calculus
integration
logarithms
real analysis
real analysis unsolved

an   integral i just made up, LOL  :D

show

$ \int_0^1\;\dfrac{\textbf dx}{\sqrt {\; \ln\;\left(\ln\; \dfrac{1}{x}\; \right)\; }}\; \equal{} \;\boxed{\sqrt \pi}$

Now tell me why the RHS of your formula contains no imaginary part.

Isn't it  $ \int_{0}^{1} \frac{dx}{\sqrt{ln(\frac{1}{x})}}\equal{}\sqrt{\pi}$ ?

If we put here $ \frac {1}{x} \equal{} e^t$, we get:
$ \int_{0}^{1} \frac {\textbf dx}{\sqrt {\ln (\frac {1}{x})}} \equal{} \int_{0}^{\infty}{e^{ \minus{} t}t^{ \minus{} \frac {1}{2}}\textbf dx} \equal{} \Gamma\left(\frac {1}{2}\right) \equal{} \sqrt {\pi}$

and if $ x>\frac{1}{e}$ then $ \ln \ln\left(\frac{1}{x}\right)<0$

oh ! sorry....  :(

you see, the computer which i use to connect to the internet is   old, and i sometimes see things on the screen that looks OK after i type (preview window), but apparently it is coming out  wrong !

you are correct, the integral i wanted to pose was

$ \int_0^1\dfrac{1}{\sqrt {\minus{}\,\ln\,x } } \;\textbf dx$

(maybe i typed in an extra $ \ln$ who knows...   :maybe: )

anyway, i was looking at $ \int_0^1x^m\cdot \left(\minus{}\,\ln\,x\right)^n\;\textbf dx\;\equal{}\;\dfrac{\Gamma(n\plus{}1)}{(m\plus{}1)^{n\plus{}1}}$

( where $ m,n>\minus{}1$ )

and thought i would ask that question. well, you guys got it right even with the wrong question.  :lol:

[quote="misan"]
you are correct, the integral i wanted to pose was

$ \int_0^1\dfrac{1}{\sqrt { \minus{} \,\ln\,x } } \;\textbf dx$

(maybe i typed in an extra $ \ln$ who knows...   :maybe: )[/quote]
After substitution $ z \equal{} \minus{} \ln x$ yields the well-known gaussian integral and the result clearly follows.

[quote="misan"]anyway, i was looking at $ \int_0^1x^m\cdot \left( \minus{} \,\ln\,x\right)^n\;\textbf dx\; \equal{} \;\dfrac{\Gamma(n \plus{} 1)}{(m \plus{} 1)^{n \plus{} 1}}$

( where $ m,n > \minus{} 1$ )[/quote]
After makin' $ z \equal{} \minus{} \ln x$ the integral becomes $ \int_0^\infty {e^{ \minus{} z(m \plus{} 1)} z^n \,dz} \equal{} \frac {{n!}} {{(m \plus{} 1)^{n \plus{} 1} }}.\quad\blacksquare$


Let $ p_i \geq 2$, $ i \equal{} 1,2, \cdots n$ be $ n$ integers such that any two of them are relatively prime. Let:

\[ P \equal{} \{ x \equal{} \sum_{i \equal{} 1}^{n} x_i \prod_{j \equal{} 1, j \neq i}^{n} p_j \mid x_i \text{is a non \minus{} negative integer}, i \equal{} 1,2, \cdots n \}
\]
Prove that the biggest integer $ M$ such that $ M \not\in P$ is greater than $ \displaystyle \frac {n \minus{} 2}{2} \cdot \prod_{i \equal{} 1}^{n} p_i$, and also find $ M$.

And the answer is $$\prod ^{n}_{i=1}p_{i}\cdot \left( n-1-\sum ^{n}_{k=1}\dfrac {1}{p_{i}}\right) $$

This is just a (pretty straightforward) generalization of problem IMO 1983/3.
(The same argument works for both problems.)

http://www.artofproblemsolving.com/community/c6h60800p366618


Αποδείξτε ότι για τους πραγματικούς αριθμούς $ a,b,c$ ισχύει

$ 4\left( \sum_{cyclic}a^{2}b^{2}\minus{}abc\sum_{cyclic}a\right)\left( \sum_{cyclic}a^{4}\minus{}
\sum_{cyclic}a^{2}b^{2}\right)\geq 3\left( \sum_{cyclic} a^{3}b\minus{}abc\sum_{cyclic}a\right)^{2}$

Δεν την έχω καλοκοιτάξει, αλλά νομίζω ότι βγαίνει με Moirhead και Holder.

Αν φέρουμε το αριστερό μέλος στο δεξί αρκεί να αποδείξουμε ότι η διακρίνουσα του τριωνύμου
$ T\equal{}3(\sum x^2y^2\minus{}xyz\sum x)r^2\minus{}3(\sum x^3y\minus{}xyz\sum x)r\plus{}\sum x^4\minus{}\sum x^2y^2$  Αρkeί νδο $ T\geq 0$

Είναι $ 3(\sum x^2y^2\minus{}xyz\sum x)\equal{}\frac{1}{2}\sum (xy\minus{}2yz\plus{}zx)^2$  ,

$ 3(\sum x^3y\minus{}xyz\sum x)r\equal{}\sum (x^2\minus{}y^2)(xy\minus{}2yz\plus{}zx)$ και $ \sum x^4\minus{}\sum x^2y^2\geq\frac{1}{2}\sum (x^2\minus{}y^2)^2$

άρα αρκεί να δείξουμε ότι $ \frac{1}{2}\sum (xy\minus{}2yz\plus{}zx)^2 r^2\minus{}r\sum (x^2\minus{}y^2)(xy\minus{}2yz\plus{}zx)\plus{}\frac{1}{2}\sum (x^2\minus{}y^2)^2\geq 0$ (1)

Τώρα παρατηρείστε ότι το μεσαίο είναι το διπλάσιο γινόμενο γινόμενο των ακραίων άρα η (1) είναι ισοδύναμη με
$ \frac{1}{2}\sum (x^2\minus{}y^2\minus{}rxy\plus{}2ryz\minus{}rzx)^2\geq 0$ , το οποίο είναι και το ζητούμενο .

Από την παραπάνω λύση βλέποντας την περίπτωση ισότητας κάποιος μπορεί να βγάλει απευθείας ότι η δοσμένη ανισότητα είναι ισοδύναμη με
$ (\sum x^2(xy\plus{}yz\minus{}2zx))^2\geq 0$    :wink:

[quote="silouan"]
Από την παραπάνω λύση βλέποντας την περίπτωση ισότητας κάποιος μπορεί να βγάλει απευθείας ότι η δοσμένη ανισότητα είναι ισοδύναμη με
$ (\sum x^2(xy \plus{} yz \minus{} 2zx))^2\geq 0$    :wink:[/quote]

 :P wow!ton elegxo ton ekanes esu sil i i mathematica??

[quote="silouan"]Δεν έκανα έλεγχο  :P  Έλυσα το σύστημα

$ \sum (x^2(kxy \plus{} kyz \minus{} lzx))^2 \equal{} 4(\sum a^2b^2 \minus{} abc\sum a)(\sum a^4 \minus{} \sum a^2b^2) \minus{} 3(\sum a^3b \minus{} abc\sum a)^2$

Αν ξέρει κανείς να τα ανοίγει συμμετρικά (βλέπε λύση silouan IMO 2008) οι πράξεις είναι πολύ σύντομες  :wink:[/quote]

exeis dikio...alla thelei ligi eksaskisi ..right?

[quote="silouan"]Δεν έκανα έλεγχο  :P  Έλυσα το σύστημα

$ \sum (x^2(kxy \plus{} kyz \minus{} lzx))^2 \equal{} 4(\sum a^2b^2 \minus{} abc\sum a)(\sum a^4 \minus{} \sum a^2b^2) \minus{} 3(\sum a^3b \minus{} abc\sum a)^2$

Αν ξέρει κανείς να τα ανοίγει συμμετρικά (βλέπε λύση silouan IMO 2008) οι πράξεις είναι πολύ σύντομες  :wink:[/quote]

Δεν καταλαβαινω πως απο τη λυση καταλαβαινουμε οτι η διακρινουσα θα εχει αυτη τη μορφη. Ισος να ειναι προφανες αλλα εχω μπερδευτει...
Μηπως μπορεις να το εξηγησεις λιγο αν εχεις χρονο?

Σας εμπλεξα λίγο  :maybe:  Λοιπόν παραπάνω έχω παρουσιάσει δύο εντελώς διαφορετικές προσεγγίσεις . Η πιο νορμάλ και εντελώς φυσιολογική είναι αυτή με τη διακρίνουσα . Η δεύτερη κάνει χρήση της άλλης ταυτότητας . Και λέω ότι την ταυτότητα μπορούμε να την πάρουμε με δύο τρόπους . Ο πρώτος είναι να λύσουμε το σύστημα με τα $ k,l$ και ο δεύτερος είναι με πονηριά (δηλαδή κοιτάμε την πρώτη λύση , βλέπουμε που έχουμε ισότητα και τους συντελεστές της ισότητας τους χρησιμοποιούμε ως όρους ενός αθροίσματος το οποίο το υψώνουμε όλο στο τετράγωνο )
Hope it is more clear now  :)   . Αλλά θα έλεγα ότι η πρώτη λύση είναι αυτή που αξίζει . Με την κατασκευή του τριωνύμου και την πολύ λογική μάζεψη του τετραγώνου  :wink:

Ναι, ομως δεν μπορω να καταλαβω στον 1ο τροπο γιατι λυνουμε το συγκεκριμενο συστημα... Γνωριζουμε στο περιπου τη μορφη που θα εχει η παρασταση οταν τη γραψουμε ετσι ωστε να φαινεται οτι ειναι θετικη και θεωρουμε το καταληλο συστημα?

Τώρα καταλάβα τι ρώτησες ... Λοιπόν κανονικά πρέπει να πάρεις κάθε όρο της μορφής $ x^3y$ με διαφορετικούς συντελεστές το καθένα από μπροστά , αλλά εδώ λόγω κυκλικότητας οι συντελεστές είναι ίδιιοι και αν δεις ποιοι όροι μένουν το απλοποιείς σιγά σιγά

[quote="silouan"]Τώρα καταλάβα τι ρώτησες ... Λοιπόν κανονικά πρέπει να πάρεις κάθε όρο της μορφής $ x^3y$ με διαφορετικούς συντελεστές ο καθένα από μπροστά , αλλά εδώ λόγω κυκλικότητας οι συντελεστές είναι ίδιιοι και αν δεις ποιοι όροι μένουν το απλοποιείς σιγά σιγά[/quote]

Α οκ, καταλαβα.
Thanx


when im practicing, im missing a lot of problems like

A jet airplane departing on, flying between two airports at an average speed of 540 miles per hour arrives 8 minutes late. departing on time and flying at an average of 480 miles per hour arrives 53 minutes late. what is the number of miles between the two airsports?

how do you solve these?

Let $ s$ miles be the distance between two airports, and $ t$ minutes be the expected time of travel. We now have the following simultaneous equations:

$ s \equal{}\frac{540}{60}*(t\plus{}8)$
$ s \equal{}\frac{480}{60}*(t\plus{}53)$

Hence we have:
$ \frac{540}{60}*(t\plus{}8) \equal{}\frac{480}{60}*(t\plus{}53)$
$ \Rightarrow 540*(t\plus{}8) \equal{} 480*(t\plus{}53)$
$ \Rightarrow 1.125(t\plus{}8) \equal{} t\plus{}53$
$ \Rightarrow 1.125t\plus{}9 \equal{} t\plus{}53$
$ \Rightarrow t \equal{} 352$

Therefore we have:
$ s \equal{}\frac{540}{60}*(352\plus{}8)$
$ \Rightarrow s \equal{} 3240$ miles

ok i worded that teribly. i suck at those types of problems. thanks thought minicon!

ok let me try these

let x be the rate and y be the time
we have
$ 120\equal{}xy$
$ 120\equal{}(x\plus{}8)(y\minus{}30)$
so
we need two numbers whose difference is 38 and whose procut is 120...
ummmm what did i do wrong?

I suck at those kind of problems, too.

Minicon, what's the solution for #2?

I got 12 minutes.

I am not confident, so can you explain it your way??

you can do that too. that saves a lot of time. thanks

person A and B are biking 1200 meters apart. they start at the same time. A travels 25 kmph, and B travels 30 kmph. what is the difference in time between when A finishes and B finishes?

1.200/30=.04. thats 2.4 minutes
1.200/25=.048. thats 2.88 minutes.
they are .48 minutes apart. thats 28.8 seconds

they r all based on the d=rt formula, with x variables and other algebra.  Try doing practice problems to improve.  For each problem, know wat u wanna find, and wat u have as x variable (sometimes they're not the same).  Also, clearly know for each equation wat is time and wats speed

I was wondering about that too, but in a MathCounts competition, if the times are like $30$, $40$, $45$, etc. minutes, I would just bash it like in minicon's solution, since trying relative speeds may take some time to think about. However, a time like $53$ minutes, probably would be quicker using relative speeds.


quadratics
algebra
number theory
modular arithmetic
quadratic formula

Find all $n \leq 100$ such that $n^{3}+n+31$ is divisible by $43$.

I have often heard that you should not use the quadratic equation in modular arithmetic, but Thomas Mildorf uses it on page 3 of http://www.tjhsst.edu/~tmildorf/math/NumberTheory.pdf. Is it only applicable to prime moduli?

