geometry
ratio
3D geometry
tetrahedron
analytic geometry
graphing lines
slope

*(Pictures attached below)
1. A circle is inscribed in a quarter-circle. Express in simplest radical form the radius of $ r$ of the circle in thers of the radius $ R$ of the quarter circle.
[hide="Answer"]$ r \equal{} (\sqrt {2} \minus{} 1)R$, or $ r \equal{} R\sqrt {2} \minus{} R$  [/hide]
2. In the diagram, circle $ Q$ is congruent to circle $ W$, and both are tangent to circle $ O$ and to each other. Circle $ S$ and circle $ T$ are congruent and are tangent to circle $ O$, to circle $ Q$ and to circle $ W$. Find the ratio of the area of smallest circle to the largest circle.
[hide="Answer"]$ \frac19$[/hide]
3. The piece below, with all indicated lengths of five units, can be folded to form a regular tetrahedron. What is the volume of the tetrahedron? Express your answer as a fraction in simplest radical form.
[hide="Answer"]$ \frac {125\sqrt {2}}{12}$[/hide]

[hide]for number one, if you draw an accurate picture with the smaller circle tangent to two radii of the larger circle, and also tangent to midway of the arc of the quarter circle, you see that R(radius of big circle)=2r(radius of little circle) + that tiny bit left. well, how can we get that "bit" into a mathematical expression? well, draw a square with 1 vertex at the center of circle r, and another vertex at the center of circle R. R=r+the diagonal of this square. the daig.=r*sqrt2 so R =r(1+sqrt2) so R/(1+sqrt2)=r and just simplify that... ill add more solutions later....[/hide]

for number 3, the volume is base*height/3, so find the area of the base(an equilateral triangle) and then find the height by playing around with the pythagorean theorem, and then multiply and divide by 3.

For number 1

R=r+r*sqrt2 b/c if u draw a line from center of circ w/ radius r straight upward it makes a square
the diagonal has 2 be r*sqrt2
the left over is r when u draw a line from the center going left and down w/ a slope of 1
so R=r+r*sqrt2
therefore r=R*sqrt2-R

ok for the one with 5 circles, the radius of the largest circle is obviously 2 times the radius of the *medium* sized circles.
then make a triangle with vertices as the centers of the 2 medium circles and 1 small circle.
then draw the altitude to the side connecting the centers of the medium circles and u get 2 congruent triangles
now use pythagorean theorem and get an equation


Let $ p$ be a prime number with $ p > 5$. Consider the set $ X \equal{} \left\{p \minus{} n^2 \mid n\in \mathbb{N} ,\ n^2 < p\right\}$.
Prove that the set $ X$ has two distinct elements $ x$ and $ y$ such that $ 1<x<y$ and $ x\mid y$.


[quote="encyclopedia"]Let $ x,y,z > 0$ prove that
\[ (xy \plus{} yz \plus{} zx)[\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2}]\le\frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \minus{} \frac {3}{4}
\]
Actually I don't know solution but I know it is true, I wait for nice solution without computer and tons calculations.[/quote]
The inequality is equivalent to
\[ \sum \frac {x}{y \plus{} z} \plus{} \sum \frac {yz}{(y \plus{} z)^2} \le \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \minus{} \frac {3}{4}
\]
It is easy to see that
\[ \sum \frac {yz}{(y \plus{} z)^2} \le \frac {3}{4}
\]
It suffices to prove that
\[ \sum \frac {x}{y \plus{} z} \plus{} \frac {3}{2} \le \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx}
\]

\[ \Leftrightarrow \sum x^2 \plus{} xyz\sum \frac {1}{y \plus{} z} \plus{} \frac {3}{2} \sum xy \le \sum x^2 \plus{} 2\sum xy
\]

\[ \Leftrightarrow xyz\sum \frac{1}{y \plus{} z} \le \frac {1}{2}\sum xy
\]
which is true because
\[ xyz\sum \frac {1}{y \plus{} z} \le \frac {1}{4}xyz\sum \left( \frac {1}{y} \plus{} \frac {1}{z}\right) \equal{} \frac {1}{2} \sum xy
\]
We have done.
:)

[quote="can_hang2007"][quote="encyclopedia"]Let $ x,y,z > 0$ prove that
\[ (xy \plus{} yz \plus{} zx)[\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2}]\le\frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \minus{} \frac {3}{4}
\]
Actually I don't know solution but I know it is true, I wait for nice solution without computer and tons calculations.[/quote]
The inequality is equivalent to
\[ \sum \frac {x}{y \plus{} z} \plus{} \sum \frac {yz}{(y \plus{} z)^2} \le \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \minus{} \frac {3}{4}
\]
It is easy to see that
\[ \sum \frac {yz}{(y \plus{} z)^2} \le \frac {3}{4}
\]
It suffices to prove that
\[ \sum \frac {x}{y \plus{} z} \plus{} \frac {3}{2} \le \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx}
\]

\[ \Leftrightarrow \sum x^2 \plus{} xyz\sum \frac {1}{y \plus{} z} \plus{} \frac {3}{2} \sum xy \le \sum x^2 \plus{} 2\sum xy
\]

\[ \Leftrightarrow xyz\sum \frac {1}{y \plus{} z} \le \frac {1}{2}\sum xy
\]
which is true because
\[ xyz\sum \frac {1}{y \plus{} z} \le \frac {1}{4}xyz\sum \left( \frac {1}{y} \plus{} \frac {1}{z}\right) \equal{} \frac {1}{2} \sum xy
\]
We have done.
:)[/quote]

Hello can_hang2007, that's very nice solution thanks  :)


function
geometry
complex analysis
complex numbers
complex analysis unsolved

hello

I have only had an introductory course in the amazing world of complex analysis
but right now I am studying a course of partial differential equations, including laplace equation

That is how I came up with this seemingly basic question, when is a function $u: \Omega\subset \mathbb{R}^2\rightarrow \mathbb{R}$ ,$\Omega$ open, the real part of an analytic function $f: \Omega\subset \mathbb{R}^2\rightarrow \mathbb{C}$
I would say if and only if it is continuously differentiable twice in $\Omega$ and has a laplacian zero everywhere in that area

[quote="fredbel6"]
I would say if and only if it is continuously differentiable twice in $\Omega$ and has a laplacian zero everywhere in that area[/quote]
That is true if $\Omega$ is simply connected. For more complicated regions, like an annulus $r<|z|<R$ you need some extra assumptions (in the mentioned annulus, for example, the function $\log|z|$ is harmonic but is not a real part of a function analytic in the entire annulus).

oo of course
btw is there a nice definition then of simple connected, i mean, no holes is a bit weird in my opinion

"Every 'loop' can be continously deformed till it get's a single point" is the normal definition for that. Is that better¿


Every road in a particular country is one-way. Every pair of cities is connected by exactly one direct route. Prove by mathematical induction that there exists a city which can be reached from every other city either directly or via at most one other city.

[hide]
The base case is obvious.  Say that the statement is true for $ n$ cities.  Partition the cities into three sets: C, which is the city we want, B, which contains all cities which connect directly to C, and A, which do not connect to C, but by assumption must connect to cities in B.

Add a new city D.  If D connects to C or a city in B, C still has the desired property.  Otherwise, C and all cities in B connect directly to D.  All cities in A connect to cities in B, and so reach D with just one intermediate city.  So D has the property we wish.

Thus the statement is true for $ n\plus{}1$ cities and the claim is true by induction.
[/hide]

are the cities connected two at a time or is every city connected to every other city?

See http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=162646 and the posts linked therein for related problems (including, actually, this one).

aadil, "connected two at a time" and "every city connected to every other city" mean exactly the same thing.


inequalities
logarithms
inequalities open

[quote="lambert"][(x+y)/2]^(x+y)<= (x^x)(y^y)

please help me with it.[/quote]

just uesing Jensen's inequality, we can prove this inequality holds for any $ x,y \in \mathbb{R}^ \plus{}$.

$ \left(\frac {x \plus{} y}{2}\right)^{x \plus{} y}\le x^xy^y$
$ \Longleftrightarrow \frac {x \plus{} y}{2}\ln{\frac {x \plus{} y}{2}}\le \frac {x\ln x \plus{} y\ln y}{2}$
let $ f(x) \equal{} x\ln x$, we have $ f''(x) \equal{} \frac {1}{x} > 0$
so ...


Μιας και ειναι ισως το πιο ωραιο στοιχειωδες προβλημα που εχω δει πιστευω πρεπει να υπαρχει στο forum

Στην περιφερεια ενος τετραγωνικου οικοπεδου κινουνται τεσσερα σκυλια ενω στο εσωτερικο του ενας λυκος.
Οταν ο λυκος βρει ενα σκυλι μονο του το τρωει ενω αν βρει δυο σκυλια αυτα τον εμποδιζουν να περασει.
Τα σκυλια κινουνται μιαμιση φορα γρηγοροτερα απο το λυκο. Να αποδειξετε οτι τα σκυλια αμα θελουν δεν αφηνουν
το λυκο να βγει απο το οικοπεδο!

Οποιος δεν το εχει δει αξιζει να το προσπαθησει :)

[quote="AlexandrosG"]Μιας και ειναι ισως το πιο ωραιο στοιχειωδες προβλημα που εχω δει πιστευω πρεπει να υπαρχει στο forum

Στην περιφερεια ενος τετραγωνικου οικοπεδου κινουνται τεσσερα σκυλια ενω στο εσωτερικο του ενας λυκος.
Οταν ο λυκος βρει ενα σκυλι μονο του το τρωει ενω αν βρει δυο σκυλια αυτα τον εμποδιζουν να περασει.
Τα σκυλια κινουνται μιαμιση φορα γρηγοροτερα απο το λυκο. Να αποδειξετε οτι τα σκυλια αμα θελουν δεν αφηνουν το λυκο να βγει απο το οικοπεδο![/quote]

Eeeep, αρχικές συνθήκες please :)

Βάλτε τα σκυλιά αρχικά στις 4 γωνίες και το λύκο στο κέντρο, αλλιώς υπάρχει περίπτωση ο λύκος να δραπετεύσει ;)

Cheerio,

Durandal 1707

PS: Και ναι, είναι πραγματικά όμορφο πρόβλημα... μου το'χε δείξει ο r_boris παλιά και μου είχε φάει όλο το γυρισμό σπίτι να το σπάσω ;)

Εχεις δικιο τις παρελειψα αλλα επιτηδες γιατι ειναι hint. :)

Δες το συνημμένο

Αυτό που δε μου άρεσε από παλιά σε αυτό το πρόβλημα είναι ότι δε λαμβάνουμε υπ'όψιν την αντίδραση των σκύλων στη κίνηση του λύκου (αυτό μάλιστα ήταν που με είχε παιδέψει όταν σκεφτόμουν το πρόβλημα για πρώτη φορά).

Π.χ. ας θεωρήσουμε ότι οι σκύλοι κινούνται με μοναδικό γνώμονα να τείνουν να ελαχιστοποιήσουν την απόσταση ανάμεσα σε αυτούς και στο λύκο. Τότε, ο λύκος μπορεί πάντα να ξεφύγει [i]ανεξαρτήτως των αρχικών συνθηκών[/i] (γιατί?).

(Επίσης, μπορείτε να γράψετε τη διαφορική εξίσωση που περιγράφει τότε την κίνηση των σκύλων?)

Άρα, για να μην ξεφύγει ο λύκος, θα πρέπει να θεωρήσουμε ότι οι σκύλοι κινούνται πιο "έξυπνα" από τον παραπάνω τρόπο, π.χ. ότι [i]το πολύ[/i] δύο σκύλοι κινούνται έτσι ώστε να τείνουν να ελαχιστοποιήσουν την απόσταση ανάμεσα σε αυτούς και στο λύκο. Αυτό κάπως το κρύβουμε κάτω απ'το χαλί με την υπόθεση ότι "οι σκύλοι συνεννοούνται τηλεπαθητικά", αλλά... meh, υπάρχει ακόμα κάτι που με ενοχλεί λίγο ;)

Cheerio,

Durandal 1707

Και εμενα κατι δεν μου καθεται. Για να λειτουργει η λυση στο συννημενο, δεν πρεπει τα σκυλια να ξερουν που θα παει λυκος εξ'αρχης? Γιατι θα μπορουσε ας πουμε ο λυκος να τραβηξει τα σκυλια καπου και τελικα αντι να βγει απο το Μ να τους τη σκασει απο καπου αλλου. Οπως λεει και ο Durandal πρεπει να λαβουμε υποψιν τον τροπο κινησης των σκυλων...

