OK, let's propose another idiot test:

1 - How do you introduce a Giraffa in a refrigirator?

[hide=""Correct Answer""]
Open the door of the refrigirator, put the Giraffa inside, close the door or the refrigirator.
This question mesures your skills to solve simple problems, and proves that you usually try to do it in a very complicated way.
[/hide]

2 - How can you introduce an Elefant in a refrigirator?
[hide=""Correct Answer""]
Open the door of the refrigirator, put the Giraffa outside the refrigirator, put the elefant inside, close the door or the refrigirator.
This question mesures your skills to think about the efects of your past actions.
[/hide]

3 - The Lion King is giving a party for the animals. All the animals go except one. What animal does not go?
[hide=""Correct Answer""]
The elefant, because it is inside the refrigirator.
This question mesures your skills to retain information in your memory.
[/hide]

If you didn't anwer the first three questions right, you have a last chance. Try harder!

4 - You have to cross a river where it is known that because there lives a very dangerous specie of crocodriles. How do you cross it?
[hide=""Correct Answer""]
You just swim across the river. All the crocodriles are in the party of the Lion King.
This question mesures your skills to learn of your mistakes.
[/hide]

You feel frustrated now. Try this test in a 4 years old kid. 90% of them will get the four questions right. Now, you can feel frustrated!!

Best,

I've seen these before, promise i didnt peek-

[hide]1. Open the door and put the giraffe in it, and close the door.
2. Open the door, take the giraffe out, put the elephant in and close the door.
3.The elephant, because he's in the refrigirator
4. Swim across, the crocodiles are at the Lion Kings party[/hide]

[hide="My answer."]Enter the refrigerator, then perform an inversion with respect to the refrigerator.  The elephant, giraffe, Lion King, crocodiles, and all other animals are now in the refrigerator, and you are out, satisfying numbers 1 and 2.  The elephant gets tired of such a cramped space, sharing a refrigerator with all the other animals, so it leaves before the party begins.  Thus, it is the one that missed out on the party, so elephant is the answer to number 3.  As for the river, all the crocodiles are in the refrigerator, so you can cross it safely.[/hide]

Am I right?  :wacko:

I think this would be more convincing if for the first two questions "the refrigerator" was used instead of "a refigerator" so that it is clear that it is a specific refigerator being reffered to.

my friends grandma told me this one before :P

Awe... I got all 4 wrong. Eh, shows you how much common sense I've got.

But here's an alternate solution:

1.) Chop up the giraffe - and stuff it in, since the giraffe, LOGICALLY, can't fit. Oh, WHAT NOW!!! Logicalness. LOL.

2.) The same process - measures your ability to repete actions.

3.) The LION. They all arrive at the party, but since they fear the lion, they all GO home, but the LION stays.

4.) Get a boat.


A geometric solid is to be built by joining the faces of 26 cubes, each having volume of one cubic inch. What is the positive difference between the number of square inches in the largest and smallest possible square areas of such a solid?

The maximum is when the sides of the solid are equal. $ 26\equiv2\pmod{3}$ so there cannot be an integer value for making a cube out of the 26 unit cubes. So the largest becomes $ 9\cdot9\cdot8\equal{}648$. The smallest is having all the unit cubes form a line 1 by 1 by 26. Therefore the solution is $ 648\minus{}26\equal{}\boxed{622}$

26 is close to 27 which is a perfect cube.

So the object with the smallest surface area is a cube missing a corner.

The surface area would be 54 sq units.

The object with the largest surface area is a long piece 1*1*26.

The surface area would be 106.

106-54 is 52.

I think that's the answer...


trigonometry
probability
real analysis
real analysis unsolved

Hi! I have a problem with convergence or divergence the series

$ \sum_{n\equal{}1}^{\infty}\frac{\sin n^2}{n}$

Please help!!!

A series like this should converge with probability 1. That is not the same as certainty.

This rests on whether $ n^2$ is uniformly distributed mod $ 2\pi$. That is very likely to be true, and very hard to prove.

[url=http://en.wikipedia.org/wiki/Weyl%27s_criterion]Weyl's criterion[/url] is your friend.  This is given as an exercise in Stein and Shakarchi and the key, which looks manageable, is the estimate

$ \left| \sum_{n\equal{}1}^{N} e^{2 \pi i f(n)} \right|^2 \le c \frac{N}{H} \sum_{h\equal{}0}^{H} \left| \sum_{n\equal{}1}^{N\minus{}h} e^{2 \pi i( f(n\plus{}h) \minus{} f(n))} \right|$

where $ c$ does not depend on $ N, H, f$ and $ H \le N$.  One then takes $ f(n) \equal{} [ \frac{n^2}{2\pi} ]$.


group theory
number theory proposed
number theory

Given $ p$ a prime number, an integer sequence is called Fibonacci-like, if $ a_{n \plus{} 2} \equiv a_{n \plus{} 1} \plus{} a_n (mod p)$, and $ 0 \leq a_n \leq p \minus{} 1$.
(a) Show that there is some minimal $ r$ such that $ a_{n \plus{} r} \equal{} a_n$ for all $ n$. Call such $ r$ to be a period.
(b) Show that periods $ r$ reaches its maximum when $ a_0 \equal{} 0,a_1 \equal{} 1$.
(c) Show that any period for any other $ a_0$ and $ a_1$" divides $ r$
--

(a) is a trivial thing. Only (c) was appeared on high school students test.

