Να βρεθουν οι πραγματικοι αριθμοι $ x,y$ που επαληθευουν την εξισωση:

$ 2^{2x^2 \minus{} 8y \plus{} 2} \plus{} 2^{2x \minus{} 8y^2 \minus{} {\frac {1}{2}}} \equal{} 2^{x^2 \minus{} 4y^2 \plus{} x \minus{} 4y \plus{} {\frac {7}{4}}}$

Σιγουρα δεν ειναι ασκηση επιπεδου ολυμπιαδων αλλα πιστευω πως θα μπορουσε να ειναι θεμα Θαλη ή   Ευκλειδη Γ Λυκειου

Βαζω και μια υποδειξη μαζι με τη λυση σε hide:

[hide]Μεταφεροντας ολους τους ορους της εξισωσης στο 1ο μελος σχηματιζεται ενα τελειο τετραγωνο, και μετα κανοντας τις πραξεις προκειπτει οτι: $ (x, y) \equal{} \left(\frac {1}{2}, \frac {1}{2}\right)$[/hide]

Ωραία ασκησούλα αν και εύκολη αφού αρκεί να παρατηρήσει κανείς καλά τους εκθέτες.

$ 2^{2x^2 \minus{} 8y \plus{} 2} \plus{} 2^{2x \minus{} 8y^2 \minus{} {\frac {1}{2}}} \equal{} 2^{x^2 \minus{} 4y^2 \plus{} x \minus{} 4y \plus{} {\frac {7}{4}}}\Longleftrightarrow$

$ (2^{x^{2} \minus{} 4y \plus{} 1})^{2} \minus{} 2\cdot2^{x^{2} \minus{} 4y^{2} \plus{} x \minus{} 4y \plus{} \frac {3}{4}} \plus{} (2^{x \minus{} 4y^{2} \minus{} \frac {1}{4}})^{2} \equal{} 0\Longleftrightarrow$

$ (2^{x^{2} \minus{} 4y \plus{} 1} \minus{} 2^{x^{2} \minus{} 4y^{2} \minus{} \frac {1}{4}})^{2} \equal{} 0\Longleftrightarrow$

$ 2^{x^{2} \minus{} 4y \plus{} 1} \equal{} 2^{x^{2} \minus{} 4y^{2} \minus{} \frac {1}{4}}\Longleftrightarrow$

$ x^{2} \minus{} x \plus{} 4y^{2} \minus{} 4y \plus{} \frac {5}{4} \equal{} 0\Longleftrightarrow$

$ 4x^{2} \minus{} 4x \plus{} 16y^{2} \minus{} 16y \plus{} 5 \equal{} 0\Longleftrightarrow$

$ (4x^{2} \minus{} 4x \plus{} 1) \plus{} (16y^{2} \minus{} 16y \plus{} 4) \equal{} 0\Longleftrightarrow$

$ (2x \minus{} 1)^{2} \plus{} (4y \minus{} 2)^{2} \equal{} 0\Longleftrightarrow$

$ x \equal{} \frac {1}{2}$ και $ y \equal{} \frac {1}{2}$

Σωστος, βαζω τωρα μια που νομιζω πως ειναι δυσκολοτερη

Να βρεθουν οι πραγματικοι αριθμοι $ x, y$ που επαληθευουν την εξισωση:

$ k^{x \plus{} y \minus{} y^2} \plus{} k^{y^2 \minus{} 2y \plus{} x} \plus{} k^{y \minus{} 2x} \equal{} 3$, οπου $ k$ τυχαιος θετικος αριθμος διαφορετικος του 1

πραγματι η ΑΜ-ΓΜ δινει λυση και στις δυο ασκησεις μας.
[hide]1)$ 2^{2x^2 \minus{} 8y \plus{} 2} \plus{} 2^{2x \minus{} 8y^2 \minus{} {\frac {1}{2}}}\geq2*\sqrt {2^{2x \minus{} 8y^2 \minus{} 1/2 \plus{} 2x^2 \minus{} 8y \plus{} 2}} \equal{} 2^{x^2 \minus{} 4y^2 \plus{} x \minus{} 4y \plus{} 7/4}$ επομενως με περιπτωση ισοτητας
εχουμε δηλαδη : $ 2x^2 \minus{} 8y \plus{} 2 \equal{} 2x^2 \minus{} 8y \plus{} 2 < \equal{} > x \equal{} 1/2 \equal{} y$
2)$ k^{x \plus{} y \minus{} y^2} \plus{} k^{y^2 \minus{} 2y \plus{} x} \plus{} k^{y \minus{} 2x}\geq3*k^0 \equal{} 3$
αρα με περιπτωση ισοτητας $ (x,y) \equal{} (3/4,3/2)$ ή $ (x,y) \equal{} (0,0)$[/hide]

για την 1 επισης μπορουμε να πουμε οτι αν διαρεσουμε τα παντα με το $ 2^{2x^2\minus{}8y\plus{}2}$
καταληγουμε σε μια σχεση $ 1\plus{}2^m\equal{}2^n$η αλλιως αρτιος=περιττος ατοπο
αρα πρεπει 2^m=1 kai 2^n=2
η αλλιως m=0,n=1
πιο συγκεκριμενα εχουμε να λυσουμε το συστημα
$ x^2\minus{}4y^2\plus{}x\minus{}4y\plus{}7/4\equal{}2x^2\minus{}8y\plus{}2\plus{}1$
$ 2x^2\minus{}8y\plus{}2\equal{}2x \minus{} 8y^2 \minus{} {\frac {1}{2}}$το οποιο εχει μονη κοινη λυση $ x\equal{}y\equal{}1/2$ :)

Athinaios ωραια η λυση σου με την ΑΜ-ΓΜ και στις 2 ασκησεις, η 2η σου λυση ομως στη 1η ασκηση νομιζω πως ειναι λανθασμενη αφου οι αριθμοι $ m, n$ μπορει να μην ειναι ακεραιοι.

[quote="Athinaios"]pragmati lathos i skepsi m.... :oops: apo sinitheia logw tis thewrias arithmwn den to skeftika swsta,[/quote]

οκ δεν πειραζει, ολοι κανουμε λαθη καμια φορα :P


A six place number is formed by repeating a three place number; for example, $ 256256$ or $ 678678$, etc. Any number of this form is always exactly divisible by:

$ \textbf{(A)}\ 7 \text{ only} \qquad\textbf{(B)}\ 11 \text{ only} \qquad\textbf{(C)}\ 13 \text{ only} \qquad\textbf{(D)}\ 101 \qquad\textbf{(E)}\ 1001$

[hide]If we express the number as $ abcabc$, where a, b, and c are digits, then we can rewrite this as $ (abc)(1001)$, so this number is divisible by $ 1001$, so $ \mathrm{ (E) }$[/hide]


let $ ABC$ be a triangle with $ \angle A\equal{}36^o$ and:

$ 2S_{\triangle ABC}\equal{}ab\sqrt{\sin^2A\plus{}\sin^2B\plus{}\sin A\sin B}$

Find the angles of $ \triangle ABC$.

[quote="sandu2508"]$ \triangle\ ABC$ with $ 2S \equal{} ab\sqrt {\sin^2A \plus{} \sin^2B \plus{} \sin A\sin B}\ \Longrightarrow\ C \equal{} 120^{\circ}\ .$ [/quote]

$ \left\|\ \begin{array}{c} 2S \equal{} ab\sin C \\
 \\
\frac {\sin A}{a} \equal{} \frac {\sin B}{b} \equal{} \frac {\sin C}{c} \\
 \\
c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos C\end{array}\ \right\|\ \Longrightarrow\ \left\|\ \begin{array}{c} \sin^2C \equal{} \sin^2A \plus{} \sin^2B \plus{} \sin A\sin B \\
 \\
\Downarrow\\\\
c^2 \equal{} a^2 \plus{} b^2 \plus{} ab \\
 \\
\Downarrow\\\\
\cos C \equal{} \minus{} \frac 12 \\
 \\
\Downarrow\\\\
\boxed {\ C \equal{} 120^{\circ}\ }\end{array}\ \right\|$


it is not 'that' goldbach conjecture....instead, it is an easy problem that is hard only to me...

We difine the set $P$ recurrently as follow.
$P$$=[m^{n}|m\geq 2 ,n\geq 2, m\notin\ P ]$

To say it another word, let$m=2$, and put the numbers of the form $m^{n}$into$P$. Then, we increase $m$ by $1$ (that is $m=3$ ), and put the numbers of the form $m^{n}$ only if $m$ is not already the member of$P$.

prove that the sum$\sum_{k\in\ P }\frac{1}{k-1}=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}..... =1$

You can find a solution in 'Concrete Mathematics' by Knuth, Graham, Patashnik (Look it up in the Exercice section of Chapter 2, then refer to the Solutions)
I will post it soon if you mind ...

sorry, but I do not have that book....I want to buy one, but my national olympiad is urgent so I cannot see that book throughly....If you give me some kindness to post the solution, I will be very appreciated...thanks for your help..^^

Enjoy : http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/goldbach.pdf   :)

How am I happy to meet you again and now I am ready to answer your lovely question. Firstly induct on the question. For n=2, answer is 10/21 and let suggest be true and for a next step it becomes always a number as an inductive step which is less than the difference of the kth and (k+1)th steps,it actually proves our problem.
Thanx.


geometry
circumcircle
perpendicular bisector
angle bisector

[i]Trying to solve a more difficult problem , I found this easy,nice and elementary result.[/i]

  Let $D$ be a point on the side $BC$ of an equilateral trinagle $ABC$. If $O$ is the circumcenter of tringle $ADC$ , prove that $BO$ bisects angle $B$

   [u]Babis[/u]

Thank you my friend ! Really sorry for the typo.I apologize.
  Now I have edited.
I please Farenhajt not to send his solution today.I'm sure he has at least one. The problem is  easy for his math-level.

  [u] Babis[/u]

As you can see, my friend, I restrained from doing so, I just corrected the typo :)

[hide]Let $M$ be the midpoint of $\overline{AC}$.
$\overline{OM}\perp\overline{AC}$ since $O$ lies on the perpendicular bisector of $\overline{AC}$.
$\overline{BM}\perp\overline{AC}$ since in equilateral triangle $ABC$, the median and altitude to one side coincide.
Therefore, $B, O, M$ are collinear.
$\overline{BOM}$ is also the angle bisector of $\angle B$, since $ABC$ is an equilateral triangle.[/hide]

Bictor,

 your nice anad very simple solution works also for this problem:

   [color=orange]Let $ABC$ an isosceles triangle with $AB = AC$ and $D$ a poiont on side $AC$.If $O$ is the circumcenter of triangle $BCD$ , prove that $AO$ is the bisector of angle $A$.[/color]


Six numbers are drawn at random, without replacement, from the set {1, 2, 3, ..., 1000}. Find the probability that a brick whose side lengths are the first three numbers can be placed inside a box with side lengths the second three numbers with the sides of the brick and box parallel.

[hide="Hint"]
The probability in question is the number of ways to, given six distinct number, break them into three sets with the property in question divided by the total number of partitions.[/hide]

[hide="solution"]
Suppose you have the six number (1, 2, 3, 4, 5, 6).  You break the set into two sets of 3 numbers each.  The first set is the brick, while the second is the box.  The second set MUST have the 6, while the first set MUST have the 1.  Once those have been assigned, there are 5 ways to partition the set into two subsets with the properties in question: $(1,2,3) (1,2,4) (1,2,5) (1,3,4) (1,3,5)$ as the items in the first set.  The total numer of ways to construct the first set is $\binom{6}{3}=20$.  So our final answer is $\frac{5}{20}=\boxed{\frac{1}{4}}$...[/hide]


i'm using geometer sketchpad to create diagrams, then making pdfs out of the gsp files using the virtual printer. after that i implement the pdf into another pdf. (a standard way to include pictures)

but I'm having a problem seeing the betas from geometer sketchpad. everything else is fine, but where there should be a beta, i see a small rectangular box.

what can i do to get beta to work correctly?

I think we had this problem too in the dark days when we used to use GSP to generate our diagrams.  I think you have to go into a graphics editor (we used to use Fireworks or Freehand) and fix it by hand.   This made us  :(

Now we use MetaPost to generate all our diagrams, and we're  :)


calculus
integration
function
real analysis
real analysis unsolved

Show that $ f(x)\equal{}cos(\frac{\pi}{x})$  for  $ 0< x\leq 1,$   $ f(0)\equal{}0$ is integrable on $ [0,1]$

I'm assuming that the question comes from a very early stage of the development of the Riemann integral, before any attempt has been made to characterize all integrable functions. I'm also assuming that some significant general integrability theorem has been proved, probably this one: if $ f$ is continuous on $ [a,b],$ then $ f$ is (Riemann) integrable on $ [a,b].$

The function in question has two significant properties that will be used.

1. It is continuous except at zero. In particular, it is continuous on $ [\delta,1]$ for any $ \delta>0.$

2. It is bounded.


[b]For every point M inside triangle ABC we define a triplet (d1, d2, d3),
where d1, d2, d3 are distances from M to the sides BC,AC,AB. Prove that
the set of points M, satisfing condition d1 ·d2 ·d3  r3, where r is the incircle’s
radius, lies inside the circle with center O and radius OI.[/b]

[quote="mufc"][b]...satisfing condition d1 ·d2 ·d3  r3, where r is the incircle’s
radius...[/b][/quote]

Please clarify your statement. (My message will be deleted after.)


1.what is the radius and center of this?

x^2 + y^2 +2x -4y = 20

2. What is the radius of this circle

x^2 + y^2 +10y + 9 = 0

a little confused for number 2- the constant term is 9, and if you move everything to the other side, the 9 is isolated and the radius is the square root of that = 3? am i correct so far?

as for the 7^2 + 10y, if you complete the square, you get a 25 - 9 on oneside, and the radius is 4. which one is the right one?

