When the speed of a rear-drive car is increasing on a horizontal road, the direction of the frictional force on the tires is

a. backward on the front tires and forward on the rear tires
b. forward on the front tires and backward on the rear tires
c. forward on all tires
d. backward on all tires
e. zero

the answers is a...can anyone explain?

i dont understand how a forward frictional force on the rear tires at their contact point with the road can make the tires roll clockwisely...wouldn't the the torque from this frictional force cause the rear tires to roll counterclockwisely?

and what does the acceleration of the car and thus the angular and tangential accelerations of the tires have to do with the direction of frictional force?...and why is the direction of frictional force on the front tires different from that of the back's?

Friction acts against the direction of motion. The rear wheels want to move clockwise, and there is a force of friction that acts against that. And this is static friction, not kinetic friction. The static friction prevents the wheels from just spinning in place.

A rear-drive car puts all its energy into the rear wheels, which "push" the car. So the front wheels actually do nothing, and just provide a frictional force backwards on the car.

oh ok so the intended motion of the rear wheels is clockwise rolling, so friction, which acts against the direction of intended motion, will give the wheels a torque that would oppose their motion by accelerating them in the counterclockwise direction, which is the same as decelerating them in the clockwise direction, which opposes the intended clockwise rolling...so I guess this means the angular acceleration given to the rear wheels by the car (by some mechanism that doesn't matter) is greater in magnitude than the angular deceleration the forward frictional force gives via the torque it produces on the wheels?

the only thing is...if the rear wheels are accelerating then aren't the forward wheels too? so then wouldn't the frictional force on the forward wheels also oppose this intented clockwise rolling of them by causing a torque (from a forward frictional force at the wheel-road contact point) that would give the front wheels counterclockwise acceleration, i.e. a clockwise deceleration, thus opposing the intended clockwise rolling of the front wheels?

this means both the rear and front wheels would have a foreard frictional force, which is wrong...why why am I wrong?

Ok, first question. By definition, the angular acceleration of the rear wheels will always be greater or equal to the angular deceleration of the frictional force. When there is a force on an object, in this case on the wheel, then frictional force $ \mu_s\ge F$. Frictional force can't actually make something go backwards, if you think about it.

I think you may be overcomplicating it a little. By rotating the rear wheels clockwise, the car is pushing back on the road in order to push itself forward, similar to the way a person will push back to properl themselves forward. Therefore, the frictional force is forward.

The forward wheels are accelerating not because of the car, but because of the frictional force. This time, there is a frictional foce backwards on the tires, causing them to turn clockwise. And this frictional force is the only torque acting on it.
It's like if you were pushing a block forwards through it center of mass. There is going to be a frictional force backwards. The car is not exerting any torque on the forward tires.

yea I guess the answer to the first question was kind of obvious.

Ah ok that makes sense: the intended motion of the point on the tire that is in contact with the road at any given instant is backwards, relative to the road, so the frictional force on that point is forward to oppose that motion...so that means there is some other force on the tire that gives a torque, greater in magnitude and opposite in direction, to the one given by the backwards frictional force in order to give the rear tires a clockwise angular acceleration right, in order for the wheel to roll forward?

yea and the front wheels aren't "trying" to do anything so the only force on them would have to be friction which has to be backwards to roll the them forward.

heh maybe I am overcomplicating this...but its definitely helping my understanding of rolling and friction.

Correct. The force that makes the rear tire move clockwise is provided by the engine and the car's inner workings and whatnot.


For an arbitrary sequence of numbers $ A\equal{}(a_0,a_1,a_2,...)$, let $ SA\equal{}(a_0,a_0\plus{}a_1,a_0\plus{}a_1\plus{}a_2,...)$ be the sequence of partial sums of the series $ a_0\plus{}a_1\plus{}a_2\plus{}...$. Is there any sequence $ A$, not constant zero, for which the sequences $ A, SA, SSA, SSSA,...$ are all convergent?

No.  Let $ A(x) \equal{} \sum_{k\equal{}0}^{\infty} a_k x^k$ be the corresponding generating function.  The generating function for the sequence of partial sums is

$ S^k A(x) \equal{} A(x) (1 \plus{} x \plus{} x^2 \plus{} ... )^k$

and if $ A(x)$ has nonzero coefficients then the coefficients of $ S^k A(x)$ clearly tend to infinity.

Get ready for one pretty intense bump.

I'm not even close to convinced by this solution.  As far as I can tell, this is a pretty hard problem, and I'd love to see a solution.

By the way, this also appeared on the Miklós Schweitzer competition in 2007.

It might make sense to move this to the analysis forum as well.

Actually, t0rajir0u's solution works perfectly well.

Notice that for any sequence $ b_0,b_1,b_2,\ldots$ we have:
$ (b_0\plus{}b_1x\plus{}b_2x^2\plus{}\cdots)(1\plus{}x\plus{}x^2\plus{}\cdots) \equal{} (b_0)\plus{}(b_0\plus{}b_1)x\plus{}(b_0\plus{}b_1\plus{}b_2)x^2\plus{}\cdots$

So, if $ A(x) \equal{}\sum_{k\equal{}0}^{\infty}a_{k}x^{k}$ is the generating function for a sequence $ A$,
then generating function for $ SA$ is: $ SA(x) \equal{} (1\plus{}x\plus{}x^2\plus{}\cdots)A(x)$

Thus, by induction, we have:
$ S^k A(x) \equal{} (1\plus{}x\plus{}x^2\plus{}\cdots)^k A(x)$
as t0rajir0u stated.

From here, it is easy to see that as $ k \to \infty$, the coefficients of $ (1\plus{}x\plus{}x^2\plus{}\cdots)^k A(x)$ approach $ \pm \infty$, and thus, do not converge.

Saying that something is "easy to see" does not make it so.  Anyhow, even if you were able to prove that, for each $ i$, the $ i$th-coefficient approaches infinity as k grows*, you would not have shown what you need to, which is that [b]there exists some $ k$[/b] such that the coefficients of $ S_k A(x)$ do not converge.

This is a hard problem!

*this only true once we've hit the first nonzero coefficient anyway--the proof isn't hard but does require some thought about polynomials.

Ah, never mind.  It seems as if the problem statement requires that $ \lim_{x \to 1} \frac{A(x)}{(1 \minus{} x)^k}$ exists for every $ k$.  I don't know enough analysis to decide whether this is reasonable or not.

It's possible to construct a function with the property you mention: $ e^{\minus{}1/(1\minus{}x^2)}$ works, for example.

But that property is insufficient.  Case in point: it seems from inspection that the coefficients of $ \frac{e^{\minus{}1/(1\minus{}x^2)}}{(1\minus{}x)^2}$ do not converge, in other words if $ A$ is the sequence of coefficient of our power series, $ SSA$ is not convergent.

Now that I look at this, I'm not sure exactly what I was thinking when I said that "the coefficients of $ (1\plus{}x\plus{}x^2\plus{}\cdots)^k A(x)$ approach $ \pm \infty$, and thus, do not converge". I might have been neglecting the $ A(x)$ part, or something like that.

Solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153929]here[/url].  The short version is that $ e^{(1 \minus{} x)^{ \minus{} 1/2}}$ produces the coefficients we want, and the name of the game is Fourier analysis.


function
floor function
algebra proposed
algebra

I can't find a link of it online, but you should be able to use the fact that the set of positive integers can be covered by a family of sequences satisfying the fibonacci recurrence $ a_n \equal{} a_{n\minus{}1} \plus{} a_{n\minus{}2}$ if you choose appropriate starting points.

[quote="tdl"]Does there exist a function $ f$ from the positive integers to the positive integers such that $ f(1) \equal{} 2, f(f(n)) \equal{} f(n) \plus{} n$ for all $ n$, and $ f(n) < f(n \plus{} 1)$ for all $ n$?[/quote]

[color=green]$ f(n) \equal{} \lfloor\dfrac{1 \plus{} \sqrt {5}}{2}.n \plus{} \frac {1}{2}\rfloor$ .[/color]


if ab+bc-ac=0     and    a-c=101 ;  a,b,c is  positive integer       Find b


algebra
polynomial
number theory unsolved
number theory

Consider the polynomial $ P(x) \equal{} x^3 \plus{} 153x^2 \minus{} 111x \plus{} 38$.
(a) Prove that there are at least nine integers $ a$ in the interval $ [1, 3^{2000}]$ for which $ P(a)$ is divisible by $ 3^{2000}$.
(b) Find the number of integers $ a$ in $ [1, 3^{2000}]$ with the property from (a).

In $ 0<x<9$ we had 2 solutions $ 3^4|f(x)$ - $ 1,4$.
1. $ x\equal{}1\plus{}9x_2, \ P(1\plus{}9x_2)\equal{}3^4(1\plus{}22x_2\plus{}156x_2^2\plus{}9x_2^3)$. Because $ 3\not |22, \ 3|156, 3|9$ for any n in $ 0<x_2<3^n$ we had exactly one solution, suth that $ 3^{n\plus{}4}|P(1\plus{}9x_2)$. Therefore number of solutions form $ x\equal{}1\plus{}9x_2<3^n$ and $ 3^n|P(x),n\ge 4$ is 9.
2. $ x\equal{}4\plus{}9x_2, \ P(4\plus{}9x_2)\equal{}3^4(23\plus{}129x_2\plus{}165x_2^2\plus{}9x_2^3)$. It give 9 solutions. Total solutions $ x<3^n,3^n|P(x),n\ge 4$ is 18.

please can someone put another solution or who can explain the first solution in detail please

[hide]Consider the polynomial $ P(x) = x^3 + 153x^2 - 111x + 38$.
(a) Prove that there are at least nine integers $ a$ in the interval $ [1, 3^{2000}]$ for which $ P(a)$ is divisible by $ 3^{2000}$.
(b) Find the number of integers $ a$ in $ [1, 3^{2000}]$ with the property from (a).[/hide]


Here it come:

$ \sqrt[9]{3\sqrt{2}\minus{}2\sqrt{5}}\sqrt[18]{3\sqrt{10}\plus{}\frac{19}{2}}$

I guess now you know why I hate radicals  :lol:

[hide="Hint:"]Notice that $ \sqrt [9]{3\sqrt {2} \minus{} 2\sqrt {5}} \equal{} \minus{} \sqrt [18]{(3\sqrt {2} \minus{} 2\sqrt {5})^2}$ (The negative sign because $ 3\sqrt{2} \minus{} 2\sqrt{5} < 0$)[/hide]

Solved.  :D
Thanks a lot gauss202  :)
I'm in a hurry now but I will put the solution for others benefit. Thanks again


calculus
derivative
calculus computations

Hi all Im new here.

Im sore for not using proper symbols for the equations as I dont have word or any math program installed on my P.C

Im trying to understand how the maclaurins series expansion of

e^(x^2)

Is solved. I have the answer but when I do the calculations I get 1 for F(o), then when I derive it I get 2xe^(x^2), when I try this for
F(o), I get 0, and this is the case for all other deritives.

This leads me to beleive the Maclaurins series expansion is simply f(x) = 1 - but my answer sheet says its 1 + x^2 + (x^4)/2 + ....

I dont get it. Is there any special rules when dealing with this case?

I get the correct expansion when doing it for e^2x and xe^x

Please help! Thanks

expand $e^x$ and replace $x$ by $x^2$ because $x$ and $x^2$ have the same value at $x=0$.

I'd prefer that student questions like this be placed over here, in the "Calculus computations and tutorials" section.

As a second issue: if you click on various parts of my post below, you'll see the TeX code used to show the formulas. Go ahead and try it out yourself with your next post. Just enclose the expression you want by dollar signs, make your best guess as to the right notation (copying what you see elsewhere) and use "Preview" to see whether it renders properly.

This question should be handled by substitution. We know that

$e^x=1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots.$

To get the series for $e^{x^2}$, just replace $x$ everywhere it appears above by $x^2$:

$e^{x^2}=1+x^2+\frac{x^4}2+\frac{x^6}{3!}+\cdots= \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}.$

Here's a question for you: if $f(x)=e^{x^2}$, then what is $f^{[10]}(0)$? (The tenth derivative evaluated at zero.)

[quote="Popee"]Is solved. I have the answer but when I do the calculations I get 1 for F(o), then when I derive it I get 2xe^(x^2), when I try this for F(o), I get 0, and this is the case for all other deritives.
[/quote]
Have you computed f''(x) yourself? it is $2e^{x^2} + 4e^{x^2}x^2$.
and we find that $f^{(n)}(0) = 0$ for odd $n$ and $f^{(n)}(0) \neq 0$ for even $n$.

Ah ive cracked it. For some reason I wasnt using the chain rule to differentiate $2e^{x^2}$ and so on!

Kent I was aware there was a simple way to do it, but im glad I figured it out. Just a silly mistake really. Its late here! I realised what id done when I got into bed and switched the light off. Just hopped out and tried it and it all works out fine.

Thanks for your help everyone! - I hope this is on my exam now.


geometry
ratio
analytic geometry
Pythagorean Theorem

This sides of a triangle are of lengths 13, 14, and 15. The altitudes of the triangle meet at point H. If line AD is the altitude ot the side of length 14, what is the ration HD:HA?

Source: AHSME 1964 #5

It is not very difficult if use Area Ratios, but what is $\text{mass point geometry}$?

Mass point geometry is another name for Barycentric coordinate, if I remember correctly.

