Θα προσπαθήσω να εξηγήσω εδώ ποιοί θετικοί ακέραιοι γράφονται σαν άθροισμα δύο ακέραιων τετραγώνων και με πόσους τρόπους μπορεί να γίνει αυτό.

Το πρώτο λήμμα είναι αρκετά απλό και η απόδειξή του αφήνεται στον αναγνώστη.

Λήμμα 1: Αν $ n \equiv 3 \bmod 4$ τότε ο ν δεν μπορεί να γραφτεί σαν άθροισμα δύο τετραγώνων.

Το επόμενο λήμμα μάλλον είναι γνωστό σε αρκετούς. Μια απόδειξή του βασίζεται σε τετραγωνικά ισοϋπόλοιπα για τα οποία ίσως να μιλήσουμε άλλη φορά.

Λήμμα 2: Αν $ p \equiv 3 \bmod 4$ πρώτος και $ p | m^2\plus{}n^2$, τότε $ p|m$ και $ p|n$. (Άρα $ p^2 | m^2\plus{}n^2$.)

Πόρισμα 3: Αν ο αριθμός n γράφεται σαν άθροισμα δύο τετραγώνων, και $ p \equiv 3 \bmod 4$ πρώτος που διαιρεί τον n τότε υπάρχει κ ώστε ο $ p^{2k}$ διαιρεί τον n αλλά ο $ p^{2k\plus{}1}$ δεν τον διαιρεί.

Η απόδειξη είναι σχετικά εύκολη και αφήνεται στον αναγνώστη.

Θα δούμε σε λίγο ότι ισχύει και το αντίστροφο. Για αρχή

Θεώρημα 4 (Fermat): Κάθε πρώτος της μορφής $ p \equiv 1 \bmod 4$ μπορεί να γραφτεί σαν άθροισμα δύο τετραγώνων

Μπορείτε να βρείτε αρκετές αποδείξεις  [url=http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares]εδώ[/url]

Παρατήρηση 5: Αν ο n και ο m μπορούν να γραφούν σαν αθροίσματα δύο τετραγώνων (επιτρέπουμε την χρησιμοποίηση του 0), τότε το ίδιο ισχύει και για τον nm

Η απόδειξη υπάρχει και στο παραπάνω link, αλλά την βάζω διότι είναι σύντομη, εξαιρετικά όμορφη και θα χρησιμεύσει αργότερα.

Απόδειξη: Αν $ n \equal{} a^2 \plus{}b^2$ και $ m \equal{} c^2 \plus{} d^2$ τότε $ nm \equal{} (ac\plus{}bd)^2 \plus{} (ad\minus{}bc)^2$.

Επειδή ο 2 γράφεται σαν άθροισμα δύο τετραγώνων, και κάθε τετράγωνο γράφεται σαν άθροισμα δύο τετραγώνων έχουμε

Θεεώρημα 6: Ο αριθμός n γράφεται σαν άθροισμα δύο τετραγώνων αν και μόνο αν όλοι οι διαιρέτες του της μορφής  $ 3 \bmod 4$ βρίσκονται σε άρτια δύναμη στην παραγοντοποίηση του n

Τώρα ξέρουμε ποιοί αριθμοί γράφονται σαν άθροισμα δύο τετραγώνων αλλά με πόσους τρόπους μπορεί να γίνει αυτό;

Θεώρημα 4β: Κάθε πρώτος της μορφής $ p \equiv 1 \bmod 4$ μπορεί να γραφτεί [b]με μοναδικό τρόπο[/b] σαν άθροισμα δύο τετραγώνων.

Για παράδειγμα, ο μόνος τρόπος να γράψουμε το 1997 ως άθροισμα δύο τετραγώνων είναι $ 1997 \equal{} 34^2 \plus{} 29^2$.

Το $ 2 \cdot 1997^2$ που μας απασχόλησε [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265010]σε άλλη άσκηση[/url] μπορεί να γραφτεί σαν $ 2 \cdot 1997^2 \equal{} (1^2 \plus{} 1^2)(29^2 \plus{} 34^2)(29^2 \plus{} 34^2)$ και χρησιμοποιώντας τον τύπο στην απόδειξη της παρατήρησης 5 βρίσκουμε $ 2 \cdot 1997^2 \equal{} (1^2 \plus{} 1^2)(29^2 \plus{} 34^2)(29^2 \plus{} 34^2) \equal{} (1^2 \plus{} 1^2)(1997^2 \plus{} 0^2) \equal{} 1997^2 \plus{} 1997^2$, $ 2 \cdot 1997^2 \equal{} (1^2 \plus{} 1^2)(29^2 \plus{} 34^2)(29^2 \plus{} 34^2) \equal{} (1^2 \plus{} 1^2)(1972^2 \plus{} 315^2) \equal{} 2287^2 \plus{} 1657^2$.

Το θεώρημα λέει πως αν δουλέψουμε όπως πιο πάνω τότε μπορούμε να βρούμε όλους τους τρόπους που γράφεται ένας αριθμός σαν άθροισμα δύο τετραγώνων. (Δεν είμαι σίγουρος αν αυτό το θεώρημα είναι του Gauss ή όχι. Μπορεί όμως να αποδειχθεί χρησιμοποιώντας Gaussian integers.)

Μπορεί κάποιος να μετρήσει με πόσους τρόπους μπορεί να γίνει αυτό αλλά είναι λίγο πολύπλοκο και θέλει αρκετή προσοχή. (Για παράδειγμα, στην άσκηση που μας ενδιέφερε είχα πει πως γίνεται με τρεις τρόπους επειδή μετρούσα το $ 1997^2 \plus{} 1997^2$ μία φορά και το  $ 2287^2 \plus{} 1657^2$ δύο φορές.)

Ελπίζω αυτά να βοηθήσουν σε παρόμοια προβλήματα.

συγνωμη για την καθυστερηση της απαντησης αλλα ειχα καιρο να μπω( ..πανελληνιες)
Το αρθρο ειναι πραγματικα ωραιο και πολυ χρησιμο!  :) Μπραβο.


Can someone do this step by step for me to learn it:

Write a costrunction for the problem. Assume that the relationship is linear.
Fixed cost, $ \$500$ ; 5 items cost $ \$3210$ to produce

[color=green][Edited by moderator to clean up non-TeX dollar sign.][/color]

[quote="ArmiAldi"]Can someone do this step by step for me to learn it:

Write a costrunction for the problem. Assume that the relationship is linear.
Fixed cost, $ 500 ; 5 items cost$3210 to produce[/quote]

JRav,

I think he means, find a general formula for the cost of $ n$ items, if there are fixed costs of $ 500$ and $ 5$ items cost $ 3210$

Answer

Clearly the 3210 contains 500 fixed costs, so 5 items cost $ 5 \times$ cost of one item $ \plus{}500$.

Therefore 1 item is $ \frac{3210\minus{}500}{5} \equal{} 542$

Therefore $ n$ items cost $ 542n\plus{}500$

Is that what he wanted?

[quote="ArmiAldi"]Want to tell u thank You Simon that is what i wanted and now what will be different if  instead   500 fixed cost we have  500 marginal cost?[/quote]

Well, what is a marginal cost?

How can we think about it?


My favorite is probably MATHCOUNTS, though I kind of like the AMC.

I'll vote for the only one I've ever tried.

Use Google.

Or read the FAQ in the AMC forum.

Ouch, my school only held the AMC 12, so I had to take it. I couldn't do the pre-calc questions :( (I'm in algebra 2)

I took the 10...

Missed half a day of school to go do math! I was happy.


inequalities
vector
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers, and let $ f(x)\equal{}\sum_{cyc} a^x(a\minus{}b)(a\minus{}c)$.

1.    Find the greatest real number $ p$ such that for all $ r>1$, $ \left(a^p\plus{}b^p\plus{}c^p\right)^{\frac{1}{p}}f(r)\geq f(r\plus{}1)$.

2.    Find the greatest real number $ p$ such that for all real nymbers $ r,k$ such that $ \frac{r}{k}\geq1$, $ \left(a^{pk}\plus{}b^{pk}\plus{}c^{pk}\right)^{\frac{1}{p}}f(r)\geq f(r\plus{}k)$.

[quote="bakerbakura"]Let $ a,b,c$ be nonnegative real numbers, and let $ f(x) \equal{} \sum_{cyc} a^x(a \minus{} b)(a \minus{} c)$.

1.    Find the greatest real number $ p$ such that for all $ r > 1$, $ \left(a^p \plus{} b^p \plus{} c^p\right)^{\frac {1}{p}}f(r)\geq f(r \plus{} 1)$.

2.    Find the greatest real number $ p$ such that for all real nymbers $ r,k$ such that $ \frac {r}{k}\geq1$, $ \left(a^{pk} \plus{} b^{pk} \plus{} c^{pk}\right)^{\frac {1}{p}}f(r)\geq f(r \plus{} k)$.[/quote]

It only makes sense to ask for the [b]smallest [/b]$ p$ in the power mean. In both cases, it is true for $ p \equal{} 1$ as just seen [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=317772]here[/url].
To show that $ p \equal{} 1$ is smallest  possible, put  $ c \equal{} 0$, then we have $ f(x) \equal{} (a \minus{} b)(a^{r \plus{} 1} \minus{} b^{r \plus{} 1})$.

Thus (1) becomes $ (a^p \plus{} b^p)^{\frac1p}\ge\frac {a^{r \plus{} 2} \minus{} b^{r \plus{} 2}}{a^{r \plus{} 1} \minus{} b^{r \plus{} 1}}$. For $ r\to1$, the RHS tends to $ a \plus{} b \minus{} \frac {ab}{a \plus{} b}$, but $ a^p \plus{} b^p\ge \left(a \plus{} b \minus{} \frac {b}{1 \plus{} \frac ab}\right)^p$ is false for $ p < 1$ if we put $ b \equal{} 1$ and make $ a$ big enough.

For (2) we get essentially the same after replacing $ a^k,b^k$ by $ a,b$ and then putting $ r \equal{} k\to\pm\infty$.

Not very nice with all those limits, but so it goes with counterexamples sometimes  :blush:

Actually, there is no power mean above.
From Jensen's Vector Inequality, $ g(p)\equal{}(a^p\plus{}b^p\plus{}c^p)^{\frac{1}{p}}$ is decreasing, so that we [b]are[/b] actually looking for the [b]greatest[/b] $ p$.

[quote="bakerbakura"]Actually, there is no power mean above.
From Jensen's Vector Inequality, $ g(p) \equal{} (a^p \plus{} b^p \plus{} c^p)^{\frac {1}{p}}$ is decreasing, so that we [b]are[/b] actually looking for the [b]greatest[/b] $ p$.[/quote]
Oh I’m sorry. I mixed it up with $ 3\left(\frac {a^p \plus{} b^p \plus{} c^p}3\right)^{\frac {1}{p}}f(r)\geq f(r \plus{} 1)$ :blush:
But I see that in fact, for $ p > 1$ the correct version doesn't hold any more.
Same counterexample: $ c \equal{} 0$, $ r\to1$, $ b \equal{} 1$. Then we get  $ (a^p \plus{} 1)^{\frac {1}{p}}\ge a \plus{} 1 \minus{} \frac {1}{a \plus{} 1}$, equivalently (after dividing by $ a$ and putting $ x \equal{} \frac1a$),
\[ 1 \plus{} x^p\ge\left(1 \plus{} \frac x{1 \plus{} x}\right)^p.\]
By Bernoulli however, $ \left(1 \plus{} \frac x{1 \plus{} x}\right)^p\ge 1 \plus{} \frac {px}{1 \plus{} x} > 1 \plus{} x > 1 \plus{} x^p$ if we choose $ 0 < x < p \minus{} 1$, i.e. $ a$ big enough.

So $ p \equal{} 1$ is the greatest possible. Similarly for the more general case with $ \frac rk$.

When we make $ c \equal{} 0, r\to1, b \equal{} 1$, then we get $ (a^p \plus{} 1)^{\frac {1}{p}}\ge a \plus{} 1 \minus{} \frac {a}{a \plus{} 1}$, not $ (a^p \plus{} 1)^{\frac {1}{p}}\ge a \plus{} 1 \minus{} \frac {1}{a \plus{} 1}$, and then after dividing by $ a$ and letting $ x\equal{}\frac{1}{a}$, this is equivalent to $ 1\plus{}x^p\geq \left(1\plus{}\frac{x^2}{x\plus{}1}\right)^p$

Solutions by computer are welcome...
Someone please solve this problem

After correction, I guess $ p_{max}\equal{}\frac{ln2}{ln3\minus{}ln2}\approx 1.7095$. This is best possible for $ (a,b,c)\equal{}(1,1,0)$ and $ r\to 1$.

But for the proof...  :maybe:  :roll:


probability
geometry
geometric transformation
homothety
ARML
trapezoid
trigonometry

[url]http://www.nycmathteam.com/wp-content/workshops_01.pdf[/url]

[hide="Tell me what you think before you read"]Personally I don't like any of them. I think I've mastered mass points and probability, especially expected value. I'm leaning towards State diagrams and Homothecy in geometry. They look most abstract and thought-provoking.[/hide]

If for some reason, everyone picks the same three, how would they decide who goes where?

[quote="Bictor717"]I think I've mastered[/quote]
wow.

well, im totally opposite.
I hate the fact that I cannot take them all.
Jan has always been a great teacher, and im sure hes going to bring something interesting, as well as other people including our Ravi B, mr.Boppana.

That's quite an impressive list of workshop-presenters there.  I don't think you could go wrong with any of them.

[quote="JBL"]That's quite an impressive list of workshop-presenters there.[/quote]indeed.