[quote="Phelpedo"]Find all $n \leq 100$ such that $n^{3}+n+31$ is divisible by $43$. [/quote]
I think you meant $n^{2}$. And I'm pretty sure you're able to use the quadratic formula, at least in prime moduli. I don't know about other moduli.
[hide="Solution, without the quadratic formula"]
\[n^{2}+n+31 \equiv n^{2}+n-12 \equiv (n+4)(n-3) \bmod 43 \implies n \in \{3,39,46,82,89\}\]
[/hide]


[quote="Interval"]Which of the following lettered items possesses
line symmetry?  Consider vertical and/or horizontal symmetry.

a. the letter F

b. an  equilateral triangle

c. the letter X

d. a square[/quote]
[hide="a"]No F does not have one.[/hide]
[hide="b"]Yes, an equilateral triangle split in half is a 30-60-90 triangle.[/hide]

[hide="c"]Yes X has one straight through it.[/hide]

[hide="d"]Yes a square has one . A square split along its diagonal makes two 45-45-90 triangles.[/hide]

[quote="Interval"]Which of the following lettered items possesses
line symmetry?  Consider vertical and/or horizontal symmetry.

a. the letter F

b. an  equilateral triangle

c. the letter X

d. a square[/quote]

B,C,D do
For the triangle, any perpendicular line that go through a corner of the triangle works
For the X, either go straight down the middle or straight across the side
Square have 4 lines of symmetry


calculus
integration
derivative
calculus computations

When using integration by parts, is there any advantage to using the du and dv?  To me, it seems much simpler to just use u, v, u', and v' (so that the formula is $ \int uv' \equal{} uv \minus{} \int u'v )$.  However, my teacher and most websites seem to prefer to use the du and dv, which appears to me to be more work and more confusing than is necessary.  So is there any reason to use du and dv instead of u' and v'?

It's just a matter of pleasure.

I do use $ \int f(t)g'(t)\,dt \equal{} f(t)g(t) \minus{} \int f'(t)g(t)\,dt,$ (*) because for me (at least) is a faster way on doing the integration by parts and saves a lot of time.

If your professor likes using $ du$ and $ dv,$ it's probably 'cause he(she) has taught so many times IBP in the same fashion that he(she) doesn't have time on changing the way on teaching it, or perhaps he(she) doesn't realize that using (*) is faster.

I personally find it less confusing rather than more confusing to use $ du$ and $ dv,$ primarily because it is the same language and the same thought processes used in making substitution, and because it creates a consistent level of bookkeeping. In my mind it's a notation that forces you to do things the right way, and that's a good thing. And that is the way I teach, because it works for me and it works for my students.

A way, way, down the road thought: Once you've defined the Riemann-Stieljes integral, then provided $ f$ and $ g$ satisfy some reasonable hypotheses,

$ \int_a^bf(x)\,dg(x)\equal{}f(b)g(b)\minus{}f(a)g(a)\minus{}\int_a^bg(x)\,df(x).$

On a side note: I sometimes see other students having trouble remembering the "formula" for integration by parts.

I then remind them that it is really just the product rule for differentiation, integrated:

$ \frac{d}{dx} (f g) \equal{} f ' g \plus{} g' f$, so $ f g' \equal{} \frac{d}{dx} ( f g ) \minus{} f' g$, which gives:

$ \int f g'  \equal{}  \int \frac{d}{dx} ( f g ) \minus{} \int f' g \equal{}  f g \minus{} \int f' g$.

"it is really just the product rule for differentiation, integrated"

That's why I prefer to use u' and v' rather than du and dv; to me; the former notation makes the connection between integration by parts and the product rule clear and obvious, while the latter notation does not.

You should never write down an integral without specifying the variable you're integrating with respect to.  The notation $ \int uv'$ is ambiguous.

[b]Edit:[/b]  More precisely, you are supposed to assume that the variable you're taking the derivative of $ v$ with respect to and the variable you're integrating with respect to are the same.  This is ambiguous, and the better notation $ \int u \, dv$ resolves the ambiguity by pointing out a more important truth: that as long as those two variables are the same variable, it doesn't matter (up to some smoothness requirement) what it is.

I agree with t0rajir0u and feel that I probably should have included that as part of my previous post.

[quote="t0rajir0u"]You should never write down an integral without specifying the variable you're integrating with respect to.  The notation $ \int uv'$ is ambiguous.[/quote]

Okay, so why not say that $ \int uv' dx \equal{} uv \minus{} \int u'v dx$ ?

Because it's still not clear whether the derivatives are with respect to $ x$.  What you should write is $ \int u \frac {dv}{dx} \, dx$, which, of course, you should really write $ \int u \, dv$.

It is maybe not clear to you why specifying variables is so important, but if you start doing a lot of changing of variables or working with more than one variable you will quickly come to see the benefit of this notation.

[quote="t0rajir0u"] It is maybe not clear to you why specifying variables is so important, but if you start doing a lot of changing of variables or working with more than one variable you will quickly come to see the benefit of this notation.[/quote]

I'll take your word on this, as I have no experience with multivariable calculus/advanced calculus (and you and Kent obviously do).  Thanks for the help and insight.

It's not even a matter of preparing yourself for multivariable calculus.  You just don't want to get into the habit of always thinking of $ x$ as the independent variable, since it could also be $ y$ or $ t$ or anything else, depending on the problem.


I registered for a hosting service this summer for a free math league our high school is coordinating ([url]http://www.rocketcitymath.org[/url]). Well, I still have about a gigabyte left, so I was hoping to put that space to good use by constructing an open content math website. Anyone who wants to can contribute articles, mock contests, proofs, etc., that will help other avid math students. If there is enough interest, I'll post an E-mail address dedicated to contributions. Thank you.

is this school math or olympiad math? :)

I'm not sure what you are asking, but if you're talking about the Rocket City Math League, it's probably around the level of normal math competitions for Algebra II / Trigonometry students or lower and is not as tough as the Olympiad. An archive of past tests is available at the website if you would like to view some of the past contests. As for contributing articles and such, I think students would be happy to see a variety of math levels covered.


geometry
circumcircle
perpendicular bisector
projective geometry
geometry solved

Given 6 arbitrary points [tex]A_1,A_2,B_1,B_2,C_1,C_2[/tex]. [tex]M_A, M_B, M_C[/tex] are the midpoints of [tex]A_1A_2, B_1B_2, C_1C_2[/tex] respectively. The perpendicular bisector of [tex]A_1A_2[/tex] and [tex]B_1B_2[/tex] meet at [tex]P_{AB}[/tex]. The same applies for [tex]P_{BC}[/tex] and [tex]P_{CA}[/tex].
Now choose two midpoints and the intersection point of the respective perpendicular bisectors and draw a circle through these three points. Do this three times, so that you'll get three circles. Proof that there's a common point of intersection

Unfortunately, this is just a complicated application of the [b]Miquel theorem[/b]:

[i]If ABC is a triangle, and X, Y, Z are three arbitrary points on its sidelines BC, CA, AB, then the circumcircles of triangles AYZ, BZX, CXY have a common point[/i].

In your case, the triangle is $P_{BC}P_{CA}P_{AB}$ and the points on its sidelines are $M_A$, $M_B$, $M_C$, respectively.

The proof of the Miquel theorem above is a simple application of the chordal angle theorem. [You can find a proof together with some further discussion at http://cut-the-knot.com/proofs/Pivot.shtml , Statement B.]

  Darij

Thanks Darij!
My poor geometric knowledge is gradually increasing. Yesterday: Carnot's theorem;Today: Miquel's Theorem; Tomorrow: Duck's Theorem!
 :D

[quote="Christian Hirsch"]Tomorrow: Duck's Theorem![/quote]

Thanks for your interest, but just to mention it, this is not the established name of that fact ;)  others would call it a particular case of Pascal's theorem with the Pascal line being the line at infinity :D

  darij


a) Prove that there does not exist any function $f: \mathbb{R}\rightarrow \mathbb{R}$ so that $(f\circ f)(x)=\{\begin{array}{cc}{II}\sqrt{2007}&\textrm{if}\;x\in\mathbb{Q}\2007,&\textrm{if}\;x \in\mathbb{R}-\mathbb{Q}\end{array}$
b) Prove that there exist an infinity of functions $g: \mathbb{R}\rightarrow \mathbb{R}$  so that $(g\circ g)(x)=\{\begin{array}{cc}{II}2007&\textrm{if}\;x\in \mathbb{Q}\\sqrt{2007},&\textrm{if}\;x \in\mathbb{R}-\mathbb{Q}\end{array}$

a) Suppose $x \in \mathbb{Q}$ and $f(x) \in \mathbb{Q}$.
Then:
- $f(x) \= q \in \mathbb{Q}$
- $\sqrt{2007}= f(q)$
- $f(\sqrt{2007}) = f(f(q) = \sqrt{2007}$
- $f(f(\sqrt{2007})) = 2007 = f(\sqrt{2007})$.
The last two lines are a contradiction.

Suppose $f(x) = r \in \mathbb{R}-\mathbb{Q}$. Then:
- $f(x) = r$
- $f(f(x)) = \sqrt{2007}= f(r)$
- $f(\sqrt{2007}) = f(f(r)) = 2007$
- $f(f(\sqrt{2007})) = 2007 = f(2007)$
- $f(2007) = f(f(2007) = \sqrt{2007}$
The last two lines are a contradiction.

b) Let r a rational number. Define f as:
$f(r) = 2007$
$f(2007) = 2007$
$f(\mathbb{Q}-\{r; 2007\}) = r$

$f(\mathbb{R}-\mathbb{Q}$.

Then, if x is rational:
- if x is r or 2007, then $f(f(x)) = f(2007) = 2007 \in \mathbb{Q}$
- otherwise, $f(f(x)) = f(r) = 2007 \in \mathbb{Q}$
If x is irrational:
- $f(f(x)) = f(\sqrt{2007}) = \sqrt{2007}$

You can choose r in infinte ways.


Demonstrati ca $\ h_{a}+h_{b}+h_{c}\leq\ 9r+\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{p}$, notatiile fiind cele obisnuite.

Ce imbunatatire!!!Acuma sa o demonstram.Avem ca $h_{a}=\frac{2S}{a}$ si analoagele.Trebuie sa demonstram ca
 $ S \left[(\sum a)\left(\sum\frac{1}{a}\right)-9\right]\leq \sum a^{2}-\sum ab$.Acuma folosim substitutiile Ravi $a=y+z,b=x+z,c=x+y$.Avem de demonstrat astfel $\sqrt{xyz\sum x}\cdot \frac{2\sum x^{3}-\sum xy(x+y)}{\prod xy(x+y)}\leq \sum x^{2}-\sum xy$.Acuma o rescriem $ 2\sqrt{xyz\sum x}\sum_{cyc}(x-y)^{2}(x+y)\leq \prod (x+y)\sum (x-y)^{2}$ echivalent cu $\sum (x-y)^{2}(x+y)[z(x+y+z)+xy-2\sqrt{xyz\sum x}] \geq 0$ echivalent cu $\sum (x-y)^{2}(x+y)(\sqrt{z(x+y+z)}-\sqrt{xy})^{2}\geq 0$.(QED) :lol:

Din $ah_{a}=bh_{b}=ch_{c}=2S$ si $r=\frac{S}{p}=\frac{2S}{a+b+c},$ avem:
$\ h_{a}+h_{b}+h_{c}\leq\ 9r+\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{p}\Longleftrightarrow$
$\Longleftrightarrow S(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\leq 9S+(a^{2}+b^{2}+c^{2}-ab-bc-ca)\Longleftrightarrow$
$\Longleftrightarrow S\bigg[\bigg(\frac{a}{b}+\frac{b}{a}-2\bigg)+\bigg(\frac{b}{c}+\frac{c}{b}-2\bigg)+\bigg(\frac{c}{a}+\frac{a}{c}-2\bigg)\bigg]\leq \frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\Longleftrightarrow$
$\Longleftrightarrow S\bigg[\frac{(a-b)^{2}}{ab}+\frac{(b-c)^{2}}{bc}+\frac{(c-a)^{2}}{ca}\bigg]\leq \frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\Longleftrightarrow$
$\Longleftrightarrow \sum \bigg[\frac{(a-b)^{2}}{ab}\bigg(S-\frac{ab}{2}\bigg)\bigg]\leq 0$, inegalitate evidenta daca tinem cont ca $S=\frac{ab\sin C}{2}\leq \frac{ab}{2}.$

[color=darkblue][b][u]Demonstratia de mai jos este oarecum similara cu demonstratia lui Marius Damian.[/u][/b]

Inmultind cu $4Rp$, inegalitatea $\sum h_{a}\le 9r+\frac{1}{p}\cdot(\sum a^{2}-\sum bc)$ devine : $2p\cdot\sum 2Rh_{a}\le 9\cdot 4Rpr+2R\cdot(2\sum a^{2}-2\sum bc)$.

Folosind formulele $\{\begin{array}{c}2Rh_{a}=bc\\\\ 4Rpr=abc\end{array}$ si identitatile cunoscute $\{\begin{array}{c}2\sum a^{2}-2\sum bc=\sum (b-c)^{2}\\\\ \sum a\cdot\sum bc=9abc+\sum a(b-c)^{2}\end{array}$ inegalitatea devine :

$\sum a\sum bc\le 9abc+2R\cdot\sum (b-c)^{2}$ $\Longleftrightarrow$ $9abc+\sum a(b-c)^{2}\le 9abc+2R\cdot\sum (b-c)^{2}$ $\Longleftrightarrow$

$\sum a(b-c)^{2}\le 2R\cdot\sum (b-c)^{2}$ $\Longleftrightarrow$ $\sum (2R-a)(b-c)^{2}\ge 0$, ceea ce este adevarat. Avem egalitate daca si numai daca $a=b=c$.[/color]


Solve.