What little I know about mass point geometry I learned from this article.

http://mathcircle.berkeley.edu/mpgeo/mpgeo.html

Hm, I don't know if there's a nicer way besides finding these lengths, but it really isn't too bad (was it one of the later questions?  are calculators allowed on AHSME's?).

draw your 13-14-15 triangle with altitude to the 14 side.  If you don't already know, its easy to show that the altitude has length 12 and divides the 14 side into lengths 5 and 9 (forming a 5-12-13 right triangle and a 9-12-15).

Now choose 1 other altitude (either work, but the one to 15 will make for simpler numbers) so since the area is 84, the length of that altitude is 168/15=56/5 and by pythatgorean theorem you can find that the segment is divided into lengths 33/5 and 42/5.

Now you have everything you need to apply basic mass points and you'll end up with 70, 55, and 99 on the three vertices, 55+99=154 on point D and thus, HD:HA= 70/154= 5/11

How do you do it with area ratios?

[quote="jelly119"]I don't know what the mass point geometry is about, but Menelaus provides a nice way for this:
[img]http://i5.photobucket.com/albums/y194/jelly119/05072501.jpg[/img]

Simple use of pythagorean gives $BD:CD=5:9$, $ AF:BF=99:70$.

Apply Menelaus's theorem to $\triangle{ABD}$ with respect to $F,H$ and $C$ yields:
\[\frac{DH}{AH}\times\frac{AF}{BF}\times\frac{BC}{DC}=1\]
\[\frac{DH}{AH}\times\frac{99}{70}\times\frac{14}{9}=1\]
\[\frac{DH}{AH}=\frac{5}{11}\][/quote]

How is it $ AF:BF=99:70$ ? Can you show ?

[quote="Hanno"]Hi.

Let ABC be a triangle with sites a,b,c and let CD be perpendicular to AB through C. Further, denote [tex]x\equiv \overline{BD}, y\equiv \overline{AD}, h\equiv\overline{CD}[/tex]. Then, by the pythagorean theorem we have [tex]x^2=a^2-h^2=a^2-b^2+y^2=a^2-b^2+c^2-2cx+x^2\Leftrightarrow x=\frac{a^2+c^2-b^2}{2c}[/tex]. Analogously we get [tex]y=\frac{b^2+c^2-a^2}{2c}[/tex], whence [tex]\frac{x}{y}=\frac{a^2+c^2-b^2}{b^2+c^2-a^2}[/tex]. Applying this to the given problem, that is, to [tex]a=13, b=15, c=14[/tex], we get [tex]\frac{AF}{BF}=\frac{99}{70}[/tex].

Hanno[/quote]

Oh thanks very much ! :D

[hide="Some random kid who just learned mass points comes along..."]

By Pythagorean, we have $15^2-CD^2=13^2-BD^2 \Rightarrow (CD,BD)=(9,5)$
Call the altitute from $B$, $BE$.
By Pythagorean, we have $14^2-CE^2=13^2-AE^2 \Rightarrow (CE,AE)=(8.4,6.6)$

Now let $a$ be the mass on $A$, etc.

$9c=5b$ and $6.6a=8.4c \Rightarrow \frac{14}{11} = a/c$ so $c$ must be a multiple of both $5$ and $11$, so $c=55$, $b=99$, $a=70$. Then $d=154$, so $\frac{70}{154}=\boxed {\frac{5}{11}}$. Yay! I did it! :D
[/hide]

[quote="erik-the-red"]I first heard of mass points only a month ago.

What is its purpose?[/quote]
If you know area ratios, then you don't need math point.

[quote="erik-the-red"]I first heard of mass points only a month ago.

What is its purpose?[/quote]
It is just a MUCH easier way of finding raios of lengths, especially triangles that is made by simuilating having weights on each point, and they have to balance out


Να λυθεί η διοφαντική εξίσωση $ 2x^3\equal{}y^2\plus{}1$ όπου $ x,y$ ακέραιοι.

Δεν ξέρω αν θα βοηθήσει αλλά $ y\equiv1(mod2)$

Καλησπέρα σε όλους και από εμένα! Είμαι νέο μέλος του forum και στέλνω πρώτη φορά.

Οι λύσεις της εξίσωσης $ 2x^3\equal{}y^2\plus{}1$ είναι: $ (x,y)\equal{}(1,1), (1,\minus{}1).$ Εξηγώ το γιατί:

Αν $ (x_{0},y_{0})$ λύση της $ 2x^3\equal{}y^2\plus{}1$ τότε προφανώς $ (2x_{0},2y_{0})$ είναι λύση της $ y^2\equal{}x^3\minus{}4$.  Η $ y^2\equal{}x^3\minus{}4$ έχει λύσεις τις $ (x,y)\equal{}(2,2), (2,\minus{}2), (5,11), (5,\minus{}11)$. Οπότε εύκολα προκύπτει το συμπέρασμα.
Εξισώσεις της μορφής $ y^2\equal{}x^3\plus{}k$ έχουν λυθεί για πάρα πολλές τιμές του $ k$ (υπάρχουν σχετικοί πίνακες σε βιβλία Αλγεβρικής Θεωρίας Αριθμών). (Για την συγκεκριμένη, έχουμε $ x^3\equal{}(y\minus{}2i)(y\plus{}2i)$ και δουλεύμε στο σώμα $ Q(i)$).

Νίκος

Γεια σου ρε Νίκο!! Καλωσήρθες στην Ομάδα μας... Είναι χαρά μας να σ'έχουμε και σένα εδώ!! Παιδιά ο Νίκος Κατσίπης ήταν συμφοιτητής μου στο μεταπτυχιακό του Πανεπιστημίου Κρήτης. Είναι εξαιρετικός μαθηματικός και πολύ γερός στη Θεωρία Αριθμών (και όχι μόνο  :wink: )! Σίγουρα θα μας βοηθήσει πολύ η συμμετοχή του!!

Να αναφέρω ότι οι εν λόγω πίνακες διοφαντικών εξισώσεων υπάρχουν σε όλα τα κλασσικά βιβλία Αλγεβρικής Θεωρίας Αριθμών. Για παράδειγμα αναφέρω το βιβλίο των Alaca - Williams "Algebraic Number Theory" στο οποίο βρίσκεται και η συγκεκριμένη κατηγορία διοφαντικών (προς το τέλος του βιβλίου).

Γιώργο μήπως έχεις κάποια άλλη λύση πιο στοιχειώδη? Σίγουρα για να λύνεται με Αλγεβρική Θεωρία Αριθμών θα θέλει αρκετή δουλειά με στοιχειώδη τρόπο! Αν έχεις λοιπόν κάποια άλλη λύση υπόψιν σου γράψε την! Πες μας επίσης και την πηγή της άσκησης αν την έχεις (βιβλίο, διαγωνισμός κτλ).

Αλέξανδρος

Έστω $ x,y\in Z$ τέτοιοι ώστε να ισχύει $ 2x^3 \equal{} y^2 \plus{} 1$. Το $ y$ είναι περιττός. Θεωρούμε στο $ Z_{[i]}$ την παραγοντοποίηση $ (y \plus{} i)(y \plus{} i) \equal{} 2x^{3}$.  $ (1)$

Κάθε κοινός διαιρέτης των $ y \plus{} i$ και $ y \minus{} i$ διαιρεί το $ (y \plus{} i) \minus{} (y \minus{} i) \equal{} 2i \equal{} (1 \plus{} i)^{2}$. Όμως το $ \delta (1 \plus{} i) \equal{} 2$ και επειδή το $ 2$ είναι πρώτος το $ 1 \plus{} i$ είναι ανάγωγο και απο την μοναδικότητα της παραγοντοποίησης στο $ Z_{[i]}$ παίρνουμε ότι κάθε κοινός διαιρέτης των $ y \plus{} i$ και $ y \minus{} i$ είναι συντροφικός με έναν απο τους $ 1,1 \plus{} i,(1 \plus{} i)^{2}$.

Eπειδή ο $ y$ είναι περιττός έχουμε $ y \plus{} i \equal{} 2k \plus{} 1 \plus{} i$ για κάποιο $ k\in Z$, άρα το $ y \plus{} i$ διαιρείται με το $ 1 \plus{} i$ αλλά όχι με το $ (1 \plus{} i)^{2}$. Επίσης το $ 1 \plus{} i$ διαιρεί το $ y \minus{} i \equal{} 2k \plus{} 1 \minus{} i$. Συνεπώς $ gcd(y \plus{} i,y \minus{} i) \equal{} i \plus{} 1$.

Οπότε τώρα έχουμε ότι $ y \plus{} i \equal{} (1 \plus{} i)(a \plus{} bi)$ και $ y \minus{} i \equal{} (1 \plus{} i)(c \plus{} di)$ όπου τα $ a \plus{} bi,c \plus{} di\in Z_{[i]}$ είναι σχετικά πρώτα στοιχεία.

Άρα απο τη σχέση $ (1)$ συνεπάγεται ότι $ ( \minus{} ix)^{3} \equal{} (a \plus{} bi)(c \plus{} di)$.

[u]Λήμμα[/u]
Έστω $ R$ μια περιοχή μοναδικής παραγοντοποίησης και $ a,b,c\in R$ με $ ab \equal{} c^{n}$, $ n > 0$. Αν τα $ a,b$ είναι σχετικά πρώτα, τότε υπάρχουν $ d,e\in R$ και αντιστρέψιμο $ u\in R$ τέτοια ώστε $ a \equal{} ud^{n}$ και $ b \equal{} u^{ \minus{} 1}e^{n}$.

Απο το παραπάνω λήμμα έπεται ότι υπάρχει αντιστρέψιμο $ u\in R$ και $ \gamma \plus{} \delta i\in Z_{[i]}$ με $ a \plus{} bi \equal{} u(\gamma \plus{} \delta i)^{3}$. Επειδή όμως κάθε αντιστρέψιμο στοιχείο του $ Z_{[i]}$ είναι τρίτη δύναμη στο $ Z_{[i]}$ μπορούμε να υποθέσουμε ότι $ u \equal{} 1$.

Άρα λοιπόν με αντικατάσταση του $ u$ έχομυε ότι
$ y \plus{} i \equal{} (1 \plus{} i)(a \plus{} bi) \equal{} (1 \plus{} i)(\gamma \plus{} \delta i)^{3} \equal{} (\gamma^{3} \minus{} 3\gamma^{2}\delta \minus{} 3\gamma \delta^{2} \plus{} \delta^{3}) \plus{} (\gamma^{3} \plus{} 3\gamma^{2}\delta \minus{} 3\gamma \delta^{2} \minus{} \delta^{3})i$.

Ως συνέπεια των παραπάνω λαμβάνουμε τελικά το σύστημα
$ y \equal{} (\gamma \plus{} \delta)(\gamma^{2} \minus{} 4\gamma\delta \plus{} \delta^{2})$

$ 1 \equal{} (\gamma \minus{} \delta)(\gamma^{2} \plus{} 4\gamma\delta \plus{} \delta^{2})$ και επειδή $ \gamma,\delta\in Z$ απο τη δεύτερη εξίσωση παίρνουμε ότι $ (\gamma ,\delta) \equal{} (1,0),(0, \minus{} 1)$. Αντικαθιστώντας στην πρώτη βρίσκουμε ότι
 $ (x,y) \equal{} (1,1),(1, \minus{} 1)$

Αλέξανδρε ο τίτλος του βιβλίου αυτού είναι ο ''μια εισαγωγή στην άλγεβρα''  απο το εθνικό και καποδιστριακό μαθηματικό τμήμα Αθηνών. :)

Μπορείς να μου εξηγήσεις λίγο τι εννοείς με το $ \delta(1 \plus{} i) \equal{} 2$ ?

Γενικά δεν έχω ασχοληθεί καθόλου με αλγεβρική θεωρία αριθμών  :oops:

Θα έρθει η ώρα της στο πανεπιστήμιο  :)

Στέλιος

Στέλιο $ \delta(a \plus{} bi) \equal{} a^2 \plus{} b^2$ όταν δουλεύεις στον δακτύλιο $ \mathbb{Z}[i]$. Είναι δηλαδή σαν να λέμε το μέτρο του μιγαδικού στο τετράγωνο. Όταν χρειάζεται να δουλέψεις σε άλλο σώμα τότε το "μέτρο" είναι διαφορετικό. Δεν υπάρχει λόγος να πούμε παραπάνω πράγματα γιατί χρειάζονται αρκετές έννοιες ανώτερης Άλγεβρας [Δακτύλιοι, σώματα, πεδία μονοσήμαντης ανάλυσης (που χρειάζονται στο πρόβλημα) και πολλά άλλα για να αναλύσουμε πλήρως την παραπάνω λύση (ας πούμε τί είναι πρώτος και τί σύνθετος στο μεγαλύτερο σύνολο? Υπάρχει η μοναδική ανάλυση σε πρώτους παράγοντες όπως υπάρχει στο σύνολο των ακεραίων? (όχι πάντα, γι'αυτό υπάρχουν τα λεγόμενες περιοχές μοναδικής ανάλυσης))]. Το σύνολο $ \mathbb{Z}[i]$ είναι το σύνολο όλων των στοιχείων $ a\plus{}bi$ όπου όμως $ a,b\in \mathbb{Z}$. Άρα έχουμε εφοδιάσει τους ακεραίους αριθμούς με το $ i$ και παίρνουμε ένα μεγαλύτερο σύνολο. Ο αριθμός $ 3$ είναι πρώτος στο σύνολο $ \mathbb{Z}[i]$ [προσπαθήστε να τον γράψετε στη μορφή $ (a\plus{}bi)(c\plus{}di)$ όπου $ a,b,c,d\in\mathbb{Z}$ και κανένας από τους δύο να μην είναι το $ \pm 1$ ή $ \pm i$ (που είναι οι μονάδες του συνόλου μας) και θα δείτε ότι θα αποτύχετε (εκεί χρειάζεται και το μέτρο στο τετράγωνο)] ενώ ο αριθμός 5 που είναι πρώτος στο $ \mathbb{Z}$ είναι "σύνθετος" στο $ \mathbb{Z}[i]$ διότι $ 5\equal{}(2\plus{}i)(2\minus{}i)$. Άλλο ερώτημα που προκύπτει είναι: Μπορεί να γραφτεί με άλλο τρόπο; Η απάντηση είναι όχι διότι ακριβώς το $ \mathbb{Z}[i]$ είναι πεδίο μονοσήμαντης ανάλυσης οπότε δεν έχουμε άλλο τρόπο γραφής του $ 5$ [ή ίσως να έχουμε πολλαπλασιασμό της παραπάνω γραφής με τις "μονάδες" του συνόλου μας (εξάλλου και στους ακεραίους που οι μονάδες είναι $ \pm 1$ η γραφή του $ 6$ ως $ 2\cdot 3$ δεν θεωρείται διαφορετική από την $ (\minus{}2)(\minus{}3)$ διότι είναι η προηγούμενη ανάλυση πολλαπλασιασμένοι με τις μονάδες των ακεραίων)]

Το όλο θέμα είναι ότι για να λύσουμε μία διοφαντική εξίσωση στους ακεραίους στους οποίους δεν μπορούμε να δουλέψουμε με στοιχειώδη μέσα, "ανεβαίνουμε" και δουλεύουμε σε ένα μεγαλύτερο σύνολο (στο συγκεκριμένο παράδειγμα στο σύνολο των ακεραίων του Gauss $ \mathbb{Z}[i]$). Όποιες λύσεις βρούμε τις βάζουμε στην αρχική και βλέπουμε αν την επαληθεύει. Προφανώς εαν η εξίσωση δεν λύνεται στο μεγαλύτερο σύνολο τότε δεν λύνεται ούτε στο μικρότερο.

