Show that for all real numbers $ a,b,c,d,$ the following inequality holds:

\[ (a\plus{}b\plus{}c\plus{}d)^2 \leq 3 (a^2 \plus{} b^2 \plus{} c^2  \plus{} d^2) \plus{} 6ab\]

I worked on this problem for about a day and couldn't do it. So please post solution!

[hide="My efforts"]
I do know that after expanding and simplifying as much as possible, AM-GM is appealing but because of $ 6ab$, it doesn't really go anywhere. Also, using Cauchy in the beginning, it tells mes that $ (a\plus{}b\plus{}c\plus{}d)^2 \leq 4(a^2\plus{}b^2\plus{}c^2\plus{}d^2)$ but I cannot find any relation between $ 4(a^2\plus{}b^2\plus{}c^2\plus{}d^2)$ and $ 3(a^2\plus{}b^2\plus{}c^2\plus{}d^2) \plus{} 6ab$ again because of pesky term of $ 6ab$. Blah, I basically got nowhere with it.  :(  [/hide]

[hide="big hint"]
$ 6 \equal{} 2*3$, so maybe if you put the pesky term inside the parenthesis...
[hide="solution"]
$ ((a\plus{}b)\plus{}c\plus{}d)^2\leq3((a\plus{}b)^2\plus{}c^2\plus{}d^2)$
[/hide][/hide]

Oh man... All I had to do was realize how it could factor and apply Cauchy like you did. Blah  :oops:

Why Cauchy? isn't it only $ AM \le QM$?

$ (a\plus{}b\plus{}c\plus{}d)^2 \le 3 (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} 6ab$
$ (a\plus{}b\plus{}c\plus{}d)^2 \le 3 ((a\plus{}b )^2 \plus{} c^2 \plus{} d^2)$
$ a\plus{}b \equal{} t \Rightarrow (t\plus{}c\plus{}d)^2 \le 3 (t^2 \plus{} c^2 \plus{} d^2)$
$ \frac{t\plus{}c\plus{}d}{3} \le \sqrt{\frac{t^2 \plus{} c^2 \plus{} d^2}{3}} \Rightarrow AM \le QM$

[quote="Alex_ITA"]Why Cauchy?[/quote]
Cauchy says $ ((a\plus{}b)\plus{}c\plus{}d) \le (1^2\plus{}1^2\plus{}1^2)((a\plus{}b)^2\plus{}c^2\plus{}d^2)$.

[quote="ghjk"][quote="pianoforte"][quote="Alex_ITA"]Why Cauchy?[/quote]
Cauchy says $ ((a \plus{} b) \plus{} c \plus{} d) \le (1^2 \plus{} 1^2 \plus{} 1^2)((a \plus{} b)^2 \plus{} c^2 \plus{} d^2)$.[/quote]
You forget the square root sign on the RHS, don't you? :wink:[/quote]
I swear I remember putting that there.

[quote="pianoforte"][/quote][quote="ghjk"][quote="pianoforte"][quote="Alex_ITA"]Why Cauchy?[/quote]
Cauchy says $ ((a + b) + c + d) \le (1^2 + 1^2 + 1^2)((a + b)^2 + c^2 + d^2)$.[/quote]
You forget the square root sign on the RHS, don't you? :wink:[/quote][quote="pianoforte"]
I swear I remember putting that there.[/quote]
OK! Here is a counter example:a=b=c=d=1/9, so the LHS=4/9>RHS=2/9, so your inequality is wrong! :wink: .Generally, your inequality is wrong with 0<a=b=c=d<2/9.

Yeah... that's because he forgot to write the $ ^2$ to square the left hand side. What he meant to write is likely:

$ ((a \plus{} b) \plus{} c \plus{} d)^2 \le (1^2 \plus{} 1^2 \plus{} 1^2)((a \plus{} b)^2 \plus{} c^2 \plus{} d^2)$

[quote=Alex_ITA]Why Cauchy? isn't it only $ AM \le QM$?

[/quote]
Doesn't matter. Cauchy implies AM-QM


Suppose that x, y, z is postive real numbers. Prove that

$ xyz(x\plus{}2)(y\plus{}2)(z\plus{}2)\le(1\plus{}\frac{2xy\plus{}2yz\plus{}2zx}{3})^3$.

This was posted two [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=274644]times[/url]


Let ABCD  be a right angled trapezoid that A  and B  are right angles.Let  point E be midpoint  of DC  and point  F belongs  [BC] line segment.BF=8, FC=2  and  [AE] is perpendicular to [FE].Find  lengts of [AD]  line  segment.

Is the length AB given? If no, then there must be a condition missing, else the problem is not uniquely defined. If yes, then denote AB = a. We will then prove that $AD=\sqrt{a^{2}+64}-2$. In fact, the quadrilateral AFCD is a trapezoid; the point E is the midpoint of its leg CD. If M is the midpoint of its other leg AF, then the segment ME is the mid-parallel of this trapezoid, so that we have $ME=\frac{1}{2}\left( AD+FC\right) $. But FC = 2; hence, $ME=\frac{1}{2}\left( AD+2\right) $. On the other hand, since AE is perpendicular to FE, the angle < AEF = 90. This means that the point E lies on the circle with diameter AF. The center of this circle must be the midpoint of the segment AF, i. e. the point M. Therefore, the point M is the center of the circle with diameter AF, which passes through the points A, F and E. Hence, MA = MF = ME. This entails AF = MA + MF = ME + ME = 2 ME, and $ME=\frac{1}{2}\cdot AF$. But since the triangle ABF is right, we can use Pythagoras and get $AF=\sqrt{AB^{2}+BF^{2}}=\sqrt{a^{2}+8^{2}}=\sqrt{a^{2}+64}$. Hence, $ME=\frac{1}{2}\cdot AF=\frac{1}{2}\cdot \sqrt{a^{2}+64}$. Comparing this with $ME=\frac{1}{2}\left( AD+2\right) $, we see that $AD+2=\sqrt{a^{2}+64}$, so that $AD=\sqrt{a^{2}+64}-2$.

But without having a given (or some other length), we cannot calculate AD !

  Darij


Πέρυσι είχα αρχίσει να γράφω ένα άρθρο πάνω στο πλήρες τετράπλευρο, το οποίο είναι και από τα αγαπημένα μου tools στη Γεωμετρία. Το είχα παρατήσει για αρκετό καιρό, αλλά μιας και η πολλή δουλειά έχει ήδη γίνει σκέφτηκα ότι είναι αμαρτία να μην το τελειώσω. Θα αναρωτιέστε προς τι αυτή η εισαγωγή... και όχι άδικα. Λοιπόν το άρθρο έχει την εξής δομή: Θεωρία(την καλύπτω σε ικανοποιητικό βαθμό από την αρχή μέχρι το τέλος με όλα τα cross references όπου κρίνεται απαραίτητο), Εφαρμογές και τέλος Προβλήματα. Αυτό που μου λείπει είναι οι ασκήσεις για την ενότητα των εφαρμογών και των προβλημάτων. Έχω προσθέσει μόνο την περυσινή άσκηση της βαλκανιάδας και σταμάτησα εκεί. Έχω αρκετές στο αρχείο μου (που apparently είναι στην Ελλάδα) κι έχω κι ένα mail-ευγενική προσφορά του Σιλουανού με κάμποσες ασκήσεις. Αν σας ενδιαφέρει το θέμα και δεν σας πειράζει να με βοηθήσετε (εννοείται ότι τα ονόματα σας θα γραφούν στα acknowledgements), θα το εκτιμούσα πάρα πολυ. Σκέφτηκα ότι το θέμα θα είναι ενδιαφέρον και για το forum και για τα παιδία που ασχολούνται με τις Ολυμπιάδες. Οπότε αν έχετε κάποια άσκηση (κλασική,καινούργια, γνωστή ή μη, θέμα διαγωνισμών -ο,τιδήποτε) και είναι πάνω στα προαναφερθέντα σας ποτρέπω να την postarete εδώ. Επίσης αν έχεται κάποιο trick,λήμμα ή κομμάτι θεωρίας πάνω σε αυτά και δεν είναι ευρέως γνωστό και θέλετε να το μοιραστείτε μαζί μας, αυτός ο χώρος ενδείκνυται.

Ακολουθήστε τους γνωστούς κανόνες για τέτοιου είδους topics:
[color=red]
[size=150][b]ΑΣΚΗΣΗ i (ΟΝΟΜΑΤΕΠΩΝΥΜΟ)[/b][/size][/color]
........

(απαντάτε με quote στην εκφώνηση κλπ)
[b][size=150][color=red]
ΛΥΣΗ (ΟΝΟΜΑΤΕΠΩΝΥΜΟ) [/color][/size][/b]
.......

όμοια [size=150][color=red][b]ΘΕΩΡΙΑ i, ΑΠΟΔΕΙΞΗ [/b][/color][/size]

Αν το ετοιμάσω σύντομα, ίσως να φανεί χρήσιμο και στα παιδιά των ομάδων!

ps. gia tous palious...nai einai ekeino pou eixa mazi mou stin ispania...kai pou tha to etoimaza, tha to ekana vivlio klp klp...nai ekei pou htan akrivos emeine!den kounithike katholou. a oute megalose...oses selides itan toses emeine to atimo!

[quote="stergiu"]Νίκο , ωραία ιδέα ! Θα το γράψεις στην Ελληνικπή ή στα αγγλικά ; Κάτι θα έχω και γω στο αρχείο μου και θα στο στείλω.

 Μπάμπης[/quote]

κ. Μπάμπη καλή ερώτηση. Ήθελα κι εγώ να αναφερθώ σε αυτό αλλά το ξέχασα. Το κείμενο είναι ήδη γραμμένο στα Αγγλικά, οπότε θα το συνεχίσω στα Αγγλικά. Αυτό σημαίνει ότι όσοι γράφουν σε αυτή την ενότητα μπορούν να γράφουν είτε στα Ελληνικά, είτε στα Αγγλικά.


articles
USAMTS
modular arithmetic
number theory
relatively prime
greatest common divisor

Just wondering, how is the McNugget Theorem (mn-m-n) formally called?

Are there any useful applications for the Mcnugget theorem?

What is the greatest amount you can't make using 5 cent coins and 8 cent coins?

5(8) - 8 - 5 = 27 cents.

Common one though,

Is there a similar theorem for the number of quantities less than or equal to a given number that cannot be made with combinations of m and n?

Btw, here's a [url=http://en.wikipedia.org/wiki/Coin_Problem]Wikipedia article[/url] I wrote. It has some more information about the McNugget formula.

[quote="math92"]I still don't understand the McNugget Theorem.
[size=75]Also, how do you write articles for Wikipedia?[/size][/quote]

What's there not to understand? If you are given the problem: "What is the largest possible value you can't make with _____s (replace with a noun) of $m$ value and _____s (replace with same noun as before) of $n$ value?" The answer is $mn-m-n$. Look at 236!'s example.

How this formula be used?

In math contest of course?  :D

I remember seeing it on Mathcounts. As for higher levels, I think this problem once appeared on USAMTS as well (well problem using this idea).

I can't remember the year..

[quote="aidan"]What if the numbers m and n are not coprime?[/quote]

I don't think there's a greatest number that can't be made, like 2 and 4, any odd number can't be made.

Let d be the greatest common divisor of m and n. Then the greatest multiple of d that cannot be made is

d(mn-m-n)

and obviously there are infinitely many non-multiples of d that can't be made.

This revised formula works for d=1 (relatively prime) too.

Is there a formula for 3 variables?[/quote]

After thinking about it for a minute, a method for three values came to me. :)

It seems that the smallest value from applying the two-variable formula to each of the three pairs of two will be nearest to the answer.  If this happens to be a multiple of the third, you can just subtract one of the values of the first smallest pair until it works, but make sure you find the greatest possibility of that.

Any more values than that could use the same method.
There is probably a complicated usage of a common divisor of two or three of the values too. :?

maybe you divide 21 and 15 by 3 to get 7 and 5, and find that 7(5) - 7 - 5 = 23, then multiply it by 3 again?  :huh:

[quote="AntonioMainenti"][quote="chess64"]What's there not to understand?[/quote]Perhaps, how we know that the theorem is true. Can anyone prove it?[/quote]

Say $p,q$ are coprime.  Then $S=\{0,p,2p,\ldots,(q-1)p\}$ is a complete residue system mod $q$.  Say that $r>(q-1)p-q$.  Clearly $r$ is congruent mod $q$ to some element of $S$, say, $ip$.  As long as $i\not=(q-1)$, we may subtract $ip$ from $r$ to get an element that is $>0$ and $0$ mod $q$.  Hence we may write $r=ip+jq$ for some $i,j>0$.

Now, if $i=q-1$, then $r-ip>-q$.  Since $r-ip$ is divisible by $q$, we conclude that $r-ip\geq 0$.  Hence we may again write $r=ip+jq$ for $i,j>0$.

It remains to show that $(q-1)p-q$ cannot be written as $ip+jq$ for $i,j>0$.  This, however, is trivial.  It is congruent to $(q-1)p$ mod $q$ and is not congruent to $ip+jq$ for $i=0,1,2,\ldots,q-2$.  Hence $i\geq q-1$.  But then $ip+jq\geq (q-1)p+jq\geq (q-1)p>(q-1)p-q$, a contradiction.  This establishes the result.