I'm going to be assigned to one, but they all look awesome!!

Bictor, you've mastered mass points?  I don't even know what a mass point is. Though "using mass points to make geometry easy" sounds cool, because geometry is one of the harder topics for me, and I like the idea of making geometry easy :)

mass points is a very cool topic- interesting and at the same time pretty easy to conceive. it does make things a lot easier if you know how to use it :)

the training?
power round was 51 parts instead of 100 ;)
team round was doable
relays crushed us

the standings (i think)
$\begin{tabular}[t]{|c|c|c|c|c|}\hline
Team&A&N&Y&C\\\hline
Power&15&15&13&16\\
Team&18&16&16&16\\
Relay 1&4&2&6&4\\
Relay 2&0&0&2&0\\\hline
Total&37&33&37&36\\\hline
\end{tabular}$

that's just plain wrong on so many levels, especially because the score is out of 100 (40-20-20-20).

<edit> hmm, based on last year's results at [url]http://www.arml.com[/url], the team round should really be worth 40, so the actual standings should be 55, 49, 53, and 52 out of 120, respectively. :P </edit>

I had a lot of fun at the training session.

The power round was insanely lengthy (just like at last year's ARML practice).  We need to find another writer of the question, or at least an editor.

What happened to you guys on the relays?  Those problems were pretty easy.

I will try to post the problems from my workshop on another thread.

From the scores, I'd think that there should be another practice session...Is there going to be one...Possibly one at the award ceremony??? Maybe with a power question that is similar to ARMls in that it doesnt have so many questions...I love 51 and all(hint 17*3), but that was too much...

A nice day well spent. Visiting City College brought back memories of my summer from 3 years ago.
The homothecy workshop was neater than I thought it would be. Mr. Rusczyk was a great speaker.

Can someone post the (other) relay questions as well? I think we were only given the solutions.

The power question was really weird. It was just super-tedious mods, after you get past the lengthy explanation. Not knowing which questions to do became a problem as well.

$\text{R1-3. Let }T=\text{TNYWR and set }K=\frac{T}{6}.\newline\text{There are }K\text{ positive integers in an arithmetic progression with common difference 1.}\newline\text{If the sum of the terms is 75, compute the value of the first term.}$

$\text{R2-3. Let }T=\text{ TNYWR. Compute the largest integer }n\text{ with }0<n\leq T\text{ such that the triangle with sides }\newline n,\;2n-3,\text{ and }3n-9\text{ has integer area.}$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35861&highlight=
I saw this a few weeks before the practice session, only since it was multiple choice, I solved it by plugging in values rather than trying it the long way. Big mistake.

That power round was ridiculous.  Especially considering how long it took for some of us to realize how leap years work... by the time I had figured it out, we only had a few minutes left.

My relay questions were pretty simple (sorry, I don't have a scanner):

R1-2. Let T = TNYWR and set K = T/15.  Square LMNP is inscribed in right triangle ABC as shown.  If PN = K, compute the product (AP)(NC).

(Diagram: Right triangle ABC, with B as the right angle; square LMNP, with L on side AB, M on side BC, and PN as part of AC, with P closer to A)

R2-2.  Let T = TNYWR.  Shown is the graph of the first quadrant portion of a relation.  The relation is symmetric over the x-axis and over the origin.  Given A(0,T), B(3T/2, 3T/2), and C(T,0), compute the area bounded by the graph.

(Diagram: Quadrant I shown, with pt A (0,T), pt B (1.5T, 1.5T) and pt C (T,0) with line segments AB and BC drawn)

Here is the first question from each relay round.

[b]R1-1. [/b] In trapezoid $ABCD$, $\ \overline{AB} \parallel \overline{CD}$ and angles $B$ and $D$ are acute.  If $\sin \angle B = \cos \angle D$, compute $\mathrm{m}\angle A - \mathrm{m}\angle B$ in degrees.

[b]R2-1.[/b]  Let $n$ be a positive integer.  If the number of integers in the domain of $y = \log(1-x)(x - n)$ equals $2n - 6$, compute $n$.

I really enjoyed the whole day!! It was so nice to meet all of you guys.  I went to the Mass Points workshop, and I followed the whole thing and learned so much.

The Power Round was definitely ridiculous, but whatever.  I was just there to have fun.  I thought the team round and the relay round were definitely doable.  However, I was the #2 relay person, and spent the first 3 mintues figuring out what TNYWR meant, so I didn't the first one.  I got the second one, though.

da team questions were ok, but da power round... it took meh more than half of da time to read those 2 pgs... n afta i finished reading, i 4got everything cuz i was really confused :?   i hope da real arml power round is not like tat  :P
newaiz, it was kool us Y team tied w/ A team... hehe  :D

da is the and 4got is forgot.
no offense, but please try not to use those fancy words here.

sry geehoon :P   i was just typing it fast... but those are not fancy words though  :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

What's an example of an abelian group $G$ and two distinct subroups $H$ and $K$ both of index 2?  ($G$ doesnt need to be cyclic)

$Z_{2}\times Z_{2}$, 3 subgroups of index 2.

[quote="m4467"]i dont understand the example.  what's G?  can you explain please?[/quote]

$G$ here is $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$, that is $\{(x,y)|x\in \mathbb{Z}_{2}\mbox{ and }y\in \mathbb{Z}_{2}\}$.

Or you can think of it as the "Klein 4 Group"


Are there two solutions for Sodoku? As in, one puzzle, but you get different numbers in every box (except the ones already filled out) I was wondering if that was possible. Is it?

It's possible to design puzzles with multiple solutions, but the standard for official sudokus is that the solution is unique.

AFAIK sudoku puzzles are generally created with a computer; it's a relatively simple thing to do, algorithm-wise. It's an exact cover problem. Essentially, you create a binary matrix that represents all the constraints in such a way that when you pick a set of rows that represents a valid grid, if you add those rows you get a row of 1s.


inequalities
logarithms
function
calculus
calculus computations

For arbiterary positive numbers $a,\ b,$ prove the following inequality. \[b\ln \frac{a}{b}\leq a-b\leq a\ln \frac{a}{b}\]

I think this is obvious due to [b]mean value theorem.[/b]
we can assume [b] a > b[/b]
then there in a c which is [b]b<c<a [/b]and [b](ln a  - ln b)/(a - b)= 1/c[/b]
so we the inequality is equal to [b]1/a < 1/c  < 1/b,[/b]
and this is obvious .

[color=darkblue][b]Lemma.[/b] $\boxed{\ t>0\Longrightarrow \ln x\le x-1\ }$ (the equation of the tangent in the point $T(1,0)$ to the graph $G_{f}$ of the concave function $f(x)=\ln x\ ,\ x>0$ is $y=x-1$).
[b]Remark.[/b] $t: =\frac{1}{t}$ $\Longrightarrow$ $\ln \frac{1}{t}\le \frac{1}{t}-1$ $\Longrightarrow$ $ln t\ge 1-\frac{1}{t}$ $\Longrightarrow$ $t\ln t\ge t-1\ .$ Therefore, $\boxed{\ t>0\Longrightarrow\frac{t-1}{t}\le \ln t\le t-1\ }\ .$

[b]Proof of the proposed inequality.[/b] $t: =\frac{a}{b}>0$  $\Longrightarrow$ $\frac{\frac{a}{b}-1}{\frac{a}{b}}\le\ln \frac{a}{b}\le \frac{a}{b}-1$ $\Longrightarrow$ $\frac{a-b}{a}\le\ln\frac{a}{b}\le\frac{a-b}{b}$ $\Longrightarrow$ $\boxed{\ b\ln\frac{a}{b}\le a-b\le a\ln\frac{a}{b}\ }\ .$[/color]

My solution is exactly same as yours, Virgil Nicula.


[quote="Interval"]How do I graph the function given below?

f(x) = pi(x^4) - sqrt{7}(x^3) + 5x - 1.[/quote]

$\text{f(}x) = \pi (x^{4})-\sqrt[7](x^{3})+5x-1$

Well unless it says sketch a graph just use a table of values and substitute simple values of $x$ for your $y$ values

[quote="Interval"]How do I graph the function given below?

f(x) = pi(x^4) - sqrt{7}(x^3) + 5x - 1.[/quote]

yes... the first thing you should always do is find he roots.  in this case that is pretty hard to do without a calculator.  next, plot some points.  try x=0,1 and -1.  Those usually are the best to try first because it gets rid of the root 7 of x^3

yeah i usually hate graphing assignments because it's a waste of time and paper

@jli:...i love halo!!!  :lol:

[quote="Interval"]How do I graph the function given below?

f(x) = pi(x^4) - sqrt{7}(x^3) + 5x - 1.[/quote]
1. find the roots by whatever ways possible. if it's not possible to find all but two roots (the last two can be done with the quadratic equation), then jus do it by making a table of points.
2. mark signs between each of the roots and at the ends of the x axis by testing points in between and at the ends.
3. test 1-2 points between each intercept, and make a smooth curve connecting them all.
make sure your leftmost side is an arrow going up and same with your rightmost side, because you are graphing an equation with an even degree and a positive leading coefficient.

I thank you all for the great tips and replies.


ratio
geometry
parallelogram
angle bisector
similar triangles
geometry unsolved

Points $D$ and $E$ are taken on sides $BC$ and $CA$ of a triangle $ BD\equal{}AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of $\angle ACB$ intersects $AD$ and $BE$ at $Q$ and $R$ respectively. Prove that $ \frac{PQ}{PR}\equal{}\frac{AD}{BE}$.

Do you use area ratios to solve this? I have a whole mess of equations with them.

[hide="Solution"]
Let $ CD \equal{} y$, $ DB \equal{} AE \equal{} x$, and $ CE \equal{} z$.  Now, Ceva's Theorem gives that $ \frac {AE}{EC}\cdot \frac {DC}{BD}\cdot \frac {BK}{AK} \equal{} 1\implies \frac {x}{z}\cdot \frac {y}{x}\cdot \frac {BK}{AK} \equal{} 1$.  This gives us that $ \frac {z}{y} \equal{} \frac {BK}{AK}$.  Notice that\[ \frac {PB}{EP} \equal{} \frac {[BPC]}{[EPC]} \equal{} \frac {[CPB]}{[APC]\cdot \frac {CE}{AC}} \equal{} \frac {[CBP]}{[APC]}\cdot \frac {1}{\frac {z}{x \plus{} z}} \equal{} \frac {BK}{AK}\cdot \frac {x \plus{} z}{z} \equal{} \frac {z}{y}\cdot \frac {x \plus{} z}{z}
\]and so\[ \frac {PB}{EP} \equal{} \frac {x \plus{} z}{y}\implies \frac {EB}{EP} \equal{} 1 \plus{} \frac {PB}{EP} \equal{} 1 \plus{} \frac {x \plus{} z}{y} \equal{} \frac {x \plus{} y \plus{} z}{y}\]which gives us that $ EP \equal{} \frac {BEy}{x \plus{} y \plus{} z}$.  Similarly, we can find that $ DP \equal{} \frac {ADz}{x \plus{} y \plus{} z}$.  $ CR$ is an angle bisector, so $ ER \equal{} \frac {EC\cdot EB}{EC \plus{} CB} \equal{} \frac {zEB}{x \plus{} y \plus{} z}$.  Similarly, $ QD \equal{} \frac {yAD}{x \plus{} y \plus{} z}$.  Hence, $ PQ \equal{} DQ \minus{} DP \equal{} \frac {AD(y\minus{}z)}{x \plus{} y \plus{} z}$ and $ PR \equal{} EP \minus{} ER \equal{} \frac {EB(y\minus{}z)}{x \plus{} y \plus{} z}$.  Thus,\[ \frac {PQ}{PR} \equal{} \frac {\frac {AD(y\minus{}z)}{x \plus{} y \plus{} z}}{\frac {EB(y\minus{}z)}{x \plus{} y \plus{} z}} \equal{} \frac {AD}{BE}\][/hide]

We define the point $ F$ in the plane, such that $ EF\parallel CRQ$ and $ AF \equal{} AE$ $ ,(1)$

So, because of $ \angle EAF \equal{} 180^{o} \minus{} 2\angle AEF \equal{} 180^{o} \minus{} \angle C,$ we conclude that $ AF\parallel BC$ $ ,(2)$

From $ (1),$ $ (2)$ and from $ AE \equal{} BD$ we conclude that $ AF\parallel \equal{} BD$ and then, $ BF\parallel \equal{} AD.$

Hence, from the similar triangles $ \bigtriangleup PQR,$ $ \bigtriangleup BFE,$ we conclude that $ \frac {PQ}{PR} \equal{} \frac {BF}{BE} \equal{} \frac {AD}{BE}$
and the proof is completed.

Kostas Vittas.

Construct the parallelogram $ AEBF$, we have $ BF\equal{}BD$, the triangle $ BFD$ is isosceles and the angle bisector of $ \angle DBF$ is perpendicular on $ CQ$, angles $ \angle BCA$ and $ \angle DBF$ being supplementary; hence $ DF\parallel{}CQR$, triangles $ \triangle ADF$ and $ \triangle PQR$ are similar, $ BE\equal{}AF$ and we are done.

Best regards,
sunken rock


geometry
circumcircle
incenter
Euler
trigonometry
geometry unsolved

R and r be circumcenter and incenter of a triangle
prove that R>2r

Dera Mathlinkers,
applied the well knowed formulas concerning the distance between the incenter and the center of the circumcircle
d² = R²-2Rr  where R, r are the ray of the circumcircle, incircle.
Sincerely
Jean-Louis

[quote="kunny"]Isn't that $ R\geq 2r?$[/quote]
yeah   :wink:

hello, using $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$, $ A\equal{}sr$ and $ A\equal{}\frac{abc}{4R}$ then you can prove
$ \frac{abc}{4A}\geq\frac{2A}{s}$.
Sonnhard.