[Μοιάζει με τις εξισώσεις με ριζικά στους πραγματικούς, στις οποίες κανονικά πρέπει να δουλέψεις στο σύνολο ορισμού τους, ΟΜΩΣ θεωρείς ότι δουλεύεις στο σύνολο των μιγαδικών αριθμών (αυτό δεν λέγεται φανερά γιατί στη Β΄ Λυκείου δεν ξέρεις μιγαδικούς), υψώνεις εις το τετράγωνο για να διώξεις τις ρίζες και μετά ΕΠΑΛΗΘΕΥΕΙΣ για να δεις εαν οι τελικές τιμές που βρήκες επαληθεύουν την αρχική. Προφανώς εαν η εξίσωση δεν λύνεται στους μιγαδικούς τότε δεν λύνεται ούτε στο σύνολο ορισμού της εξίσωσης].

Περισσότερα στο μάθημα Αλγεβρικής Θεωρίας Αριθμών (νομίζω μόνο στο Μαθηματικό Θεσσαλονίκης είναι υποχρεωτικό μάθημα 3ου ή 4ου έτους και μόνο στην κατεύθυνση των καθαρών μαθηματικών).

Ελπίζω να σε βοήθησα έστω και λίγο Στέλιο.  (μπορείς αν θέλεις να μάθεις περισσότερα να ανατρέξεις στο βιβλίο που αναφέρω παραπάνω και να διαβάσεις μερικά πράγματα. Θεώρησε κάποια από αυτά δεδομένα (δεν μπορείς να μάθεις όλη την Άλγεβρα από μερικές σελίδες) και προχώρα στα παραδείγματα τα οποία είναι ΠΟΛΥ αναλυτικά γραμμένα.

Αλέξανδρος

Αλέξανδρε,

σε ευχαριστώ πάρα πολύ για τον πολύτιμο χρόνο που διέθεσες και μου εξήγησες μερικά πράγματα :)

Σίγουρα τώρα τα έχω καταλάβει καλλίτερα !!

Με την πρώτη ευκαιρία θα αγοράσω το βιβλίο που προτείνεις! Είμαι σίγουρος ότι θα 'ναι πολύ καλό!

Στέλιος


One considers the positive integers $a < b \leq c < d $ such that $ad=bc$ and $\sqrt d  - \sqrt a \leq 1 $.

Prove that $a$ is a perfect square.

Okay, I will post my soln, since I think this is a very nice problem.

From $a-d > c-b $ we obtain that $ ( a-d)^2 > (b-c)^2 \Rightarrow (a-d)^2+4ad > (c-b)^2 + 4bc $ and thus $(a+d)^2 > (b+c)^2 \Rightarrow a+d \geq b+c+1. $

Suppose first that $\sqrt d - \sqrt a = 1$. Then $ a+d - 2\sqrt {ad} = 1 \Leftrightarrow $ $ a+d = 1 + 2 \sqrt {bc } \leq 1 + b+c \Rightarrow b=c$ and $a+d=2b+1$. But then $ad=b^2$, so $\gcd (a,d)=1$ because otherwise, if $p\ | \ \gcd(a,d)$ and $p$ is a prime we would have $ p\ | \ b$ and thus $p \not | 2b+1$ contradiction. So, each of $a$ and $d$ are perfect squares.

Now, suppose that $\sqrt d - \sqrt a <1$. Then $a+d < 1+ 2\sqrt {ad} \leq 1+b+c \leq a+d $ which is a contradiction.

Sorry I ment $|a-d| > |b-c|$ which is obvious given that $a<b\leq c<d$.

[quote=spider_boy]:? Where do we get thata-d>c-b,and also a-d is negative,but c-b is positive!
and we can not square both sides because  a-d is negative! :?[/quote]

The solution still works after squaring regardless of sign because of absolute value, but yes that is an error.

Nice problem!


Se considera cinci puncte conciclice.Sa se arate ca perpendiculareleduse din centrele de greutate ale triunghiurilor formate cu oricare trei puncte,pe coarda determinata de celalalte doua,sunt concurente. ;)  :P

Vad ca problema este neinteresanta,poate mult prea simpla. :oops:

Intr-adevar, e mult prea simpla: Daca $O$ este centrul cercului, $A$, $B$, $C$, $D$, $E$ - cele cinci puncte, iar $M$ punctul din plan pentru care
\[
\overline{OM}=\overline{OA}+\overline{OB}+\overline{OC}+\overline{OD}+\overline{OE}\,,
\]
atunci $M$ este punctul de concurenta al dreptelor date.

Apropo, dupa semnele de punctuatie se lasa de obicei spatii.

Problema asta a fost data la olimpiada locala in Constanta la clasa a noua .

Intr-adevar,a fost propusa de "cineva" care in ultimii trei ani la olimpiada din constanta a dat probleme cunoscute, cel
 putin la geometrie vectoriala,dar ce sa faci,asta e! :(

:? chiar inspectorul le-a propus   :rotfl:

Din partea mea inspectorul asta sa fie sanatos.Poate sa fie cine vrea el, problema este ca ele sunt cunoscute. :(

cred totusi ca destul de trist e ca doar o persoana a luat punctaj maxim la ea  :(

Intr-adevar, e chiar pacat, avand in vedere simplitatea problemei. Dar, ma rog, tu de unde stii? ;)

Eu abia acum am invatat geometrie vectoriala asa ca inca (vectorial nu am facut-o) dar am gasit o solutie sintetica.

Solutie sintetica??! Suna interesant. Poti sa o postezi? :?

Fie cele cinci puncte de pe cerc A1 , A2 , A3 , A4 , A5 . Luam toate triunghiurile cu vf in A1 si ducem din centrele lor perpendiculare pe coardele formate de celelalte doua puncte .Daca consideram omotetia de centru A1 si raport 3/2 care imi transforma centrele de greutate ale triunghiurilor cu vf in A1 in mijloacele laturilor patrulaterului A2A3A4A5 ; cum omotetia imi transforma drepte concurente in drepte concurente => daca demonstram ca perpendicularele duse din mijloacele laturilor unui patrulater inscriptibil pe laturile opuse sunt concurente atunci problema este rezolvata(aproape).Se poate verifica ca punctul de intersectie al acestora este simetricul centrului cercului circ. patr. fata de mijlocul segmentului EF unde E si F sunt mijloacele diagonalelor patrulaterului c.c.t.d.Prin repetarea procedeului cu A2 , A3 , A4 , A5 , vom obtine ca dreptele cerute(15 la numar) sunt concurente patru cate patru ceea ce inseamna ca sunt concurente toate intre ele .

Imi cer scuze pt intarziere dar eu am monitoru stricat si acum scriu problema de la un prieten .


Let a; b; c be real numbers. Prove that

(ab + bc + ca-1)^2 le (a^2 + 1)(b^2 + 1)(c^2 + 1)

[hide]After expanding it becomes equivalent to $ (a+b+c-abc)^{2}\geq 0$.[/hide]


Ok my problem is simple, but id just like to make sure i got it right.  A right-triangle has the following measurements:

X = 50
                |
                ||
                ||||
                ||||||
                ||||||||        Z = ?
                ||||||||||
                ||||||||||||
                ||||||||||||||
                ||||||||||||||||
                ||||||||||||||||||

                     Y = 60  <----

Z 2 = X 2  + Y 2
Z 2  = 50 2  + 60 2
Z =  :sqrt: 6100
Z = 78

And the actual problem is that the height (X) is getting taller at a rate of 10 meters per minute, what is the rate in which Y is shrinking?

    Well this is how I did the problem:

1.) Particular Information:

 X = 50
 Y = 60
 Dx/Dt = 10 meters/minute
 Dy/Dt = ? when X = 50

2.) Steps in solving the problem:
 X 2  + Y 2  = 78
 2x (dx/dt) + 2y(dy/dt) = 0             <------- Derivative
 2x (dx/dt) = -2y(dy/dt)                  <------- DID I SET IT UP RIGHT?
 2 (50) (10) = -2 (60) (dy/dt)
 1000 = -120 (dy/dt)
 dy/dt = -8.333...

Shrinking at rate of 8.333... meters per minute

IS THAT CORRECT?

No picture really needed for this one. Obviously we have $x^{2} + y^{2} = z^{2}$. $Z = 78$ by Pythagoren Theorem using $x= 50$ and $y=60$.

$x^{2} + y{2} = 78$
$\frac{dx}{dt} = 10$

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

Substituting in 50 for x and 60 for y and 10 for dx/dt we'll end up getting:

$\frac{dy}{dt} = \frac{-1000}{120} = \frac{-25}{3} \frac{m}{min.}$

So yeah, you're right.


A boy begins to solve a problem at the time between 4:00 and 5:00 p.m. when the clock's hands are together. He finishes when the minute hand is exactly opposite the hour hand. How many minutes it take him to solve the problem, and when does he finish it ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

[hide="Solution"]At 4:00 pm the hour hand is at 20 minutes. It moves $ \frac {1}{12}x$ minutes as the minute hand moves x minutes to met it. Then $ 20x\cdot\frac {1}{12}x \equal{} x$, thus $ x \equal{} 21\frac {9}{11}$. Thus the boy starts the problem at 4:21$ \frac {9}{11}$ pm. Again, starting at 4:00 pm, the hour hand moves $ \frac {1}{12}x$ minutes from 20 as the minute hand moves x minutes pass the hour hand by 30 minutes. Then $ 20 \plus{} \frac {1}{12}x \plus{} 30 \equal{} x$, and $ x \equal{} 54\frac {6}{11}$. Thus the boy finishes at 4:54$ \frac {6}{11}$ pm, and the problem takes him 32 $ \frac {8}{11}$ minutes to solve it.[/hide]


[quote="Drunken_Math"]Prove that a 17-gon can or cannot be divided into parallelograms.[/quote]

A convex 17-agon can't, and a concave 17-agon can.  This has already been posted, but I'll post a hint:

[hide]Show that in order to divide a convex polygon into parallelograms, the number of sides must be even.[/hide]


Let $ n$ and $ a$ be positive integers greater than 1 and $ (n,a)\equal{}1$. Show that $ Z_n^*$ and the set
\[ S\equal{}\{1,a,a^2,\ldots,a^{\varphi(n)\minus{}1}\}
\] are equal.

This is not possible unless $ n \equal{} 1, 2, 4, p^k, 2p^k$ for $ p$ an odd prime and is only true for $ a$ a [url=http://en.wikipedia.org/wiki/Primitive_root_modulo_n]primitive root[/url] $ \bmod n$; in the cases that they exist, there are $ \varphi(\varphi(n))$ of them.


[quote="gauss202"]Look at this remarkable article that I found! http://www.cco.caltech.edu/~mamikon/VisualCalc.html[/quote]

Interesting. Thanks for sharing the link.

Wow! Apostol needs to write another book now.

wowowowowowow. This guy = fields medal pronto.

I think he's a bit too old for a fields medal if he's been teaching for 50 years  :)


induction
linear algebra
linear algebra unsolved

Let $A,B\in\mathcal M_{2}(R)$ and $AB-BA=A$. Show that $AB^{n}A=O_{2}$ for any $n\in\mathbb{N}^{*}$

One can use the fact that $A^2=0$ and an easy induction (I think).

I have one question. What's Caley consequence?  :|


Okay guys,

If we don't want to go to Carowinds next year, which I presume is the case because 26 USD for less than five rides is incredibly outrageous (26 USD for less than 25ish rides is pretty bad, but hey, I'll take what I can get), then we need to propose a viable alternative.