$ sin(2\theta)\equal{}\sqrt{2}cos(\theta)$

\[ \sin 2x \equal{} \sqrt{2} \cos x \iff 2 \sin x \cos x \equal{} \sqrt{2} \cos x.\]

If $ \cos x \neq 0$,
\[ \sin x \equal{} \frac{1}{\sqrt{2}} \iff x \equal{} \frac{1}{4}\pi, \frac{3}{4}\pi.\]

If $ \cos x \equal{} 0$,
\[ \cos x \equal{} 0 \iff x \equal{} \minus{}\frac{1}{2}\pi, \frac{1}{2}\pi.\]

Thus $ x \equal{} \frac{1}{4}\pi, \frac{3}{4}\pi, \pm \frac{1}{2}\pi$.


Given triangle $ ABC$, with circumcenter $ O$, let the $ A$-symmedian hit $ BC$ at $ S$.  Let the parallel through $ S$ to $ AO$ hit the $ A$-altitude at $ X$.  Prove that $ OX\perp AS$.

Let the A-symmedian cut  the circumcircle again at $ T.$ Tangents to $ (O)$ at $ A,T$ meet at a point $ V$ on $ BC,$ i.e. the center of the A-Apollonius circle of $\triangle ABC.$ Let $ OY \perp AS$ and define $ X'\equiv OY \cap AH_a.$  From the cyclic quadrilaterals $ AYH_aV$ and $ YSH_aX'$ we get  $ \angle YSX' \equal{} \angle YH_aA \equal{} \angle YVA \equal{} \angle OAY $ $ \Longrightarrow$ $ OA \parallel SX'$ $\Longrightarrow$ $ X \equiv X'.$

My solution:

Let $ Z $ be the projection of $ X $ on $ AS $ .
Let the tangent of $ \odot (ABC) $ through $ A $ intersect $ BC $ at $ Y $ .

It suffices to prove $ O \in XZ $ .

From $ SX \parallel AO \Longrightarrow SX \perp AY $ ,
so $ X $ is the orthocenter of $ \triangle AYS \Longrightarrow Y \in XZ $ .

Since $ AS $ is the polar of $ Y $ WRT $ \odot (O) $ ,
so we get $ YO \perp AS \Longrightarrow O, X, Y, Z $ are collinear .

Q.E.D

[quote="buratinogigle"]An extension

Let $O$ be circumcenter of triangle $ABC$. $P$ is a point on $OA$. $(K)$ is circumcircle of triangle $PBC$. $PQ$ is diameter of $(K)$. $AQ$ cuts $BC$ at $S$. $T$ is a point such that $AT\perp BC$ and $ST\parallel AO$. Prove that $PT$ and $AS$ intersect on $(K)$.[/quote]

Reformulation

Let $O$ be circumcenter of triangle $ABC$. $S$ is a point on $BC$;  $T$ is a point such that $AT\perp BC$ and $ST\parallel AO$ ; $P$ a point on $AO$  sucht that $PT \bot AS$ and $ PT \cap AS \doteq F$.  Prove that  $B,C,F$ and $P$ are cyclic

Proof

$ BC \cap TP \doteq Z $ and  $T$ is orthocenter of $\Delta AZS$ then $TS \bot AZ$ ie. $ AO \bot AZ \Rightarrow ZP.ZF \doteq AZ^2$ and $AZ$ is tangent to circumcircle of $ \Delta ABC$ ie. $ZB.ZC \doteq AZ^2$
$ ZP.ZF \doteq ZB.ZC \doteq AZ^2 \Rightarrow B,C,F$ and $P$ are cyclic.
Done!

[quote="buratinogigle"]An extension

Let $O$ be circumcenter of triangle $ABC$. $P$ is a point on $OA$. $(K)$ is circumcircle of triangle $PBC$. $PQ$ is diameter of $(K)$. $AQ$ cuts $BC$ at $S$. $T$ is a point such that $AT\perp BC$ and $ST\parallel AO$. Prove that $PT$ and $AS$ intersect on $(K)$.[/quote]
My solution:

[b]Lemma[/b]:

Let $ O $ be the circumcenter of $ \triangle ABC $ and $ P $ be a point on $ AO $ .
Let $ Q $ be the antipode of $ P $ in $ \odot (PBC) $ and $ M $ be the projection of $ P $ on $ BC $ .

Then $ AM, AQ $ are isogonal conjugate of $ \angle BAC $ .

[b]Proof[/b]:

Let $ R $ be the isogonal conjugate of $ Q $ WRT $ \triangle ABC $ .
Let $ Y, Z $ be the projection of $ P $ on $ AC, AB $, respectively .
Let $ B^*, C^* $ be the projection of  $ R $ on $ AC, AB $, respectively .

It suffices to prove $ A, M, R $ are collinear .

Since $ PM, PQ $ are isogonal conjugate of $ \angle BPC $ ,
so we get $ \angle C^*BR=\angle QBC=\angle BPM=\angle BZM \Longrightarrow MZ \parallel RB $ .
Similarly we can prove $ MY \parallel RC $ .

From $ P \in AO $ we get $ YZ \parallel CB $ ,
so $ \triangle MYZ $ and $ \triangle RCB $ are homothetic with center $ A \Longrightarrow R \in AM $ .
____________________________________________________________
[b]Back to the main problem[/b]:

Let the tangent of $ \odot (ABC) $ through $ A $ intersect $ BC $ at $ X $ .
Let $ Y $ be the projection of $ T $ on $ AS $ and $ M $ be the projection of $ P $ on $ BC $ .

From $ ST \parallel AO \Longrightarrow ST \perp AX $ ,
so $ T $ is the orthocenter of $ \triangle AXS \Longrightarrow Y \in XT $ .
Since $ A, P, M, X $ lie on a circle with diameter $ XP $ ,
so combine with the lemma we get $ \angle MXY=\angle TAS=\angle MAP=\angle MXP $ ,
hence $ P, T, X, Y $ are collinear $ \Longrightarrow \angle QYP=90^{\circ} $ . i.e. $ Y \equiv AS \cap PT \in \odot (K) $

Q.E.D


Find the value of the following expression
$ \sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}....\sin\frac{13\pi}{14}$

[quote="RAGHAV GROVER"]Find the value of the following expression
$ \sin\frac {\pi}{14}\sin\frac {3\pi}{14}\sin\frac {5\pi}{14}....\sin\frac {13\pi}{14}$[/quote]

[hide]
Since $ \sin \theta=\cos({\pi\over2}-\theta)$

Then $ \sin\frac {k\pi}{14}=\cos({7\pi\over14}-\frac {k\pi}{14})$;$ \quad k=1,3,5\cdots13$

$ \therefore \sin\frac {\pi}{14}\sin\frac {3\pi}{14}\sin\frac {5\pi}{14}....\sin\frac {13\pi}{14}=\cos\frac {3\pi}{7}\cos\frac {2\pi}{7}\cos\frac {\pi}{7}....\cos\frac {-3\pi}{7}$
$ =\left(\cos\frac {3\pi}{7}\cos\frac {2\pi}{7}\cos\frac {\pi}{7}\right)^2$

And

$ \cos\frac {3\pi}{7}\cos\frac {2\pi}{7}\cos\frac {\pi}{7}$

$ ={\cos\frac {3\pi}{7}\cos\frac {2\pi}{7}\cdot 2\sin \frac {\pi}{7}\cos\frac {\pi}{7}\over 2\sin \frac {\pi}{7}}$

$ ={\cos\frac {3\pi}{7}\cos\frac {2\pi}{7}\sin \frac {2\pi}{7}\over 2\sin \frac {\pi}{7}}$

$ ={\cos\frac {3\pi}{7}\cdot2\cos\frac {2\pi}{7}\sin \frac {2\pi}{7}\over 2^2\sin \frac {\pi}{7}}$

$ ={\cos\frac {3\pi}{7}\sin \frac {4\pi}{7}\over 2^2\sin \frac {\pi}{7}}$

$ ={\cos\frac {3\pi}{7}\sin \frac {3\pi}{7}\over 2^2\sin \frac {\pi}{7}}$

$ ={2\cos\frac {3\pi}{7}\sin \frac {3\pi}{7}\over 2^3\sin \frac {\pi}{7}}$

$ = {\sin \frac {6\pi}{7}\over 2^3\sin \frac {\pi}{7}}={\sin \frac {\pi}{7}\over 2^3\sin \frac {\pi}{7}}= {1\over 2^3}$

$ \therefore \sin\frac {\pi}{14}\sin\frac {3\pi}{14}\sin\frac {5\pi}{14}....\sin\frac {13\pi}{14}=  \left({1\over 2^3}\right)^2= {1\over 2^6}$

[/hide]

[quote="Kouichi Nakagawa"][hide="Hint"]$ m$ is integer numbers such that $ m\ge2$.
\[ \prod_{i \equal{} 1}^{m \minus{} 1}\sin\frac {i \pi}{m} \equal{} \frac {m}{2^{m \minus{} 1}}
\]
[/hide][/quote]
[hide="Using formula"]
$ \prod_{i \equal{} 1}^{13}\sin\frac {i\pi}{14} \equal{} \left(\prod_{i \equal{} 1}^{6}\sin\frac {i\pi}{7}\right) \times \left(\prod_{i \equal{} 1}^{7}\sin\frac {(2i \minus{} 1)\pi}{14}\right)$

So,
$ \prod_{i \equal{} 1}^{13}\sin\frac {i\pi}{14} \equal{} \frac {14}{2^{13}}$
$ \prod_{i \equal{} 1}^{6}\sin\frac {i\pi}{7} \equal{} \frac {7}{2^{6}}$

Thus, $ \prod_{i \equal{} 1}^{7}\sin\frac {(2i \minus{} 1)\pi}{14} \equal{} \left(\prod_{i \equal{} 1}^{13}\sin\frac {i\pi}{14}\right)/\left(\prod_{i \equal{} 1}^{6}\sin\frac {i\pi}{7}\right) \equal{} \frac {\frac {14}{2^{13}}}{\frac {7}{2^{6}}} \equal{} \frac {2}{2^7} \equal{} \frac {1}{2^6} \equal{} \boxed{\frac {1}{64}}$.[/hide]

[quote="Kouichi Nakagawa"][hide="Hint"]$ m$ is integer numbers such that $ m\ge2$.
\[ \prod_{i \equal{} 1}^{m \minus{} 1}\sin\frac {i \pi}{m} \equal{} \frac {m}{2^{m \minus{} 1}}
\]
[/hide][/quote]

Thanks for ur help ,ur solution is very cute and short can u plz tell me the same  formula for cosine also :)

What is the name of this formula , i have never seen this one ,plz tell me so that i can search for its proof.. :!:

Use complex numbers! Although, it may not look like it, it can easily be evaluated using complex numbers.
[hide]Hint: z^14=1.[/hide]

Hey,Kouichi Nakagawa can u evaluate the following by the
formula,you have stated
$ \sin\frac{5\pi}{14}\sin\frac{3\pi}{14}\sin\frac{\pi}{14}$. :roll:

Thanks in advance!!

I'm not Kouichi Nakagawa, but, here's what I'd do...

[hide]Let's say $ x \equal{}  \sin\frac{5\pi}{14}\sin\frac{3\pi}{14}\sin\frac{\pi}{14}$. Note that since $ \sin y \equal{} \sin(\pi \minus{} y)$, we get that $ x^2 \sin{\frac{7\pi}{14}} \equal{} x^2 \equal{} \sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}....\sin\frac{13\pi}{14} \equal{} \frac{1}{64}$. $ x$ must be positive, so $ x \equal{} \frac{1}{8}$.

And Google calculator agrees with me. =D
[/hide]

I'm sure you can do it easily with complex numbers too. Anyone want to give it a shot?

[quote="electron-proton"]Can anyone tell me the link on which the above formula is stated (except AOPS)??[/quote]

You can look this book inside: Challenging Mathematical Problems With Elementary Solutions vol.2 written by A.M. Yaglom & I.M. Yaglom

See Page 23-24!

[url]http://www.flipkart.com/challenging-mathematical-problems-elementary-solutions/0486655377-gyw3fulx8d#previewbook[/url]

How to solve
$ \sin\frac{\pi}{30}.\sin\frac{7\pi}{30}.\sin\frac{11\pi}{30}.\sin\frac{17\pi}{30}$
by the above mentioned formula..

I am not posting the problem in the new topic because this question is directly linked to these discussions..

[quote="RAGHAV GROVER"]Kunny is there a same type of formula for cosine also??[/quote]

Yes, here is:

$ \boxed{\prod_{j \equal{} 1}^{n \minus{} 1} \cos \frac {2\pi}{n}j \equal{} \frac {\sqrt{n}}{2^{n \minus{} 1}}}$


Determine the number of ways to arrange the letters of the word ELEVEN

The solution needs a little more for those wanting to understand why, something like:
"If we treat the three E's as distinct, there are 6x5x4x3x2x1 ways to rearrange the six letters.  In each arrangement there are 3x2x1 ways to arrange the three E's, so to correct for overcounting we divide 6! by 3!."

Well the official solution just says There are $ 6!$ ways to arrange the letters, but we divide by $ 3!$ for overcounting to get $ \frac{6!}{3!}\equal{}120$ or something like that.  Mr. Batterson showed WHY you divide by $ 3!$.


geometry
rectangle
real analysis
real analysis unsolved

Ok. This is a simple question, yet it managed to get me thinking for a couple of days.
I hope someone here might be able to solve it.