You want "is divisible by $ r$."  This is true of any periodic sequence, so I don't know why it's specialized to this problem.  The relevant structure here is that of the cyclic group $ C_r$.

I meant in (c) "period for any other $ a_0$ and $ a_1$" divides $ r$.


inequalities
function
integration
inequalities unsolved

let f,g be real functions such that f(x),g(x)
\[ \in
\]
I
\[ \left( {\forall x \in \left[ {a,b} \right]} \right)
\]
and
\[ f \prec g
\]
. if  f and g are nonincreasing then for every convex function
\[ \begin{array}{l} \phi : I \to R \\
\int_a^b {\phi \left( {f\left( x \right)} \right)} dx \le \int_a^b {\phi \left( {g\left(x \right)} \right)} dx \\
\end{array}
\]


geometry
geometric transformation
reflection
trigonometry
function
AMC

1. If x is acute and tan(2x) = (cos(x) - sin(x))/(cos(x) + sin(x)), compute |tan(4x)|.

2. For any set S, let |S| denote the number of elements of S and n(S) denote the number of subsets of S, counting S itself and the empty set.  If n(A) + n(B) + n(C) = n(Union(A, B, C)) and |A| = |B| = 100, what is the minimum possible value of |Intersection(A, B, C)|?

Note: Edited to reflect typos in question 2.

[quote="JBL"]
2. For any set S, let |S| denote the number of elements of S and n(S) denote the number of subsets of S, counting S itself and the empty set.  If n(A) = n(B) = n(Union(A, B, C)) and |A| = |B| = 100, what is the minimum possible value of |Intersection(A, B, C)|?[/quote]

Shouldn't that be n(A) + n(B) + n(C) = n(Union(A, B, C))?   I did that AMC problem just the other day, unless you're thinking of something else.

JBL wrote:1. If x is acute and tan(2x) = (cos(x) - sin(x))/(cos(x) + sin(x)), compute |tan(4x)|.

[hide]2tanx / (1+tan^2(x) = 2sinx*cosx

2sinx*cosx = (cos(x) - sin(x))/(cos(x) + sin(x))

2sinx*cos^2(x)+2sin^2(x)*cosx = (cos(x) - sin(x))

(2sinx*cos^2(x)+2sin^2(x)*cosx+sinx)/cosx= 1

2sinxcosx + 2sin^2x + tanx = 1

sin(2x) + tan(x) = cos(2x)

And here, I get stuck.[/hide]

Mystic points out that I typoed in the first line of that solution, which is doubtlessly part of the problem, if not all of it.

Quote:1. If x is acute and tan(2x) = (cos(x) - sin(x))/(cos(x) + sin(x)), compute |tan(4x)|.

My Solution

[hide]

tan(2x)^2 = (cos(x)-sin(x))^2 / (cos(x)+sin(x))^2

simplifying gives:

(1-cos(4x)) / (1+cos(4x)) = (1-sin(2x)) / (1+sin(2x))

(1-cos(4x))(1+sin(2x)) = (1+cos(4x))(1-sin(2x))

simplifying:

sin(2x) = cos(4x)

sin(2x) = sin(90-4x)

x=15

|tan(15*4)| =  root(3)

Is this the right answer?[/hide]

For #1 you might also try [hide]multiplying top and bottom by (cos x - sin x) and it simplifies nicely into trig functions of (2x). [/hide]

Yes, I also got [hide] root 3 [/hide].

yey Linda for the pretty solution ...

(By the way, in case anyone was wondering, I'm Linda's (wyra's) personaly cheerleader :)

Thanks, Alison -- typos with the plus signs and all.  Are you 100% sure it's not n(A) + n(B) = n(C) = n(Union(A, B, C))?

[quote="JBL"]
Thanks, Alison -- typos with the plus signs and all.  Are you 100% sure it's not n(A) + n(B) = n(C) = n(Union(A, B, C))?[/quote]

Pretty much.  It's interesting both ways, though.

Joel... a cheerleader?  I feel honored.

#2. (not very sure) For readability, let A n B mean Intersection(A,B), A u B mean Union(A,B).

[hide]n(A) = 2|A| for any set A. Then 2|A| + 2|B| + 2|C| = 2|A u B u C|

2100 + 2100 + 2|C| = 2|A u B u C|

2101 + 2|C| = 2|A u B u C|

|C| = 101 and |A u B u C| = 102

|A n B| = |A| + |B| - |A u B| for any sets A, B (?)