I thus suggest Dave and Buster's in Concord, NC.  The Dave and Buster's location listing [url=http://www.daveandbusters.com/Locations/default.aspx]here[/url] gives its Saturday hours as 11:30 AM - 2:00 AM, so we could definitely hang out there for a while (i.e., about 4 hours at the max, but less if we wanted to get home earlier).  The establishment is basically a giant arcade.  It's a little north of the center of Charlotte, so Charlotte kids could easily be dropped off there.  Plus it's quite close to I-85 ([url=http://maps.google.com/maps?f=q&hl=en&geocode=&q=+8361+Concord+Mills+Blvd+Concord+,+NC+28027&sll=37.0625,-95.677068&sspn=33.160552,76.992187&ie=UTF8&ll=35.263562,-81.287842&spn=4.269412,9.624023&z=7]map[/url]) so it's not too terribly out of our way.

Any other suggestions?

I checked the website... Are you sure about this? It looked like a casino to me.

I know that the Dave and Buster's in Georgia have an age limit after 5pm or so.

@bismbard: Although it may look like a casino, it's actually a giant arcade.  I know that there are a lot of similarities.  For example,
[list]
you spend money at both places;
you can't win back hardly any money at either place;
you allegedly have fun at both places;
they're buildings;
weird people inhabit both;
they're for-profit enterprises;
they have various ways to lose money;
the food is ridiculously overpriced because you're a captive audience;
etc.
[/list]

@gtf: I don't know anything about an age limit after 5 pm.  I know that I went to one with Siemens people in NYC in 12/07, and a freshman was allowed in, but other than that I have no idea.  I wouldn't be too surprised either way.  You could always call the NC one and find out.


$8x$, the rhombus must be a square to be inscribed in a circle.

[quote="4everwise"]A rhombus is inscribed inside a circle. The length of one diagonal of the rhombus is 8x. What is the length of the other diagonal.[/quote]

To be inscribed in a circle, it must be a square, so the answer is $8x$

[hide]so if you have a rhombus in a circle, doesn't that mean that it is a square?  So if the diagonal is 8x and the rhombus is a square then the other diagonal is 8x[/hide]


If $a\geq b\geq 0$ and $t\geq 1$ prove $(a-b)^{t}\leq a^{t}-b^{t}$

Let's notice that for $t\geqslant 1$ the function $f(x)=x^{t}$ is convex for $x\geqslant 0$, since $f''(x)=t\cdot (t-1)\cdot x^{t-2}$. Then let's consider two cases:
[b]1.[/b] $a\geqslant 2b$
We have $(a,0)\succ (a-b,b)$, because $a\geqslant a-b$ and $a=a-b+b$.
[b]2.[/b] $a\leqslant 2b$
We have $(a,0)\succ (b,a-b)$, because $a\geqslant b$ and $a=b+b-a$.
In both cases we have $(a,0)\succ (a-b,b)$, so from the Karamata inequality we get $a^{t}\geqslant (a-b)^{t}+b^{t}\iff a^{t}-b^{t}\geqslant (a-b)^{t}$.


vector
combinatorics proposed
combinatorics

Here is the problem :
Let $n$ be an integer, and $E$ a set of $n$ elements.
Let $A_i, i=1..n$  some subsets of $E$  such that
$i\neq j \Longrightarrow A_i \neq A_j$. Add the
condition $card( A_i \cap A_j) =1$  when $i\neq j$.
Prove that $E= \bigcup_{i=1}^{n} A_{i}$.

Fabrice

I think it can be shown that if $A_1,A_2,\ldots,A_m$ are distinct subsets of a set with $n$ elements s.t. the intersection of any two has cardinal $1$, then $m\le n$. Applied here, it gives the result.

Actually, let $A_i$ denote the characteristic vectors of the sets with the same name (over $\mathbb Q$). Assume we can find integers $a_i$ s.t. $\sum a_iA_i=0$. We partition the indices of the non-null $a_i$'s in two classes, according to whether $a_i$ is positive or negative. Denote the classes of indices $I,J$, respectively. We have $\sum_{i\in I}a_iA_i=\sum_{j\in J}(-a_j)A_j\ (*)$. Assume $\sum a_i>\sum(-a_j)\ (**)$ (if we have equality in $(**)$, we deal with it in a similar manner). Now take the scalar product between equality $(*)$ and $A_{i_0}$ for some $i_0\in I$. We will get a contradiction.

[quote="grobber"]Actually, let $A_i$ denote the characteristic vectors of the sets with the same name (over $\mathbb Q$).[/quote]

Could you please remind me what a "characteristic vector" is ?

Fabrice

In this case, it's a vector with $n$ entries, having $1$ as the $j$'th entry if $j$ belongs to $A_i$, and $0$ otherwise (so it's a vector with elements $0,1$). I mentioned that it's over $\mathbb Q$ because sometimes, such vectors are considered to have elements in $\mathbb Z_2$, for example, in questions involving parity. In this situation, we deal with real entries $0$ and $1$, not the elements of some other field.

You will find related problems and informations in the attached file :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=caruso&t=13941

Pierre.


Harvard
college
MIT
Princeton
Stanford
geometry
3D geometry

Take a look at it here:
http://ed.sjtu.edu.cn/rank/2004/top500list.htm

though is is strange: Ecole Polytechnique is 207th !!!

My university ranks first  ;)  :P  :lol:  :)  :D !!! However, I don't accept the idea of ranking universities; such ranks can sometimes be quite delusive (for instance, yes, Ecole Polytechnique certainly does not deserve to be placed at 207th place, and the top 3 in my opinion should have been made of Harvard, MIT, Princeton (in no particular order) --- although this is only my (subjective) opinion ... ). Anyway, I don't accept such statistics, and I am convinced that prestige should not be the decisive factor in our choice of college.

--Vesselin

My university ranks 47th.  :)  I think it always depends what the rankinks are based on and what priority which section got assigned to. Mostly they are focussed on math / nature sciences. I think you should rather make a ranking based on department. For instance Carnegie Mellon is ranked first in Software Engineering world wide. But I also like Stanford, Princeton etc.   :)

How is the highest Indian Institute of Technology only 404th?  On [i]60 minutes[/i], it was decribed as "Harvard, MIT and Princeton together."

[quote]How is the highest Indian Institute of Technology only 404th? On 60 minutes, it was decribed as "Harvard, MIT and Princeton together."[/quote]

Like all rankings and comparisons, both this ranking and the 60 minute description are very subjective.

And with this in mind, let me make my complaint :)

The University of Michigan is far far superior to U Wisconsin, no joke. Its also superior to all the UCs except possibly Berkeley.

[quote="theone853"]UC San Diego got ranked unusually high[/quote]

I agree, and that's not the only one. Any school ranked higher on the world list than any well-accepted US list is ranked "unusually high".

The fact that no univerisity in China (or for that matter the rest of Asia except Japan) is ranked in the top 100 is quite unusual.

The fact that the US has 17 of the top 20...*sigh*...dont get me started. Dont believe what these people tell you.

[quote]Tetrahedron, do you want to go to U of M?[/quote]

It wouldn't be my first choice, but ultimately, I wouldn't mind going. Being in state, UM isn't very competitive.

[quote="MithsApprentice"]

I agree, and that's not the only one. Any school ranked higher on the world list than any well-accepted US list is ranked "unusually high".

The fact that no univerisity in China (or for that matter the rest of Asia except Japan) is ranked in the top 100 is quite unusual.

[/quote]

No doubt that several of the American TOP universities deserved their rank and are really amazing.

But I also completely agree about Asia. Asia is completely underestimated. There are lots of TOP universities in China. And when I see that Saint Petersburg is ranked 302th-403th and Moscow just 66th I just can laugh. They have lots of top students, including lots of IMO gold medal winners and perfect scorers !!!

I also feel that universities as Korea Advanced Inst. of Science & Tech (302-403), National University of Singapore (101-152), Pohang University of Science & Tech., Korea (302-403) are ranked ridiculously low.

In general I feel that the [url=http://ed.sjtu.edu.cn/rank/2004/Methodology.htm]"Ranking Methodology"[/url]  cannot be really objective. It includes citation indices that are rather spread in the Western hemisphere such as Science Citation Index-expanded and Social Science Citation Index.

The same goes for articles published in Nature and Science. The people who gets awarded Nobel Prizes and Fields Medals also heavily depends on who decides about it. I also want to point out that just few of the used [url=http://ed.sjtu.edu.cn/rank/2004/Resources.htm]resources[/url] are Asian ones.

[quote="MithsApprentice"]

The fact that no univerisity in China (or for that matter the rest of Asia except Japan) is ranked in the top 100 is quite unusual.
[/quote]

Especially since some Chinese university is doing this...

I also think it's weird that they would rank themselves 404-502.

Well, the rankings of educational facilities are totally different than those of the quality of students that attend. My friend already started graduate school in physics at Yale University. He got decent scores during undergraduate and did well on the GRE (well, not of IPhO caliber but still...) Anyways, thinking that he was well prepared, to his surprise, a few of his classmates from Taiwan and China had backgrounds exceeding his own! They told him that Chinese Universities had tougher programs than those offered in US. Currently, he is still studying physics, but he was surprised by the fact that Chinese schools are a lot harder. Same with European schools I believe. BTW, my friend and I graduated from Rutgers (we went to same school).

I think many of you are missing the point -- this is a ranking of [i]Universities[/i], not undergraduate colleges.  That means that it's not primarily about you (or me) as undergraduates -- no matter how brilliant I may think myself, I and my friends are not the reason that Harvard is #1 on that list.  In fact, you'll note that not a single one of the criteria they use would be affected by any change in the undergraduate composition of the school.  China does not have particularly well-established research universities (although they're working on it), so it's not incredibly surprising that they don't have any universities ranked highly.  IIT isn't a "well-rounded" university -- no matter what quality engineers it produces, it simply isn't a major center of reasearch in a large number of fields and (without basis) I doubt it sends many of its students into academia which forfeits a whole bunch of points immediately.  (Actually, I don't even know what IIT is like as a research university.)

By the way, Vesselin, what year at Harvard are you?

[quote="JBL"]I think many of you are missing the point -- this is a ranking of [i]Universities[/i], not undergraduate colleges. [/quote]Exactly. That is why no Romanian University for example appears in the first 500 :(

The ranking procedure is ridiculous. Of course it is very difficult for a school not in the US to be ranked near the top.

The ranking is ridiculous, of course, but I'm happy because my school (well, beginning next year, I'm in high school now) is number 10 in the world.

Magnara, you got into U Chicago?  Very nice!

China's trying to catch up. but i think there should be some China's
Universities rank in the top 100.  :( as a matter of fact, the best ranks (202-203) really got me started.

[quote]

Y ?[/quote]

Hey I don't see (see the picture on the right) my  school in the list at all!  I was told that the "Guide Dog School"  is the best Guide Dog School in the world.

From newspaper without comments ...
[quote]LONDON, England - 03/02/05 - When it comes to ranking
institutions of higher learning worldwide, US
universities - particularly those in California -
dominate.

That's the word from a new list of the world's most
prestigious and famous universities created by The
Times of London annual Higher Education Supplement.

Led by Harvard University, fully seven of the top 10
schools are in the US, with three of those leading
institutions located in California.

The University of California - Berkeley ranked 2nd on
the international list, while the California Institute
of Technology (CalTech) in Pasadena and Stanford
University in Palo Alto came in at 4th and 7th,
respectively.

Oxford and Cambridge are the only British universities
to make the cut in the Top 10, although two other
schools - the London School of Economics and Imperial
College came in at 11th and 14th, respectively.

Rounding out the Top Ten are the Massachusetts
Institute of Technology (3rd); Yale University (8th);
Princeton University (9th); and ETH Zurich (10th).

The Times Higher Education World University Rankings
are based on the opinions of 1,300 academics in 88
countries, as well as the latest measures of research
excellence and teaching.

Of the top 200 universities, all are located in 29
countries. Leading the list are the US with 62, the
United Kingdom with 30, Germany with 17, and Australia
with 14.[/quote]

my current math teacher used to be a university professor, and he told us that these rankings can be very very misleading. It is all about name recognition. When groups compile these lists they just go to universtities and hand out lists to the professors of colleges and ask thhem to list them in order of how good they are. Well of course not every professor will spend lots of time researching each college and trying to make the results as accurate as possible. THey are busy people, they have research projects they havelectures. So many just look at the names and put them down based on recognition. E.g. "oh i have  a know a very intelligent prof at UM. Ill rank them 20."
of course some profs will spend adequate time in choosing, but most just write them down quickly and w/o much thought.

so i dont know how accurate these lists may be.... :?

The ranking of the Canadian Universities doesn't coincide with the Maclean's magazine rankings(not that it should). An entire issue of Maclean's is devoted to the university rankings every year--it's always a hot topic.

Maclean's ranks Best Overall:

                                              [list]1. University of Waterloo
                                              2. University of Toronto
                                              3. Queen's University
                                              4. McGill University[/list]

[quote]Carnegie Mellon is ranked first in Software Engineering world wide. [/quote]
orl, can you direct me to the source of this quote,please?


there was a problem on the 2005 CPSC, about how many possible moves are there for the first three moves in a game chess. i got the answer and but i was wondering that if someone could post some proof of it(hopefully nothing illegal would happen becasue the contest ended yesterday!)
[hide]my answer was 8904[/hide]

can someone moves this to the pre-olympiad section?
 ;)

Amirhtlusa,

     This was a pretty hard/intricate problem... the answer turns out to be 5362.  White's first move has 20 options (16 pawn moves and four knight moves).  Black's first move is not influenced by any of white's moves, so black has a similar 20 options.  Thus, there are 400 possible board positions after the second move (Black's first move).  For each of these positions, the number of possible moves for white must be analyzed.  Typically, white has around 20 options, but can occasionally have many more due to revealed queen and bishop possibilities.  Counting all the possibilities for white's third move should give a total of 8902.  However, this is not the number of unique board positions!  There are many positions that can be reached through two different sets of white moves (e.g. a white pawn advances, then a different one does), so these must be subtracted once each.  There are also four ways to move a knight and then return it to its original position, so three of these must be subtracted.  There are also two ways for each knight to end up four squares in front of its initial position.  After all of this is taken into account, there are only 5362 unique positions after three moves.