[quote="aidan"]I believe there is a proof in the website chess64 provided. Very complex though. By the way, how bout for m = 21 and n = 15? What is the largest number that can't be made?[/quote]

Any number that isn't a multiple of 3 (meaning that the number $\ne0\text{(mod 3)}$) can't be made...

Does this mean for two numbers $m$ and $n$ with $k=gcd(m,n)$, the numbers that can't be made are $\ne 0 \pmod {k}$?


limit
ratio
geometry
floor function
probability
expected value

Anybody solving any of the following problems will recieve four times the number of points listed after the question given divided by the number of hours after it was posted.  (Half that if it's wrong, the other half recieved if it's fixed.)  I will post more and more questions.  Hide answer in one thing and solution in another.  A second solution to a problem will recieve half the number of points they would have gotten normally.  The first one to 100 points wins!!

1) 10/31, 9:30 AM
    991191 / 882282 = 1.123440125...
    999111991191 / 888222882282 = 1.124843788...
    As you add more digits, this sequence converges to what       number? (70 pts.)

2) 10/31, 9:30 AM
    100 trillion purely random integers from one to 100 are split into 50 trillion pairs of 2.  For each pair, if the sum of the squares of the two numbers is greater than 10000, add 1 to a counter 'd'.  Otherwise, add one to a counter 'c'.  After all 50 trillion pairs have run through this process, compute the expected value of 4c over c + d. (60 pts.)

More problems coming soon!

Do you need a solution, or just an anwer?  :blush:

I'll solve number 1.

[hide="Answer."]$\frac{9}{8}$.[/hide]
[hide="Solution."]Just remark that \[\begin{aligned}\underbrace{999\cdots 9}_{n\ \text{times}}\underbrace{111\cdots 1}_{n\ \text{times}}& = \sum_{i=1}^{n}9\underbrace{000\cdots 0}_{n-1\ \text{times}}1\underbrace{000\cdots 0}_{n-i\ \text{times}}\\ & = 9\underbrace{000\cdots 0}_{n-1\ \text{times}}1\cdot\sum_{i=1}^{n}10^{n-i}= \left(9\cdot 10^{n}+1\right)\left(\frac{10^{n}-1}{9}\right)\end{aligned}\]Use the same technique to see that \[\underbrace{888\cdots 8}_{n\ \text{times}}\underbrace{222\cdots 2}_{n\ \text{times}}= \left(8\cdot 10^{n}+2\right)\left(\frac{10^{n}-1}{9}\right)\]So we want to calculate \[\begin{aligned}\lim_{n\to\infty}\frac{\underbrace{999\cdots 9}_{n\ \text{times}}\underbrace{111\cdots 1}_{n\ \text{times}}}{\underbrace{888\cdots 8}_{n\ \text{times}}\underbrace{222\cdots 2}_{n\ \text{times}}}& = \lim_{n\to\infty}\frac{\left(9\cdot 10^{n}+1\right)\left(\frac{10^{n}-1}{9}\right)}{\left(8\cdot 10^{n}+2\right)\left(\frac{10^{n}-1}{9}\right)}= \lim_{n\to\infty}\frac{9\cdot 10^{n}+1}{8\cdot 10^{n}+2}\\ & = \lim_{n\to\infty}\frac{9+\frac{1}{10^{n}}}{8+\frac{2}{10^{n}}}= \frac{9+0}{8+0}= \boxed{\frac{9}{8}}\end{aligned}\][/hide]

Kurt Godel: 2*(70/9.5) pts = approximately 15 pts
(The question was asking for things of the form 99991111999111991191 / 88882222888222882282, not
99991111 / 88882222)

3) 11/01 9:00 AM
There are 19 houses on Elm Street.  A mailman notices that no two consecutive houses have no mail and no three consecutive houses all have mail.  How many ways is this possible? (50 pts.)

[quote="miyomiyo"]Kurt Godel: 2*(70/9.5) pts = approximately 15 pts
(The question was asking for things of the form 99991111999111991191 / 88882222888222882282, not
99991111 / 88882222)[/quote]

The two forms converge to the same number.  His argument relies on the fact that as we continue adding more and more digits, the number of $9$s and $8$s goes to infinity and dominate the limit behavior.  The other terms are irrelevant.

[hide="Problem 2: Answer"] $\pi$ [/hide] [hide="Problem 2: Solution"] This is the Monte Carlo Method.  We are approximating the value of $\pi$ by finding four times the ratio of the area of a quarter-circle to the area of the square circumscribed around it (which is $4 \cdot \frac{\pi}{4}$) by plotting random lattice points.

http://en.wikipedia.org/wiki/Monte_Carlo_method [/hide]

[quote="t0rajir0u"][hide="Problem 2: Answer"] $\pi$ [/hide] [hide="Problem 2: Solution"] This is the Monte Carlo Method.  We are approximating the value of $\pi$ by finding four times the ratio of the area of a quarter-circle to the area of the square circumscribed around it (which is $4 \cdot \frac{\pi}{4}$) by plotting random lattice points.[/hide][/quote]
I thought of it. But they are integer numbers from $1$ to $100$ and using Monte Carlo you are assuming real numbers from $0$ to $100$. Can you tell how good is your approximation?
[hide="Exact result using a spreadsheet"]$3.1016$
Using
$d=\sum_{i=1}^{100}\lfloor\sqrt{100^{2}-i^{2}}\rfloor=7754$[/hide]

[quote="lordWings"]they are integer numbers from $1$ to $100$ and using Monte Carlo you are assuming real numbers from $0$ to $100$. [/hide][/quote]

Imagine that each point "stands for" the lattice square to the left and below it.

Whenever an integer falls into the circle we use in the Monte Carlo method, its corresponding lattice square also falls within that circle, so the [b]expected value[/b] of the two methods should be the same.  Moreover, the areas they represent (by this scheme) are also the same.

$a, b, c$ integers?  Reals?  Positive integers?  Positive reals?

[quote="miyomiyo"][quote="cincodemayo5590"][quote="miyomiyo"]Kurt Godel gets the remaining half (15 pts)
t0rajir0u gets 60 / 6 pts (10 pts)

4) 8:30 PM
Find the values of a, b, and c so that a + b + c = 1, a + c is the absolute minimum possible, abc is the maximum, and b is minimized.  (100 pts)

I changed it so that first one to 40 pts wins.[/quote]

[hide="Umm..."] The question is a bit confusing as to which is the most important of the three criteria... if you want $a+c$ to be the smallest possible (first) then the other criteria don't make sense as we're forced into $b=1$. Similarly if you want $abc$ maximized first then we're forced into $a=b=c=\frac13$ by AM-GM, and if you want $b$ minimized first then we're forced into $a+c=1$ and $abc=0$. Keep in mind that if $a,b,c$ can be negative reals then the problem changes.

Is there some kind of weighting system for this? One sum that should be maximized? (ie. for the criteria you gave it would make sense to ask, "Maximize $abc-(a+c)-b$", or something along those lines.)[/hide][/quote]

The order in which the details were given is the order in which I want them to be weighted.[/quote]

Weighted how?

[hide="Since right now, I would say"]
$a+c=0 \Rightarrow b=1$[/hide]


function
inequalities proposed
inequalities

find the minimum of the following function:
$f(x)=\sqrt{2x^{2}+2x+1}+\sqrt{2x^{2}-2x+1}$

This page tends to formulize things kunny said see here:

[quote="kunny"]Let $A(0,-1),\ B(0,1),$ For $P(x,\ x),\ f(x)=\overline{AP}+\overline{BP}.$[/quote]

Your answer is correct, navid. :)
In my solution, when $A,\ B,\ P$ are collinar, i.e. $P(0,\ 0),\ f(x)$ is minimized.


calculus
integration
calculus computations

Can somebody solve this? I tried, and tried, ... and failed :(

ps: don't know those codes you use, sorry but new here
      did some work in paint

I deleated my post,because my solution is wrong basically.

kunny

Thanks for helping me!

But is it possible that the end of your first line should be: (see below)

Anyhow, thanks alot, wasn't sure what method to use

sorry, I was mistaken. I will edit.

kunny

Here's an alternative approach:

Let $u=\sqrt{x},\,x=u^2,dx=2u\,du.$

$\int\sqrt{x+\sqrt{x}}\,dx=\int2u\sqrt{u^2+u}\,du$

Now complete the square and let $v=u+\frac12$, obtaining

$\int(2v-1)\sqrt{v^2-\frac14}\,dv.$

$\int2v\sqrt{v^2-\frac14}\,dv=\frac23\left(v^2-\frac14\right)^{\frac32}.$

$-\int\sqrt{v^2-\frac14}\,dv$ can be handled by standard trigonometric (or hyperbolic) substitutions, or for that matter, you can look it up in the ubiquitous tables of integrals in the backs of calculus books.

[quote="fedja"][quote="kunny"]sorry, I was mistaken. I will edit.
[/quote] Alas, then you have to change all the following lines as well and the answer will not be so simple anymore...[/quote]

Yes.you are right.

After some calculation I ended up with
$\displaystyle \frac {(40x + 10\sqrt{x} - 15)\cdot \sqrt{x +
\sqrt{x}}}{60} + \frac {\sinh^{-1} (x^{\frac14})}{4}$

Anyone able to confirm?


In a store with large barrels of jellybeans, Susan scooped out 1 lb, 7 oz of red jellybeans and 12 oz of white jellybeans. Given that she wants to buy enough black jellybeans to make her purchase a whole number total of pounds, what is the smallest number of ounces of black jellybeans she can buy?

16 oz=lb.

She has 23+12=35 oz. She needs the next highest multiple of 16 which is 48, 48-35=13

she has 23+12 jellybeans already=35, 48 is the next factor of 16.

48-35=13.

Also note that the number of pounds isn't necessary in this problem.

We can just take 12+7 = 19, which is 1 lb 3 oz.

Then we need 16-3 = 13 ounces more.


geometry
inequalities
inequalities proposed

Given triangle $ABC$ area $S$ three median $m_{a},m_{b},m_{c}$ prove that
$(b+c)m_{a}+(c+a)m_{b}+(a+b)m_{c}\ge 12S$
$am_{b}+bm_{c}+cm_{a}\ge 6S$

$am_{b}+bm_{c}+cm_{a}\geq ah_{b}+bh_{c}+ch_{a}=2\left(\frac{aS}{b}+\frac{bS}{c}+\frac{cS}{a}\right)\geq 6S$ from Am-Gm :!:  :P

[color=purple]With medians? gemath if you like i think you should kill this problem [/color]
  $\frac{{m_{a}}^{2}}{{h_{a}}^{2}}+\frac{{m_{b}}^{2}}{{h_{b}}^{2}}+\frac{{m_{c}}^{2}}{{h_{c}}^{2}}\leq \frac{3R^{2}}{4r^{2}}$

It is only corollary that
$\frac{R}{2r}\ge\frac{m_{a}}{h_{a}}$ however it is quite hard!

ok Albanian Eagle your solution is so simple, I didn't think so, but my problem is special case of the inequality I posted before
Triangle $ABC$ area $S$ for all $M$ then
$bMA+cMB+aMC\ge 4S$
Now I can think it isn't too trivial! :D
And I still have three problems with median that unsolve

http://www.mathlinks.ro/Forum/viewtopic.php?t=113643
http://www.mathlinks.ro/Forum/viewtopic.php?t=109127
http://www.mathlinks.ro/Forum/viewtopic.php?t=118204

Try it if you have pleasant! :)

[quote="gemath"]
Triangle $ABC$ area $S$ for all $M$ then
$bMA+cMB+aMC\ge 4S$
[/quote]
If $M$ in triangle $ABC$, we have ${MA.BC\geq 2(S_{MAB}+S_{MCA}})$,...

If $M$ not in triangle $ABC$ then ...  :maybe:

[quote="N.T.TUAN"]
If $M$ in triangle $ABC$, we have ${MA.BC\geq 2(S_{MAB}+S_{MCA}})$,...

If $M$ not in triangle $ABC$ then ...  :maybe:[/quote]
It only work for $aMA+bMB+cMC\ge 4S$ but mine is $bMA+cMB+aMC\ge 4S$
 :D  :D
By the way there is a theorem said
$xMA+yMB+zMC\ge t$ is true for all $M$ in triangle then it is true for all $M$ in plane


probability
abstract algebra
superior algebra
superior algebra unsolved

Consider $ (G,*)$ is a group . s.t. $ |G|\equal{}n$ , where n is +ve integer.

Let $ S\equal{}\{(x,y) \in G \times G , xy\equal{}yx\}$.

if we define the probability of G as follows :

$ P(G)\equal{}\frac{|S|}{|G\times G|}$ ,

Prove that $ P(G)\leq \frac{5}{8}$ whenever $ G$ is not an abelian group.