Θα ελεγα οτι συνθηκη ικανη και αναγκαια ειναι οτι [i]τα σκυλια μπορουν να κινουνται με τετοιο τροπο ωστε σε οποιαδηποτε θεση μεσα στο οικοπεδο (του σκυλου) αν εκεινος να αποφασισει να "επιχειρησει εξοδο" τοτε τουλαχιστον 2 σκυλια να μπορουν να φτασουν στο ιδιο σημειο με αυτον σε τουλαχιστον τον ιδιο χρονο[/i].

Και αμα θελετε ριξτε και μια ματια εδω http://www.mathlinks.ro/Forum/viewtopic.php?search_id=174773530&t=34424 (δεν ξερω αν ειναι σωστη η λυση γιατι με τις εξετασεις δεν εχω προλαβει να την διαβασω :mad: ...)

Edit: Με προλαβες Δημητρη. Ωραια λυση :clap2:

[quote="Demetres"]Εγώ νομίζω πως δεν χρειάζεται τα σκυλιά να συνεννοούνται τηλεπαθητικά:

Βάλτε τον λύκο στο (0,0) και τα σκυλιά στα Α = (0,1), Β = (1,0), Γ = (0,-1) και Δ = (-1,0). Ο σκύλος στο Α θα κινείται ώστε όταν ο λύκος φτάσει στην πλευρά ΑΒ τότε θα βρίσκεται και αυτός στο ίδιο σημείο, όταν ο λύκος φτάσει στην πλευρά ΑΔ τότε θα βρίσκεται και αυτός στο ίδιο σημείο. Δεν το ενδιαφέρει αν ο λύκος πάει να ξεφύγει από το ΒΓ η ΓΔ. Ομοίως και για τους άλλους 4 σκύλους. Αν βρω τέτοια στρατηγική τελείωσα επειδή πάντα θα υπάρχουν (τουλάχιστον) δύο σκύλοι σε όποιο σημείο πάει να ξεφύγει ο λύκος.

Στρατηγική για τον σκύλο Α: Όταν ο λύκος βρίσκεται στο (χ,ψ), ο σκύλος βρίσκεται στο (χ,1-|x|). Επειδή η ταχύτητά του είναι τουλάχιστον 1.5 φορά την ταχύτητα του λύκου, αυτή η στρατηγική δουλεύει.[/quote]

Ναι, ακριβώς αυτό εννοώ όταν λέω ότι πρέπει τα σκυλιά να μη λειτουργούν "μυωπικά" και ότι πρέπει όταν βγαίνει ο λύκος από κάποια περιοχή να μην ασχολούνται μαζί του - δηλ. να μην προσπαθούν απλώς να μειώσουν τυφλά την απόσταση με το λύκο αλλά να έχουν κάποια όρια στο πόσο θα τον κυνηγήσουν.

Όπως λες κι εσύ, αυτή η στρατηγική δουλεύει. Το "τηλεπαθητικό" σχόλιο πήγαινε στη συνήθη διατύπωση του προβλήματος, π.χ. σε αυτήν του r_boris. Αυτό που κάπως με ενοχλεί είναι ότι στο αρχικό πρόβλημα δε γίνεται καμία αναφορά στο πόσο "έξυπνοι" θεωρούνται οι σκύλοι οπότε το βρίσκω λίγο διφορούμενο, that's all ;)

Cheerio,

Durandal 1707


Okay, I have two questions regarding to the TI-89 Titanium. :help:

Question 1: How would I get the cube root of 125 on the TI-89?

Question 2: How would I get the floating point answer in the form of a  fraction?

Thanks. :roll:

Answer the first: 125^(1/3)

Answer the second: Make sure the calculator's not in APPROX mode.  If in AUTO, make sure you didn't use non-fractions in your expression; 0.25*3 will never return a fraction in APPROX.  Make it (1/4)*3

If you want to get the fraction equivalent to a decimal, for example 0.125, you can change it to EXACT mode and just type in 0.125 and press enter.


Oh, you added a less than 30 option for me...good...

...shut up.

[quote="the future"]lol you posted it twice.

are you allowed to post questions?

i want to see what the number 30 on sprint and number 8 on target were like[/quote]

yes well the problems are up on the mathcounts website and there is a topice about this on the mc forum and it hasn't been deleted so i am pretty sure we can discuss it.

lol i just looked at number 30 and i missed it..

I got 1099/350+1100/350.

lol its suppose to be -1099/350 (bad handwriting)...

Funny how if you just look at the answer I got and then the problem, you clearly can tell that wasnt a realistic answer...too bad i wanst aware of that.

Didnt do number 8 on target but i looked at it, itsnt that kinda easy for an 8 on target?

The entire test was easy for Mathcounts. There were tons of perfect scorers, too.


Hi,
Does anyone have the solution or instructor resource manual to Paul Zeitz's Art and Craft of Problem Solving? If you do please email it to me at fdasdfasdf at gmail.com.

 thanks

Do a search for this.  It's a been discussed before.

Locked.


probability
analytic geometry
graphing lines
slope

well, I was going through the 2008 National Sprint round and i came across 2 problems that were fairly tricky for me (?).
[b]
27. Three players each have a red card, blue card, and green card. The players play a game that consists of three rounds. In each round, each player randomly turns over one of his/her cards without replacement. What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds?[/b]
Well, that's one boring game alright.

AND
[b]
22. To the right is an example of a square grid of dots with three dots along each side. In a similar square grid of dots with 8 dots along each side, how many distinct lines pass through the bottom left corner and at least
one other dot of the grid?[/b]

sorry, no pic

so anyone who can help me will receive three helping dollars [hide="cough cough"](which can be spent on virtually nothing)  :) [/hide]

For #27, I think it is 1/18 (correct me if I am wrong.)

Here's the explanation (If that is right...)
[hide]
Let's take the first round first, the first person can flip over any of the cards (so his choice doesn't matter), the second person can only flip over 2 of the possible 3 cards, thus with probability 2/3. The third person has only one choice for the last card to flip, with probability 1/3. Thus the probability for a successful round 1 is 2/9.

For round 2 (each of the people have 2 cards left), the first person starting can pick any of the two (so his choice doesn't matter), the second person has only 1 out of the 2 that is going to yield success. Similarly, the last person also is going to pick the last remaining card (the color which hasn't been picked yet), with probability 1/2. Thus the probability for success in this round is 1/4.

In round three, each of the players have only one card left. Since both, rounds 1 and 2 have to be successful, we have to multiply the probabilities. So the answer is 2/9 * 1/4 =  1/18 (Basically because, for each of the 2/9 times round 1 is successful, round 2 is successful 1/4 of those times.)
[/hide]

[hide="#27"]WOW I am stupid. I didnt read the part on how you dont replace your cards, so I just did (2/9)^3.

Yeah, thats right then Limac. You could do: (# of ways to succeed on round 1)/(# of total ways to do round 1)x(# of ways to succeed on round 2)/(# of total ways to do round 2)x(# of ways to succeed on round 3)/(# of total ways to do round 3).

Round 1 you win 3x2x1 times out of a total of 3x3x3 times
Round 2 you win 2x1x1 times out of a total of 2x2x2 times.
Round 3 you only have 1 card left for everyone, so if you succeeded on round 1 and 2, you will succeed on round 3, so its 1.

for a total of (6)/(27)x(2)/(8)x(1)/(1) =$ \boxed{1/18}$
[/hide]

[hide="#22"]As for number 22, its not that hard if you just list the possible slopes systematically, take out the fractions that equals a fraction listed before it (ie. for 3/2 and 6/4 we would need to take out 6/4 since its the same slope) then add 2 (you will see why). The slopes are:
1/1 1/2 1/3 1/4 1/5 1/6 1/7
2/1 2/3 2/5 2/7
3/1 3/2 3/4 3/5 3/7
4/1 4/3 4/5 4/7
5/1 5/2 5/3 5/4 5/6 5/7
6/1 6/5 6/7
7/1 7/2 7/3 7/4 7/5 7/6

But then we need to add 2 because we didnt list the 2 where the slope is undefined (follows y=7 when plotted one a grid) or when the slope is 0.

S0 there are a total of 35+2=$ \boxed{37}$[/hide]

[quote="mathlearner2012"]You are not allowed to post answers. If you want to find the answer, there is another site.[/quote]

I see you're new. If you are referring to this rule:

[quote="AIME15"]Some guidelines that you should follow when you are asking for help:

[b][u]What You Tried[/u][/b]
[list][*]Post what you attempted, even if it may not have worked.  We are not here to give you solutions, we are here to help you learn.
[*]Be detailed.[/list]
[/quote]

It has a LOT of flexibility. Even if the person doesnt post what he tried, we will usually still answer the question.  :)

Wow I have like ten posts that tells me AIME15 said this rule. And your right, I am fairly new.


Can someone give an explanation onto how they did this problem

CD must equal the perimeter of the base.  We need to find the radius so we can find volume:

$ 2\pi r\equal{}10\pi, r\equal{}5$

$ V\equal{}\pi r^2 h\equal{}(\pi)(5^2)(6)\equal{}150\pi$

We know $ h\equal{}6$ b/c AD=6

oh thanks i did it the other way with 10pi as the height and 6 as the circumference  :rotfl:

CD is the circumfrence of the circle, since it is 10pi, the radius must be 5.

5^2pi=25pi which times height =150.

Out of curiosity, where did you get that problem? Mathcounts doesn't do multiple choice.

[quote="neurosurgeon1118"]Its from KCATM its a local competition that we have every year in Kansas last year i won 1st place individual for 6th Grade.[/quote]

Sorry, but it isn't a annual competition for Kansas.  Do you even know why it's called KCATM :Kansas City Association of Teachers of Mathematics.  YOu can clearly see its says KANSAS CITY!!

but nowhere does it say that you can't post questions from it in the forum.  When asking about the competition itself, that goes somewhere else.


Is anyone in upstate New York (especially Rochester Area) interested in joining my HMMT team?

Also, do you think there is any ethical issue with having my friend (who will be 18 by that time) driving the team to Boston (Cambridge)?

[quote="frankthetank"]Also, do you think there is any ethical issue with having my friend (who will be 18 by that time) driving the team to Boston (Cambridge)?[/quote]

Not ethical, but maybe insurance-wise.

This is how Albany Area does it.  (Yay for carpooling!)  Although that wouldn't work as well with the longer drive from out west.

[quote="frankthetank"]Maybe I could have everyone arrange their own transportation.[/quote]

Do we have to have a coach come along with us?


This was already asked and solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=36404]here[/url]


Given a polyhedron with $ n$ edges ($ n \geq 8$), and a plane which intersects the polyhedron at its edges. The plane dose not pass through any vertex or edge of the polyhedron. Prove that:
(1) There exists at most $ n \minus{} 2$ edges on which the plane intersects.
(2) For each $ n \geq 8$, there exists a polyhedron and a plane, suth that the plane intersects $ n \minus{} 2$ edges of the polyhedron at its interior.

Is this one too difficult or too boring? No one replays to this in a week.

Choose a face of the polyhedron.  If the plane intersects 3 edges of this face, it must be the plane of the face, a contradiction.  Thus, we can find an edge of this face through which the plane does not pass.  Choose a face of the polyhedron not containing this edge.  Then again there is some edge of this face through which the plane does not pass.  Thus, the plane cuts at most $ n \minus{} 2$ edges.

Unless I've done something wrong, this seems to indicate that the second part is false (and indeed, that the only polyhedron for which this estimate is sharp is a tetrahedron).

Unfortunately your "solution" is wrong. You've ignored that these intersection point could lie on a single line. Here is an example.
In fact I've known that $ n\minus{}2$ is truely available for $ n \geq 8$ and $ n\equal{}6$.

Yes, you're right, I did not consider polyhedra with concave faces.