- 3,
- root 14,
- 4,
- root 34

Moved.
furia, try completing the square.  :)

[quote="K81o7"][quote="furia"]
2. What is the radius of this circle

x^2 + y^2 +10y + 9 = 0

a little confused for number 2- the constant term is 9, and if you move everything to the other side, the 9 is isolated and the radius is the square root of that = 3? am i correct so far?

as for the 7^2 + 10y, if you complete the square, you get a 25 - 9 on oneside, and the radius is 4. which one is the right one?
[/quote]

If you move the 9 over to isolate it, you get $x^2+y^2+10y=-9$.  But $\sqrt{-9}$ is NOT equal to 3...You, of course, do have to complete the square to get $x^2+(y+5)^2=4^2$[/quote]

yes, but if you move everything to the other side, the 9 will be positive

I don't quite understand what you mean by "move everything to the other side"... Anyway, whenever you move something to the other side, its sign changes. In this case, when you move $9$ to the right side, we have $x^2+y^2+10x=-9$, as K81o7 wrote. If you move everything to the other side, though, every single sign changes, so we get $0=-x^2-y^2-10x+9$. But this is essentially the same as the original equation because dividing (or multiplying) the new equation by $-1$ gives the original equation.

Also, the constant term ($9$ in this example) doesn't have to be $r^2$. In fact, the constant doesn't mean anything before you complete the square.


function
real analysis
real analysis unsolved

Let f be a derivable, strictly increasing and bijective function from the set of positive real numbers to the same set. Suppose that $\frac{f'(x)}{x^2}$ tends to infinity when x tends to infinity. Prove that the adherence of the sequence $ f^{-1}(n)\cdot \{f^{-1}(n)\}$ includes the set of positive real numbers.

[i]Edited by Myth[/i]

I don't understand. Do you mean closure of set $f^{-1}(n)\{f^{-1}(n)\}$ is $[0,+\infty)$?


algebra
polynomial
inequalities
inequalities proposed

I don't follow.  We have $ M_3 \ge A_3 \ge G_3 \ge H_3$ by Power Mean.

[quote="Guest"][quote="fjwxcsl"]Sorry!
$ A_3\geq{\frac {M_3 + G_3 + H_3}{3},A_3\geq{\frac {2M_3 + G_3 + H_3}{4}, A_3\geq{\frac {7M_3 + 3 G_3 + 3H_3}{13},......}}}$

If $ A_3\geq{\frac {kM_3 + G_3 + H_3}{k + 2},}$then $ k_{max} = ?$[/quote]
The best possible value of $ k$ is the greatest root of the following equation:

$ 1728\ k^{15} - 63936\ k^{14} + 855936\ k^{13} - 4508416\ k^{12} + 1529136\ k^{11} + 53628816\ k^{10} - 45390080\ k^9 - 286818432\ k^8 + 148772532\ k^7 + 787201932\ k^6 + 66791448\ k^5 - 1115075232\ k^4 - 911934927\ k^3 - 7355637\ k^2 + 242300916\ k + 69074694 = 0,$

that is, $ k_{max} = 2.388282179\cdots.$[/quote]
Can you explain how you got that?
My progress was: By the uvw-theorem it's easy to prove that we only have to check the case where $ x_3 = 0$ and the case where $ x_2=x_3=1$. The case $ x_3=0$, is quite easy, and it holds whenever $ k \le 1 + \sqrt{3}$.
In the case where $ (x_1,x_2,x_3) = (t,1,1)$ I was a bit more lost.. I inserted $ (t,1,1)$ and after squaring (with my computer!) I got a polynomial in $ t$ and $ k$ where $ t$ was of degree $ 10$ and $ k$ was of degree $ 2$. I realised that $ (t-1)^2$ was a root and hence I divided it out. But from that point I was pretty lost. :(

[quote="Mathias_DK"][/quote][quote="Guest"][/quote][quote="Mathias_DK"][quote="Guest"][quote="fjwxcsl"]Sorry!
$ A_3\geq{\frac {M_3 + G_3 + H_3}{3},A_3\geq{\frac {2M_3 + G_3 + H_3}{4}, A_3\geq{\frac {7M_3 + 3 G_3 + 3H_3}{13},......}}}$

If $ A_3\geq{\frac {kM_3 + G_3 + H_3}{k + 2},}$then $ k_{max} = ?$[/quote]
The best possible value of $ k$ is the greatest root of the following equation:

$ 1728\ k^{15} - 63936\ k^{14} + 855936\ k^{13} - 4508416\ k^{12} + 1529136\ k^{11} + 53628816\ k^{10} - 45390080\ k^9 - 286818432\ k^8 + 148772532\ k^7 + 787201932\ k^6 + 66791448\ k^5 - 1115075232\ k^4 - 911934927\ k^3 - 7355637\ k^2 + 242300916\ k + 69074694 = 0,$

that is, $ k_{max} = 2.388282179\cdots.$[/quote]
Can you explain how you got that? :([/quote][quote="Guest"]
I am sorry but I employ a package called "BOTTEMA" which works with Maple.[/quote][quote="Mathias_DK"]
Cheater ! :P :P :P[/quote]
You are an uneducated person! Perhaps you were never taught that do not animadvert on what beyond your knowledge.

Come on, that was just a joke of him.
Perhaps you were never taught to understand jokes...  :huh:

[quote="Mathias_DK"][quote="Guest"]You are an uneducated person! Perhaps you were never taught that do not animadvert on what beyond your knowledge.[/quote]
Where I come from the " :P " smiley is used to indicate, that the thing one just wrote was meant in a humorous context - and thus not animadverting.
You should really be better at insulting people, though. "Uneducated" is not really that harsh a word. You should at least say "stupid" :P[/quote]
I am very sorry indeed for my misunderstanding.

[quote="fjwxcsl"]Suppose

   $ A_3 \equal{} \frac {x_1 \plus{} x_2 \plus{} x_3}{3},G_3 \equal{} \sqrt [3]{x_1x_2x_3},M_3 \equal{} \sqrt {\frac {x_1^2 \plus{} x_2^2 \plus{} x_3^2}{3}},H_3 \equal{} \frac {3}{\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3}}$

  $ ({x_1,x_2,x_3} > 0),$

prove that

  $ A_3\ge{\frac {M_3 \plus{} G_3 \plus{} H_3}{3}}.$[/quote]

Replace $ (x_1;x_2;x_3)$ by $ (x;y;z)$

Inequality is equivalent with

$ x \plus{} y \plus{} z\ge \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3}} \plus{} \sqrt [3]{xyz} \plus{} \frac {3xyz}{xy \plus{} yz \plus{} zx}$

Assume $ xyz \equal{} 1,x \plus{} y \plus{} z \equal{} p,xy \plus{} yz \plus{} zx \equal{} q$, we have

$ RHS \le \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3}} \plus{} 2 \equal{} \sqrt {\frac {p^2 \minus{} 2q}{3}} \plus{} 2$

We will prove

$ p \minus{} 2\ge \sqrt {\frac {p^2 \minus{} 2q}{3}}$

$ \Leftrightarrow 3(p \minus{} 2)^2\ge p^2 \minus{} 2q$

$ \Leftrightarrow 2(p \minus{} 3)^2 \plus{} 2(q \minus{} 3)\ge 0$

Which true by AM-GM

[quote="gigaman"]

Inequality is equivalent with

$ ... \plus{} \frac {3xyz}{x \plus{} y \plus{} z}$

[/quote]

this must be $ \frac{3xyz}{xy\plus{}yz\plus{}zx}$

however, it's a nice proof

[quote="hungnsl"][quote="gigaman"]

Inequality is equivalent with

$ ... \plus{} \frac {3xyz}{x \plus{} y \plus{} z}$

[/quote]

this must be $ \frac {3xyz}{xy \plus{} yz \plus{} zx}$

however, it's a nice proof[/quote]

Sorry, I just fix it :D


AMC
USA(J)MO
USAMO
AIME
email
AMC 12
AMC 10

[quote="AMCDirector"]
The lowest AIME score among these students was 6 (144.0 + 10*6) so now we look at all students in 10th grade and below who got a 6 or better on the AIME.  There are 468 of them, so we need to use the USAMO Index to discriminate among them.

The 156 highest USAMO indices among them are at Index 202.5 and higher.

[/quote]

Thanks!

Meaning if you are a <=10 grade, you needed a 6 on the AIME with a index (AMC + AIME*10) score of 202.5

Oh wow, I greatly predicted this wrong.
So the index was still a 6, but there were so many of them many people that got 6s,7s,8s, or even 9s (maybe 10s) on AIME for <=10th graders still got left out.

I bet there were already more than 170 people that got a 7+ on AIME for <=10 grade, but the AMC 12 takers really brought this thing down, which is very very unfortunate for the people that got like a 120/8 or even 132/7 (and very lucky for someone I know that got a 144/6)

But seriously, i expect (and might agree) with some of the complaining thats going to be be happening on about this "index" for <=10 graders.

The big thing is that, should the floor for <=10 graders be decided based on the AMC 12 takers scores??  The problem is that, there were probalby already more than 160 or so people that were <= 10th grade that got a 7+ on the AIME (of course i dont know this for sure, but from that stats, its most liekly the case), so why did we have to make the more confusing by lowering even 1 more to 6+ (and thus greatly increasing the amount of people, and thus greatly increasing the index for <=10th graders).

Whoa, that's a really close cutoff for the floor score.  Sorry to those who didn't make it.

Thanks to the AMC for releasing the cutoffs so early.

This is very interesting.
What surprised me the most, when you look at Mr. Dunbar's post, is the mean score of the AIMEs. I mean, it's almost double the mean scores of 2004-2006.
Ooh, and I just noticed, this REALLY sucks for my friend who got a 132/7. Ouch, seriously.

[quote="the future"]Actually the 12 scorer who got 144+6*10 for the 204 index (and just made the normal index) and set the floor in 6 will be absoultely HATED by those who got 7s and 8s on the AIME (like a 132/7 or 120/8) but still didnt make USAMO.

Because of that 144/6, it probalby bumped this <=10th index up by like 10 points and thus made the AMC 10 WAY MORE SIGNIFICANT than the AIME.[/quote]

Keeping in mind what Mr. Dunbar posted elsewhere about the USAMO index being a better predictor of USAMO performance than AIME score alone, if the "floor" requirement for 10th grade and younger was eliminated, then the selection of approximately 160 young students would be chosen strictly upon their Index score.  This would yield an identical selection of 156 students at the Index of 202.5.

[quote="Math Geek"]In the words of Billy Dorminy, yippeeyahoo!!!!!  Yay, I made AIME as the lowest AIME scorer!! :rotfl:

But seriously, wow, I feel so special!  Fate smiled upon me today, cuz I thought I epik phailed and had no chance of making USAMO whatsoever!  Fortunately, I made the index cutoff by 1.5 points!!!!!

Okay, I think that's enough exclamation points to express my joy.  Back to studying for USAMO, I guess.

PS: How embarassing to be that 144+6*10 person. :blush:[/quote]

Same here! Woot! I also got a 144/6 because of careless mistakes on AIME...But now I feel really refreshed!  :lol: God bless that 144/6 person!

EDIT: Maybe I should not be too excited...barely making it with floor and 1.5 points above (if it's changed back to 7....)...yeah, anyways I should start on USAMO, even if I didn't qualify at the end...

[quote="Anaxerzia"]Man a 6 index is really surprising! This AIME was waaay to easy for that... IMO we know not to trust AoPS polls for determining Indices/Floor :P

EDIT: And WOW! I cannot believe someone got a 144/6 for teh AMC12/AIME!! Thats messed up lol[/quote]

I think there was a 150/6 last year who set the floor :).

Thanks to AoPS for the USAMO expansions recently.

[quote="Math Geek"][quote="the future"]Actually the 12 scorer who got 144+6*10 for the 204 index (and just made the normal index) and set the floor in 6 will be absoultely HATED by those who got 7s and 8s on the AIME (like a 132/7 or 120/8) but still didnt make USAMO.

Because of that 144/6, it probalby bumped this <=10th index up by like 10 points and thus made the AMC 10 WAY MORE SIGNIFICANT than the AIME.

Anyways, did you guys receive your official emails yet?[/quote]

Wait, I thought we were talking about grades 10 and below?  Cuz I'm 8th grade and took the 10 and got 144 and 6 on the AIME.  Does that apply to the 6 floor score and >=202.5 score to qualify, or are all my hopes and dreams going to suddenly be destroyed by the next post?[/quote]

yes you made it if you got 144/6.

What I am saying is that the reason it is a 6 (and not 7) is because some person that took the 12 got a 144/6 on the 12/AIME.  And this is good news for people like you but really bad news for people that got 7s or 8s on AIMEs with not so high AMC 10 scores.

ANyways, the names are out on the AMC website.  You should check.

Ah man this [b]really[/b] stinks with a 117/8.  If only I hadn't made that careless error on #2 on the AMC 10A, I would've made it with a 203.

[quote="beta"][quote="the future"]Actually the 12 scorer who got 144+6*10 for the 204 index (and just made the normal index) and set the floor in 6 will be absoultely HATED by those who got 7s and 8s on the AIME (like a 132/7 or 120/8) but still didnt make USAMO.

Because of that 144/6, it probalby bumped this <=10th index up by like 10 points and thus made the AMC 10 WAY MORE SIGNIFICANT than the AIME.

Anyways, did you guys receive your official emails yet?[/quote]

Mr.Dunbar said he's still expecting about a few more qualifiers, maybe that will bump the current index cutoff to 204.5 and maybe change the floor to 7[/quote]

Any additional qualifiers will qualify by the posted qualification scores.  The indices and floor will not change.

Elizabeth Claassen
AMC Senior Accounting Clerk
(and resident number cruncher)

Yeah, I know a freshman who got 126/7.  I feel pretty bad for him right now.

[quote="ProtestanT"][quote="i_like_pie"]127.5/7 :cursing:

No further comments...[/quote]

I guess that's a consequence of them making it 330 of those that took AMC12s and only 160 of those who took AMC10s[/quote]
Actually, its 330 AMC12ers, and in the 160 <=10th graders, there are a good deal of AMC12ers too.

I need to work on not bubbling things wrong, I managed to lose 10% of my grade in math by bubbling stuff wrong on tests, and now I miss usamo because I didn't bubble 2 answers correctly.

even though I took the AIME unofficially, i also got 132/7 :lol:

The statement above would have been typed with much more passion had I taken this for real.

but good luck to those who've qualified!

I might as well continue the trend by adding my 132/7 (although I took the AMC12 since I'm a junior).

[quote="Lazarus"]132/10 blah[/quote]

-__-... note the long list of depressing messages above yours daesun =P.  And then... yours xD...

Lots of 132/7s.