 :D

I'm pretty sure there was another thread (with a more detailed proof), but at least I found http://www.mathlinks.ro/viewtopic.php?t=18800 .


calculus
integration
trigonometry
derivative
calculus computations

Compare $ \displaystyle f(\theta) \equal{} \int_0^1 (x \plus{} \sin \theta)^2\ dx$ and $ \ g(\theta) \equal{} \int_0^1 (x \plus{} \cos \theta)^2\ dx$ for $ 0\leqq \theta \leqq 2\pi .$

hi
here's my answer:
0 <= x < pi/4   :   f(theta) < g(theta)
pi/4 <x < 3pi/4 :   f(theta) > g(theta)
3pi/4 <x <5pi/4 :   f(theta) > g(theta)
5pi/4 < x< 7pi/4 :  f(theta) > g(theta)
7pi/4 < x <2pi   :  f(theta) < g(theta)
At all other points f(theta) = g(theta)
Hope it is correct :)
Shyam

What's your result of $ f(\theta)\minus{}g(\theta)?$

We only need to compare $ f,g$ (not necessarily evaluate each one). The difference $ f\minus{}g$ is
\[ \int_0^1 (x\plus{}\sin \theta )^2\minus{}(x\plus{}\cos \theta )^2\ dx\equal{}(\sin \theta \minus{}\cos \theta )\int_0^1 2x\plus{}\cos \theta \plus{}\sin \theta \ dx\]
\[ \equal{}(\sin \theta \minus{}\cos \theta )(1\plus{}\cos \theta \plus{}\sin \theta )\]
This difference is equal to 0 at $ \theta \equal{}\frac{\pi}{4},\pi ,\frac{5\pi}{4},\frac{3\pi}{2}$, since $ \sin \theta \equal{}\cos \theta$ when $ \tan \theta \equal{}1$, and $ 1\plus{}\cos \theta \plus{}\sin \theta \equal{}0$ when $ \cos \left( \theta \plus{}\frac{\pi}{4}\right) \equal{}\minus{}\frac{1}{\sqrt{2}}$. All zeroes are single zeros (there is no $ \theta$ for which both factors have a zero, or where a factor has a zero with zero derivative) so the sign switches between zeroes. $ f\minus{}g$ is negative at $ \theta \equal{}0$, so we get
\[ 0\leq \theta <\frac{\pi}{4},\pi <\theta <\frac{5\pi}{4},\frac{3\pi}{2}<\theta \leq 2\pi \Rightarrow f<g\]
\[ \frac{\pi}{4}<\theta <\pi, \frac{5\pi}{4}<\theta <\frac{3\pi}{2}\Rightarrow f>g\]
\[ \theta \equal{}\frac{\pi}{4},\pi ,\frac{5\pi}{4},\frac{3\pi}{2}\Rightarrow f\equal{}g\]


ratio
geometry
trigonometry
incenter
homothety
angle bisector
INMO

Let $I$ be the triangle incenter and $\Gamma \equiv (I)$ the incircle We follow the usual construction of the circle $\Gamma' \equiv (J)$ tangent to the lines $b = CA, c = AB$ and externally tangent to the incircle $(I)$ with the help of an expansion transformation. Consider the circles $(I), (J)$ and the tangents $b, c$ directed. For example, let the circle $(I)$ be directed clockwise, i.e., its radius $-r < 0$ is negative. Then the circle $(J)$ then has to be directed counter-clockwise with a positive radius $r_0 > 0$. The directed lines $b, c$ are tangents of the directed circle $(J)$, but not of the directed circle $(I)$. Perform an expansion transformation by the distance $r > 0$. The directed circle $(I)$ is carried into a circle of zero radius, i.e., into the point $I$. The directed lines $b, c$ are carried into the directed lines $b', c'$ parallel to $b, c$, respectively, but shifted by the distace $r$ to the left when looking down their directions. The directed circle $(J)$ is carried into the directed circle $(J')$ with the same center $J$ and radius $r_0 + r$, tangent to the directed lines $b', c'$, and passing through the point $I$. To construct the center $J$ of the circle $(J')$, we draw an arbitrary circle tangent to the lines $b', c'$ centered on the bisector of the angle formed by the lines $b', c'$, which is identical with the bisector of the angle formed by the lines $b, c$, i.e., with the internal bisector $AI$ of the $\angle A$. To simplify things, we choose the circle $(A)$ centered at the point $A$, which has radius $r$, i.e., it is congruent to the circle $(I)$. This circle is centrally similar to the desired circle $(J')$ with the external homothety center at the intersection $A' \equiv b' \cap c'$ of the lines $b', c'$. Using this homothety, the tangency points $U, K$ of the circles $(A), (J')$ with the line $b'$ and the intersections $V, I$ of these 2 circles with the angle bisector $AI$ are centrally cimilar with the same homothety coefficient. Hence, the lines $UV \parallel KI$ are parallel. This allows to find the tangency point $K$ of the circle $(J')$ with the line $b'$. The center $J$ of this circle is simply the intersection of a normal to the line $b'$ at the tangency point $K$ with the angle bisector $AI$. Knowing the common center $J$ of the circles $(J'), (J)$, we can easily construct the desired circle $(J)$.

Let $S$ be the tangency point of the incircle $(I)$ with the triangle side $CA$ and let $P, Q$ be the intersections of the angle bisector $AI$ with the incircle, so that the points $A, P, I, Q$ follow on the angle bisector $AI$ in this order. Obviously, the point $P$ is the tangency point of the circles $(I), (J)$. The isosceles triangles $\triangle AUV \sim \triangle JKI$ are centrally similar with the homothety center $A'$, hence, their corresponding sides are parallel. The isosceles triangles $\triangle AUV \cong \triangle ISQ$ and their sides are parallel. The angles $\angle QIS = \angle IJK$, $\angle ISQ = \angle JKI$ are easily calculated:

$\angle IJK = \angle QIS = \pi - \angle AIS = \pi - \left(\frac \pi 2 - \frac{\angle A}{2}\right) = \frac{\pi + \angle A}{2}$

$\angle JKI = \angle ISQ = \frac{\pi - \angle QIS}{2} = \frac{\pi - \angle A}{4}$

Let $S'$ be the intersection of the normal $IS \perp b, b'$ with the line $b'$. From the right angle triangle $\triangle IS'K$ wit the side $IS' = 2r$ and the angle $\angle KIS' = \angle ISQ = \frac{\pi - \angle A}{4}$, we get

$IK = \frac{IS'}{\cos \widehat{KIS'}} = \frac{2r}{\cos \frac{\pi - \widehat A}{4}}$

and from the isosceles triangle $\triangle JKI$ with the sides $JI = JK = r + r_0$ and the base angles $\angle JKI = \angle JIK = \frac{\pi - \angle A}{4}$, we get

$r + r_0 = JK = \frac{IK}{2 \cos \widehat{JKI}} = \frac{IK}{2 \cos \frac{\pi - \widehat A}{4}} = \frac{r}{\cos^2 \frac{\pi - \widehat A}{4}}$

$r_0 = \frac{r}{\cos^2 \frac{\pi - \widehat A}{4}} - r = r\ \frac{1 - \cos^2 \frac{\pi - \widehat A}{4}}{\cos^2 \frac{\pi - \widehat A}{4}} = r\ \frac{\sin^2 \frac{\pi - \widehat A}{4}}{\cos^2 \frac{\pi - \widehat A}{4}} = r\ \tan^2 \frac{\pi - \widehat A}{4}$

[quote][color=darkred]Let $\Gamma (J,\rho )$ be a circle which is tangent externally to the incircle $w=C(I,r)$ of $\triangle ABC$ and to the rays $(AB$ , $(AC$ . Find the ratio $\frac {\rho}{r}$ .[/color][/quote][b][u]Proof[/u].[/b] In the interior of the angle $\widehat {BAC}$ exist two circles,  $\Gamma_1(J_1,\rho_1 )$ and $\Gamma_2(J_2,\rho_2 )$ with mentioned properties, where $\rho_1<r<\rho_2$ and $\rho_1\rho_2=r^2$ . Prove easily that $\boxed{\frac {s-a}{r}=\frac {2\sqrt {r\rho}}{|\rho -r|}}\ (1)$ , where $a+b+c=2s$ . The relation $(1)$ is equivalently with $(s-a)^2\cdot\rho^2-2r[(s-a)^2+2r^2]\cdot\rho+r^2(s-a)^2=0$ $\iff$ $\frac {\rho}{r}\in\{\frac {(s-a)^2+2r^2\pm 2r\sqrt{(s-a)^2+r^2}}{(s-a)^2}\}$ . Since $IA^2=(s-a)^2+r^2$ obtain $\frac {\rho}{r}\in \{\frac {IA^2+r^2\pm 2r\cdot IA}{(s-a)^2}\}$ $\iff$ $\frac {\rho}{r}\in\{(\frac {IA\pm r}{s-a})^2\}$ . Since $\frac {IA\pm r}{s-a}=\frac {1\pm \frac {r}{IA}}{\frac {s-a}{IA}}=$ $\frac {1\pm \sin\frac A2}{\cos\frac A2}=$ $\frac {1\pm \cos\frac {\pi -A}{2}}{\sin \frac {\pi -A}{2}}$. In conclusion, $\{\begin{array}{c}
\rho_1=r\cdot \tan^2\frac {\pi -A}{4}\\\\
\rho_2=r\cdot\cot^2\frac {\pi -A}{4}\end{array}$ $\iff$ $\frac {\rho}{r}\in \{\tan^2\frac {\pi -A}{4}\ ,\ \cot^2\frac {\pi -A}{4}\}$ .

[quote][color=darkred][b][u]Extension[/u].[/b] Let $w=C(I,r)$ , $w_1=C(I_1,r_1)$ , $w_2=C(I_2,r_2)$ be three circles which are internally to the angle $\widehat{BAC}$ , are tangent to
rays $(AB$ , $(AC$ and the circle $w$ is externally tangent to the circles $w_1$ , $w_2$ . Then $\{r_1,r_2\}=\{r\cdot \tan^2\frac {\pi -A}{4}\ ,\ r\cdot\cot^2\frac {\pi -A}{4}\}$ .[/color][/quote]

[quote="Rushil"]Let $ABC$ be a triangle and a circle $\Gamma'$ be drawn lying outside the triangle, touching its incircle $\Gamma$  externally, and also the two sides $AB$ and $AC$. Show that the ratio of the radii of the circles $\Gamma'$ and $\Gamma$ is equal to $\tan^ 2 { \left( \dfrac{ \pi - A }{4} \right) }.$[/quote]

Shouldn't the question say that $\Gamma'$ be drawn lying inside the triangle??


topology
function
geometry
advanced fields
advanced fields unsolved

let $C$ be the unit circle in the complex plane and $D$ be the disc it bounds. let $x,y\in D-C$. give a homemorphism from D to D which interchanges x and y and fixes all the points on the circle.

Consider a special case: $y=-x$. The homeomorphism $f(re^{i\theta})=re^{i\theta+g(r)}$ will do the trick if $g$ is a continuous function such that $g(|x|)=\pi$ and $g(1)=0$.

In the general case, find a fractional linear transformation $h$ such that $h(y)=-h(x)$, and use $h^{-1}\circ f\circ h$.

well i dont no much of complex analysis. isnt there any other way? can you give me some geometric picture? i first tried to get a homoemorphism without the restrictions on the circle, ie just interchanging the two points, but even this i wasnt able to get. :(

 thanks for any kind of help

ok now i have a vague idea of what mlok is saying.  dont you mean $f(re^{i\theta})=re^{i\theta+ig(r)}$? anyway i still dont no what this map means geometry wise. can someone help?


function
integration
algebra
domain
real analysis
topology
real analysis unsolved

I read but I can not understand, so please to help me.
Let $ w(x), x\in\mathbb{R}^n$, be the function such that:
$ w\in C_0^\infty(\mathbb{R}^n), w(x)\geq 0, w(x) \equal{} 0$ if $ |x| > 1$, and $ \int_{\mathbb{R}^n}w(x)dx \equal{} 1$.
Let $ u_p(x) \equal{} \frac {1}{p^n}\int_{\mathbb{R}^n}w(\frac {x \minus{} y}{p})u(y)dy$.
And $ u_p$ is called a mollification of $ u(x)$.
Let $ \Omega '$ be a bounded domain, $ \Omega '\subset\subset\Omega$.
Why $ u_p\in C_0^\infty(\Omega)$ if $ p < dist(\Omega ',\partial\Omega)$ ???
Please help me to solve that as soon as posible. Thanks alot

your question is simple, let's see..

