field of number theory!).

Indeed :)

On the other hand, to rescue this thread, it is easy to solve this equation for any given $m,n$ with $(m,n) > 1$. For example:

Solve in integers $1+x+\cdots+x^{9}= 1+y+y^{3}$.

Same for the other problem: show that $1+x+\cdots+x^{md}=y^{d}$ has a finite number of solutions $(x,y,m)$ for any fixed $d>1$.

You should ask T.N.Shorey for the complete answer guy  :D, i agree with harazi that he's very well-known arround this type of equations  :wink:

[quote="N.T.TUAN"]Thanks, use Google I also have  :
[1]Y. Bugeaud and T. N. Shorey, On the Diophantine Equation $\frac{x^{m}-1}{x-1}=\frac{y^{n}-1}{y-1}$, Pac. J. Math. 207(1) (2002),61-75.

[2]R. Goormaghtigh, Nombres tel que $A=1+x+...+x^{m}=1+y+...+y^{n}$, L'Intermediare des Mathematici Monatsh. Math. 3 (1892), 265-284.

[3]Yu. Nesterenko and T. N. Shorey, On the Diophantine Equation $\frac{x^{m}-1}{x-1}=\frac{y^{n}-1}{y-1}$, Acta Arith. 83 (1998), 381-399.

Now i think that my friend joke for me :P[/quote]
N.T.TUAN, Could you tell me what did you input to get so many result.
Could you tell me some tips about use google.I find nothing by using google about this equaiton. :blush: Thanks

Keyword : T.N. Shorey

We'll have all. An example math.tifr.res.in/~shorey/


First of all, and before posting the sense of this topic, I would like to know, nowadays, all communist countries in the world. Then, I'll post my question (just to make sure it's not useless).

[quote="bubka"]According to the 4th Communist International, there is no communist country in the world.

The countries where the communists have made maximum progress are Nepal and Peru.[/quote]

and what about, for example, Cuba?

Well I would say these five :

Laos, Vietnam, North Korea, People's Republic of China and Cuba

However, some say Turkmenistan is also communist as the ruling Party and the president were communists in the Soviet era.  Belarus is also said to be communist sometimes.

I know my answer will be criticized anyway, there is no pleasing when answering this question.

So let me put it this way : you mean countries where everyone works and gets the same wealth as everyone else?
None.

[quote="fredbel6"]Laos, Vietnam, North Korea, People's Republic of China and Cuba[/quote]

Thanks fredbel6.

Know, if there're 200 countries, more or less, non-communist, and only 5 of them communist, why is communism accepted! What about big countries, like USA, threatening this 5 countries to stop being like that? I know it sounds ridiculous, but, I think, USA can "make" (by focre) Cuba being non-communist, and they can move Castro from power (like they did with Saddam). Now, why don't they do it?

[quote="José"]I think, USA can "make" (by focre) Cuba being non-communist, and they can move Castro from power (like they did with Saddam). Now, why don't they do it?[/quote]

Because it turned out so well with Iraq, of course ...  :roll:

In fact, while we're at it, why don't we just smash international terrorism, poverty, hunger, and that annoying kid in my building who plays his stereo too loud?

[quote="JBL"] and that annoying kid in my building who plays his stereo too loud?[/quote]

Jajajaja poor you.

I mean, I personally think that communism is bad for a country, really bad. So, I'd like to find a way to stop communism everywhere.

The point is if all of us (people that live in a country without communism) think the same about it

Jose you mean well obviously, but I think experience is of great importance here. How old are you?  I don't mean to humiliate you at all : when I was 16 I laughed at what I thought about the Gulf war that time, now I am 21 and I laugh at what I thought when I was sixteen.

When I will be 30, I will laugh at what I think now probably.

First of all there also other noncommunist countries that the USA doesn't get along with like Zimbabwe and Myanmar (and Belarus if you don't consider them communist).

Second of all : USA doesn't want to spread democracy :

[img]http://upload.wikimedia.org/wikipedia/en/c/cf/Allende-Pinochet.jpg[/img]

Also remember their support of the contras (for whom they sold weapons to Iran!) against the Sandinistas in Nicaragua.  And Saddam?  Where to begin if you think it's all about ousting a dictator, maybe with this picture :

[img]http://www.gwu.edu/~nsarchiv/NSAEBB/NSAEBB82/handshake300.jpg[/img]

Third point : what do you want the USA to do?  Invade the country?  Invading a country like China is impossible as they have nuclear weapons and the technology to launc them (and USA doesn't like to attack countries that can really hit back).  The same goes for North Korea as the PROC is likely to intervene on their behalf.  Vietnam : they have been there, done that, it didn't work, they only made Laos communist in the progress!
And as for Cuba, it enjoyed Soviet protection in the Cold War, I am not completely aware of their current relations with the PROC for instance, but they do have good relations with Bolivia and Venezuela now.  And as I said, US'ers like to think of war as sending some troops overseas, watching some bombardments and weeping over a few military casualties, Cuba on the other hand is not THAT far from mainland USA.

Oh well, if you think Iraq is a success, regardless of the maybe 100,000 civilian casualties, war with Cuba might be good...

[quote]Jajajaja poor you. [/quote]

Please don't react to JBL's comment in that way.  He wasn't lamenting about his current situation, he was pointing out that hitting everyone that displeases you won't make anyone better.

I guess I'll stop here for now.

[quote="akalra1"]Bush doesn't notice that you can't solve problems with force
there has to be another way, but Bush and his great brains haven't found it yet[/quote]

Well as it is offtopic I will hide it , but :
[hide]
[quote="Antibush site Idleworm"]We've got: Israeli Jews fighting Lebanese Shi'a and Palestinian Sunnis; Palestinian Fatah militants who've stopped fighting Hamas militants, but only because they're both fighting the Israelis; Saudi Sunni fundamentalists issuing fatwas against Hezbollah Shi'a fundamentalists; Egyptian Sunni fundamentalists backing those same Hezbollah Shi'a fundamentalists; Iraqi Sunnis killing Iraqi Shi'a and vice versa; Iraqi Shi'a (the Mahdi Army) jousting with Iraqi Shi'a (the Badr Brigade); Iraqi Kurds trying to push Sunni Arabs and both Sunni and Shi'a Turkomen out of Kirkuk; Turks threatening to invade Kurdistan; Iranians allegedly shelling Kurdistan, Syrian Kurds rebelling against Syrian Allawites who are despised by Syria's Sunni majority but allied with the Lebanese Shi'a who are hated and feared by the House of Saud and its Sunni fundamentalist minions. Oh, and American and Israeli neocons threatening to bomb both Syria and Iran.
Don't worry - President Bush will sort it out. He just found out about the whole Sunni/Shia thing back in 2003. I wonder what he'll do if he finds out that the Iranians aren't Arabs...[/quote][/hide]

I don't mean attacking them. What happened with Irak was terrible.
But I think communism has to stop now.