[color=darkblue][b][u]Try to present an another proof for the following problem :[/u][/b][/color]

[color=darkred]Let $ABC$ be a triangle and a point $L$ in the its plane. Denote the points :

$\|\begin{array}{c}D\in BC\ ,\ LD\perp BC\\\ E\in CA\ ,\ LE\perp CA\\\ F\in AB\ ,\ LF\perp AB\end{array}\|\begin{array}{c}M\in BC\ ,\ MB=MC\\\ N\in CA\ ,\ NC=NA\\\ P\in AB\ ,\ PA=PB\end{array}\|\begin{array}{c}U\in EF\ ,\ MU\perp EF\\\ V\in FD\ ,\ NV\perp FD\\\ W\in DE\ ,\ PW\perp DE\end{array}\|\ .$

Prove that the lines $MU$ , $NV$ , $PW$ are concurrently.[/color]

[color=darkblue][b][u]Remark.[/u][/b] The triangle $DEF$ is orthologically to the triangle $MNP\ ,$ i.e. there is a point $L$

so that $\{\begin{array}{c}LD\perp NP\\\ LE\perp PM\\\ LF\perp MN\end{array}\ .$ Therefore, and the triangle $MNP$ is orthologically to the triangle $DEF\ ,$ i.e. exists a point $S$ so that $\{\begin{array}{c}SM\perp EF\\\ SN\perp FD\\ SP\perp DE\end{array}\ .$

[hide="A metrical proof (with the Carnot-s lemma) of the proposed problem."]
I'll apply the [b][u]Carnot's lemma[/u][/b] for the point $L$ and the triangle $ABC\ :$

$(FB^{2}-FA^{2})+(EA^{2}-EC^{2})+(DC^{2}-DB^{2})=0\ \ (*)\ .$ Therefore,

$\{\begin{array}{c}MU\perp EF\Longleftrightarrow UF^{2}-UE^{2}=MF^{2}-ME^{2}\\\ \mathrm{theorem\ of\ the\ median\ }\Longrightarrow \{\begin{array}{c}4\cdot MF^{2}=2\cdot(FB^{2}+FC^{2})-a^{2}\\\ 4\cdot ME^{2}=2\cdot(EB^{2}+EC^{2})-a^{2}\end{array}\end{array}\|$ $\Longrightarrow$

$2(UF^{2}-UE^{2})=FB^{2}+FC^{2}-EB^{2}-EC^{2}\ \ (1)\ .$ Analogously,

$2(VD^{2}-VF^{2})=DC^{2}+DA^{2}-FC^{2}-FA^{2}\ \ (2)\ ,$

$2(WE^{2}-WD^{2})=EA^{2}+EB^{2}-DA^{2}-DB^{2}\ \ (3)\ .$

Using the relation $(*)\ ,$ from the sum of the relations $(1)$ , $(2)$ , $(3)$ obtain

$(UF^{2}-UE^{2})+(WE^{2}-WD^{2})+(VD^{2}-VF^{2})=0\ ,$

i.e. (again the [b][u]Carnot's lemma[/u][/b]) the lines $MU$ , $NV$ , $PW$ are concurrently.[/color][/hide]

I have an exam tomorrow, and I put it over far too long.  :|
So I'm really sorry, but I have no time to elaborate my possible solution, maybe I will post it later:

Because $DEF$ is a pedal triangle, we can find all it's angles. Now, we can prove the statement by using Ceva in trigonometric form on triangle $MNP$, which is the medial triangle (so we know all the angles, and we can make the computation)

I think it will work, but I have no time to make sure right now.

It was also discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=123843]here[/url].


[quote="April"]Let $\ a,b,c>0,a^{2}+b^{2}+c^{2}=1$. Prove:
$\ A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{1}{ab+bc+ca}\preceq\frac{1}{2}$[/quote]
$\iff\sum(\frac{a}{b+c}-\frac{a^{2}}{ab+bc+ca})\leq\frac{1}{2}$
$\iff\frac{abc}{ab+bc+ca}.\sum{\frac{1}{b+c}\leq\frac{1}{2}}$
Use $\frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b}$ we have it's true.


There are six barrels, containing 15 gallon, 8 gallons, 17 gallons, 13 gallons, 19 gallons and 31 gallons.  Each barrel contains either oil or vinegar.  The oil sells for twice as much per gallon as the vinegar.  A customer buys 14 dollars worth of each, leaving one barrel.  Which barrel was left?

It must be put in systematic list "dictionary order"

Thanks for any help.

[hide]
add the numbers together to get 103. Since he bought twice as much vinegar as oil, the amount he actually bought must be divisible by 3. so we need to subtract the amount in 1 barrel to get a number divisible by 3. we see that 13, 19, and 31 work. Since he bought them for 28 total, the result needs to be divisible by 28, else it would not make sense to by a fraction of a gallon. we see that 103-19=84, which is divible be 28. thus 19 is the amount left behind.[/hide]


function
topology
inequalities
advanced fields
advanced fields unsolved

Is it easy to find the functions of $L^p([0;1],\lambda)$ , $1< p <\infty$, that are extreme points of the unit ball ?
(Such functions exist because the unit ball of $L^p$ is compact for the weak topology, and the weak topology defines a locally convex topology. So krein-Millmann shows there are such functions)

PS: question motivated by Rainwater theorem used in this recent post http://www.mathlinks.ro/Forum/viewtopic.php?t=85772  :) )

Don't they are the points with norm $1$ in $L^p([0,1])$ for $1<p<\infty$ ?
At least we can easilty show that all such points are extreme points: if we suppose the contrary for some $f$ for which $||f||_p=1$ we have $f= \lambda g+ (1-\lambda)h$, then by Minkowski inequality $||\lambda g+ (1-\lambda)h||_p \leq \lambda ||g||_p + (1-\lambda) ||h||_p \leq 1 =||f||$, so the there must actually be an equality , which means that $g=kh$ for some constant $k$. And it is clear that then $f=g=h$.


A row of 8 light switches controls different banks of lights in a school gymnasium. In preparation for a school dance, students decide that they want either one or two banks of lights on during the dance. How many possible ways are there of setting these 8 switches so that either one or two banks of lights are on?

This is simply $ \binom{8}{1} \plus{} \binom{8}{2} \equal{} \boxed{36}$.


Once you open eg Preolympiad section, you see that, that is as an attachment. In this case there are 156 pages.

And now it is very hard to get to page numbered say 70. You would have to click to "page 4", then "page 7", "page 10"... and that would waste user's time...

Therefore I suggest it to be "clickable" the "Goto page" text. And after that the user will be asked for number of page he/she would like to see.

It will be more convenient that way.

conceive

Rather than click through 3 pages at a time, you can alter the URL directly:

[url=http://www.artofproblemsolving.com/Forum/index.php?f=138&ppage=50&sort=lastpost&order=DESC&start=50]http://www.artofproblemsolving.com/F...&order=DESC&start=50[/url]

is the url of the 2nd page of Round Table, given my settings.  To get to the 3rd page given my settings, I just change it to

[url=http://www.artofproblemsolving.com/Forum/index.php?f=138&ppage=50&sort=lastpost&order=DESC&start=100]http://www.artofproblemsolving.com/F...&order=DESC&start=100[/url]

There used to be the goto page option, which worked exactly as you suggested. It disappeared a couple of years ago and I've been "lobbying" Valentin to put it back since then :( I'm happy to see that I'm not the only one who is too lazy to practice the "times 50" table every day :P

Guys you've got to understand that there are tons of upgrades going on at all times (even if you don't see anything on the page) and this is not very high on our priorities list. I will try to solve this problem though when I'm coming back from Vietnam :)

Ok, the option was added. Push F5 (refresh) once to make it work if you have the pages in cache.


calculus
integration
algebra
function
domain
vector
abstract algebra

Let $ \mathfrak{m}$ be a maximal ideal in an integral domain $ R$. Suppose $ \mathfrak{m}^n \subset \mathfrak{m}^{n\minus{}1}$ is proper and $ \mathfrak{m}^{n\minus{}1} / \mathfrak{m}^n \cong R/\mathfrak{m}$ for any $ n \geq 2$.

(i) If $ u \in R$ and $ u \notin \mathfrak{m}$, show that image of $ u$ in $ R/\mathfrak{m}^n$ is a unit.

(ii) If $ p \in \mathfrak{m}$ but $ p \notin \mathfrak{m}^2$, show that $ R/\mathfrak{m}^n$ is a principal ideal domain with only a finite number of ideals $ (p), \cdots, (p^{n\minus{}1})$.

maximal ideals in $ R/\mathfrak{m}^n$ correspond to maximal ideals of $ R$ containing $ \mathfrak{m}^n$, i.e. containing $ \mathfrak{m}$. thus $ R/\mathfrak{m}^n$ is local with maximal ideal $ \mathfrak{m} R/\mathfrak{m}^n$. therefore (i).

in the situation of (ii) we see that (i) imples that multiplication with $ p$ yields a injective homomorphism $ R/\mathfrak{m} \to R/\mathfrak{m}^2$. thus we get an exact sequence of $ R/\mathfrak{m}$-vector spaces

$ 0 \to R/\mathfrak{m} \to \mathfrak{m}/\mathfrak{m}^2 \to \mathfrak{m}/(p) \to 0$.

by assumption, $ \dim_{R/\mathfrak{m}}(\mathfrak{m}/\mathfrak{m}^2) \equal{} 1$, so that $ \mathfrak{m} \equal{} (p)$.

now let $ \mathfrak{a}$ be an ideal of $ R/\mathfrak{m}^n \equal{} R/p^n$. if it is not proper, it's generated by $ 1$. otherwise it lies in the maximal ideal $ p R/p^n \cong R/p^{n \minus{} 1}$ and corresponds to an ideal $ \mathfrak{b}$ of $ R/p^{n \minus{} 1}$ such that $ \mathfrak{a} \equal{} p \mathfrak{b}$. continue with $ \mathfrak{b}$ in the same fashion as with $ \mathfrak{a}$. this shows that the ideals of $ R/p^n$ are $ (0) \equal{} (p^n) \subseteq (p^{n \minus{} 1}) \subseteq ... \subseteq (p) \subseteq (1) \equal{} R/p^n$. in particular, every ideal is principal, but $ R/p^n$ is not a principle ideal [i]domain[/i].

(iii) Let $ k < n$ and $ M$ any $ R/\mathfrak{m}^n$-module. Show that $ \mathfrak{m}^k M \subset M$ is a $ R/\mathfrak{m}^n$-submodule in $ M$ and $ \phi: R/\mathfrak{m}^n \times M/\mathfrak{m}^k M \rightarrow M/\mathfrak{m}^k M$ factors through $ \psi: R/\mathfrak{m}^k \times M/\mathfrak{m}^k M \rightarrow M/\mathfrak{m}^k M$.

(iv) Finally, show that any finitely generated $ R/\mathfrak{m}^n$-module can be represented as a (finite) direct sum of $ R/\mathfrak{m}^k$ with $ k \leq n$ and the number of terms in the representation is independent of the choice of splitting.

is this is a homework you want to get solved from us?!

who is P in (iii)?

All the $ P$ should have been replaced by $ \mathfrak{m}$. This is just a slight revision of the problem I found in Chapter 12 of Dummit and Foote. I want to emphasize that this is not a homework problem. Why would I post a homework problem in here? Anyway, thank you so much!


Prove that there are no numbers x,y>0 for which x*2^y+y*2^(-x)=x+y.

[b] Solution [/b]:

This is done with my favourite theorems in the world :D :D :D
[b]Lagrange theorem[/b]
if x=y we get that x=y=0 which is impossible because x,y>0.
lets consider that there exist x and y>0 with x<>y for which x*2^y+y*2^(-x)=x+y [b](1)[/b]
Let y=x1-x2, x=x2-x3 where x1>x2>x3>0.
From [b](1) [/b] we have that:
[2^(x1-x2)-1]/[1-2^(x3-x2)]=(x1-x2)/(x2-x3) which can be written like:
(2^x1-2^x2)/(x1-x2)=(2^x2-2^x3)/(x2-x3) now we apply Lagrange and we get that:
(2^x1-2^x2)/(x1-x2)=2^(alpha1)*ln2 where alpha1 is in (x2,x1)
(2^x2-2^x3)/(x2-x3)=2^(alpha2)*ln2 where alpha2 is in (x2,x1)
so we need to have 2^(alpha1)*ln2=2^(alpha2)*ln2 which gives us alpha1=alpha2 which is absurd! :D :D
[b]cheers! [/b] :D :D

It's as easy as pie

Let f(x)=x2y+y2-x-x-y.
f'(x)=2y-y.2-xln 2-1>2y-y.ln 2-1>0 since
y.ln 2+1 is a tangent to convex curve 2y in point (0,1).
Thus f(0)=0 and f is an increasing function ==> f(x)>0 for all x>0.


calculus
derivative
calculus computations

how do i differentiate arcsin sqroot theta

$ \frac{d}{d\theta}(\arcsin(\sqrt{\theta}))\equal{} \frac{1}{\sqrt{1\minus{}\sqrt{\theta}^2}}\cdot \frac{1}{2}\theta^{\minus{}1/2}\equal{} \frac{1}{2\sqrt{\theta}\sqrt{1\minus{}\theta}}$

i.e. use the definition of derivative of $ \arcsin (x)$ and the chain rule.


calculus
integration
calculus computations

For a positive constant number $ t$, let denote $ D$ the region surrounded by the curve $ y \equal{} e^{x}$, the line $ x \equal{} t$, the $ x$ axis and the $ y$ axis. Let $ V_x,\ V_y$ be the volumes of the solid obtained by rotating $ D$ about the $ x$ axis and the $ y$ axis respectively. Compare the size of $ V_x,\ V_y.$

[quote="IndoChina"]Kunny... do you mean rotating about the x-axis and the y-axis... ?

You wrote... "[u]rotating  about the axis[/u] and the y axis respectively. Compare the size of..."[/quote]

Sorry, I made a typo.