Euler  :)

You can use $ \cos \frac{B\minus{}C}{2}\ge \sqrt{\frac{2r}{R}}$

Or prove the stronger one : $ 2r\le \frac{2\sqrt{3}p}{9}\le R$


Harvard
college
calculus
MIT
real analysis
integration
function

I have been doing a lot of reading regarding math 55 at Harvard University and I have one question. Does MIT's course, 18.014-18.024, calculus with theory, compare to Math 55 in terms of rigor and difficulty. Any feed back will be appreciated.

In my humble opinion, MIT's course that you mentioned might be like Harvard's Math 25, but Math 55 is in a class by itself.

The 18.0x4 series mainly just covers calculus.  Math 55 covers calculus, linear algebra and samples other topics as I understand it.

Also, I agree with what tokenadult said.  Math 55 [i]seems[/i] to be very unique.

From what I've heard about Math 55, it's not too different from Stanford's 50H series.  50H sounds a little less intense, though; no 40-hour weekly problem sets for us.

I think Stanford's course also is better analogized to Math 25 rather than Math 55.

[quote="joml88"][quote="drunner2007"]but you can take math 55 while you're at MIT right? (if you can fit it in your schedule)[/quote]

Yes, you [i]can[/i].  But I don't know anyone who has.  One thing to consider is the amount of group studying that you'll have to do to survive the class.  Either you'll be spending a LOT of time over at Harvard or you'll have to get a couple other kids at MIT to take it with you and even then you might still spend a lot of time at Harvard.[/quote]

did you know anyone over at MIT that was thinking about it?

darktreb and a couple other MOPpers were going to do it but ended up aborting that plan.  I think some MIT professors highly discouraged it for some reason.  There's also the hassle of getting to Harvard every day and the fact that their school schedule is different from MIT's.

Thanks for the update on the MIT students who had considered taking Harvard's Math 55--I was wondering how that plan would turn out.

What is usually the highest math class incoming freshmen can take at MIT then?  MIT's Analysis has three levels (A, B, and C), which would seem similar to Harvard having three intro classes Math 23, 25, and 55 (although I don't know how the subject matter would compare).  Also, they MIT's Analysis uses Rudin just like Math 25 and 55 do (I don't' know about Math 23).

I think, and I'm pretty sure I learned this from AoPSer and current Harvard student JBL, the policy at MIT is to let freshmen take ANY math course they are ready for. Some MIT freshmen come in with astounding background in college-level mathematics, and are permitted to register for amazingly hard upper-division courses.

I too would like to know what current MIT students have observed about what actual courses some MIT freshmen take.

As far as I know, MIT does let its freshmen take whatever they want in math, although I suppose that individual professors teaching individual classes might discourage freshmen from taking them.  One MIT student my year took math 55 at Harvard.

By the way, as a general proposition: math 55 is [i]not[/i] worth becoming obsessed over.  It won't make a significant difference in your life after freshman year of college that you took math 55 instead of, say, Sally's Analysis at Chicago or 18.100b (I hope that number is right) at MIT or any of the other classes people have mentioned.  It's a very challenging math course, but it's not anything more than that.

Also, analogizing to math 25 is slightly misleading: 25 jumps around in difficulty quite a bit, depending on the year and the professor, from just above the upper edge of 23 to approaching the lower edge of 55.  Math 25 does not necessarily use Rudin (although it may have in the past 2 or 3 years).

18.100B is generally the class that freshman take for math if they have a couple math credits going in and want a challenge.  The class is one semester.  towersfreak2006 took it this semester so he might have something to say about it.  Freshmen can take a few more math classes if they feel that one's not enough.  I believe Nimish took 18.100B plus two other maths this semester.  Just to drop another name, billzhao took it this semester also.

In general MIT lets students have a lot of freedom with what they take.  You don't need much more than professor approval.  The only thing that isn't flexible is the credit limit for freshmen which might be a little annoying but does save your sanity your first semester.

I justo took 18.014 and its easy but very nice... and the professor this year was aaaaawsome.

my other math class was 18.510 .. logic.. and i really liked it though it was hard, because it is very different to what every high school olympiad student has seen.. but its beautiful and powerfull

[quote="quantum leap"]MIT's Analysis has three levels (A, B, and C), which would seem similar to Harvard having three intro classes Math 23, 25, and 55 (although I don't know how the subject matter would compare).[/quote]Here is the given description for the three flavors of 18.100:

"Three options offered, each covering fundamentals of mathematical analysis: convergence of sequences and series, continuity, differentiability, Riemann integral, sequences and series of functions, uniformity, interchange of limit operations. Each option shows the utility of abstract concepts and teaches understanding and construction of proofs. Option A chooses less abstract definitions and proofs, and gives applications where possible. Option B is more demanding and is for students with more mathematical maturity; it places more emphasis on point-set topology and n-space, whereas Option A is concerned primarily with the real line. Option C is a 15-unit (4-0-11) variant of Option B, with further instruction and practice in written and oral communication."

The math department (as well as most others, I think) requires two CI-M classes. 18.100C is a CI-M, a communication intensive (class) in your major. This means you will probably be expected to revise and submit your proofs possibly several times. Even though I usually did my problem sets in LaTeX, so they would be easily editable, I considered myself experienced enough in proofwriting that I would not benefit much from the extra hours demanded by 18.100C. Also, freshmen are generally limited to 54 units, so the extra 4 units might take up important slots. 18.100B definitely focuses on topology, which I hear many people say is the hardest subject in the class.

[quote="quantum leap"]What is usually the highest math class incoming freshmen can take at MIT then?[/quote]
Of course it depends on your background. The somewhat loose sequence of classes is roughly:
18.01x Calculus I or BC Calculus -> 18.02x Calculus II-> 18.03x Differential Equations -> 18.06 or 18.700 Linear Algebra
From what I can tell, the theoretical math track after that is:
18.100 Analysis I, 18.701 Algebra I -> 18.101 Analysis II, 18.103 Fourier Analysis, 18.702 Algebra II, 18.901 Topology
I don't know how applied math majors typcially go about, but they do need to take 18.310 and 18.311 Applied Math, and 18.04 or 18.112 Complex Variables.

Of the classes listed above, I know there were freshmen last semester in at least 18.01, 18.02, 18.03, 18.700, 18.100, 18.701, 18.310.

If you plan to do Putnam, a good class is Richard Stanley's problem solving seminar. I wasn't in that, but I'm sure several AoPSers were.

Also, remember that first semester students are on pass-no record. I took a theoretical math class and an applied math class to get a feeling for each. Namely, my math classes last semester consisted of 18.100B, 18.310, and 18.091, the last of which served as a companion class to 18.310.


[quote="Nbach"]We write again this problem in other way: Find all function $ f: Z_{+}\mapsto Z_{+}$ such that:
                $ f(f(x))+f(x) =2x \forall x \in Z_{+}$
Solution: Denote $ f_{n}(x) = f(f(....f(x)...))$ ( where  n  times of f ) . Note that $ f_{0}(x) =x$ and $ f_{1}(x)=f(x)$
        We easily see that $ f_{n+1}(x) =-f_{n}(x)+2.f_{n-1}(x)$
      Consider equation $ x^{2}+x-2 =0$ .It has two roots are $ x_{1}=1$ and $ x_{2}=-2$
Hence $ f_{n}(x) = A+B.(-2)^{n}$ where A,B will be defined by $ f_{0}(x)$ and $ f_{1}(x)$
We'll get $ 3A= 2x+f(x)$ and $ 3B=x-f(x)$ . Let $ n$ be a sufficently large odd number.Because $ f_{n}(x) > 0$ ,hence $ B=0$ .We get $ f(x) =x$.[/quote]
I haven`t seen this problem there yet. So, great solution :)

An other solution . Let $ h(x)$ be the function defined by $ h(x) = \frac{f(x)}x$
The initial equation becomes $ h(x) = \frac 2{1+h(f(x))}$ . Manifestly $ h$ is bounded . Let $ I,S$ be its bounds : $ 0\leq I=inf (h(x)) \leq S=sup (f(x))$ .
$ h(f(x)) \leq S \Longrightarrow h(x)\geq \frac 2{1+S}\Longrightarrow I\geq \frac 2{1+S}\Longrightarrow I \geq 2-IS$  ;  $ h(f(x)) \geq I \Longrightarrow h(x)\leq \frac 2{1+I}\Longrightarrow S\leq \frac 2{1+I}\Longrightarrow S \leq 2-IS$
So  $ S\leq I$ . But $ 0 \leq I\leq S$ ,  so $ 0\leq I=S=2-IS \Longrightarrow I=S=1 \Longrightarrow f(x) =x$
 :cool:

a somehow similar one with the similar idea is the leningrad 1981 9th grade:
[b]in the sequence $ 1<a_{1}<a_{2}<...<a_{n}<...$ from natural numbers,for every natural $ n$ we have:
$ a_{n+a_{n}}=2a_{n}$
prove that there exists natural number $ c$ s.t for every natural number $ n$ we have $ a_{n}=n+c$[/b]


The function f(x) is not defined for x=0.  It has the property that for all nonzero real numbers x, f(x) + 2f(1/x)=3x.  Find all values of a such that f(a) = f(-a)

i've sorta figured out that this function is not one-to-one and it is a fraction so far.  Any hints or solutions would be great! :lol:

When does $ f(x) \equal{} f(1/x)$?
$ x \equal{} \frac {1}{x}$
$ x^{2} \minus{} 1 \equal{} 0$
$ x \equal{} \pm 1$

Edit: Sorry  :blush: $ f(1)\equal{}1$ and $ f(\minus{}1)\equal{}\minus{}1$

Setting $ x \minus{} \minus{} > \frac {1}{x}$ we take:

$ f \left(\frac {1}{x} \right) \plus{} 2f(x) \equal{} \frac {3}{x}$.

So $ f(x) \equal{} \frac {2}{x} \minus{} x,\forall x \in \mathbb{R^*}$.

So we need:
$ \frac {2}{a} \minus{} a \equal{} \minus{} \frac {2}{a} \plus{} a \Longleftrightarrow \frac {2}{a} \equal{} a \Longleftrightarrow a \equal{} \sqrt {2}$ or $ a \equal{} \minus{} \sqrt {2}$.

Dimitris


How can I have a set of equations aligned (like inside an align), but have left-aligned text in between some of them without disrupting their alignment?

Use \intertext as in
\begin{align}n = 1\\
\intertext{but sometimes we have}
n = 0
\end{align}


Find all continuous function $f: \mathbb{R}\to\mathbb{R}$ such that $f(x+f(y+f(z)))=f(x)+f(f(y))+f(f(f(z))) \forall x,y,z\in\mathbb{R}$.

(*)  $f(x+f(y+f(z)))=f(x)+f(f(y))+f(f(f(z)))$

Let's assume $f\not\equiv 0$ and set $g(x): =f(x)-f(0)$.

Then
(1) $g(x+f(y))=g(x)+g(f(y))$ and $g(nf(y))=ng(f(y))$ for all $x,y\in\mathbb{R},n\in\mathbb{Z}$:
$x=0$ in (*) gives $f(f(y+f(z)))=f(0)+f(f(y))+f(f(f(z)))$ and
$f(x+f(y+f(z)))=f(x)+f(f(y))+f(f(f(z)))=f(x)+f(f(y+f(z)))-f(0)$.
So $g(x+f(y+f(z)))=g(x)+g(f(y+f(z)))$ for all $x,y,z\in\mathbb{R}$.
Replacing $y+f(z)$ by $y$ gives the first result.
Then $g((n+1)f(y))=g(nf(y)+f(y))=g(nf(y))+g(f(y))$ for all $n\in\mathbb{Z}$.
The rest follows by induction.

Now comes the ugly part of my proof.
(2) For all $x\in\mathbb{R},n\in\mathbb{N}$ there exists some $y\in\mathbb{R}$ with $f(x)=nf(y)$:
First $x=y=0$ in (*) gives $f(0)=-f(f(0))$.
Then by the intermediate value theorem there exists
some $z\in\mathbb{R}$ between $0$ and $f(0)$ with $f(z)=0$.
Then ${1\over n}f(x)$ lies between $f(x)$ and $0=f(z)$ for all $x\in\mathbb{R},n\in\mathbb{N}$.
So again by the intermediate value theorem there exists
some $y\in\mathbb{R}$ between $x$ and $z$ with $f(y)={1\over n}f(x)$.

Now take $t\in\mathbb{R}$ with $f(t)\not =0$ and set $a={g(f(t))\over f(t)}, b=f(0)$. Then
(3) $f(x)=ax+b$ for all $x\in\mathbb{R}$:
For all $m\in\mathbb{Z},n\in\mathbb{N}$ there exists some $y\in\mathbb{R}$ with $f(t)=nf(y)$ by (2).
And then $g({m\over n}f(t))=g(mf(y))=mg(f(y))={m\over n}g(nf(y))=a{m\over n}f(t)$ by (1)
and $f({m\over n}f(t))=a{m\over n}f(t)+b$.
But $\{{m\over n}f(t): m\in\mathbb{Z},n\in\mathbb{N}\}$ is dense in $\mathbb{R}$, as $f(t)\not =0$, and the result follows.