Well as said many times before, communism in its pure form hasn't been established anywhere, and countries like Myanmar which aren't communist may be worse in terms of for instance freedom of speech and human rights than Vietnam( I don't know, haven't visited either)

Restricting myself to Cuba for this one post :

What the USA DOES is imposing severe restrictions on travelling to Cuba for its citizens (my uncle, a Belgian who dropped his nationaly to adopt that of the USA, claims it comes down on basically no US'er being allowed to visit Cuba)(I invite better informed members to correct me/ elaborate) and an embargo, making it illegal for US companies to invest in Cuba.
I think this mainly affects the normal population and doesn't really force the leaders to give power back to the people. (Just like the embargo imposed on Iraq after the first Gulf War...)

So José, I think we basically agree then, you oppose invasion, but want something to be done.  But what?

One point though : I cannot evaluate Cuba's economy and other matters.  It is often said though that Cuba has a very decent healthcare system.

[hide="Offtopic reply to fredbel6"]It's hindsight, but if an ethnically divided Iraq with a civil war in progress is the best that could have been done, there would have been a much less painful way of getting there - in 1991.

To review: Iraq invaded and conquered Kuwait. The U.S., leading a substantial coalition (including very important support from both Saudi Arabia and Turkey) launched a counterattack that drove Iraq out of Kuwait and destroyed that part of the Iraqi army that had been in Kuwait, along with an unhealthy fraction of Iraq's infrastructure. At the end of that war, President George H.W. Bush encouraged Iraqis to rebel against Saddam Hussein - and just such a rebellion happened, concentrated in the major Shi'a cities of the south. Hussein sent in his Republican Guard troops (largely spared from the debacle in Kuwait) and attack helicopters and brutally crushed that rebellion. The U.S.-led alliance did not act to prevent this. Why not? Perhaps in an attempt to leave a Iraq (well, Iraq less Kurdistan) as a unified country; perhaps out of concern over Iranian influence.

That's the "what if": what if the U.S. had ordered the Iraqi government to ground its helicopters and leave its heavily-armored troops in their bases, promising to shoot down the helicopters and to destroy from the air any moving army on the highways? Given the U.S. and British air power available, that wouldn't have been an idle threat. Facing only light and police weapons, the rebellion might have become a Shi'a entity controlling southern Iraq and dividing Baghdad - an entity in which the most respected ayatollahs would likely have had more voice than the young hotheads. Is that any worse than the current situation? And all without any U.S. occupying forces as targets.[/hide]

[quote="bubka"]I have recently come to know that Castro is a traitor. USA has nothing to fear from him. The Cuban wing of the international is right now working against him. The same applies for the other countries named by fredbel6.

By the way, the left party is in power in Germany as well.[/quote]

Doesn't it strike you as rather odd to simultaneously declare Cuba non-communist and the center-right Christian Democratic Party to be the "left party" in Germany?

Jose, if communism is as bad as you say it is, then the countries' people will soon realize it and should switch expect maybe in the case of North Korea. China is a special case of communism, decades ago they were definitely more solid communist in terms of economy and general every life but now their economy is becoming slightly more captialist everyday but politically, their ways are still pretty much communist.

Anyway, the US would be a fool to attack places like China or Vietnam or Laos. China is the 2nd biggest loaner of money to the US at the moment with its huge debt and almost everything in the US is made with China because of the huge amount of trade. If items of China stop coming in, prices of everything will surge even more than they have now because of oil, and the economy will be in shambles, maybe even as bad as it was during the Great Depression because the rising cost of goods will cause hyperinflation to set in. Lastly, attacking any country right now especially Vietnam and Laos is idiotic for the government considering the highly unpopular and non-progressing physical war they're fighting in Iraq and Afghantistan and the decades of failed attempts to get peace in the Middle East. North Korea and Iran continue to not care about the US' warnings and deadlines are just passing as the countries stall and in the case of North Korea, start testing its missiles.

[quote="Ignite168"]and almost everything in the US is made with China because of the huge amount of trade.[/quote]

I know facts can be troublesome and all, but China is currently the third largest trading parter of the United States, after Canada and Mexico.  (Fourth, if you count "the EU" as a single unit instead of 25 different countries.)

US won't attack anyone other than potential communists or islamic fundamentalists.

Why do you think they created the base in Afghanistan? To attack Nepal of course. Oil and stuff are secondary.

US is not in any hurry to stop the war in Lebanon, for it knows that, while Israel is still a far cry form being a real threat, Lebanon can become the base of an Islamic terrorist group quickly.

It is important to note that, in the above discussions, I am not taking any sides.

[quote="fredbel6"]Uhm, Bubka, the USA has always had an eye on oil from the Caspian sea.  Pumping it through Pakistan is definitely possible, Iran is out of question.  Now that Afghanistan is a friend the oil can flow.
But wait, what is that weird country at the Caspian sea?
Turkmenistan : considered ultradictatorial, and ruled by former communists.
[img]http://www.ringnebula.com/Oil/pipeline.jpg[/img]

Nepal isn't even communist (yet).[/quote]

what about iran?

I said nepal is the closest to communist rule. No country has communism.

[quote="akalra1"]
what about iran?[/quote]

What is your question?

couldn't they just pump fuel trhough iran?

:|  Iran?!

Of course not! Iran is their "enemy", they're dangerously close to attacking the country at this point.

Ever since the shah was ousted from power by the ayatollahs, Iran has no diplomatic relations with the USA.

[quote="bubka"]They will attack iran soon, don't worry. They are getiing restless. Sensing trouble I presume. What, eh?  :wink:[/quote]
Dare I say- I mean literally- them and what army?  Afghanistan and Iraq do provide an interesting base to launch attacks but they can hardly control the situation in those two countries as it is and the war is already costing a fortune.
I think most things indicate no war is about to happen with Iran.  Remember their treats to shut off the Persian gulf?