[b]Def.[/b] Define $ \phi \in C^\infty(\mathbb R^N)$ by

$ \phi \left( x \right): \equal{} \left\{\begin{gathered} Ce^{\frac {1}{{\left| x \right|^2 \minus{} 1}}} \quad \operatorname{if} \left| x \right| \leqslant 1, \hfill \\
0 \quad \operatorname{if} \left| x \right| > 1. \hfill \\
\end{gathered} \right.$

The constant $ C > 0$ selected so that
\[ \int\limits_{\mathbb{R}^N } {\phi \left( x \right)dx} \equal{} 1.
\]
[b]Def.[/b] For $ \varepsilon > 0$, set
\[ \phi _\varepsilon \left( x \right): \equal{} \frac {1} {{\varepsilon ^N }}\phi \left( {\frac {x} {\varepsilon }} \right).
\]
We call $ \phi _\varepsilon$ the standard mollifier. The function $ \phi _\varepsilon$ are $ C^\infty$ and satisfy
\[ \int\limits_{\mathbb{R}^N } {\phi _\varepsilon \left( x \right)dx} \equal{} 1
\]
and
\[ \operatorname{supp} \left( {\phi _\varepsilon } \right) \subset B\left( {0,\varepsilon } \right)
\]
[b]Def.[/b] If $ f : U \to \mathbb R$  is locally integrable, define its mollification $ f^\varepsilon : \equal{} \phi _\varepsilon *f$, that is,
\[ f^\varepsilon \left( x \right) \equal{} \int\limits_{\mathbb{R}^N } {\phi _\varepsilon \left( {x \minus{} y} \right)f\left( y \right)dy} \equal{} \int\limits_{B\left( {0,\varepsilon } \right)} {\phi _\varepsilon \left( y \right)f\left( {x \minus{} y} \right)dy} .
\]
[b]Def.[/b] Assume that $ U$ is open and $ \varepsilon > 0$. Denoting
\[ U_\varepsilon : \equal{} \left\{ {x \in U: \operatorname{dist}\left( {x,\partial U} \right) > \varepsilon } \right\}
\]

We shall prove that $ f^\varepsilon \in C^\infty(U_\varepsilon)$. Indeed, fix $ x \in U_\varepsilon$ and $ h$ so small that $ x \plus{} he_i \in U_\varepsilon$. Then
\[ \frac {{f^\varepsilon \left( {x \plus{} he_i } \right) \minus{} f^\varepsilon \left( x \right)}} {h} \equal{} \frac {1} {{\varepsilon ^N }}\int\limits_V {\frac {1} {h}\left( {\phi \left( {\frac {{x \plus{} he_i \minus{} y}} {\varepsilon }} \right) \minus{} \phi _\varepsilon \left( {\frac {{x \minus{} y}} {\varepsilon }} \right)} \right)f\left( y \right)dy}
\]
for some open set $ V \subset \subset U$. As
\[ \frac {1} {h}\left( {\phi \left( {\frac {{x \plus{} he_i \minus{} y}} {\varepsilon }} \right) \minus{} \phi _\varepsilon \left( {\frac {{x \minus{} y}} {\varepsilon }} \right)} \right) \to \frac {1} {\varepsilon }\frac {{\partial \phi }} {{\partial x_i }}\left( {\frac {{x \minus{} y}} {\varepsilon }} \right)\]
uniformly on $ V$, $ \frac {{\partial f^\varepsilon }}{{\partial x_i }}$ exists and equals
\[ \int\limits_U {\frac {{\partial \phi_\varepsilon }} {{\partial x_i }}\left( {\frac {{x \minus{} y}} {\varepsilon }} \right)f\left( y \right)dy}.\]
A similar argument shows that $ D^\alpha f^\varepsilon \left( x \right)$ exists and
\[ D^\alpha f^\varepsilon \left( x \right) \equal{} \int\limits_U {D^\alpha \phi _\varepsilon \left( {\frac {{x \minus{} y}} {\varepsilon }} \right)f\left( y \right)dy}\]
for each multiindex $ \alpha$.

Note the close relationship between this and the original problem asked in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169412]this recent topic[/url].

Combine the two, and we get that for an integrable (or even locally integrable) $ f,$ we can find a sequence of $ C^{\infty}$ functions that converge to $ f$ at every point of the Lebesgue set of $ f.$

[quote="bookworm_vn"][quote="kidmath"]Let $ \Omega '$ be a bounded domain, $ \Omega '\subset\subset\Omega$. Why $ u_p\in C_0^\infty(\Omega)$ if $ p < dist(\Omega ',\partial\Omega)$ ???[/quote]

In your question, since $ p < dist(\Omega ',\partial\Omega)$, then
\[ \left\{ {x \in \Omega : dist\left( {x,\partial \Omega } \right) > p} \right\} \equal{} : \Omega _p \supset \Omega '.
\]
Therefore
\[ f \in C^\infty \left( {\Omega '} \right).
\]
[/quote]
I still understand clearly. Please show me.


A bullet of mass .0157 kg and speed $ v_b$ passes completely through a pendulum bob of mass 1.3 kg. The bullet emerges with a speed $ \frac{v_b}{2}$. The pendulum bob is suspended by a flexible cord of length .96 m and negligible mass. What is the minimum value of $ v$ such that the pendulum bob will swing through a complete vertical circle?

Let's apply conservation of linear momentum to the colision between the bullet (of mass m) and the bob (of mass M) - $ v_p$ will be the bob's velocity after the bullet have emerged from it:

$ m v_b \equal{} M v_p \plus{} m \frac{v_b}{2} \equal{}> v_p \equal{} \frac{m v_b}{2M}$.

In order for the bob to describe a complete vertical circle, its minimum velocity at the top of the circle verifies (T is tension of the cord, and L its lenght)

$ T \equal{} 0 \equal{}> F_c \equal{} P \equal{}> M \frac{v_{min}^2}{L} \equal{} Mg \equal{}> v_{min} \equal{} \sqrt{2gL}$.

Let's now apply conservation of mechanical energy to the lowest and highest points of the bob's trajectory:

$ \frac{1}{2} M v_p^2 \equal{} Mg2L \plus{} \frac{1}{2} M v_{min}^2 \equal{}> \frac{m^2 v_b^2}{4M^2} \equal{} 4gL \plus{} 2gL \equal{} 6gL$

$ \equal{}> v_b \equal{} \frac{M \sqrt{24gL}}{m}$.

Substituting by the given data we get $ v_b \equal{} 1244\,\, m/s$.

In the highest point of the trajectory, Fc = T + P (where Fc in the centripetal force): the minimum velocity there corresponds to T = 0.

But $ F_c$ is in units of N but momentum $ P$ is in units N*seconds?

P is not a momentum, but the weight of the bob.


Edit: Solution removed. Part of math summer camp application.

[quote="mmahdi"]hi
oh you guys...don't you see the generalization of "cs".it's:
$({a_1}^2+{a_2}^2+...+{a_n}^2)({b_1}^2+{b_2}^2+...+{a_n}^2)({c_1}^2+{c_2}^2+...+{c_n}^2)\geq({a_1}{b_1}+...+{a_n}{b_n})({a_1}{c_1}+...+{a_n}{c_n})({b_1}{c_1}+...+{b_n}{c_n})$
ok wasn't it nice?.try to prove it by the original method of "CS".then you will find a nice strategy for using"CS" and then you can easily prove your inequality.bye[/quote]
 :o What's the purpose of your post?
$(a_1^2+a_2^2+\cdots a_n^2)(b_1^2+b_2^2+\cdots b_n^2)\geq (a_1b_1+a_2b_2+\cdots a_nb_n)^2$
$(b_1^2+b_2^2+\cdots b_n^2)(c_1^2+c_2^2+\cdots c_n^2)\geq (b_1c_1+b_2c_2+\cdots b_nc_n)^2$
$(c_1^2+c_2^2+\cdots c_n^2)(a_1^2+a_2^2+\cdots a_n^2)\geq (c_1a_1+c_2a_2+\cdots c_na_n)^2$
Multiplying these three inequalities gives the desired inequality.

[b]paladin8[/b] mentioned from another post this is problem #1 of 2006 SIMUW application quiz with deadline 2/15/2006. Lock/delete solutions... [b]wizard[/b] should respect the [b]Honor[/b] code


Consider the number 3025. If we split the digits into 30 and 25, add these two numbers and square the result, we get 3025 back. Find another four digit number, all four digits distinct, which has the same property.

[hide]

9801[/hide]

I used guess and check.

The answer is 9801.

The solution can be found in the following way. Represent the four digit number as 1000a + 100b + 10c + d. Assume, m = 10a, n = 10c. Then, the representation is changed as 100m + 100b + n + d. Then 99(m+b) + (m+n) + (b+d).

According to problem defiinition the first two digits of the four digit number and the second two digit number are added and squared. Representing this based on our form of equation, it is ((10a + b) + (10c + d)) power 2. Now, it can be represented as ((m+n) + (b+d)) power 2. Equate the above two forms based on the problem definition.

99(m+b) + (m+n) + (b+d) = ((m+n)+(b+d))power2. Assume, m+b = P, (m+n+b+d) = X. Then,

X power 2 = X + 99P. Solve for X, we get X = (1 + SQRT(1 + 396P))/2. If you find a suitable "P" such that (1 + 396P) is a perfect square, we get the answer.

teehari,

thank you for taking the
time and effort to give
(almost) all of the work,
but leaving the last bit
for the reader to finish
up on his or her own,
if wanted.

[quote="Jean-Louis"]Find another four digit number, [b]all four digits distinct,[/b] which has the same property.[/quote] I need to add text, because this post is only quote tags.

the question was a four digit number with all the four digits to be distinct. In 2025 the number "2" repeats twice.


Two semicircles are tangent to middle circle, and both semicircles and middle circle are tangent to the horizontal line as shown. If $ PQ \equal{} QR \equal{} RS \equal{} 24,$ then find the length of radius $ r$.

[img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1999Number3.jpg[/img]

[hide]Call the center of the middle circle $ O$.  Since $ \bigtriangleup OPS$ is isosceles with leg length $ r + 24$, and since $ \overline{PS} \perp \overline{OT}$, $ \bigtriangleup POT \cong %Error. "bigtriangle" is a bad command.
SOT$ by AAS.  $ \bigtriangleup POT$ is right and $ PT = \frac {3}{2}\cdot24 = 36$, $ OT = r$ and the hypotenuse length $ OP = r + 24$.  Now $ 36^2 + r^2 = (r + 24)^2$.  Simplifying both sides we get $ r^2 + 1296 = r^2 + 48r + 576$, and subtracting common terms we get $ 720 = 48r$, from which we find $ r = 15$.[/hide]

EDIT: $ \overline{PT}$ changed to $ \overline{OT}$.
EDIT2: OP fixed the problem.

 :blush: Sorry, I just read the rules that said to leave these for less experienced problem solvers.  Well, I hid it, so anyone who wants to do this won't be spoiled, and... I guess if they can't figure it out, they have a solution.

[quote="syckls"][hide]Call the center of the middle circle $ O$.  Since $ \bigtriangleup OPS$ is isosceles with leg length $ r + 24$, and since $ \overline{PS} \perp \overline{OT}$, $ \bigtriangleup POT \cong %Error. "bigtriangle" is a bad command.
SOT$ by AAS.  $ \bigtriangleup POT$ is right and $ PT = \frac {3}{2}\cdot24 = 36$, $ OT = r$ and the hypotenuse length $ OP = r + 24$.  Now $ 36^2 + r^2 = (r + 24)^2$.  Simplifying both sides we get $ r^2 + 1296 = r^2 + 48r + 576$, and subtracting common terms we get $ 720 = 48r$, from which we find $ r = 15$.[/hide][/quote]
I did same thing as he but when
$ 36^2 + r^2 = (r + 24)^2$, we can solve this little faster and without big numbers.

$ 36^2 + r^2 = r^2 + 48r +24^2$
$ 36^2 - 24^2=48r$
$ (36-24)(36+24)=48r$
$ 12*60=48r$
$ r=15$

Yup. That's what I did as well.

In problems like this (especially with AIME), avoid actually multiplying out the numbers when you have to divide (so save time and get more accuracy).


algebra
polynomial
superior algebra
superior algebra solved

Prove that x^4 + 15x^3 + 7 is irreducible in Q[x].

The easiest way, I think, is to prove that it is also irreducible $\mod 2$, since that's just checking of a few cases.

i understand mod p irreduciblity test i just don't get how mod p irreducibility test con be used to test members  of Q[x] for irreducibility.  do you apply the test in the exact same way?  I tried that and i worked because i get mod 2 of f(x) = x^4 + x^3 +1 and since f(0) = 1 and f(1) = 1 we can conclude f(x) is irreducible over Z(2).  and that proves Q but what about Q[x].  the book don't seem to explain that.  would appricate greatly any help
thanks,
gap

What do you mean with "that proves Q"¿

You must use:
A monic polynomial with integer coefficients is irreducible in $\mathbb{Q}[x]$ iff it is irreducible in $\mathbb{Z}[x]$.

Because it's the case here, you can apply the $\mod p$ test.
But to show the irreducibility $\mod 2$ it's not enough to look at $f(0),f(1)$ only.


geometry
rectangle
ratio
similar triangles

In the diagram, $ D$ is the midpoint of $ \overline{EC}$, and the area of $ \Delta EDF$ is 4 square centimeters. What is the number of square centimeters in the area of rectangle ABCD?

[asy]pair A,B,C,D,E,F;
A = (0,0); B=(3,0); C = (3,2); D = (0,2); E = (-3,2);
F = D/2;
draw(E--B); draw(A--E--C--B--A--D);
label("A",A,S); label("B",B,SE); label("C",C,NE);
label("D",D,N); label("E",E,NW); label("F",F,NE);[/asy]

Note that $ \triangle FDE\cong\triangle FAB\implies[FDE] \equal{} [FAB]\implies[ABCD] \equal{} [FAB] \plus{} [DCBF]$
$ \equal{} [FDE] \plus{} [DCBF] \equal{} [CBE] \equal{} (2)^2\cdot4 \equal{} \boxed{16}$, where the last step comes from the fact that the square of the ratio of sides of similar triangles is the ratio of their respective areas.

An easier way to solve this is to see and realize that, since the area of $\Delta EDF$ is 4 sq. centimeters, and since $\frac{\overline{ED} \times \overline{DF}}{2}$ $=$ $\frac{\overline{EF} \times \overline{FA}}{2}$, you know that $\Delta EDA=8$.

You also see that $\Delta EDA=\frac{1}{2} \text{ABCD}!!!$

Therefore, the area of rectangle $\text{ABCD}=\Delta EDA \times{2} \rightarrow \boxed{16}.$


this was the bonus on my calc test yesterday...

for what $p$ does: $\sum^\infty_{n=1} \frac {1}{n^p} (1+\frac{1}{2^p}+\frac{1}{3^p}+...+\frac{1}{n^p})$ converge?

whats the right answer?