138/9  :P
i was pretty sure i would make it

[quote="dnkywin"]150/8 (AMC 10)  :P . Three people in my school (including me) made USAMO [b]AND[/b] nats in mathcounts.  :D  :D[/quote]

144/9 AMC 10.
Would have made nats if not for losing 7 or 8 points because of careless mistakes...

im confused, how come i didnt make it with a 135/7?

i thought that was enough to make it...

[quote="jrshoch"]Discussions such as this one arise every year.  Two years ago I was intensely interested in the debates, as I was a freshman who missed the index by less than 2 points but had an AIME score either one or two above the floor, i don't remember.  That year they ended up eliminating the index because the teacher manual actually advised taking the AMC12 in hopes of qualifying for USAMO; this misdirection was accounted for in that year, and has since been remedied by editing the teacher's manual.
[/quote]

Yeah dude I launched that whole rebellion thing and you can still look up the thread if you wish! (Letter Regarding the Non-Preferencing...)

Anyway so the AMC decided to remedy the situation by substituting a bunch of meaningless words in the Teachers' Manual with DIFFERENT meaningless words, and problem solved!

But I think the point is, I meant to spark a discussion, not just indulge my megalomania and qualify myself (and 56 other guys, mind you) into the USAMO via shouting "Rule 7" ad infinitum.  I sought to avoid such a debacle in future years by calling for a substantive re-thinking of the tie-breaking policy.

This did not happen(!)  I really, really cannot tell you why.

But dudes, dudes if you were on the borderline (say a 132 on the AMC 12 and a 7) and unfairly cheated out of the USAMO, you should not worry that much, because you have a bunch more years, and (take it from me) you were probably not that likely to destroy the USAMO anyway.  The best thing to do is just keep practicing and learn to think about hard problems, like in hard problems movie, which was a terrible movie, but whatever.  That way, you will be well-poised to EAT the USAMO 2009.  I wish you all good luck.

Sorry about the randomness, but: my friend made 144 on amc10 and 2 on aime. I just wanted to emphasize the big roles a "bad day" can play, bc he couldve gotten much better on aime.........

just wondering: is it bad to post on a dead thread?

It is indeed a very bad thing to do. If I had mod powers, I would reprimand you even more.

Don't revive old non-mathematical threads unless there's something new and interesting to say. Threads die for a reason; if you don't have anything new to say, people still won't be interested.


Problem:

Calculate sum of series $ S \equal{} \sum_{k\geq 0} \frac {k}{2^k}$.

What I've tried:

If we take $ f(k) \equal{} \frac {k}{2^k}$
we have
$ f(k \plus{} 1) \equal{} \frac {1}{2}f(k) \plus{} \frac {1}{2^{k \plus{} 1}}$
if we sum on both sides of this we get
$ S \minus{} 1 \equal{} \frac {S}{2} \plus{} 1$
so we get $ S \equal{} 4$

I'm probably wrong so I'd like to ask how to calculate this.

[quote="spx2"]$ S \minus{} 1 \equal{} \frac {S}{2} \plus{} 1$[/quote]

You made a small mistake here.  The first term of the series is not $ 1$, it is $ 0$, so what you want is:

$ S \minus{} 0 \equal{} \frac {S}{2} \plus{} 1$, hence $ S \equal{} 2$.

Another technique for doing this kind of series is to look at $ g(x) \equal{} \sum_{k\geq 0} \frac {k}{2^k} x^k$ and compute $ g'(x)$, but what you did works just fine.

2(S-1) = S/2 +1

S-1 = S/4 + 1/2

3/4*S = 3/2

S = 2

The sum should equal 2.

Essentially they're the same. F(k) = k/2k. The first term (k = 0) is equal to 0.
For F(k+1), the first time is the 2nd term of F(k), which makes them equal since the addition of 0 doesn't change the answer.

So you could also say F(k+1) = F(k) summation wise.... therefore
S = S/2 + 1,  not S-1 on the left hand side.

Aww, he wrote his answer as I was mine xD

[quote="LydianRain"]

Another technique for doing this kind of series is to look at $ g(x) \equal{} \sum_{k\geq 0} \frac {k}{2^k} x^k$ and compute $ g'(x)$, but what you did works just fine.[/quote]
 I think you mean :
define for $ |x|<1$ function
$ g(x) \equal{} \sum_{k\equal{}0}^{\infty}x^{k}$
on one hand : $ xg'(x)|_{x\equal{}1/2}\equal{}your sum$
on other hand $ g(x)\equal{}\frac{1}{1\minus{}x}$

Well yes, what I wrote was rather ridiculous, wasn't it?  :blush:


inequalities
induction
absolute value
inequalities proposed

Let $a_1,a_2,\ldots,a_n$ be real numbers such that
\[ a_1^2+a_2^2+\cdots+a_n^2=1.  \]
Prove that
\[ 1+n\geq(a_1+a_2+\cdots+a_n)(a_1+a_2+\cdots+a_n+a_1^3+a_2^3+\cdots+a_n^3).  \]

Oh, sorry I posted it in the wrong box. Mod, please move it.

You meant "suffices" I'm sure.

This is not "user-friendly" in the sense that it has an overabundance
of sigmas, sub i's, sub j's, and excessive absolute value symbols whereas the original problem had none of that.

"But it is obvious by" is not a helpful mathematical phrase at all.  By whose measure?

an induction as an alternative, showing the steps,  BUT I do not want the symbols that the alleged solver used above (i.e. sigmas, sub i's, sub j's, and absolute value symbols).


linear algebra
matrix
linear algebra unsolved

Find all positive integers $ n$ for which there exist matrices $ A,B,C\in M_{n}(\mathbb{Z})$ such that $ ABC \plus{} BCA \plus{} CAB \equal{} I.$ Here $ I$ is the identity matrix. (V. Brayman)

[hide]Since tr(ABC) = tr(CAB) = tr(BCA), we have tr(ABC + BCA + CAB) = tr(I) <=> tr(ABC) + tr(BCA) + tr(CAB) = tr(I) <=> 3 tr(ABC) = n <=> tr(ABC) = n/3. Since each of the matrices A, B, C is in Z, so will ABC and so tr(ABC) is an integer. Therefore, n must be a multiple of 3.[/hide]

I believe that it is not enough what Carcul said.It's only a necessary condition, but it's not sufficient.

For the converse, it's sufficient to give an example for $ n\equal{}3$:
$ c\equal{}\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix},b\equal{}\begin{pmatrix}0&0&0\\0&0&0\\0&1&0\end{pmatrix},a\equal{}\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$.

Assume $ A_1,B_1,C_1$ are solutions when $ n\equal{}3$. Then clearly
$ A_k\equal{} \begin{pmatrix}A_1 & 0 &...& 0 \\
0 & A_1 &...& 0 \\...&...&...&...\\
0 & 0 &...& A_1\end{pmatrix}$
$ B_k\equal{} \begin{pmatrix}B_1 & 0 &...& 0 \\
0 & B_1 &...& 0 \\...&...&...&...\\
0 & 0 &...& B_1\end{pmatrix}$
$ C_k\equal{} \begin{pmatrix}C_1 & 0 &...& 0 \\
0 & C_1 &...& 0 \\...&...&...&...\\
0 & 0 &...& C_1\end{pmatrix}$
is a solution when $ n\equal{}3k$.
As loup blanc wrote for $ n\equal{}3$
$ C_1 \equal{} \begin{pmatrix}0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0\end{pmatrix},B_1 \equal{} \begin{pmatrix}0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 1 & 0\end{pmatrix},A_1 \equal{} \begin{pmatrix}0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0\end{pmatrix}$ is a solution.

So $ n \equiv 0 (mod 3)$ is necessary and sufficient condition.


Solve    2sinx cos40° = sin(x+20°)   when 0°<x<90°.

This should probably moved to a different forum.

There should ABSOLUTELY be no trig in any forum in Middle School, let alone Classroom. (just a note, not too serious)  :D

I was reading the excellent article PDF "An Intriguing Geometry Problem" by Tom Rike (BMC) and...page 4, second solution
of  "Franklin's kite problem" ([b]kunny[/b] the 30th May 2005) finish with this equation ...wich
[b]I can not solve ![/b]
Wich forum is appropriate ? :ninja:

This is already in the right place.  So other people can discuss it now.  If the moderators here don't think this should be here, they can move it if they want.  :)


geometry
3D geometry
tetrahedron
sphere
IMO
incenter
IMO 1962

The tetrahedron $SABC$ has the following property: there exist five spheres, each tangent to the edges $SA, SB, SC, BC, CA, AB,$ or to their extensions.

a) Prove that the tetrahedron $SABC$ is regular.

b) Prove conversely that for every regular tetrahedron five such spheres exist.

a) Let $\mathcal{S}$ be the sphere tangent to the edges from the inside, and let $\mathcal{S}_X$ be the sphere tangent to the edges of the face opposite to vertex $X$, and the extensions of the edges from $X$.
Consider say $\mathcal{S}$ and $\mathcal{S}_S$. If we cut these spheres by plane $ABC$, in both cases we get a circle on plane $ABC$ tangent to the segments $AB,BC,CA$ - this is the incircle of $\triangle ABC$. Hence $\mathcal{S}$ and $\mathcal{S}_S$ touches the edges $AB,BC,CA$ in the same points.
Now cut the two spheres by plane $SAB$. We get the incircle and $S$-excircle of $\triangle SAB$. We have proven that $\mathcal{S}$ and $\mathcal{S}_S$ touch $AB$ in the same point, so the incircle and $S$-excircle of $\triangle SAB$ touch $AB$ at the same point. This implies that $SA=SB$.
Similarly, we get that any two edges are the same length, therefore the tetrahedron is indeed regular.

b) Look at the regular tetrahedron $SABC$.
By symmetry, the center $O$ of the regular tetrahedron $SABC$ is of equal distance $d$ from all its edges, so taking $\mathcal{S}(O,d)$ works.
Let the incenter of say $\triangle SAB$ be $I$, the $S$-excenter of $\triangle SAB$ be $J$. Consider the homothety with center $S$ that maps $I$ to $J$. This homothety maps the incircle $\omega$ of $\triangle SAB$ to the $S$-excircle of $\triangle SAB$; since $\omega\subset \mathcal{S}$, we have $\omega'\subset\mathcal{S}'$ and therefore $\mathcal{S}'$ also touches edge $AB$. By rotational symmetry, $\mathcal{S}'$ also touches edges $BC$ and $CA$. Moreover, because $\mathcal{S}$ touched edges $SA,SB,SC$, $\mathcal{S}'$ will touch the extensions of $SA,SB,SC$. This proves that $\mathcal{S}_S=\mathcal{S}'$ is up for the job.
By symmetry, so do $\mathcal{S}_A,\mathcal{S}_B,\mathcal{S}_C$ exist.


geometry
incenter
inequalities
cyclic quadrilateral
national olympiad

Here are the problems from INMO held on 5th Feb 2006

(You can also view the problems at http://www.mathlinks.ro/Forum/resources-78-46-2006.html )

1)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73403]Problem 1[/url]

In a non equilateral triangle ABC the sides a,b,c form an Arithmetic Progression.Iis the incentre and o is the orthocentre of the triangle.Prove that
(1) IO is perpendicular to BI.
(2)If BI meets Ac in K, and D,E are the midpoints of BC,BA respectively. Prove that I is the circumcentre of triangle DKE.

2)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73399]Problem 2[/url]

Prove that for every positive integer n there exists a unique ordered pair (a,b) of positive integers such that
\[ n = \frac{1}{2}(a + b - 1)(a + b - 2) + a  \]

3)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73400]Problem 3[/url]

Let X denote the set of all triples\[ (a,b,c)  \]
of integers.
Define \[ f: X \to X  \]
by\[ f(a,b,c) = (\sum a ,\sum {ab} ,abc)  \]

Find all triples\[ (a,b,c)  \]
such that\[ f(f(a,b,c)) = (a,b,c)  \]

4)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73401]Problem 4[/url]

Some 46 squares are randomly chosen from a 9*9 chess board and coloured [color=red]red[/color].Show that there exists a 2*2 block of 4 squares of which atleast three are coloured [color=red]red[/color].

5)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73404]Problem 5[/url]

In a cyclic quadrilateral ABCD, AB=a,BC=b,CD=c,\[ \angle ABC = 120^\circ  \]
and\[ \angle ABD = 30^\circ  \]
Prove that
(1)\[ c \ge a + b  \]
(2)\[ |\sqrt{c + a} - \sqrt{c + b} | = \sqrt{c - a - b}  \]

6)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73405]Problem 6[/url]

(a)Prove that if n is a integer such that \[ n \ge 4011^2  \]
then there exists an integer l such that \[ n < l^2 < (1 + \frac{1}{{2005}})n  \]

(b) Find the smallest positive integer M for which whenever an integer n is such that \[ n \ge M  \]
then ther exists an integer l such that \[ n < l^2 < (1 + \frac{1}{{2005}})n  \]

(this means can you sharpen the inequality \[ n \ge 4011^2  \])

Thanks! Problems were introduced in the resources section and can be found at http://www.mathlinks.ro/Forum/resources-78-46-2006.html :)


Let $f_{n}$ be a sequence of continous functions on $[a,b]$. The functions have the property that for any $x\in [a,b]$ we have that $\{f_{n}(x)\}$ forms a non-increasing sequence.

Show that if $f_{n}\to f$ pointwise, and $f$ is continous on $[a,b]$ then actually $f_{n}\to f$ uniformly.

It is not that hard to prove. But nethertheless it is interesting.

Have a look at Theorem 7.13 in Blue Rudin, and indeed it is interesting.

Well, it is too classical. WLOG, $f=0$. Fix $\varepsilon>0$. For every point $x$ consider $n(x)$ such that $f_{n(x)}(x)<\frac \varepsilon 2$. Let $U(x)$ be the neighborhood of $x$ in which $f_{n(x)}<\varepsilon$. Since $[a,b]$ is compact, we can choose finitely many such neiborhoods $U(x_{j})$ that cover $[a,b]$. Then $f_{n}<\varepsilon$ on $[a,b]$ for all $n>\max_{j}n(x_{j})$.  :sleeping:

It is a beautiful problem if you see it for the first time, though :).