[quote="Xevarion"]I think if Iran gets enough of the rest of the world pissed off by making nukes, the US might attack them. Especially if Republicans think they won't win the 2008 presidential election anyway.[/quote]
Aren't the october 2006 elections an obstruction already?  And now already they are not getting along perfectly anymore with the Iraqi and Afghan government, and they would like to use them to launch their attacks from both sides.

Point for all of you the governments in the "communist" countries is actually a dictatorship. Communism is when everone gets the same pay and is equal. That can't happen in the world because nearly everone is greedy and wants more money.

Also if Iran makes nukes America won't attack Iran as Iran could wipe out Isreal with the nuke. Note North Korea it has a confirmed nuke which pissed everyone off but no one attacked it.

North Korea doesn't have oil or muslims.

[quote="bubka"]North Africa has muslims but no oil.
(Incidentally as far as I know, the scene is hotting up for the communists in africa as well ...)[/quote]

Uhm you don't consider Nigeria,Chad,Algeria, the Western Sahara or Libya  North African?

Which communist groups in Africa are yo speaking of?  Because Laurent Kabila supposedly was a Marxist too...

No no not the revisionists. I mean the international.

DARN I THOUGHT I KILLED THIS TOPIC

[quote="bubka"][quote="junggi"][quote="José"]But I think communism has to stop now.[/quote]

but what can you do about it?[/quote]

Join the counter-recolutionaries perhaps?  :wink:[/quote]
ok
can I be a 5star general?
hey this could look good on my college app 4 years from now
"I served as general of the counter-revolutionaries"
yeah.

[quote="Ars"]Well it depends in a society of angels communism is the best but here with human greed socialism is.[/quote]

My thoughts exactly, I would use a modified version of socialism that has many capitalist viewpoints thrown in there

Since I am tired of people who understand neither socialism nor communism talking about them both, I will go back to junggi.

[quote="junggi"]DARN I THOUGHT I KILLED THIS TOPIC

[/quote][quote="bubka"][quote="junggi"][quote="José"]But I think communism has to stop now.[/quote]

but what can you do about it?[/quote]

Join the counter-recolutionaries perhaps?  :wink:[/quote][quote="junggi"]
ok
can I be a 5star general?
hey this could look good on my college app 4 years from now
"I served as general of the counter-revolutionaries"
yeah.[/quote]

It will work only if the counter revolutionaries win - which is unlikely  :(

[quote="bubka"]Since I am tired of people who understand neither socialism nor communism talking about them both, I will go back to junggi.

[/quote][quote="junggi"]DARN I THOUGHT I KILLED THIS TOPIC

[/quote][quote="bubka"][quote="junggi"][quote="José"]But I think communism has to stop now.[/quote]

but what can you do about it?[/quote]

Join the counter-recolutionaries perhaps?  :wink:[/quote][quote="junggi"]
ok
can I be a 5star general?
hey this could look good on my college app 4 years from now
"I served as general of the counter-revolutionaries"
yeah.[/quote][quote="bubka"]

It will work only if the counter revolutionaries win - which is unlikely  :([/quote]
can I still be the 5star general? :maybe:

[quote="junggi"][/quote][quote="bubka"]Since I am tired of people who understand neither socialism nor communism talking about them both, I will go back to junggi.

[/quote][quote="junggi"]DARN I THOUGHT I KILLED THIS TOPIC

[/quote][quote="bubka"][quote="junggi"][quote="José"]But I think communism has to stop now.[/quote]

but what can you do about it?[/quote]

Join the counter-recolutionaries perhaps?  :wink:[/quote][quote="junggi"]
ok
can I be a 5star general?
hey this could look good on my college app 4 years from now
"I served as general of the counter-revolutionaries"
yeah.[/quote][quote="bubka"]

It will work only if the counter revolutionaries win - which is unlikely  :([/quote][quote="junggi"]
can I still be the 5star general? :maybe:[/quote]

Ummm ... probably. Joining them is better than not joining anybody, in any case. I suppose there is no harm in trying. But you must work your way up the army. Five stars is not an easy thing, you know.  :|


I made a [url=http://en.wikipedia.org/wiki/Mass_point_geometry]mass points article on Wikipedia[/url] and while writing it learned some new things about mass points, which is involved in the problem below.  I'm interested in seeing how many people know of this method.

In triangle $ ABC$, $ D$, $ E$, and $ F$ are on $ BC$, $ CA$, and $ AB$ respectively so that $ AE \equal{} AF \equal{} CD \equal{} 2$, $ BD \equal{} CE \equal{} 3$, and $ BF \equal{} 5$. If $ DE$ and $ CF$ intersect at $ O$, compute $ \tfrac{OD}{OE}$ and $ \tfrac{OC}{OF}$.

This can be solved using splitting mass points:

Assign mass $ 3$ to $ A$, mass $ 2$ to $ C$, $ 5$ to $ E$, then there are two values for $ B$, $ 6/5$ for side $ AB$ and $ 4/3$ for side $ BC$. Then $ D$ has mass $ 2\plus{}4/3\equal{}10/3$, so $ OE: OD\equal{}10/3: 5$; $ F$ has mass $ 3\plus{}6/5\equal{}21/5$, so $ OC: OF\equal{}21/5: 2$

Does anyone have a solution without using mass points geometry? Also does anyone have a proof of the technique of splitting mass points?


limit
induction
calculus
calculus computations

Let $ (x_n)$ a sequence, $ x_0\equal{}0$ and $ x_n\equal{}\sqrt{n^2\plus{}x_{n\minus{}1}}, n \geq 1$. Find the limit $ \lim_{n \to \infty}(x_n\minus{}n)$.