This converges for $p>1$ and diverges for $p\le 1.$

For $p>1,$ there is a number $C$ (which depends on $p$) such that $\left(1+\frac1{2^p}+\frac1{3^p}+\cdots+\frac1{n^p}\right)\le C.$

Hence the original series converges by comparison to $\sum_n\frac C{n^p}.$

For $p\le1, \left(1+\frac1{2^p}+\frac1{3^p}+\cdots+\frac1{n^p}\right)\ge 1.$

The original series then diverges by comparison to $\sum_n\frac1{n^p}.$


how come people rarely post in the carolinas sections? looks like the entire section just died out. joe, but something fun and attractive here so it will be "happening".

eh, if you care (I don't see the point of all these legions of forums, but I digress) and another state agrees, you could probably ask a site moderator to merge with that other state to create a greater base

What do you want me to do?  I can't make the forum more active if there is hardly anyone from the Carolinas to be active...

If someone would like to post problems here more often, it would be more active.

ouch, x has a lot to do with n.

so f(x) is a constant!!!!!!

Arnav, have you really been banned for the NC forum?

(and UNC lost! ha!)


symmetry
superior algebra
superior algebra unsolved

Let $ L|K$ be a Galois extension of fields such that $ Gal(L|K)\equal{}<\sigma>$ and $ |Gal(L|K)|\equal{}a_1 a_2$, where $ a_1$ and $ a_2$ are relatively prime.  Let $ F_i$ be the fixed field of $ \sigma^{a_i}$ , for $ i\equal{}1,2$.   Then,

[b](1)[/b] if $ F_i\equal{}K(\alpha_i)$, then $ \sigma^{a_1 b}$ acts trivially on $ \alpha_1 \plus{} \alpha_1$ if and only if $ a_2 | b$.

[b](2)[/b] $ a_i$  divides $ |Gal(K(\alpha_1 \plus{} \alpha_2)|K)|$, for $ i\equal{}1,2.$

[b](3)[/b] $ L\equal{}K(\alpha_1 \plus{}\alpha_2)$.

Danken!

(1) $ \sigma^{a_1 b}(\alpha_1 \plus{} \alpha_2) \equal{} \alpha_1 \plus{} \sigma^{a_1 b}(\alpha_2)$ equals $ \alpha_1 \plus{} \alpha_2$ iff $ \sigma^{a_1 b}$ fixes $ \alpha_2$ iff $ \sigma^{a_1 b} \in \langle \sigma^{a_2} \rangle$ iff $ \sigma^{a_1 b a_1} \equal{} 1$ iff $ a_1 a_2 | a_1 b a_1$ iff $ a_2 | b$.
(2) $ \sigma^{a_1}$ is an element of order $ a_2$ in $ Gal(K(\alpha_1\plus{}\alpha_2)/K)$ by (1), thus $ a_2$ divides the order of the group. by symmetry, also $ a_1$ does.
(3) from (2) we get $ |K(\alpha_1 \plus{} \alpha_2) : K|\equal{}|L : K|$, qed.


Source: Massachusetts Mathematics League March 2003

Given a heptagon where no two diagonals are parallel, and no three intersect at one point. How many points in the interior of the heptagon are determined by the intersections of the diagonals?

[hide]
I get $35$ from drawing it out

[/hide]

It would be nicer to post a solution, so that it can be applied to maybe... a 71-gon?  :lol:

[hide]If we draw a heptagon and two diagonals so as to produce one point of intersection, it is easy to see that 4 vertices determine one point of intersection. Therefore the solution is $_7C_4$, which equals 35. [/hide]

However, i'm pretty sure there is another way to solve this...

can you explain yours in more detail? I don't get what you did.

[hide]To determine an intersection point, two diagonals must be drawn. To determine a diagonal, 2 vertices must be chosen. Therefore, it takes 4 vertices to determine a single intersection point. Since every 4 vertices produce a unique intersection point, all you have to do is $_7C_4$[/hide]

Do we need another way to solve it?  I think your method is best...


[quote="mifriend"]Legs of right triangle are 8 and 15. find altitude to the hypotenuse?[/quote]

[hide]Use $[ABC]=\frac{a\cdot h_{a}}{2}$.
You know $[ABC]=60$ and $a=17$, so solving, $h_{a}=\boxed{\frac{120}{17}}$[/hide]

hmm isn't this too easy for intermediate?

[hide="Beautiful Question even though its quite easy for Intermediate"]Just get the area 60. If the hypotenuse is  17 by pythagorean, then the answer would be $\frac{120}{17}$?[/hide]

edit: fixed typo

[hide="solution"]The hypotenuse has length 17, and the area of the triangle is 60. Therefore, the altitude to the hypotenuse is 120/17[/hide]


Find all natural k such that equation:  x(x+k) = y(y+1) has solution in positive integers.

All the integers $k \geq 4$ and $k=1$.

For $k=1$ just choose $x=y$. Now suppose that $k>1$ and that there exist $x,y$ such that $x(x+k) = y(y+1)$. (1)
Then $y>x$, so let $a = y-x \geq 1$.

The equation (1) becomes :
$x(x+k) = (x+a)(x+a+1)$ that is $kx = (2a+1)x + a^2 + a$. (2)
Since $a \geq 1$, we deduce that $kx \geq 3x + 2 > 3x$ thus $k \geq 4$.
In another hand, (2) can be rewritten as :
$k = 2a+1 + \frac {a(a+1)} x$.
For $x = a(a+1)$ it leads to $k = 2a+2$, which proves that each even integer > 2 is admissible. Conversely, for $k = 2a+2$, let's choose $x = a(a+1)$ and $y = x + a$ to have a solution of (1).

Now, let's note that for each $a$ the number $a(a+1) = 2b$ is even. Let's choose $x = b$, thus $k = 2a+3$, which proves that each odd integer $k > 3$ is a solution, and we are done.

Pierre.


Let $ Q$ be the first Fibonacci number with six different perfect square factors. How many prime factors of $ 2002$ are not factors of $ Q$?

[hide]I believe that the first Fibonacci number with 6 different square factors is 144:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, [b]144[/b], 233...

It's divisible by 1, 4, 9, 16, 36, and 144.

2002 = 2 x 7 x 11 x 13

So, the prime factors of 2002 are 2, 7, 11, and 13.

Three factors, 7, 11, and 13, are NOT factors of 144.

Therefore, the answer is [b]3[/b].[/hide]

Sorry 'bout that. It's hidden now!

I believe you are correct, just don't forget to hide your answers/solutions. :wink:


Can you change the font in a LaTeX document?

Yes, but not easily. :)

I should amend that: in some cases it is quite easy.  For example, if you want to redo your whole document in Palatino, just put
\usepackage{pxfonts}
at the beginning (before \begin{document}).

If you want to change just a particular portion of your document to a particular font, you can use the \usefont command, which is not exactly what I would call user-friendly.  For example:
\[ \text{Sample in Computer Modern \usefont{T1}{pnc}{m}{n}Sample in New Century Schoolbook \usefont{T1}{put}{b}{sl}Sample in Utopia Bold-Slanted} \]
It is a bit of a pain.  I'd check out a good LaTeX reference book (such as Chapter 7 of [i]The LaTeX Companion[/i]) or do some web-searching.


I have a probability question which I am struggling to answer.

You have a bucket of 5 apples with 1 rotten apple and 4 good apples and you randomly select 2 apples at the same time.  What is the probability that 1 of the 2 apples selected is rotten?

I've been told the answer is 40% or 2/5.

Is there a formula to use to calculate?  I was looking into taking 1/5 + 1/4 (the odds of pulling the rotten apple in a concsecutive, dependent series of events), but obviously the result ends up being 45% and not 40%.   It can't be 1/5 + 1/5, despite the right answer, because the formula couldn't be applied if you had 3 rotton apples (e.g. 3/5 + 3/5)... so any guidance (like the exact formula) would be fantastic!!

Thanks

I like to do these problems by looking at the different cases.

There are 10 total possibilities for which apples you will select (1 + 2 + 3 + 4). There are 4 possibilities for getting a rotten apple. Therefore, the answer is 40%.

Generalization: The probability of getting $c$ "different" items by removing $b$ items from $a$ total items is $\frac{b}{ac}.$

Instead of pulling out your apples at the exact same time, pull one out and then pull a second one out, without replacing the first apple. This doesn't change the problem at all.

So whats the probability of pulling out a rotten apple on the first apple? 1/5
And whats the probability of pulling out a rotten apple on the second apple? 1/4
However, be careful. Because the events are not independent. What we are really looking for in the second quetion is the probability that we pull out a rotten apple, given that we havent already found a rotten apple on the first try. So, whats the probability of not finding a rotten apple on the first try? 4/5
Thus we get 1/5+4/5*1/4=2/5


rotation
geometry
geometric transformation

Here is a nice problem, and I wanted to see how you would proceed.

Suppose a stick of uniform mass density, and length $ l$ falls from a height $ h$ above a the surface of a table.  At the begining, the stick is exactly horizontal. The edge of the stick hits the edge of the table in a perfectly elastic collision. What is the velocity of the center of mass of the stick just after the collision?  What is the angular velocity of the stick just after the collision?

Thank you.

PS: An additional interesting part that I thought for this problem, is what if the corner of the table touches the stick at a length x away from its edge? What if the collision what not table edge-stick edge, but point on stick to table edge?

Can someone please verify if my solution is correct?

Solution

[hide]

Conservation Energy:
$ Mgh \equal{} \frac 12 M v_{cm} ^2 \plus{} \frac {M l^2}{24} \omega ^2$
$ (1) 24 gh \equal{} 12 (v_{cm})^2 \plus{} (\omega l)^2$

Linear and angular momentum change.  The edge of the table will impart some angular momentum relative to the CM, while substracting some linear momentum (the table exerts a force upward)

$ F \Delta t \equal{} M \Delta v \Longleftrightarrow v_{cm} \equal{} \sqrt {2gh} \minus{} \frac FM \Delta t$
$ F \Delta t \frac l2 \equal{} I \omega$
$ (2) v_{cm} \equal{} \sqrt {2gh} \minus{} \frac {\omega l}{6}$

Am I on the right path? My goal is to substitute (2) into (1) for $ \omega$ in terms of $ v_{cm}$, but I don't want to solve that nasty equation until I'm certain my way of solving it is correct

[/hide]

I really appreciate the help.

Thanks.

The velocity of the end of the rod that hits the edge of the table shall be $ \minus{}\sqrt{2gh}$ after collision. I wonder if the rod necessarily rotates about the C.M.

The rotation doesn't need to be from the center of mass.  The motion of any extended body can be represented by translational motion and rotational motion by the CM.

Total KE, once again, can be represented as translational KE + rotational KE (internal vibrations don't count so far).


Solve the following system:
$ x\plus{}[y]\plus{}<z>\equal{}200$
$ <x>\plus{}y\plus{}[z]\equal{}190.1$
$ [x]\plus{}<y>\plus{}z\equal{}178.8$
here, [x] represents the greatest integer less than or equal to x and <x> is the fractional part of x greater than 0.

[quote="hell_ever"]Solve the following system:
$ x + [y] + < z > [] = 200$
$ < x > + y + [z] = 190.1$
$ [x] + < y > + z = 178.8$
here, [x] represents the greatest integer less than or equal to x and <x> is the fractional part of x greater than 0.[/quote]

I like this problem.  :)

[hide="Solution"]Adding all of these, we get:
\[ x + [x] + < x > + y + [y] + < y > + z + [z] + < z > = 568.9
\]
Since $ [x] + < x > = x$, we have:
\[ 2(x + y + z) = 568.9\Rightarrow x + y + z = 284.45
\]
Now subtract the first equation from this, and we have:
\[ y - [y] + z - < z > = 84.45
\]
Since $ y - [y] = < y >$, we have:
\[ < y > + [z] = 84.45
\]
Thus $ < y > = 0.45$, and $ [z] = 84$.

Substituting $ [z] = 84$ into the second equation, we get:
$ < x > + y + 84 = 190.1$
$ \Rightarrow < x > + y = 106.1$
$ \Rightarrow < x > + [y] + 0.45 = 106.1$
$ \Rightarrow < x > + [y] = 105.65$
So $ [y] = 105$, $ < x > = 0.65$

Substituting $ < y > = 0.45$ into the third equation, we get:
$ [x] + z = 178.35$
$ \Rightarrow [x] + < z > + [z] = 178.35$
$ \Rightarrow [x] + < z > = 94.35$
So $ [x] = 94$, $ < z > = 0.35$.

Finally, we can conclude:
\begin{align*} x = [x] + < x > &= 94.65 \\
y = [y] + < y > &= 105.45 \\
z = [z] + < z > &= 84.35 \\
\end{align*}[/hide]

Unless I made a calculation error  :maybe:


geometry
3D geometry
sphere
geometric transformation

Sphere X with radius 3/4 sits on top of sphere Y with radius 1. They are tangent, and the point on X which is diametrically opposite of the tangent point is marked with a dot. The dot is currently 2.5 units away from the center of Y.

Sphere X is rolled down one side of Y for 90 degrees, and now touches its equator. Then X is rolled along Y's equator 90 degrees to the right. Last, X is rolled back up to the top, 90 degrees. The spheres are in the same position now, but the dot has moved. How far is it from the dot to the center of sphere Y?