I have just edited it.

kunny

hello, we get $V_x$ as $V_x=\int_{0}^{t}e^{2x}\,dx=\frac{\pi}{2}\left(e^{2t}-1\right)$
Sonnhard.

hello, we get $V_y$ as $V_y=\pi\int_{0}^{t}x^2e^{x}\,dx=\pi\left(-2+2e^{t}-2e^{t}t+e^{t}t^2\right)$.
Sonnhard.


Can someone please help me wit this Question !!!

We have to Prove that :->

You have to prove that  $ 991 | 111...1$

so write  $ 111...1$ in another form and use FLT.

[quote="cdymdcool"][quote="mathson"][hide="Solution"]
$ 1982|222 \ldots 22 \Leftrightarrow 991|111 \ldots 11$
now we want to prove $ 991|111 \ldots 11$.
but note that: $ 111 \ldots 11 = \frac {10^{1980} - 1}{9}$. Of course :
$ 991|111 \ldots 11 \Leftrightarrow 991|\frac {10^{1980} - 1}{9} \Leftrightarrow 991|10^{1980} - 1$ because $ \gcd (991 , 9) = 1$.
now we know: $ 10^{\phi (991) } - 1 \equiv 0 \pmod {991} \Rightarrow 10^{990} \equiv 1 \pmod {991}$.
$ 10^{1980} - 1 = 10^{2 \times 990} - 1 \equiv 0 \pmod {991}$. QED[/hide][/quote]

Why because $ \gcd (991 , 9) = 1$, so you cancel the 991 and 9.  Did you use the eludcean algorithm in the second part?[/quote]

we can write: $ 10^{1980}-1 = 9a \Rightarrow 991|\frac{9a}{9}\Rightarrow 991|a$.
but $ \gcd(991,9)=1 \, \textrm{ then }\, 991|9a$.

$ 991$ is a prime number  :D


function
complex analysis
real analysis
real analysis solved

Find all functions f, entires such that equation $ f(z)=a $ have a finite number of solutions, for all $ a\in C. $

Piard's great theorem states that if f is an entire function which is not a polynom, then $f^{-1}(x)$ is an infinite set for every $x\in \mathbb{C}$ with at most one exception.

So your functions are polynomials.

You are right! I found the same solution

how did you prove it? Picard is not that trivial, after all...

just interested.


[quote="hanschinaman"]a,b,c,d>0 with a+b+c+d=4,prove
$ 5(abc+abd+acd+bcd) \leq 8abcd+12$[/quote]
It's my own inequality.  :lol:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5972
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71736


Hi! there,

I use Latex for my thesis. I need to plot some graphics in Latex. May I ask which GUI software is your recommended please? I know Tpx, Matagraf...

I do not know which one is the best. Thx for your suggestion! :rotfl:

[quote="stevem"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url][/quote]
Asymptote is best because you can insert it in the forum with [asy]and[/asy] tags.

[quote="aopsfan"][quote="stevem"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url][/quote]
Asymptote is best because you can insert it in the forum with [code][asy]
and
[/asy][/code] tags.[/quote]

Added code tags.


[quote="Absolut"]OK i agree that $f(n)=n$ for all natural numbers n and the number 0. But how can we prove that f(n)=nfor all real numbers? Can you explain me[/quote]

Let $f(x)=z$ then $f(f(x))=z=f(z)$ so if $x\in Imf$ then  $f(x)=x$

i didn`t understand nothing.. i`m grade 9! how can i get this problem

[quote="cckek"][quote="Absolut"]Find all functions f:R-R, such that f(1)=1, and f(xy+f(x))=xf(y)+f(x), for all x,y-real[/quote]

$y=1\implies f(x+f(x))=x+f(x), x=1 \implies f(y+1)=f(y)+1\implies f(0)=0\implies f(f(x))=f(x)\implies f(x)=x$[/quote]

Actually you proved that $f(x)=x$ only for $x\in Im(f)$. Any way ..

$y=\frac{1}{x}\Longrightarrow f(1+f(x))=xf(\frac{1}{x})+f(x)$
But $f(1+f(x))=1+f(f(x))=1+f(x)$ , so $1+f(x)=xf(\frac{1}{x})+f(x)\Longrightarrow f(\frac{1}{x})=\frac{1}{x}$
Finally $f(x)=x$ for all $x\in R$ .. More general, you can take any $K$ (commutative field) in place of $R$, and the proof is the same ..
:cool:


ARML
AMC
AIME
induction
inequalities
number theory
least common multiple

1. Compute the largest prime factor of
3(3(3(3(3(3(3(3(3(3(3 + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1.
2. Assuming the expression converges, determine the largest integer n with n under or equal to 9, 000, 000 for
which$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+...}}}}$  is rational.
3. The sequence 1, 3, 5, 7, 9, 15, 17, . . . consists of the positive integers whose binary representation
reads the same left to right as right to left. (Such as 9, which is written as 10012 in binary.)
Find the number of 1’s in the binary representation of the 1717th term of the sequence.
4. What is the largest positive integer n for which $n^3 + 100$ is divisible by $n + 10$?
5. Let m and n be positive integers with 1 under or equal to m < n. In their decimal representations, the last
three digits of $1978^m$ are equal, respectively to the last three digits of $1978^n$. Find m and n
such that m + n has its least value.
6. Find all positive integers a, n such that
$a^{n+1} - (a + 1)^n = 2001.$
7. Find with proof all primes that are one more than the square of a Fibonacci number.

[hide="1"]
the expression is equivalent to $\frac{3^{12}-1}{2}$, so we have to find the largest prime factor of $3^{12}-1$, lets say it's $p$, then $3^{12} \equiv 1 (\bmod\ p)$, and by fermat's theorem we see that $3^{p-1} \equiv 1 (\bmod\ p)$ also.  So we have $12k = p-1$, and $p = 12k+1$.  and I don't know what to do now...
[/hide]

[quote="PenguinIntegral"]Do you mean 73? That is what the answer in the .pdf is.[/quote]

I just realize that I was wrong and changed it :blush: any way to continue from that?

to make up here's number 2:
[hide]
using common infinite recursion technique and letting $k$ be the expression, we have $k^2 - k - n = 0$.  Looking at the determinant we see that $4n+1$ must be a perfect square.  Since this is odd, and all squares leaves remainder of 1 mod 4, if $x$ is this odd square, then $x < 4*9000000 + 1 = 36000001$.  this nasty thing is a little bit more than $6000^2$,  so we have $x = 5999^2$ and solve it from there we get $n = 8997000$.
[/hide]

Yeah, shobhit. I thought AoPS stole one of the ARML tests when I first looked at it.

I've been racking my brain trying to solve 5,6 and 7. 7 is induction I think, but you have to slpit it up and it gets hard.

[quote="molina"][hide="2"]

              Do $\large{a_i\,=\,\underbrace{\sqrt{n+\sqrt{n+\sqrt{n+...}}}}_{i\,\text r{times}}}$

             Assuming the convergence :

           $\large{a_i^2\,=\,a_i+n\,\Longrightarrow{\,\Delta\,=\,1+4n}}$

           How $\large{a_i}$ is rational , $\large{\exists\,k\,\in\mathbb{Z_+}\,|\,1+4n\,=\,k^2\,}$

           Now it  easy...[/hide][/quote]
Complete solution?

[quote="PenguinIntegral"]I'm confused by your use of the ".". What operation does it stand for?[/quote]

It means multiplication.

[quote="molina"][hide="4"]$\large{(n+10)|(n^3+10^2)\,\Longrightarrow{\,(n+10)|[(n+10)(n^2+10n+10^2)-(10^3-10)]\,\Longrightarrow{\,(n+10)|10.99.101}}}$

      Then $\large{n\,=\,10(99.101-1)}$[/hide][/quote]

I believe you got the identity wrong.  it is $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
[hide="the rest"]
 continuing you should get that $n+10 | 900$, which implies that $n = 890$.
[/hide]
I edited in the solution for number 2 above in my second post too.  Anyone want to try number 1??

[quote="pkerichang"][quote="joml88"][hide="Number 1"]As Eric said, it's equal to $\frac{3^{12}-1}2$.  You can see this by multiplying it out (not literally) to get $3^{12}+3^{11}+\cdots+1$ and using the formula for the sum of a geometric series.

So just factor it:

\begin{eqnarray*} 3^{12}-1 &=& (3^6+1)(3^6-1)\\ &=& (3^2+1)(3^4-3^2+1)(3^3+1)(3^3-1)\\ &=& 10\cdot 73\cdot 28\cdot 26 \end{eqnarray*}

So the answer is 73.[/hide][/quote]
uuuuugggggggghhhhhh!!!! should have thought of that....Those factoring tactics always trick me...
anyone can give me tips on where to look for factoring stuff???[/quote]
Everywhere. You should factor everything, absolutely everything. Make it a habit.

At least that is what the admins always say.

[quote="PenguinIntegral"]
3. The sequence 1, 3, 5, 7, 9, 15, 17, . . . consists of the positive integers whose binary representation
reads the same left to right as right to left. (Such as 9, which is written as   [b]10012[/b] in binary.)
Find the number of 1’s in the binary representation of the 1717th term of the sequence.
[/quote]
You mean $1001_2$?
[hide="3"]
Each number is of the form $(2k+1)_10$.  The $1717th$ term is $2(1717)-1=3433$
Since $3433=2048+1024+256+64+32+8+1=110101101001_2$ there are $\boxed{7}$  $1$'s.[/hide]
[hide="2"]  Solving for the expression, $v$ yields ${v=\frac{1+\sqrt{1+4n}}{2}}$.  Checking to see which values of $n$ give rational answers, we find $2,6,12,...$ work, all of the form $a(a+1)$, [hide="check yourself"]plug in $a(a+1)$ for $n$ to see this always works[/hide].  $n=a(a+1)\leq9,000,000 (=3000^2).$  $a=2999$ satisfies the inequality, so $n=2999(3000)=\boxed{8,997,000}$[/hide]
[hide="spoiler"]Yeah, I saw #1 on this postest on a previous year's individual arml questions, and a question really similar to #2 on an arml team round.  #5's in AaCoPS and is from a past aime.[/hide]

I need to find solutions for 5 and 6. 7 isn't really that hard.

I think you can brute force with 5, but I want a nice shortcut...

[quote] think you can brute force with 5, but I want a nice shortcut...[/quote]
FWIW
[hide]$1978 = 2 \cdot 989$   and  notice that  (last three digit to be same ) $1000=8 \cdot 125$
Now since $\phi(125) =100$  and $\phi (8) = 4$ and LCM of 100 and 4 is 100.

$989^{100} \equiv 1 \bmod 125$  and $989^{100} = 1 \bmod 8 \implies 989^{100} \equiv 1 \bmod 1000$
(We also have to show that 100 is smallest such number, see the note below)

(and $2^3=8$  so)

  $1978^3 \equiv 1978^{103} \bmod 1000$

To show that $103+3$ is smallest such value, all you have to show that  order of $979$   is not less than 100.
This is easy to show because $989 \equiv -11 \bmod 125$ and one can easily check that that order has to be a  mulitple of 4 (checking the last 2 digits) and   checking for $(-11)^{20}$  we see that it would not work.[/hide]

You have to notice, quantum leap, that neither 11 nor 13 is in the sequence and that your solution doesn't read the same in both directions (it's not a palindrome).

Now for my attempt at a solution: [hide="3"]Firstly, the first and last digit of each binary number must be 1.

For those members of the sequence with even numbers of digits in their binary representations, the right half must be the mirror image of the left half, hence it depends on the left half. So the left half starts at 1000...0 and goes up by increments of 1 to 1111...1. So if there are $k$ digits in the binary sequence, there are $2^{k/2 - 1}$ numbers in whose binary representation contains $k$ digits.

For those members whose binary representations consist of an odd number of digits, the right half also mirrors the left half. If there are $k$ digits in one of these, then there are $2^{(k+1)/2 - 1}$ with $k$ digits. There is 1 number with 1 digit, 1 with 2, 2 with 3, 2 with 4, 4 with 5, 4 with 6 etc. The number of terms with any number of digits, even or odd, is $2^{\lceil k/2\rceil-1}$.

Now we need to figure out how many digits the 1717th term has. So first we'll find out what which groups of numbers of digits have been filled: $2^1 + 2^2 + 2^3 +...+ 2^n < 1717$, where $n$ is the half of the last even group that's been filled. So $2^{n+1} - 2 < 1717$ and $n = 10$. The next 1024 terms after the 1022th have odd numbers of digits, including the 1717th. So the 1717th term has 21 digits.