[quote="fedja"]
It is a beautiful problem if you see it for the first time, though :).[/quote]

Is this "Didi's Theorem"? The proof I know of uses the Bolzano-Weierstrass Theorem.

And am I right to say it provides the sufficient conditions for the converse of the following theorem to be true: "If $f_{n}$ are continous on a closed interval and it converges uniformly to $f$ then $f$ is continous on the closed interval"?


[quote="zdyzhj"][img]http://bbs.math.org.cn/attachment.php?aid=1430&k=2500e9e65b62becd8f9e1cfc946cba87&t=1252291493&fid=20&noupdate=yes&sid=f3ff9C06oG6eOlTTWKIA3v%2B7r4eRJcJ6RAI203Hea5jDCsg[/img]

I have solved the problem![/quote]

 :huh: I can't read anything


The numbers a, b, c and d are positive. Prove that among the inequalities
a + b < c + d;
(a + b)(c + d) < ab + cd;
(a + b)cd < ab(c + d)
at least one is false.

bump... I don't know how to do this.  I was hoping someone would provide an answer so I could learn.

hint: [hide]these are a classic type of problems. the way u solve most of them is by contradiction. assume all are true. then, use the two least "complex" statements and incorporate them into the most complex statement. finally, obtain a contradiction. this way u could easily solve this prtoblem.[/hide]

srulikbd wrote:GeoMar wrote:[hide]
Consider p(x)=(x-a)(x-b)(x+c)(x+d)=0, whose roots are a,b,-c,-d
Expand that out to
x^4+(c+d-a-b)x^3+(cd+ab-ac-bd-ad-bc)x^2+(abc+abd-acd-bcd)x+(abcd)

If all three inequalities are true, p(x) has all positive coefficients, however such a polynomial cannot have positive roots (a,b), so at least one inequality cannot be true
[/hide]
I didn't understand...does someone can explain me or give a good solution to this nice question? thanks a lot

GeoMars method Consider the polynomial  If all the inequalities were true,  would only have positive coefficients, so  if . As , we know that  for all  and therefore,  cannot be a root of . This is the contradiction.


geometry
rectangle
perimeter
pigeonhole principle
combinatorics unsolved
combinatorics

You are given $1999$ rectangles, with their sides not exceeding $1998$. Prove that among these $1999$ rectangles there are three rectangles, say $A,B,C$ so that $A$ will fit inside $B$ and $B$ will fit inside $C$.

my pitiful attempt (edit: assuming the sides must be integer lengths as something such as (.1,.9) (.01,.99) (.001, .999)... wouldn't satisfy the above relation)

[hide]Let S(x) denote the number of retanges those perimeters equal twice of x.

The Maximum of S(x) is when x=1999, where there are 999 rectangles.  Only using rectanges from this set, we can make 999*2=1998 rectangles in which no three satisfy the relation given in the problem.  But by pigeonhole principle, when you add the 1999th rectange there are three rectanges that satisfy the above relation.

If you only use rectanges from a different set S(x), there are even less rectangles to choose from, resulting in the same result above.

If you choose rectanges from a mix of different sets, the maximum number of rectanges we can make that do not satisfy the above relation consists only of rectanges where the set S(x) is greatest.  If we use rectangles from other sets than where S(x) is greatest than the rectangle from the set can be contained by more than 1 of the rectangles in the set where S(x) is greatest.  This leads to even less rectanges when can choose before three rectanges satisfy the above relation.

Since S(x) is greatest when x=1999 and 2S(x)<1999, there is always three rectangles if we choose 1999 rectangles that have side lengths not exceeding 1998 in which A will fit inside B and B will fit inside C[/hide]


search
algebra
polynomial
national olympiad

could anyone post the problems of vietnam tst 2008???
 :D
would be glad to see them :!:

some problems of the tst 2008 of vietnam are already posted in different forums
i think 4 of them are there you could try a search as tst 2008 vietnam
 :lol:

vietnameese people were supposed to to be kind and helping
iwonder what is taking them so much time :(

Sorry my bad english. I can't translate all problems.
[size=150]
[b]Day 2:[/b][/size]

[b]Problem 1:[/b] Let m, n be positive integer numbers. Prove that
$ 6m \mid (2m\plus{}3)^n\plus{}1$ if and only if $ 4m \mid 3^n\plus{}1$

[size=150]
[b]Day 1:[/b][/size]
[b]Problem 2:[/b] Find all m be positive integer numbers such that exist three polynomials with real coefficients $ P(x), Q(x), R(x,y)$ to satisfy the following condition:
If $ \forall a,b \in R, a^m\minus{}b^2\equal{}0$ we always have $ P(R(a,b))\equal{}a$ and $ Q(R(a,b))\equal{}b$

I'll update problems.

[quote="gauravpatil"][quote="SOUMYASHANT NAYAK"]From this year's onwards,
none of the questions are going to be set from the ISL... :wink:  :P  :rotfl:

because people using AOPS get unfair advantage and limited but still some access to the ISL of previous year through the TST of other countries....
And I am going to complain abt it....

Some people tried to misuse aops for solving the postal sets and response by BJV was immediate....[/quote]
i dont care somyashant whether they are or not from isl
i just want problems to solve  :P
i dont need the isl for cheating and nor did i know that the tst's werre taken from the isl
i am just interested in new problems to solve (and not the same old ones you fool  :P  :rotfl:  :rotfl: )for i am kind of out of problems (most of the ones on the forum are already seen)
i personelly feel that i am far better than you at ethics :wink: so rest assured i myself will tell BJV that i have seen these problems beforehand :P
thanks a lot Nyugen[/quote]

I am not blaming you...
But earlier the same used to be the case...

[quote="April"][quote="Valentin Vornicu"]I don't think they had the NMO yet.[/quote]
It was posted here: http://www.mathlinks.ro/viewtopic.php?t=185641

[quote="NguyenDungTN"]Sorry my bad English. I can't translate all problems... I'll update problems.[/quote]
It's not necessary: http://www.mathlinks.ro/viewtopic.php?t=201709[/quote]

Sorry, I haven't seen it.

[hide="To April"] Anh dạo này hay online nhỉ :D Chúc anh đạt kết quả cao trong kì thi TN và ĐH[/hide]

How do you get your own country's TST?

I have never seen my own country's TST.


geometry
3D geometry
sphere
incenter
conics
ellipse
Asymptote

I need two specific cones and they are:
A cone that is a $ 30^{\circ}$ sector of a circle with radius 12.
Also a cone with base radius of 7, height of 24 and a sphere inscribed in it.
Also a cylinder with height 4 and radius.

Thanks in advance!

You need to read the documentation on graph3 package. I'm not really good with 3D graphics myself, so I stick to the basic routines like [b]surface()[/b] that allows you to draw any parametrized surface. The main problem with your request is that you want the sphere to be visible inside the cone and there is no perfect way to do it as far as I know. The code below is some compromise: it draws the sphere in solid color and the cone as a mesh. Note also, that I had to take special care of what part of the cone is in front of the sphere and what part is behind. If the resulting picture is not satisfactory, think of how you would draw it on paper and tell what exactly you want. I added the $ x$ and $ y$ axis just to show how the projection is set.
[asy]import olympiad;
import graph3;
size(250);
currentprojection=orthographic(1,1,1);
draw((-10,0,0)--(10,0,0));
draw((0,-10,0)--(0,10,0));
label("$x$",(10,0,0),S);
label("$y$",(0,10,0),S);
real p=3.1415926;
triple f(pair a)
{
real x=a.x,y=a.y;
return (0,0,24)-(7*x*cos(y),7*x*sin(y),24*x);
}
pair A=(-7,0),B=(7,0),C=(0,24);
pair I=incenter(A,B,C);
real r=inradius(A,B,C);
triple g(pair a)
{
real x=a.x,y=a.y;
return (0,0,I.y)+r*(cos(x)*cos(y),sin(x)*cos(y),sin(y));
}
add(surface(f,(0,-pi/4),(1,3*p/4),60,nullpen,blue));
add(surface(g,(-p,-p),(p,p),30,red));
add(surface(f,(0,3*pi/4),(1,7*p/4),60,nullpen,blue));[/asy]
As to the other pictures, it should be clear how to handle them after you figure out how the code works but if you have difficulties, do not hesitate to ask :)

I didn't really need to have them made into a solid or whatever.
All I really need is a normal cone and (maybe) an elipse or something.
I'm trying to recreate an image on one of my challenge sets and the sphere is looks just like a circle even though the cone looks 3D

You mean just some quick picture like this?
[asy]import olympiad;
size(200);
pair A=(-7,0),B=(7,0),C=(0,24);
draw(incircle(A,B,C),blue);
draw(ellipse((0,0),7,2),red);
draw(A--C--B,red);[/asy]

The olympiad package (available on this site at [url]http://www.artofproblemsolving.com/Forum/latexrender/asy/olympiad.asy[/url]) I used it to avoid the computations related to the circle; the [b]incircle()[/b] command from that package returns the circle inscribed into the triangle with given 3 vertices.

sorry for double posting but i figured it out.  :)

Ugg triple posting.
Anyway so far I have this:
[asy]import olympiad;
size(200);
pair A=(-7,0),B=(7,0),C=(0,24);
draw(incircle(A,B,C),blue);
draw(ellipse((0,0),7,2),red);
draw(A--C--B,red);
draw((0,0)--(0,24));
draw((-7,0)--(7,0));[/asy]
But how can I draw an elipse so that is fits nicely above the 'sphere' yet still be tangent to the cone. Basically what i mean is the elipse on the bottom is nice and tucked in the cone yet still touching the sphere. How can i make another one, just higher up and still nicely fitting?
I know the radius of the sphere is $ 5\frac{15}{31}$ (that was the actual answer to the question asked) but thats all I really know.

Like this?
[asy]import olympiad;
size(200);
pair A=(-7,0),B=(7,0),C=(0,24);
draw(incircle(A,B,C),blue);
draw(ellipse((0,0),7,2),red);
draw(A--C--B,red);
draw((0,0)--(0,24));
draw((-7,0)--(7,0));
pair I=incenter(A,B,C), X=foot(I,A,C),Y=foot(I,B,C);
real r=abs(X-Y)/2;
draw(ellipse((X+Y)/2,r,2/7*r),blue);[/asy]

Actually a little higher, like the very top of the sphere.

[code]import solids;
size(250);
currentprojection=orthographic(1,1,1);

draw((-10,0,0)--(10,0,0));
draw((0,-10,0)--(0,10,0));
label("$x$",(10,0,0),S);
label("$y$",(0,10,0),S);

revolution((7,0,0)--(0,0,24),Z,-45,135).fill(green+opacity(.2));
fill(circle((0,0,0),7),gray+opacity(.5));
sphere((0,0,5.25),5.25).fill(red);
fill(circle((0,0,10.5),7*13.5/24),gray+opacity(.3));
revolution((7,0,0)--(0,0,24),Z,135,315).fill(lightgray+opacity(.2));[/code]

For some reason, it won't render properly on AoPS. Here's the pdf file I got on my computer though. It's pretty nice. :D

[quote="undefined117"]
For some reason, it won't render properly on AoPS.[/quote]
Older versions do not understand "lightgray" as an alternative spelling of "lightgrey". Anyway, there is one more problem with that approach. If you do it the standard way producing an .eps file and using the standard GSviewer, you just get this:
[asy]import solids;
size(250);
currentprojection=orthographic(1,1,1);

draw((-10,0,0)--(10,0,0));
draw((0,-10,0)--(0,10,0));
label("$x$",(10,0,0),S);
label("$y$",(0,10,0),S);
revolution((7,0,0)--(0,0,24),Z,-45,135).fill(green+opacity(.2));
fill(circle((0,0,0),7),gray+opacity(.5));
sphere((0,0,5.25),5.25).fill(red);
fill(circle((0,0,10.5),7*13.5/24),gray+opacity(.3));
revolution((7,0,0)--(0,0,24),Z,135,315).fill(lightgrey+opacity(.2));[/asy]
To get your picture, you seem to have to produce a .pdf file, which can be achieved by changing settings.tex to "pdflatex" for example, but, for security reason, playing with settings variables is disabled on AoPS. So, alas, no such nice pictures for now :(.


HI,

If a television  set were laid on its back, wold electrons be able to move upward against the force of gravity?
What potential difference, acting over a distance of 3 cm would be needed to balance the downward force of gravity so that an electron would remain stationary? Assume that the electric field is uniform.

We simply need-
   $eVd$ = $mg$
This gives $V$ = $\frac{mg}{ed}$
Substituting the value given -

$V$ = $2 nV$ (approximately)
 and i guess thats a very small $P.D$
 By the way why do you want to put the television to rest on its back.Poor thing , alrready made to overwork ,does is deserve such a torture.

In these scenarios ($e^{-}$ moving in potentials) gravity is usually negligible.


ratio
analytic geometry
vector
projective geometry

Let $(E,M,I,N)$ be an harmonic division with ratios $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$.