BTW Blender is useful for making 3d math figures  :)

[hide]This may not be correct but I got $\frac {\sqrt {37}}{4}$ because when the smaller sphere moved up, it traveled $\frac 14$ of the larger sphere's circumference, which is $\frac 12 \pi$. $\frac 12 \pi$ is $\frac 13$ of the smaller sphere's circumference, so the dot moved $120^{\circ}$ downward. Then,  noticing the equilateral triangle and by Pythagorean's theorem we get $\frac {\sqrt {37}}{4}$. It's hard to explain.[/hide]

Sorry. What I said first was definitely wrong. I redid it and got this.
[hide]$\sqrt {(\frac {20 - 3 \sqrt 3}{8})^2 + (\frac {9}{16})^2} = \frac {\sqrt {1317 - 120\sqrt 3}}{16}$. Uhhh explaining how I got that is rather difficult.[/hide]

Nice problem!

[hide]Not sure if I did this right, but from the first rolling, you get a change of 210 degrees in the xz plane (90 translation + 120 rolling).  The second rolling brings about another 210 degrres in the xy plane.  Then another 210 in the yz plane.  Then you just have to do some solid geometry.  I think it comes out to 58/36 + 105/64.[/hide]


inequalities
quadratics
inequalities unsolved

if  $ a,b,c$  are nonnegative reals prove that:
$ \frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(a\plus{}c)^2}\geq \frac{1}{a^2\plus{}bc}$

[quote="DIMITRIS T"]if  $ a,b,c$  are nonnegative reals prove that:
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(a \plus{} c)^2}\geq \frac {1}{a^2 \plus{} bc}$[/quote]

One way to prove this neat result is straightforward.
Just multiply out and you get

$ a^4  \minus{} 2a^2bc \plus{} bc(b^2  \minus{} bc \plus{} c^2 ) \ge 0$

Now look at the discriminant

$ (bc)^2  \minus{} bc(b^2  \minus{} bc \plus{} c^2 ) \equal{} \minus{} bc(b \minus{} c)^2 \le 0$

and the proof is complete.

thnks!!!
is there any solution using BCS?

Yes, we can use Cauchy  :D.
$ \frac{c^2}{c^2(a\plus{}b)^2}\plus{}\frac{b^2}{b^2(a\plus{}c)^2} \ge \frac{1}{a^2\plus{}bc}$
$ \frac{(b\plus{}c)^2}{c^2(a\plus{}b)^2\plus{}b^2(a\plus{}c)^2} \ge \frac{1}{a^2\plus{}bc}$ (Cauchy)
After expanding, this simplifies to
$ 2a^2\plus{}b^2\plus{}c^2 \ge 2ab\plus{}2ac$
$ (a^2\plus{}b^2)\plus{}(a^2\plus{}c^2) \ge 2ab\plus{}2ac$
True by AM-GM on each parentheses

[quote="DIMITRIS T"]if  $ a,b,c$  are nonnegative reals prove that:
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(a \plus{} c)^2}\geq \frac {1}{a^2 \plus{} bc}$[/quote]
By Cauchy,
$ (a^2 \plus{} bc)(b \plus{} c)\ge b(a \plus{} c)^2$
$ \Leftrightarrow \frac {1}{(a \plus{} c)^2}\ge \frac {b}{(b\plus{} c)(a^2 \plus{} bc)}$
$ (a^2 \plus{} bc)(c \plus{} b)\ge c(a \plus{} b)^2$
$ \Leftrightarrow \frac {1}{(a \plus{} b)^2}\ge \frac {c}{(b \plus{} c)(a^2 \plus{} bc)}$

then, add two inequality, done!

PS.Thank you, DIMITRIS T.

thanks to all!!!
especially the last solution is very clever!

PS:NextPeace in the last two inequallities I think you made a typo (but the idea is nice :))

[quote="auj"][quote="DIMITRIS T"]if  $ a,b,c$  are nonnegative reals prove that:
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(a \plus{} c)^2}\geq \frac {1}{a^2 \plus{} bc}$[/quote]

One way to prove this neat result is straightforward.
Just multiply out and you get

$ a^4 \minus{} 2a^2bc \plus{} bc(b^2 \minus{} bc \plus{} c^2 ) \ge 0$

... [/quote]

As a matter of fact, there is an even easier way to verify this inequality (not using the discriminant).
Indeed,

$ a^4 \minus{} 2a^2bc \plus{} bc(b^2 \minus{} bc \plus{} c^2 ) \ge 0$

$ <\equal{}>$

$ (a^2 \minus{} bc)^2 \plus{} bc(b \minus{} c)^2 \ge 0$

Well the discriminant in the quadratic formula is derived by completing the square, so that's kind of the same.


What are the special properties of the 4k+3 primes? :wink:

is $(\frac{x}{p})$ the Legendre symbol? :maybe:

are there any other? :wink:

Yes.

Just take any property of $4k+1$ primes and "invert" it (and look at $2$).
E.g. $4k+3$ primes never divide $x^{2}+1$.

They are the only primes that remain prime in $ \mathbb{Z}[i]$. In fact I think they're the only integers that remain prime in $ \mathbb{Z}[i]$

There is no difference between primes remaining prime and integers remaining prime ;)


Warning: bad language
Warning:Really, really bad language in the lyrics
Warning: Do not read the lyrics, bad for little kids. shoo.
[url=http://www.wired.com/news/culture/0,1284,67970,00.html]Evil. Pure Evil.[/url]

[quote]My flow is so intense that I will overflow your buffer,
Corrupt your stack pointer makin' all your data suffer.
I've got saturated edges but your flow is sparser,
Real gangstas sip on Yacc; instead you generate a parser.[/quote]

[quote="Treething"]lmaonade[/quote]
Quoted for trutheness.

Also, I cannot access your blog....

[quote="Derek"]I'd love to hear somesome rap the digits of pi[/quote]
I can do a chant, but not a rap.

[quote="Sly Si"]Wasn't that done for the ARML song contest last spring?[/quote]
yep at Penn State

wow. i just downloaded some of that geeksta rap.  :D  awesomeness.


I think it's false. Try x=y=z=3. But your last inequality with $ (x\minus{}y)^4$ was very hard, I couldn't get it!  :lol:

Maybe it's this

Rearrangement, Chebyshev, Nesbitt, AM-GM gives

$ \sum\frac {x^6}{(y^2 \plus{} z^2)^2} \ge \frac {1}{3}\left(\sum\frac {x^3}{y^2 \plus{} z^2}\right)^2 \ge \frac {1}{3}\left(\frac {1}{3}(x \plus{} y \plus{} z)\left(\sum\frac {x^2}{y^2 \plus{} z^2}\right)\right)^2 \ge \frac {1}{3}\left(\frac {x \plus{} y \plus{} z}{2}\right)^2 \ge \frac {3\sqrt [3]{(xyz)^2}}{4} \equal{} \frac {27}{4}$

My bad , is like ocha said : $ \sum_{cyc}\frac{x^6}{(y^2\plus{}z^2)^2}\ge \frac{27}{4}$ , with $ xyz\ge x^2\plus{}y^2\plus{}z^2$ .

[hide="Another solution"]By condition : $ xyz\ge x^2\plus{}y^2\plus{}z^2\ge 3\sqrt[3]{x^2y^2z^2}\Rightarrow xyz\ge 27$

Use AM-GM : $ \frac{x^6}{(y^2\plus{}z^2)^2}\plus{}\frac{y^2\plus{}z^2}{8}\plus{}\frac{y^2\plus{}z^2}{8}\ge \frac{3x^2}{4}$

So we obtain $ LHS \ge \frac{x^2\plus{}y^2\plus{}z^2}{4}\ge \frac{3\sqrt[3]{x^2y^2z^2}}{4}\ge \frac{27}{4}$[/hide]

What inequality you reffer dgreen ?


function
trigonometry
algebra unsolved
algebra

[quote="irantst"]problem 1)Find all f:N---->N such that for any imteger m,n :
               f(m+n)|m+n
problem 2)Find all f:N---->N such that for any imteger m,n :
                m+n|f(m+n)[/quote]

There are surely some errors in these problems :

problem 1) so $ f(x)|x$ and any function which associates to a natural number any of its divisor is solution.

problem 2) so $ x|f(x)$ and any function which associates to a natural number any of its multiple is solution (for exemple $ f(x) \equal{} xg(x)$ for any $ g(x)$)

[quote="mahdlbelly"]question:
problem 1)Find all f:N---->N such that for any imteger m,n :
f(m)+f(n)|m+n
problem 2)Find all f:N---->N such that for any imteger m,n :
m+n|f(m)+f(n)[/quote]

Why do you say " I made a big mistake in writing". Are you a double identity of the original poster irantst ?

maybe he/she said that because he/she had been agreed that it's from iran's exms
if you doubt you can watch his/her subject descripction
but he/she is right
I(irantst) made a mistake and problem is like mahdlbelly explained before
excuse me pco for mistake  from my side
it was a  misunderstanding

I said that because before that I said(my previous posts) it is from iran's exams and it means I accept the problem is true and i will send the soloution
but after pco said that i understood that i made a big mistake for declaring it is true
it means at first irantst after that me made mistakes!
bye
 :wink:

[quote="mahdlbelly"]question:
problem 1)Find all f:N---->N such that for any imteger m,n :
f(m)+f(n)|m+n
[/quote]

$ f(n)\plus{}f(n)|n\plus{}n$ and so $ 1\le f(n)\le n$ $ \forall n$

Let then a prime $ p>n$

$ f(n)\plus{}f(p\minus{}n)|p$ and since $ f(n)\plus{}f(p\minus{}n)>1$, we get $ f(n)\plus{}f(p\minus{}n)\equal{}p$. But since $ 1\le f(n)\le n$ and $ 1\le f(p\minus{}n)\le p\minus{}n$, we get $ f(n)\equal{}n$ which, indeed is a solution

Hence the answer : $ \boxed{f(n)\equal{}n}$ $ \forall n$

[quote="mahdlbelly"]m+n|f(m)+f(n)[/quote]

For this one, I have infinitely many solutions (including trivial $ f(n)\equal{}an$) but up to now am unable to establish if the family of solutions I got is the whole set of solution.

So i'll wait till tomorrow when you will be less busy and when you will post your own general solution.

[quote="irantst"][quote="pco"][quote="mahdlbelly"]m+n|f(m)+f(n)[/quote]

For this one, I have infinitely many solutions (including trivial $ f(n) \equal{} an$) but up to now am unable to establish if the family of solutions I got is the whole set of solution.

So i'll wait till tomorrow when you will be less busy and when you will post your own general solution.[/quote]

I found f(x)=-f(-x) so f(x) is an odd function
I tried some odd functions and all of them was true like:(except sin, tan,... )
i)f(x)=ax
ii)f(x)=a(x^(2k+1))         like f(x)=x^3,x^5,x^7,.....
so I think the soloution is f(x)=a(x^(2k+1))

I should say that the problem 1 was iran's question but I made this question(problem 2 ) myself.[/quote]

$ f(n)$ is from $ \mathbb N\to\mathbb N$ so it's impossible to speak about $ f(\minus{}x)$ and $ \sin, \tan, ..$ are meaningless.

There are many other functions than $ f(x)\equal{}ax^{2k\plus{}1}$

And since mahdlbelly will give us his general solution tomorrow, I suggest you to be a little bit patient.

[quote="mahdlbelly"]hi
I know the soloution
I will send it tommorow ,now I'm bust
bye[/quote]

Thanks for giving us your solution.

[quote="irantst"]hi pco
Please send your soloution
please
I want to know the answer dear PCO
bye :(  :?:[/quote]

As I previously said, I dont have.
I Just have a family of solutions (larger than yours) and am unable to know if this is the unique family.

I stopped searching when I discovered that mahdlbelly fully solved the problem and I'm waiting for his solution (he promised for yesterday, so I hope he'll soon post it).


Find all primes $p$ for which $5p+1$ is a perfect square. Give convincing reasons that you've found all of them and there can't be any more.

I think this may be better suited as an easy intermediate problem, although it would be fine if the Getting Started mods decide to leave it here as a tough getting started Number Theory problem.

Here is my solution, in spoiler:

[hide]

We have, for some integer n:

5p + 1 = n^2

5p = n^2 - 1

5p = (n+1)(n-1)

Since 5 and p are each prime, we must have one of the following:

5 = n + 1 and p = n - 1

5 = n - 1 and p = n + 1

The first set of equations yields n = 4 and p = 3, and the second set yields n = 6 and p = 7. Thus the only solutions are p = 3 or 7. Testing these in the original expression, we find that they each yield perfect squares.

[/hide]

I think that putting it here was why I danced around with that "Give convincing reasons ..." language rather than using the "p" word.

Of course JSRosen knows he's right, but just so the rest of you know: the correct solution is in his post. Try to do it without looking at that.

I agree with JSRosen3 as to the level, and with Kent Merryfield as to the appropriate way to sneak the "p word" into a problem that would fit in Getting Started level. Demanding a tad bit less formality emboldens the beginning problem solvers to try reasoning their way to a solution.

I haven't looked at JSRosen's solution. Here's mine for what its worth:

[hide]5p+1=x^2

5p=(x+1)(x-1)

since p is prime, either x+1 or x-1 is equal to 5. so x either equals 4 or 6. this gives us p=3 and p=7.[/hide]

hopefully its correct...

kool_dudy wrote:I'm interested in this type of problem.  Does anyone have another example?

I have one but I don't think I am allowed to show it at the moment.

Try these:

1. Let n be a positive integer. Find all values of n such that 1 + 3 + 5 + 7 + ... + (2n-3) + (2n-1) is a prime number.

2. Find all primes p that can be expressed in the form 1 + 2 + 3 + ... + n, where n is some positive integer.

3. (a bit more challenging) An nonconstant arithmetic sequence has all terms positive integers, and the sum of all the terms is a prime number. [hide]Prove[/hide]   that the sequence has at most two terms.