$ \frac{x_n^2\minus{}n^2}{x_n\plus{}n}\equal{}\frac{x_{n\minus{}1}}{x_n\plus{}n}\equal{}x_n\minus{}n\equal{}\frac{x_{n\minus{}1}\minus{}(n\minus{}1)\plus{}(n\minus{}1)}{x_n\minus{}n\plus{}2n}$

$ y_n\equal{}x_n\minus{}n$

$ y_n\equal{}\frac{y_{n\minus{}1}\plus{}n\minus{}1}{y_n\plus{}2n} \to \lim_{n \to \infty} y_n\equal{}\frac{1}{2}$

[quote="kenn4000"]$ y_n \equal{} \frac {y_{n \minus{} 1} \plus{} n \minus{} 1}{y_n \plus{} 2n} \to \lim_{n \to \infty} y_n \equal{} \frac {1}{2}$[/quote]

Why?

assume $ \lim_{n \to \infty} y_n \equal{} L$

$ L \equal{} \lim_{n \to \infty} \frac {y_{n \minus{} 1} \plus{} n \minus{} 1}{y_n \plus{} 2n}$

$ \equal{} \frac {\lim_{n \to \infty}\frac{y_{n \minus{} 1}}{n} \plus{} 1\minus{}\frac{1}{n} }{\lim_{n \to \infty}\frac{y_n}{n} \plus{} 2}\equal{}\frac{1}{2}$

[quote="kenn4000"]assume $ \lim_{n \to \infty} y_n \equal{} L$

$ L \equal{} \lim_{n \to \infty} \frac {y_{n \minus{} 1} \plus{} n \minus{} 1}{y_n \plus{} 2n}$

$ \equal{} \frac {\lim_{n \to \infty}\frac {y_{n \minus{} 1}}{n} \plus{} 1 \minus{} \frac {1}{n} }{\lim_{n \to \infty}\frac {y_n}{n} \plus{} 2} \equal{} \frac {1}{2}$[/quote]

Warning , you have to  prove that  $ y_n$ is convergent  !  :D
First of all you observe (for ex. by induction)  that $ n \leq x_n  \leq n\plus{}1$ , so $ y_n \equal{} x_n\minus{}n$ is bounded , so it's convergent because $ \displaystyle \lim_{n \to \infty} \frac { \frac{y_{n \minus{} 1}}{n} \plus{} 1\minus{}\frac{1}{n} }{ \frac{y_n}{n} \plus{} 2}\equal{}\frac{1}{2}$
 :cool:


algorithm
number theory
greatest common divisor
relatively prime
Euclidean algorithm
number theory unsolved

[quote="lightrhee"]This is the first problem of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422487]the second round of 19th Korean Mathematics Olympiad[/url].
1.
For two positive integers a and b, which are relatively prime, find all integer that can be the great common divisor of $a+b$ and $\frac{a^{2005}+b^{2005}}{a+b}$.[/quote]

Let's use the Euclidean algorithm to find the GCD($a+b$,$\frac{a^{2005}+b^{2005}}{a+b}$)$=$
($a+b$,$a^{2004}-a^{2003}b+$...$+b^{2004}$)$=$($a+b$,$-2a^{2003}b+$...$+b^{2004}$)$=$...$=$($a+b$,$2005b^{2004}$)$=$($a+b$,$2005$)
As $2005=5*401$ (primes) $\Rightarrow$ the required integers are $1$ , $5$ , $401$ , $2005$
(We can take $a+b$ equal $1$ , $5$ , $401$ , $2005$ respectively and $b=1$)

Another solution:

Let $d$ be the greatest common divisor.

Since $d|a+b$, $a\equiv -b$ (mod $d$).

We also have $a^{2004} - a^{2003}b + a^{2002}b^2 - \ldots{} + b^{2004} \equiv 0$ (mod $d$)

Thus $b^{2004} + b^{2004} + \ldots{} + b^{2004} \equiv 0$ (mod $d$)

$2005b^{2004} \equiv 0$ (mod $d$).

So $d|2005b^{2004}$ and also $d|2005a^{2004}$. Since $a \perp b$, $d|2005$, and so we can have $d=1,5,401,2005$.

Again, this can be realized by taking $b=1$ and $a=1,4,400,2004$.

[quote="ZetaX"][quote="11235813"]Trash Problem.[/quote]
Why¿  :huh:
Well, it's easy, and I have seen it as homework for students, but why is it trash¿[/quote]

SO,I guess Korean students bad at math.

No idea how many rounds they have, but the german olympiad has four rounds, and the problems in round two or three are not harder...

this problem generalized form is merely an easy lemma thought to students in olympiads


Given the four expressions for positive reals xj, j = 1,2,3,4 :

E1 = ( \sum xj 2 ) 2
E2 = 16x1x2x3x4
E3 = ( \sum xj)( \sum xj 3 )
E4 = 4( \sum xj 4 )

Order the expressions above in ascending order.

E4 \geq E3 by Chebyshef.
E3 \geq E1 by Cauchy.
E1 \geq E2 by AM-GM.
So E4 \geq E3 \geq E1 \geq E2.

I think that E2<=E1<=E3<=E4 because
1)E2<=E1:
We use the MA-MG inequality for the numbers x1 2 ,x2 2 ,x3 2 ,x4 2.
So x1 2 +x2 2 +x3 2 +x4 2 >=4 \sqrt (x1x2x3x4).So (x1 2 +x2 2 +x3 2 +x4 2 )  2 >=16x1x2x3x4.
2)E1<=E3:
This is the Cauchy-Buniakovsky-Schwartz inequality.
3)E3<=E4:
For this one we use the equality:
(x1+x2+x3+x4)(x1^3+x2^3+x3^3+x4^3)=
x1^4+x2^4+x3^4+x4^4+sum(xi^3*xj) where i<>j.
We need now to demonstrate that sum(xi^3*xj)(with i<>j)<=3(x1^4+x2^4+x3^4+x4^4) which is obvious using the inequality
a1*b1+a2*b2+....an*bn<=a1*c1+a2*c2+...an*cn where c1<=c2<=...<=cn and b1,....,bn is a permutation of {c1,...cn}.
I think this is called the arangemets inequality.

[quote="Pungaru Marius"]I think this is called the arangemets inequality.[/quote]
actually it's called [b]re[/b]-arrangement inequality :D

actually, I think it's all just the same with what arne wrote...


geometry
3D geometry
graph theory
combinatorics solved
combinatorics

Prove that  an equilateral triangle can't be partitioned into pairwise distinct equilateral traingles.
(Note that a square CAN be partitioned into incongruent squares, there is an example in "Graph theory - An introduction course" written by B.Bollobas)

Are you sure a square can be divided in squares of diferent sizes? For example, I know for sure that a cube can't be divided into cubes of diferent sizes and seems to me that a similar argument works for squares.

First, an answer to Dzeta :
Yes, it is possible to dissect a square into smaller squares each of different sizes.
The number of squares used in such a dissection is called the order of the dissection. The lowest possible order is 21, and has been found with computer by Duijvestijn in 1978.
It is difficult to give it here, but the squares on its boundary and from the left-upper one and in counterclock direction have respective sides :
50 (left-up), 35, 27 (right-up), 19, 24, 42 (right down), 37, 33 (left-down), 29.