$Z$ is a point $\in (MI)$ which divides the segments $MI,NE$ in the same ratio $\frac{ZM}{ZI}= \frac{ZN}{ZE}= \mu$

Then $\boxed{\mu = \frac{1}{2}\left(\frac{1}{\lambda}-\lambda\right)}$

First of all, you probably meant
[quote="pontios"]Let $(E,M,I,N)$ be an harmonic division with ratios $\frac{\overline{EM}}{\overline{EN}}=-\frac{\overline{IM}}{\overline{IN}}= \lambda$ ...[/quote]

Next,
[hide="The solution"]
I'll use coordinates on the line $MN$: $M(0)$, $N(1)$, $E(e)$, $I(j)$, $Z(z)$.
From $\frac{0-e}{1-e}= \lambda$ we get that $e = \frac{\lambda}{\lambda-1}$, whilst from $-\frac{0-j}{1-j}= \lambda$ we get that $j = \frac{\lambda}{\lambda+1}$.
Now, we know that $\frac{0-z}{j-z}= \mu = \frac{1-z}{e-z}$, so $-z (e-z) = (1-z)(j-z)$, i.e. $z = \frac{j}{1+j-e}$.
Thus,
\begin{eqnarray*}\mu &=& \frac{z}{z-j}\\ \ &=& \frac{\frac{j}{1+j-e}}{\frac{j}{1+j-e}-j}= \frac{\frac{j}{1+j-e}}{\frac{-j^{2}+ej}{1+j-e}}\\ \ &=& \frac{j}{-j^{2}+ej}= \frac1{-j+e}\\ \ &=& \frac1{-\frac{\lambda}{\lambda+1}+\frac{\lambda}{\lambda-1}}= \frac1{\frac{\lambda \left( 1-\lambda \right)+\lambda \left( \lambda+1 \right)}{\lambda^{2}-1}}\\ \ &=& \frac{\lambda^{2}-1}{\lambda-\lambda^{2}+\lambda^{2}+\lambda}= \frac{\lambda^{2}-1}{2 \lambda}\\ \ &=& \frac12 \left( \lambda-\frac1{\lambda}\right) .\end{eqnarray*}
[/hide]

[quote="perfect_radio"]First of all, you probably meant
[quote="pontios"]Let $(E,M,I,N)$ be an harmonic division with ratios $\frac{\overline{EM}}{\overline{EN}}=-\frac{\overline{IM}}{\overline{IN}}= \lambda$ ...[/quote][/quote]

Actually, I was not thinking of vectors, but [u]segments[/u]

[hide="Here is my solution"]
$ \begin{array}{cc}ZM = \mu ZI \\ ZN = \mu ZE \end{array} \}\Rightarrow (ZN+ZM) = \mu (ZE+ZI) \Rightarrow MN = \mu EI \Rightarrow$

$\mu = \frac{MN}{EI}\ \ (1)$

$\frac{EM}{EN}= \lambda \Rightarrow \frac{EM}{MN}= \frac{\lambda}{1-\lambda}$

$\frac{IM}{IN}= \lambda \Rightarrow \frac{MI}{MN}= \frac{\lambda}{\lambda+1}$

$\frac{EI}{MN}= \frac{EM}{MN}+\frac{MI}{MN}= \lambda (\frac{1}{1-\lambda}+\frac{1}{\lambda+1}) = \frac{2\lambda}{1-\lambda^{2}}$

So from $(1) \Rightarrow \mu = \frac{1-\lambda^{2}}{2\lambda}= \frac{1}{2}(\frac{1}{\lambda}-\lambda)$[/hide]

[hide="Replying to pontios"][quote="pontios"]Actually, I was not thinking of vectors, but [u]segments[/u][/quote]

Actually, I was thinking of directed (i.e. with sign) segments. That's what the bar over $EM$ simbolizes (you can't divide vectors, you know).

For example, [url=http://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml]here[/url] it is said that
[quote="Cut The Knot"]The cross-ratio (ABCD) of four collinear points A, B, C, D is defined as the "double ratio":

(1)	(ABCD) = CA/CB : DA/DB,

where all the segments are thought to be [b]signed[/b].[/quote]

As you could see, you and I obtained different answers (our answers add up to $0$). That probably happened because you said that $ZN+ZM = MN,$ and that's not always true when dealing with undirected segments.

Anyway, in such problems it is always useful to keep in mind the sign (and thus use directed segments), because otherwise $\frac{AM}{BM}= k$ represents two $M$ points.[/hide]


[hide="Question"]$\text{How do you get to mathmode in latex?}$[/hide]

Offtopic: Please do not use text in $\LaTeX$ tags anymore. Thanks.

I'm not sure if this is the math mode but you can use the dollars sign to open and close what I think is mathmode.

ledurt


A number $n$ is said to be "insolent" if the sum of the digits of $3^n$ is more or equal than the sum of the digits of $3^{n+1}$. For example, $2$ and $11$ are insolent....

Prove that there are infinitely many insolent numbers.

[quote="Megus"]You can take $n$ such that $3^n=10^k-1$ for some $k$ (there exist $\infty$ such $n$) and try to show that $3^{n+1}$ has sum of digits smaller than sum of digits of $3^n$.[/quote]
I think equation $3^n=10^k-1$ had only one solution $n=2,k=1$.
Let $s_n$ is the summ of digits $3^n$. If "incolent" numbers is finetly, then for n>N $s_{n+1}>s_n\Longrightarrow s_{n+1}=s_n+9$, because $s_k=0(mod \ 9)$ for k>1. It give contradition with $s_n<9*(n*lg(3)+1)$ (number of digits is n[lg(3)+1]).

[quote="Rust"][quote="Megus"]You can take $n$ such that $3^n=10^k-1$ for some $k$ (there exist $\infty$ such $n$) and try to show that $3^{n+1}$ has sum of digits smaller than sum of digits of $3^n$.[/quote]
I think equation $3^n=10^k-1$ had only one solution $n=2,k=1$.
Let $s_n$ is the summ of digits $3^n$. If "incolent" numbers is finetly, then for n>N $s_{n+1}>s_n\Longrightarrow s_{n+1}=s_n+9$, because $s_k=0(mod \ 9)$ for k>1. It give contradition with $s_n<9*(lg(3)+1)$.[/quote]

Sorry, can you explain it with more detail????

[quote="Arne"]Posted before many times. See, for example, [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=S%283%5En%29&t=659]this very old post of mine.[/url][/quote]

Oh sorry, I found it in an Ibero TST......


How do I align the equal signs? For example, [quote]\[ \frac {1}{2} \minus{} \frac {1}{3} \equal{} \frac {3}{x}\newline \frac {1}{6} \equal{} \frac {3}{x}\newline 18 \equal{} x\newline \boxed{180} \equal{} 10x
\][/quote] looks quite ugly. Thanks.

ThinkFlow

The align environment is built to do exactly what you want:
\begin{align}
\frac {1}{2} - \frac {1}{3} & = \frac {3}{x} \\
\frac {1}{6} & = \frac {3}{x} \\
18 & = x \\
\boxed{180} & = 10x
\end{align}
or without equation numbers:
\[ \begin{align*}
\frac {1}{2} - \frac {1}{3} & = \frac {3}{x} \\
\frac {1}{6} & = \frac {3}{x} \\
18 & = x \\
\boxed{180} & = 10x
\end{align*}*
\]
Note the & to line up whatever symbol follows, in this case =
\\ is simpler to type than \newline.
In a document you must have \usepackage{amsmath} in the preamble ie between \documentclass and \begin{document}

Hi ThinkFlow,

perhaps some of those tutorials and documentation could be useful for you: [url=http://texblog.net/latex-link-archive/mathematics/]Mathematics with LaTeX[/url].
The amsmath documentation is a must, the Mathmode document is a very rich source.

Stefan

Thank you. Another question: how do you write 21st (as in 21st century) without resorting to math mode? Specifically, I write 21st as $ 21^{st}$ in my documents, but I'm somewhat unhappy with the st being slanted. Is there an alternative?

That only works in maths mode which is not what is wanted.
[quote="ThinkFlow"]how do you write 21st (as in 21st century) without resorting to math mode?[/quote]

stevem: after searching for a similar command for subscripts, I found that ^ and _ work outside of math mode too. Thus, to type 21st, I wouldn't need to type 21\textsuperscript{st}, just 21^{st} without dollar signs.

Just thought this would be a useful note.

ThinkFlow

EDIT: Sorry, I must have misread the documentary of the ^ and _ commands. The above doesn't work. I apologize for posting this. :(


induction
combinatorics unsolved
combinatorics

Let $ n > 2$ be an integer. Suppose that $ a_{1},a_{2},...,a_{n}$ are real numbers such that $ k_{i} \equal{} \frac {a_{i \minus{} 1} \plus{} a_{i \plus{} 1}}{a_{i}}$ is a positive integer for all $ i$(Here $ a_{0} \equal{} a_{n},a_{n \plus{} 1} \equal{} a_{1}$). Prove that $ 2n\leq k_{1} \plus{} k_{2} \plus{} ... \plus{} k_{n}\leq 3n$

[quote="kihe_freety5"]Let $ n > 2$ be an integer. Suppose that $ a_{1},a_{2},...,a_{n}$ are real numbers such that $ k_{i} \equal{} \frac {a_{i \minus{} 1} \plus{} a_{i \plus{} 1}}{a_{i}}$ is a positive integer for all $ i$(Here $ a_{0} \equal{} a_{n},a_{n \plus{} 1} \equal{} a_{1}$). Prove that $ 2n\leq a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}\leq 3n$[/quote]

Are you sure there is no other condition?  If $ \{a_i \}$ satisfies the condition then $ \{\minus{}a_i\}$ also satisfies the condition. ???

I've seen this problem on the site before, but can't find it. Anyone know where it is?

Use induction. For $ n\equal{}3$ it is easy to show. Now assume its true for $ n\minus{}1$
We can assume $ a_n$ is the biggest of all $ a_i$'s and it can be assumed to be positive. Then there are only two cases for $ k_n$ to be a positive integer.
$ (1)$ If $ a_1\plus{}a_{n\minus{}1}\equal{}2a_n$ Then it turns out that all $ a_i$ are equal which satisfy inequlity.
$ (2)$ If $ a_1\plus{}a_{n\minus{}1}\equal{}2a_n$. Then replace $ a_n$ with $ a_1\plus{}a_{n\minus{}1}$. It gives
$ 2n\minus{}3 \leq k_1\plus{}...k_{n\minus{}1} \leq 3(n\minus{}1)$ which is true by induction hypothesis.

[hide]I think $ k_1\plus{}...k_{n} \leq 3n\minus{}1$ is also true[/hide]

oh, I have found the solution , the solution is in this file, It is problem 227


topology
calculus
advanced fields
advanced fields unsolved

Let $ M,N$ be smooth manifolds, $ M$ connected. Let $ f: M\rightarrow N$ be a smooth map such that $ f_*\equal{}0$ at all points of $ M$, where $ f_*$ is the differential of $ f$. Prove that $ f$ is constant.

Thanks.

Try to show that $ A(p_0) \equal{} \{p\in M\mid f(p) \equal{} f(p_0)\}$ for some point $ p_0\in M$  is a non-void open and closed subset of $ M$.

Or try using the Fundamental Theorem of Calculus along a path in M.  Of course, you cannot integrate f directly, so try something sneakier.


inequalities
function
integration
trigonometry
real analysis
real analysis unsolved

let $f: [0,1] \to R$ be a continuous function satisfying $xf(y)+yf(x) \le 1$ for every $x,y \in [0,1]$
a)show that $\int_0^1 f(x)dx \le \frac{\pi}{4}$.b)find a fuction such  for equality occurs.

I can only prove $\int_0^1 f(x)dx \le 1$, :(
somebady help me.thanks

[quote="alekk"]try the substitutions $x=sin(u)$ and $x=cos(u)$ ..[/quote]
thanks,can you tell me when equality occurs?

sorry for my stupid question.
then $f(x)=\cos {\arcsin x}$
sorry :)


Let $P_{n}$ denote the number of paths in the coordinate plane traveling from $(0, 0)$ to $(n, 0)$ with three kinds of moves: [i]upstep[/i] $u = [1, 1]$, [i]downstep[/i] $d = [1,-1]$, and [i]flatstep[/i] $f = [1, 0]$ with the path always staying above the line $y = 0.$ Let $C_{n}= \frac{1}{n+1}\binom{2n}{n}$ be the $n^{th}$ Catalan number. Prove that

$P_{n}= \sum_{i = 0}^\infty \binom{n}{2i}C_{i}$ and $C_{n}= \sum_{i = 0}^{2n}(-1)^{i}\binom{2n}{i}P_{2n-i}.$

[hide="Solution to Part 1"]

Let a path string, $S_{k}$, denote a string of $u, d, f$ corresponding to upsteps, downsteps, and flatsteps of length $k$ which successfully travels from $(0, 0)$ to $(n, 0)$ without passing below $y = 0.$ Also, let each entry of a path string be a slot. Lastly, denote $u_{k}, d_{k}, f_{k}$ to be the number of upsteps, downsteps, and flatsteps, respectively, in $S_{k}.$

Note that in our situation, all such path strings are in the form $S_{n},$ so all our path strings have $n$ slots. Since the starting and ending $y$ values are the same, the number of upsteps must equal the number of downsteps.

Let us observe the case when there are $2k$ downsteps and upsteps totally. Thus, there are $\binom{n}{2k}$ ways to choose the slots in which the upsteps and the downsteps appear. Now, we must arrange the downsteps and upsteps in such a way that $d_{n}= u_{n}$ and a greater number of upsteps preceed downsteps, as the path is always above $y = 0$. Note that a bijection exists between this and the number of ways to binary bracket $k$ letters. The number of binary brackets of $k$ letters is just the $k^{th}$ Catalan number. We then place the flatsteps in the rest of the slots. Thus, there are a total of $\sum_{k = 0}^\infty \binom{n}{2k}C_{k}$ ways to get an $S_{n}.$

[/hide]

[quote="Karth"] Now, we must arrange the downsteps and upsteps in such a way that $d_{n}= u_{n}$ and a greater number of upsteps preceed downsteps, as the path is always above $y = 0$. Note that a bijection exists between this and the number of ways to binary bracket $k$ letters. The number of binary brackets of $k$ letters is just the $k^{th}$ Catalan number.
[/hide][/quote]

Could you please explain that a bit more? What's the binary bracket thing? What is the bijection?

[hide="As for part 2"]
$\binom{2n}{0}P_{2n}$ is just $P_{2n}$

$\binom{2n}{1}P_{2n-1}$ counts paths with atleast one flatstep. First it picks one slot for the flat step, then it creates a path for the other 2n-1 slots. However, this overcounts paths with atleast two flats steps.

$\binom{2n}{2}P_{2n-2}$ counts paths with atleast two flatsteps, but it overcounts those with atleast 3 flatsteps

... and so on

Using the principle of inclusion-exclusion, you can count the number of paths with 2n slots without any flatsteps. That's exactly what the sum is doing. And this is the nth Catalan number.
[/hide]

Here's a link that demonstrates binary bracketing  :)

[url]http://mathworld.wolfram.com/BinaryBracketing.html[/url]


Let be a set with n elements.  How many ways can you select two non-intersecting sets from S (without regard to order)?

Does this mean that if I select two empty sets, they are still non-intersecting?  Or are empty sets consider to be intersecting?

I would think they're non-intersecting, since their intersection is still the empty set.