To minimize |A n B n C| we should maximize |A u B|.. I think. Let |A u B| = 102.

|A n B n C| = |A n B| + |C| - |(A n B) u C|. Minimal |A n B| = 98; maximal |(A n B) u C| = 102. Both conditions are possible simultaneously.

|A n B n C| = 98 + 101 - 102 = 97? Heh.. eh..?[/hide]

I like it for sure up to [hide]|A u B u C| = 102.  Then it's easier (and more verifiable) than what you did:  there are 102 elements between the three of them.  1 of them isn't in C, 2 of them aren't in A, and 2 of them aren't in B.  This means that at most 5 elements are not in one of A, B, or C.  So at least 97 are in all of them.[/hide]

Ah! That's much nicer. I get it now, thanks.

"To minimize |A n B n C| we should maximize |A u B|.. I think. Let |A u B| = 102."

I think this amounts to the same thing, basically.  I'm wondering if your method would work equally well in all cases if we just picked numbers for |A|, |B|, |C| and |A u B u C|


I'm sorry to double post, but could someone please give a solution? Thanks so much.

[quote="ragnarok23"]In the figure, circle K has diameter ABl circle L is tangent to circle K and to AB at K;l and circle M is tangent to circle K, circle L, and AB. The ratio of the area of circle K to the area of circle M is equal to $\frac{A}{1}.$ Find A.[/quote]

[hide]Hint:
Draw segments connecting the centers of the circles and find equations....[/hide]

[hide]Answer: 16[/hide]


USAMTS
induction
number theory
greatest common divisor
combinatorics unsolved
combinatorics

I posted this problem few days ago in USAMTS forum for USAMTS practice but this appears little bit too hard.

Problem was in:

http://artofproblemsolving.com/Forum/viewtopic.php?t=40771

[hide="I tried to.."]

I tried to expand $\binom{n}{1} = n, \binom{n}{2} = \frac{1}{2} n(n-1)$ and so on but this didn't get me far. I thought about induction but I don't seem to get even close to the answer.

BTW, I do not know the answer. [/hide]

So can anyone help? Appreciated. (I need full-proof not just "do that" and "do this" and "this will appear. QED")

I will use the two following well-known results :

a) Lucas' theorem :

Let $p$ be a prime.
Let $n = n_qp^q + \cdots + n_1p+n_0$ be the expansion of the integer $n$ in base $p$ (so that $n_q \neq 0$).
For $k=1, \cdots, n-1$, let $k=k_qp^q + \cdots + k_1p+k_0$ be the expansion of $k$ in base $p$, which may have been completed so as to have exactly $q+1$ digits.

Then $C_n^k = \prod_{i=0}^q C_{n_i}^{k_i} \mod [p].$

In particular, $C_n^k$ is divisible by $p$ if and only if $n_i < k_i$ for some $i$.

b) Legendre's theorem :
With the above notations, the exponent of $p$ in the prime decomposition of $n!$ is $\nu _p (n!) = \frac {n- s_p(n)} {p-1}$, where $s_p(n) = n_0+ \cdots + n_q$, the sum of the digits of $n$ in base $p$.

Now, if $n = p^a$ for some prime $p$.
Then $n_q=1$ and $n_i = 0$ for all $i<q$. Thus, for each $k \in \{1, \cdots , n-1 \}$ there is some $i<q$ such that $k_i > 0 = n_i$. Using a), it follows that $p$ divides $C_n^k$. Thus $p$ divides the desired $gcd$, say $g$.
In another hand, using b) for $k=p^{a-1}$, we easily deduce that $\nu_p (C_n^k) = \nu_p(p^a!) - \nu_p(p^{a-1}! )- \nu_p ( (p^a-p^{a-1})!) = \frac {p^a-1} {p-1} - \frac {p^{a-1} - 1} {p-1} - \frac {p^a-p^{a-1} - p + 1} {p-1} = 1.$
Therefore, $p^2$ does not divides $g$.

At last, since $g$ has to divides $C_n^1 = p^a$, it is clear that no other prime divides $g$, so that $g=p.$

Now, assume that $n$ is not the power of a prime.
Let $p$ be any prime. It follows that either there is some $j<q$ such that $n_j \neq 0$ or $n_q >1.$
In the first case, consider $k$ such that $k_i = n_i$ for all $i$ except that $k_j =n_j-1$. In the second case, consider $k$ such that $k_i = n_i$ except $k_q = n_q-1.$
In both cases, using a), we deduce that $C_n^k \neq 0 \mod [p].$
It means that $p$ does not divides $g$. Since $p$ is arbitrary chosen, we deduce that $g=1.$

Pierre.


inequalities
function
induction
algebra unsolved
algebra

Let $ \{a_n\}^{\infty}_1$ be a sequence of real numbers such that $ a_1 \equal{} 2,$ and \[ a_{n\plus{}1} \equal{} a^2_n \minus{} a_n \plus{} 1, \forall n \in \mathbb{N}.\] Prove that \[ 1 \minus{} \frac{1}{2003^{2003}} < \sum^{2003}_{i\equal{}1} \frac{1}{a_i} < 1.\]

[quote="orl"]Let $ \{a_n\}^{\infty}_1$ be a sequence of real numbers such that $ a_1 \equal{} 2,$ and
\[ a_{n \plus{} 1} \equal{} a^2_n \minus{} a_n \plus{} 1, \forall n \in \mathbb{N}.
\]
Prove that
\[ 1 \minus{} \frac {1}{2003^{2003}} < \sum^{2003}_{i \equal{} 1} \frac {1}{a_i} < 1.
\]
[/quote]
$ a_{n \plus{} 1} \equal{} a^2_n \minus{} a_n \plus{} 1$ thus $ \frac{1}{a_n}\equal{}\frac{1}{a_n\minus{}1}\minus{}\frac{1}{a_{n\plus{}1}\minus{}1}$
hence $ \sum^{2003}_{i \equal{} 1} \frac {1}{a_i}\equal{}\frac{1}{a_1\minus{}1}\minus{}\frac{1}{a_{2004}\minus{}1}$
The last is easy :)

How is the left side of the inequality true?