Let $ (X_i,\mathcal{M}_i), i\equal{}1,2,3$ be three measurable spaces. We can define the product measure in more than on way. For instance, $ \mathcal{M}_1 \otimes \mathcal{M}_2 \otimes \mathcal{M}_3$ is the $ \sigma$-algebra generated by the mesurable rectangles $ A_1 \times A_2 \times A_3$ where $ A_i \in \mathcal{M}_i, i\equal{}1,2,3$. Another way would be to define $ \left( \mathcal{M}_1 \otimes \mathcal{M}_2 \right) \otimes \mathcal{M}_3$ as the $ \sigma$-algebra generated by measurable rectangles $ B \times A$ where $ B \in \mathcal{M}_1 \otimes \mathcal{M}_2$ and $ A \in \mathcal{M}_3$.
Now Folland claims that $ \mathcal{M}_1 \otimes \mathcal{M}_2 \otimes \mathcal{M}_3 \equal{} \left( \mathcal{M}_1 \otimes \mathcal{M}_2 \right) \otimes \mathcal{M}_3$.
I can see why $ \mathcal{M}_1 \otimes \mathcal{M}_2 \otimes \mathcal{M}_3 \subseteq \left( \mathcal{M}_1 \otimes \mathcal{M}_2 \right) \otimes \mathcal{M}_3$, since any mesurable rectangles $ A_1 \times A_2 \times A_3$ where $ A_i \in \mathcal{M}_i, i\equal{}1,2,3$ is trivially in $ \left( \mathcal{M}_1 \otimes \mathcal{M}_2 \right) \otimes \mathcal{M}_3$.
I can't even start to imagine the other direction.

Well, just consider the sets $ A\subset X_1\times X_2$ such that $ A\times B\subset M_1\otimes M_2\otimes M_3$ for all $ B\in M_3$. It is a sibma-algebra and conatins all rectangles. ;)

Hi fedja,
I have thought along similar lines, but was unable to show that it was closed under complements.
But to tell you the truth, I think the family you have devised is easily shown to be a $ \sigma$-algebra.

Thank you for the help.

[quote="karitaru"]
I have thought along similar lines, but was unable to show that it was closed under complements.[/quote]
$ [(X_1\times X_2)\setminus A]\times B\equal{}(X_1\times X_2\times B)\setminus (A\times B)$


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

A chessboard is the set $ C \equal{} \{(i,j),1\leq i,j\leq 8\}$. We put $ 21$ $ 3\times 1$ tiles on chessboard such that no two tiles overlaps. Determine those points in $ C$ that can not be covered by some tile.

Would you give one example of placing 21 of $ 3\times1$ dominoes ? ( I believe we is allowed to use  $ \pi/2$ rotation of it too ie., $ 1\times 3$ domino). I could not find one and I feel there does not exist one such arrangement.

[quote="srikanth"]Would you give one example of placing 21 of $ 3\times1$ dominoes ?[/quote]
Good point! I haven't found a proof that it is possible or impossible. That should be proven first.
[quote="srikanth"]( I believe we is allowed to use  $ \pi/2$ rotation of it too ie., $ 1\times 3$ domino).[/quote]
Yes. You can use also $ 1\times 3$ dominoes.

[quote="MeKnowsNothing"]Is the following approach correct? One can number the squares of the board as
12312312
31231231
23123123
12312312
31231231
23123123
12312312
31231231
Now every tile covers three different numbers. So after putting $ 21$ tiles on the board we have one square with number $ 1$ uncovered. On the other hand, one can cover the board as
21321321
13213213
32132132
21321321
13213213
32132132
21321321
13213213
Again one $ 1$ stays uncovered. Now the result is those position $ a_{i,j}$ on the board which has $ 1$ in both numberings.[/quote]

This is what I did.  In total there are four squares which cannot be covered.

Generalization:  For each integer $ n>3$ with $ n$ not a multiple of 3, find the number of ways to deleta a square from an $ n^2$ chessboard so we can tile it completely with $ 1$ x $ 3$ tiles.


[quote="Fabiano"]Give a,b,c >0
Show that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {{9({a^2} \plus{} {b^2} \plus{} {c^2})}}{{{{(a \plus{} b \plus{} c)}^2}}}$[/quote]

We have stronger Inequality
$ 2(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})\plus{}1 \ge \frac{21(a^2\plus{}b^2\plus{}c^2}{(a\plus{}b\plus{}c)^2}$
SOS can slove it :)


I am interested in studying Mathematics and Computer Science.
Does anyone know if Caltech has a good math and computer science department?

And what are the top ratings for mathematics instead?

I guess MIT, Caltech, Harvard, Princeton, Stanford. Others?

Thanks for the answer.

UC Berkeley's graduate math program is consistently in the top 5, if not number one on a some occasions

[quote="XT"]
I guess MIT, Caltech, Harvard, Princeton, Stanford. Others?
[/quote]
Chicago

[quote="tokenadult"]Yes, it does. Ordinarily, Caltech is just outside the top five ratings for computer science, but that is partly because of the very theoretical orientation of the department. Caltech's CS department is academic, not job training.[/quote]

As an alum of Caltech CS:

Everything at Caltech is theoretical.  It's not that good if you plan to work in industry (though a lot of cs grads end up at Microsoft/Yahoo/Google/nVidia/etc), but I find that it teaches you a lot of more general skills that let you pick up things very quickly, so you end up being able to outpace the standard software engineers and have a deeper understanding of the problems.  Of course, Caltech's also so small that there aren't any co-op programs, so most CS grads got their job offers as a result of summer internships.  (I believe)

As far as math goes, I don't think there's a huge difference in the quality among the top schools.  I'm willing to bet their curricula are similar, since the math is likely to be theoretical anyway.  The only major negative I would mention about Caltech is that because it's so small, you won't get a diverse selection of faculty working on all sorts of problems, there might only be one or two professors working in any particular subfield.  This could potentially be a factor depending on what you want to study.

I would say that most people who end up at Caltech chose because the environment is unique, not because Caltech is a top-school.  (I don't think many people slip by our admissions who can't handle any other top school.)


Find integer roots of the equation : $x^{10} + y^{10} -z^{10} = 1999$

Hmmm... I presume you meant $x^{10}+y^{10} - z^{10} = 1999$, am I wrong?

Pierre.

According to Fermat's little theorem, $n^{10} = 0$ or $1$ modulo $11$.
Therefore $x^{10}+y^{10}-z^{10} = 0,1,2,10 \mod [11]$. Since $1999 = 8 \mod [11]$ there is no solution at the given equation.

Pierre.


Ok.. the way I see it, the main problem with this merger is that it makes the forum selection page abhorrently convoluted.
My suggestions on how to correct it:
Take away all the links to the subforums under advanced and college.
Then on the next page, where the links to the subforums are, take away all those subforum links. Now, everything on that topic will be in that topic, and not in 5 other possible places.
Now get rid of cultural exchange. Aops has had that in the round table (in addition to the hundred billion posts on the current administration).
Next, lets get rid of the news flashes area and the polls section. Thats quite an extra feature that was already used everywhere else on the site.
Also, I think it would be a better design if instead of posting a bunch of separate continents in that one section (the one with Non-US Americas, Africa/Australia, Asia, Europe), it was just combined into "Other Continents".
And a shout-out to all those people who have a knack for picking on others' english grammar, hold back on it for a while.
I think thats pretty much it. If I think of anything else, I'll post it.

Well,  they were fine combined into one before AoPS merged.  But now we have all kinds of people from different countries.  I don't think that it is fair if we combine it into one.  I think rep's idea is fine.  Don't take each one away but maybe make subforums.  I don't know how many people are from which continent so it also kind of depends on that.

[quote="jelyman"]My suggestions on how to correct it:
Take away all the links to the subforums under advanced and college.
Then on the next page, where the links to the subforums are, take away all those subforum links. Now, everything on that topic will be in that topic, and not in 5 other possible places.[/quote]

I would rather propose a systematic subdivision of the maths sections into "Getting Started", "Intermediate Level" and "Advanced Corner". Each of these sections should be subdivided into the subforums "Algebra", "Combinatorics", "Geometry", etc. etc.. Actually, this is because many of us are beginners in one field and professionals in another one, hence it is better if everybody has the appropriate category for each field. The material currently crowded in the "Advanced Problems (Temporary)" forum should be moved into the other categories, namely each thread into the appropriate subforum (I know this is a lot of work, but it should be worth doing it).

[quote="jelyman"]Next, lets get rid of the news flashes area and the polls section.[/quote]

Indeed this is right.

[quote="jelyman"]Also, I think it would be a better design if instead of posting a bunch of separate continents in that one section (the one with Non-US Americas, Africa/Australia, Asia, Europe), it was just combined into "Other Continents".[/quote]

Yes. Take the subforum "National Olympiads" out of the "IMO" forum, and join it will the Americas, Africa/Australia, Asia, Europe etc. olympiad ones.

  Darij

One idea that we have discussed is letting students select which forums show up on their main page when they log in.  This ways students can select to see only the 4 or 10 or 15 they normally look at, which they can click a button to bring up the full list when they feel like it.

It is still undetermined how much work this would take, but any feedback on this idea would be appreciated.

it's just my opinion, but i like these sub-subforum links. they don't take much place and if you've got a modem which has just 56kbit and every site needs about 5 or 10 seconds to load, it is quite useful you don't have to click through the thousands of subforums, but can go directly to the forum you want to go to.

Peter

Yes I like the sub-subforums too. Cuz I also have a 56 kbit connection. And BTW I would like to have the text mode back (even if it's not the most important thing).

Concerning the old AoPS ranks, I don't know what they were like, but if they were better than the MathLinks rank then I have nothing against using them.

  Darij

Yeah, that'd definitely be nice. I think most community members don't use all the forums, so it would be very useful.

[quote="Rep123max"]That would be awesome to let us choose what forums show up when we log on. Cause I really only look at about 6 forums.[/quote]

We're looking into this.  Not clear yet how hard it will be to do.

[quote="jelyman"]Hmm. It seems as if I worded that section wrong. I didnt meant to say get rid of all of the subforums completely. What I meant was that on the main forum page, get rid of those little subforums sections that u can click on in each of the advacned and college boxes. Then on the next page (when u click either the advanced or college section), dont get rid of all those links, but get rid of all of those sub-sub forums. Just combine all of the double-subs into one sub. Get it? Heh.[/quote] You can do that if you use the MathLinks skin (profile -> style -> MathLinks) and then hit the options and you can customize anything.
These options will soon be available on the AoPS skin soon (Richards' working on them).

Here's what I'd like to see (or something similar)

[u]Problem Solving[/u]
[b]-Getting Started[/b] (as is)
[b]-Intermediate Topics[/b] (as is)
[b]-Advanced Problems[/b]
[i]Perhaps leave the sub-forums under advanced problems, but I don't think the sub-sub-forum classification is necessary; I feel it creates entirely too much clicking and is confusing. Instead of having Unsolved Problems, Proposed & Own Problems, Open Questions, Solved Problems, Theorems and Formulas sub-sub-forums, just combine it all under the respective sub-forums (i.e. once someone clicked on "algeba," it would look just like the rest of the forums, with all the "algebra" topics listed normally) The old "Advanced Problems" could be filed under those topics[/i]
[b]-College Playground[/b]
[i]Again, leave the sub-forums, ditch the sub-sub-forums. There also appears to be a "main index" link with a globe next to it that takes you back to the forum index. That seems unnecessary. Mathematics Contests for Undergraduate Students should perhaps be moved to "Other Problem Solving Topics[/i]
[b]-Other Problem Solving Topics[/b] (as is)

[u]US Contests and Programs[/u] (as is)

[u]International Contests & Programs[/u]
[i]This can be consolidated a lot. The Europe, Asia, and Africa & Australia forums have one post each. The Americas outside the US forum has some posts, but they could probably be moved to "Other Problem Solving Topics" or something similar. As for the Mathlinks Contest and Download forums, I think that perhaps a better idea would be to have a seperate webpage for these forums. The mathlinks contest could be on a page similar to the "Resources" page on AoPS, with a link off to the far left. It doesn't really need to be a forum at all, probably. The download forum could be similar, on a seperate page listing the olympiads, etc that are available for download (in a civilized, ordered manner), along with a way for members to send new files to the site maintainers to add to the page. It wouldn't have to be a forum either. The idea sounds like a lot of work, but it seems like it should be possible to script some PHP that would help out considerably in the process. (Unfortunately, I don't know PHP, so I can't help.) Doing those things would allow you to eliminate this entire section, without losing any content.[/i]
----
What happens to the rest of the sections doesn't matter much, although the Announcements section seems pointless.

I hope this post was of some help.

[quote="Valentin Vornicu"] These options will soon be available on the AoPS skin soon (Richards' working on them).[/quote]

Done.  Try the Options button above - you can set it to see no subforums listed.

I just noticed a subforum under the LaTeX forum. That seems like it can go, since it serves the same purpose as the LaTeX forum and the preview button, essentially.

[quote="confuted"]Here's what I'd like to see (or something similar)

Perhaps leave the sub-forums under advanced problems, but I don't think the sub-sub-forum classification is necessary; I feel it creates entirely too much clicking and is confusing. Instead of having Unsolved Problems, Proposed & Own Problems, Open Questions, Solved Problems, Theorems and Formulas sub-sub-forums, just combine it all under the respective sub-forums (i.e. once someone clicked on "algeba," it would look just like the rest of the forums, with all the "algebra" topics listed normally) The old "Advanced Problems" could be filed under those topics
[/quote]

That would be a really pain-in-the-ass for people searching only to solve the (yet) unsolved problems. Experience shows that for Olympiad problems that's the best way to do it.
[quote="confuted"] [i]Again, leave the sub-forums, ditch the sub-sub-forums. There also appears to be a "main index" link with a globe next to it that takes you back to the forum index. That seems unnecessary. Mathematics Contests for Undergraduate Students should perhaps be moved to "Other Problem Solving Topics[/i]
[/quote]

The main index link, which you don't see unless you go into the problem solving category, is there because most of the Forum links on MathLinks skin are designed to get you there (skip all the other parts) and then if still users want to get into the other places, they would use that link.