Then, use 4, 25, 16, 9, 7, 18, 2, 15, 17, 6, 11, 8 to fill the interior.

Have fun with this puzzle  :D

Pierre.

Now, let's answer the original problem :

Suppose, for a contradiction, that a tiling of the equilateral triangle T is possible with n  \geq  2  pairwise incongruent triangles.

- First, it is easy to see that if T' and T'' are two triangles from this tiling, then  they have parallel sides :
Since there is  finite number of tiles, it is possible to draw a line segment from an interior point of T' to an interior point of T''. Each point where this segment intersects the boundary of some tile is then a common point to the sides of two neighbour triangles. Then these two triangles have a parallel side. Since they are equilateral, it ensures us that their three sides are parallels. Since it is true for any two neighbour triangles intersected by the line segment, it is trus for T' and T''.
And then, using tiles which have common sides with T, we deduce that they all have their sides parallel to the sides of T.

Now, we say that triangle T' stings triangle T'' if one vertex of T' is on the boundary of T'' but is not a vertex of T''.
Let E be the set whose elements are T and each of the tiles.

Lemma 1.
Let ABC be in E. If the side [AB] is not entirely contained in the side of another element of E then there exists a triangle which stings ABC.

Proof.
Either ABC is T, and the result is clear, or [AB] is not in a side from T. In that case, since the tiles are pairwise incongruent :
- A is a vertex of another tile, with vertex M in ]AB[, or
- A is on the sides but not the vertex of another tile with vertex M in ]AB[.
In both cases, we get the result.

Lemma 2.
Let ABC be in E. The number of triangles which sting ABC is greater than or equal to the number of triangles stinged by ABC.

Proof.
Let p be the number of elements of E which are stinged by ABC. Then p = 0,1,2,3.
- If p = 0, the result is obvious.
- If p = 1 : Suppose that ABC stings S in A. Then [AB] and [AC] are not sides of T. Thus : Either [AB] or [AC] satisfies the statement of the lemma 1, and the result holds. Or [AB] and [AC] are entirely contained in sides of two triangles bigger than ABC, and in that case [BC] satisfies the statement of the lemma 1, and we are done.
- if p = 2 : Suppose that ABC stings T' and T'' at A and B respectively. Thus [AB] satisfies the statement of lemma 1. Moreover, if [BC] does not satisfies the statement then [BC] is entirely contained in a side of a triangle bigger than ABC, and so [AC] satisfies the statement of lemma 1. Thus ABC is stinged at least two times..
- If p = 3, then clearly the three sides of ABC satisfy the statement of the lemma 1, and the result holds.

Now, we end the proof of the initial problem.... :
The total number of points where the elements of E sting other elements of E is equal to the total number of points where the elements of E are stinged by other elements of E. From lemma 2, this forces each element ABC of E to sting exactly the same number of elements of E that there are elements of E which sting ABC. In particular, for ABC = T. But T does not sting any element of E, so it is not stinged, which is impossible since there is at least 2 tiles....A contradiction.

Pierre.


geometry
Locus problems
polygon
rotation
Locus
IMO
IMO 1986

Let $A,B$ be adjacent vertices of a regular $n$-gon ($n\ge5$) with center $O$. A triangle $XYZ$, which is congruent to and initially coincides with $OAB$, moves in the plane in such a way that $Y$ and $Z$ each trace out the whole boundary of the polygon, with $X$ remaining inside the polygon. Find the locus of $X$.

Let $ C \not \equal{} A$ the vertex which is adjacent to $ B$. While $ XYZ$ moves from $ OAB$ to $ OBC$, it is easy to see $ XYBZ$ is cyclic. Thus $ X$ lies on the bisector of $ \angle YBZ \equal{} \angle ABC$. Moreover, $ X$ is the intersection of a circle passing through $ B$ (the circumcircle of $ XYBZ$) and with a fixed radius (the radius is a function of $ \triangle XYZ$). Therefore $ X$ varies in a line segment ended in $ O$. When $ Y$ and $ Z$ pass through the other sides, we get as locus $ n$ distinct line segments, each passing throught $ O$ and contained in $ OV$ (but not in $ \vec{OV}$) for some vertex $ V$ of the polygon. Each two of these lines are obtained one from another by a rotation with center $ O$.


function
calculus
derivative
trigonometry
calculus computations

Let f''(x) + f(x) = 0 and

f'(0) = f(0) = 0

Prove that f(x)=f'(x) = 0 for every x.

How to prove that f'' is constant? Or f' is constant?

Multiply both sides of the first equation by $ f'$ to get $ f'f''\plus{}ff'\equal{}0$.  Therefore, $ [(f')^{2}\plus{}f^{2}]'\equal{}2(f'f"\plus{}ff')\equal{}0$, so $ (f')^{2}\plus{}f^{2}$ must be a constant function.  Since $ f(0)\equal{}0$ and $ f'(0)\equal{}0$, this constant function must be 0.  Therefore $ (f'(x))^{2}\plus{}(f(x))^{2}\equal{}0$ for all $ x$, which of course is possible if and only if $ f(x)\equal{}0$ for all $ x$.

[hide="Exercise"]If $ f$ has a second derivative everywhere and $ f''\plus{}f\equal{}0$, $ f(0)\equal{}a$, and $ f(0)\equal{}b$, show that $ f(x)\equal{}b\sin{x}\plus{}a\cos{x}$.[/hide]

hello, setting $ f(x)=e^{\lambda x}$ we get all solutions of your equation in the form
$ f(x)=C_1\sin(x)+C_2\cos(x)$
From here we find
$ f^{'}(x)=C_1\cos(x)-C_2\sin(x)$
Now you can calculate $ C_1$ and $ C_2$ with the condition $ f^{'}(0)=f(0)=0$
Sonnhard.

The existence-uniqueness theorem for differential equations can be used here.

This is a properly posed ODE with an initial condition; the solution must be (at least locally) unique. Since $ f(x)\equal{}0$ is a solution by inspection, the existence-uniqueness theorem says there are no others.

JRav's approach can be thought of as motivated by physics. The quantity $ (f')^2\plus{}f^2$ is (a constant multiple) of the total energy of the system, with $ (f')^2$ being a multiple of the kinetic energy and $ f^2$ a multiple of the potential energy.