Why do you say it trivial sir????
I havenot seen any thing that yoy say trivial?

[quote="arkanm"]LHS inequality is saying that $1-\frac{1}{2003^{2003}}<\frac{1}{a_1-1}-\frac{1}{a_{2004}-1}\Leftrightarrow a_{2004}-1>2003^{2003}$. Now from the recursion observe that $a_{n+1}-1=a_n(a_n-1)$, hence by repeatedly applying the recursion to $a_n-1$ we get $a_{n+1}-1=(a_1-1)\prod_{i=2}^na_i=\prod_{i=2}^na_i$. It suffices to show that $\prod_{i=2}^{2003}a_i>2003^{2003}$, which is trivial.[/quote]
First, your indexing is wrong; it should be $a_{n+1} - 1 = \prod_{i=1}^na_i$; but that is mostly irrelevant. What is relevant is your claim that $\prod_{i=1}^{2003} a_i> 2003^{2003}$ is trivial, which[b] it isn't[/b] (at least at first sight).

We want to show that $a_{n+1} > n^n + 1$. This can easily be checked for the first few small values of $n$. The function $f(x) = x^2 - x + 1$ is increasing on $[1,\infty)$, and we have $a_{n+2} = f(a_{n+1})$, so $a_{n+2} = f(a_{n+1}) > f(n^n+1) = n^{2n} +n^n +1$, which we would like to be at least $(n+1)^{n+1} +1$. But $n^{2n} > (n+1)^{n+1}$ is equivalent to $\dfrac {n^n}{n+1} > \left (\dfrac {n+1} {n}\right )^n = \left (1 + \dfrac {1} {n}\right )^n$.
Now, $LHS > 3$ for $n\geq 3$, and $RHS < 3$ (from the known proof for $\textrm{e}$), so induction does the job.

Funny problem :P The LHS bound is pretty stupid and somewhat of a red herring.

Notice that $$a_{n+1} = a_n(a_n-1) \iff \frac 1{a_n} = \frac 1{a_n-1} - \frac 1{a_{n+1}-1}.$$ As a result $$\frac 1{a_1}+\frac 1{a_2}+\cdots + \frac 1{a_{2003}} = 1 - \frac 1{a_{2004} - 1}.$$ The first display yields further that $a_{n+1}-1 > (a_n-1)^2$, so recursing $$a_{2004} - 1 > (a_2-1)^{2^{2002}} = 2^{2^{2002}} > (2^{2003})^{2003} > 2003^{2003},$$ and obviously $a_{2004} - 1 > 0$, proving both bounds.


[quote="parinaz.sa"]bacheha biyayn yek marathone hendese rah bendazimm man ba yek soalle sade shoroo mikonam shomaha edamash bedin!!!
$1$.dar mosalase $ABC$noghtehaye $K,L$bar zele'haye $AB,BC$tori ekhtiar mishavand ke $\mid AK\mid=\mid KL\mid=\mid LC\mid$az noghteye barkhorde khathaye $AL ,CK$khate rastii be movazate nimsaze zaviyeye $B$ rasm mishavad ta khate $AB$ra dar noghteye $M$ghate konad sabet konid:$\mid AM\mid=\mid BC\mid$.[/quote]

lozoomi nadare ke $KL$ ba $AK=LC$ mosavi bashad ,faghat $AK=LC$ kafie.

[quote="creative"]harfe Amir.S, ro tayid mikonam.
hamoon AK = LC, kafiye.

ba neveshtane, 2 ta menelaos hal mishe.

man maratone hendese ro payam, chera kesi dige soal nemige!?

man chon latex, balad nistam nemitoonam dorost soal bedam! :blush:[/quote]
valla taghsire man nist jozve farziate masalle bood vaelle khodam ham midoonam hal mishee man faghat masalaro kamel neveshtamm ba $latex $ham kari nadareee kar kardan faghat do tarafe har chi mikhay benevisi  $ bezar va symbolasham az latex help begirrr!!!vali kodayii payeye marathone hendese bashin hendese az hame bahaltare!! :wink:

noghtehaye M ,N bar zelhaye AB,CD az chaharzeliye mohaddabe ABCD tori ekhtiar shodeand ke anha ra be yek nesbat (ba ehtesab az rashaye A,C) taghsim mikonand.ba vasl kardane in noghteha be hameye rashaye chaharzeli ABCD ,an ra be shesh mosalas va yek chaharzeli  taghsim mikonim,sabet konid masahate chaharzeli  barabar ast ba majmooe masahathaye do moslalase mojaver be zelhaye AD ,BC.
creative, latexa bikhial, soaleta benevis