[quote="confuted"] This can be consolidated a lot. The Europe, Asia, and Africa & Australia forums have one post each. The Americas outside the US forum has some posts, but they could probably be moved to "Other Problem Solving Topics" or something similar.
[/quote] At the beginning the US Contest & Programs had only one post per forum. It's not a real criterion, having more or less posts (at least for now).

[quote="confuted"]As for the Mathlinks Contest and Download forums, I think that perhaps a better idea would be to have a seperate webpage for these forums. The mathlinks contest could be on a page similar to the "Resources" page on AoPS, with a link off to the far left. It doesn't really need to be a forum at all, probably. The download forum could be similar, on a seperate page listing the olympiads, etc that are available for download (in a civilized, ordered manner), along with a way for members to send new files to the site maintainers to add to the page. It wouldn't have to be a forum either. The idea sounds like a lot of work, but it seems like it should be possible to script some PHP that would help out considerably in the process. (Unfortunately, I don't know PHP, so I can't help.) Doing those things would allow you to eliminate this entire section, without losing any content. [/quote]You can post yourself any materials you want. That's the purpose of those files being in the forum. Many people use them already. It's also a good way to track what people are downloading.

[quote="confuted"] What happens to the rest of the sections doesn't matter much, although the Announcements section seems pointless.
[/quote] It only seems.

[quote="confuted"] I hope this post was of some help.[/quote] Glad to be of assistance in clearing some of the issues. :)


Isn't it true that the limit of the sequence a_n=n^2*|sinn| doesn't exist? But how do you prove that?

It is really hard if we want to show n^2.|sin n| does not coverge to infinity.
The question n^2|sin n| has not finite limit is triavial, because for every N, we can find n > N such that |sinn| > 1/2.


geometry
circumcircle
geometric transformation
analytic geometry
homothety
symmetry

Let $O$ be the circumcenter of an acute-angled triangle $ABC$, let $T$ be the circumcenter of the triangle $AOC$, and let $M$ be the midpoint of the segment $AC$. We take a point $D$ on the side $AB$ and a point $E$ on the side $BC$ that satisfy $\angle BDM = \angle BEM = \angle ABC$. Show that the straight lines $BT$ and $DE$ are perpendicular.

Assume the circumcircle of $\triangle AOC$ intersects the sides $AB$ and $BC$ again at $A'$ and $C'$ respectively. The foot of the perpendicular from $T$ to $BC$ is the midpt of $CC'$. It's easy to show (quick angle chase) that $AC'||DM||A\"C$, where $A\"$ is the reflection of $A$ in $D$. All of this means that $DM$ cuts $BC$ in the midpoint of $CC'$, which is also the foot of the perpendicular from $T$ to $BC$. In the same way we show that $EM$ cuts $AB$ in the midpt of $AA'$, which is the foot of the perpendicular from $T$ to $AB$.

Let the perpendicular from $T$ to $AB$ cut $BC$ in $C_1$. Similarly we define $A_1$. Let $X$ and $Y$ be the midpts of $AA'$ and $CC'$ respectively. The figures $BXEC_1$ and $BYDA_1$ are obviously similar, so $A_1C_1||DE$ (#).

In $\triangle BA_1C_1$ the point $T$ is the orthocenter, so $BT\perp A_1C_1$ (##).

From (#) and (##) we get the desired conclusion.

Consider a homothety $h$ with center in centroid of $\triangle ABC$, s.t. $h(M)=B$. Denote
$\angle ABC=\alpha$
$F=h(D)$ and $G=h(E)$.
It is easy that
$\angle BFC=\angle BCF=\angle BGA=\angle BAG=\angle TCO=\angle TOC=\angle TAO=\angle TOA=\alpha$.
Thus
$\triangle CFB \sim \triangle COT$ and $\triangle AGB \sim \triangle AOT$, so
$\triangle CFO \sim \triangle CBT$ and $\triangle AGO \sim \triangle ABT$. Hence
$OF=\frac{TB\cdot OC}{TC}=\frac{TB\cdot OA}{TA}=OG$ and
$\angle FOG=\angle OCT+\angle OAT=2\alpha$.
Let $k$ be a bisector of $\angle FOG$. We have
$\angle(OF, BT)=\angle OCT=\alpha=\angle(OF, k)$.
Thus $BT \parallel k$ and $k\perp FG$ because $\triangle OFG$ is isocles
triangle, so $BT\perp FG$. By homothety $DE\parallel FG$ and in the end $BT\perp DE$. :)   :D

[quote="Thanhliem"]we have BE=RsinA+2RsinCcosB
             BD=RsinC+2sinAcosB
 DE 2 =R 2 [(sinC) 2+6sinAsinCcosB +(sinC) 2 -8sinAsinC(cosB)^3]
 then[ 1/(sinDEB) 2] -1=
[(cos(A-B)) 2 /(sinC+2sinAcosB) 2 ]
 then tg TBC=[cosBcos(A-B)]/[sin(A-B)cosB-sinA] then tg TBC=cotgDEB => we prove[/quote]

Can you involve a little bit more detail?

[quote="orl"][color=darkred]Let $ C(O)$ , $ C(T)$ be the circumcircles of an acute $ \triangle ABC$ and $ \triangle AOC$ respectively. Let $ M$ be the midpoint of $ [AC]$ .

Consider the points $ D\in [AB]$ and $ E\in [BC]$ which satisfy $ \widehat {BDM}\equiv\widehat {BEM}\equiv\widehat {ABC}$ . Show that $ BT\perp DE$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u][/b] (similar with the nice [b]Grobber[/b]'s proof). I"ll use the [b]Grobber[/b]'s notations. Denote the points $ \left\{\begin{array}{c} A'\in BA\ ,\ A'B = A'C \\
 \\
C'\in BC\ ,\ C'B = C'A\end{array}\right\|$ .

Observe that $\left \{\begin{array}{c} m(\widehat {AOC}) = 2B \\
 \\
m(\widehat {AA'C}) = m(\widehat {AC'C}) = 180^{\circ} - 2B\end{array}\right\|$ , i.e. the points $ A'$ , $ C'$ belong to the circle $ C(T)$ .

Denote the midpoints $ X$ , $ Y$ of the segments $ [AA']$ , $ [CC']$ respectively. Observe that $ \left\{\begin{array}{c} MX\parallel A'C\implies E\in XM \\
 \\
MY\parallel C'A\implies D\in MY\end{array}\right\|$ .

$\left \{\begin{array}{c} TX\perp BA \\
 \\
TY\perp BC\end{array}\right\|\implies$ the quadrilateral $ BXTY$ is ciclically (the circumcenter is the midpoint of the segement $ [BT]$ ).

From the relations $ m(\widehat {XDY}) = m(\widehat {XEY}) = 180^{\circ} - B$ obtain that the quadrilateral $ DEYX$ is cyclically.

Thus, the perpendicular line $ d$ from the point $ B$ to the line $ DE$ passes through the center of the circumcircle of $ \triangle BXY$

(and of the quadrilateral $ BXTY$ ) , i.e. the midpoint of the segment $ [BT]$ . Thus, $ d\equiv BT$ . In conclusion, $ BT\perp DE$ .[/color]

[hide="Solution"]
Let $ \angle ABC\equal{}\alpha$.  Letting $ DM$ and $ EM$ meet $ AB$ and $ BC$ at $ X$ and $ Y$ respectively, we see that $ \angle XDB\equal{}\angle XBD\equal{}\alpha$.  Thus, $ \angle MXA\equal{}\angle XDB\plus{}\angle XBD\equal{}2\alpha\equal{}\angle AOC\equal{}\frac{\angle ATC}{2}\equal{}\angle ATM$.  It follows that $ TXAM$ is cyclic, so $ TX\perp AB$ since $ TM\perp AC$.  Similarly, $ TY\perp BC$, so $ BYTX$ is cyclic.  Now, notice that $ \angle TYM\equal{}\angle TCA\equal{}90\minus{}\angle AOC\equal{}90\minus{}2\alpha$.  Letting $ \angle TYX\equal{}\beta$, we see that $ \angle TBX\equal{}\beta$, so $ \angle TBY\equal{}\alpha\minus{}\beta$.  Furthermore, $ \angle XDY\equal{}\alpha\equal{}\angle YEX$, so $ YXED$ is cyclic, so $ \angle XDE\equal{}\angle XYM\equal{}\angle TYX\plus{}\angle TYM\equal{}90\minus{}2\alpha\plus{}\beta$.  Yet, $ \angle XDB\equal{}\alpha$, so $ \angle BDE\equal{}\angle XDB\plus{}\angle XDE\equal{}90\minus{}\alpha\plus{}\beta$.  It follows that $ \angle TBD\equal{}\alpha\minus{}\beta\equal{}90\minus{}(90\minus{}\alpha\plus{}\beta)\equal{}90\minus{}\angle BDE$, so $ BT\perp DE$. [/hide]

[quote="iandrei"]Let N be the midpoint of BC and P be the midpoint of AC. Let D' be the intersection of the perpendicular bisector of MN with AB. It is easy to show that D' is the midpoint of BD. Let E' be the intersection of the perpendicular bisector of MP with BC. E' is the midpoint of BE.

Set up a coordinate system with the center in O and such that the circumcenter of ABC is the unit circle. Then, by a quick and easy computation we obtain : (we have denoted by lowercase leters the affixes of points denoted by uppercase letters)

d = 2a + b + c - ab/c
e = a + b + 2c - bc/a
t = ac/(a+c)

We must prove that (d-e)/(b-t) is an imaginary complex number. But :

d-e = (ab+bc-ac)(c-a)/ac
b-t = (ab+bc-ac)/(a+c)

Therefore z = (d-e)/(b-t) = (c-a)(c+a)/ac. It is easy to see that if z' is the conjugate of z , z+z'=0. Whence the conclusion.[/quote]
It's a pretty minor thing, but I think it's worth pointing out: it should be
\[2d=2a+c-\frac{ab}{c}\]and
\[2e=a+2c-\frac{bc}{a}.\]

[hide="Solution"]
We do not use isometric coordinates.

Let $BD\cap EM=N$ and $BE\cap DM=P$.  By trivial angle chasing, $\angle BNE=\angle ATM=\angle MTC=\angle BPD$.  Then $\angle ANT=\pi-\angle TMA=\dfrac{\pi}{2}$, and $\angle CPT=\pi-\angle CMT=\dfrac{\pi}{2}$.  Thus, quadrilateral $BNTP$ is cyclic.  Since $\angle NDP=\pi-\angle ABC=\angle NEP$, quadrilateral $NDEP$ is also cyclic.  Thus, we have $\angle DBF+\angle BDF=\angle NBT+\angle BPN=\angle NBT+\angle NTB=\dfrac{\pi}{2}$, so $BT \perp DE$ as desired.

QED
[/hide]

Let $N,P,L$ be midpoints of $BC,BA$ and the center of the 9-point circle of $\triangle ABC$. $S$ is the foot of perpendicular of $B$ on $AC$. Since the composition of inversion with center $B$ and radius $\frac{BA\cdot BC}{2}$ and symmetry WRT the internal bisector of $\angle ABC$ pictures $S,N,P$ to $O,A,C$ then it pictures $\odot SNP$ to $\odot AOC$ so $\angle CBT=\angle ABL$(1). Note that $MN||BD$ and $\angle MDB=\angle NBD$ then $MNBD$ is an isosceles trapezoid so since $LM=LN$ then $LB=LD$ Likewise $LB=LE$ so $L$ is the circumcenter of $\triangle BDE$. So from (1) $BT\perp DE$

Let the circumcircle of AOC intersects AB and CB at S and P,respectively.Now,we easy have that MD/ME=BS/BP=BC/BA(Let R and Q be the midpoints of BC and BA,then MD=MQ and MR=ME and the conclusion follows).Now,it is easy by angle chasing that <STP=2<ABC,now let point N be such that BR=BN and <RBN=<STP and <SBN=3<ABC.Now,SBN is similar to MDE,so we need to prove that(this is reduced by trivial angle chase) <PNS=<PBT,but this is obvious aplying spiral homothety with center P,so BPN is similar to STP,from which we have that SPN is similar to BPT,so we are finished.

Let $F$ be the point such that $\triangle FBA \sim \triangle FCB$. It is well-known that $BF$ is a symmedian in triangle $ABC$ and points $A,F,O,C$ lie on a circle. Let $E'$ be the intersection of lines $AB$ and $CF,$ and $D'$ be the intersection of lines $AF$ and $BC$.

Apply inversion at $B$ of radius $\sqrt{\frac{1}{2}\cdot AB\cdot BC}$ followed by reflection in the internal bisector of angle $ABC$. As the circle $(BDM)$ passes through the midpoint of $BC$ and $M \rightarrow F,$ we conclude that $D \rightarrow D'$. Similarly, $E \rightarrow E'$ and we obtain $D'E' \parallel DE$.

Finally, we note that $$\angle D'BF=\angle D'CB \Longrightarrow D'B^2=D'F\cdot D'C.$$ Similarly, $E'B^2=E'F\cdot E'A$. Thus, the line $D'E'$ is the radical axis of $B$ and the circle $(AFC)$. The conclusion $BT \perp DE$ follows.

[b]Note:[/b] The fact that $B$ acts as a "point circle" was crucial to finish the proof.

Solved with [b]p_square, Aayuu[/b]

Let $Z$ denote the reflection of $O$ across $AC$. Observe that $AMZD$ and $CMZE$ are cyclic. So, it suffices to show $BZ$ and $BT$ are isogonal since $BZ$ is the diameter of $(BDZE)$.