[hide="Large hint"] Each element on $ S$ is either in the first non-intersecting subset, the second non-intersecting subset, or in neither. ;) [/hide]

[quote="firecricket91"]ok thanks, thats all i needed because the answer would correspond with that statement.

you guys can still do the problem if you want.[/quote]

[hide]For distinguishable sets A,B, every n elements can either be in A, B, or neither (C). So there are $ 3^n$ such sets. However, since A and B are indistinguishable, we divide by two for every set pair of solutions such that their elements are just switched. This applies to every solution except (empty,empty,S). So add 1 to account for that, divide by two, $ (3^n\plus{}1)/2$?[/hide]

yea thats right..

thats almost like an old AIME problem but it was the union of the two sets is S and there was a number for n..but its the same trick and the same answer in terms of n


geometry
3D geometry
pyramid
calculus
integration
calculus computations

Derive the formula for the volume of a truncated pyramid in terms of two base areas (in at the top and one at the bottom) and slant height.
http://answers.yahoo.com/question/index;_ylt=ApwDLF5H3T8kCLRKlMXH6I3ty6IX;_ylv=3?qid=20080603140008AAKbcrQ&show=7#profile-info-y8CshceTaa

Also derive the formula in terms of two base areas (upper and lower) and height (not slant height this time).

Also try using integral calculus.

EDIT: The link you give does not give the derivation, but only the formula.

See this: [url]http://mathworld.wolfram.com/PyramidalFrustum.html[/url]

Also, the basic forum is not for integral calculus. See the calculus section for help in that area.

One of the very oldest mathematical documents known (an Egyptian scroll known as the Moscow Papyrus, which dates from perhaps 1800 BC) contains a problem which apparently involves the volume of a frustrum of a square pyramid  - and the answer is correct. Since the document contains no justification, we don't really know why they divided by 3.


trigonometry
analytic geometry
geometry
graphing lines
slope

In triangle $ ABC$ denote the middlepoint $ M$ of the side $ [BC]$ and the point $ N\in (BC)$ for which $ 3\cdot NC\equal{}BC$. Prove that $ AB\perp AC$ $ \Longleftrightarrow$ $ \tan\widehat {BAM}\cdot\tan\widehat {CAN}\equal{}\frac 12\ .$

[hide]If part:

Draw BD that is perpendicular to AB and parallel to AC.
Extend AM until it touches BD at point E.
Extend AN until it touches BD at point F.  Extend AC so that the normal from F to AC is F'.
$ \tan{BAM} \equal{} BE/AB$ and $ \tan{CAN} \equal{} AB/AF' \equal{} AB/BF$
Multiplying removes the AB, so BE/BF.

We now use coordinate geometry to solve.
Set A as the origin.  B is (0,a) and C is (b,0).  M is (b/2, a/2) and N is (2b/3, a/3).
Note that the slope of MA is a/b but the slope of NA is a/2b.  Now, since BD is perpendicular to AB and parallel to AC, BD is the line y=a.  The intersection of BD with MA is at point E or (b,a), but the intersection of BD with NA is at point F or (2b,a).  Thus, the distance BE=b, and BF=2b so BE/BF=1/2.

Only if part:

We again use coordinate geometry.
Set A as the origin.  B is (0,a), but C is (b,c), where all the points are in Quadrant I.  We need to show c=0.  M is (b/2, (a+c)/2) and N is (2b/3, (a+2c)/3).  Now, draw altitudes from M and N to AB and BC, respectively.  Thus, $ \tan{BAM} \equal{} \frac {b}{a \plus{} c}$, and we use tangent subtraction identity from trignometry to solve for the angle between AC and AN, since we can't yet assume c=0.  $ \tan{CAN} \equal{} \frac{\frac {a \plus{} 2c}{2b}\minus{}\frac{2c}{2b}}{1\plus{}\frac{a\plus{}2c}{2b}(c/b)}$, which simplifies to $ \frac{ab}{2b^2\plus{}ac\plus{}2c^2}$.  We know that $ \frac {b}{a\plus{}c} * \frac{ab}{2b^2\plus{}ac\plus{}2c^2} \equal{} 1/2$  Cross multiplication gives $ 2ab^2\equal{}(a\plus{}c)(2b^2\plus{}ac\plus{}2c^2)$.  We subtract to get $ 0\equal{}c(a^2\plus{}2ac\plus{}2b^2\plus{}ac\plus{}2c^2)$.  We oriented this so $ a,b >\equal{}0$, but b isn't 0 or B and C are collinear, and a isn't 0 or A and B are now the same point.  In both cases, we no longer have a triangle.  However, if neither point is 0, the expression in the parenthesis must be positive; thus, c must be 0.[/hide]


If $ q_{1}(x)$ and $ r_{1}$ are the quotient and remainder, respectively, when the polynomial $ x^{8}$ is divided by $ x \plus{} \frac {1}{2}$,and if $ q_{2}(x)$and $ r_{2}$ are the quotient and remainder, respectively, when $ q_{1}(x)$ is divided by $ x \plus{} \frac {1}{2}$, then $ r_{2}$ equals?

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

I know the basic way to approach this problem but I dont understand on step.

I started with

$ x^{8} \equal{} (x \plus{} \frac {1}{2})q_{1}(x) \plus{} r_{1}$ and

$ q_{1}(x) \equal{} (x \plus{} \frac {1}{2})q_{2}(x) \plus{} r_{2}$

so for the first one I plugged in $ x \equal{} \minus{} \frac {1}{2}$.  Then I got $ \frac {1}{256} \equal{} r_{1}$, then plugging that for $ r_{1}$ and rearranging I got
$ q_{1}(x) \equal{} \frac {x^{8} \minus{} \frac {1}{256}}{x \plus{} \frac {1}{2}} \equal{} \frac {(x^{4} \plus{} \frac {1}{16})(x^{2} \plus{} \frac {1}{4})(x \plus{} \frac {1}{2})(x \minus{} \frac {1}{2})}{x \plus{} \frac {1}{2}}$

Now for this $ q_{1}(x)$, am I allowed to cancel out the $ x \plus{} \frac {1}{2}$?  Because if I am for the next step, in the second equation, I would plug in $ x \equal{} \minus{} \frac {1}{2}$ again and easily get $ r_{2}$.  However, if I am not allowed to cancel out the top and bottom, how am I supposed to find $ r_{2}$?

The reason I was thinking you cannot cross out is because if you do, you would have to restrict the fact that x cannot equal $ \minus{} \frac {1}{2}$ or $ q_{1}(x)$will be undefined?  So am I allowed to make that move by canceling out, and not having to restrict it to not equaling $ \minus{} \frac {1}{2}$?

When can you just cancel out the top and bottom without changing the equation? Is it only when you are solving for it when it equals 0 that you have to restrict it?

If I am allowed to cancel out, then I would just find $ q_1( \minus{} \frac {1}{2}) \equal{} \minus{} \frac {1}{16} \equal{} r_{2}$ because $ q_{2}(x)$ will be gone since $ \minus{} \frac {1}{2} \plus{} \frac {1}{2} \equal{} 0$

[quote="Tenoreoz"]$ q_{1}(x)$ is undefined at $ x \equal{} \minus{} \frac {1}{2}$ so you cannot plug it into the second equation to find $ r_{2}$. However, here's something you can do:

$ q_{1}(x) \equal{} (x^4 \plus{} \frac {1}{16})(x^2 \plus{} \frac {1}{4})(x \minus{} \frac {1}{2})$ for all $ x \neq \minus{} \frac {1}{2}$. This means that $ q_{2}(x)$ is also defined only for all $ x \neq \minus{} \frac {1}{2}$. (This is why you can't plug $ \minus{} \frac {1}{2}$ in).

The problem now suffices to finding the remainder, or constant $ r_{2}$, when $ q_{1}(x)$ is divided by $ x \plus{} \frac {1}{2}$, so we wish to find the remainder of
\[ \frac {(x^4 \plus{} \frac {1}{16})(x^2 \plus{} \frac {1}{4})(x \minus{} \frac {1}{2})}{x \plus{} \frac {1}{2}}
\]
Now you can just use long division to find it.

The answer I got when I did it was [hide]$ r_{2} \equal{} \minus{} \frac {1}{16}$
[/hide][/quote]

Thanks for your reply.

But however, lets say you were allowed to cancel out:  then it would be

$ q_{1}(x) \equal{} (x \plus{} \frac {1}{2})q_{2}(x) \plus{} r_{2}$ for the second equation
using remainder theorm-same eqaution you have,just written differently
then we would have

$ (x^4 \plus{} \frac {1}{16})(x^2 \plus{} \frac {1}{4})(x \minus{} \frac {1}{2}) \equal{} (x \plus{} \frac {1}{2})q_{2}(x) \plus{} r_{2}$.

But lets say we plug in $ x \equal{} \minus{} \frac {1}{2}$, we would have $ \minus{} \frac {1}{16} \equal{} r_{2}$, getting the same answer you got from long division.  So I dont understand why you were allowed to sub back in $ x \equal{} \minus{} \frac {1}{2}$ in this particular equation or is this just a special case or is this always going to be true?


AMC is one of those competitions that has speed as a major factor in it.  I've noticed that even though I might be able to do the problems without a time limit, I can't do it within the allotted 3 minutes.  This is probably because in order to solve one of the harder problems in the test, you must be fairly familiar with the tactic that you should use to solve it.  AIME on the other hand gives you 12 min/problem.  Even so, like the AMC, you still have to be quite familiar with the problem to solve it within the given time limit.  The USAMO seems to give a large quantity of time per problem, but I've never solved a single one of them.  How do I prepare for such a competition?  Where do I start?  By the way, I think this is an awesome site.  Keep up the great work!!!

-interesting_move

"The USAMO seems to give a large quantity of time per problem, but I've never solved a single one of them. How do I prepare for such a competition? Where do I start?"

Our goal to answer that question with a resounding "You can start right here!"

The Art of Problem Solving books have been around for many years.  They cover much of what is needed to learn how to crack USAMO problems.

The Math Jams might be of interest to you.  The harder ones will deal with material that would be useful on the USAMO.

Eventually we plan to offer online classes for material of that level.

You might also try starting threads about various USAMO problems and see what you learn from them!

Thank you very much for that information!

-interesting_move  :D

Do you recommend I find a school teacher or mentor?  I'm pretty much just teaching this material to myself.

Whoa dude, wasn't this thread on the last page of the forum?
Maybe WOOT next year will help. I'm doing it next year; I've heard it helps a lot with harder AIME and USAMO. ACoPS by Zeitz and Problem Solving Strategies help too, I hear.
But before that, I'm just gonna finish off AoPS 2 before September.


function
algebra
domain
integration
inequalities
complex analysis
complex analysis unsolved

Suppose that the fuction $ f$ is analytic in a domain $ \Omega$, and that
\[ \left| {f\left( z \right) \minus{} 1} \right| < 1\qquad\forall z \in \Omega
\]
Prove that for any closed contour $ \gamma$ in $ \Omega$
\[ \int\limits_\gamma  {\frac{{f'\left( z \right)}}
{{f\left( z \right)}}dz}  \equal{} 0.
\]

Can you solve it?  :D

the condition implies that $ \log(f(z))$ is an analytic function in $ \Omega$, so ...

Use inequality and Rouche's theorem to conclude that $ f$ doesnt have a root inside $ \Omega$ hence $ \frac{1}{f}$ is analytic inside $ \Omega$ as well as $ f'$.


A suitcase lock has 3 dials with the digits $ 0, 1, 2,..., 9$ on each.
How many different settings are possible if all three digits have
to be different?

For the first digit, there are $ 9 \minus{} 0 \plus{} 1 \equal{} 10$ possibilities. No matter what digit we choose, there will be 9 choices left for the second and 8 for the third. Thus, our answer is $ 10 \times 9 \times 8 \equal{} \boxed{720}$.

[hide=Solution]Classic Factorial problem.
For the first dial, there are $10$ options, for the second dial, there are $9$ options, and for the third dial, there are $8$ options.
Thus there are
$$10 \times 9 \times 8$$
$$ \boxed{720 \: \textrm{diffrent settings}}$$
$\blacksquare$ [/hide]

[hide=Sol]10 cases for the first digit. 9 cases for the second digit. 8 cases for the 3rd digit. Ans = 10*9*8 = 720[/hide]


vector
rotation
geometry
geometric transformation

We have triangle ABC and A1,B1,C1 the projections of G(weight point) on its sides. Prove that a:^2: GA1+b :^2: GB1+c :^2: GC1=0; GA1,GB1,GC1 are vectors.

What about when we move G onto a vertex?  Then the sum is two zero vectors plus a non-zero vector, which definitely isn't zero.

Edit:  thanks, Ravi.  Ignore this comment.

By "weight point", did the problem mean the centroid (aka center of mass, aka intersection of medians)?

Oh, well that makes a lot more sense!  Figures why he chose G, then.

Okay, so we can multiply everything by three to convert GA1, GB1 and GC1 into ha, hb and hc, the altitudes to the sides a, b and c respectively.  So we want to show a :^2: ha + b :^2: hb + c :^2: hb = 0.  Rotate the three vectors 90 degrees.  Then ha will be parallel to BC, hb will be parallel to CA, and hc will be parallel to AB.  The length of a :^2: ha is a*2K, where K is the area of the triangle and similarly for the other sides.  Thus, under this rotation we have three vectors parallel to the (directed) sides of the triangle whose lengths are proportional to those of the sides of the triangle.  This means we can rearrange them into the form of a triangle similar to the original, so the result of travelling all three is a return to where we started, so their sum is 0.