And now $0=f(0)+f(f(0))=(a+2)b$.
So the solutions of (*) are $f(x)=cx,x\in\mathbb{R}$ and $f(x)=c-2x,x\in\mathbb{R}$ for any $c\in\mathbb{R}$. (this includes $f\equiv 0$)

In solving this problem , I know so you. But prove $a=-\infty$ and $A=+\infty$ is not easy.
[quote="Baronov"]Assume first f(0) = 0.
Then put z=0 we get :
      f(x + f(y)) = f(x) + f(f(y)). for all x,y.

now let $a=inf_{x\in \mathbb{R}}f(x).$ and$A=inf_{x\in\mathbb{R}}f(x)$
we want to prove than $a=-\infty$ and $A=\infty$
if we prove this than since f is continous then f is surjective on $\mathbb{R}$ and we have f(x+y) = f(x)+f(y) for all x,y.and since f is continous then f(x)=cx, for some constant c and these functions are solutions for all $c\in\mathbb{R}$.
It is not so hard to prove but now i don't have time to write it down.

And it remains to prove that f(0)=0.[/quote]

Thank for help of you. But now i will check your solution :D
[quote="olorin"]
(1) $g(x+f(y))=g(x)+g(f(y))$ [/quote]
Why, olorin?  :maybe:

[quote="N.T.TUAN"]Why, olorin?  :maybe:[/quote]
??? That's what I proved after the ":".
Here
[quote="olorin"]$g(x+f(y))=g(x)+g(f(y))$ for all $x,y\in\mathbb{R}$:
$x=0$ in (*) gives $f(f(y+f(z)))=f(0)+f(f(y))+f(f(f(z)))$ and
$f(x+f(y+f(z)))=f(x)+f(f(y))+f(f(f(z)))=f(x)+f(f(y+f(z)))-f(0)$.
So $g(x+f(y+f(z)))=g(x)+g(f(y+f(z)))$ for all $x,y,z\in\mathbb{R}$.
Replacing $y+f(z)$ by $y$ gives the first result.
[/quote]
Please tell me what you don't understand.

:lol: sorry!
[quote="olorin"]
(2) For all $x\in\mathbb{R},n\in\mathbb{N}$ there exists some $y\in\mathbb{R}$ with $f(x)=nf(y)$:
First $x=y=0$ in (*) gives $f(0)=-f(f(0))$.
Then by the intermediate value theorem there exists
some $z\in\mathbb{R}$ between $0$ and $f(0)$ with $f(z)=0$.
Then ${1\over n}f(x)$ lies between $f(x)$ and $0=f(z)$ for all $x\in\mathbb{R},n\in\mathbb{N}$.
So again by the intermediate value theorem there exists
some $y\in\mathbb{R}$ between $x$ and $z$ with $f(y)={1\over n}f(x)$.

Now take $t\in\mathbb{R}$ with $f(t)\not =0$ and set $a={g(f(t))\over f(t)}, b=f(0)$. Then
(3) $f(x)=ax+b$ for all $x\in\mathbb{R}$:
For all $m\in\mathbb{Z},n\in\mathbb{N}$ there exists some $y\in\mathbb{R}$ with $f(t)=nf(y)$ by (2).
And then $g({m\over n}f(t))=g(mf(y))=mg(f(y))={m\over n}g(nf(y))=a{m\over n}f(t)$ by (1)
and $f({m\over n}f(t))=a{m\over n}f(t)+b$.
But $\{{m\over n}f(t): m\in\mathbb{Z},n\in\mathbb{N}\}$ is dense in $\mathbb{R}$, as $f(t)\not =0$, and the result follows.
[/quote]

Very nice! Thanks again!


Bob plays a game where, for some number $ n$, he chooses a random integer between 0 and $ n\minus{}1$, inclusive. If Bob plays this game for each of the first four prime numbers, what is the probability that the sum of the numbers he gets is greater than 0?

Let's do this by complementary counting.

For n=2, the chances of picking 0 is 1/2
For n=3, the chances of picking 0 is 1/3
For n=5, the chances of picking 0 is 1/5
For n=7, the chances of picking 0 is 1/7

Thus the probability of getting 0 as a sum is 1/210, so the probability that the sum is larger than 0 is 209/210


algebra
polynomial
probability
quadratics
system of equations

Hi

Solve the following equation in real numbers x:

$ x^{4}\plus{}6x^{3}\minus{}27x^{2}\minus{}44x\plus{}68\equal{}0$.

I don't know if this has "nice" solutions.

I guess you'd first try the rational zeros ($ \pm 1, \pm 2, \pm 4, \pm 17, \pm 34, \pm 68$)

none of which work...

A general quartic formula exists but it's really ugly~

did u use the intermediate value theorem of polynomials to find the upper and lower bound?

[quote="mathgeniuse^ln(x)"]Hi

Solve the following equation in real numbers x:

$ x^{4} \plus{} 6x^{3} \minus{} 27x^{2} \minus{} 44x \plus{} 68 \equal{} 0$.

I don't know if this has "nice" solutions.[/quote]
My answer is not! Since we easily check that this equation doesn't have any rational solutions, now we jut only hope that it can be factored into both polynomial degree 2 with integer coefficients(the same as rational coefficients when clearing all denominators). So, we assume that there exists such 2 polynomials degree 2 like this:(x^2+ax+b)(x^2+cx+d)=x^4+x^3(c+a)+x^2(d+ac+b)+x(ad+bc)+bd(a,b,c,d are integers)
Matching those coefficients with the original coefficients to solve the following systems:
bd=68
ad+bc=-44
d+ac+b=-27
c+a=6
Considering that this system is inconsistent(doesn't have any integer solutions), so the polynomial above can't be factor into both polynomial degree 2 with integer coefficients. Therefore, it's an irreducible polynomial! Thus, a nice solution is nearly impossible to find(I don't know whether it exists a trigonometric solution like x=tg2a or x=tg3a,etc but the algebraic approach is impossible as proved above).Applying quartic formula seems to be the best way to solve this problem! :)

idea:

1) substitute $ x \equal{} z \minus{} 3/2$ and expand (z^3 term drops out)
2) move the rest of the terms except $ z^4$ to the RHS and hope they form a perfect square

ok... perhaps not such a great idea.  If the RHS is a perfect square, 3) take the plus or minus square roots of both sides, yeilding two quadratics.  Knowing my luck, though, the probability of this working is about... $ 0$.  If you want exact answers, quartic formula (Giant mess...).  Otherwise, just approximate them with a calculator: $ \minus{} 2.145925$, $ 1.0525418$, $ 3.5570875$, and $ \minus{} 8.463704$

yeah I didn't think it would be a perfect square, and I was too tired to try for myself.  Anyway, I hope you are not expected to find the exact values of the x-intercepts, because the only way looks like the quartic formula, which is almost impossible to use because of its size.

[quote="mathgeniuse^ln(x)"]Hi

Solve the following equation in real numbers x:

$ x^{4} \plus{} 6x^{3} \minus{} 27x^{2} \minus{} 44x \plus{} 68 \equal{} 0$.

I don't know if this has "nice" solutions.[/quote]

You are correct: this equation does not have "nice" solutions.

I checked it out with an online utility:

[url]http://www.akiti.ca/Quad4Deg.html[/url]

The following roots were computed:

1.05254179198807
-2.145924982682401
3.557087484476999
-8.463704293782667

Woah... what a coincidence... I was attempting to solve this exact quartic equation for the Olymon problems in June (I already sent my solutions)... I just used an online utility to approximate the roots. Interesting.  :o

EDIT: My 199th Post!!! :D

check out [url]http://mathworld.wolfram.com/QuarticEquation.html[/url]

in short, to solve a quartic $ x^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e \equal{} 0$
1) substitute $ x \equal{} z \minus{} b/4$.  We now have an equation of the form $ z^4 \plus{} pz^2 \plus{} qz \plus{} r \equal{} 0$
2) now add and subtract $ z^2u \plus{} zu^2/4$ to the equation.  Define $ u$ such that the equation can be factored as $ P^2 \minus{} Q^2 \equal{} (P \minus{} Q)(P \plus{} Q)$ where $ P \equal{} z^2 \plus{} u/2$.
3) Find a value for $ u$ such that $ q^2 \equal{} 4(u \minus{} p)(1/4u^2 \minus{} r)$ (Yes, solve a cubic).  Then $ Q \equal{} z\sqrt {u \minus{} p} \minus{} \frac {q}{2\sqrt {u \minus{} p}}$.
4) solve the quadratics $ P \plus{} Q \equal{} 0$ and $ P \minus{} Q \equal{} 0$ for $ z$.
5) back-substitute to find $ x$.

To solve a cubic $ x^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} 0$
1) substitute $ x \equal{} z \minus{} b/3$ to get $ z^3 \plus{} pz \plus{} q \equal{} 0$
2) substitute $ z \equal{} w \minus{} \frac {p}{3w}$
3) solve the resulting 6th degree quadratic type for $ w$
4) backsubsitute to find $ z$ and then $ x$.

Work the system of equations given by the coefficients; you can write the quartic as the product as two quadratics with a leading coefficient of one.

$ p(x)=(x^2+ax+b)(x^2+cx+d)$

I have b as the conjugate of d with $ b,d=\frac{-3\pm i\sqrt{13319}}{14}$
and a is the same to c but $ a,c=3\pm \sqrt{\frac{274}{7}}$.


abstract algebra
quadratics
Gauss
Galois Theory
superior algebra
superior algebra unsolved

K be a splitting field of x^n-1 over the rationals Q

for which n does x^2+1 split in K ?

perhaps more generally : every subextensions of degree two corresponds to adjoining a root of a squarefree integer
is there a way to determine those integers?

I am familiar with the fundamental theorem of Galois theory, the fact that for primes p you adjoin the square root of (-1)^(p-1/2)*p, and that the galois group of such extensions is isomorphic to the group of units in Z mod n

If $K_n$ is the splitting field of $X^n-1$ over $\mathbb Q$ then the roots of unity in $K_n$ are just the $n$-th roots if $n$ is even, or the $2n$-th roots if $n$ is odd.  Hence $X^2+1$ splits in $K_n$ iff $n \equiv 0 \bmod 4$.

We define the [i]conductor[/i] of an abelian extension $F$ of $\mathbb Q$ to be the least $n$ such that $F \subseteq K_n$.  It is a theorem of class field theory that the conductor of a quadratic extension is equal to its discriminant.

We can see this in one direction by considering the primes that ramify in $K_n$: they are exactly the primes dividing $n$ (where we take $n$ odd if possible).  So any quadratic field in $K_n$ must have discriminant dividing $n$.  In the other direction, consider the Gauss sum.


I could easily find the answers to the 2009 AMC 8 problems. However, I did not manage to find the problems so that I could check my answers. Does anybody know a website that has all of the AMC 8 problems or a place on AoPS where the problems are listed?

[quote="eqjj168"]well i do have a copy of the test booklet...[/quote]

Could you post them or send them to me? Thanks.

it seems that it is being posted right now...

I am posting all that don't have a diagram right now, and I will try to make the diagrams for the remaining problems.

EDIT: All up except for 3, 4, 7, 9, 12, 24, and 25, which have diagrams.

Could you send the AMC8 problems or post it?  Thanks

On the forum. Look for subject descriptions that say "2009 AMC 8 Problem _". I couldn't post 3, 4, 7, 9, 24, or 25 because they had diagrams I didn't know how to make (and I am too lazy to learn how).

Update: I am posting some more problems. I will edit with updates. Right now, there is just 3, 4, 7, 9, 12, and 25 left. Here are the links to each problem:

[hide="AMC 8 2009"]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314287]Problem 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314289]Problem 2[/url]
Problem 3
Problem 4
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314290]Problem 5[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314292]Problem 6[/url]
Problem 7
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314293]Problem 8[/url]
Problem 9
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314299]Problem 10[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314294]Problem 11[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315239]Problem 12[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314295]Problem 13[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314298]Problem 14[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314302]Problem 15[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314310]Problem 16[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314311]Problem 17[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314326]Problem 18[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314312]Problem 19[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314315]Problem 20[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314316]Problem 21[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314303]Problem 22[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314317]Problem 23[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315205]Problem 24[/url]
Problem 25[/hide]

EDIT: Added #12 as well.


AMC
USA(J)MO
USAMO
number theory proposed
number theory

[quote="ith_power"]Let $ a,b$ be two distinct odd natural numbers.Define a Sequence $ { < a_n > }_{n\ge 0}$ like following:
$ a_1 \equal{} a \\
a_2 \equal{} b \\
a_n \equal{} \text{largest odd divisor of }(a_{n \minus{} 1} \plus{} a_{n \minus{} 2})$.
Prove that there exists a natural number $ N$ such that $ a_n \equal{} gcd(a,b) \forall n\ge N$.[/quote]

Nice :)

[hide="My solution"]
Let $ b_n$ the greatest power of two in $ a_{n\minus{}1}\plus{}a_{n\minus{}2}$. We have $ a_{n\minus{}1}\plus{}a_{n\minus{}2}\equal{}2^{b_n}a_n$ and $ b_n\geq 1$ $ \forall n$ (since $ a_i$ are all odds, and so $ a_i\plus{}a_{i\plus{}1}$ all even).

So $ \sum_{k\equal{}1}^n(a_k\plus{}a_{k\plus{}1})\equal{}\sum_{k\equal{}1}^n2^{b_{k\plus{}2}}a_{k\plus{}2}$ and so $ a_1\plus{}2a_2\equal{}\sum_{k\equal{}3}^n(2^{b_k}\minus{}2)a_k$ $ \plus{}(2^{b_{k\plus{}1}}\minus{}1)a_{k\plus{}1}\plus{}2^{b_{k\plus{}2}}a_{k\plus{}2}$

Since LHS is fixed, we need to have $ b_k\equal{}1$ $ \forall k>N$ and the equality above becomes $ c\equal{}a_{k\plus{}1}\plus{}2a_{k\plus{}2}$ $ \forall k>N$ and so $ a_k\equal{}A$ constant for $ k>N\plus{}1$

From $ a_{n\minus{}1}\plus{}a_{n\minus{}2}\equal{}2^{b_n}a_n$, we see that $ \gcd(a_{n\minus{}1},a_{n\minus{}2})\equal{}\gcd(a_{n\minus{}1},a_n)$ and so $ \gcd(a_1,a_2)\equal{}\gcd(A,A)\equal{}A$

Q.E.D.[/hide]

Mr. PCo, I am in a place where there the internet is not fast and nice.