     By the way, if your school participated in the 2005 Collaborative Problem-Solving Contest, your coach should have detailed solutions to all problems which we e-mailed last Thursday.

yeah, we got the answers...
got to tell ya, i liked that problem....

>got to tell ya, i liked that problem....

     Thanks!  We love getting feedback on our contests, especially when it's positive.  :-)  This year's contest was extremely popular... I think teams are becoming better at working together and we're getting better at assembling sets of problems with universal appeal.  Hopefully next year will be even better!


The sum of two positive numbers multiplied by the first of the two is 84. The difference of the two multiplied by the second is 16. Find the largest and smallest possible sums for the two numbers.

Subtracting, x^2 + y^2 = 68. Letting one of the numbers be 0, the other is Sqrt(68) = 2 * Sqrt(17), which is the smallest sum. Letting x = y, they are both equal to Sqrt(34), so the largest is 2*Sqrt(34).

[color=cyan]Well, there is clearly a difference of opinion.  Fierytycoon, can you show that those represent the largest and smallest?  Also, have you considered all cases of the question?[/color]

Whoops, my bad. Ok, so x^2 + y^2 = (2*Sqrt(17))^2 is a circle of radius 2*Sqrt(17). So to minimize the sum...  :?:

I was relying on the title, which was "Number Theory #11." Number theory deals with properties of rational integers. sqrt(17) isn't a rational integer. Proof:

Assume, for purposes of contradiction, that a/b = sqrt(17) and (a,b)=1. Squaring and cross-multiplying, a^2 = b^2 * 17, so 17 divides a^2. 17 is a prime, so 17 divides a. Then let a = 17t. The equation becomes 243t^2 = 17b^2, so b^2 = 17t^2, and 17 divides b. But then 17 divides both a and b, which contradicts our hypothesis that (a,b)=1! Therefore sqrt(17) cannot be expressed as a rational number a/b.

A direct corollary is that Fierytycoon's answer is wrong.

Don't let the title fool you. The title need not have anything at all to do with the problem. If it doesn't say you're working in integers or rationals, you shouldn't do so.

In that case, his lower bound is correct but the upper bound is 10sqrt(2).

[color=cyan]A wonderful proof of the irrationality of the square root of 17.  Unfortunately, the question was mislabeled.  It's the result of making darkquantum break up my 15-question post all on his own.

If I titled a question "The answer is 17," would you just put down 17 as your answer?  Or might you actually do it?

So, anyhow, lets make it two questions:

1)  do it for integers.
2)  do it for reals.
[/color]

Affirmative.

We should have never trusted him alone with such a difficult, demanding task.


Let $d(n)$ be the number of divisors of $n$. Show that:

$\sum_{k=1}^{n+1}\lfloor \frac{n+1}{k}\rfloor-\sum_{k=1}^{n}\lfloor \frac{n}{k}\rfloor = d(n+1)$

Sorry, my bad - calculation error. Post deleted.

Do anyone can try solve this problem? I gave up.

Dunno if it's right:
$\lfloor \frac{n+1}{k} \rfloor = \lfloor \frac{n}{k} \rfloor$ when $n \not\equiv-1 \pmod k $. And, if $n \equiv -1 \pmod k$ then $\lfloor \frac{n+1}{k} \rfloor = \lfloor \frac{n}{k} \rfloor +1$. So, when $k \mid n+1$ we have that $\lfloor \frac{n+1}{k} \rfloor -\lfloor \frac{n}{k}\rfloor = 1 $. When $k \nmid n+1$ we have that $\lfloor \frac{n+1}{k} \rfloor -\lfloor \frac{n}{k}\rfloor = 0 $.
Thus, the sum is equal to the number of divisors of $n+1$.


Prof. Merryfield, could you please review the statement. The [b]7[/b] there seems to me to be out of place, should be (I guess) [b]y[/b]. By the way, that calenar is quite extrange, some of the problems are quite easy, and others are very trickly!!

[quote="Kent Merryfield"]¿Cuántos números de cuatro cifras se pueden formar con los dígitos 1, 2, 3, y 4, de manera que los dígitos se puedan repetir y ningún dígito tenga uno más grande a su derecha?[/quote]

Si he entendido bien el problema, la siguiente es mi solucion:

[hide="Mi Solucion"]
Bueno che, si no se pueden repetir digitos, pues entonces se tienen que usar todos una unica vez. Ademas, si ninguno puede tener digitos mas grandes a su derecha, pues el 4 tiene que ser el primero (de otra manera, el primero de los digitos tendria a 4, que es mas grande que el mismo, a su derecha), por el mismo criterio, el 3 tiene que estar en la segunda posicion, y el 2 en la tercera y el 1 en la cuarta, de modo que se tiene solo al numero [b]4321[/b], osea, la respuesta es 1.
[/hide]

Ciao

David - I should proofread very carefully, since I don't read Spanish. (Recognizing words isn't too hard, but that's not all there is to reading a language.) I've made the correction: "y", not "7."

However, I think your answer is wrong. Isn't the condition that the digits don't increase from left to right? That's not the same thing as decrease. The answer in the back of the calendar makes clear that that is the interpretation; 4422 would be allowed, among others.

Yes, the problems on this calendar vary widely; some are too easy or elementary for me to even bother posting; some are quite challenging.

Ok, es sencillo. Consideraremos el numero al extremo derecho. Si es 4, entonces el numero es puros 4 (porque no puede haber un numero menor a su izquierda). Si es 3, puede tener solo 3's y 4's a su izquierda, entonces hay a lo mas un "escalon" (primero viene un bloque de 3's y luego un bloque de 4's), y el luegar de este "escalon" se puede escoger de 4 maneras. Si el numero termina en 2 tenemos a lo mas 2 escalones (donde podrian estar en el mismo espacio), entonces se pueden escoger de ${5 \choose 3}$ maneras. Si termina en 1, tiene a lo mas 3 escalones que se escojen de ${6 \choose 3}$ maneras. Con esto, sumando, obtenemos
$1+4+{5 \choose 3}+{6\choose 3}=35$
Y este es el resultado deseado

EDIT: Este problema lo corregi, debido a un error en la manera de contar

Prof. Merryfield, sorry, I terribly misread. First I understood that the digits could not be repited, and it is all the oposite, and maybe some kind of "selective dyslexia", because I understood left instead of right!  :blush:  :oops:

Leonardo, Prof. Merryfield esta en lo correcto. Siguiendo tu idea de "escalones" me parece que lo resolvi, de hecho, tengo una solucion para el caso general, sin embargo no lo tengo como solucion cerrada, si no en terminos de una sumatoria. Tengo pereza de ponerme a simplificarla, la voy a proponer en la parte de Combi a ver quien la simplifica.

[hide="My Lazy Solution"]Siguiendo la idea de Leonardo: 56[/hide]

Ciao,

Hmmm, todavia no veo cual es mi error  :?

Uy, claro, que tonto. Ya vi...., la voy a editar

Podemos poner la solucion general. Supongase que se quiere el numero de secuencias crecientes de longitud $n$ cuyos elementos estan en el conjunto $\{1,2,\dots,n\}$. Bueno, despues del error que cometi por una cosa estupida (cuatro numeros tienen tres espacios en el medio, no cuatro, ay que tipo mas baboso soy a veces!!), ya tengo una solucion para este. Recuerden que dice ser crecientes, no estrictamente crecientes.

Ciao,

David

Para este, procedemos igual, la generalizacion es evidente. Vemos que si el numero termina en $k$, entonces tiene a lo mas $n-k$ escalones, que se pueden escoger de ${2n-k-1\choose n-1}$. Sumando sobre $k$ obtenemos
$\sum_{k=1}^{n}{2n-k-1\choose n-1}={2n-1\choose n}$

Y obtenemos lo que queriamos.

Con esto, ya no veo como extender mas este problema, al menos de que le agreguemos condiciones adicionales ;)


The sum of the recipricols of three distinct natural numbers is $\frac{31}{30}$. What is the least possible product of these three numbers?

30, the numbers are 2, 3, and 5

It just made sense to test them out because 2 x 3 x 5 = 30. Educated guess and check, if you will.

lol the best way isn't to guess haha.  you can do it a bit more systematically like this:

look carefully at 31/30.  it's a bit over 1.  we have 3 fractions.  can it be true that they are all less than or equal to 1/3?  no.  why not?  because if they were all 1/3, even THEN they would just sum to 1.  that means one of them has to be greater than 1/3, the only choices being 1/2 and 1.  obviously you can't have BOTH of those, so let's consider the 2 cases.

[b][u]case 1: you choose 1/2.[/u][/b]
then the remaining two numbers have to sum to 31/30 - 1/2 = 31/30 - 15/30 = 16/30 = 8/15.  now can both of those numbers be less than or equal to 1/4? no, because then at most their sum would be 1/2, which is less than 8/15.  so one of them HAS to be 1/3.  subtract 1/3 from 8/15 to get 1/5.

thus in the case that we choose 1/2, we have only the combination 1/2, 1/3, 1/5 for a total product of 30.

[u][b]case2: you choose 1/1 = 1.[/b][/u]
then the remaining 2 numbesr have to sum to 31/30 - 1 = 31/30 - 30/30 = 1/30.  note that whatever we decompose 1/30 into, the denominators will be greater and that our product will eventually exceed what it was in case 1.  so this track is done.

thus the best is case 1: the product is 30.

I was playing around with 1/1 for about a thousand minutes before I realized that it won't work with natural numbers.  :blush:

I solved it using what i will call logic but what could be considered "guess-and-check" the factors of 30 are $1$,  $2$, $3$, $5$, $6$, $15$, and $30$. We definitely can't use 1, and in order to get anywhere close to a sum of 31/30 you [u]have[/u] to use 1/2 and 1/3. and even if this logic is wrong...it worked pretty well  :D

yeah that's basically a good way of solving them for these quick contests where time is really critical, but I just wanted to show a more formal approach that would work for general fractions.

the thing, mathcounts tends to stress a lot of speed and what not.  as you go into high school and if you get exposed to contests like AIME, and more importantly USAMO etc, you realize that you get like an hour or more per problem.  in those cases you really have to understand your fundamentals and know how to prove and really set your answer in stone.  because then you can't just give an answer -- you need to show an explanation too and even hints of guesswork earn you 0 points (even if you got the answer right!)


algebra
polynomial
number theory
least common multiple
number theory proposed

Prove that for any integers $a_1,...,a_n$ the number ${\frac{LCM(a_1,...,a_n)}{a_1a_2...a_n}}\cdot\prod_{i<j}{(a_i-a_j)}$ is an integer divisible by $ 1! 2!...(n-2)!$. Moreover, there exists n-uplets for which the two numbers are equal, so the result is the best.

I think there is an not-so-old exercise from X oral examination which was asking for a proof that $\prod _{i<j} (a_i-a_j)$ is divisible by $1!2! \cdots (n-1)!$.
As far as I'm remembering the solution was quite boring. I think there is a more elegant way using Van der Monde determinants. Moubi is probably aware of it (it appeared in the RMS).

Pierre.

That X oral exercise is in fact the classical exercise I was reffering to. But it is much older than oral X. It appeared very long time ago in AMM and continues to appear in olympiads. It was posted and received three solutions: one from Myth, using multivariable polynomials, one from Valiowk, counting exponents and one from me, with determinants. I find it a very nice problem, not a boring one.

I was talking about the solution given in the RMS.

Pierre.

Ah, thus this problem appeared before in RMS? I saw it in the last issue only and there was no solution.

[b]Hint:[/b] Denote $L = \text{LCM} (a_1,\dots,a_n)$. Then the following determinant with integer entries
\[\left|
\begin{array}{cccc}
\frac{L}{a_1}  & \frac{L}{a_2} & \cdots & \frac{L}{a_n} \\
1 & 1 & \cdots & 1 \\
a_1-1 \choose 1 & a_2-1 \choose 1 & \cdots & a_n-1 \choose 1\\
\vdots & & & \vdots \\
a_1-1 \choose n-2 & a_2-1 \choose n-2 & \cdots & a_n-1 \choose n-2\\
\end{array}
\right|
\] equals (up to a sign)
\[\frac{\text{LCM}(a_1,\dots,a_n)}{a_1 a_2 \cdots a_n}\cdot
\frac{\prod_{i<j}{(a_i-a_j)}}{1! 2! \cdots (n-2)!}.\]
Alin.

Yup, this was also my solution. It follows exactly the solution for the classical problem. What about for a case of equality to show that the result is optimal? ;)

Choose $a_n=(n!)^2,a_i=(n!)^2+i,\forall i\in\{1,2,...,n-1\}$
Then $lcm (a_1,a_2,...,a_n)=n!na_1...a_{n-1}$ because the numbers $a_n,\frac {a_i} {i},i=1,2,...,n-1$ are pairwise coprime.So we easily get the desired equality.
I want to add that,this is really excellent problem .


if x/y=2/3, then find the values of (x+y)/y and (x+2y)/4y

[hide]$\frac{x+y}{y}=1+\frac{x}{y} = 1\frac{2}{3}$

$ \frac{x+2y}{4y}=\frac{1}{4}(\frac{x}{y})+\frac{2y}{4y} = \frac{1}{6}+\frac{1}{2}=\frac{2}{3}$

That's how to do it algebraically. You can assume x=2 and y=3 also. [/hide]

you are correct. for the second one, you could've also substituted 3x for 2y, get 4x/4y and then get 2/3. nice job!