$10000000000000000001_2$ is the 1023rd term, and using the second digit up to and including the middle digit, we will count to $1717-1023 = 494 = 0111110000_2$. So the 1717th term has $\fbox{12}$ ones. I hope.
[/hide]

[quote="PenguinIntegral"]6. Find all positive integers a, n such that
$a^{n+1} - (a + 1)^n = 2001.$
[/quote]

[hide="partial solution"]look in mod a,

$0\equiv 2002\mod a$

$2002=2*11*7*13$

now it is just case work, noting that for each value of $a$, there can be at most one solution for $n$ because exponentials are strictly increasing or decreasing[/hide]

[quote="Altheman"]
[hide="partial solution"]look in mod a,

$0\equiv 2002\mod a$

$2002=2*11*7*13$

now it is just case work, noting that for each value of $a$, there can be at most one solution for $n$ because exponentials are strictly increasing or decreasing[/hide][/quote]
A little more,  that is if you really need it...
[hide]
Check mod 3, and you will see $a$  can not be 0 or 2 mod 3, so has to be 1 mod 3 this makes n even, and same logic as above, ($a+1$) must be a factor of 2002.

(So $a$  must be 1 or 13,  no dice for 1 so it must be 13.)
(A little more work and one can see the Only solution is n=2)
[/hide]

BTW, Altheman - You have to be a little careful when  you say

  [quote] ... noting that for each value of $a$, there can be at most one solution for $n$ because exponentials are strictly increasing or decreasing [/quote]

Actually as n increases  ($a^{n+1} - (a+1)^n$)  increases, reaches a maximum and then decreases, (becomes zero and then goes to negative infinity)  so  there can be [b]two [/b]  possible values to check.

(One is lucky here that when a=13,  only n=2 is an integer)


hello, write the term in the following form
$\frac{7-6\left(\frac{6}{7}\right)^{n}}{1-\left(\frac{6}{7}\right)^{n}}$
regards,Sonnhard.


geometry
3D geometry
pyramid
analytic geometry

Sue arranges ping-pong balls in the form of a square-based pyramid. Sue's pyramid has five square layers, each square layer has sides one ping-pong ball longer than the layer above it, and the top layer is one ball. How many ping-pong balls did she use?

So, with five layers, the bottom will be $ 5^2$, then $ 4^2$, $ 3^2$ and so on. Adding 25+16+9+4+1=55

[quote="isabella2296"]Or we can just do the sum of the first 10 numbers. $ 10 \times 11 \equal{} 110 \div 2 \equal{} \boxed{55}$.[/quote]

im pretty sure thats just a coincidence. try it with 6 layers.

$ 1\plus{}4\plus{}9\plus{}...\plus{}n^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$

ill answer this in a series of steps

[hide="this is my explanation"]1.we know the following fact(s)

 I.we know that there is only 1 ball at the top

 II.and there are 5 layers

 III. the lower you go the more balls there will be

2.now that we have the information we get this

 the 1st layer-->1x1
 the 2nd layer-->2x2
 the 3rd layer-->3x3
 the 4th layer-->4x4
 the 5th layer-->5*5

3.getting these coordinates we get the #'s 1,4,9,16,25 for the layers 1,2,3,4,5 layer respectively

4.adding the #'s 1+4+9+16+25=55

5.[hide="ANSWER"]55[/hide][/hide]

[hide="."][size=200][color=#FF4000]PLEASE RATE MY POST THANK YOU SO MUCH!![/color][/size][/hide]

[size=50]please be fair and please PM if you think there is a better way to solve this thank you so much!![/size]

:bruce:
may the force of AoPS be with you.


limit
trigonometry
FTW
calculus
derivative
calculus computations

Find the limits:
[b]1.[/b]$\mathop {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{\sin (x - \frac{\pi }{3})}}{{1 - 2\cos x}}$
[b]2.[/b]$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} tg2x \bullet tg(\frac{\pi }{4} - x)$
[b]3. [/b]$\mathop {\lim }\limits_{x \to 4} \frac{{x\sqrt x - 8}}{{4 - x}}$

I don't think these go in Getting Started but oh well here I'll try.

[hide="number 1"]
mr. l'hopital ftw

cos(x-pi/3) / (1 + 2sinx)

1/(1+sqrt(3))
[/hide]

[hide="hint for 3 cuz i am way too lazy"]
factor
[/hide]

Yeah, calculus is not allowed in GS. [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]This[/url] is probably the best place to post this type of problem.

[hide="Number 3"]
Yeah, you can factor the numerator as Treething said, but it's faster to use I'Hopital's rule.

$\lim_{x\to 4} \frac{x\sqrt{x}-8}{4-x} =\lim_{x\to 4} \frac{\frac{3}{2}\sqrt{x}}{-1} =-\frac{3}{2}\cdot 2=-3$[/hide]

[quote]
number 1
[/quote]
Are you sure that L'Hopital's Rule was applied properly to the denominator? Remember, the derivative of a constant is 0.

[hide]
1. Use L'Hopital's Rule. Taking the derivative of both numerator and denominator, we get:
cos(x-pi/3)/(2sinx)
Plugging in pi/3 to this equation, we get 1/(sq. root 3) = [b](sq. root 3)/3[/b]

2. Simply plug in the x-value of the limit into the expression. On the right side of the product sign the pi/4 terms cancel out, giving us an answer of [b]0[/b]

3. The equation can be rewritten as
(x^(3/2)-8)/(4-x)
Use L'Hopital's Rule again, as plugging 4 into this expression also gives us 0/0. Therefore, we get
((3/2)x^(1/2))/-1 = -(3/2)x^(1/2). Plugging 8 into this expression, we get an answer of [b]-3[/b]. [/hide]

[hide]1.$\lim_{x\rightarrow \frac{\pi}{3}} \frac{\sin \left(x-\frac{\pi}{3}\right)}{1-2\cos x}=-\lim_{x\rightarrow \frac{\pi}{3}} \frac{\frac{\sin \left(x-\frac{\pi}{3}\right)}{x-\frac{\pi}{3}}}{\frac{f(x)-f\left(\frac{\pi}{3}\right)}{x-\frac{\pi}{3}}},$ where $f(x)=2\cos x.$

2. You can set $x-\frac{\pi}{4}=\theta.$

3. $\lim_{x\rightarrow 4} \frac{x\sqrt{x}-8}{4-x}=-\lim_{x\rightarrow 4} \frac{g(x)-g(4)}{x-4},$ where $g(x)=x\sqrt{x}.$[/hide]


Find all x fulfilling the inequality:

1.

$ |cosx|^{1\plus{}sinx\plus{}cosx} \geq 1$

2.

$ \sqrt {sinx} \plus{} \sqrt {cosx} > 1$

1. Equavalent to $ 1\plus{}sinx\plus{}cosx\le 0$ or $ (2n\plus{}1)\pi \le x\le (2n\plus{}\frac 32 )\pi$ ($ 0^0\equal{}1$).
2. Equavalent to $ 2n\pi <x<(2n\plus{}\frac 12 )\pi$.


induction
number theory proposed
number theory

Show that for every positive integer $ n$ the number $ (\sqrt{2}\minus{}1)^n$ is of the form $ \sqrt{m}\minus{}\sqrt{m\minus{}1}$ for some $ m \in \mathbb{N}$.

[hide]Call $ m_n \equal{} \frac {(\sqrt {2} \plus{} 1)^{2n} \plus{} (\sqrt {2} \minus{} 1)^{2n} \plus{} 2}{4}$.  Clearly, $ \sqrt {m_n} \equal{} \frac {(\sqrt {2} \plus{} 1)^n \plus{} (\sqrt {2} \minus{} 1)^n}{2}$ because $ (\sqrt {2} \plus{} 1)(\sqrt {2} \minus{} 1) \equal{} 1$, and clearly also, $ \sqrt {m_n \minus{} 1} \equal{} \frac {(\sqrt {2} \plus{} 1)^n \minus{} (\sqrt {2} \minus{} 1)^n}{2}$, leading to $ (\sqrt {2} \minus{} 1)^n \equal{} \sqrt {m_n} \minus{} \sqrt {m_n \minus{} 1}$.  It remains only to be proved that $ m_n$ is an integer.  Call $ y_n \equal{} (\sqrt {2} \plus{} 1)^{2n} \plus{} (\sqrt {2} \minus{} 1)^{2n} \equal{} (3 \plus{} 2\sqrt {2})^n \plus{} (3 \minus{} 2\sqrt {2})^n$, it suffices to show that it is an even integer which is not a multiple of $ 4$ for all non-negative integers $ n$.  For $ n \equal{} 0$ the result is clearly true with $ y_0 \equal{} 2$, and for $ n \equal{} 1$ the result is clearly true with $ y_1 \equal{} 6$.  If the result is true for $ n \minus{} 1,n$, then
\[ y_{n \plus{} 1} \minus{} 6y_n \equal{} (3 \plus{} 2\sqrt {2})^{n \minus{} 1}(9 \plus{} 8 \plus{} 12\sqrt {2} \minus{} 18 \minus{} 12\sqrt {2}) \plus{} (3 \minus{} 2\sqrt {2})^{n \minus{} 1}(9 \plus{} 8 \minus{} 12\sqrt {2} \minus{} 18 \plus{} 12\sqrt {2}) \equal{} \minus{} y_{n \minus{} 1},
\]
and since $ y_n,y_{n \minus{} 1}$ are even integers which are not multiples of $ 4$, then so is $ y_{n \plus{} 1} \equal{} 6y_n \minus{} y_{n \minus{} 1}$; it is clearly even, and if it was a multiple of $ 4$, then so would be $ 6y_n$, hence $ y_{n\minus{}1}$, contradiction.[/hide]

I conjecture that $ (\sqrt{N}\minus{}\sqrt{N\minus{}a})^k$ can be written as $ \sqrt{M}\minus{}\sqrt{M\minus{}a^k}$ for some $ M\in\mathbb{N}$. Can someone prove this?

[quote="Johan Gunardi"]I conjecture that $ (\sqrt {N} \minus{} \sqrt {N \minus{} a})^k$ can be written as $ \sqrt {M} \minus{} \sqrt {M \minus{} a^k}$ for some $ M\in\mathbb{N}$. Can someone prove this?[/quote]

The general form for $ M_k$ would then be

[hide]$ M_k \equal{} \frac {(\sqrt {N} \plus{} \sqrt {N \minus{} a})^{2k} \plus{} (\sqrt {N} \minus{} \sqrt {N \minus{} a})^{2k} \plus{} 2a^k}{4}$[/hide]

(see my post above), and then if $ a$ is even, we need to show that

[hide]$ y_k \equal{} (\sqrt {N} \plus{} \sqrt {N \minus{} a})^{2k} \plus{} (\sqrt {N} \minus{} \sqrt {N \minus{} a})^{2k}$ is a multiple of $ 4$ for all $ k\geq1$,
[/hide]

whereas if $ a$ is odd, we need to show that

[hide]$ y_k \equal{} (\sqrt {N} \plus{} \sqrt {N \minus{} a})^{2k} \plus{} (\sqrt {N} \minus{} \sqrt {N \minus{} a})^{2k}$ is an even number which is not a multiple of $ 4$ for all $ k\geq1$,
[/hide]

clearly true since

[hide]it satisfies the recursive equation $ y_{k \plus{} 1} \equal{} 2(2N \minus{} a)y_k \minus{} a^2y_{k \minus{} 1}$, and $ y_1 \equal{} 4N \minus{} 2a$, $ y_2 \equal{} 16N(N \minus{} a) \plus{} 2a^2$, hence always even, and multiple of $ 4$ or not iff $ a$ is even or odd, respectively, and this property is carried out from $ y_{k \minus{} 1},y_k$ into $ y_{k \plus{} 1}$, induction, done![/hide]


Are you going to do it if Treething makes a mock Mathcount (Sprint and Target, State/National Level)?

I would, but I'd already know the answers to be 69  :rotfl: .

I'd do it if the answers were not 69 for all of them.

[quote="i_like_pie"][quote="SplashD"]I would, but all the answers would end up to be 69 and there's really no point[/quote]
I hope the reason isn't the one I'm thinking of. :|[/quote]

Well, it's Treething's favorite number, not mine.

[quote="i_like_pie"][quote="SplashD"]I would, but all the answers would end up to be 69 and there's really no point[/quote]
I hope the reason isn't the one I'm thinking of. :|[/quote]
It is.


It's really easy but I'm just not completely sure

(x/3)
------
x+3

Can it be simplified?

Hint: [hide]Dividing is the same as multiplying by the recripocal.  In other words

(a/b)

------- = (a/b)(d/c)

(c/d)

[/hide]

I already got the answer which was x/(x(x+3))

And please remember tomorrow 6:00 EST, okay?

Do you mean x/(3*(x+3)) ?