So, I am not able to see your solution as well as post my solution. Atleast, let me see your solution please. Thanks.

Indian TSTs are sadly decreasing their standard.

It's also 1993 USAMO, #4:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58207[/url]

In fact this can be done usinf infinite descent on $ max{a_i,a_{i\plus{}1}}$,which is decreasing.
@Agr_94_Math: I agree with you,But TST2 problems are hard.

@above here is the
[hide=Solution]
Without loss of generality assume that \(\gcd(a,b) = 1\) (otherwise just divide all terms by the \(\gcd\)). For the sake of contradiction, assume that \(a_n > 1\) for all \(n\). Then we have
\[a_n = \frac{a_{n-1} + a_{n-2}}{2^{\nu_2(a_{n-1} + a_{n-2})}} \leq \frac{a_{n-1} + a_{n-2}}{2} < \max(a_{n-1},a_{n-2})\]
This is impossible; then \((a_n)\) is a decreasing sequence of positive integers but \(a_n > 1\) for all \(n\).
[/hide]


Show that if p is an odd prime and 0 < k < p, then (p-k)!(k-1)! :equiv: (-1)k (mod p).

I think I can do this, but only relying on Wilson's theorem, which I don't remember how to prove:

[hide](k - 1)! = 1 * 2 * 3 * ... * (k - 1) :equiv: -(-1) * -(-2) * ... * -(1 - k) :equiv: (-1)k - 1*(p - 1)*(p - 2)*...*(p - k + 1).

Thus (p - k)!*(k - 1)! :equiv: (-1)k - 1*(p - 1)! :equiv: -1k by Wilson's theorem.[/hide]

Trying to Tex it, a little at a time:

[tex]\displaystyle{(k - 1)!=1\cdot2\cdot3 ... \cdot(k - 1)\equiv-(-1)\cdot-(-2)\cdot...\cdot-(1-k)\equiv}[/tex][tex]\displaystyle{(-1)^{k-1}\cdot(p-1)\cdot(p-2)\cdot...\cdot(p-k+1)}[/tex]

Thus [tex](p-k)!\cdot(k-1)!\equiv(-1)^{k-1}\cdot(p-1)!\equiv(-1)^k(\mod p)}[/tex]

I think the multiplication dot is \cdot .

[tex]a \cdot b[/tex]

The proof to Wilson's theorem is basically that every residue x mod p has a unique y such that xy = 1 mod p. We can continue to pair off until we are left with 1 mod p and -1 mod p, which multiply out to -1 mod p.


What is the maximum number of colors, so that I can paint $ 8\times 8$ chessbord, so that each $ 1 \times 1$ square shares an edge with at least two squares with the same color as it?

Same color as it, or same color as each other?

[hide="Solution"]The answer is 16. Note that 16 is possible: for each of the colors, color a 2x2 square completely that color. You can split the board into 16 2x2 squaes.

Now, note that if we use a color, we must use it at least 4 times. Why? Because, lets say a square is colored red. Then two of the squares adjacent to it are also colored red. But those two squares can't be adjacent to each other, so there must be another square colored red, or at least 4 reds. Therefore, 16 is the maximum possible number.
[/hide]

Two follow-ups:

1) The solution given above works for any even-sided board.  What about boards with an odd side-length?

2) What about if each square must be adjacent to two squares that have the same color as each other, but not necessarily the same color as the selected square?


If $a=-2$, the largest number in the set $\{-3a, 4a, \frac{24}{a}, a^2, 1\}$ is

$A)\quad -3a \qquad B)\quad 4a \qquad C)\quad \frac{24}{a} \qquad D)\quad a^2 \qquad E)\quad 1$

[hide]
Just substitute $a$.  You should immediately see that it's $-3a$, which is $\boxed{A}$.
[/hide]

[hide]a is 6
b is -8
c is -12
d is 4
e is 1

a is the largest.[/hide]

If you plug in the variable, A is the largest one.  :D


Let $a,b,c$ be such real numbers that $a+b+c=2$ and $ab+bc+ca=1$. Prove that $\max (a,b,c)-\min (a,b,c)\leq \sqrt{\frac{4}{3}}$.

Let $a\leq b\leq c,\ \ a,\ b,\ c$ are the roots of the cubic equation $x(x-1)^{2}=k$ where $k=abc.$ By the graph of $y=x(x-1)^{2}$, we have $0\leq a\leq \frac{1}{3}\leq b\leq 1 \leq c\leq \frac{4}{3}.$


Find the next letter in the pattern:

O, T, T, F, F, S, S...

Here are the next three, in spoiler:

[hide]E, N, T [/hide]

This one just takes a little bit of lateral thinking.

Anyway I didn't look at your spoiler tolkenadult and I got E for the next letter.

one, two, three, four, five, six, seven, eight, nine, ten, eleven...

Isnt E(for Ether Bunny) correct?? This time, SAY IF IM CORRECT!!! WE WANT TO KNOW!!

I HAVE TO GUESS!! :(


I need some help with my science homework.

Balance the chemicla equations by adding coeffecients:

A) HCl + NaOH --> H 2 O + NaCl

B) Fe 2 O 3 + C --> Fe + CO2

C) SO2 + O2 --> SO3

Please explain how you got your answer because I am thoroughly confuzzled. Thank you in advance.

<edit> TYPO!!!>

Because of the law of conservation of mass, the same number of atoms of a certain element in the reactants must equal the number of atoms of that element in the product.  Note that a coefficient before a compound distributes over the compound, and something like 4[H2] indicates 8 hydrogen atoms.

The first equation can never be balanced, since there is no sodium or oxygen in the products...

2[Fe 2 O 3] + 3C --> 4Fe + 3CO2
Now, on both sides of the equation, we have 4 Fe, 6 O, and 3 C.

2[SO2] + O2 --> 2[SO3]
2 S, 6 O on either side.

I hope that helped.  =)  I'm not a chem expert, as this is my first year taking it, but I'm pretty sure what I did is correct.

[quote="MathFiend"]um...you kinda lost me. My science teacher wasn't at school today, so we had a sub, so no one explained this. I don't get what "balancing" is, or how to do it...I kno WHAT the law on conservation of energy is, but I don't wuite understand it...I'm so confuzzled...[/quote]

OK, I'll try to explain it a little more.  My book actually gives a pretty good explanation... Basically, forget about chemistry for a moment and pretend you're sending bicycle parts to some place that will build the bike.  To construct a bike, you need some wheels (W), a frame (F), pedals (P), and handlebars (H).  So, the left side of the equation will look something like W+F+P+H.  The right side of the equation will be the finished bike, which will look something like WFPH.  But to actually make the bike, you need more than one of each of these.  In fact, you need two wheels, one frame, two pedals, and one set of handlebars.  So the formula for the finished bike is W2FP2H (those twos are subscripts).  This will be the right side of the equation.  However, if all you sent to the factory was one wheel, one frame, one pedal, and one handlebar, they couldn't construct the bike.  So, on the left side of the equation, you'll need two wheels and two pedals.  Thus, the finished and balanced equation is
2W+F+2P+H --> W2FP2H.
The same applies in a chemical reaction - you need the same number of atoms on the left side as on the right side.  It doesn't seem reasonable to have more atoms at the start of a reaction than at the end of a reaction, or vice versa.

I hope this helped... Good luck =)

ooh! I get it kinda now!!! Thank you...that illustration made it much easier.

No problem.. once you get the basic idea, balancing should become pretty easy, as you are a math person =)

for the first one, HCl and NaOH are strong acids and bases, which mean they form a salt (NaCl, table salt!) and water.... so the NaCl is aqueous and the HCl and NaOH are aqueous, but the water is liquid... so if you do a net ionic equation, you get
H+(aq) plus Cl-(aq) plus Na+(aq) plus OH-(aq) which yields H2O(liq) plus Cl-(aq) plus Na+(aq)
the Cl- and the Na+ cancel out, leaving the formation of water.... er... theres no H2Cl formed, so that question is a little... wrong... yeah, i will ask my ap chem teacher about it but i know she will look at me like im crazy, ^THATS what happens when you mix a strong acid and a strong base... are you sure you copied it right?^

I LOVE THESE THINGS!!!

My teacher made me do ten worksheets full of them. One had 51, 197, 288 and all sorts of weird numbers that I had to form three variable equations for - that took me a while.

However, if you know the basics behind the different charges of compounds and how certain compounds interact the process should be much easier. You also need an open mind - realize chemicals change molecular structure through reactions, so OH could separate into O2 and Hydrogen combined with something else.

The whole process is just really fun.

[quote="white_horse_king88"]
The whole process is just really fun.[/quote]

Yeah, once you understand what's going on, chem is really cool.  Writing and balancing equations can be quite satisfying, too =)

Caroline gave a nice explanation on why equations have to be balanced.  However, I don't think anyone's gone through the steps of balancing these equations, so I'll do that.

For the first one,

[quote="MathFiend"]B) Fe 2 O 3 + C --> Fe + CO2[/quote]

The first thing we do is count up the number of atoms on each side of the unbalanced equation.

Left side:
2 Fe
3 O
1 C

Right side:
1 Fe
2 O
1 C

We want the number of atoms on either side to be the same.  Since the number of carbons is the same on both sides, we only need to change the number of irons (Fe) and oxygens (O).

To make the number of Fe atoms equal, we would multiply whatever right-side compound containing Fe by 2.

First, let's try making the number of O atoms equal.  To do this, we need twice as many O atoms on the left side, and 3 times as many O atoms on the right side.  Thus, we need to multiply Fe 2O 3 by 2 and CO2 by 3.

Doing this produces the equation 2Fe 2 O3 + C --> Fe + 3CO2.

We count up the number of atoms again.  Now we have

Left side
Fe - 4
O - 6
C - 1

Right side
Fe - 1
O - 6
C - 3

Now we try and equalize the number of Fe molecules.  To do this, we need four times as many Fe atoms on the right side.  So we multiply Fe by 4, producing 2Fe 2 O3 + C --> 4Fe + 3CO2.

Again, we count up the number of atoms.

Left side
Fe - 4
O - 6
C - 1

Right side
Fe - 4
O - 6
C - 3

Almost there!  Now, we need to triple the number of C atoms on the left side.  So...we multiply the C by 3, producing 2Fe 2 O3 + 3C --> 4Fe + 3CO2.

Are we done?  Well, we count up the number of atoms, and indeed find an equal number:

Left side
Fe - 4
O - 6
C - 3

Right side
Fe - 4
O - 6
C - 3

So this equation is balanced.

Try doing this on equation C:

[quote]C) SO2 + O2 --> SO3[/quote]

I fixed the first one...I made a typo on accident...

Apparently, A was supposed to be HCl + NaOH --> H2O + NaCl.  Here is how I would solve it:

Count the number of atoms in the unbalanced equation.

HCl + NaOH --> H2O + NaCl

Left side:
H - 2
Cl - 1
Na - 1
O - 1

Right side:
H - 2
Cl - 1
Na - 1
O - 1

Wait a second - there are the same number on both sides.  This equation is already balanced.  We were done before we even started!

C is not the same as A - my bad.

SO2  + O2 --> SO3

Right now, we have
left side:
S - 1
O - 4

right side:
S - 1
O - 3

We need to make the number of Os equal.

Our instinct is to multiply the number of Os by 4 on the right and by 3 on the left.  But we don't have to do that, because we can actually get our oxygens from two different places on the left side.  Instead, try multiplying SO3 on the right by 2.  Then, we have

left side:
S - 1
O - 4

right side:
S - 2
O - 6

Now, let's try and balance the S atoms.  We multiply the SO2 by 2, resulting in 2SO2 + O2 --> 2SO3

left side:
S - 2
O - 6

right side:
S - 2
O - 6

And by some luck, this turns out to be balanced.  So we're done.

huh??? u like totally lost me on that last statement...they don't equal...

I'll assume MathFanatic made a counting error (I'm very familiar with those ;) )... as I said up there somewhere,
2[SO2] + O2 --> 2[SO3]
2 S, 6 O on either side.

Yep.  Was in the process of correcting it when you posted.

[size=150]THANK YOU[/size] [size=200]SOOOOOOOOO[/size] [size=150]MUCH YOU GUYS!!![/size]

I never could have done it without you...

No problem - good luck with chem/science in the future =)

Thanks. I really understand it now. There is, however one problem on a worksheet that is making me rather angry. The problem is:

C3H5(NO3)3 --> CO2 + N2 + H2O + O2

Start with the simplest elements (those that appear once on each side). Then play around with the more complicated one.

Left: C[size=75]3[/size]+H[size=75]5[/size]+N[size=75]3[/size]+O[size=75]9[/size]
Right: C+O[size=75]5[/size]+N[size=75]2[/size]+H[size=75]2[/size]
so it's 3C+5H+3N+9O and C+5O+2N+2H
You need to add 2C+3H+N+4O to the right side to make it right

Gee... guys - shouldn't we tell MathFiend [i]how[i] to do it? Not give the answer?