(15+x)(20+x) < 900
x^2 + 35x - 600 < 0
then use the quadratic formula to find the positive root.

Is there anything that says that the deck can't be exactly 900 square feet?

[quote="archimedes1"](15+x)(30+x) < 900
x^2 + 35x - 600 < 0
then use the quadratic formula to find the positive root.[/quote]

Archimedes1, $(15+x)(30+x)=x^{2}+45x+450$. Did you mean $(15+x)(20+x)$?

I think the deck's area can be exactly 900 square feet. The problem just says the area can't be [i]greater[/i] than that.

Thanks for pointing that out.
"Not greater than" means "less than or equal to."
It's sort of like positive and nonnegative.

[quote="sharkman"]A homeowner wants to increase the size of a rectangular deck that now measures 15 feet by 20 feet but the local building code laws state that a deck cannot be > 900 square feet.  If the length and width are to be increased by the same amount, x, find, to the nearest tenth, the maximum number of feet by which the length of the deck may be legally increased.

I got 12 feet but the textbook tells me that the correct answer is 12.6 feet.

How do I get 12.6 feet as the right answer?[/quote]
[hide]We want to maximize the size of the deck, so $(x+15)(x+20)=900$.
\[x^{2}+35x+300=900 \]
\[x^{2}+35x-600=0 \]
\[x=\frac{-35\pm\sqrt{1225+2400}}{2}\]
\[\boxed{x\approx12.6}\][/hide]


Hi, I need help with some questions

Here it is:

Year 9 Maths teacher Mr Stan Deliver has his class practicing mental arithmetic. He gets each student to choose a set of consecutive numbers, none of which ends in zero. The student is to add these numbers and then divide the total by the last digit of each number. For each whole number answer a student earns a house point.
For example, Andy chooses 15, 16, 17 with a total of 48, and calculates 48 ÷ 5, 48 ÷ 6, 48 ÷ 7. Just one of these divisions given a whole number, so he wins one house point.
a) Show that if Andy chooses eight consecutive numbers, he wins at least one house point.
b) Belinda chooses five consecutive numbers and wins five house points.
What is the smallest total her five numbers can have? Explain.
c) Caroline chooses six consecutive numbers. Explain why she wins no more than two house points.

this one I've done a) and I found out that the 5 consecutive numbers for b) are 22,23,24,25,26 but I'm stuck with the explanation and I dunno how to do c)

For c) , all the possible sums are:
$21 = (1+2+3+4+5+6)$ , $27 = (2+3+4+5+6+7)$, $33 = (3+4+5+6+7+8)$, $39 = (4+5+6+7+8+9)$.

The factors of 21 are: 1,3,7,21. Only 1 and 3 work.
The factors of 27 are: 1,3,9,21. Only 3 works.
The factors of 33 are: 1,3,11,33. Only 3 works.
The factors of 39 are: 1,3,13,39. None work.

This is only by brute force. A better strategy can used by observing patterns.

Thank you very much. I'll think on an explanation for that. For b this is what I get:

Because we need 5 consecutive numbers so:

all the last digits possible: 1-5, 2-6, 3-7, 4-8, 5-9

sums are: 15, 20, 25, 30, 35
if I have those + 10 (11-15,...) the sums are: 65, 70, 75, 80, 85 respectively
if I have those + 20 (21-25,...) the sums are:115, 120, 125, 130, 135

so from here I get that whenever they + 10s their sum increases by 50 and I just did this randomly and I got 120 which are divisible to all last digits from 2-6 so I'm confident that 22,23,24,25,26 are the answer but I'm not sure if this is the right way to do it because I'm confused as well.

Is there a sort of pattern between the answers of b) ?.

Thanks, I know how to do it now. Pretty much I analyzed the structure of the sums of 1-5, 2-6, 3-7, 4-8, 5-9 and from here I got that only 2-6 and 4-8 are possible since other ends up with 5 as => not divisible by even numbers and I used trial and error which got me to 120 with 22,23,24,25,26


Find the solution set for $ (x\minus{}1)(x\minus{}4) \plus{} 1< 2(x\minus{}4)$.

irreal answers ( forgot the word in English)   are (7+3i)/2  and (7-3i)/2

 :maybe:

hello, the given inequality is equivalent with $ \minus{}x^2\plus{}7x\minus{}13>0$. This is not fulfilled for any real $ x$.
Sonnhard.

[quote="mathwizarddude"]Find the solution set for $ (x \minus{} 1)(x \minus{} 4) \plus{} 1 < 2(x \minus{} 4)$.[/quote]
$ (x \minus{} 1)(x \minus{} 4) \plus{} 1 < 2(x \minus{} 4)$
$ \minus{}x^2\plus{}7x\minus{}13>0$
$ \minus{}\left( x \minus{} \frac{7}{2} \right)^2 \minus{} \frac{3}{4}>0$

Thus we get no solutions at all (the inequality is impossible).

[quote="mathwizarddude"]Find the solution set for $ (x \minus{} 1)(x \minus{} 4) \plus{} 1 < 2(x \minus{} 4)$.[/quote]

The problem is very easy :D I think that!!!
$ (x \minus{} 1)(x \minus{} 4) \plus{} 1 < 2(x \minus{} 4) \leftrightarrow x^2  \minus{} 5x \plus{} 4 \plus{} 1 < 2x \minus{} 8$

$ \leftrightarrow x^2  \minus{} 7x \plus{} 13 < 0$

$ \Delta  \equal{} 49 \minus{} 4.13 < 0 \Rightarrow LEFT > 0$

So that problem isnot have solution! :D

Inequalities are not well-defined for complex numbers, unless you're doing something like taking the norm of both sides (in which case it is not a total ordering).

OK, let's define $ z > a, z \in \mathbb{C}, a \in \mathbb{R}$ to be $ |z| > a.$
Sorry about the original question!  :D

[quote="mathwizarddude"]You can have solutions with complex numbers.[/quote]

I think you should show us solve that problem in complex number after you post that problem!  :?:

It's a nice exercise to prove that there is no way to define the analogue of a "positive number" on the complex numbers; in other words, there is no subset $ \mathbb{C}^{\plus{}}$ of $ \mathbb{C}$ such that

- $ \forall z \in \mathbb{C} : z \equal{} 0, z \in \mathbb{C}^{\plus{}} \text{ or } \minus{}z \in \mathbb{C}^{\plus{}}$
- $ \mathbb{C}^{\plus{}}$ is closed under multiplication and addition.

(I hope I phrased that correctly.)  Since the inequality sign $ a < b$ over the reals is usually defined to mean $ b \minus{} a \in \mathbb{R}^{\plus{}}$, there is no way to talk about inequalities in $ \mathbb{C}$ without losing one or more of the familiar properties of an inequality.  None of the alternate formulations here are particularly meaningful.

[quote="t0rajir0u"](I hope I phrased that correctly.)[/quote]  Yes -- an equivalent phrasing is to say that there is no total order $ \preceq$ such that $ \mathbb{C}$ satisfies the [i]ordered field axioms[/i] $ a \preceq b \Longrightarrow a \plus{} c \preceq b \plus{} c$ and $ 0 \preceq a, 0\preceq b \Longrightarrow 0 \preceq ab$.  The multiplication part of either definition is key, since the method of proof is to show that $ 0 \preceq a^2$ for any $ a$ in an ordered field $ F$.


I will try to be as quick as possible and put the questions online
please check this post periodically untill the subjects appear

It would have been even more fitting to bump this thread last summer when the IMO was (again) in Japan. :)


Let n be an integer and p be a prime with p ≠3. Then if the equation n ≡x3+ y3mod p has a solution so also does the equation n≡  x3+ y3mod p^α  for every  α ≥ 1.

Proof. Let n ≡x3+ y3mod p. Assume that p not divide x. (If p | x and p | y then p | n, so we can use x = 1
and y =-1.)

Why assume p not divide x?
What happen if I assume p|x and p not divide y?
Can I use x = 2 and x = -2? :(

I assume you are reading this proof from some book or somewhere - this is only the very start of the proof. At the moment its just divided it into two cases - a) both x and y are divisible by p and b) at least one of x, y (without loss of generality, x) is not divisible by p, and then proven case a.

If you read further I guess it will prove case b too..

Just thought it was mildly interesting that this seems to be either Fermat's Little Theorem (I think it was this one...) or something closely related.  I don't understand what x, y, and the 3's are though.


probability
logarithms
LaTeX
probability and stats

[i]Preamble:[/i]

Suppose we have a random mapping  [i]f[/i]   from a set  [i]D[/i]  of size  [i]d[/i]  to a set  [i]R[/i]  of size [i]r[/i]. We also consider that [i]d > r[/i]  (there are more elements to be mapped in [i]D[/i]  than there are available values in [i]R[/i] ). This means that some elements from [i]D[/i] will be mapped to the same values in [i]R[/i]. We assume that the way [i]f[/i] maps elements is random, with each value in [i]R[/i] having the same probability of being "chosen".

[i]Statement of the problem in mathematical terms:[/i]

Let [i]h[/i] be a particular fixed value from [i]R[/i]. How large does set [i]D[/i] need to be in order to contain some element [i]e[/i] such that [i]f(e) = h[/i]?

[i]Statementof the problem in computational terms:[/i]

Suppose elements in set [i]D[/i] can be generated computationally using a brute-force approach. Initially, [i]D[/i] contains no elements and in each new step a new value is added to the set. Let [i]h[/i] be a particular fixed value from [i]R[/i]. What is the computational effort (in terms of time complexity) that we need to spend in order to generate a set that contains some element [i]e[/i] so that [i]f(e) = h[/i]?

Dilemma :)

I have managed to compute the probability that, in step [i]k[/i] we have one such element. The formula I've got is:
[i]1 - (1 - 1/r) ^ k = p[/i]

However, I am unable to reach a definite conclusion about the computational effort that's needed. Would you reckon it's simply
[i]ln(p - 1) / ln(1 - 1/r)[/i], which is effectively [i]O(ln(p))[/i]? (as the other term is constant)

If you can help clarify things for me, please do so. =]

That should be $ \ln (1 \minus{} p)$ for $ p$ near 1, which you can't replace with $ \ln p$.  And you should be aware that the constant $ \frac {1}{\ln (1 \minus{} 1/r)}$ is negative.  Otherwise, looks fine.

To learn about using LaTeX to make your formulas look nice on the forum, see [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


[quote="TTsphn"]Let p is a prime number
Find all $ (a,b,c)\in N$ satisfy
$ \Large\ a!b! \equal{}{p^{c}}c!$[/quote]
Do you mean find all $ (a,b,c)$ for which all $ p$ suffice that equation? If so, there is only one solution: $ (1,1,1)$.

If you mean find all $ (p,a,b,c)$, then there are probably many solutions, but I have only found one: $ (2,2,4,3)$.

Its obviously that we must find  all $ (p,a,b,c)$.The solutions are trivial..... :!:

[quote="TTsphn"]Let p is a prime number
Find all $ (a,b,c)\in N$ satisfy
$ \Large\ a!b! \equal{}{p^{c}}c!$[/quote]

Also $ (2,1,2,1)$ and $ (2,2,1,1)$...

Let $ p$ be a prime number.  Find all $ (a,b,c)\in\mathbb{N}$ satisfy
\[ a!b!\equal{}p^{c}c!.\]
[quote="Αρχιμήδης 6"]There are not no trivial solutions.[/quote]
First of all, what do you mean by not no trivial solutions? ("there are not no" = "there are"?)  The remark is not clear, and to me it seems that the problem is [b]not [/b]trivial.  I am not sure what you were thinking.  Please explain.
Roughly speaking, for "large" primes, it appears somewhat trivial.  But, for the "smaller" primes this does not appear as easy.

[quote="i_like_pie"]
Do you mean find all (a,b,c) for which all p suffice that equation? If so, there is only one solution: (1,1,1). [/quote]
Note that $ 1!1!\neq p^{1}1!$ for all primes $ p$.

Baz i mean that solutions are ''small'' and we can find no hard and fast...Can sb find a ''big'' solution of (p,a,b,c)?

[quote="TTsphn"]Find all solution $ (p,a,b,c)$
Let $ f(c) : N\minus{}\to N$
Exist $ k\in R$
$ \frac{f(c)}{c}< k$
Then the equation
$ a!b! \equal{} p^{f(c)}c!$ has finite solution.[/quote]

I don't understand this?  What does $ f(c)/c$ being bounded have to do with anything?  If you are trying to somewhat generalize the original problem of this post, then why doesn't someone prove that first.  It may not be too difficult, but explicitly stating all the solutions takes time.  Doing this rigorously is not easy.

If $ c\to\infty$ then $ \frac{f(c)}{c}<k$
For Example
$ f(c)\equal{}\frac{c}{2}$

[quote="TTsphn"]If $ c\to\infty$ then $ \frac{f(c)}{c}< k$
For Example $ f(c) \equal{}\frac{c}{2}$[/quote]
TTsphn,  my question is how does this relate to there being finitely many solutions to $ a!b! \equal{} p^{f(c)}c!$ in positive integers $ a,b,c,p$ ($ p$ prime)?

P.S.  What's up with the bad rating for these posts?