Yes I will remember.


algebra
polynomial
quadratics
induction
Galois Theory
number theory proposed
number theory

a) Show that there exist polynomials that don't have rational roots, but have a root mod $n$ for any positive integer $n$.
 b) What is the possible degree of such a polynomial (b) is very difficult, IMHO and I don't have any elementary solution).

a)Well-known example: http://www.mathlinks.ro/Forum/viewtopic.php?t=22166
(confused polynomial in one variable with polynomial in 3 variables)

There was a similar topic some time ago concerning reducible instead of having roots ( http://www.mathlinks.ro/Forum/viewtopic.php?t=64273 ).

a) We can take $ (x^{2} \minus{} 2)(x^{2} \minus{} 17)(x^{2} \minus{} 34)$:
It has a solution $ \mod$ all odd primes $ p$ since at least one of the numbers $ 2,17,34$ is a square $ \mod p$. And this solution then raises to a solution to any power of $ p$ by e.g. Hensel's lemma.
Similar, it has a solution $ \mod$ any power of two.
By the Chinese remainder theorem, we finally get roots $ \mod$ all integers.

b) I also don't see an elementary way, but here's the one you probably had in mind:
Claim: the smallest degree is $ 5$.
An example is $ (x^{2} \plus{} 3)(x^{3} \minus{} 7)$: if $ \minus{} 3$ is a quadratic residue $ \mod p$ (odd $ \neq 3$), then we are done (first factor!) by Hensel.
If $ 3$ is not a qudratic residue, then $ p \equiv 2 \mod 3$ and all residues are cubes. Thus we are also done then.
If $ p \equal{} 2,3$, we are done by the second factor.

Now let's prove that there is none with smaller degree.

Let $ p(x)$ be such a polynomial with zeros $ S \equal{} \{a_{1},a_{2},...,a_{n}\}$ and let $ G$ be it's Galois group.
By Chebotarev's density theorem, we get that every $ g \in G$ must have a fixed point on it's operation on $ S$ (otherwise we take a $ g$ without fixed point and we get a lot of primes for that $ g$ is the Frobenius automorphism, thus it can't have linear factors).

This already excludes irreducible polynomials of prime degrees $ q$ (their Galois group will always have a cycle of length $ q$).
In fact, it excludes ever irreducible polynomial:
An irreducible polynomial causes $ G$ to act transitively on $ S$, giving the known $ \sum_{g \in G}\mathrm{Fix}(g) \equal{} |G|$. But $ \mathrm{Fix}(e) \equal{} |S| > 1$, thus there is a $ g \in G$ with $ \mathrm{Fix}(g) \equal{} 0$.

More general:
Let $ p(x) \equal{} p_{1}(x) p_{2}(x) ... p_{m}(x)$ (all factors irreducible with degree $ > 1$). Then we can't have that for all nonempty subsets $ A,B$ of $ \{1,2,...,m\}$ that the splitting fields of $ \prod_{a \in A}p_{a}(x)$ and $ \prod_{b \in B}p_{b}(x)$ have trivial intersection:
We have to show that there is an element from the Galois group without fixed points.
The condition tells us that the Galois group of $ p$ is the direct product of the Galois groups of the $ p_{i}$, which induction have an element without fixed point as we've shown before.
Thus we simply get an element without fixed point for the whole group by taking one for each factor.

But back to the initial problem:

We have to exclude a polynomial of type $ p(x)q(x)$, where the factors are irreducible of degree $ 2$.
If their splitting fields are disjoint, we are done by the previous. If it is the same, the nontrivial element from the Galois group has no fixed points.

PS: If we have a polynomial having the desired property (no rational roots, but roots $ \mod n$), we can always add a factor like $ (x^2\minus{}2)$ and get one with degree increased by $ 2$. As there are such polynomials of degree $ 5$ and $ 6$, there are such polynomials for all degrees $ \geq 5$ (and as shown, none for degrees $ <5$).

@above At the example of a degree of a polynominal you should think of (mod 2)


search
inequalities proposed
inequalities

[quote="Marius Damian"]From $x,y,z>0$ and $x(1-y)+y(1-z)=z(x-1)$, results at least one of $x,y,z$ is $\geq 1$.
Let $x \geq 1$. Then $x^{\dfrac {1}{y}} \geq 1$.
Hence $x^{\dfrac{1}{y}}+y^{\dfrac{1}{z}}+z^{\dfrac{1}{x}}\geq 1$.[/quote]

What a beautiful solution!! I amazed you. :first:

[quote="nickolas"]The  initial problem  should  be : $x,y,z>0$  and  $x(1-y)+y(1-z)+z(1-x)=0$. Prove that ¨
$x^{\frac{1}{y}}y^{\frac{1}{z}}z^{\frac{1}{x}}\geq1$
Am i right  Silouan?[/quote]

Yes I was inspired by this. This problem also posted by my but it is still unsolved.
Just search for <<like another one>> posted by me.


This problem was given to us by Dr. Alipour
I solved it in 5 minutes.
Suppose we have $n$ points in plane.
Suppose the bigest and smallest distance between points are $M$ and $m$
Prove:
\[ \displaystyle \frac Mm \geq \frac {\sqrt n - 1}{\sqrt 2}\]

i am suspicious...
i think it's almost trivial...
anyway, let's try to solve it.
if we draw circles of ray m/2 with center in every point, they won't intersect.
but all circles lie in a circle of ray (M+m)/2.
of course n*pi*m/4 <= pi(M+m)/4, which is M/m >= n-1.
but n-1 = (sqrt(n) + 1)(sqrt(n) - 1) >= (sqrtn - 1)/sqrt2.
am i totally wrong?

Following the lines of the answer of Ma_go, the result may be shapened to $ \frac M m > \frac {\sqrt {3}} 2 (\sqrt {n} - 1)$ by considering the disk $D$ with smallest radius $R$ which encloses the $n$ points.
It is a classical result that some two of the $n$ points form a diameter of $D$ or some three of the $n$ points lies on $D$ and are the vertices of an acute triangle. From that it is easy to deduce Jung's theorem which states that if the diameter of the set of the $n$ points is $M$ then $R \leq \frac M { \sqrt {3} }$, which leads to the above inequality.

Pierre.

Well my solution is very simple.
Consider the longest segment $AB$ and draw $L_1$ and $L_2$ perpendicular to $AB$ from $A$ and $B$.All points are in the region between $L_1$ and $L_2$ projection of any segment on $L_1$ has lengh at most $AB$ so there is a square with side length $AB$ that contain all points...


[b]3.[/b]  Suppose that $ x,y\in[0,\infty)$ satisfy $ 2x \plus{} y \equal{} 6$. Find the maximum and the minimum of $ 4x^2 \plus{} 3xy \plus{} y^2 \minus{} 6x \minus{} 3y.$

[b]4.[/b]  Suppose that $ x,y\in\mathbb R$ satisfy $ 4x^2 \minus{} 5xy \plus{} 4y^2 \equal{} 5$. Find the maximum and the minimum of $ x^2 \plus{} y^2$.

[quote="ifai"][b]3.[/b]  Suppose that $ x,y\in[0,\infty)$ satisfies $ 2x \plus{} y \equal{} 6$. Find the maximum and the minimum of $ 4x^2 \plus{} 3xy \plus{} y^2 \minus{} 6x \minus{} 3y.$[/quote]

$ P \equal{} 4x^2 \plus{} 3xy \plus{} y^2 \minus{} 6x \minus{} 3y \equal{} (2x \plus{} y)^2 \minus{} 3(2x \plus{} y) \minus{} xy$
$ \equal{} 18 \minus{} xy \equal{} 18 \minus{} 2x(3 \minus{} x)$ with $ 0\leq x\leq 3$ , yielding $ 18 \minus{} 2\cdot \frac {9}{4}\leq P\leq 18 \minus{} 2\cdot 0\Longleftrightarrow \boxed{\frac {27}{2}\leq P\leq 18}$

[quote="ifai"]4.[/b]  Suppose that $ x,y\in\mathbb R$ satisfies $ 4x^2 \minus{} 5xy \plus{} 4y^2 \equal{} 5$. Find the maximum and the minimum of $ x^2 \plus{} y^2$.[/quote]

Let $ x^2 \plus{} y^2 \equal{} r^2$,  $ 4x^2 \minus{} 5xy \plus{} 4y^2 \equal{} 5\Longleftrightarrow xy \equal{} \frac {4}{5}r^2 \minus{} 1$, since $ |xy|\leq \frac {1}{2}(x^2 \plus{} y^2)$,

we have $ \left|\frac {4}{5}r^2 \minus{} 1\right|\Longleftrightarrow \frac {1}{2}r^2\Longleftrightarrow \minus{} \frac {1}{2}r^2\leq \frac {4}{5}r^2 \minus{} 1\leq \frac {1}{2}r^2$,

yielding $ \boxed{\frac {10}{13}\leq x^2 \plus{} y^2\leq \frac {10}{3}}$

Alternative Solution:

Let $ x\equal{}r\cos \theta ,\ y\equal{}r\sin \theta$, $ 4x^2 \minus{} 5xy \plus{} 4y^2 \equal{} 5\Longleftrightarrow 4r^2\minus{}5r^2\sin \theta \cos \theta \equal{}5$

$ \therefore r^2\equal{}\frac{5}{4\minus{}5\sin \theta \cos \theta}\equal{}\frac{5}{4\minus{}\frac{5}{2}\sin 2\theta}$ Since $ \minus{}1\leq \sin 2\theta \leq 1$, yielding $ \boxed{\frac {10}{13}\leq x^2 \plus{} y^2\leq \frac {10}{3}}$


Hi everyone,

Just a post about a new "Mathcounts-style" middle-school contest on the web -- [url]http://www.NonemptySet.com[/url]. A couple friends and I put this together over the past few days...  two divisions (one at Chapter/State difficulty, one at State/National difficulty), 30 questions per round, one hour-long round every month, and a couple small prizes ;)

If anyone is interested, please give this a try. Our website is mostly functional, we've set up a java program to check answers, and we're ready for takers :) It's our first try at this, so there may be a couple rough spots, but we're hoping this will go well and be a good warm-up for the upcoming MathCounts season.

Also, we'd certainly welcome any feedback about the website,  format, question difficulties, articles, etc. from all you experienced problem solvers =)

Thanks!

[quote="SplashD"]Wow, this is pretty cool.

Can you submit your answers and registration information at the same time?[/quote]

Thanks :) We'd prefer if you could register first -- that way the whole ID number setup thing is easier to manage, and we can also get a feel for the number of participants we have early on.

On an unrelated note: there were typos on 14 and 19 of the Division A round last night, but they've since been fixed -- for anyone who might have printed the contest out before ~3:30est today, though, it'd be great if you could re-print now or check over the corrected versions online. We're also sending this notice out with the registration e-mails, and we'll definitely do a better proofread next time :oops:


limit
logarithms
integration
calculus
calculus computations

How did you do the third line, and evaluate the integral?

[quote="red_dog"]Let $ x_{n}=\frac{1}{n^{4}}\prod_{i=1}^{2n}(n^{2}+i^{2})^{\frac{1}{n}}$
Then $ \ln x_{n}=\ln\prod_{i=1}^{2n}(n^{2}+i^{2})^{\frac{1}{n}}-4\ln n=\sum_{i=1}^{2n}\ln(n^{2}+i^{2})^{\frac{1}{n}}-4\ln n$
$ =\sum_{i=1}^{2n}\;\frac{1}{n}\left(2\ln n+\ln\left[1+\left(\frac{i}{n}\right)^{2}\right]\right)-4\ln n=\sum_{i=1}^{2n}\frac{1}{n}\ln\left[1+\left(\frac{i}{n}\right)^{2}\right]$
$ \textbf{and\;\;\;so},\;\;\;\lim_{n\to\infty}\ln x_{n}=\int_{0}^{2}\ln (1+x^{2})\;\textbf dx=2\ln 5-4+2\tan^{-1}2$
$ \textbf{therefore},\;\;\;\lim_{n\to\infty}x_{n}=\textbf e^{(2\ln 5-4+2\tan^{-1}2)}=\boxed{25\;\textbf e^{(2\tan^{-1}2-4)}}$[/quote]

nicely done, red_dog !  :) however, i have [b]fixed [/b]the setting up
of the integral above... recall we have the index $ i$ running
from $ 1$ to $ 2n$.... so the limits of the integral will run from
$ 0$ to $ 2$.... do you agree ? :| the [i]final [/i]answer will be different
from the one you posted, obviously.

Sorry for my mistake and thank you for correcting me.


Hello!
I love maths and am thinking of studying it. But I still have some doubts. How big is the demand for mathematicians? And also how well paid are the jobs? I like working in a team with people. Is this mostly the case as a mathematician, or do you work secluded on your own? Is it quite likely that you'll end up in a field fairly unrelated to maths, like business consulting?
 appreciate help...