Anyway - you may end up adding a coefficient to C3H5(NO3)3.:wink:
Not as simple as "You need to add 2C+3H+N+4O to the right side".

yeah...I already know HOW to do it though. It simply wasn't cooperating when I tried to add coeffecients here and there. I did, however, figure out the answer right before I went to science class this morning. :)

You might get a reaction thats nearly impossible to balance by hand in any normal amount of time. Then you get to use a more fun method: Splitting into half reactions, balancing the charge of each of those and then combining them again. This is what we did in my ap class, so that we can balance anything in a couple minutes.

I used to love these in school.  Little number theory problems.

For this:

[quote]C3H5(NO3)3 --> CO2 + N2 + H2O + O2[/quote]

I'd start by sticking an x in as the coeff of the thing on the left, then you can find the others in terms of x fairly quickly; then make everything integers.  Sometimes you'll need a couple dummy variables, but I don't think I ever ran into one that couldn't be solved this way pretty quickly.

Interesting, I never thought to do it like that.  I usually just play w/ the coefficients of all of them till I get it.

I only ever treated them as number theory problems when I was trying to prove them unsolvable. (That happened once on a test in one of my chemistry classes in high school.)

I've always wanted to balance an equation in a really mathy way, but my teacher inevitably gives us equations that are simple to balance; the coefficients usually range from 1-4.


inequalities
absolute value
analytic geometry

Okay I'm sorta having trouble with graphing absolute value inequalities... They've been bugging me. Could some one help me? And also, I mean on a Cartesian Plane, not like number line wise. Thanks!

sorry I can't help I'm learning this lesson tomorrow so maybe dayafter.

Say you are given the inequality $ |x|\plus{}|y|\le5$. We have to separate this into four cases and find the extreme points. i.e., the points where $ x$ is the largest and $ y$ is the smallest, and the points where $ x$ is the smallest and $ y$ is the largest.
Case 1: $ x\plus{}y\le5$
The maximum points are $ (0,5),(5,0)$

Case 2: $ \minus{}x\plus{}y\le5$
The maximum points are $ (0,5),(\minus{}5,0)$

Case 3: $ x\minus{}y\le5$
The maximum points are $ (0,\minus{}5),(5,0)$

Case 4: $ \minus{}x\minus{}y\le5$
The maximum points are $ (0,\minus{}5),(\minus{}5,0)$

When we graph these points,

[asy]size(200);
import graph;

Label f;
f.p=fontsize(6);

xaxis(-7, 7, Ticks(f, 10, 2));
yaxis(-7, 7, Ticks(f, 10, 2));

dot("(0, 5)", (0, 5), black);
dot("(0, -5)", (0, -5), black);
dot("(5, 0)", (5, 0), black);
dot("(-5, 0)", (-5, 0), black);

draw((5, 0)--(0, -5));
draw((0, 5)--(5, 0));
draw((-5, 0)--(0, 5));
draw((0, -5)--(-5, 0));[/asy]

From this, it becomes obvious that the shape of the graph is a square. The region we want is the region inside the square.

Would we shade it in? I don't recall our teacher shading it in... I'm just asking...

Well if it's an inequality it has to be shaded in, but if the equation was $ |x|\plus{}|y|\equal{}5$, then the resulting figure would just be the square.

Okay...
So say the equation was |x+y|>3, would we have -(x+y)>3 and x+y>3, or would we need another two? As in -(x+y)<3 and x+y<3... I failz...

but remember the hobbits wouldn't understand this

never mind I learned this before.

eh our teacher never covered the case where $ |x\plus{}y|\le\text{some number}$ so I don't expect it to be on the test. (if you're curious, you can try to graph it, don't ask me, I won't tell you :P)

Lolz... Well duh... Meh I'll do them anyways  :P

basically it's just absolute value inequalities.
just solve those than you go on to graphing them.


function
combinatorics proposed
combinatorics

Let $S=\{1,2,\cdots,n\}$ where $n\in {\Bbb N}\setminus \{0\}$. How many functions $f:S\to S$ are there such that

\[
x+f(f(f(f(x))))=n+1\qquad \hbox{for all} \,\, x\in S.
\]

Michael

we make a couple of simple observations first;
a) $f$ is a permutation:indeed,if $f(x)=f(y)$ then $f^{(4)}(x)=f^{(4)}(y)$ and so $n+1-x=n+1-y$ forcing $x=y$ so $f$ is $1-1$,hence onto.
b) $f^{(8)}=I$,i.e.$f^{(8)}(x)=x\ \forall x \in S$.again simple since $f^{(4)}(f^{(4)}(x))+f^{(4)}(x)=n+1 \Rightarrow f^{(8)}(x)=x$.
so graphically if we have a directed graph with vertex set $S$ and edge set $\{(x,f(x))|x \in S\}$ then we see that this graph is the union of directed cycles whose lengths are divisors of $8$;but again if $x \neq y$ and $f(x)=y,f(y)=x$ then $f^{(4)}(x)+x=n+1 \Rightarrow x=(n+1)/2$;likewise $y=(n+1)/2,\rightarrow\leftarrow$.similarly if there exists a $4$-cycle,i.e.we have $x\neq y\neq z\neq w\neq x,f(x)=y,f(y)=z,f(z)=w,f(w)=x$ then we again get$x=y=z=w=(n+1)/2\rightarrow\leftarrow$.
hence every cycle is an $8$-cycle or is a $1$-cycle;in the case of the $1$-cycle it is unique with $x=(n+1)/2$;every other cycle has to be an $8$-cycle.
hence if $a_n$ denotes the number of such functions we have
    $a_n=0$ if $n \neq 0,1(mod\  8)$.
to count the number now is fairly simple.if $n=8k$ then each $8$-cycle consists of $4$ pairs of 'oppositely located' numbers with the numbers opposite each other being $i,n+1-i$ respectively.
so for $n=8k,a_n=\frac{(4k)!}{(4!)^k} (2^3\cdot 3!)^k=2^{n/8}n!$.
and if $n=8k+1$ then once again all the numbers except $x=4k+1$ occur in $8$-cycles while $x=4k+1$ is a fixed point.so once again $a_n=2^{(n-1)/8}(n-1)!$.


a triangle with sides of 36, 40, 52 has one of its sides increased by x, a second by 2x, and a third by 3x where x is a positive integer. if the newly formed triangle is a right triangle, find the sum of all possible distinct perimeters of the newly formed triangle.

aren't there like 6 cases and within each case 3 cases? is there a better way?

not all of the 6 cases will have 3 sub-cases,
but I don't see any way of doing this other than casework

[hide]
scale down the triangle by a factor of 4 to simplify the arithmetic
9, 10, 13 triangle

case 1:
$(9+x)^{2}+(10+2x)^{2}= (13+3x)^{2}$
$181+58x+5x^{2}= 169+78x+9x^{2}$
$4x^{2}+20x+12 = 0$
$x^{2}+5x+3 = 0$ no integer solutions

partial solution: will continue
[/hide]

so anyone have a solution? the answer is 448.

due to lack of responses, i think that means there is no elegant solution. so i will brute force and see where i get.


function
real analysis
real analysis unsolved

Let D be the set of points of R^2 such |p| is less than or equal to 1. Let  f : D => D be a  function such that for any y of D there is at least one x of D such f(x) = y and            |f(p) - f(q)| is less than or equal to |p - q| for any points p, q of D.                           Show that  |f(p) - f(q)| = |p - q|.
(  |(x,y)| = sqrt(x^2 + y^2)  )


Does anybody know how to prove Euler's result that for any triangle ABC
OH=sqrt(R^2 - 4Rp) where O is the circumcenter, H the orthocenter, R is the circumradius and p is the inradius of the orthic triangle?

1. H-the orthocenter is the incenter of the orthic triangle.
2. O*- the circumcenter of the orthic triangle is the midpoint of OH and its circumradius is $\frac{R}{2}$.
So $O^{*}H^{2}= \frac{R^{2}}{4}-2 \frac{R}{2}p$ by Euler.
And the result follows.


Brian has 100 feet of fencing. He wants to enclose a play area for his puppy. If the play area must be a quadrilateral, what is the maximum number of square feet he can enclose?

[hide="my answer"]Squares have larger areas then rectangles of the same perimeter so the side of the lay area will be 25feet[/hide]

[quote="aznmonkey1992"][hide]25 on each side...i think....[/hide][/quote]

Good, because the greatest area for a quadrilateral for a given perimeter is a square (I hope everyone knows this).

Now, what would be the area?

[hide="answer"]$625ft^2$[/hide]would be the answer i believe.

Here's a question then:

what is the least possible area if the sides have to be integers?

[hide]i think the least possible area is 49ft^2.  it occurs when the sides are a rectangle with length 49 and width 1.[/hide]

correct, what about the smallest integer area?  :D

no. first he asked for smallest area with integer sides. now its smallest integer area with any positive real sides. :)

[quote="h78reg"]well, the smallest integer area is 1 when the length is some large number and the width is some small number. :)[/quote]

Actually it can be 0 if it's a straight line.  ;)

[quote="h78reg"]but, then brian's puppy can't play anywhere. :lol:[/quote]

However, math is more important than the puppy.  :rotfl:
The smallest area is a straight line (or integerly, a thin rectangle), the largest is a square. That's all I want to point out.

[quote="236factorial"]
what is the least possible area if the sides have to be integers?[/quote]
[hide]you can have a sides of 49 and 1 so the area is 49 sq. feet[/hide]

[hide="1"]
Since square's areas are greater than rectangles, the maximum area is 25*25=625.[/hide]

[hide="2"]
49, if one side has to be 1(49*1)
But if one side can be as small as it can, than the answer is 0(49*0)
I don't know which of these answers are correct. :fool:
But if I had to have only one answer, I'd go with 49*1=49[/hide]


BTW does anyone know where the AMC forum (IF there is one) is???

Anyway, my question relates the question 16 of AMC 12 B of 2006...

"Regular hexagon ABCDEF has vertices A and C at (0,0) and (7,1), respectively. What is its area?

A) $ 20\sqrt{3}$
B) $ 22\sqrt{3}$
C) $ 25\sqrt{3}$
D) $ 27\sqrt{3}$
E) $ 50$"

Here is my question...

Can anyone do a "Geoalgebra" diagram of this question because I have a hard time visualizing this?

What I mean is...

If one wanted to find the side length which we'll call "x", let's use the x-coordinates. We know that $ \frac{3}{2} \cdot x \equal{} 7 \Longrightarrow x \equal{} \frac{14}{3}$

BUT if we use the y-coordinates, then we get $ \frac{\sqrt{3}}{2} \cdot x \equal{} 7$, from which we get $ x \equal{} \frac{14}{\sqrt{3}}$

Isn't this a contradiction?

HELLO!!!!
[hide="EDIT: FORGOT HIDE AND HELLO"]
Imagine connecting A C and center O
You are creating a isoceles triangle with angles 30 30 120
The length of AC is simply $ \sqrt {1^2 \plus{} 7^2} \equal{} \sqrt {50}$
Then we find AO, and we know since it is regular hexagon, it will equal the side length.
$ \sqrt {50}/(2*sin(60) \equal{} \frac {\sqrt {150}}3$
and then using $ S \equal{} \frac {\sqrt 3} 4 a^2$ of triangle AOB
we find that the area of the hexagon is C $ 25\sqrt 3$
[/hide]
Greetings from Roooooooooooooooooooooobiks

Hello
And I have no idea where 3/2x=7 comes from while looking at the x-cord
Maybe I have a bad time visualing  :blush:

Rooobiks

Forget coordinates. Look at the attached picture. Hint: the area of an equilateral triangle is $ \frac{\sqrt{3}}{4} s^2$.

[hide="Solution"]The area of the hexagon is twice that of the orange triangle. Thus, the area of the hexagon is $ 2 \cdot \frac{\sqrt{3}}{4} \cdot(AC)^2 \equal{} \frac{\sqrt{3}}{2} (7^2\plus{}1^2) \equal{} \boxed{\textbf{(C)} \; 25 \sqrt{3}}$[/hide]

WoW!!!!!!!!!

Thanks so much TZF and everyone else!!!!!

BTW, TZF, which program did you use to generate that diagram (I want to learn to make prettiful pictures too  :D  )

I'm quite sure such diagrams can be made with any picture-processing software, like Microsoft Paint and GIMP.


Use the angle bisector theorem to find the lengths of the two segments created when the angle bisector cuts the third side. Then you have all three sides and the lengths of the two segments cut by the angle bisector, and you apply Stewart's Theorem from there

Given triangle $ ABC$ and $ AD$ is the angle bisector.
By applying Stewart's Theorem, we get
$ AD^{2}\equal{} AB.AC\minus{}DB.DC$

This also can be proven by applying the Law of Cosine.


Hi people,

I desperately need full MathCounts practice tests. The thing is, I have nothing to offer.  :(  If anybody has:

a) Free tests;
b) A online resource besides MATHCOUNTS or Elias Saab,

please PM me!

Thanks a lot in advance,
biblioman

I can give you some. What do you need?

(Also, I don't know if you meant [url=http://www.mnspe.org/mathcounts/coach.htm]this[/url] by MATHCOUNTS, but it is out there too.)

You could also ask for some in the MATHCOUNTS Resources Here! sticky.

Hi PowerofPi,

Anything is fine, the later the better. Thank you for the generosity! Also, the resource is great!