Not the most elegant write-up, but I like the proof itself.


calculus
integration
logarithms
calculus computations

My answer is $ \frac{2\sqrt[1/2]{3}}{3}\minus{}\frac{pi}{3}$

Ok, nice and easy one! :)

but [b]the_wonder[/b], i think you made a really small mistake! or maybe i've made a mistake, who knows! :P

let $ u\equal{}e^{x}\minus{}1$ then $ \text{d}u\equal{}e^{x}\text{d}x \Longrightarrow \text{d}u\equal{}(u\plus{}1)\text{d}x \Longrightarrow \frac{\text{d}u}{u\plus{}1}\equal{}\text{d}x$, and the integral will be:

$ I\equal{}\int^{\ln\frac{4}{3}}_{0}\sqrt{e^{x}\minus{}1}\text{d}x\equal{}\int^{\frac{1}{3}}_{0}\frac{\sqrt{u}}{u\plus{}1}\text{d}u$

Now let $ u\equal{}t^{2}$ then $ \text{d}u\equal{}2t\text{d}t$ and we'll have:

$ I\equal{}\int^{\frac{\sqrt{3}}{3}}_{0}\frac{t\cdot2t}{1\plus{}t^{2}}\text{d}t\equal{}2\int^{\frac{\sqrt{3}}{3}}_{0}\frac{1\plus{}t^{2}\minus{}1}{1\plus{}t^{2}}\text{d}t$

$ \equal{}2\int^{\frac{\sqrt{3}}{3}}_{0}\left(1\minus{}\frac{1}{1\plus{}t^{2}}\right)\text{d}t\equal{}2\left[t\minus{}\arctan t\right]^{\frac{\sqrt{3}}{3}}_{0}$

$ \equal{}2\frac{\sqrt{3}}{3}\minus{}\frac{\pi}{6}$

 :)


Find all 2-variable polynomials P such that

1)if P is of n-th degree, $ P(tx,ty) \equal{} t^n \cdot P(x,y)$ for all real t,x,y and natural n

2)$ P(a \plus{} b,c) \plus{} P(a \plus{} c,b) \plus{} P(b \plus{} c,a) \equal{} 0$ for all real a,b,c

3)$ P(1,0) \equal{} 0$


What year in this century, when its digits are written in reverse
order, is a year in the next century?

We need a $ 2$ in the first and last digits, a $ 0$ in the second digit and a $ 1$ in the third digit.  Thus, our number is $ \boxed{2012}$.

I meant that in the upcoming centuries, the answer will change.

@above That is not true. If this question was written in the 22nd century, then 2122 would be the answer.

and in the 23rd century, 2232 is the answer.

20th century, 1902 is the answer.


I was seriously wondering if there are any mathematics or physics or chemistry or computerscience or any other related contests for undergraduate students in india.

If any of you have any info about any such contest, then please post - it would be very helpful.
Thank you in adavnce.

Come on guys Reply.Seniors esp, u guys ought to be replying

CS-related competitions include the Topcoder contests, ACM ICPC and Google Code Jam.

@mamama:Thanks.

Are there any core science or maths contests?

Watch this forum shortly for details on the online math contest of IIT Madras.

[quote="Ram"]Seniors esp, u guys ought to be replying[/quote]
I don't recollect anyone here who's already in college  :huh:


limit
trigonometry
algebra unsolved
algebra

find limit:
 1)  $ \lim_{x \to 0}\frac{\tan( \sin{x}) - \sin( \tan{x})}{x^7} $
 2)  $ \lim_{x \to 0}\frac{1- \cos{x}\cdot \sqrt{\cos{2x}}\cdot \sqrt[3]{\cos{3x}}}{1- \cos{x}\cdot \cos{\sqrt{2x}}\cdot \cos{\sqrt[3]{3x}}} $

the Taylor series of tan(sinx)-sin(tanx) is indeed (1/30)x^7 + O(x^8). Then the limit of problem a is 1/30.

The series expandsion of \[ 1-cos(x).\sqrt{cos(2x)}\sqrt[3]{cos(3x)}\] is \[ 0 + O(x)\], but the series expandsion of \[ 1-cos(x).cos(\sqrt{2x}).cos(\sqrt[3]{3x})\] is \[  \frac{\sqrt[3]{9}}{2}x^{\frac23}+O(x) \].
Therefore, the limit of b is 0

sorry, is there any way else to solve the problems. thank you


These counts were done with the command
[code]
grep "^[^ ] A" 2007qual.txt -c
[/code]
where the A was any capital letter.

I did not change the file, so "Brady Zarathustra" is counted as a Z and not a B. Change it mentally if you so desire.

[hide="The statistics"]
[code]
A:	14
B:	26
C:	41
D:	8
E:	3
F:	13
G:	19
H:	38
I:	4
J:	17
K:	28
L:	47
M:	26
N:	8
O:	4
P:	19
Q:	1
R:	13
S:	51
T:	20
U:	0
V:	3
W:	39
X:	12
Y:	20
Z:	31
[/code]

Winner: S
2nd: L
3rd: C
[/hide]

So, what do you guys think of the last names?

Yes!!  S wins!!!! :first:

I'm surprised to see L and C as second and third.

J wins for first name.
S wins for last name.
I bet N wins for middle name.

[quote="Kent Merryfield"][quote="dasherm"]I'm surprised to see L and C as second and third.[/quote]
Remember how many Chinese names there are on this list. C for Chen, Chan, Chang, L for Lee, Li, Liang. Note also the high number of W (Wang, etc.) and Z (Zhang, etc.).[/quote]

Also many Asians have H for last name (Huang, etc), hence why H is also up there.

[quote="dasherm"]Yeah, I was expecting W or Z to beat L.  That's what I meant.[/quote]

Well, L has Lee, Le, and Li, for example, which represents a very wide range of Asians. However, W and Z seem to be more location-specific. If this isn't true, please correct me, but that's how it seems to me.

I wouldn't know.   :|

btw, it should be noted that, once again, U was the only letter without a USAMO qualifier.  (Too bad they got rid of the one-per-letter rule  :wink: )

My chances are still slim
even
with "H" having 38

dude just about the only correlation with IMO team is L

that is pretty shady

LOUCHE

go 'v'.  3 whole people.  but other than 'q', that's the least of any letter that HAD usamo qualifiers...

[quote="Kent Merryfield"][quote="dasherm"]I'm surprised to see L and C as second and third.[/quote]
Remember how many Chinese names there are on this list. C for Chen, Chan, Chang, L for Lee, Li, Liang. Note also the high number of W (Wang, etc.) and Z (Zhang, etc.).[/quote]

And by Wang, etc., you mean Wang and Wu.
There were 15 Wangs this year.
And there were 14 Wus this year, despite Wu being a less common name.  We are definitely in power (I'm Jeffrey).

This leads to the question, "Which last name bred the most USAMO qualifiers this year?"
Zhang had 12 qualifiers.
Chen had 12 qualifiers.
Kim had 9 qualifiers.
Lee had 9 qualifiers.
Only 2 Chans and 1 Chang.

This leads me to think Wang and Wu might be #1 and #2.  And yet W only comes in 4th place... Oh well, Wangs and Wus are definitely holding it up.

[quote="Aznwithabrain"]Don't forget:

Huang's got 6
and
Li's got 8 (Li is the most common last name in China)(Check on Wikipedia)[/quote]

Heh but Wang and Wu are still on top.

[quote="ra5249"]Too bad Tian isn't a very common chinese last name.  :wink:[/quote]
Well, we have a Tian here in Southern California.

yay for H!

too bad hoch is german..... and nobody else qualified with that name...

so surprised to see "z" having 31 ppl qualified.
i can't think of any common non-asian surname with z

Every single "Z" name is obviously Chinese. Actually, Chinese names are probably a majority of all USAMO qualifiers.

It is interesting to see more Thai names than Vietnamese names- there's not a single Nguyen.

[quote="bloodysloth"][quote="Aznwithabrain"]Wu is a Taiwanese last name right?

I think Li, Zhang, and Chen are the most popular Chinese last names (according to Wikipedia)[/quote]

Wu is Chinese

14 qualifiers; where else would we be from? =P[/quote]

Taiwan. That's why he's asking if Wu is mainland or Taiwanese.


Hi every body,

Can any one tell me how to divide page into many column using $\LaTeX$?
And if some one can give me a link of a COMPLETE tutorial of LaTeX that will be better.

Thanks

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59948&sid=52c0b3832b9d85594c6a3feed5cc2bfe]www.artofproblemsolving.com/Forum/viewtopic.php?t=59948&sid=52c0b3832b9d85594c6a3feed5cc2bfe[/url]

This isn't a complete tutorial but once you have the name of the package, look it up google or ctan to get a pdf document about the package in question.

And personnaly, I don't think there's such thing as a "complete" tutorial.  There's always something else.


Let $a,b,c>0$.Prove that $ a^b +b^c+c^a \geq \frac{3}{2}$.

 When does equality hold? ;)

Is this true?  Set $a=1$ so it becomes $b^c+c \geq \frac{1}{2}$, but take for instance $c=\frac{1}{10}, b = \frac{1}{10^{10}}$... unless I'm missing something  :P


[quote="radio"]Let $A,B$ be two fixed points and $r$ a fixed positive number. Find the locus of points $P$ such that $\frac{AP}{PB}=r$[/quote]

It isnt the right place to post this well known theorem!Open any geometry book and read the proof!

Can someone show me where to find the proof? Maybe the name of one such book.


[hide="Solution?"]
Factor each equation. \[ \begin{align}x(x^2-3y^2)&=1\\y(3x^2-y^2)&=0\\(x+y)^2(x-y)^2&=-\frac{1}{2}\\4xy(x-y)(x+y)&=-\frac{\sqrt3}{2}\end{align}.\] From $ (2),$ we can immediately tell that either $ y=0$ or $ 3x^2=y^2.$ Checking $ y=0$ we find that it doesn't work, as $ (4)$ would not be true. So that means \[ 3x^2=y^2\newline y=x\sqrt{3}.\] Plugging this into $ (1)$ gives the new equation \begin{align*}x[x^2-3(x\sqrt{3})^2]&=1\\x(x^2-9x^2)&=1\\-8x^3&=1\\x&=-\frac{1}{2}.\end{align*}Testing in $ (3)$ (along with the value of $ y$ that corresponds with it), we find that it works, so the solution pair is $ \boxed{\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)}.$ [/hide]

[hide="Hint"] Let $ z \equal{} x \plus{} yi$ and compute $ z^3$ and $ z^4$. [/hide]

Adding first & second :

\[ (x\minus{}y)^3\equal{}1\]

adding third & fourth:

\[ (x\minus{}y)^4\equal{}\frac{1}{2}(\sqrt{3}\minus{}1)\]

.. next ?

Obel1x, check again.  The signs are wrong for what you've written.

Okay since no one is replying to my "solution" I'm assuming it's wrong? But I checked the pair in all four equations and it seems to satisfy all of them. Unless I'm missing something obvious?

Your solution is fine, but the hint is a little faster.

[hide="A version of the hint that does not involve complex numbers"] What is $ x(x^3 \minus{} 3xy^2) \minus{} y(3x^2 y \minus{} y^3)$?  What is $ y(x^3 \minus{} 3xy^2) \plus{} x(3x^2y \minus{} y^3)$? [/hide]


Find all real polynomials $p(x)$ such that
\[p^{2}(x)+2p(x)p\left(\frac{1}{x}\right)+p^{2}\left(\frac{1}{x}\right) = p(x^{2})p\left(\frac{1}{x^{2}}\right)\]
For all non-zero real $x$.

Very nice, thank you very much!  :wink:  I will NOT post my solution because it is very long and bad  :(

If it was wrong, can anyone gives a solution?

[quote=ThE-dArK-lOrD]No, it wasn't wrong.[/quote]

could you point it out why?

[quote=reveryu]Bump!

I think the reason is it’s polynomial but can anyone gives me mathematically explanation?[/quote]

I suppose that you're asking about this line:

[quote=olorin]
$(q(x)+(-x^{-1})^{n}q(x^{-1}))^{2}=(x+1)^{2n}(-x^{-2})^{n}q(x^{2})q(x^{-2})$
[/quote]

Consider the polynomial  $x^{N}(\mathrm{LHS}-\mathrm{RHS})$ (for some big enough $N$ so that it's a polynomial)
This polynomial is zero for any $x\neq 1$, so it must be the zero polynomial. Hence the above equation holds even when $x=1$.

[quote=talkon][quote=reveryu]Bump!

I think the reason is it’s polynomial but can anyone gives me mathematically explanation?[/quote]

I suppose that you're asking about this line:

[quote=olorin]
$(q(x)+(-x^{-1})^{n}q(x^{-1}))^{2}=(x+1)^{2n}(-x^{-2})^{n}q(x^{2})q(x^{-2})$
[/quote]

Consider the polynomial  $x^{2n}(\mathrm{LHS}-\mathrm{RHS})$.
This polynomial is zero for any $x\neq 1$, so it must be the zero polynomial. Hence the above equation holds even when $x=1$.[/quote]

Thank you!
I was thinking about that.


ratio
trigonometry
projective geometry
geometry unsolved
geometry

Let $ ABC$ be a triangle and $ (I)$ be its incircle. Let $ D$, $ E$, $ F$ be the tangency points of $ (I)$ with the sides $ BC$, $ CA$, $ AB$, respectively. $ M$, $ N$, $ P$ lie on $ EF$, $ FD$, $ DE$, respectively such that three lines $ DM$, $ EN$, $ FP$ are concurrent at one point, let it's $ J$. Denote by $ X$, $ Y$, $ Z$ be the second intersections of $ DM$, $ EN$, $ FP$ with the circle $ (I)$. Then, prove that

$ (a).$ $ AM$, $ BN$, $ CP$ are concurrent at $ K$.

$ (b).$ $ AX$, $ BY$, $ CZ$ are concurrent at $ L$.

$ (c).$ $ J$, $ K$ and $ L$ are collinear.

$ (a)$ follows from taking ratios using Ceva and Trig Ceva:

By Ceva,
$ \frac{DP}{PE}\frac{EM}{MF}\frac{FN}{ND}\equal{}1$
But because $ \triangle{BFD}$ is isosceles, $ \frac{FN}{ND}\equal{}\frac{\sin{\angle{FBN}}}{\sin{\angle{NBD}}}$
Applying the same thing to the two other triangles, we get an expression which, by trig Ceva, gives the desired result.

$ (b)$ is Steinbart, right?  Maybe I'm wrong...but if it is, it follows from Trig Ceva as well.

Part a,b can be proved by using Trig Céva (like the above posts). In part c we can use Desargues' theorem. Apply the theorem with two triangles $ BNY$ and $ CPZ$, we need the concurrence of $ BC,YZ,NP$, which is true and can be proved by using Pascal's theorem with the hexagon $ YZFDDE$

[quote="nktp"]b) Let $ \frac {sin\widehat{XAF}}{sin\widehat{XAE}} \equal{} \frac {S_{XAF}}{S_{XAE}} \equal{} \frac {XF.AF.sin\widehat{XFA}}{XE.AE.sin\widehat{XEA}}$
$ \equal{}\frac {XF.sin\widehat{XFA}}{XE.sin\widehat{XEA}} \equal{} \frac {sin^2\widehat{XDF}}{sin^2\widehat{XDE}}$
Similarly with $ 2$ other equation and multiply them we have result by trig Xeva[/quote]

can I know why

$ \frac {XF.sin\widehat{XFA}}{XE.sin\widehat{XEA}} \equal{} \frac {sin^2\widehat{XDF}}{sin^2\widehat{XDE}}$

?