Basically studying math you do not learn any "concrete/specific" skill you rather learn problem-solving and handle complex, abstract matters. I mean if you like to stay in academia just follow the topic you enjoy most and are best at.

[url=http://www.courses-careers.com/graduate-careers/articles/mathematics.htm]Here[/url] is an example for actuary and banking. I mean basically any job where you need some quantitative and/or modelling skills. Statistical data analysis, software development and project management are ver popular too. As you see in industry math is mostly applied within some domain (application field) which can often be learnt much quicker than the other way around!

Politics also needs problem-solving. Former IMO participant [url=http://www.imo-official.org/participant_r.aspx?id=9974]Reinhard Höppner[/url] of the German Democratic Republic was minister president of one of Germany's states, i.e. Saxony-Anhalt, from 1994-2002.

As you know German you might know 2008 is [url=http://www.jahr-der-mathematik.de/]Germany's math year.[/url] Very recently there was the [url=http://www.math.uni-rostock.de/staunt/]STAUNT!-seminar[/url] where former German mathletes gave presentations about their short (or long) career since then.


geometry
geometric transformation
reflection
symmetry
incenter
geometry unsolved

The interior bisector of the angle $A$ of the triangle $ABC$ cuts the incircle $c$ at two points; the one closer to $A$ is denoted by $O_A$. Define the points $O_B$ and $O_C$ similarly on the internal bisectors of the angles $B$ and $C$, respectively. The circle $c_A$ is drawn about $O_A$ and it is touching the sides $AB$ and $CA$. Similarly, $c_B$ is about $O_B$ and it is touching the sides $BC$ and $AB$, finally the circle $c_C$ is about $O_C$ and it is touching the sides $CA$ and $BC$. Taken any two of the circles $c_A$, $c_B$ and $c_C$ consider their common external tangents different from the corresponding sides of the triangle, respectively.
Prove that these three tangents are concurrent.

I'll just sketch it:

We can show that all three tangents pass through the incircle of $X_aX_bX_c$.

Let $X_a,X_b,X_c$ be the contacts between the incircle of $ABC$ and $BC,CA,AB$ respectively. Also, let $X_{uv}$ be the intersection between $X_uX_v$ and $O_vO_t$ for all permutations $(u,v,t)$ of $(a,b,c)$.

Now show by an angle chase, that, for example, the common tangent of $c_B,c_C$ passes through $X_{ba}$. This is easy to do, since $c_C$ is also tangent to $X_bX_a$ and thus to $X_{ba}X_a$, so the tangent fro $X_{ba}$ to $c_C$ is the reflection of $X_{ba}X_a$ in $O_aO_c$, and from here it's easy. By symmetry, the common tangent of $c_B,c_C$ also passes through $X_{ca}$.

After this, prove that $X_{ba}X_{ca}$ passes through the incenter of $X_aX_bX_c$. This setting has appeared before in many problems on the forum.


Ramsey Theory
combinatorics unsolved
combinatorics

what is ramsey number please tell me about it

In http://www.math.aau.dk/research/reports/R-99-2012.ps David Cariolaro proved that $R(3,6)\leq 18$. I have not yet found a graph to prove that $R(3,6)>17$ but I believe that is very straightforward.

How can I open the file with ps ?I have not such a program...

Install [url=http://www.cs.wisc.edu/~ghost/]Ghostview[/url] in your computer.

To finish the proof that $R(3,6)>17$, see [url=http://math.stackexchange.com/questions/3333/how-to-construct-17-vertexed-graph-for-ramsey-number-r3-6-18]this[/url] link.


The wheel has radius $ x $ in, the gear has radius $ y $ in, and the circle connected to the gear by a chain and is connected to the pedals has radius $ z $. If the biker makes $ c $ revolutions of the pedals per second, what is the speed of the bike in mph?

Yes, please - I don't see what you're saying.

[hide="Im soo dumb"]
When in $1$ sec $y$ moves $2{\pi}y$,
$x$ moves $2{\pi}x$
But when $z$ moves $2%Error. "pizc" is a bad command.
$
$x$ also moves the same amount.
So,
$\dfrac{2{\pi}y}{2{\pi}zc}=\dfrac{2{\pi}x}{a}$
where we need to find $a$,
So speed=$\dfrac{2{\pi}xzc}{y}$
[/hide]

[quote="riddler"][hide="Im soo dumb"]
When in $1$ sec $y$ moves $2{\pi}y$,
$x$ moves $2{\pi}x$
But when $z$ moves $2%Error. "pizc" is a bad command.
$
$x$ also moves the same amount.
So,
$\frac{2{\pi}y}{2{\pi}zc}=\frac{2{\pi}x}{a}$
where we need to find $a$,
So speed=$\frac{2{\pi}xzc}{y}$
[/hide][/quote]
good, I just wanted it mph

[quote="royalfire"][u]20xz(pi)c miles[/u]
76y hours
sorry LaTeX is giving me problems  :oops:[/quote]

I don't know where 20 and 76 came from, but here's the LaTeX'ed version:

$\frac{20xz \pi c \text{ miles}}{76y \text{ hours}}$

EDIT: ...Uh, LaTeX is giving me problems too.


geometry
3D geometry
tetrahedron
octahedron
icosahedron
topology
combinatorics proposed

Some edges are painted in red. We say that a coloring of this kind is [i]good[/i], if for each vertex of the polyhedron, there exists an edge which concurs in that vertex and is not painted red. Moreover, we say that a coloring where some of the edges of a regular polyhedron is [i]completely good[/i], if in addition to being [i]good[/i], no face of the polyhedron has all its edges painted red. What regular polyhedrons is  equal the maximum number of edges that can be painted in a [i]good[/i] color and a [i]completely good[/i]? Explain your answer.

Here are the maximums for a few regular polyhedra:
[hide]Tetrahedron: 4
Cube: 8
Octahedron: 8
Explanation: for each of these, I can visually see the coloring in my head to prove it works. To prove they are the maximum possible values, I provide the following arguments:
Tetrahedron: Each edge corresponds to 2 vertices, so there must be 2 uncolored edges for the 4 vertices
Cube: Each edge corresponds to 2 vertices, so there must be 4 uncolored edges for the 8 vertices
Octahedrom: Each edge corresponds to 2 faces, so there must be 4 uncolored edges for the 8 faces[/hide]
I did not find a solution for an icosahedron or dodecahedron.

Not that interesting a problem - since quite complicated definitions, and only $5$ objects to apply to! I would prefer some more general statement for all convex polyhedra, like the classical

[color=#0000FF]Theorem. If all edges of a convex polyhedron are oriented, and no vertex is such that all incident edges have same orientation, then at least one facet of the polyhedron (in fact at least two) is such that its border is an oriented cycle![/color]

Notice that this result is not anymore true if the "polyhedron" is toroidal (i.e. its surface is homeomorphic with a torus). Think of a $\mathbb{Z}_m \times \mathbb{Z}_n$ toroidal lattice, oriented "up" and "to the right".


logarithms
calculus
derivative
integration
calculus computations

I'm on vacation right now, and this computer doesn't have adobe reader, so I cannot figure out how to use latex.  I'll try my best to make it easy to understand, though.

I need to find the antiderivative of:

(dx)/(x^1/3) which is:

1/(x^1/3) dx

Is the antiderivative ln(x^1/3) ?  That's what I thought it would be, but there seems to be something missing?

What you're forgetting is the chain rule.

$\frac{d}{dx}\ln\left(x^{1/3}\right)=\frac1{x^{1/3}}\frac{d}{dx}\left(x^{1/3}\right)$

$=\frac1{x^{1/3}}\cdot \frac13x^{-2/3}=\frac1{3x}.$

So that's not the antiderivative. (And notice that I could have done that derivative calculation more efficiently by first noting that $\ln\left(x^{1/3}\right)=\frac13\ln x.$)

Your mistake is a very common one among beginning calculus students, but it is also a serious mistake. Two points:

1. Do [b]not[/b] say that $\int\frac1{\rm{something}}\,dx=\ln(\rm{something}).$

You must make a substitution, and if the substitution gets you to $\int\frac{du}u,$ [b]then[/b] you have a logarithm.

Integration by substitution is "the chain rule, read backwards."

2. Don't forget the general rule for the integration of powers:

$\int x^m\,dx=\frac{x^{m+1}}{m+1}+C$ for every $m$ [b]except for[/b] $m=-1.$

The $-1$ power is special, and gives the logarithm.

---

$\int\frac{dx}{x^{1/3}}=\int x^{-1/3}\,dx=\frac1{\frac23}x^{2/3}+C =\frac32x^{2/3}+C$


$ x,y,z$ is positive real numbers. Prove that
$ \sum \frac{x^2}{y^3\plus{}z^3\plus{}xyz}\ge \sqrt[3]{\frac{x^3\plus{}y^3\plus{}z^3}{x^3y^3\plus{}y^3z^3\plus{}z^3x^3}}$

[quote="nguoivn"]My first hint is:
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \geq\ \sqrt [3]{\frac {x^3 \plus{} y^3 \plus{} z^3}{x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3}}$
My second hint is:
$ \sum \frac {x^2}{y^3 \plus{} z^3 \plus{} xyz}\ge \ \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$
But both of them seem not true  :wink:[/quote]

What do you mean?  :maybe:


Find the exact value of $\sin{18}$ (18 in degrees).


conics
hyperbola
Asymptote
function
algebra
rational function

1) I've derived the hyperbola formula $ \frac{y^{2}}{a^{2}}-\frac{x^{2}}{c^{2}-a^{2}}= 1$ from the hyperbola's constant definition. However, I cannot find a way to prove that $ c^{2}-a^{2}= b^{2}$ where $ a$ is half the vertical axis, b is half the horizontal axis and c is the distance from either focii to the center. How on Earth do you prove this?

2) Also, how do prove that the function's asymptotes are $ y = \pm \frac{a}{b}x$ ?

3) For the rational function $ y = \frac{a}{x}$, what are the co-ordinates of the focii?

What eqn is that?

I thought hyperbola is

$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1$

I typed up soln. to find asymp. but when I submit it asked for login.. and now its deleted!!

But this is how you do it

Make $ y^{2}$ the subject from the above eqn

$ y^{2}$ = $ \frac{b^{2}}{a^{2}}(x^{2}-a^{2})$

$ \div$ $ x^{2}$ then examine what happens to $ \frac{y^{2}}{x^{2}}$ as x approaches infinity then make y the subject and you should get your eqn

I gave the formula for a vertical hyperbola while you gave the one for a horizontal one. I'll consider what you said though, thanks  :lol:

Edit: If the hyperbola is written in the form

$ y = \pm a \sqrt{\frac{x^{2}}{b^{2}}+1}$ and we have our possible asymptote $ y = \pm \frac{a}{b}x$

now suppose that  $ \pm a \sqrt{\frac{x^{2}}{b^{2}}+1 }= \pm \frac{a}{b}x$

$ \frac{a^{2}x^{2}}{b^{2}}+a^{2}= \frac{a^{2}x^{2}}{b^{2}}$

$ a^{2}= 0$

proof by contradiction.

The problem is, this doesn't prove that the asymptote isn't simply any random locus of points. For instance, I could say the asymptotes are y = 0 and y = -0.0001. The hyperbola never touches these lines, so by this proof method the lines y = 0 and y = -0.00001 could very well be asymptotes (even though they aren't).

Suffice to say, I'm still stuck  :wallbash_red:

I understand now. For the statement

$ \frac{y^{2}}{x^{2}}= \frac{b^{2}}{a^{2}}(1-\frac{a^{2}}{x^{2}})$

if $ \frac{a^{2}}{x^{2}}$ = 0 then $ y = \frac{b}{a}x$. However, $ \frac{a^{2}}{x^{2}}\neq 0$ because the numerator has a preset value $ > 0$.

Thanks a lot for the help  :D

Does anyone have an answer for 1 and 3  :huh:

Yes your correct that $ a>0$ so $ \frac{a^{2}}{x^{2}}$ never equals 0. But the value is so small that it is CONSIDERED AS 0.

for example $ \frac{10}{999999999}$ is approaching 0 so it is considered 0

But what about $ \frac{y^{2}}{x^{2}}$ isnt that also approaching 0??
sry Im stuck with 1 and 3 :D


While running to a bus stop by my high school, i thought of this question:

Say bus $A$ comes every twenty minutes and bus $B$ comes every 25 minutes (on time -- like that's ever going to happen in real life). I get out of the school at $5$-o'clock, and while walking the five minute walk to the stop I do not see the bus pass by. I have no idea at what time the bus will come (I forgot my schedule), so I wait. If i can take either to get home, whats the probability that my wait will be less than $5$ minutes?