I looked at them and I'm still looking.

Looks hard to me but it's just to me.

I think i solved the 2. :

p=1+2+3+...+n =(n+1)n/2=(n+1) (n/2)

If it's prime, then: n+1=1 OR n/2=1   n integer, so n/2=1 and n=2

So this is true for [b]p=3[/b]. FOr p different from 3, then (n+1) (n/2) is a way to write p, with n+1 and n/2 different from 1? so p is not prime.

HUGO

I'll  try to solve the 1., i think i have to work with the same strategy.

Yes, i don't show the proof of the 1., but it's very similar with the 2.

Now it's time to work on the 3.

Let Un be an arithmetic sequence: Un=U0+nr

The sum of the n first terms is: (U0+nr)U0 (n/2) = ab with a= (U0+nr)U0
and b=(n/2)

ab is prime, so a=1 and b=p, OR a=p and b=1.

B=1 for n=2, then p=(U0+2r)U0

If n>2, then b different from 1 and a different from one too. (postive terms and so on...)

So p is prime for n=2

The manner of the solution is more appropriate for the Intermediate forum, so I will move this thread now. I leave threads in Getting Started if they get answered by real beginners, otherwise I tend to move them up.

Do you mean the original problem, or the three that I posted? If you mean the three that I posted, I can split them off to a new thread and move the original back to Getting Started. I think the original can still be in Getting Started.

Hmm...

[hide]

1 + 2 + 3 + 4 +...+ n is equal to (n)(n+1)/2

So (n)(n+1)/2 must be prime, or more specifically, odd, as two does not work.

(n)(n+1)/2 must be an integer, while (n)(n+1)/4 may not be an integer.

(n+1) is odd, so it will not affect the divisibility-by-two-ness of n.

So n must be able to be expressed as 2(s+1), where s is an integer.

(n+1)(s+1) must be prime. So n+1 must be prime, and s+1 must be 1, as n+1 cannot be 1, because if it was, then n would be 0, which could not happen.

If s is zero, then n = 2 (see above) and so n+1 = 3.

2(2+1)/2 = 2(3)/2 = 3.

3 is the only prime number that can be expressed in that form.

The next one:

If the arithmetic sequence has an odd number of terms, it would look something like this:

x-an, x-(a-1)n, x-(a-2)n,...,x,..., x+(a-1)n, x+an

The first and last terms, second and second-to-last terms, third and third-to-last terms, etc. all the way down to the terms right to the terms next to the x term will average out to be x.

My mom is forcing me to go to sleep (it's 9:44 <_< ), so I can't finish the rest of the proof. Sorry 'bout that.

[/hide]


Find a four-digit base-ten number abcd ( a is nonzero) which is equal

a^a + b^b + c^c + d^d

Unique answer: 3435

Since 6^6, 7^7, 8^8, 9^9 > 9999, the number cannot contain any digits greater than 5. This leaves (5)(6)(6)(6) = 1080 possibilities—small enough to be checked by computer or (as I did) programmable calculator.


geometry
circumcircle
trigonometry
geometric transformation
reflection
power of a point
radical axis

Since $\angle PAT+\angle PBT=\pi$ (a simple angle chase), it means that $H$ lies on the circumcircle of $APT$. Proving that the tangents in $P,T$ to $(APT)$ meet on $AH$ means, basicly, proving that $PATH$ is a harmonic quadrilateral, and this ammounts to showing that $HP\cdot AT=HT\cdot AP$, or $\frac{BT}{BP}=\frac{AT}{AP}\ (*)$.

Let $A'$ be the symmetric of $A$ wrt the midpoint of $PT$ (note that $A'$ lies on $AB$). $(*)$ is now reduced to showing that $PA'\cdot PB=TA'\cdot TB$, and this follows from the fact that $AB$ is the median of $BPT$, so the areas of $BPA',BTA'$ are equal, and $\sin(\angle BPA')=\sin(\angle BTA')$.

Or look at http://www.mathlinks.ro/Forum/viewtopic.php?t=18258 (even with the reference to Vietnam 2001...).

  darij

Sorry for reviving this topic.

Is there a way to prove this using symmedian technique?
I know that AS is the symmedian of A, but I don't know what to do from there.

Daniel

I think i've found one.
Solution:
[hide] $AS$ is the symmedian of $\triangle APT$
Let $AB\cap PT=R$ So by power of point and radical exsis theorem $PR=RT$. It implies $AR$ is median of $\triangle APT$. Let $AS\cap  c(APT)=H'$. Then $\angle H'PT=\angle H'AT= \angle PAR =\angle BPR$. Same way $\angle H'TP= \angle BTP$ and so $H'=H$

[/hide]

[quote="DanielKang"]Thanks! What is radical axis theorem?[/quote]
Check the link: http://en.wikipedia.org/wiki/Radical_axis

(And it's not a direct theorem, just a property of Radical_axis)

[hide="Quick"]Let $M$ be the midpoint of $PT$. Since $M$ has the same power WRT the two circles, it lies on their radical axis $AB$. Angle chasing yields $H$ is the intersection of the $A$-symmedian of triangle $APT$ with $\odot(APT)$, so we're done. $\blacksquare$

Notice that $B$ is the same point as $K$ [url=http://www.artofproblemsolving.com/Forum/blog.php?u=90347&b=95047]here[/url].[/hide]

$ AS $ is symmedian of $ APT $ (well-known lemma). It is easy to see that $ H $ is on the circumcircle of $ APT $ We need to show that $ \angle APS = \angle PHS $ which is true since $ APS $ is similar to $ PHS $ due to $ \angle SPH = \angle PTH = \angle PAH $ and a common angle.

Dear Mathlinkers,
AB is also the A-median of APT... which offers a new way...
Sincerely
Jean-Louis

[hide="Solution"]
Note that by Angle-by-Tangent, $\angle PTB=\angle BAT$ and $\angle TPB=\angle PAB$. Since $H$ is the reflection of $B$ across $PT$, it follows that $H$ is on the circumcircle of $APT$. It suffices to show that $HTAP$ is harmonic. Denote $BH\cap PT=X$. In $\Delta PBX$, we find that $\frac{\sin{\angle TPB}}{BX}=\frac{\sin\angle PXB}{PB}$, and in $\Delta BTX$, we find that $\frac{\sin{\angle PTB}}{BX}=\frac{\sin{\angle TBX}}{BT}$. Dividing the two, we find that $\frac{\sin{\angle TPB}}{\sin{\angle PTB}}=\frac{PB}{BT}=\frac{PH}{HT}$. Now denote $AB\cap PT=M$. Similarly, we use $\Delta PAM$ and $\Delta AMT$ to find that $\frac{\sin\angle PAM}{\sin\angle TAM}=\frac{PA}{AT}$. But we know from before that $\angle PAM=\angle TPB$ and $\angle TAM=\angle PTB$, so $\frac{PH}{HT}=\frac{PA}{AT}$; $APHT$ is harmonic, and we may conclude. $\blacksquare$[/hide]

$B$ is the $A-$ HM point of $\Delta APT$ it is well known that reflection of HM point on the opposite side lies on symmedian coming from the corresponding vertex so done $\blacksquare$

Dear Mathlinkers,

http://www.mathlinks.ro/Forum/viewtopic.php?t=18258

Sincerely
Jean-Louis

As $\angle PHT=\angle PBT=\angle PBA+\angle TBA=\angle PTA+\angle TPA=180^{\circ}-\angle PAT \implies H$ lies on $(APT)$.
Clearly, $HS$ is $H-$symmedian of $\Delta HPT$, it reduces to prove $HA$ is $H-$symmedian of $\Delta HPT$.

As $P$ and $T$ are tangent to the two circles and $AB$ is radical axis, $AB$ will bisect $PT$, say at $X$.
This reduces to prove $\angle AHT=\angle XHP$.

This is now just an angle chase,$$\angle AHT=\angle APT=\angle PBX=\angle XHP$$
as desired. $\blacksquare$

Since, $\odot (APT)$ is the reflection of $\odot (BPT)$ over $PT$, hence, $H$ lies on $\odot (APT)$, since, $ATHP$ is harmonic $\implies$ $A,S,H$ collinear

[quote=orl]In the plane let two circles be given which intersect at two points $A, B$; Let $PT$ be one of the  two common tangent line of these circles ($P, T$ are points of tangency). Tangents at $P$ and $T$ of the circumcircle of triangle $APT$ meet each other at $S$. Let $H$ be a point symmetric to $B$ under $PT$. Show that $A, S, H$ are collinear.[/quote]

Solution:
$AS$ is the symmedian

Define the point: $M=AB \cap PQ$

$AM$ is median in $\triangle{APQ}$ because the Radical Axis bisects the common tangent

We have to show that $H$ is on $AS$
$PF$: we have $\angle PBQ = \angle PHQ$
$\angle PAB = \angle BPQ$ and $\angle BAQ = \angle BQP$
$\angle PBQ = 180 - \angle BPQ - \angle BQP$
$\angle PBQ = 180 - \angle PAB - \angle BAQ$
$\angle PBQ = 180 - \angle PAQ = \angle PHQ$
$\implies P,A,Q,H$ are concyclic

By symmetry we have :
$\angle HPQ = \angle BPQ = \angle HAQ = \angle PAB$
By the definition of the symmedian $H$ must lie on $AS$
$Therefore$, $A , H , S$ are collinear
And we won a pizza :P

Observe that $AS$ is symmedian of $\triangle APT$ and that $B$ is $A$ Humpty point of $\triangle APT$. Thus we get the following equivalent problem:
[quote]Given $\triangle ABC$ with orthocenter $H$ and $A$ Humpty point $H_A$. Let $D$ reflection of $H_A$ over $BC$. Prove that $D$ lies on $A$ symmedian. [/quote]

It is well -known that $H_A$ lies on $\odot(BHC)$. Consequently we have that:
$$ \angle BHC = \angle BH_AC = \angle BDC = 180^{\circ} - \angle A$$
Thus $D$ lies on $\odot(ABC)$. Now let $E$ be reflection of $H$ over line $BC$, which clearly lies on $\odot(ABC)$. Now observe that $EHH_AD$ is isosceles trapezoid by symmetry. On another hand by radical axis theorem on $\odot(ABC)$, $\odot(BHH_AC)$ and $\odot(EHH_AD)$ we have that lines $ED$, $HH_A$ and $BC$ are concurrent point $X_A$ with $(X_A, AH \cap BC;B,C) =-1$. To finish the problem note that:
$$ -1=(X_A,AH\cap BC;B,C) \overset{E} = (D,A;B,C)$$
This implies that $ABCD$ is harmonic quadrilateral and consequently $D$ lies on $A$ symmedian.

Observe that $\angle PBA=\angle TPA$ and $\angle TBA=\angle PTA$, so
\[\angle PAT=180-(\angle PTA+\angle TPA)=180-(\angle TBA+\angle PBA)=180-\angle PBT=180-\angle PHT,\]
which implies that $H$ lies on circle $(PAT)$.

Clearly $HS$ is the $H$-symmedian of $\triangle HPT$. Let $AB$ intersect $PT$ at $C$. By the radical axis theorem, $C$ is the midpoint of $PT$. Note that $$\angle PHC=\angle PBC=\angle PBA=\angle TPA=\angle THA,$$
so $A$ lies on the $H$-symmedian of $\triangle PHT$. Hence, $A, S, H$ are collinear.

Let $\omega=(ATP).$ Notice $$\measuredangle TAP=\measuredangle TPA+\measuredangle ATP=\measuredangle TBP=\measuredangle THP$$ so $H$ lies on $\omega.$ Letting $C$ be the reflection of $B$ over $M,$ the midpoint of $\overline{PT},$ we see $$\measuredangle PCT=\measuredangle TBP=\measuredangle PAT$$ so $C$ also lies on $\omega.$ In addition, by homothety at $B,$ we see that $\overline{PT}\parallel\overline{CH}.$ Hence, $CHTP$ is a cyclic isosceles trapezoid so $\angle TAH=\angle MAP$ and $H$ lies on the $A$-symmedian of $\triangle ATP.$ But $S$ also lies on the $A$-symmedian so we are done. $\square$

Let $SA$ meet $APT$ at $K$. Let $AB$ meet $PT$ at $M$. $AS$ is symmedian of $PKT$ and $M$ is midpoint of $PT$ so $\angle TKA = \angle PKM$ and we have $\angle TKA = \angle TPA = \angle MBP$ so $K$ is $H$ so $S,A,H$ are collinear.

Long time i dont see 2 circles joining problems.
[b][color=#f00]Claim 1:[/color][/b] $H$ lies on $(APT)$
[b][color=#f00]Proof:[/color][/b] By angle chasing
$$\angle PHT=\angle PBT=\angle PBA+\angle BAT=\angle TPA+\angle PTA=180-\angle PAT \implies H \in (TAP)$$
[b][color=#f00]Claim 2:[/color][/b] Let $AB \cap PT=M$ then $M$ is midpoint of $PT$.
[b][color=#f00]Proof:[/color][/b] By PoP
$$MP^2=MA \cdot MB=MT^2 \implies MP=MT \implies M \; \text{midpoint of} \; PT$$
[b][color=#f00]Finishing:[/color][/b] Now by angle chasing
$$\angle MHT=\angle MBT=\angle PTA=\angle PHA \implies HA \; \text{symedian of} \; \triangle PHT$$
Since $HA$ is symedian on $\triangle PHT$ we have that it passes through $S$ hence we are done  :blush:

Let $\angle PAB=\theta_1,\angle QAB=\theta_2$, $\alpha=\angle PAQ=\theta_1+\theta_2$, $\angle APQ=\beta$, and $\angle AQP=\gamma$.