My english quite bad so I can't talk about it.
Suppose the equation has infinite  solution  then $ \frac{f(c)}{c}<k$
You can solve this equation when $ f(c)\equal{}\frac{c}{2}$


Let $ ABCD$ be a convex quadrilateral, $ I$, $ J$ be the midpoints of diagonals $ AC$ and $ BD$ respectively. Denote $ E \equal{} AJ \cap BI, F \equal{} CJ \cap DI$.
Let $ H, K$ be the midpoints of $ AB, CD$. Prove that $ EF \parallel{} HK$


real analysis
calculus
derivative
integration
limit
function
real analysis unsolved

a measurable $E\subset R$ is said to have density $d$ at $x$ if the $Lim_{h\rightarrow 0}\frac{m(E\cap [x-h,x+h])}{2h}=d$.define $\phi(E)$ the set at which $E$ has density $1$.prove that $m(E\Delta \phi(E))=0$.

This is an immediate corollary of the Lebesgue theorem on the differentiation of the integral: if $f$ is locally integrable, then

$\lim_{h\to0}\frac1{2h}\int_{x-h}^{x+h}f=f(x)$ almost everywhere.

Just apply this theorem to the characteristic function of $E.$

Do you want a proof of the differentiation of the integral theorem? Or are you trying to prove the density without quoting that theorem?

[quote="Kent Merryfield"]This is an immediate corollary of the Lebesgue theorem on the differention of the integral...[/quote]
where might I find the statement of that theorem? do you know if it is in bartle or royden?

(by the way, a proof that does not rely on the theorem that kent mentioned is in "Measure and Category" by Oxtoby page 17)

I have a 2nd edition of Royden: the key chapter is Chapter 5, Differentiation. The result is never quite stated in exactly that way, but the gist of it is in that chapter.

I don't have a copy of that Bartle, so I can't help you there. I know it's a skinny book, so I don't know what's included and what isn't.

In Wheeden and Zygmund, it's given in $\mathbb{R}^n$ generality as Thoerem 7.2 (p.98) and the proof and several corollaries (including the statement on points of density that started this topic) last for the next 10 pages. This proof proceeds by way of the Hardy-Littlewood maximal function, which is my preferred proof.

Somewhere, somehow, there has to be a "covering lemma" (Vitali, or some other) in the proof of this theorem.

I'll use Vitali's Theorem ([url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=vitali&t=23976]discussed on the forum[/url]), as Kent suggested. I think we should restate the problem first, since it seems a bit awkward :):

Let $E$ be a measurable subset of the real line with positive measure, and let $M$ be the set of points of $E$ where the density is $1$. Then $m(E\setminus M)=0$.

For $x\in E$ and $h>0$, define $\phi(x,h)=\frac{m(E\cap[x-h,x+h])}{2h}$. Let $A_t\ (0<t<1)$ be the set of points $x$ of $E$ which satisfy $\liminf_{h\to 0}\phi(x,h)<t$. It's easy to see that $A_t$ is measurable. If we fix an arbitrary $t\in(0,1)$ and prove that $m(A_t)=0$, we're done. We can also assume WLOG that all the measures involved are finite, i.e. $m(E)<\infty$. In fact, we can assume that $E$ is bounded.

For every $x\in A_t$, we can find a sequence $(h^x_n)_{n\ge 1}$ which decreases towards $0$ such that $\phi(x,h^x_n)<t,\ \forall n\ge 1$. Let $I^x_n=[x-h^x_n,x+h^x_n],\ \forall x\in A_t,\ \forall n\ge 1$. If we assume that $A_t$ has positive measure, we'll be able to find an interval $I$ such that $m(A_t\cap I)>tm(I)$. Now $\Delta'$, the subcollection of $\Delta=(I^x_n)_{x,n}$ consisting of those $I^x_n$ which are contained in $I$, Vitali-covers $A_t\cap I$. This means that we can find a disjoint sequence $I_1,I_2,\ldots\in \Delta'$ such that $m((A_t\cap I)\setminus\bigcup_{j\ge 1}I_j)=0$, according to Vitali's Theorem. However, we have $m(A_t\cap I_j)<tm(I_j)$ and the $I_j$'s are disjoint, so $tm(I)<m(A_t\cap I)<t\sum_jm(I_j)\le tm(I)$, contradiction.


geometry
geometric transformation
rotation
reflection
abstract algebra
group theory
superior algebra

let $ ABC$ be  equilateral triangle ,Consider the rotation $ r$  with center $ O$ and angle  $ \frac{2\pi}{3},A \to B ,B \to C ,C \to A$ and $ s$ be the reflection with respect to the line $ OA$,wich map $ A$ to $ A, B$ to $ C$ and $ C$ to $ A$ .prove that :
{$ id,r,r^2,s,rs,r^2s$}is a group .it is an abelian group ?

have a look [url=http://members.tripod.com/~dogschool/trianglegroup.html]here[/url] for example...


Many people have been complaining that you lost your badges, you lost your videos, etc. There's actually nothing you can do about that. When it happened to me, all I did was wait for a few days, and restart my computer.

There still might be bugs in the system. I think you're stuff will come back if you just wait for a while.

(I'm hope I'm not wrong; I said all that stuff confidently; if it is wrong then I will be very, very embarrassed :blush: )

Actually, when you lose your badges, it means you are answering all those questions wrong.  You then lose your badges.

I just remembered, when I lost everything, I waited a few days, restarted my computer, AND did a few problems. So basically the cure (not exactly, you won't get everything back), is to do a couple of problems CORRECTlY.

Restarting your computer will do nothing to help it, unless your browser isn't rendering images properly.


geometry
3D geometry
probability
pyramid
AMC

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

$ \textbf{(A)}\ \frac{1}{4} \qquad
\textbf{(B)}\ \frac{3}{8} \qquad
\textbf{(C)}\ \frac{4}{7} \qquad
\textbf{(D)}\ \frac{5}{7} \qquad
\textbf{(E)}\ \frac{3}{4}$

[hide]
There are $ \binom{8}{3}\equal{}56$ ways to choose the three vertices.  The plane contains points iff it does not coincide with a face of the cube.  For each face, there are $ 4$ ways to choose three points from it.  Thus the number of faces not containing interior points is $ 6(4)\equal{}24$.  So the desired probability is

$ \frac{56\minus{}24}{56}\equal{}\frac{4}{7}\equal{}C$.
[/hide]

There are (8)(7)(6)/3! = 56 triples of vertices of the cube. Any three points forming a plane with points inside the cube must have at least one diagonal on a side of the cube. Take the first case in which there is only one diagonal on a side of the cube. To maintain this, there are only two possible points for each diagonal on the opposite side of the cube. Since there are 2 diagonals per side of the cube and 6 sides, there are 12 diagonals. Therefore, there are (2)(12) = 24 possible triples with one diagonal. When there are two diagonals, one can infer that there must be a third when adjoined with the third vertice. Since each possible figure consisting of the three diagonals on the sides of the cube forms a pyramid centered around a single vertice, and there are 8 vertices, there must be 8 possible triples with 3 diagonals. Therefore (24+8)/56 = 4/7.

[hide = Solution]

There are $\binom{8}{3}$ possible combinations of vertices.

Also, there are two cases for successful combinations.

[b]Case 1:[/b] a triple in which the cross-section forms an equilateral triangle. There are obviously $8$ triples that satisfy the condition

[b]Case 2: [/b]a triple which forms a right triangle with an edge as one of its legs joined with an opposite vertex. For each edge, there are $2$ vertices that are valid. Hence, there are $2 \cdot 12 = 24$ valid triples.

Our probability is

$$P(A) = \frac{8+24}{\binom{8}{3}} = \frac{32(3)(2)}{8 \cdot 7 \cdot 6} = \frac{4}{7} \implies \boxed{\textbf{(C)}}$$

[/hide]


Source: 360 PMC
Solve the system of equations
$2\sin x + 3\cos y = 3$
$3\sin y + 2\cos x = 4$.

[quote="pkerichang"][hide]
Squaring both equations and adding together we have:
$4(\sin^2{x} + \cos^2{x}) + 9(\sin^2{y} + \cos^2{y}) + 12(\cos{x}\sin{y} + \cos{y}\sin{x}) = 25$, or $12\sin{(x+y)} = 12$, which implies that $x+y = 90$, and $\cos{y} = \sin{x}$ and $\cos{x} = \sin{y}$, plug and solve we have that $x$ and $y$ are the two non-right angles of a 3-4-5 triangle, with $y > x$.
[/hide][/quote]

I see how $x+y=90$, because $\sin{x+y}=1$, but shere did you get $\cos{y} = \sin{x}$ and $\cos{x} = \sin{y}$?

Notice that since $x+y=90$, then $y=90-x$. Also, $\cos{x}=\sin(90-x)$, or in other words $\cos{x}=\sin{y}$


Two circles $ k_1$ and $ k_2$ intersect each other at point $ A$ and $ B$. The line tangent to $ k_1$ at $ A$ intersects $ k_2$ at point $ C$ ,and the line tangent to $ k_2$ at $ B$ intersects $ k_1$ at a point $ D$. Prove that $ \frac{BC}{AD}\equal{}\frac{AC^2}{BD^2}$

$ \angle ADB\equal{}\angle BAC$ and $ \angle DBA\equal{}\angle ACB$ $ \Rightarrow \triangle ADB \sim \triangle BAC$

$ \Rightarrow \frac {AD} {BA}\equal{}\frac {DB} {AC}\equal{}\frac {AB} {BC} \Rightarrow \frac {AD} {BA} \cdot \frac {AB} {BC}\equal{}\frac {DB} {AC} \cdot \frac {DB} {AC} \Rightarrow \frac {BC} {AD}\equal{}\frac {AC^2} {BD^2}$


geometry
analytic geometry
Euler
geometric transformation
reflection
projective geometry
perpendicular bisector

let $ ABCD$ be a convex quadrilateral such that $ <DAB\equal{}<ABC\equal{}<BCD$ .let $ H$ and $ O$ be the orthocenter and cicumcenter of $ ABC$.prove that $ D,O,H$ collinear

We can use barycentric coordinates WRT ABC, though I believe there must be an easier approach. Using Conway formula for $ \theta \equal{} \alpha\minus{}\beta$  and $ \phi\equal{}\alpha\minus{}\gamma,$ we compute the coordinates of $ D$ WRT $\triangle ABC.$ Then we verify that $D$ satisfies the equation of the Euler line $e$ of $\triangle ABC,$ namely

$e \equiv S_A(S_B \minus{} S_C)x \plus{} S_B(S_C \minus{} S_A)y \plus{} S_C(S_A \minus{} S_B)z \equal{} 0$

What is conway's formula? Can you explan it to me, please?

The Conway formula gives the barycentric coordinates of a point $P$ defined by the [i]balanced angles[/i] $\angle PBC=\theta$ and $\angle PCB=\varphi.$ Keeping in mind the notation $S_\theta=S \cot \theta$

\[ P \equiv (-a^2:S_C+S_{\varphi}:S_B+S_\theta) \]

Let $ K, M$ be midpoints of $ BC, AB$ and $ G \equiv AK \cap CM$ centroid of $ \triangle ABC$ and $ OGH$ its Euler line. $ CD$ is reflection of $ AB$ in the perpendicular bisector $ OK$ of $ BC.$ Similarly, $ AD$ is a reflection of $ BC$  is the perpendicular bisector $ OM$ of $ AB.$ $ AB, CD$ intersect at $ X \in OK$ and $ AD, BC$ intersect at $ Y \in OM.$ By Pappus theorem for the lines $ XAM, YCK,$ the intersections $ O \equiv XK \cap YM, G \equiv AK \cap CM, D \equiv AY \cap CX$ are collinear.


We are now accepting enrollment for our AMC/AIME Problem Series course which begins on January 4, 2004.  The course will be on Sunday evenings from 7:30-9:00PM EST.  The first 6 weeks will cover challenging problems from the AMC-10 and AMC-12, and the last 6 weeks will focus on the harder AIME problems.  The instructor for the class is Mathew Crawford, who scored a perfect score on both of these exams when he was a student in the mid 1990s.

If you have any questions about the course, please post them here.

Will this be a good course for last minute prepare for the exams? My goal is to make it to USAMO this year and I've been doing my own work for it but I'd like to know if it would be a good idea to sign up for this course to help.  I was thinking of doing the Algebraic Problem Solving Course but this sounds very enticing.  Just wondering...

Both the Algebraic Problem Solving and the AMC/AIME class will help with qualifying for the USAMO.  Most of the Algebra class is AMC/AIME level.  The AMC/AIME class is, of course, all AMC/AIME level, and is designed specifically for a student who hopes to make the USAMO.  Given the timing of the classes, you could well take both - the classes only overlap for a few weeks.

Yeah, I have time to think about it.  I definitely want to do the AMC but a class on Algebraic Problem Solving would be good for me I think.  I might do both.

This is an emergency!! My parents are not home right now and won't be back until 6:00 PM. I don't have any amount of money to enroll for the class. The post office closes at 5 and will be closed tomorrow. How will I ever make my payment for AIME problem solving course to you on time? What should I do?

It would be a good idea.... but my dad wouldn't let me use credit card online. Well, I guess I just have to convince him then.

I was unable to send the payments to AIME/AMC problem solving class and the deadline was yesterday. I wonder if i'm allowed to mail my payment tomorrow when the post office is open and start the class a lesson late. Help needed.

Sorry I didn't answer your earlier posts - I was on the road.  I have enrolled you in the class; you can mail the entry fee by printing out the entry form [url=http://www.artofproblemsolving.com/Classes/AoPSClasses.pdf]here[/url] and filling out the requisite information.  Please let me know if you have any more questions, and enjoy the course!