Thanks,
biblioman

EDIT: I would like Nationals a lot, but States are good too.
ANOTHER EDIT: Well, I was kicked out of the Mathcounts Resources forums.  :mad:


topology
group theory
abstract algebra
set theory
superior algebra
superior algebra unsolved

Hi,

I was reading the Hatcher book about algebraic topology more precisely the introduction of the free product of groups. The definition given in the book doesn't seem very rigorous as it uses expressions that weren't introduced before, e.g. "reduced word". One can think of a recursive definition of a (reduced) word of length $ n$ but my instructor often places emphasis on the fact that one should just take the unions of sets if one is aware of the common superset where the union is being taken. This limitation to the underlying set theory has other serious disadvantages: not only the difficulty in constructing the coproduct in the category of groups but also constructing the coproduct in the category of sets (disjoint union).

So the question may be formulated as: Is it possible to construct coproducts in the category of groups or sets without taking the union of sets (without superset)?

laxton

You really can't do anything without taking the union of sets.  What exactly do you mean?  Union has a clear meaning in ZF.

This is the definition of union given by the instructor:
Let $ A$ be a set and $ B,C\subseteq A$. Then $ B\cup C$ is defined to be the subset of $ A$ satisfying the following property $ x\in B\cup C\Leftrightarrow x\in B\vee x\in C$

If I remember correctly ZF requires unions of arbitrary sets without known superset ($ A$). If I want to define the free product of two groups then I'd have to take the union of all sets $ S_n$ where $ S_n$ denotes the set of reduced words of length $ n$. What is the common superset $ A$ in this case?

laxton

It's the set that the [url=http://en.wikipedia.org/wiki/Axiom_of_union]axiom of union[/url] asserts exists.

So what is a rigorous but handy definition of the free product?

laxton

The free product is the [url=http://en.wikipedia.org/wiki/Coproduct]coproduct[/url] in the category of groups.  This is the only important property of the free product.  What you are asking for is not a definition, but a construction, and I don't understand what you don't like about the construction you already have.

What I don't like about the construction I already have is the usage of mathematical terms which weren't defined before e.g. "words of length $ m$" or "cancellation rules". Probably the best way of making clear what I want is to start a construction:
Let $ (G_{\alpha})_{\alpha\in I}$ be a family of groups. Define $ G: \equal{}\dot\bigcup_{\alpha\in I}G_{\alpha}$ to be the set-theoretic coproduct/disjoint union of the family $ (G_{\alpha})_{\alpha\in I}$ (I accepted that it is necessary to admit arbitrary unions and that this implies the existence of coproducts in the category of sets/disjoint unions). A word of length $ m\geq 0$ is an element of $ G^m$. A word $ (g_1,\dots,g_m)\in G^m$ is called reduced iff the following conditions hold:
1) $ g_i\in G_{\alpha_i}$ and $ g_{i\plus{}1}\in G_{\alpha_{i\plus{}1}}$ for some $ \alpha_i,\alpha_{i\plus{}1}\in I$ implies $ \alpha_i\neq\alpha_{i\plus{}1}$ for all $ i\equal{}1,\dots,m\minus{}1$ and
2) $ g_i\in G_{\alpha_i}$ for an $ \alpha_i\in I$ implies $ g_i\neq 1_{G_{\alpha_i}}$.
The set of reduced words of length $ m$ is called $ G_r^m$. Then as a set $ \ast_{\alpha\in I}G_{\alpha}$ equals the union $ \bigcup_{i\in\mathbb{N}}G_r^m$. Now we want to define the product in $ \ast_{\alpha\in I}G_{\alpha}$:
Let $ (g_1,\dots,g_m),(h_1,\dots,h_n)\in\ast_{\alpha\in I}G_{\alpha}$. The juxtaposition of these two words is defined to be the following (not necessarily reduced) word of length $ m\plus{}n$: $ (g_1,\dots,g_m,h_1,\dots,h_n)\in G^{m\plus{}n}$.

Questions: How can I define the proper product in $ \ast_{\alpha\in I}G_{\alpha}$ without making use of the "algorithmic definition" of "reducing" the word according to given "cancellation rules"? This has to be kept as simple as possible as it remains to prove associativity (existence of neutral and inverse elements is clear, isn't it?) and the universal property. I'm open to other constructions of the free group if they admit an easier proof of these properties but I insist on this level of formalism altough I can understand it's repulsive nature.

laxton

What problem do you have with the algorithmic definition?  Do you not trust the definition of the free group on a set, either?

One technique to prove existence which skirts proving associativity, etc., is a common one.  You let each of the groups $ G_\alpha$ act on $ \ast_{\alpha\in I}G_{\alpha}$, and show this is a faithful action, so all of these groups embed in the permutation group on $ \ast_{\alpha\in I}G_{\alpha}$.  Now the free product will be the subgroup of the permutation group generated by these images.  This is the route taken by Massey in "A Basic Course in Algebraic Topology", section III.4.

Steve


[quote="Syler"]Real numbers $ x_{1},x_{2},..,x_{n}$ are positive. Prove the inequality:

$ \frac {x_{1}}{x_{2} \plus{} x_{3}} \plus{} \frac {x_{2}}{x_{3} \plus{} x_{4}} \plus{} ... \plus{} \frac {x_{n \minus{} 1}}{x_{n} \plus{} x_{1}} \plus{} \frac {x_{n}}{x_{1} \plus{} x_{2}} > (\sqrt {2} \minus{} 1)n$[/quote]
Q.E.D $ \Leftrightarrow \sum \frac{a_1\plus{}a_2\plus{}a_3}{a_2\plus{}a_3}\geq \sqrt{2}n$
we need to prove that:$ \prod  \frac{a_1\plus{}a_2\plus{}a_3}{a_2\plus{}a_3}\geq {\left( \sqrt{2} \right)}^{n}$(by AM-GM)
we have:$ \prod {\left( {a_1\plus{}a_2\plus{}a_3}\right)}^{2}\geq \prod {\left( a_1\plus{}\frac{a_2}{2}\plus{}\frac{a_2}{2}\plus{}a_3\right)}^{2}$
$ \geq 4\prod \left( a_1\plus{}\frac{a_2}{2}\right)\left( \frac{a_2}{2}\plus{}a_3\right)\equal{}\prod \left( \left( 2a_1\plus{}a_2\right)\left( 2a_3\plus{}a_2\right) \right)$
$ \geq \prod \left(2{\left( a_1\plus{}a_2\right)}^{2} \right)\equal{}{2}^{n}{\left( \prod\left( a_1\plus{}a_2\right) \right)}^{2}$
So we have Q.E.D :)

[quote="trungk42sp"][quote="Syler"]Real numbers $ x_{1},x_{2},..,x_{n}$ are positive. Prove the inequality:

$ \frac {x_{1}}{x_{2} \plus{} x_{3}} \plus{} \frac {x_{2}}{x_{3} \plus{} x_{4}} \plus{} ... \plus{} \frac {x_{n \minus{} 1}}{x_{n} \plus{} x_{1}} \plus{} \frac {x_{n}}{x_{1} \plus{} x_{2}} > (\sqrt {2} \minus{} 1)n$[/quote]
Q.E.D $ \Leftrightarrow \sum \frac {a_1 \plus{} a_2 \plus{} a_3}{a_2 \plus{} a_3}\geq \sqrt {2}n$
we need to prove that:$ \prod \frac {a_1 \plus{} a_2 \plus{} a_3}{a_2 \plus{} a_3}\geq {\left( \sqrt {2} \right)}^{n}$(by AM-GM)
we have:$ \prod {\left( {a_1 \plus{} a_2 \plus{} a_3}\right)}^{2}\geq \prod {\left( a_1 \plus{} \frac {a_2}{2} \plus{} \frac {a_2}{2} \plus{} a_3\right)}^{2}$
$ \geq 4\prod \left( a_1 \plus{} \frac {a_2}{2}\right)\left( \frac {a_2}{2} \plus{} a_3\right) \equal{} \prod \left( \left( 2a_1 \plus{} a_2\right)\left( 2a_3 \plus{} a_2\right) \right)$
$ \geq \prod \left(2{\left( a_1 \plus{} a_2\right)}^{2} \right) \equal{} {2}^{n}{\left( \prod\left( a_1 \plus{} a_2\right) \right)}^{2}$
So we have Q.E.D :)[/quote]

Thank you.
This is my solution:
Let the LHS be S.
$ S\geq\ (\sqrt{2}\minus{}1)n| *(\sqrt {2}\plus{}1)$
  $ S(\sqrt{2}\plus{}1)\geq\ n$
    $ \frac{S}{n}\geq\ \frac{1}{\sqrt{2}\plus{}1}$
    It enough to prove that:
$ \prod\frac{x_{1}\plus{}x_{2}\plus{}x_{3}}{x_{2}\plus{}x_{3}}\geq\ \frac{1}{(\sqrt{2}\plus{}1)^n} (AM\minus{}GM)$
It is obvious that $ \prod(x_{1}\plus{}x_{2}\plus{}x_{3})\geq\ \prod(x_{2}\plus{}x_{3})$
 and $ (\sqrt{2}\plus{}1)^n\geq\ 1$
then our inequality is true.


[quote="hendrata01"]Let $a,b,c,d$ be real numbers that satisfy:
$\frac{1}{a+1} + \frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1} = 2$ and $a,b,c,d > -\frac{1}{2}$
Then show that:
$(a^3+b^3+c^3+d^3) + 11(a+b+c+d) \geq 5(a^2+b^2+c^2+d^2) + 28$[/quote]
$(a^3+b^3+c^3+d^3) + 11(a+b+c+d) \geq 5(a^2+b^2+c^2+d^2) + 28\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^3-5a^2+11a-7)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(a^3-5a^2+11a-7+16(\frac{1}{a+1}-\frac{1}{2})\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-1)^4}{a+1}\geq0.$  :)


Will there be an Intermediate Number Theory?

Yes.  We expect to have an intermediate number theory class some time next year.  Join the mailing list at Stay Informed & you'll get an email as soon as the class is listed.


number theory
least common multiple
function
algorithm
vector

Pairs

Let there given a strict positive integer N. You must determine how much pairs of strict positive integers have the least common multiple equal to N.

Date IN

In the in folder "pairs.in" there will be the number N.

Date OUT

In the out folder "pairs.out"  you will atach the determined number.

Restrictions and advices:

    * 0 < N < 231
    * The pairs (a,b) si (b,a) are considered the same thing.

Example

perechi.in
12

perechi.out
8

Explanation: The pairs which have the least common multiple equal to 12 are: (1,12), (2,12), (3,12), (4,12), (6,12), (12,12), (3,4), (4,6)

Well, if N were a bit larger you could solve it by looking at the prime factors, but with such a tiny N a brute force method of checking the LCM of all pairs up to (N,N) will run near instantly..

Here's my solution

[code]
int Number_Of_Pairs (int N) {
int i,j;
int sum=0;

	for (i=1; i<=N; i++) if (N%i==0)
	for (j=i; j<=N; j++) if (N%j==0)
		if (GCD(i,j)==1) sum++;

return (sum);
}

int GCD(int A, int B) {

	if (A==B) return (A); else
	if (A<B) return (GCD(A,B-A)); else
	return (GCD(A-B,B));

}
[/code]

[u][color=green]Explanation[/color][/u]
If $i,j$ are two divisors of $N$ such that $(i,j)=1$ then the least common multiple of $\frac{N}{i}$ and $\frac{N}{j}$ is the number $N$

Sweet! I like this little trick. Oh by the way, you can use GCD(A, B % A) instead of GCD(A, B - A) and as a matter of fact, there are also function __gcd(a, b) and __gcd<T>(a, b) defined in <algorithm> I think. (You shouldn't use them though since they are undocumented functions.)

Actually I know this problem from a contest and the correct restriction is
$N<2^{31}$

That's more like it :). With this restriction you would have to do TripleM's factorization:
[code]
#include <cstdio>
#include <vector>

using namespace std;

vector<int> exponents = vector<int>();

int main() {
    int n;
    scanf("%d", &n);
    int result = 1;
    for (int i = 2; i * i <= n; i ++) {
        int power = 0;                  // What is the highest power of i that divides n?
        while (n % i == 0) {
            n /= i;
            power ++;
        }
        result *= 2 * power + 1;
    }
    if (n > 1) {                        // The last divisor of n is a prime in first power.
        result *= 3;
    }
    result = (result + 1) / 2;          // We counted most of the solutions twice.
    printf("%d\n", result);
    return 0;
}
[/code]


Let $ a,b$ be positive real numbers such that $ a > b$ and $ a^2 \plus{} b^2 \equal{} 1$. Let $ n > m$ be natural numbers. Prove that

$ a^m \minus{} b^m\geq \frac {2}{3}(a^n \minus{} b^n)$

[quote="radio"]Let $ a,b$ be positive real numbers such that $ a > b$ and $ a^2 \plus{} b^2 \equal{} 1$. Let $ n > m$ be natural numbers. Prove that

$ a^m \minus{} b^m\geq \frac {3}{2}(a^n \minus{} b^n)$[/quote]
Why $ \frac 32$?
Let $ a\equal{}1,b\equal{}0$ we get $ a^m\minus{}b^m\equal{}1\equal{}a^n\minus{}b^n$.

Well, I'm terribly sorry, Rust. It must be $ \frac{2}{3}$! I'll edit it.