Let $T'$ be the isogonal conjugate of $T$ in $\triangle ABC$, so it suffices to show $Z,B,T'$ are collinear or that $BT'$ goes through the circumcenter of $\triangle AHC$, or show that $T'$ lies on the euler liner of $\triangle AHC$.

So we have reduced the problem to the following.

[quote=Rephrased problem]
Let $ABDC$ be a quadrilateral with $\angle A = \angle B = \angle C$. Show that $D$ lies on the Euler line of $\triangle ABC$.
[/quote]

To do this, let $X = AB \cap CD$ and $Y = BD \cap AC$. Let $P,Q$ denote midpoints of $AB,AC$ and let $G$ be the centroid. Note that $YB = YA$ and $XA = XC$.

Black magic, by Pappus on $XBP$ and $YCQ$, $D$ lies on $OG$, the Euler line, done. $\blacksquare$

Let rays $BA,BC$ meet $\odot (AOC)$ again at $S,R$ respectively; $DM\cap BC=N,DE\cap BA=K.$ Observe that $RA\parallel MD$ (analogously $SC\parallel ME$), since $\angle RAB=\pi-\angle ABR-\angle ARB=\angle ABR=\angle MDB.$ Hence by the Thales $N,K$ are midpoints of $CR,AS.$ Next $\angle BKT=90^{\circ}=\angle BNT,\angle KDN=\angle KEN,$ so $BNTK,DENK$ are cyclic.
Thus the diameter $BT$ of $\odot (BNK)$ is perpendicular to $DE,$ as desired.

Let $P$ and $Q$ be the feet of the altitudes from $M$ to $\overline{BA}$ and $\overline{BC}$ respectively. Then it is clear that $D$ is the reflection of the midpoint of $\overline{BA}$ over $P$, and a similar fact is true for $E$. Now use complex numbers with $(ABC)$ as the unit circle, setting $A=a,B=1,C=c$, so $M=\tfrac{a+c}{2}$. We can compute $P=\tfrac{3}{4}a+\tfrac{1}{4}c-\tfrac{a}{4c}+\tfrac{1}{4}$ and $Q=\tfrac{3}{4}c+\tfrac{1}{4}a-\tfrac{c}{4a}+\tfrac{1}{4}$, hence $D=a+\tfrac{1}{2}c-\tfrac{a}{2c}$ and $E=c+\tfrac{1}{2}a-\tfrac{c}{2a}$. Furthermore, the circumcenter formula yields $t=\tfrac{ac}{a+c}$. We wish to prove that
$$\frac{2(d-e)}{t-1}=\frac{a-c-\frac{a}{c}+\frac{c}{a}}{\frac{ac-a-c}{a+c}}=\frac{(a^2-c^2)(1-\frac{a+c}{ac})}{ac-a-c}=\frac{a^2-c^2}{ac}$$
is imaginary, which is clear since its conjugate is its negative.


algebra
polynomial
Rational Root Theorem

solve the natural solution  system   $ \left\{\begin{array}{l}

  x^4\plus{}y\equal{}17 \\

  x\plus{}y^4\equal{}3

\end{array}\right.$  :roll:   :wink:

I dnt know how to formally show this is the only real solution, but the only positive integer solution is (2,1)

When you're solving polynomials of a high degree for integer roots, it is often useful to use the:
[hide="One of many possible hints"]Rational Root Theorem[/hide]
[hide="One of many possible solutions"]$ x \equal{} 3 \minus{} y^4 \implies (3 \minus{} y^4)^4 \plus{} y \minus{} 17 \equal{} 0$

Substituting $ y \equal{} 0$ into the LHS reveals that the constant term is $ 64$, so $ y|64$.

Thus $ 17 \minus{} x^4 \equal{} y \geq \minus{} 64 \implies x^4 \leq 81 \implies |x| \leq 3$

Testing these seven values, the only solution is $ (x,y) \equal{} (2,1)$.[/hide]


The probability of wind tommorrow is 2/3, and the probability of rain tommorow is 2/5. If they are independent, what is the probability of wind but not rain tommorrow.

how would u do this problem?

If the probability of an event is p, then the probability of that event [b]not[/b] occuring is 1-p. So the probability of there being no rain is 1- 2/5.

The probability of two independent results occuring= probability of one * probability of the other.

[hide="solution"]Probability of no rain = 3/5, so probability of wind but no rain = 3/5 * 2/3 = 6/15 = 2/5[/hide]

So you want wind, which has probability $\frac{2}{3}$. You don't want rain, which has probability $1-\frac{2}{5}$ = $\frac{3}{5}$. We have to multiply these since they are independent:
$(\frac{2}{3})(\frac{3}{5})=\frac{2}{5}$
The answer is $\frac{2}{5}$.

Could you please hide your answers?  Thank You

[hide]The answer is

$\frac{2}{3}*(1-\frac{2}{5})=\frac{2}{5}$[/hide]


Euler
geometry
3D geometry
tetrahedron
analytic geometry
geometry unsolved

For any tetrahedron, Is following true?

Let $R$ be the radii of circumsphere, and $r$ be the radii of insphere of a tetrahedron.

Then $R^2 - 3Rr = OI^2$

No!!
For instance, consider an isoseles tetrahedron, the coordinates of the vertices being
(a,0,0); (0,b,0); (0,0,c); (a,b,c)
We have O = I and (R/r)^2 = (a^2+b^2+c^2)(1/a^2+1/b^2+1/c^2)

Oh, You're right.

Then Could you show me the extention form of Euler theorem in a tetrahedron?  :D

I think that a relation involving only r, R and OI cannot exist.


سلام دوباره.
امروز رو چه كرديد؟
من بد ندادم.
20 ta zadam ke fekr konam 2tash ghalate.

Kilid ro kasi nadare?

[quote="shoki"]salam man ke kelid nadaram...
vali motaassefane gand zadam (be onvane yek 3-vomi) va faghat 20 ta ro zadam...
kode daftarcham 2 hast...
1-d,2-b,3-a,4-( :D ,fekr konam befahmid manzuram chie ...[hide="click konid ta befahmid"] nazadam  :rotfl: [/hide])
5-c,6-e,7-b,8-ghalat zadam (soale be in asuni ro  :rotfl: ),10-a,11-a,12-a,13-e,14-e,15-d,16-fekr mikonam soal ghalat bashe manzuram ine ke javab tuye gozineha nist badan tozih midam chera (age mishe kasi ke javab dade bege chijuri daravorde)
17-c,18-a,19-a,23-c,24-b,25-(nazadam vali fekr konam a beshe chon hatman katar az 20 hast)
28-b,29-c
mibinid cheghadr gand zadam  :) ...[/quote]

thx.

rasti soale kashia ro ($ 1388 \times 1388$) che joori hal mikonid?

hal nakardam har chand ke sare emtehan fahmidam taye jahayisho vali ....(hesssesh nabud :D )
baraye un soal ta jayi ke midunam faghat kafiye in mozu ro esbat konid ke do ta az char ta adadi ke avale kar mizarim va gharare majmueshun bar 6 bakhshpazir bashe barabar hastan va baraye do taye dige ham ye khassiyate dige darim(be peymaneye 3 hamneheshtand) ...
vali be har hal...
soalate marhale aval (nemigam hamash vali) 99% asunand (manzuram ine ke chize fogholaddei nadarand ) vali baraye man... nemidunam chera.... vali vaght kam miyaram :rotfl:  :rotfl:
kelid ro ham tu site imo09 ta ye chand saate dige mizaran

Salam !! be nazare man ke kheyli sakht nabood. man [b]63%[/b] zadam .(az ye dovommi bishtar entezar nadashte bashin  :!: )
fek mikonam kaf az [b]30%[/b] kamtar nabashe (khodam doost daram kamtar bashe chon chand ta az doostam bad dadan  :( ) vali az [b]40%[/b] ghatan bishtar nist.
Bye !

[quote="navidj"]na . soal ghalat nist .
javabesh mishe 116 . halesh ham ine ke avalan $ x \plus{} y | 5x$ va $ x \plus{} y | 5y$ .
hala $ d \equal{} (x,y)$ va $ d$ ro az 2 taraf sade kon natije mide $ x` \plus{} y` | 5$ pas 2 halat darim :
1 - $ x` \equal{} 1$ va $ y` \equal{} 4$ dar in soorat chon $ y < 101$ yani $ d \leq 25$ yani $ 25$ ta javab darim  ke be elate vojode halat $ x` \equal{} 4$ va $ y` \equal{} 1$ dar ye 2 zarb mishe .
2 - $ x` \equal{} 2$  va $ y` \equal{} 3$ mesle ghabli to in halat $ 33$ ta javab darim ke dar 2 zarb mishe va mishe $ 66$ .
pas javab e soal mishe $ 50 \plus{} 66 \equal{} 116$[/quote]
Eyval manam haminjoori hal kardam  :lol: ...
bebakhshid [b]navidj[/b] shoma chandomi ?

injoor ke maloomev hamatoon balaye 60,70% zadin. mane badbakht....

agha man 55% zadam.torokhoda begid ghabool misham ya na,daram az esteres mimiram.

albate bachehamoon bazia kheili kharab kardan,manam omidvaram kaf paiin bashe.

man ke injuri halesh kardam be dast miarim $ x\plus{}y |5y$
pas yani $ x\plus{}y\equal{}\frac{5y}{k}$
chon $ k$ hadeaksar $ 4$ mitune bashe (va agar na $ x$ gheyre tabii mishe) $ 4$ ta halat darim
1-$ k\equal{}1$ be ma midahad $ (y,4y) ; (4y,y)$ ya be ebarate digar $ 50$ ta zoj.
2-$ k\equal{}2$ be ma midahad $ (y,\frac{3}{2}y) ; (\frac{3}{2}y,y)$ ke kolan yani $ 66\equal{}2 \times 33$.dalil an ast ke avalan $ \frac{3}{2}y \le 100$ ke natije midahad $ y \le 66$ va saniyan $ y$ bayad zoj bashad yani az beyne adade zoje az $ 1$ ta $ 66$ entekhab shavad pas $ 33$ entekhab baraye $ y$ mojud hast pas dar kol $ 66$ zoj be dast midahad.
3-$ k\equal{}3$ ke ba estedlali moshabeh zojhaye $ (y,\frac{2}{3}y) ; (\frac{2}{3}y,y)$ ra be ma midahad yani $ 66\equal{}33 \times 2$ ta zoj midahad.
4-$ k\equal{}4$ ke be dast midahad $ (y,\frac{y}{4}) ; (\frac{y}{4},y)$ ama agar yekami aghl ra dar inja be kar andazim mifahmium ke in zojha dar halate 1 yani $ k\equal{}1$ she morde shodeand.
pas dar kol $ 66\plus{}66\plus{}50$ ta zoj darim yani $ 182$ ta zoj.
montazere javabetun hastam ke be man begin koja eshtebah kardam mamnun misham...

Oon soale ke ye 4zeli mohaddab bood javabesh chi mishe ? man sabet kardam $ 6$ mishe vali bazia migoftan etela`at kafi nist !!
Bayad sabet koni $ B,C,D$ rooye dayere ee be markaze $ A$ hastan . Dalilesh ine ke $ \frac{110}{2} \plus{} 125\equal{}180$

[quote="mahanmath"][quote="navidj"]Man 3vomam .
Chetor mage ???[/quote]
etela`at omoomi !! (hamoon fooozoooli)[/quote]
 :rotfl:   :rotfl:   :rotfl:
Javabe jalebi bood ... .
Be nazaretoon kaf emsal chande ?
Man ke fekr nemikonam kaf riazi emsal balaye 12 ya 12.5 bere nazare shoma chiye ?

[quote="mahanmath"]Oon soale ke ye 4zeli mohaddab bood javabesh chi mishe ? man sabet kardam $ 6$ mishe vali bazia migoftan etela`at kafi nist !!
Bayad sabet koni $ B,C,D$ rooye dayere ee be markaze $ A$ hastan . Dalilesh ine ke $ \frac {110}{2} \plus{} 125 \equal{} 180$[/quote]

javabesh mishe 6.yani bachehaye ma migan 6.

vaaaaaaaaaayyyyyyyyyyyyyyyy fahmidam koja eshtebah kardam halate 2,3 yeki bud khakkkkkkkkkkkkk bar saram ajab eshtebahiiiiiiiiiiiiiiiiiiiiii :rotfl:  :rotfl:

[quote="amparvardi"][quote="navidj"][quote="amparvardi"]vali age kaf 11 bashe ke hame ghaboolan.mage kasi zire 11 ham zade?[/quote]

Parsal ham emtahan oonghadr sakht nabood ke KAF BESHE 8 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]

HARFETOON kamelan matine.
rasti shoma parsal chand ta zadin?(va ghabool shodin?)
emsaal chi?
yeki az bachehaye ma dasht az esteres mimord.behesh begam kaf 11e?[/quote]

man parsal 16.75 shodam (18 ta zadam 1 ghalat) . Emsal ham 17.75 shodam (19 ta zadam 1 ghalat) .
(Man kolan Test Olympiad ro khob nemizanam  :D )
raje be doostetoon hanooz nemishe goft kaf 11 e . vali dar hamoon hoodood hast .

[quote="shoki"]vaaaaaaaaaaaaaaaayyyyyyyyyyy eeeeeeyyyyyyyyyyyyy       khhhhhhhhhhhhhhhhhooooooooodaaaaaaaaaaaaaa ye ghalate dige  :bomb:
ajab sutii dadam....
maaaaaaaaaaamannnnnnnnnnnnnnnnnn che ghadr bad dadam........ :wallbash_red:[/quote]

na baba chera roohie mANFI midin?