[hide]If the time you have to wait for the buses is uniformly random distributed, the probability that you have to wait less than five minutes for each bus is $\frac{5}{20}$ and $\frac{5}{15}$.
So the probability that you'll have to wait less than five minutes is
$1-(1-\frac{5}{20})(1-\frac{5}{15})=\frac{1}{2}\equiv 50\%$.
So, five minutes is the median.[/hide]


probability
number theory
modular arithmetic
\/closed

I talked my parents into letting me spend my summer studying along with a part time job, so I sighned up for the intermediate number theory and intermediate counting classes.

I want to get the most out of these classes, so here was my plan for getting ready:

-finish up intro to counting book ( on second last chapter )
-review all problems I didn't get the first time around in the counting book
-do all the problems in aops 1 on counting and probility

go-back over the number theory section in aops 1

read through and do the number theory and counting sections in aops 2, do as many problems as i can, try and get a feel for what the courses will be exploring.

while my number theory is not bad ( the aops 1 stuff is not too bad ), I plan on grabbing the intro to number theory book as soon as it comes out, and just doing it on the side to make sure I don't miss any number theory fundmentals

this is my plan, any comments/suggestions on how to prepare best?

Hmm, I think you don't have to prepare quite as much as you're doing. Just make sure that you know enough material to completely understand the course (look at the pre-test). Otherwise, don't busy yourself by trying to learn every little thing about number theory or counting and probability you can prior to the course.

From what I've seen, I think you should know all the basics of modular arithmetic and factorization/integer-type stuff for the Intermediate Number Theory course, and you should know the basic probability facts and know about permutations and combinations for the Intermediate Counting and Probability course.


Which state do you think will be the most improved team? Also, what are your rankings for all 57 teams?

West Virginia will be the most improved team.  We're going to go from 25th to 1st in 1 year.  BTW, Leet, what does 'ftw' mean...

1. Nats people really shouldn't waste time on such idle speculation.
2. We already are wondering about how teams will rank, we needn't wonder who will be most improved in a separate thread.

Ergo, 3. Locked.


Hi.

I know how to multiple matricies and vectors together when i am dealing with two.

However, given
[code]cAb
[/code]
where c and b are vectors and A is a matrix, would you calculate cA first and then multiply by b, or calculate Ab then mulitply by c, or is there some other way of multiplying these three at once

cheers

Matrix multiplication is associative so it doesn't matter which way you multiply them, either one will work.


function
algebra
polynomial
real analysis
real analysis solved

Let $n\geq 2$. Find all real continuous functions f such that $\sum_{i=1}^{n}{f(x_i)}-\sum_{1\leq i<j\leq n}{f(x_i+x_j)}+...+(-1)^{n-1}f(x_1+x_2+...+x_n)=0$ for all $x_1,...,x_n$ real numbers.

Answer: polynomials of degree less than $n$ which are zero at 0.

First, by setting all $x_i=0$, $f(0)=0$. Rewrite the formula as
$f(0)-\sum_{i=1}^{n}{f(x_i)}+\sum_{1\leq i<j\leq n}{f(x_i+x_j)}+...+(-1)^nf(x_1+x_2+...+x_n)=0$
By setting $x_2=x_3=\dots=x_n=1,\, x_1=k$, we get that the $(n-1)$-th difference of the sequence $\{f(k)\}$ is constant. Therefore, the restriction of $f$ to the integers is a polynomial of degree less than $n$. Interpolating this sequence with rational numbers, we have that $f(\frac km)$ is also a polynomial of degree less than $n$; since it is equal to the previous polynomial infinitely often, it is the same polynomial.

Now $f(x)=p(x)$ for some polynomial $p$ of degree less than $n$ and all rational $x$. By continuity, $f(x)=p(x)$ everywhere.
To prove that all such polynomials work, note that $p(x+y)-p(x)=yq(x,y)$ for some polynomial $q$ of degree 1 less than the degree of $p$. The expression $f(0)-\sum_{i=1}^{n}{f(x_i)}+\sum_{1\leq i<j\leq n}{f(x_i+x_j)}+...+(-1)^nf(x_1+x_2+...+x_n)$ is a polynomial of degree less than $n$ in all of these variables, and by the above it is divisible by $x_1x_2x_3\dots x_n$, so it is zero.


Given that $ A$ is the set of positive integers, $ B$ is the set of whole numbers less than or equal to $ 20$, and $ C$ is the set of even integers, what is the sum of the numbers in $ A \cap B \cap C$?

The $ \cap$ symbol represents an intersection, or the items in common between two lists. The only numbers in common between the three sets given are the positive even numbers $ 20$ and below, which are $ 2$, $ 4$, $ 6$, $ 8$, $ 10$, $ 12$, $ 14$, $ 16$, $ 18$, and $ 20$. Adding them up, $ 2\plus{}4\plus{}6\plus{}8\plus{}10\plus{}12\plus{}14\plus{}16\plus{}18\plus{}20\equal{}2(1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9\plus{}10)\equal{}2(\frac{10(10\plus{}1)}{2})\equal{}10(11)\equal{}\fbox{110}$.


The sum of the first four terms of an arithmetic sequence is $ 10$. If the fifth term is $ 5$, what is the sixth term?

Call the second term $ x$ and the common difference $ d$.  Then, the fifth term would be $ x \plus{} 3d$, and the sum of the first four terms would be $ 4x \plus{} 2d$.  So we have

$ 4x \plus{} 2d \equal{} 10$
$ x \plus{} 3d \equal{} 5$

Now, we solve, and we get

$ 10d \equal{} 10$
$ d \equal{} 1$

Add that to $ 5$, and we get $ \boxed{6}$.

Actually thats wrong :wink: ...

I just realized that the sequence is: $ 1,2,3,4,5,6,...$
So the $ 6$th term is $ 6$.

Yeah, sorry, edited while you posted... :wink:


probability geometry rectangle trigonometry perimeter AMC

Could someone please start a  solutions page for the AMC E? I really need to find the answers, but I don't understand.

1. The sum of the digits of the year 1999 is 28. Let abcd be the next smallest (i.e. next earliest) year where the sum of the digits equals 28 ( a, b, c, and d are digits, not necessarily distinct). Compute  a*b*c*d.

Well, you know a = 2. You want to minimize b, so, maximize c and d for now. You have 2,9,9, and b. 28-9-9-2 = 8. So, abcd = 2899. 2*8*9*9 = 1296. (A)

2. You know the formula for sum of interior angles is 180(n-2). SO 180(n-2) = 2700. Thus 180n = 3060, and n = 17. (E)

3. This requires too much explaining of my behalf. XD. Someone else will be glad to help.

4. You can factor this as a difference of squares. By doing so, you get 17, 241, 13, 5, 3, and 2. Those summed = 286. (A)

5.This is just conditional probability. Probability (3 consecutive rolls are prime and same) / P(3 consecutive rolls are prime). You get this to be (1/72) / (1/8) = 1/9. (D)

6. My solution is horrible, but after some trial and error, I realized that most of the terms cancel, so you are left with 16 as the remaining sum of coefficients. (C)

7. This is block walking. 4 by 3 rectangle represented. So... you have C(7,3) or C(7,4) which is 7 paths choose 3 up or choose 4 right. C(7,3) = 35 (A)

That's all for now...

19)  Label the triangle ABC with AB=AC=10, BC=12, M be the midpt of BC, BMA is a 3-4-5 triangle, MA=8. Let the center of the incircle be O, the radius of the incircle=[ABC]/semi-perimeter=3 so MO=3, AO=5. Let the circle tangent to AB, AC, and circle O have center at P and tangent to AC at point D. APD is a 3-4-5 triangle since it's similar to AMC. AP=8-6-r=2-r, PD=r, and (2-r)/r=5/3
solve for r, we get r=3/4

20) Let f(x)+x^2=g(x),
since g(x) will also be a quartic and have leading coefficient of 1, it can have at most 4 roots.
so all roots to g(x) must be (-1, 2, -3, 4)
g(x)=(x+1)(x-2)(x+3)(x-4)

g(1)=24=f(1)+1,     f(1)=23

I ve got some time so I ll post some solutions.

8.
$10584000=2^63^35^37^2$

The key in this question is that they are teenagers.  Looking at the 7's they can only be pared with a single six or else they wont fall in the teens.  So two ages are 14 and 14.  This leaves $2^43^35^3$.  The 5's can only go with single 3's.  This makes 15, 15, and 15.  Finally the 2's all go together to make 16.

$14+14+15+15+15+16=89$  E

9.  You can just look at the answer choices and see that 4 are obviosly too big.
.1001=1/2+...  1/2 is bigger than 1/5 to start with
.0101=1/4+...   1/4 is bigger than 1/5
.00111=1/8+1/16+1/32+....=.21875+(something)   too big
.01101=1/4+... too big

this leaves B which is too big

Thanks guys. I'm starting to understand some of the problems. Pleeeeeeeeeeeeeeeeease keep the solutions coming. Thanks :) .

10. In order to go to the 7th game, out of the first six games, Team A must win 3 and Team B must win 3.  There are $\binom{6}{3} = 20$ ways to choose which three games Team A will win, and the probability of any given way occurring is $\left( \frac{3}{4} \right)^3 \left( \frac{1}{4} \right)^3 = \frac{27}{4096}$.  So the answer is $\frac{27}{4096} \cdot 20 = \frac{135}{1024}$.  (C)

11. Dividing both sides of the first equation by $\cos x$, we get $\tan x = 3$.  Draw a right triangle with legs 1 and 3.  Clearly, the hypotenuse will have length 10, and so $\sin x = \frac{3}{\sqrt{10}}$ and $\cos x = \frac{1}{\sqrt{10}}$; $\sin x \cos x = \frac{3}{10}$. (E)

12. Rearranging, we have $x^2 - y^2 + yz - xz = 0$, or $(x-y)(x+y) - (x-y)z = 0$, or $(x-y)(x+y-z) = 0$.  This will occur either if $x-y = 0$ or $x+y-z=0$; that is, $x=y$ or $x+y=z$.  In the first case, we can have (x,y,z) = (1,1,1-5), (2,2,1-5), ... (5,5,1-5), for a total of 25 solutions.  The second case contributes 8 more solutions: (1,2,3), (1,3,4), (1,4,5), (2,3,5), and the other four solutions where x and y are switched around.  In all, there are 25+8 = 33 solutions.

13. $\frac{2}{x} = \frac{y}{3} = \frac{x}{y}$.  Using the left half of the equation, we get $y = \frac{6}{x}$.  Subsituting this for $y$ in the right half ($\frac{y}{3} = \frac{x}{y}$), we have $\frac{6}{3x} = \frac{x^2}{6}$, or $3x^3 = 36$, or $x^3 = 12$.  (B)

14.  See diagram.


Let ABC be an acute triangle and let M be the
midpoint of BC. There exists a unique interior point N such that
ABN = BAM and ACN = CAM. Prove that BAN =
CAM.

Let Ax be the symedian of triangle ABC. N' lies on Ax such that $ \angle ABN'\equal{}\angle MAB\equal{}\angle N'AC$.Let F, G be the projection of N' onto $ AB, AC$ then $ \frac{AB}{AC}\equal{}\frac{FN'}{GN'}\equal{}\frac{BN'}{AN'}$ so $ \Delta ABN'\sim \Delta CAN'$
We get $ \angle N'CA\equal{}\angle N'AB\equal{}\angle MAC$
Therefore $ N'\equiv N$. We are done!


Hey guys, I am in search for the 2000 , 2002, and 2004 School Competitions. Can you send it to me if you have the solutions/answer key/ or test for it ? PM me and I will give you further information for contact.

No, sorry, I don't have any of those. I wish I could help you, but maybe somebody may post a site of their location or type up the answer key quickly and send it to you. Good luck and I hope you do find those! Ciao.


When I printed the PDF the margins were huge.  How do I fix this?

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#layout]Click here to read about margins[/url].

You might want to spend an hour or two looking around the pages in the guide to get to know what's there - you'll probably learn a lot of methods to do things you'll want to do in the future.

Good point, sorry I probably should have read the guide in detail before I posted questions, my bad.


hi, please vote again for the speed of the site :)

Damn, just after I voted for "good for a maths site", the next page needed 1 minute to load! ([ Page generation time: 55.6705s (PHP: 67% - SQL: 33%) ])

  dg

well there seems to be a small bug with the latex environment ... we'll see if it can be fixed .. :) :P

Wow, tonight I saw that MathLinks was being moved to a more powerful server... and I must say that the result is great: [i]much[/i] shorter loading time! Two thumbs up, dudes!   :cool:

indeed we have moved to a new location. I expect all simple page generation times under 1 second :)

is it so ?

people with Cable, DSL or T1/T3 connections should feel it especially.

oh well it's 2 sec

anyway I care if it's 1 or 2 sec

as long as it's not 10+ seconds as it's been some while ago :(
I'll be happy :D