salam... man kheyli vaght bood mathliks nayoomade boodam
hint:
oon nesbataye barax rooye 2 zele rooberoo mikhad bege ke oon 2noghte baa mahale barkhord e aghtaar e 4zeli ,hamkhat and
bad papus ham mige ke oon 2 ta rase rooberooyie 4 zelie va mahal barkhor e 4 zeli ham khat an..........
1kam befekr,nashod halesho mizaaram

agha behsan, be hich vajh nemitooni natije begiri ke oon 2 noghte ba mahale barkhorde aghtar hamkhate. baraye darke behtar kafie 4 zel ee ro tori bekeshi ke AB va CD ba ham ye zavieye kheili tond besaze. :)
be har hal hallet dasht ghashang mishod ke nashod dige.

agha oveis mer30
man dishab soaalo avalesho khoondam ba 1 soale Dge eshtebaah gereftam
soal in bood: 1 noghtei rooye MN begir.(meseK)..ke nesbatesh hamoon nesbate oon 2 ta baashe
saabet kon ke nesbate S(KBC)/S(KDA) ye meghdare saabete(ba taghire m,n)

ama hint e oon:soale ghashangie(nemidoonam hal kardi ya na age hal nakardi)
az M be movaazate CD rasm kon ta biaad yeki az zelhaye oon mosallase ke ye zelesh AD ast ro ghat kone
alan S e oon mosallase ke tooye 4zelie va zavieye mishtarak dare ba oon mosallase barabare...!
bara N ham hamin kaar ro bokon va bad 2 ta mosallas be vojood miaad ke bayad begi baraabaran....( sharmande vaaght nadaram shekl bekesham vali age hal nashod begoo taro tamiiz raahe hal ro bezaaram)

oon soale ke dadam ,alan too safeye 1 e ham fekr konid....paarsaal dadam bere IMO vali chon 2 raahe hal daasht too short list naraft(raahe hallesho ta 3-4 rooz Dge mizaaram)....movafagh bashi

bache ha in soalo taaze tarh kardam...va mikhaam bebinam khoobe ya na.....lotfan hal va time e hal shodanetoono benevisid

dar mosallase ABC ,T vasate kamane BC(rooye dayere mohoti) va A1 rooberoo ghotrie A ast......az A1 be I(markaze dayere mohati) vasl konid ta dayera ro dar P ghat kone. khate PT, BC ro dar A2 ghat mikone....
B2 , C2 ro be shekle moshabeh peyda konid.
sabet konid AA2.BB2,CC2  hamresand

ba enekas T(enekas rooye T vasate kamane BC be k=TB^2=TC^2) natije mishe ke:
1/I va B va C sabet mimoonand va BC be kamane ABC va barax tabdil mishe.
paye dayereye mohati(X) mire be X1 rooye dayere
ama chon IXC=IXpayenimsaj=90 darnatije zaviyeye IX1A ham 90 daraje mishe.....yani X1I az rooberoo ghotrie A migzare pas A2,B2,C2 paye dayereye mohatian ke midoonim AA2,BB2,CC2 hamresand


Find all integers $ x$ for which the number $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ is a perfect square.

For positive even $ x$, we have $ \left(x^2\plus{}\frac12x\right)^2\ge x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\le\left(x^2\plus{}\frac12x\plus{}\frac32\right)^2$. For positive odd $ x>3$, we have $ \left(x^2\plus{}\frac12x\minus{}\frac12\right)^2\le x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\le\left(x^2\plus{}\frac12x\plus{}\frac12\right)^2$. $ x\equal{}1$ yields $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}5$ which is not a perfect square. If $ x\equal{}3$, $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}11^2$. So the only solution for positive $ x$ is $ x\equal{}3$.

$ x\equal{}0$ is a solution. Now we shall find the solutions for negative $ x$. Let $ y\equal{}\minus{}x$. If $ y$ is even, we have $ \left(y^2\minus{}\frac12y\plus{}1\right)^2\le y^4\minus{}y^3\plus{}y^2\minus{}y\plus{}1\le\left(y^2\minus{}\frac12y\right)^2$. If $ y>1$ is odd, we have $ \left(y^2\minus{}\frac12y\plus{}\frac12\right)^2\le y^4\minus{}y^3\plus{}y^2\minus{}y\plus{}1\le\left(y^2\minus{}\frac12y\minus{}\frac12\right)^2$. $ y\equal{}1$ is clearly a solution. So the only solution for negative $ x$ is $ \minus{}1$.

Answer: $ \minus{}1,0,3$.


[quote="kchande"]A pyramid has a rectangular base 10 cm by 25 cm and four triangular faces: two opposing faces both with altitude 13 cm and the other two faces  with altitudes of 20 cm and 15 cm. What is the volume of the pyramid. I know how to find volume of regular pyramid. But this is irregular pyramid.  How do we find volume of such a pyramid.