,migan kaf 11e.shomam ke sad dar sad bishtar az 15 ta dorost zadid.
dige ghametoon chie??

be nazaretoon kaf hala 11 hast??

manam ye ghalate tablo daram ke nemidoonam chera eshtebaah zadam.
soale  5 kode1,hamoon 100 ta danesh amooze alef o b.
nemidoonam chera javabesh mishe 49,man zadam 25!!!
age momkene rahetoon ro begid.

[quote="shoki"]man ke osulan kaf ro khub nemitunam takhmin bezanam(behtare begam ke hich vaght takhmini nazadam).
khob kafiye masalan $ 49$ ta $ 51$ begirid baraye kelase b(manzuram kelase ba miyangine kamtare)va yeki ham 0 va 50 ta 50 begirid baraye kelase a.
khahid did ke dorost dar miyad.[/quote]

ey lanat.

velesh dige mohem nist.ba ye soal be jaii nemiresim.

[quote="stvs_f"]Dustan guya kasi soale akharo fek nakarde!
kasy hast hal karde bashe?
haddeaksar tedade davayer........[/quote]

javabesh mishe 4.vali mese in ke bachehaye ma ba 6 ta ham dar avorde boodan.
nemidoonam.
miporsam :!:  :!:  :!:

[quote="stvs_f"]Dustan guya kasi soale akharo fek nakarde!
kasy hast hal karde bashe?
haddeaksar tedade davayer........[/quote]

javabesh gozineye akhar fekr konam beshe.

[quote="amparvardi"]migam in soale 23 daftarche 1 (hamoon chand jomleii ba daraje 5) chi mishe/?
man har chi fekr kardam nafahmidam bayad che konam.[/quote]

mishe 48 vali man eshtebah zadam :blush:

hi to all
on soale daraje 5 kheili ghashang bud!
man halidam
taghsimesh ke mikoni bedast miad zaribr x^5 ba zaribe x^2 va zaribe x^4 ba x va x^3 ham ba x^0 barabare ka ba halat bandie adade 1 ta 6 kolan 48 joor mishe ja gozari kard!
soale akharo begid chi mishe??

[quote="amparvardi"][quote="mahanmath"][quote="MJ GEO"]salam.soal akhar mishe binahyat masalan mitunid khat ro dayere be shoae binahayat dar nazar begirid(omidvaram manzooramo resoonde basham :!: )[/quote]

Man ke nafahmidam !!( kheily khengam  :blush: ) age mishe bishtar tozih bede.

amma mishod injoori in masalaro hal kard ke ye kore ro dar nazar begirid bad dayere haye azimasho rasm konid ...  :!:[/quote]

dayerehaye chisho rasm konim??[/quote]
dayere haye azime (mesle ostova too koreye zamin) ...(azime = عظیمه)

Dar har soorat mamnoon.

[url]http://www.IMO09.blogfa.com[/url]  javabha ro zade,age mikhaid berid bebinid.bahal javab dadan.


analytic geometry
geometry
conics
perpendicular bisector
radical axis
geometry proposed
BritishMathematicalOlympiad

Let $ \Delta ABC$ be a triangle and $ (O)$ be a circle which intersects the sides $ BC,CA,AB$ at $ \{A_1,A_2\},\{B_1,B_2\},\{C_1,C_2\}.$ Also the intersection of the tangents at $ A_1,A_2$ to $ (O)$ is $ A_3.$ The two other points ($ B_3,C_3$) are defined similarly. Prove the concurrency of $ AA_3,BB_3,CC_3.$ I know a solution which is really ugly and I wanted to know if there exists other solutions. (Sorry if it has been posted before, I don't know the source).

[b]Theorem:[/b] $(O)$ is an arbitrary circle in the plane of $ \triangle ABC.$ $P,Q,R$ are the poles of the sidelines $ BC,CA,AB$ with respect to $ (O).$ Then $AP,BQ,CR$ concur.

[b]Proof:[/b] $ \delta(P,a)$ stands for the distance from $ P$ to the sideline $ BC$  and $ \delta(P,b),$  $ \delta(P,c)$ are defined similarly. By [url=http://mathworld.wolfram.com/SalmonsTheorem.html]1st Salmon's theorem[/url] we have the following relations:

$ \frac {\overline{RO}}{\overline{PO}} = \frac {\delta(R,a)}{\delta(P,c)} \ , \ \frac {\overline{QO}}{\overline{PO}} = \frac {\delta(Q,a)}{\delta(P,b)} \ \Longrightarrow \ \frac {\delta(P,c)}{\delta(P,b)} = \frac {\delta(R,a)}{\delta(Q,a)} \cdot \frac {\overline{QO}}{\overline{RO}} \ \ (1)$

By cyclic exchange we have the relations:

$ \frac {\delta(Q,a)}{\delta(Q,c)} = \frac {\delta(P,b)}{\delta(R,b)} \cdot \frac {\overline{RO}}{\overline{PO}} \ . \ (2) \ , \ \frac {\delta(R,b)}{\delta(R,a)} = \frac {\delta(Q,c)}{\delta(P,c)} \cdot \frac {\overline{PO}}{\overline{QO}} \  \ (3) $

Multiplying $ (1),(2)$ and $ (3)$ together yields:

$ \frac {\delta(P,c)}{\delta(P,b)} \cdot \frac {\delta(Q,a)}{\delta(Q,c)} \cdot \frac {\delta(R,b)}{\delta(R,a)} = \frac {\delta(R,a)}{\delta(Q,a)} \cdot \frac {\delta(P,b)}{\delta(R,b)} \cdot \frac {\delta(Q,c)}{\delta(P,c)}\cdot \frac {\overline{QO}}{\overline{RO}} \cdot \frac {\overline{RO}}{\overline{PO}} \cdot \frac {\overline{PO}}{\overline{QO}}$

$ \frac {\delta(P,c)}{\delta(P,b)} \cdot \frac {\delta(Q,a)}{\delta(Q,c)} \cdot \frac {\delta(R,b)}{\delta(R,a)}  = 1$ $ \Longrightarrow$ $AP,BQ,CR$ concur by Ceva's theorem.

I didn't check the computations but your idea was really NICE
ALL OF MY CONGRATS :D  :D
That was a really nice idea to use this theorem (and what a useful theorem it is! :) )

Thanks for your kind words dear shoki, indeed it is a very useful theorem but surprinsingly unknown as well. The problem is really nice. I was working on a solution since several days ago when a friend of mine sent me the generalization via private message and then I realized that you submitted it earlier.

We have an interesting concurrency when the subject circle concides with the 9-point circle.

Given a triangle $ \triangle ABC$ and its 9-nine point circle $ (N).$ $ P,Q,R$  are the poles of the sidelines $ BC,CA,AB$  with respect to $ (N).$ Then $ AP,BQ,CR$ concur at the isotomic conjugate of $ X_{1078}.$

[quote]Given a triangle $ \triangle ABC$ and its 9-nine point circle $ (N).$ $ P,Q,R$  are the poles of the sidelines $ BC,CA,AB$  with respect to $ (N).$ Then $ AP,BQ,CR$ concur at the isotomic conjugate of $ X_{1078}.$[/quote]
The pole $ P$ of $ BC$ WRT $ (N)$ is the intersection of the tangent of $ (N)$ at the midpoint of $ BC$ and the perpendicular dropped from $N$ to $ BC.$ Its anticomplement $P'$ is then the intersection of the tangent $ \tau_a$ of $ (O)$ at $ A$ and the perpendicular bisector $ \ell_a$ of $ BC.$ Using barycentric coordinates with respect to $ \triangle ABC,$ we have:

$ \ell_a \equiv (b^2 \minus{} c^2)x \plus{} a^2y \minus{} a^2z \equal{} 0 \ , \ \tau_a \equiv c^2y \plus{} b^2z \equal{} 0$

$ P' \equiv \ell_a \cap \tau_a \equiv{} (a^2(b^2 \plus{} c^2): b^2(c^2 \minus{} b^2): c^2(b^2 \minus{} c^2))$

Therefore, $ P(2b^2c^2 \minus{} b^4 \minus{} c^4: a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \minus{} c^4: a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \minus{} b^4)$

Coordinates of $ Q,R$ are found  by cyclic permutation. Thus, $ AP,BQ,CR$ concur at

$ \left (\frac {1}{b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \minus{} a^4}: \frac {1}{b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \minus{} b^4}:\frac {1}{b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \minus{} c^4} \right )$

Which is the isotomic conjugate of $ X_{1078} \equiv (a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \minus{} a^4)$

[quote="Luis González"][b]Theorem:[/b] $(O)$ is an arbitrary circle in the plane of $ \triangle ABC.$ $P,Q,R$ are the poles of the sidelines $ BC,CA,AB$ with respect to $ (O).$ Then $AP,BQ,CR$ concur.
[/quote]
Here is a synthetic solution for this theorem:-
Its enough to prove that intersection of the polars of $ A,B,C $ wrt $ (O) $ with $ BC,CA,AB $ are collinear. Suppose, $ A',B',C' $ are the inverse points of $ A,B,C $ wrt $ (O) $. $ A_1,B_1,C_1 $ are located on $ BC,CA,AB $ such that $ A_1A'\perp OA $ and similar for the others. We have to prove that $ A_1,B_1,C_1 $ are collinear. Suppose, $ AA_2,BB_2,CC_2 $ are the altitudes of $ ABC $. Consider the circles $ \odot AA'A_2,\odot BB'B_2,\odot CC'C_2 $. Clearly, the orthocenter $ H $ of $ ABC $ wrt these circles has same power and so has $ O $. So these circles are co-axial. Their centers are midpoints of $ AA_1,BB_1,CC_1 $ and so they are collinear and this implies $ A_1,B_1,C_1 $ are collinear. So done.

A friend of mine from New Zealand( Malcolm Granville,  a former participant of IMO) remarked to me that he discovered this theorem while working on a special case of Nine Point Circle which appeared in some Romanian TST, as also remarked by Luis in his second post! The generalized statement, as he puts, is the following:

Given a triangle $ ABC $ and a conic $ \omega $. Let the poles of the lines of the sides be $ A' $, $ B' $ and $ C'  $ corresponding  to $ BC $  , $ AB $ and $ AC $. Then, the lines $ AA' $ , $ BB' $ and $ CC' $ are concurrent.

Proof: Let the points   $ AB \cap A'B' =   Z  $ ,  $ AA'  \cap BB'  = P $ ,      $ AB \cap   CC'  =   U  $ and   $  A' B'  \cap CC'   =  U'    $  be in the projective plane.

We have  $  ( A, B  ; Z , U ) = ( CA  , CB  ; CZ  ,  CU ) $ and taking the poles of the lines in the last bundle, we have  $  ( CA, CB ; CZ ,  CU )  =  (  B' , A' ; U' , Z  )  = ( A' , B' , Z , U' )  =  ( PA', PB' , PZ, PU' ) $ and so the perspectivity from $ P $ sends $ A $, $ B $ , $ Z $ and $ U $ to $ A' $ , $ B' $ , $ Z $ and $ U ' $  respectively or $ P $, $ U $ and $ U' $  are  collinear which means $ P \in  CC' $.

The existence of Gergonne and Symmedian Point can be proved by this theorem, as observed by him.

[quote="RSM"]Clearly, the orthocenter $ H $ of $ ABC $ wrt these circles has same power and so has $ O $. So these circles are co-axial.[/quote]
I'm not understanding this part. $H$ can  be the radical center of those three circles but how can you claim that those three circles are co axial?

[quote="RSM"]I wrote that, both $ H $ and $ O $ have same power wrt the 3 circles. So clearly, they are co-axial.[/quote]
Opps. Sorry i didn't notice that before :oops_sign:  :oops:

We can prove a stronger result when $A, B, C$ are inscribed in a conic:

Let $\triangle ABC$ be inscribed in a conic $\mathcal{C}$ and let $M$ be a variable point on $\mathcal{C}.$ Let $P, Q, R$ be the poles of $BC, CA, AB$ WRT $\mathcal{C}$, and let $\tau$ be the trilinear polar of $M$ WRT $\triangle ABC.$ Then $AP, BQ, CR, \tau$ concur.

[b]Proof:[/b] Let $\triangle M_AM_BM_C$ be the cevian triangle of $M$ WRT $\triangle ABC$ and denote $D \equiv BC \cap M_BM_C.$ Let $Y, Z$ be the poles of $BQ, CR$ and denote $X \equiv QR \cap YZ.$

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By Brokard's Theorem for quadrilateral $ABCM$, $\triangle M_AM_BM_C$ is self-polar WRT $\mathcal{C}.$ Hence the poles of $P, M_B, M_C$ concur (at $M_A$), so $P, M_B, M_C$ are collinear.

Note that the polar of $D$ is the intersection of the polars of $BC$ and $M_BM_C$, which is precisely line $PM_A.$ As $D \in \tau$, we get that $PM_A$ passes through the pole $\mathcal{T}$ of $\tau$ WRT $\mathcal{C}.$ Similarly, $PM_A, QM_B, RM_C$ are concurrent at $\mathcal{T}.$ From \[ Q(X, Y; Z, M_B) = (A, Y; C, M_B) \stackrel{P}{=} (A, B; Z, M_C) = R(X, Y; Z, M_C), \] we get that $\mathcal{T} \equiv QM_B \cap RM_C \in YZ.$ Hence, the polars $BQ, CR, \tau$ of $Y, Z, \mathcal{T}$ are concurrent. Similarly we get $AP, BQ, CR, \tau$ are concurrent. $\square$

Essentially the same solution as posted by multiple users, but I still feel like posting it :)

Solution: Let $O$ be the centre of the circle $(A_1B_1C_1)$. Clearly, $A_3$ lies on the perpendicular bisector of $A_1A_2$. Thus, the perpendiculars from $A_3$ to $BC$, and the like, are concurrent at $O$. Notice that $A$ lies on the polar $B_3$ and $C_3$ since $A=B_1B_2\cap C_1C_2$ and by duality, $B_3C_3$ is the polar line of $A$ wrt $(O)$. Thus, perpendiculars from $A$ to $B_3C_3$ and other lines defined similarly, are concurrent at point $O$. Hence, triangles $ABC$ and $A_3B_3C_3$ are orthologic and both the centers of orthology coincide. By the Sondat's theorem, the above pair of triangles is perspective and the result follows.