By circumcircle tangency, $\angle BPQ=\theta_1$ and $\angle BQP=\theta_2$, so $\angle PBQ=180-\alpha$. This means that $\angle PHQ=180-\alpha$ as well, so $H$ lies on $(APQ)$. Thus, it suffices to show that $APHQ$ is a harmonic quadrilateral. From law of sines on $\triangle HPQ$, we have $$\frac{HQ}{HP}=\frac{\sin\theta_1}{\sin\theta_2}$$

Note that $AB$ is a median in $\triangle APQ$ since the radical axis intersects $PQ$ at its midpoint. Thus, by ratio lemma, $$\frac{\sin \theta_1}{\sin \theta_2}=\frac{AQ}{AP}=\frac{\sin\beta}{\sin\gamma}.$$

Since $\frac{AQ}{AP}=\frac{\sin \beta}{\sin \gamma}$ by law of sines, so $$\frac{AQ}{AP}=\frac{HQ}{HP},$$ thus $APHQ$ is harmonic as desired.

[b]remark:[/b] This problem was stated in the very unnatural reference frame of circles $(ABP)$ and $(AQB)$ as the "main objects". A much more natural reference frame for this problem is $\triangle APQ$, and defining the two circles as e.g. the circle tangent to $PQ$ at $P$ passing through $A$, which motivates the solutions more.


[url=http://scientificamerican.com/article.cfm?chanID=sa006&articleID=00010347-101C-14C1-8F9E83414B7F4945&pageNumber=4&catID=2]Scientific American article[/url]

This article implies that mathematical ability (or any kind of ability for that matter) is not at all genetic.  According to the article, "what matters is not experience per se but "effortful study," which entails continually tackling challenges that lie just beyond one's competence."

So this is why AoPS is so adept at producing mathematical geniuses!

I don't mean to bring up the whole Nature vs. Nurture debate; this article just strikes me because it reminds me so much of what AoPS is doing.

I wasn't the only one struck by that article, then  :P

The point it makes seems obvious when viewed from one angle, but very deep from another.


a, b ,c are all positive

prove: (a/b+b/c+c/a)^2≥(a+b+c)(1/a+1/b+1/c)


superior algebra
superior algebra solved

Suppose that [b]$R$[/b] is a commutative ring with unity and [b]$I$[/b] is a [i]proper[/i] ideal of [b]$R$[/b] with the condition that

      if [b]$r \in R$[/b] and $r \not \in I$, then $r$ is a unit.

Show that [b]$I$[/b] is a [i]maximal[/i] ideal of [b]$R$[/b] and that [b]$I$[/b] is [i]unique[/i].

Suppose there exists ideal $M$, s.t. $I \subset M \subset R$, then if any $x \in M\I$ is unit then there exists $x^{-1}$ in $R$ and since $M$ is ideal we obtain $1=xx^{-1} \in M$ and $M=R$.
The same argument works for uniqueness:if there exists another [b]proper[/b] ideal $I_2 \neq I_1$, then
since any $x\in I_2\I_1$ is unit and $1=xx^{-1} \in I_2 \Rightarrow I_2=R$ .


is the answer... [hide]342.9[/hide]

If it is, then I will post how I got it

No, it's not. That's what we put down...

[quote="Ubemaya"]A water container with a circular bottom tapers linearly to a circular top with a radius one-half that of the base. A mark halfway up the side say "200 gallons." How many gallons, to the nearest tenth, does the water container hold?[/hide][/quote]
[hide]Extend to make a cone.
The volume of the cone is (37*200)/64=(37*25)/8
The volume of that weird thing is that times $\frac{3}{4}$ so our answer is $\frac{111 \cdot 25}{8}=346.9$ to the nearest tenth (unless I'm mistaken).[/hide]

[hide] I got 248.5 the first time, 351.5 the second time, and 135.6 the third time. I don't think third times the charm[/hide]

Okay, I just figured out a solution, with massive hints from solution posters, especially bpms.

First extend the thingy to make a cone. The area of it that ISN'T the 200 gallon part is $\frac{3}{4}^{3}$, so the area of the 200 gallon part is $\frac{37}{64}$ of it. The area of the entire half-cone is $1-\frac{1}{4}^{3}$. So the answer is \[200 \times \frac{64}{37}\times \frac{7}{8}= 302.7\].

Yes!  That's what I got too.  I found the volume of the weird things, left them in terms of pi(r)^2, and used proportions.


Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, $n^p -p$ is not divisible by $q$.

Let $q\neq p$ be a prime and let $g$ be a primitive root modulo $q$. Then $g^{a}\equiv p\pmod q$ for some positive integer $a$.

We now see that $n^{p}\equiv p\pmod q$ has no solution in $n$ if and only if $bp\equiv a\pmod{(q-1)}$ has no solution $b$, since there must be a $b$ so that $g^{b}\equiv n\pmod q$.

However, $bp\equiv a \pmod{(q-1)}$ certainly has a solution in $b$ if $p$ has an multiplicative inverse modulo $q-1$, ie if $\gcd(p,q-1)=1$, so $p\mid q-1$ is a necessary condition.

Now, if $n^{p}\equiv p\pmod q$ for some $n$, then it follows that
\[p^{\frac{q-1}{p}}\equiv n^{p\cdot\frac{q-1}{p}}\equiv n^{q-1}\equiv 1\pmod q,\]

so let's prove that there exist a prime so that
[list]
[*] $q\equiv 1 \pmod p$
[*] $p^{\frac{q-1}{p}}\not\equiv 1\pmod q$, that is, $ord_{q}(p)\nmid \frac{q-1}{p}$.[/list]

We shall now prove that there exist a prime $q$ which satisfies
[list]
[*] $q\equiv 1\pmod p$
[*] $ord_{q}(p)=p$
[*] $p\nmid \frac{q-1}{p}$, that is, $p^{2}\nmid q-1$[/list]

Since $p$ is prime, we have $ord_{q}(p)=p$ if and only if
[list]
[*] $p^{p}\equiv 1\pmod q$
[*] $p^{1}\not\equiv 1\pmod q$[/list]
which is equivalent to
\[p^{p-1}+\ldots+p^{2}+p+1\equiv 0 \pmod q.\]

Let $q_{1}\cdot\ldots\cdot q_{l}= p^{p-1}+\ldots+p^{2}+p+1$, where $q_{i}$ are prime numbers. By construction, every $q_{i}$ satisfies $p\mid q_{i}-1$, however $q_{i}\equiv 1\pmod{p^{2}}$ cannot hold for every $q_{i}$ since this would imply $p+1\equiv 1\pmod{p^{2}}$, a contradiction.

Hence, we have found a $q$ which satisfies all conditions. $\Box$

At first consider that:
\[\frac{p^p-1}{p-1}\equiv{p^{p-1}+p^{p-2}+\cdots+p^2+p+1}\equiv{p+1}\not\equiv{0}~~~\mbox{(mod p^2)}\]
Now if all of the prime factors of $\frac{p^p-1}{p-1}$ like $q$ be equivalent $1$ modulo $p^2$, then we have:
\[ \frac{p^p-1}{p-1}\equiv {1}\not\equiv{p+1}~~~\mbox{(mod p^2)}\]
Which is a contradiction. So there exists a prime $q$ such that $q|\frac{p^p-1}{p-1}$ and $q\not\equiv{1}~~\mbox{(mod p^2)}$

We claim that this $q$ convince all of the properties of the problem.

Now we check two cases:

[size=150]Case 1:[/size]

\[q|p-1~~\Longrightarrow~~{p}\equiv{1}~~\Longrightarrow~~{p^p}\equiv{1}~~~\mbox{(mod q)}\]

[size=150]Case 1:[/size]

\[{q}\nmid{p-1}~~\Longrightarrow~~gcd(p-1,q)=1~~\Longrightarrow~~{p^p-1}\equiv{0}~~\Longrightarrow~~{p^p}\equiv{1}~~~\mbox{(mod q)}\]
So we always have $p^p\equiv{0}~~\mbox{(mod q)}$.

Now suppose that $q|n^p-p$, which means:
\[{n^p}\equiv{p}~~\Longrightarrow~~{n^{p^2}}\equiv{p^p}\equiv{1}~~~\mbox{(mod q)}\]
Now let $ord_{q}{(n)}=d$. We have:
\[d|p^2~~\Longrightarrow~~\left\{\begin{array}{cc}{a)}d=1\\{b)}d=p\\{c)}d=p^2\end{array}\right.\]
If $d=1$ then we have $n\equiv{1}~~\mbox{(mod q)}$, so:

\[1\equiv{n^p}\equiv{p}~~~\mbox{(mod q)}~~\Longrightarrow~~{p^{p-1}+p^{p-2}+\cdots+p^2+p+1}\equiv{p}~~~\mbox{(mod q)}\]
But we have:
\[\frac{p^p-1}{p-1}\equiv{p^{p-1}+p^{p-2}+\cdots+p^2+p+1}\equiv{0}~~~\mbox{(mod q)}\]
So we should have $p\equiv0~~\mbox{(mod q)}$ which is a contradiction.

If $d=p$ then we have $1\equiv{n^p}\equiv{p}~~\mbox{(mod q)}$ which follows the previous case.

And at last if $d=p^2$, it means that $p^2|\varphi{(q)}=q-1$ which is a contradiction since we supposed that $q\not\equiv{1}~~\mbox{(mod p^2)}$.

So we are done. :lol:
[hide="PS"]This is not my own solution and I've only made some changes in a proof which I'd read it in a book. :) [/hide]

Thank you all for the problem and answers.

Is there some example of the using this result?


A sequence of positive integers $ a_1, a_2, a_3, ...$ is defined as follows: $ a_1 \equal{} 1, a_2 \equal{} 3, a_3
\equal{} 2, a_{4n} \equal{} 2a_{2n}, a_{4n\plus{}1} \equal{} 2a_{2n} \plus{} 1,$
$ a_{4n\plus{}2} \equal{} 2a_{2n\plus{}1} \plus{} 1, a_{4n\plus{}3} \equal{} 2a_{2n\plus{}1}$. Show that the sequence is a permutation of the positive integers.


Given A,B, and C are digits (ie. AB is a two digit number), if AA^B = ABA and A+B+C =AB then AA x C= ?

How would you do this problem?

Is it 88, its easy to see that if its in the for ABA, then it has to be 11^2, so 121, now 1+2+c=11, C=8, so 11*8=88?

The smallest two digit number is 10, and when we cube this, we get a 4 digit number.  Therefore B must be equal to 2.

$32^{2}=1024$, so AA must be either 11 or 22.  It is easy to check that it can only be 11, in which case A=1, B=2, and C=AB-A-B=12-1-2=9, so $AA\times C=11\times 9=99$.


A bag contains only blue and green marbles. If three green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be $ \frac{2}{5}$. If, instead, seven more blue marbles are added to the bag, the probability of obtaining a blue marble would be $ \frac{5}{8}$. What was the number of blue marbles in the bag before any changes were made?

$ \frac{g\minus{}3}{b\plus{}g\minus{}3} \equal{} \frac{2}{5}$

Cross multiply

$ 5g\minus{}15 \equal{} 2b\plus{}2g\minus{}6$

$ 3g\minus{}2b \equal{} 9$

$ \frac{b\plus{}7}{b\plus{}g\plus{}7} \equal{} \frac{5}{8}$

Cross multiply

$ 8b\plus{}56 \equal{} 5b\plus{}5g\plus{}35$

$ 21 \equal{} 5g\minus{}3b$

Now we have a system of equations. We wish to cancel the green marbles so multiply the first equation by 5 and the second by 3.

$ 15g\minus{}10b \equal{} 45$

$ 15g\minus{}9b \equal{} 61$
________________(-)

$ \minus{}b \equal{} \minus{}16$

$ \boxed{b\equal{}16}$

The answer is 18. What he did wrong above was that 21*3 is 63, not 61.


Consider the two functions \[f(x) = x^2+2bx+1\quad\text{and}\quad g(x) = 2a(x+b),\] where the variable $x$ and the constants $a$ and $b$ are real numbers. Each such pair of the constants $a$ and $b$ may be considered as a point $(a,b)$ in an $ab-$plane. Let $S$ be the set of such points $(a,b)$ for which the graphs of $y = f(x)$ and $y = g(x)$ do NOT intersect (in the $xy-$ plane.). The area of $S$ is

$\textbf{(A)} \ 1 \qquad \textbf{(B)} \ \pi \qquad \textbf{(C)} \ 4 \qquad \textbf{(D)} \ 4 \pi \qquad \textbf{(E)} \ \text{infinite}$

[hide]$\boxed{B}$
We need $x^2+2bx+1=2ax+2ab$ to not have any real solutions, implying that the discriminant is less than or equal to zero. Actually calculating the discriminant and simplifying, we get $a^2+b^2 <1$, which describes a circle of area $\pi$.
[/hide]

[quote="deej21"]Correct answer, but it should be pointed out the discriminant must be $<0$, not $\le 0$.  If it equals 0, there is still one real root.[/quote]

ah. that's right. my bad. i'll fix it in the solution.