Thanks,
Karan.[/quote]

[hide]It is impossible for the two triangles both with height 13 cm to have a base shared with a side of the rectangle side with length 10, so it must be the side with length 25.

Drawing the height of the pyramid from the top point, you have a right triangle with side lengths, 13, 5, and the height of the pyramid. Solving for the height of the pyramid, you find that is' equal to 12.

So the volume of the pyramid is $\frac{12*250}{3}=250*4=1000$ No matter what shape the pyramid is, as long as it's a pyramid, the volume of it will be it's altitude times its base all divided by 3. [/hide]


What are the most important (and frequent) inequalities you use in a contest? Some important ones are:
AM-GM, Cauchy-Schwarz, and Chebyshev

However, I've seen many inequalities that I've never even heard of before. :what?:  Can someone please tell me (in your opinion) which inequalities would be worth looking over?

Any help would be appreciated.  :)


[quote="covery"]In Isosceles triangle ABC, Line AC is congruent to line CB. IF D is a point on base line AB lying between A and B, line CD is drawn what?

i didnt know how to make the line over the letters, and i didnt know the congruent symbol. i coulndt find these on the keyboard.[/quote]

What is your qusestion?

  darij

would the answer be straight? it sounds like what you are asking


[hide="Hint"] The first step is to rewrite as $ (x^2 \plus{} y^2)^2 \plus{} 3(xy)^2 \equal{} z^2$.  :) [/hide]

[quote="radio"]Find all integer solutions of the equation $ x^4 \plus{} 5x^2y^2 \plus{} y^4 \equal{} z^2$.[/quote]
[hide="some progress I guess"]
mod 5, the LHS is 2, 1, or 0. It can't be 2, because that isn't a quadratic residue.
If it is zero, then it's not too hard to see that 25|z and we have infinite dissent.
If it is 1, then either x or y is a mult of 5.
[/hide]
[hide="by using hint of t0rajir0u"]
Mod 3 gives us an infinite dissent if z=0 mod 3. If $ z^{2}\equal{}1 \mod 3$ then $ x^2\plus{}y^2 \equiv 1 \mod 3$ implying that one of them is a mult of three. Plugging back into original tells us that the other must also be a mult of three if one is, setting up another infinite dissent.[/hide]
Shockingly efficient :|
As a side note, how can you tell whether or not a porblem such as this requires algebraic manipulation or an all out modular attack? (I tried the second and got what I thought was a promising result, but the porblem was nowhere near done, whereas there were only three lines of work needed after your manipulation)

[quote="t0rajir0u"]I see no contradiction if $ z^2 \equiv 1 \bmod 3$.  As for your other question, mods are more likely to be useful if the problem involves more than one exponent (for example, exponents $ 2$ and $ 5$ would suggest $ \bmod 11$), whereas algebra is more likely to be useful if the problem is quadratic or otherwise has a particularly nice form.[/quote]
Whoops, it looks like I made a careless mistake for that case  :oops:

Why would $ x^{2}$ lead to mod 11?

[b]Proposition:[/b]  There are $ \frac{p\minus{}1}{\gcd(k, p\minus{}1)}$ distinct nonzero $ k^{th}$ powers $ \bmod p$ for $ p$ prime.

In particular, if $ k | p\minus{}1$ there are $ \frac{p\minus{}1}{k}$ such powers.  In other words, there are $ 5$ squares and $ 2$ fifth powers $ \bmod 11$, which suggests that for example $ x^2 \plus{} y^5 \bmod 11$ does not take on all residues $ \bmod 11$.

[b]Exercise:[/b]  Find all integer solutions to $ x^3 \plus{} y^4 \equal{} 19^{19}$.

[hide="The correct mod to work in is..."] $ 13$. :) [/hide]

[quote="SimonM"][quote="t0rajir0u"][b]Proposition:[/b]  There are $ \frac {p \minus{} 1}{\gcd(k, p \minus{} 1)}$ distinct nonzero $ k^{th}$ powers $ \bmod p$ for $ p$ prime.

In particular, if $ k | p \minus{} 1$ there are $ \frac {p \minus{} 1}{k}$ such powers.  In other words, there are $ 5$ squares and $ 2$ fifth powers $ \bmod 11$, which suggests that for example $ x^2 \plus{} y^5 \bmod 11$ does not take on all residues $ \bmod 11$.

[b]Exercise:[/b]  Find all integer solutions to $ x^3 \plus{} y^4 \equal{} 19^{19}$.
[/quote]

[hide]
Are there no solutions *tentative* (I created a table, but am not sure how accurate the table is
[/hide]

Also, could someone provide a hint towards proving what t0rajir0u has proposed[/quote]

residues of $ x^3$ mod 13 are -5, -1, 0, 1, 5
residues of $ y^4$ mod 13 are 0, 1, 3, 9

$ 19^{19}$ is equivalent to 7(13)

clearly we have no solution


[quote="Combinatoricsman"]Why are these things such as P vs NP and Riemann Hypothesis under our names in our posts?[/quote]

Just in case this doesn't get added to FAQ.. that's an indication of our rank.  You'll notice that everyone with, say, two green question marks all have the same theorem (Hodge Conjecture), and all have between 50-100 posts (I think..).  I didn't come up with the system, but I'd guess it's just to make the ranks a little more mathy =D

Okay.