[b]Comment[/b] The result also holds when the circle is replaced by any conic. The proof goes by applying a projective transformation sending the conic to a circle and proceeding as above.


The letters x and y represent rectangular coordinates. Write each equation using polar coordinates (r, theta).

(1) x^2 + y^2 = x

(2) 4x^2y = 1

(3) y = -3

A = theta

1. x^2 + y^2 = x
r^2 = r cos A
r = cos A

2. 4x^2*y = 1 (is this what you meant by 4x^2y?)
r^2*cos^2 A = 1/(r sin A)
r^3*cos^2 A = 1/sin A
r^3 = 1/sin A/cos^2 A.
r^3 = 1/(cos^2 A * sin A) = sec^2 A *csc A

(I'm not too sure if this one's right)

3. y = -3
r sin A = -3
sin A = -3/r

Sorry if this is wrong--haven't done trig for a while


With 4 digits  5, 5, 5  and 1, and the operator + - *  /  can you present the number 24 ? Remeber $5*5-1^{5}$ is not allowed. Also you can't have $55$ in your solution.

  Good luck  :D  :D  :D

[quote="funcia"][quote="rnwang2"][hide]
$5\cdot\left(5-\frac{1}{5}\right)$[/hide][/quote]
is there any method to do this or is it guess and check?
JW[/quote]
It's basically just guess and check.

[quote="lingomaniac88"][quote="Gibbenergy"][quote="rnwang2"][hide]
$5\cdot\left(5-\frac{1}{5}\right)$[/hide][/quote]

 Very nice sols. I think it is the only way too.  :D[/quote]

*plays clip of Dr. Cox singing "wrong wrong wrong wrong..."*

[hide]$5-5+5-1$[/hide][/quote]
what the heck?  that equals 4 :maybe:

[quote="lingomaniac88"][quote="Gibbenergy"][quote="rnwang2"][hide]
$5\cdot\left(5-\frac{1}{5}\right)$[/hide][/quote]

 Very nice sols. I think it is the only way too.  :D[/quote]

*plays clip of Dr. Cox singing "wrong wrong wrong wrong..."*

[hide]$5-5+5-1$[/hide][/quote]
uhh that's 4

I have a disease.  I can't see the number 2.  :rotfl:

Huh?  Did I type something after the word "number"?  What is it?  I don't see it.  I swear I did, but I can't see it!

[quote="mathgeniuse^ln(x)"]Another way of doing it is this

$(5-\frac{1}{5})(5)$[/quote]
Uh that's the same as rnwang's solution.

[quote="Gibbenergy"][quote="rnwang2"][hide]
$5\cdot\left(5-\frac{1}{5}\right)$[/hide][/quote]

Very nice sols. I think it is the only way too.  :D[/quote]

The following webpage can search for the solutions in this type of questions: :lol:
[url]http://www.mikekong.net/Maths/Quizzes/24game.php[/url]

I have entered (5,5,5,1), and the only outcome is the solution given by rnwang2, so I believe that (5,5,5,1) is the only way. :)


inequalities
function
integration
calculus
real analysis
real analysis unsolved

let $f,g,h: \mathbb{R}\to \mathbb{R}^{+}$ be $3$ positive continuous functions and $\lambda \in [0;1]$. Suppose that for every $x,y \in \mathbb{R}$ the following inequality holds:
$h(\lambda x+(1-\lambda ) y) \geq f^{\lambda}(x) g^{1-\lambda}(y)$
Show that
$\int_{\mathbb{R}}h \geq (\int_{\mathbb{R}}f)^{\lambda}(\int_{\mathbb{R}}g)^{1-\lambda}$

This inequality is true for function from $\mathbb R$ to $\mathbb R$. For 1-dimensional situation it's collorary of Brunn-Minkowski inequality.
Actually wlog $\| f \|_{\infty}= \| g \|_{\infty}= 1$. For an arbitrary $a \in [0,1)$ we have relation
$\{ h \geq a \}\supset \lambda \{ f \geq a \}+(1-\lambda) \{ g \geq a \}$. Therefore $\mu \{ h \geq a \}\geq \lambda \mu \{ f \geq a \}+(1-\lambda) \mu \{ g \geq a \}$. The last step is integration over $a \in [0,1]$.


Erika, who is 14 years old, flips a fair coin whose sides are labeled 10 and 20, and then she adds the number on the top of the flipped coin to the number she rolls on a standard die. What is the probability that the sum equals her age in years? Express your answer as a common fraction.

The only way to get 14 is to get 10 on the coin and 4 on the die.  This happens with probability $ \frac{1}{2}\cdot \frac{1}{6}\equal{}\boxed{\frac{1}{12}}$


So here's a problem I've been thinking about, but I've not been able to settle yet. Let's say you have a convex polygon with n sides, where n >= 4 and then we cover that polygon with a smaller homothetic copy. Is it true that the non-covered region will always be nonconvex? Prove whether this is true or not.

If the smaller copy does not have to be in the interior of the original larger polygon, then no, the statement is not true and the non-covered region can be convex. One example is the pentagon with just one edge of the smaller copy lying inside the original; this can be done homothetically.

If the smaller copy must be contained in the larger original, then the statement is true. Since the smaller copy can share at most two (consecutive) edges with the original, the copy will have at least two consecutive edges, neither one which shares edges with the original. Since the polygon is convex, the region outside of those two edges is concave.

Cover a hexagon (with appropraite shape, say regular hexagon) with 4 copy at four coner with two opposite coner ignored, then remained areas are 2 triangulars and 3 quadrangles.


lithium heptylide with 1-bromo 3-chloro propane ofcourse with reasoning :)

[hide="Answer"]Sn2 reaction, with bromide being a better leaving group than chloride.[/hide]

yes that's perfect but i am 'worried' about the aswer pretty much the reason we were discussing in the other thread today...
why doesn't the strong base heptylide capture the acidic $ \beta$ hydrogen $ \beta$ to both chlorine and bromine :?:

Because that's a secondary carbon from which the base would have to approach (with also the two big halogens encumbering the way). SN2 at a primary and electropositive carbon is much faster.

would the answer be the same if i had acetylide ions :maybe:

can we take this as a thumb rule that the alkenylide ions mostly behave as nucleophilles with halides,tosylates that is Saytzeff Leaving Groups :)

Yes, if they are primary. Otherwise, elimination will occur.

I think this is corey-houser reaction. :)

yes somewhat but not corey house itself i suppose corey house specifically refers to alkane chain expansion :maybe:

The Corey-Posner, Whitesides-House reaction refers to a completely different thing: the reaction between Gilman reagents and organic halides.


Take each car separately.  First, you have $v_{0}=\frac{220}{9}\,\textrm{m}/\textrm{s}$ and $a=-8\,\textrm{m}/\textrm{s}^{2}$.  Now use the equation \[v=v_{0}+at=0,\] solve for $t$, then plug this into the equation \[d=v_{0}t+\frac{1}{2}at^{2}\] to get the distance travelled.  Doing the same thing for both cars will give you $d_{1}$ and $d_{2}$.  If $d_{1}+d_{2}<85$ then the cars will not collide.  Otherwise they will.

[quote="arlene de la cruz"]the maximum deceleration of a car on a dry road is about 8.0 m/s2. if two cars are moving head on toward each other at 88 kph, and thier drivers apply thier brakes when they are 85 m apart, will they collide? if so, at what relative speed? if not. how far apart will they be when they stop?[/quote]

Apply the formula: $v_{f}^{2}= v_{i}^{2}+2a\Delta x$ in the following way.

$v_{f}= 0,$

$v_{i}= 88 \times 10^{3}\text{ m/hr}\times\frac{\text{ 1 hour}}{3.6 \times 10^{3}\text{ seconds}}= 24.45 \text{ m/s}$

$a =-8.0\, m/s^{2}$

$\therefore \Delta x = \frac{(0-(24.45)^{2})}{2(-8.0)}= 37.35\text{ m}$

${\Delta x < \frac{85 \text{ meters}}{2}}$, so yes the cars will stop before they collide.

$\text{separation}= 85 \text{ m}-2(37.35\text{ m}) = 10.3 \text{ m}$


geometry probability AMC AIME USA(J)MO USAMO algebra \/closed

How many people from WOOT are taking it next year?

I'm trying to convince my parents to let me take WOOT this year...I think I'll probably be able to, but I have to convince them that it's worth the money.

I want to know if WOOT will be more difficult and intense than last year before I decide whether I want to join or not. The problem sets last year were short and many had too many easyish problems. Also does anyone know about the difficulty of the problems that will be done in class? I hope WOOT is more difficult this year and also I think there is some repetition which isn't nice.

We do not intend for there to be any repetition of problems in 2009-10 from 2008-09.  As for the difficulty of the problems, I think you're in the minority -- we had more students complaining about the problems being too hard than too easy.  That said, there will be some more very hard problems, as we will attempt to increase the variance of the difficulty of the material.

I should add that we will re-use some handouts from 2008-09, but these will be in addition to the 8+ handouts that we will offer for 2009-10 that were not part of 2008-09 WOOT.  (That is, we'll have more than 8 handouts for the course.)

Overall will there be more message board problems this year because there wasn't enough challenging message board problems (there was like one or two really hard ones I couldn't get but I didn't really like that.)

We'll make sure there are plenty of challenging message board problems.

[quote="thing1"]Will there be Olympiad classes this upcoming year? I think those are generally more challenging than the WOOT lectures. If not will WOOT be more challenging this year? Last year many of the classes only did easy problems while the message board problems were often hard.[/quote]

As you'll note from the WOOT pages, we have restructured the course somewhat, by grouping classes roughly by subject area.  Our intent here is to allow us to drill deeper into each subject area, pushing more of the introductory material into handouts, and allowing us to do more of the challenging material in class.  Of course, this only helps students who will actually read the handouts and do some of the problems therein!

We might offer the Olympiad Geometry class next spring, as we did this year, but we haven't yet decided.  Most of the material from our old Olympiad Problem Solving course has been integrated into WOOT and the Intermediate Counting & Probability course.

Is there any way we can buy the course materials or take the course without feedback for cheaper? I would still be willing to pay reasonably.

So does the Intermediate Counting and Probability class cover a lot of AIME/USAMO-level material?  Is it also targeted toward WOOT-level people?  Does it cover material not included in WOOT?

Yes, it has a lot of AIME - beginning USAMO material.  It is targeted towards a similar crowd as beginners in WOOT.  It has a little topical overlap with WOOT, but will not use the same problems.

Some of the sample problems like all the polynomial ones are really simple or stuff thats in the AOPS text. Same goes for many other sections but the graph theory has one from 2008 USAMO so I'm wondering what the level of difficulty of WOOT will be for this year? Will the classes be as challenging as the advanced part of the practice olympiads?

Yes, some will.  That is part of the reason we re-structured WOOT the way we have, to allow more depth in fewer areas each year rather than just touching on many areas.  We'll still give some exposure to areas outside the lectures, through handouts and problems on the message board, but rather than try to have classes on 12-16 different areas, we'll have classes on fewer areas, but cover them to greater depth (i.e. to greater complexity and difficulty) by having more than one class per area for most areas we study.

[quote="blackbelt14253"]Hmm...
Will WOOT still be offered to all MOPpers for free next year?[/quote]

The 2009 MOPpers will be invited to 2009-2010 WOOT for free.

[quote="veezbo"]I've got a question. I REALLY REALLY want to take WOOT next year, because all the topics seem interesting, and I definitely want to improve my proof-writing skills.

There are three things though:

1. My average score on the AIME is precisely 5 now, and I am unsure whether this will be enough for WOOT (because I have not seen any incredibly increase in score)

2. Put simply, I find it extremely difficult to write a proof. Usually, what ends up happening is that I think I have this great solution, but when I try and put on paper, it either doesn't work, or there is some case that I missed and renders the whole thing useless.

3. My problem solving and mathematical ability...unfortunately is not that high. I am mostly good at finding patterns and trying something out for a while before I can come to a legitimate conclusion. This type of thinking is also very difficult to write down on paper in the form of a proof.
In other words, I can solve problems requiring trivial observations, but for something more complex (like nearly any olympiad problem), something always goes wrong (to this day, I haven't solved one).

Do you still think WOOT would be a valid choice for me? One reason I want to take it next year is because that is 10th grade for me, and it will be the year I can put most time into it, as Junior year is 3 times more hectic.[/quote]

It depends on what you do between now and September.  If you do nothing, then you would likely be better off with the Intermediate classes than WOOT.  If you spend the next 2-3 months working on math, then you will likely be ready for WOOT.  Some resources to consider: from AoPS, Interm Algebra, Interm Counting & Probability, AoPS Vol 2.  From Amazon: Art and Craft of Problem Solving by Paul Zeitz, Problem Solving Strategies by Arthur Engel.  (I recommend the latter two after you've done either both Interm books or AoPS Vol 2.)  You don't have to finish all of these books prior to WOOT, but you should spend a significant amount of time with them.

As for proof writing, we will be including mathematical writing as part of the curriculum in WOOT this year.