[quote="JRav"]Multiply both sides of the first equation by $ f'$ to get $ f'f'' \plus{} ff' \equal{} 0$.  Therefore, $ [(f')^{2} \plus{} f^{2}]' \equal{} 2(f'f" \plus{} ff') \equal{} 0$, so $ (f')^{2} \plus{} f^{2}$ must be a constant function.  Since $ f(0) \equal{} 0$ and $ f'(0) \equal{} 0$, this constant function must be 0.  Therefore $ (f'(x))^{2} \plus{} (f(x))^{2} \equal{} 0$ for all $ x$, which of course is possible if and only if $ f(x) \equal{} 0$ for all $ x$.

[hide="Exercise"]If $ f$ has a second derivative everywhere and $ f'' \plus{} f \equal{} 0$, $ f(0) \equal{} a$, and $ f(0) \equal{} b$, show that $ f(x) \equal{} b\sin{x} \plus{} a\cos{x}$.[/hide][/quote]

Thank you for the direct solution.

[quote="Kent Merryfield"]The existence-uniqueness theorem for differential equations can be used here.

This is a properly posed ODE with an initial condition; the solution must be (at least locally) unique. Since $ f(x) \equal{} 0$ is a solution by inspection, the existence-uniqueness theorem says there are no others.

JRav's approach can be thought of as motivated by physics. The quantity $ (f')^2 \plus{} f^2$ is (a constant multiple) of the total energy of the system, with $ (f')^2$ being a multiple of the kinetic energy and $ f^2$ a multiple of the potential energy.[/quote]

We did not cover any DE. This was given in Analysis class, when we cover the Main Theorem of Calculus.

[quote="Dr Sonnhard Graubner"]hello, setting $ f(x) = e^{\lambda x}$ we get all solutions of your equation in the form
$ f(x) = C_1\sin(x) + C_2\cos(x)$
From here we find
$ f^{'}(x) = C_1\cos(x) - C_2\sin(x)$
Now you can calculate $ C_1$ and $ C_2$ with the condition $ f^{'}(0) = f(0) = 0$
Sonnhard.[/quote]

[quote="Dr Sonnhard Graubner"]hello, what kind of question do you have?
Sonnhard.[/quote]

Sorry, I had to restart my computer in the middle of the posting and it seems some got lost.

But the good news is that I worked out the problem I asked.
I didin't know how to prove that all solutions have the form A.cos(x) + B.sin(x) but I did it now.

Thank you for your attention and thanks to all replies!


search
greatest common divisor
combinatorics unsolved
combinatorics

On the segment $ [0,N]$, the endpoints and two more points are marked so that the lengths of the three segments into which the segment $ [0,N]$ is divided are integers with the greatest common divisor $ 1$. If $ A$ and $ B$ are two marked points whose distance is a multiple of $ 3$, then we can divide the segment $ AB$ into three equal parts, mark one of the division points and unmark one of the points $ A,B$. Is it true that we can mark any given integer point on the segment $ AB$ with several such operations?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=797233841&t=209640 is exactly the same problem, http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=797233841&t=205176 is a similar-looking problem that maybe implies the desired result?  (I searched for "mark* operation*" to find these threads.)


inequalities
geometry
geometric transformation
reflection
homothety
circumcircle
geometry unsolved

Each vertex of a triangle is reﬂected in the opposite side. Prove that the
area of the triangle determined by the three points of reﬂection is less than
ﬁve times the area of the original triangle

It was a problem from competition in Hungary 1986

Reflection triangle $\triangle A'B'C'$ of $\triangle ABC$ is homothetic to the pedal triangle $\triangle A_0B_0C_0$ of its 9-point center $N$ through the homothety with center $G$ and factor $4.$ For a proof see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=353274]here[/url], so $[A'B'C']=16[A_0B_0C_0].$ Hence, we shall prove $16[A_0B_0C_0] <5[ABC].$ Let $O$ be the circumcenter of $\triangle ABC.$ By Euler's theorem for $\triangle A_0B_0C_0,$ we have

$\frac{[A_0B_0C_0]}{[ABC]}= \frac{|p(N,(O))|}{4R^2} \Longrightarrow  4\cdot |p(N,(O))|< 5R^2$

Assume that $N$ lies outside  $(O),$ otherwise the inequality is trivial. Then $4(ON^2-R^2)<5R^2$ $\Longrightarrow$ $ON^2 < \frac{_9}{^4}R^2.$ Since $ON=\frac{_3}{^2}OG,$ it follows that $\frac{_9}{^4}OG^2< \frac{_9}{^4}R^2$ $\Longrightarrow$ $OG <R,$ which is clearly true since $G$ always lies inside $(O).$


logarithms
trigonometry
geometry
geometric transformation
reflection
calculus
calculus computations

another question i asked in class on $ \Gamma(x)$.... no, no..... don't thank me, thank Carcul !!  :rotfl:

prove, for $ |x| < 1$,

$ \sum_{n = 1}^\infty\;\dfrac{(\zeta(2n + 1) - 1)}{2n + 1}\;x^{2n + 1} \quad =$
\[ \boxed{(1 - \gamma)x - \ln\sqrt {\dfrac{\sin\,(\pi\,x)}{\pi\,x}} - \ln\,(\Gamma(1 + x)) - \ln\sqrt\dfrac{1 + x}{1 - x}}}
\]

Hint:
by using $ \Gamma(z) \equal{} \frac {n!n^{z}}{z(z \plus{} 1)...(z \plus{} n)}$ we can easy find serie for $ \ln \Gamma(z)$ and $ \ln \Gamma(z \plus{} 1)$ by using Stirling ($ \ln n! \equal{} ...$)  formula and  taylor formula for
$ \ln (1 \plus{} x)$ when $ |x| < 1$
we can also easy calculate serie for $ \ln \sqrt {\frac {1 \plus{} x}{1 \minus{} x}}$ just take $ ... \equal{} \frac {1}{2}(\ln (1 \plus{} x) \minus{} \ln (1 \minus{} x))$ and use taylor formula .
our aim is to calculate serie for $ \ln \sqrt {\frac {\sin \pi z}{\pi z}}$ it's easy by using  Euler's reflection formula. I mean :
$ \Gamma(1 \minus{} z)\Gamma(z) \equal{} \frac {\pi}{\sin \pi z}$ just take $ \ln$.
After just   substitute , also few  :lol:  calculation and we are done .


geometry
trapezoid
geometric transformation
homothety
power of a point
radical axis
geometry unsolved

(V.Yasinsky, Ukraine) Suppose $ X$ and $ Y$ are the common points of two circles $ \omega_1$ and $ \omega_2$. The third circle $ \omega$ is internally tangent to $ \omega_1$ and $ \omega_2$ in $ P$ and $ Q$ respectively. Segment $ XY$ intersects $ \omega$ in points $ M$ and $ N$. Rays $ PM$ and $ PN$ intersect $ \omega_1$ in points $ A$ and $ D$; rays $ QM$ and $ QN$ intersect $ \omega_2$ in points $ B$ and $ C$ respectively. Prove that $ AB \equal{} CD$.