I looked at the FAQ, and it explained the ranks, but I didn't know that that's what those were.

Thanks.


When Galileo carried out his experiment (the inclined plane one). He found that the acceleration due to gravity is about $ 7 ms^{\minus{}2}$. Why did he find such results?

My teacher said it has something to do with how once you increase the angle of inclination, the bal starts rolling as well. I have no idea what that means......... I am hopeless at Physics (I only took it to understand Chemisty better  :D  joking).

Seriously though can someone explain? Thanks.

simply speaking....yes it's due to the friction that makes the ball roll as ur teacher said

So why did he specify the fact that it only happens for large angles of inclination?


According to the "Cyclic inequality of degree 3-" theorem, I proposed and proved in my book a very very general inequality. I think it is super hard, too.

$\bigstar$ (hungkhtn's, SII book)
Let $a,b,c$ be non-negative real numbers. For each $k,l \ge 0$, denote
\[F=\frac{a}{ka+lb+c}+\frac{b}{kb+lc+a}+\frac{c}{kc+la+b}. \]
Prove that
\[\min F=\min\lt\{\frac{3}{k+l+1}\ ; \ \frac{1}{k}\ ; \frac{2}{k+\sqrt{l}}\rt\}\ ; \]

\[\sup F=\max\lt\{\frac{3}{k+l+1}\ ; \frac{1}{k}\ ; \frac{2}{k+\sqrt{l}}\rt\}\ ; \]
For some particular cases, we have

Given non-negative real numbers $a,b,c$. For all $k\ge 1$, prove that
\[\frac{3}{k^{2}+k+1}\le \frac{a}{ka+k^{2}b+c}+\frac{b}{kb+k^{2}c+a}+\frac{c}{kc+k^{2}a+b}\le \frac{1}{k}. \]
$\bigstar$
Given non-negative real numbers $a,b,c$. For all $k\ge 0$, prove that
\[\frac{a}{ka+(2k-1)b+c}+\frac{b}{kb+(2k-1)c+a}+\frac{c}{kc+(2k-1)a+b}\ge \frac{1}{k}. \]
$\bigstar$
Given non-negative real numbers $a,b,c$. For all $k\ge 0$, prove that
\[\frac{(k^{2}-k+1)a}{(2k^{2}-3k+2)a+k^{2}b+c}+\frac{(k^{2}-k+1)b}{(2k^{2}-3k+2)b+k^{2}c+a}+\frac{(k^{2}-k+1)c}{(2k^{2}-3k+2)c+k^{2}a+b}\le 1. \]
In the last inequality, by letting $k=3$, we obtain the following result

$\bigstar$
Given non-negative real numbers $a,b,c$, prove that
\[\frac{a}{11a+9b+c}+\frac{b}{11b+9c+a}+\frac{c}{11c+9a+b}\le \frac{1}{7}. \]
It is an inequality of mine solved by THMM, notice that the equality can holds for $a=b=c$ or $(a,b,c)\sim(3,1,0)$.

Letting $k=2$, we have
\[\frac{a}{4a+4b+c}+\frac{a}{4b+4c+a}+\frac{c}{4c+4a+b}\le \frac{1}{3}. \]
This is also a very nice inequality of mine a long time ago (and... The Soul).

I will post the Solution next 2 months.

For the initial of the Soul,  it is easy to prove by Cauchy-Schwarz. So It is not so necessary to post.

[quote="nayel"][quote="The soul of rock"]Prove that for any $a,b,c>0$, $k\ge1$, we always have
$\frac{a+b}{k(a+b)+c}+\frac{b+c}{k(b+c)+a}+\frac{c+a}{k(c+a)+c}\ge \frac{3}{2k+1}$[/quote]

as arqady said, $\sum\frac{a+b}{k(a+b)+c}\ge \frac{4(\sum a)^{2}}{k\sum(a+b)^{2}+2\sum bc}\ge \frac{2}{k}\iff 2k\sum bc\ge \sum bc$

and clearly, $\frac{2}{k}>\frac{3}{2k+1}$.[/quote]
nayel, $\frac{4(\sum a)^{2}}{k\sum(a+b)^{2}+2\sum bc}\ge \frac{2}{k}\Leftrightarrow2k\sum bc\ge 2\sum bc$
and there is much easier way. :wink:

[quote="arqady"][quote="The soul of rock"]Prove that for any $a,b,c>0$, $k\ge1$, we always have
$\frac{a+b}{k(a+b)+c}+\frac{b+c}{k(b+c)+a}+\frac{c+a}{k(c+a)+c}\ge \frac{3}{2k+1}$[/quote]
For $a,b,c>0$ and $k\geq1$ holds:
$\frac{2}{k}\leq\frac{a+b}{k(a+b)+c}+\frac{b+c}{k(b+c)+a}+\frac{c+a}{k(c+a)+c}\leq\frac{6}{2k+1}.$ :wink:[/quote]
I'm sorry, this is my mistake on typing. My inequality looks like yours!
$\frac{a+b}{k(a+b)+c}+\frac{b+c}{k(b+c)+a}+\frac{c+a}{k(c+a)+c}\leq\frac{6}{2k+1}.$
This inequality is easy by using Schur's!