Let $ a \leq b \leq c$. Prove that:
$ 3a \leq a \plus{} b \plus{} c \pm \sqrt {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca} \leq 3c$

[quote="ngoclam1088"]Let $ a \leq b \leq c$. Prove that:
$ 3a \leq a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca} \leq 3c$[/quote]
The first is obvious:
$ 3a \leq a \plus{} b \plus{} c  \le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}$
For the second, it is equivalent to:
$ 2c\minus{}a\minus{}b \ge \sqrt {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca} \equal{} \sqrt {\frac{1}{2}((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)}$
Then let $ c\minus{}b \equal{} x \ge 0$ and $ b\minus{}a \equal{} y \ge 0$.
The we have $ 2c \minus{}a\minus{}b \equal{} 2y \plus{} x \ge \sqrt{\frac{1}{2}(x^2\plus{}y^2\plus{}(x\plus{}y)^2)} \equal{} \sqrt{x^2\plus{}y^2\plus{}xy}$ $ \iff$
$ 4y^2 \plus{} 4xy \plus{} x^2 \ge x^2 \plus{} y2 \plus{} xy$ $ \iff$ $ 3y(x\plus{}y) \ge 0$ which is obvious. Equality iff $ y\equal{}0$ which is $ b\equal{}c$. QED.

[quote="ngoclam1088"]Let $ a \leq b \leq c$. Prove that:
$ 3a \leq a + b + c - \sqrt {a^2 + b^2 + c^2 - ab - bc - ca}$[/quote]

We can normalize to $ a + b + c = 1$ with $ a \leq b \leq c$. Then the inequality becomes
$ 1 - 3a \& \geq \sqrt {1 - 3(ab + bc + ca)} = \sqrt {1 - 3bc - 3a + 3a^2}$
and after squaring both sides
$ 2a^2 + bc \geq a \Leftrightarrow 2a^2 + bc \geq a(a + b + c) \Leftrightarrow a^2 + bc \geq ab + ac$
But the last inequality is obvious, because $ (a - b)(a - c) \geq 0$.

Nice! FelixD  :lol:
how do we can normalize to $ a\plus{}b\plus{}c\equal{}1$ with $ a \le b \le c$

The condition $ a \leq b \leq c$ is given. And we can normalize, because the inequality is homogeneous, in i.e. set $ a \equal{} a'l$, $ b \equal{} b'l$ and $ c \equal{} c'l$. Then the inequality becomes
$ l(3a') \leq l(a' \plus{} b' \plus{} c' \minus{} \sqrt {{a'}^2 \plus{} {b'}^2 \plus{} {c'}^2 \minus{} a'b' \minus{} b'c' \minus{} c'a'})$
That means, that we can choose $ l$ arbitrarily. So if we set $ l \equal{} \frac {1}{a' \plus{} b' \plus{} c'}$, then $ a \plus{} b \plus{} c \equal{} a'l \plus{} b'l \plus{} c'l \equal{} l(a' \plus{} b' \plus{} c') \equal{} \frac {1}{a' \plus{} b' \plus{} c'}\cdot(a' \plus{} b' \plus{} c') \equal{} 1$.
Moreover, if $ a \leq b \leq c$, then $ a'l \leq b'l \leq c'l \Leftrightarrow a' \leq b' \leq c'$. :wink:


inequalities
function
integration
vector
triangle inequality
Functional Analysis
topology

I have found the following thm in two different texts.
If $ X$ is a finite measure space(with positive measure).and $ 1 \leq p<q<\infty$,then $ L^{q}\subset L^{p}$.
In the proof,the writer applies holder's inequality to the functions $ |f|^{p}$ and$ 1$.
My question is that It seems that the same argument works for $ 0 < p<q<\infty$.
What goes wrong?

Holder's inequality is not true for $ 0 < p < q < 1$. Furthermore, the "norm" that goes with the $ L^p$ space violates the triangle inequality.

$ \int_{X}|f|^{p}d\mu \equal{} \int_{X}|f|^{p} \cdot 1 d\mu \leq (\int_{X}|f|^{pq/p}d\mu)^{p/q}\cdot(\int_{X}d\mu)^{1 \minus{} p/q}$
$ \equal{} (\int_{X}|f|^{q}d\mu)^{p/q}\cdot\mu(X)^{(1 \minus{} p/q)}$
Here $ 1 < q/p$.

What goes wrong?

[quote="thefuturebird"]laipou, are you still trying to write this proof for $ 0 < p < q < \infty$ when you ask "what goes wrong?" what are you asking about?[/quote]
I found this thm in two different texts.
They all assume $ 1 \leq p < q < \infty$.
I don't know why this thm fails when  $ 0 < p < q < \infty$.
Need someone to point out which step fails when  $ 0 < p < q < \infty$.

It's what Tyl said. Look at Hölder's inequality, it's only applies when 1< p < q

http://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality

(In fact, Hölder's inequality has this version for 0< p < q ... but it is *reversed*.) (EDIT, STRIKE THIS:You use Hölder's inequality in the step you showed us. So, we are saying you can't do that step unless  1< p < q.)

[quote="thefuturebird"]It's what Tyl said. Look at Hölder's inequality, it's only applies when 1< p < q

http://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality

(In fact, Hölder's inequality has this version for 0< p < q ... but it is *reversed*.) You use Hölder's inequality in the step you showed us. So, we are saying you can't do that step unless  1< p < q.[/quote]

Actually,"$ p$" here is $ q/p$,which is $ >1$.

[quote="Soarer"]The problem is as what Tyl said - the triangle inequality for $ L^p$ is false unless $ p \ge 1$. So (I guess) it is not quite meaningful to study $ L^p$ for $ p < 1$. I am not sure how your book defines $ L^p$, but wikipedia defines $ L^p$ for $ p \ge 1$ only. (http://en.wikipedia.org/wiki/Lp_space#Lp_spaces)

What you have done is okay though.[/quote]

Here $ L^{p}$ consists of equavalent class of functions(a vector space).
I'm not saying that $ \parallel{}\cdot\parallel{}_p$ must be a norm.
I just want to make sure $ L^{q}\subset L^{p}$ is true for$ 0 < p<q<\infty$.

Over a finite measure space and with $ 0<p<q\le\infty,$ $ L^q\subset L^p.$

Nothing at all goes wrong with the proof of the containment when $ p$ or $ q$ are less then $ 1.$ That's the answer to laipou's question.

There's a long-ingrained habit of only thinking about $ L^p$ for $ p\ge 1,$ but that has nothing to do with this particular result.

In $ L^p$ for $ p\ge1,$ the triangle inequality for the $ p$-norm is true - so it is a norm. $ L^p$ is a Banach space, and the dual of $ L^p$ for $ 1\le p<\infty$ is $ L^{p'},$ where $ \frac1{p}\plus{}\frac1{p'}\equal{}1.$

So what is $ L^p$ for $ 0<p<1?$ It's a vector space and a complete metric space, with $ \|f\minus{}g\|_p^p$ as the metric, but that metric is not a norm. We can't use the $ p$-norm as a norm because the triangle inequality fails, and for the same reason, the unit ball in this metric space is not convex. Over a non-atomic measure, there are no nontrivial bounded linear functionals - in other words, there's no usable dual space. Because it's not a Banach space and may not have a dual, the motivations for studying $ L^p$ for $ p<1$ are fairly weak.

Without Holder you can argue $ \int |f|^p d\mu \equal{}  \int_{\{|f|\leq 1\} }|f|^p d\mu \plus{}  \int_{\{|f| > 1\}} |f|^p d\mu \leq \mu ({\{|f|\leq 1\}}) \plus{} \int_{\{|f| > 1\}} |f|^q d\mu$.


Zach has chosen five numbers from the set $ \{1, 2, 3, 4, 5, 6, 7\}$. If he told Claudia what the product of the chosen numbers was, that would not be enough information for Claudia to figure out whether the sum of the chosen numbers was even or odd. What is the product of the chosen numbers?
------
Really I wonder how do get it? There is many probabilities  :huh: .

[hide]Clearly we are looking for an ambiguous case.  Considering the product of the two unchosen numbers (since this is equivalent to telling the product of the five numbers Zach has chosen), we see that 12 can be obtained by {3, 4} and {2, 6}, and 6 by {1, 6} and {2, 3}.

However, in the second case, the sum of the remaining five numbers is odd for both {1, 6} and {2, 3}.  Hence the product of the two remaining numbers must be 12, and so the product of the numbers Zach has chosen is:

$ \frac{1*2*3*4*5*6*7}{12} \equal{} 420$
[/hide]

I still didn't understand even with the solution maybe becuase I didn't understand the question itself. What is the needing of the colored section of the question?
[quote="HMMT 2004"]Zach has chosen five numbers from the set $ \{1, 2, 3, 4, 5, 6, 7\}$. If he told Claudia what the product of the chosen numbers was, [color=red]that would not be enough information for Claudia to figure out whether the sum of the chosen numbers was even or odd.[/color] What is the product of the chosen numbers?[/quote]

It means that there are at [i]least[/i] two ways of obtaining the product that Zach has told Claudia.  More importantly, out of these ways, at least one of them has a sum that is odd, and at at least one of them has a sum that is even.

If Zach had told Claudia that he had chosen 5 numbers which had a product of 840, then she would know that his choice had either been {2, 3, 4, 5, 7} or {1, 4, 5, 6, 7}.  However, in either case, the sum is odd.

If Zach had told Claudia that he had chosen 5 numbers which had a product of 420, then his choice would either have been {1, 2, 5, 6, 7} or {1, 3, 4, 5, 7}.
Unfortunately the first sum is odd while the second one is even, so Claudia would not have enough information to tell whether the sum was even or odd, just from knowing the product alone.


Trên mặt phẳng cho 5 điểm $ A, B, C, P, Q$ sao cho không có 3 điểm nào thẳng hàng.
Chứng minh rằng:$ AB\plus{}BC\plus{}CA\plus{}PQ<AP\plus{}AQ\plus{}BP\plus{}BQ\plus{}CP\plus{}CQ$

[quote="ntkhang-vn"]Trên mặt phẳng cho 5 điểm $ A, B, C, P, Q$ sao cho không có 3 điểm nào thẳng hàng.
Chứng minh rằng:$ AB \plus{} BC \plus{} CA \plus{} PQ < AP \plus{} AQ \plus{} BP \plus{} BQ \plus{} CP \plus{} CQ$[/quote]
Kho day Khang!Cho moi nguoi xem ban giai ra sao di??

[quote="ntkhang-vn"]Trên mặt phẳng cho 5 điểm $ A, B, C, P, Q$ sao cho không có 3 điểm nào thẳng hàng.
Chứng minh rằng:$ AB \plus{} BC \plus{} CA \plus{} PQ < AP \plus{} AQ \plus{} BP \plus{} BQ \plus{} CP \plus{} CQ$[/quote]
bài gì lạ vậy. Bạn tự sáng tác hả


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AMC 8

Has anyone received the MOEMS 2008 [b]Dr. George Lenchner Award [/b]because of the perfect score?

I have, but the school hasn't given it to me yet. I got one in 6th grade, and I couldn't compete in 7th grade due to conflicts with orchestra, and I got (will get) it again in 8th grade. :D

Of course, that doesn't show anything about how I did on the AMC/AIME/USAMO. (You don't even know if I did the AIME, let alone USAMO :D )

No, I got a 24 this year because I stink. And I got a 23 last year. Hopefully I can get a 25 next year at last.

I got gipped. On one of the problems, my math teacher took off a point since I put it as a percentage. Break of 3 year George Lenchner streak :mad:

Yongyi, when you get the award this month, please let me know so I can ask my teacher.  Thanks!

Yongyi, I checked the MOEMS web http://www.moems.org/newsletter_files/HonorRoll0708.pdf, I could not find your name on the perfect score list.  I know your last name bcause of MathCounts.  You need to check with your teacher.

Hmm... odd. Weird. AAAAA! This means that my teacher messed up in inputting the scores and I won't get the award! Sad! But wait, she said she had the medallion. Maybe the list isn't complete. :)

ugh i got 24 for the 4th straight year and in two of the four years i got the first problem on one of the tests wrong

See this:[url]http://www.moems.org/enroll.htm[/url]
I'm not sure, but I think you have to sign up as a school.

Darn, I'm too old to participate...9th grade is too old, right?

Yes. Plus, if you're in 9th grade, you should focus more on the AMCs.[/img]

haha thats weird...
i remember my teacher telling me you can use the points to buy prizes...
ROTFL
and im sure its MOEMS bcuz it says it on the trophy.
weird..

There was this organization called school-plus that offered MOEMS out of school. I got a 24 :( but I'm not trying again because it was a careless mistake and I finished most of the tests in 5 minutes. The people there said I was the best person they had which is kind of pathetic. They also have never heard of MathCounts.

Seriously? They've never heard of MATHCOUNTS? That's sad....

Well, neither have I, but they're the people at MOEMS. If your business is in math competitions, then you SHOULD know about MATHCOUNTS...

Who said that they were from MOEMS? All that Countdown King said was that some organization's people hadn't heard of Mathcounts.

Dude no offense to anyone here but quite frankly [b]I think[/b] that MOEMS is a test of how little stupid mistkaes you can make. You have like 30 minutes to do each round or some insanely large amount of time to do AMC 8 level questions.

That's your opinion.

Don't insult people like that. I find it offensive because it's not that easy for me. Sure, it's easier than say, MATHCOUNTS, but it isn't what you make it out to be.

Yeah. I'm sorry. When I was in fourth grade, it was what introduced me to contest math. But as a rule of thumb, try doing some Mathcounts problems from the handbook @isabella. You'll find MOEMS really easy. ;-)

I made my sister do that this year and all but the first round, she perfect scored

[quote="isabella2296"]My school doesn't do MOEMS. How can I participate?  :maybe:[/quote]
There are some homeschool groups that might offer it. Perhaps you can contact MOEMS and they can tell you if there is a local one.

Hi. I just received my awards today and my coach said that the school's scores (John Glenn Middle School) got messed up. Therefore, they had to fix it at the last minute and as a result, JGMS wasn't in the list for highest team achievement and I wasn't in the list for perfect scores (on the MOEMS website) even though the sheet said I scored 25.