A,B,P,Q are concyclic
L tangent to ⊙O at M , L1 tangent to ⊙O1 at A , L2 tangent to ⊙O2 at B
Then L1∥L2∥L
But ∠BAP=∠BQP=∠PML => AB∥L
=> AB=L1=L2
=> AB is the common tangent of ⊙O1 and ⊙O2
Analogously,CD is the common tangent of ⊙O1 and ⊙O2
=> AB=CD

M being on radical axis of w1 and w2, we get AM.PM=BM.MQ, therefore ABPQ is cyclic, so  ÐBAM=ÐBQP ( 1 ). From MPNQ cyclic we get ÐPMN=ÐPQN ( 2 ). The circles w and w1 are homothetic at P, hence AD||AD, so ÐMAD=ÐPMN ( 3 ). From (1), (2) and (3) we get ÐBAD = ÐMQN. ( 4 )
Analogously we have QPCD cyclic and get ÐADC = ÐMQN ( 5 ), also w and w2 homothetic and BC||MN, hence AD||BC and, with (4) and (5) ABCD is an isosceles trapezoid, done. (Here the sign Ð means "angle").

Best regards,
sunken rock

I guess this is similar to sunken rock's solution, except worded differently.

Consider the homothety sending $\omega$ to $\omega_1$.  This sends $M$ to $A$ and $N$ to $D$, and as such $AD\parallel MN$.  Similarly, $BC\parallel MN$ and therefore $AD\parallel BC$.

Now, since $XY$ is the radical axis of $\omega_1$ and $\omega_2$, we have $AM\times MP=BM\times MQ$, so $AQPB$ is cyclic.  Similarly, $DQPC$ is cyclic.  Hence, by angle chasing with what we know, we find \begin{align*}\angle DAP=\angle NMP=\angle NQP=\angle NDC,\qquad \angle BAM=\angle MQP=\angle MNP=\angle ADP.\end{align*}  Therefore $\angle BAD=\angle ADC$, so $\triangle BAD=\triangle CDA\implies AB=CD$ as desired.  $\blacksquare$

Solution that avoids messy angle chasing (too lazy to do it :)):

Let O_1 and O_2 be the centers of w_1 and w_2, respectively.

Consider the homothety that sends w to w_1, which sends MN to AD. Thus, AD || MN. Similarly, BC || MN so AD || BC || XY.

Now note that O_1O_2 is perpendicular to XY, and since AD and BC are parallel to XY, O_1O_2 is perpendicular to both AD and BC. Now, we consider a reflection of the diagram across O_1O_2. Let A be mapped to A'. Clearly, A' is a point on w_1. Since AD is perpendicular to O_1O_2, A' must be a point on AD. However, the only point that satisfies these conditions is D, so A is mapped to D. Similarly, B is mapped to C. Finally, since reflection does not change lengths, AB = CD. QED


i think 1001 to the 999th power sounds bigger.

But Farenhajt is always right.  :lol:


AMC
AIME
USA(J)MO
USAMO
geometry
trigonometry
geometric transformation

I got 11 - all of the first 12 except for #5.

I got 9. I really want to say FAIL but when I look at it, it's not really that fail at all.

I guess I wanted to say FAIL because I could've gotten a 12.

[color=darkred]Edit: I answered 1-9, 11, 12, 14. I got 7, 9, and 14 wrong, so I got 1-6, 8, 11, 12 right.[/color]

9 (1-5, 7-8, 11, 14)

I missed 6 because i said 4^4=64, so my answer was (256-64) too big >.>
And then I don't know combinatorics/counting whatsoever so I had no chance and made absolutely no progress on 9 and 10.

8:
1-3, for 4 i thought 3009/17 was 144..., 5,7, for 8 i forgot to subtract 10*2^0, for 9 i overcounted..., 10-12

Got 11

Missed Interplanetary Council and the Prize Money one

Didn't do Circle in the Triangle or the last one

10, I got all of the first 11 except #5.

9. Gah i cant believe i got number 1 wrong, and number 10 wrong cuz i thought it was MVE instead of MEV.

3, go me!!! Yay, I'm terrible, but my goal was 2. Got #1,6,9

Attempted the first 11, got 6. I'll just say that I made two more careless mistakes in addition to the three I made on the AMC 12A.

This just isn't my year...

8... +4 from last year, yay.
Made USAMO at least ><
<- got 1 :lol:
1-3, 6-8, 13-14

Wait! How do you know what you got?

Are the answers posted? :help:

Sorry if I missed something  :ewpu:

11 here...I got all the non-geometry problems except #13 and none of the geometry problems except the analytic one (#11). (I'm really bad at geometry)

I tried to grind through #5 with Angle Bisector Theorem and Law of Cosines, that wasted a good 45 minutes (I got it to an expression for the answer that I later solved with a calculator in <10 seconds, but would have taken a good 15-30 minutes by hand) that I probably should have spent on #13, oh well.

First 14.
I observed all the things on number 15 except for the 60 degree.

6. I placed an emphasis on going slowly and trying to avoid careless errors. But....

On #14, I tried to verify my solution worked for the equation a+16b+81c+256d=4745 and added wrong several times, until I finally added once and got it to come out right. Of course, I had actually been adding right except the last time, and my mistake came from finding integer solutions to the equation 51c+211d=1950 instead of 50c+210d=1950.

On #11, which I did toward the end, I found the answer 600, but then thought "What? An AIME answer can't be a nice round number!" and thought some more about the distinct condition, and, aided by a poor visualization, decided that reflections about the line from the origin to the midpoint of the quadrant I section of the line y=2009-41x yielded non-distinct triangles, meaning each triangle not reflected onto itself by that transformation would be double-counted, so I changed my answer to 324. Oops...

On #6, I both added wrong and missed that there was a solution for 1.

And, on #1, I totally failed, forgetting that the sequence could have a nonintegral common ratio.