I got lost there. =D tq

Great works, dear [b]PDatK40SP[/b]. Thank you very much.

Best regards

Thank you, April. Your problem is very nice :)
$ EJ,FJ$ cut $ BC$ at $ Q,R$
We have:
$ (JNYQ) \equal{} D(XFYD) \equal{} E(XFYD)\equal{} (XMJD)$ $ \equal{} F(XEZD) \equal{} D(XEZD) \equal{} (JPZR)$
So $ (BJ,BK,BL,BC) \equal{} (CJ,CK,CL,CB),$ or $ J,K,L$ are collinear

[quote="nktp"][quote="zestforlife"][quote="nktp"]b) Let $ \frac {sin\widehat{XAF}}{sin\widehat{XAE}} \equal{} \frac {S_{XAF}}{S_{XAE}} \equal{} \frac {XF.AF.sin\widehat{XFA}}{XE.AE.sin\widehat{XEA}}$
$ \equal{} \frac {XF.sin\widehat{XFA}}{XE.sin\widehat{XEA}} \equal{} \frac {sin^2\widehat{XDF}}{sin^2\widehat{XDE}}$
Similarly with $ 2$ other equation and multiply them we have result by trig Xeva[/quote]

can I know why

$ \frac {XF.sin\widehat{XFA}}{XE.sin\widehat{XEA}} \equal{} \frac {sin^2\widehat{XDF}}{sin^2\widehat{XDE}}$

?

I got lost there. =D tq[/quote]

Because:
$ \widehat{XFA} \equal{} \widehat{XDF},\widehat{XEA} \equal{} \widehat{XDE}$ ($ AE,AF$ tangent with $ (I)$)
$ XF \equal{} 2rsin\widehat{XDF},XE \equal{} 2rsin\widehat{XDE}$[/quote]

Ah... Understood. Thank you!   :)


this is an interesting example for using AM-Gm[/hide]


floor function
number theory
greatest common divisor
induction
relatively prime
strong induction

Prove that if $ \gcd(a,b)\equal{}1$, then $ \gcd(2^a\minus{}1,2^b\minus{}1)\equal{}1$.

[hide]Let's write this in binary, with the following notation:
$ 1_3 0_1 1_4 0_2 1_2 = 111011110011$
$ [1_2 0_1]_2 = 110110$
$ m \% n = m - \left \lfloor{\frac{m}{n} \right \rfloor n = \left \{ \frac{m}{n} \right \} \cdot n}$

[b]Solution:[/b] Consider $ m,n$ with $ m > n$. Then, $ 2^m - a = 1_m$ and $ 2^n-1 = 1_n$.

Note that $ 1_m =  [1_1 0_{n-1}] _{\lfloor m / n \rfloor -1} 1 _1 0_{m - \; m \% n} \cdot 1_n + 1_{m \% n}$
[b]Ex:[/b] $ 1_{10} = 10010010 \cdot 111 + 1$

Thus, $ \text{GCD} [2^m, 2^n]$
$ = \text{GCD} [1_n, \; [1_1 0_{n-1}] _{\lfloor m / n \rfloor -1} 1 _1 0_{m - \; m \% n} \cdot 1_n + 1_{m \% n}]$
$ = \text{GCD} [1_n, \; 1_{m \% n}]$

In other words, these GCD's reduce in the same manner that GCD's of integers reduce! We can just use the above property recursively on original inputs $ (a,b)$ to find that $ \text{GCD} [1_a, \; 1_b] = 1_{\text{GCD}[a,b]}$[/hide]

$ \gcd(2^a \minus{} 1,2^b \minus{} 1) \equal{} \gcd(2^a \minus{} 1,2^a \minus{} 2^b)$
$ \equal{} \gcd(2^a \minus{} 1,2^b(2^{a \minus{} b} \minus{} 1)) \gcd(2^a \minus{} 1, 2^{a \minus{} b} \minus{} 1)$, then you can keep continuing this, subtracting the smaller from the larger. If a and b are relatively prime, then one of the exponents must end up being 1 and we are done, otherwise they may end up being the same, and we can only get 0 as the exponent (Euclidean Algorithm).

Proceed by strong induction on $ a\plus{}b\equal{}n$.

Base $ n\equal{}2\equal{}1\plus{}1$ is trivial...Suppose $ a>b$. (If $ a\equal{}b$, then $ (a,b)\equal{}a\equal{}1$.)

$ (2^a\minus{}1,2^b\minus{}1)\equal{}(2^a\minus{}2^b,2^b\minus{}1)\equal{}(2^b(2^{a\minus{}b}\minus{}1),2^b\minus{}1)\equal{}(2^{a\minus{}b}\minus{}1,2^b\minus{}1)\equal{}1$ by inductive hypothesis because $ (a,b)\equal{}1\implies (a\minus{}b,b)\equal{}1$ and $ a\plus{}b>(a\minus{}b)\plus{}b$.

actually, that's well-known fact that $ gcd(a^b \minus{} 1,a^c \minus{} 1) \equal{} a^{gcd(b,c)} \minus{} 1$. The result follows (it is simple to prove that fact, just proceed as is shown in the first posts).

[quote="TZF"][hide]Let's write this in binary, with the following notation:
$ 1_3 0_1 1_4 0_2 1_2 = 111011110011$
$ [1_2 0_1]_2 = 110110$
$ m \% n = m - \left \lfloor{\frac {m}{n} \right \rfloor n = \left \{\frac {m}{n} \right \} \cdot n}$

[b]Solution:[/b] Consider $ m,n$ with $ m > n$. Then, $ 2^m - a = 1_m$ and $ 2^n - 1 = 1_n$.

Note that $ 1_m = [1_1 0_{n - 1}] _{\lfloor m / n \rfloor - 1} 1 _1 0_{m - \; m \% n} \cdot 1_n + 1_{m \% n}$
[b]Ex:[/b] $ 1_{10} = 10010010 \cdot 111 + 1$

Thus, $ \text{GCD} [2^m, 2^n]$
$ = \text{GCD} [1_n, \; [1_1 0_{n - 1}] _{\lfloor m / n \rfloor - 1} 1 _1 0_{m - \; m \% n} \cdot 1_n + 1_{m \% n}]$
$ = \text{GCD} [1_n, \; 1_{m \% n}]$

In other words, these GCD's reduce in the same manner that GCD's of integers reduce! We can just use the above property recursively on original inputs $ (a,b)$ to find that $ \text{GCD} [1_a, \; 1_b] = 1_{\text{GCD}[a,b]}$[/hide]
 [/quote]
Are the RHS's of

$ 1_3 0_1 1_4 0_2 1_2 = 111011110011$
$ [1_2 0_1]_2 = 110110$

in binary?
(BTW, your notation is a bit different from the one know - $ a_b$ means the number a written in base b)

If they are indeed in binary, then according to your notation, $ 1_n = \underbrace{1111...1111}_{n \ times} (binary)= 2^n + 2^{n-1} + 2^{n-2} + ... + 2^2 + 2 + 1$, and if this is indeed equal to $ 2^n-1$, then we would have $ 2^{n-1} + 2^{n-2} + ... + 2^2 + 2 + 1 = -1$, which isn't true.


find the following
lim[1-cosx cos2x cos3x...cosnx]/x^2  as x goes to 0

Applying l'Hopital yields the limit being $ \lim_{x \to 0} \frac {1} {2x} \sum_{k\equal{}1}^n \left ( k\sin(kx) \prod_{i\neq k} \cos(ix) \right ) \equal{} \frac {1} {2} \sum_{k\equal{}1}^n k^2 \equal{} \frac {n(n\plus{}1)(2n\plus{}1)} {12}$.

thanks. if we want lim{1-(cosx)^2 (cos2x)^2......(cosnx)^n]/x^2 as x goes to 0

Assuming that is $ \lim_{x \to 0} \frac {1 \minus{} \prod_{k\equal{}1}^n (cos(kx))^k} {x^2}$,

applying l'Hopital yields the limit being $ \lim_{x \to 0} \frac {1} {2x} \sum_{k\equal{}1}^n \left ( k^2\sin(kx) \cos(kx)^{k\minus{}1}\prod_{i\neq k} \cos(ix)^i \right ) \equal{} \frac {1} {2} \sum_{k\equal{}1}^n k^3 \equal{} \frac {n^2(n\plus{}1)^2} {8}$.


For $ a,b,c$ positive reals prove that

$ \sum \frac {a^2 \plus{} b^2}{c(a \plus{} b)} \geq (\sum a^2)\sqrt {\frac {3}{abc(a \plus{} b \plus{} c)}}$

i think that : write it in the form :
$ \sum \frac{x(a\plus{}b)}{y\plus{}z} \geq \sqrt{3(ab\plus{}bc\plus{}ca)}$
    but very hard.
we have :$ \frac{b^2\plus{}c^2}{a(b\plus{}c)} \plus{}1 \equal{}\frac{bc}{ab\plus{}ac}.(\frac{c\plus{}a}{b}\plus{}\frac{a\plus{}b}{c})$.
     who can help to finish it.

[quote="blackmouse"][quote="Hong Quy"]i think that : write it in the form :
$ \sum \frac {x(a \plus{} b)}{y \plus{} z} \geq \sqrt {3(ab \plus{} bc \plus{} ca)}$

[/quote]
Is it true?[/quote]
can find it at : http://reflections.awesomemath.org/currentissue.html
 it's sure true

Hi, my old friend Manlio. :)
The solution is
\[ \sum\frac{b^2+c^2}{a(b+c)} \ge (\sum a^2)\sqrt{\frac{3}{abc\sum a}}
\]
\[ \Leftrightarrow \sum \frac{b^2+c^2}{a(b+c)} \ge  (\sum a^2) \frac{\sqrt{ 3abc\sum a}}{abc\sum a}
\]
\[ \Leftarrow \sum \frac{b^2+c^2}{a(b+c)} \ge \frac{(\sum a^2)(\sum ab)}{abc\sum a}
\]
\[ \Leftrightarrow abc (\sum a) (\sum \frac{b^2+c^2}{a(b+c)}) \ge (\sum a^2)( \sum ab)
\]
\[ \Leftrightarrow \frac{1}{2}abc\sum \frac{(a-b)^2}{a+b} \ge 0
\]
Which is true. :)

Some ideas from that article, obvious the above is weaker than
\[ x(b \plus{} c) \plus{} y(c \plus{} a) \plus{} z(a \plus{} b)\ge 2\sqrt {(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)}
\]
by let $ x\to\frac {x}{y \plus{} z}$ a short solution for it:
\[ x(b \plus{} c) \plus{} y(c \plus{} a) \plus{} z(a \plus{} b) \equal{} (x \plus{} y \plus{} z)(a \plus{} b \plus{} c) \minus{} (ax \plus{} by \plus{} cz) \equal{} \sqrt {[x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} yz \plus{} zx)][a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)]} \minus{} (ax \plus{} by \plus{} cz)\ge 2\sqrt {(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)} \plus{} \sqrt {(x^2 \plus{} y^2 \plus{} z^2)(a^2 \plus{} b^2 \plus{} c^2)} \minus{} (ax \plus{} by \plus{} xz)\ge 2\sqrt {(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)}
\]

[quote="Hong Quy"]i think that : write it in the form :
$ \sum \frac {x(a \plus{} b)}{y \plus{} z} \geq \sqrt {3(ab \plus{} bc \plus{} ca)}$
    but very hard.
we have :$ \frac {b^2 \plus{} c^2}{a(b \plus{} c)} \plus{} 1 \equal{} \frac {bc}{ab \plus{} ac}.(\frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c})$.
     who can help to finish it.[/quote]

Write it as

$ \sum \frac{a^2\plus{}b^2}{c(a\plus{}b)}\equal{} \sum \frac{c}{a\plus{}b}(\frac{a^2}{c^2}\plus{}\frac{b^2}{c^2})$ and then apply

$ \sum \frac {c(A \plus{} B)}{a \plus{} b} \geq \sqrt {3(AB \plus{} BC \plus{} CA)}$  :)

Thank you very much for your nice solution, can_hang2007 :)

Your nice solutions are always welcome :)
 :)

some problems like this problem
$ \sum\frac{ab(a^3\plus{}b^3)}{a^2\plus{}b^2} \geq \sqrt{3abc(a^3\plus{}b^3\plus{}c^3)}$
  or
$ \sum ab\frac{a\plus{}c}{b\plus{}c} \geq \sqrt{3abc(a\plus{}b\plus{}c)}$


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calculus
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real analysis theorems

help me!
 :?:  :)
let sequence  :{fn} ,fo=f1=1,fn+1=fn+fn-1.find lim fn+1/fn.

It is quite easy .
Use the general explit of $ f_n$.

[[Fibonacci number]]
[[Binet's formula]]
[[Phi]]
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=262179577&t=151189

To make it a little more of a calculus problem:

Let $ a$ and $ b$ be arbitrary positive real numbers. Let $ x_0\equal{}a,$ $ x_1\equal{}b,$ and after that, $ x_{n\plus{}1}\equal{}x_n\plus{}x_{n\minus{}1}.$

Show that $ \lim_{n\to\infty}\frac{x_{n\plus{}1}}{x_n}$ exists and has the same value independent of $ a$ and $ b.$

[quote="Kent Merryfield"]To make it a little more of a calculus problem:

Let $ a$ and $ b$ be arbitrary positive real numbers. Let $ x_0 \equal{} a,$ $ x_1 \equal{} b,$ and after that, $ x_{n \plus{} 1} \equal{} x_n \plus{} x_{n \minus{} 1}.$

Show that $ \lim_{n\to\infty}\frac {x_{n \plus{} 1}}{x_n}$ exists and has the same value independent of $ a$ and $ b.$[/quote]

observe $ x_n \equal{} fibonnaci_n*a \plus{} fibonnaci_{n \plus{} 1}*b$ and when you apply binet's formula,  a+b simply cancels out for $ \frac{x_n}{x_{n\plus{}1}}$. and then if fibonnaci has limit, then this one has.