I have a question abou the payment for the class. I know the class has already started, but as you can see, I 'm paying the tuition late. Since I currently don't have pdf installed (and for some reasons it can't be downloaded), I can't get the information about class ID and the amount of tuition. Can anyone please supply the information?

Problem solved. Please disregard the last post.


[quote="balan razvan"][color=darkred]Let $ ABCD$ be a circumscriptible quadrilateral.Prove that the inscribed circles of the triangles $ ABC$ and $ ACD$ are tangent.[/color][/quote]Nice problem!
[u][b]Solution:[/b][/u]
Here, the Incircles either meet at one point (i.e. tangent) or they never meet.
But they meet, $ AC$ at some points. Let, the incircles of $ \triangle ABC$ and $ \triangle ACD$ meets at $ X$ and $ X'$ respectively.
Let,$ AB = a,BC = b,CD = c,DA = d,CA = x$
Now, we use directed length, and get that, $ |XA| = \frac {x + a - b}{2}$ and $ |X'A| = \frac {x + d - c}{2}$
So, we need to prove that, $ x + a - b = x + d - c \Leftrightarrow a - b = d - c$
As, $ ABCD$ is an circumscribed one,
Here,$ a + c = b + d\Leftrightarrow a - b = d - c$
So, $ XA = X'A$ and $ X \equiv X'$


didn't want to make another topic so:

are there prizes for achieving certain cumulative scores?

and if so, what?

[quote="avshockeyw"]the rounds tend to get harder and harder, right?[/quote]

The scores seem to show that, but by my looking at past years problems, it's very slight.


If $a_{1}, a_{2},..., a_{n}$ are $n$ distinct odd natural numbers not divisible by any prime greater than 5, show that
$\frac1{a_{1}}+\frac1{a_{2}}+\frac1{a_{3}}+...+\frac1{a_{n}}< 2$.

A similar problem has been posted, but I couldn't find it. The idea is the following
[hide]

let $S$ be our subset of the inverses of odd natural numbers. Now consider

$U=\left\{ \left( 1+\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{n}}+... \right) \left( 1+\frac{1}{5}+\frac{1}{5^{2}}+...+\frac{1}{5^{n}}+... \right) \right \}$

Clearly, $S \supset U$. But the sum of the element of $U$ is equal to $\frac{ \left( \frac{1}{3}\right)^{n+1}-1}{\frac{1}{3}-1}\cdot \frac{ \left( \frac{1}{5}\right)^{n+1}-1}{\frac{1}{5}-1}$

but when $n \rightarrow+\infty$, the sum of the elements of $U \rightarrow \frac{15}{8}<2$

sorry for my english[/hide]


function
induction
calculus
calculus computations

Build the Wronskian. It's essentially a Vandermonde determinant.

[quote="mayhem"]Show that, functions $ e^{k_1t},e^{k_2t},e^{k_3t},...,e^{k_nt}$ are linerly independent. for $ i \neq j , k_i \neq k_j$[/quote]

Suppose $ \sum_{m\equal{}1}^{n} c_me^{k_mt}\equal{}0$ for t in some interval. Then $ e^{k_1t}(c_1\plus{}\sum_{m\equal{}2}^{n} c_me^{(k_m\minus{}k_1)t}) \equal{} 0$ and therefore the term in parentheses equals $ 0$. So differentiate that term and use the induction hypothesis (which I never stated, but you get the idea).


vector
linear algebra
matrix
superior algebra
superior algebra solved

Let R be a ring of 2 x 2 matrices over reals; suppose that I is an ideal of R. Show that I = (0) or I = R.
Please answer the whole question...Thanks.

I move it to the college playground forum.
Please be carefull when posting.

Pierre.

It is not so hard I think. I will write the matrices in the vector form $(a,b,c,d)$.

Step 1: Let $(a,b,c,d)$ be a matrix of $I$. Then we have $(x,0,0,0)(a,b,c,d)(1,0,0,0)=(xa,0,0,0)\in I$ for all real $x$.

Step 2: We consider again a matrix $a,b,c,d$ in $I$. Similiar to step 1 we have $(0,0,0,x)(a,b,c,d)(0,0,0,1)=(0,0,0,xd)\in I$ for all real $x$.

Step 3: For each $(a,b,c,d)\in I$ we have $(0,1,1,0)(a,b,c,d)=(c,d,a,b)\in I$.

Step 4: For each $(a,b,c,d)\in I$ we have $(a,b,c,d)(0,1,1,0)=(b,a,d,c)\in I$.

Step 5: From the previous steps we conclude that for each position in a matrix in $I$ we can have all the real numbers(case 1) or only $0$ (case 2). From steps 3 and 4 we get that we have case 1 for all positions in the matrix or case 2 for all positions in the matrix. Thus, $I$ has the form $(\Bbb R,\Bbb R,\Bbb R,\Bbb R)$ or $(0,0,0,0)$. So $I=\{0\}$ or $I=R$.

Misha


Which chapters or topics covered in AoPS V2 and Intermediate Counting and Probability are most relevant to the AMC 10?

Going through chapter 19 (Probability) in AoPS volume 2 and then answering the problems at the end of the chapter will further your ability to answer probability problems...or at least it did for me.

Is the intermediate counting book mainly too in depth and high level for AMC 10 problems?

yes. they designed it to be from early aime to late aime level. master aops vol 1. aops v2 has stuff that's pretty much all beyond amc 10 and can be done without, it's just nice to know some extra tricks. :D
honestly, i think it would be in your best interest to work through the amc 10 tests and then see where you need to further your knowledge.
other than that, if you think you are really lacking in information, i would probably get to work. :wink:

I would think Inter. C+P is a lot harder than the AMC 10, except maybe for the very simplest problems.


A pair of positive integers $ (x, y)$ satisfies the equation $ 31x \plus{}29y \equal{} 1125$. What is $ x\plus{}y$?

Take $ \pmod{29}$ to get $ 2x\equiv23\equiv\minus{}6\pmod{29}\implies x\equiv\minus{}3\equiv26\pmod{29}$. It is clear that the only solution satisfying all conditions is $ x\equal{}26\implies y\equal{}11\implies x\plus{}y\equal{}\boxed{37}$.

A quick way to do this without finding x and y is to set up an inequality.

When y=0, x=36.3..... meaning the sum is 36.3.....

When x=0, y=38.8..... meaning the sum is 38.8.....

The sum will increase as x decreases, so as x and y are integers, it is 37 or 38.

31 and 29 are both odd, so to get an odd number, the total number of x's and y's must also be odd, making our answer 37.


problem 1
let a belongs to R* , a^3 + 1/ a^3 =18
find a^4 +1/a^4


inequalities
trigonometry
LaTeX
inequalities solved

Let ABC be a triangle.
Prove that
[tex] \frac 94 \geq \sin ^2  A+\sin ^2B+\sin ^2 C+ \frac 13 \sum \left(\frac {m_a-m_b}c \right) 2 . [/tex]

[color=red][Admin Edit: Made LaTeX compliant][/color]

Are you sure, MinhThang?
  Take a=3,b=4,c=5. :D

i have tried a lot but unsuccessful
Can you post the solution Minh Thang?


[quote="GameBot"]How many degrees are in the measure of the smaller angle formed by the hour and minute hands of a clock when the time is 7 p.m.?[/quote]

At 7 p.m., the hour hand is exactly on the 7 and the minute hand is exactly on the 12. There are 5 "intervals" of 5 minutes each between the 7 and 12 on a clock face, and 12 "intervals" of 5 minutes each in the whole clock face, so $ \frac{5 \cdot 5}{5 \cdot 12} \equal{} \frac{5}{12}$ of the entire clock face is the region between the 7 and 12. The whole clock face measures $ 360^\circ$, since it's a circle, and 5/12 of that is the angle we're looking for. Thus our answer is $ \frac{5}{12} \times 360 \equal{} \boxed{150^\circ}$.

sorry if this is the same
what i did was this

a full circle is 360 degrees

there are 12 sections on the clock so each angle is worth

$\frac{360}{12}$ $=$ 30

and between 7 and the 12 there are  5 hours so we get the equation

5*30=150

so the answer is $\boxed{150}$

[size=50]sorry if this will be considred SPAM[/size]


inequalities
function
logarithms
inequalities proposed

Let $a,b,c,d$ be positive numbers such that $abc=d^{3}$. Prove that
$(a+b-d)^{2}+(b+c-d)^{2}+(c+a-d)^{2}\ge$
$\ge (a-b+d)(c-b+d)+(b-c+d)(a-c+d)+(b-a+d)(c-a+d)$

The inequality is equivalent to
\[\sum a^{2}+3\sum ab-4d\sum a \ge 0 \]
Setting $x=\frac{a}{d},y=\frac{b}{d},z=\frac{c}{d}$ then $xyz=1$ and we have to prove
\[\sum \left( x^{2}+\frac{3}{x}-4x \right) \ge 0 \]
Consider the function $f(x)=x^{2}+\frac{3}{x}-4x+5\ln x$, we have
\[f^{\prime}(x)=\frac{(x-1)(x^{2}-2x+3)}{x^{2}}\]
\[f^{\prime}(x)=0\Leftrightarrow x=1 \]
Then we can easily check that $f(x) \ge 0$ $\forall x>0$, hence
\[x^{2}+\frac{3}{x}-4x \ge-5\ln x \qquad \forall x>0 \]
Therefore
\[\sum \left( x^{2}+\frac{3}{x}-4x \right) \ge-5\sum \ln x=0 \]
We are done. :)

It's not nice.
This is my solution
With same denote as toanhocmuonmau, we have to prove that
$x^{2}+y^{2}+z^{2}+3xy+3yz+3zx\ge 4(x+y+z)$
Or
$(x+y+z)^{2}+xy+yz+zx\ge 4(x+y+z)$
But
$(x+y+z)^{2}+xy+yz+zx\ge (x+y+z)^{2}+3 \ge 4(x+y+z)$
or
$(x+y+z-1)(x+y+z-3)\ge 0$
Which is straightforward from Am-Gm inequality.


Suppose $P(x)$ is a polynomial such that $P(1)=1$ and
\[\frac{P(2x)}{P(x+1)}= 8-\frac{56}{x+7}\]
for all real $x$ for which both sides are defined. Find $P(-1)$.

I got that $P(x)$ must have infinitely many solutions...???
[hide]
Cross-multiply out to get
\[(x+7)P(2x) = 8xP(x+1) \]
Since $x=0$ is a root of the RHS, it must be a root of the LHS. Since it is not a root of $x+7$, it must be a root of $P(2x)$. Thus, $P(0)=0$ and we replace $P(x) = xQ(x)$.
\[(x+7)(2x)Q(2x) = 8x(x+1)Q(x+1) \]
Similarly, $-2$ is a root of $Q(x)$, so $Q(x)=(x+2)R(x)$. Then
\[(x+7)(2x)(2x+4)R(2x) = 8x(x+1)(x+3)R(x+1) \]
Once again, $-6$ is a root of $R(x)$, so $R(x) =(x+6)S(x)$. Then
\[(x+7)(2x)(2x+4)(2x+12)S(2x) = 8x(x+1)(x+3)(x+7)S(x+1) \]
Now $-14$ is a root of $S(x)$, etc. So it looks like we can construct an infinite chain of roots... :huh: [/hide]

Paladin pointed out a mistake, PLUS you can upper bound the degree of the polynomial.

[quote="paladin8"][quote="Phelpedo"]Similarly, $-2$ is a root of $Q(x)$[/quote]

Eh? How did you get this?[/quote]

-2 is a root because if we plug $x=-1$ into
\[(x+7)(2x)Q(2x) = 8x(x+1)Q(x+1) \]
We get $(-2+7)(-4)Q(-2) = 0$.

But actually the problem was sorted out.


hi again!

also, I want to create tables with a specific width of columns.  I used

\begin{tabular}{p{10cm} p{5cm}}
...
\end{tabular}

But I want to have to second column right aligned.

thank

ledurt

Don't know how to do this without using the [b]array[/b] package.

If you use the package, then you can do something like
\begin{tabular}{>{\raggedright\hspace{0pt}\arraybackslash}p{10cm}%
>{\raggedleft\hspace{0pt}\arraybackslash}p{5cm}}
... text here ...
\end{tabular}

I stole this example from the [i]LaTeX Companion[/i], page 247.  I have no idea why you need the \hspace{0pt}, but it's in the example.

thanks, it worked perfectly!

in return, I did some test on thoses command and I have the same result with:

\begin{tabular}{|p{10cm}|%
>{\raggedleft\arraybackslash}p{5cm}|}
...
\end{tabular}

the command \arraybackslash seems to tell to array that this is the last column of the table because without it, I get errors of :

"Extra aligment tab"...bla bla, the usual when u mess up the & in a tabular

is the > is use to introduce a command in the package array?

thanks again

ledurt


[img]http://img69.imageshack.us/img69/1426/problemhs.jpg[/img]

The answer is: NO!

but how to show that by the following counterexample?
thanks~

counterexample:
$ f_{n}(x)\equal{}\sqrt{n}$ , $ 0\leq x\leq\frac{1}{n}$
$ f_{n}(x)\equal{}0,\frac{1}{n}< x\leq1$

$ n\geq1$

For $ m>n$, $ \|f_n\minus{}f_m\|\equal{}\sqrt{1\plus{}\frac{n}{m}\minus{}2\sqrt{\frac{n}{m}}}$. So if you fix $ n$ and let $ m\to \infty$, you get $ \|f_n\minus{}f_m\| \to 1$. So this sequence has no convergent subsequence.