But hey, I did manage to catch a careless error on #13.

So, in summary, 10 questions answered, with one wrong due to careless error, two wrong due to small errors, and one wrong due to a FUBAR. I find it quite funny that the careless error was on #14 and the FUBAR  on #1...

Still, on many practice tests (and last year's real AIME) I'd answer 9 questions and get 5 wrong due to careless errors, so this an improvement on that. Of course, I'd usually only give myself a little more than 1.5 hours on practice tests...

Anyway, I should say: The problems I got right were 2-4, 7-8, and 13. I'm quite embarrassed at not getting #5 and #9. #5 especially because I found the parallel lines and looked for a while at how to use Angle-Bisector Theorem (and then spent a little time [albeit not 45 minutes] looking at how to use Stewart's Theorem and the Angle Bisector Theorem to find the lengths of AP and PQ where Q is the intersection of AP with BC).

Yay I doubled my score from last year! 12 (1-9,11,12,14). I didn't make any stupid mistake this time (spent a lot of time checking)  :)

[quote="azndemon43"]About when can we expect official scores to return?[/quote]
Next week maybe?

Missed #1 like the majority of people...  :wallbash_red:  :wallbash_red:  :wallbash_red:
even worse, I actually spent time to check it too...

[quote="gauss1181"]I absolutely failed with a 6..........................................................................=(
I got #1, 9, and 14 wrong even though they were so easy....if I hadn't made stupid computational errors I woulda just gotten a 9 and made USAMO.....................................................................................
dang it dang it dang it dang it I ABSOLUTELY HATE MYSELF!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Got #2, 3, 4, 6, 7, 8 right.

[size=200]DARN I EPICLY PHAILED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! im gonna smack my head on the wall[/size]

 :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:[/quote]

sigh...9. I was hoping for a >1 point improvement from last year...

@gauss: How do you post that many emoticons? did they change the "3 emoticons per post" rule?

i got 6 right..fail
ok so i solved 11
then i found out two of the solutions were wrong
that wasn't too bad..still could've gotten a 9
but then....three arithmetic errors (especially #7 and #8)
sigh

Ehhh... this was a really bad AIME for me.  I never got #5, somehow got 1/4 of the right answer on #9, and my guess on #14 was less than 513... Oh yeah, and I failed at adding on #1, #2, and #8, so I got #3, #4, #6, #7, #11 for a 5, worse than last year  :(

hmm I got my lowest score since 8th grade. maybe it is useful to not set sig goals (oh darn I will fail my sig goal pretty miserably :( )

meh....9
3. i reversed everything somehow, and got 1/6
6. 4^4=64
7. i misread as a_n-a_n-1 instead of a_n+1-a_n...boohoo

Im so mad...should have had 12...:(

This was the first time I took the AIME, so I only got a 4. But I only answered 4 of them!! :rotfl:

I got 5, pretty good for someone that never saw AIME problems until this February.  I messed up on 12, I couldn't see how to do it.  I even tried Ceva, but I got the 1=1, which is obvious.  Messed up on one, got 807 instead of 840.    :(

[quote="ytrewq"]10 :D !
Twice what I got last year!
I half guessed on number 13 and got it right :lol: ![/quote]
one of my classmates guessed on one of them and got it one off xD
too bad for him D:

[quote="Illos"]Got 7, counted the evens case wrong on number 11 for some reason and got 576 instead of 600. My main problem was that I spent a loooooong time on number 4 since I failed to realize 3009 was divisible by 17. A disappointing performance for my first and last AIME.[/quote]

haha i got 7 too, also because i missed 11...but i got 624, 24 more than the answer rather than 24 less :(

I got 8, 1-8 besides 5 and 11.  I'm happy with my score, +2 from last year, this gets me to USAMO (I hope at least...), could have solved 5, 9-10, but don't really feel bad about not getting them

I only got a 3 (4, 5, and 6), but this is my first time taking the AIME and I had no time to prepare at all so it's three more than I was expecting. On number 1 I forgot that the common ratio can be less than one so I had 469 as the largest instead of 964, and on the one with the differences between elements in the set (I forget which number that one was) I forgot all about 10^10 for some reason. All the others I got wrong based on faulty logic or not knowing how to do them.

I'm surprised that not many people here got number 5; it was the only one I was absolutely sure of. Has anyone here heard of mass points? It made the problem ridiculously easy.

does anyone know links to articles about mass points BESIDES the link to the berkeley math circle one?


if $ \sqrt{b} \plus{}\sqrt{c} ,\sqrt{c} \plus{}\sqrt{a} ,\sqrt{a} \plus{}\sqrt{b}$ are in Harmonic progression,then $ a,b,c$ are in??????????

As $ \sqrt{b} \plus{}\sqrt{c} ,\sqrt{c} \plus{}\sqrt{a} ,\sqrt{a} \plus{}\sqrt{b}$ are in HP, it follows that

$ \frac{1}{\sqrt{b} \plus{}\sqrt{c}}, \frac{1}{\sqrt{c} \plus{}\sqrt{a}},\frac{1}{\sqrt{a} \plus{}\sqrt{b}}$ are in AP.

Hence

$ \frac{1}{\sqrt{c} \plus{}\sqrt{a}} \minus{} \frac{1}{\sqrt{a} \plus{}\sqrt{b}} \equal{}   \frac{1}{\sqrt{b} \plus{}\sqrt{c}} \minus{} \frac{1}{\sqrt{c} \plus{}\sqrt{a}}$

$ \frac{\sqrt{b} \minus{} \sqrt{c}}{\sqrt{a} \plus{}\sqrt{b}} \equal{} \frac{\sqrt{a} \minus{} \sqrt{b}}{\sqrt{b} \plus{}\sqrt{c}}$

$ (\sqrt{b} \minus{} \sqrt{c}).(\sqrt{b} \plus{}\sqrt{c}) \equal{} (\sqrt{a} \plus{}\sqrt{b}).(\sqrt{a} \minus{} \sqrt{b})$

$ b\minus{}c \equal{} a \minus{} b$

OR $ b \equal{} \frac{a\plus{}c}{2}$  :)

omg
how dumb i am

i missed out this simple question in quiz
simple calculation mistake

thank you very much :)

@rt_08
1,4,6,8,[b]3[/b]???????
are you sure

YA ITS THREE !
AND THATS WHERE I FUMBLED!

then its beyond my reach

btw,i'll try again