Asymptote
function
geometry
Support
rotation
perpendicular bisector

This is just to announce that [color=blue]cse5[/color] has been updated a bit from its original version. I sent Valentin the newest code. If you want to download it, check that the version is [b]4.1[/b], not [b]4[/b] (there is a small problem with version 4 that I discovered a bit too late: it has all the functions below working but [b]IP()[/b] is not 100% safe because the solution for the infamous intersection of the sides of a triangle with its incircle problem that worked in Asymptote 1.33 no longer worked in Asymptote 1.42, so I had to adjust the code to take care of that. I'm still not completely satisfied with how I did [b]IP()[/b] but the only way to improve it I see is to rewrite the Asymptote intersection routine from scratch and this is hardly what I want to do).

The most important changes are slight modifications in [b]D()[/b], [b]L()[/b], and [b]MP()[/b] and addition of 4 new functions: [b]WP()[/b], [b]MC()[/b], [b]MA()[/b], and [b]CC()[/b]. Below are the details with examples of usage:

1) The main drawing command [b]D()[/b] now supports arrows when drawing paths. Just use them the same way as in [b]draw()[/b]. For instance,
[color=red]
D((0,0)--(1,1),green,MidArrow);
[/color]
results in
[asy]size(100);
D((0,0)--(1,1),green,MidArrow);[/asy]
The arrowbar argument should be used in the last position in [b]D()[/b].
Also, when drawing points and labeling them simultaneously, you no longer need to (but still can) write [b]MP(...,D(),...)[/b]. Instead it is possible to write just [b]D()[/b] and use the usual format of [b]MP()[/b] in the parentheses except the fontsize and pen for the label should now preceed the pair and the pair specifying the direction of the label should be the very last argument. For instance,
[color=red]
D(D(45,"A",8,blue,(0,0),W)--D(45,"B",16,red,(1,1),black,E),green,MidArrow);
[/color]
results in
[asy]size(100);
D(D(45,"A",8,blue,(0,0),W)--D(45,"B",16,red,(1,1),black,E),green,MidArrow);[/asy]
This comes especially handy when initializing the pairs. Now, if you want to define, dot, and label a pair at once, all you need to do is to write [b]pair A=D("A",(1,2));[/b]. If you do not want to dot the pair but only to label it, [b]MP()[/b] does the job. If you just want to dot it without labeling, just use [b]D()[/b] as before without the label arguments.

2) The line command [b]L(A,B,r,s)[/b] now also allows to draw a line that stretches beyond $ A$ and $ B$ by $ r$ and $ s$ units respectively (so it is possible to use [i]absolute[/i] sizes rather than [i]relative[/i] ones. All one needs to do for that is to write [b]L(1,A,B,r,s)[/b]. Also there are [b]L(2,A,B,r,s)[/b] and [b]L(3,A,B,r,s)[/b] that return the perpendicular bisector to $ AB$ (the difference is again that [b]L(2,...)[/b] uses relative lengths and [b]L(3,...)[/b] absolute ones to extend the bisector in both directions. Thus
[color=red]
pair A=D((0,0)), B=D((3,3));
D(L(1,A,B,1)); D(L(2,A,B,1,2),orange);
[/color]
results in
[asy]pair A=D((0,0)), B=D((3,3));
D(L(1,A,B,1)); D(L(2,A,B,1,2),orange);[/asy]

3) The labeling command [b]MP()[/b] allows a real argument that tells by how many degrees the label has to be rotated. It should be used in the very first position. So, if you want your labels to be rotated by 45 degrees, say, you can write
[color=red]
pair A=D((0,0)), B=D((3,3));
D(MP(45,"A",A)--MP(45,"B",B));
[/color]
resulting in
[asy]pair A=D((0,0)), B=D((3,3));
D(MP(45,"A",A)--MP(45,"B",B));[/asy]

4) The [b]WP()[/b] or [b]WayPoint()[/b] is essentially equivalent to [b]waypoint()[/b] in the olympiad package except it can be also used as [b]midpoint()[/b]; for that you just onit the second parameter. So, writing
[color=red]
path P=D((0,0)..(1,2)..(3,1)..(4,6)..(5,7));
for (int k=0;k<9;++k) D(WP(P,k/8));
[/color]
results in
[asy]path P=D((0,0)..(1,2)..(3,1)..(4,6)..(5,7));
for (int k=0;k<9;++k) D(WP(P,k/8));[/asy]

5) The curve marking command [b]MC()[/b] or [b]MarkCurve()[/b] is essentially a combination of [b]MP()[/b] and [b]WP()[/b]. More precisely, writing [b]MC([color=green]a[/color],[color=red]s[/color],[color=green]f[/color],[color=red]P[/color],[color=green]t,B,p[/color])[/b] is equivalent to writing [b]MP(a,s,WP(P,t),f,B,p)[/b] (the arguments that need to be present are red and the optional arguments are green) except the defaults for positioning the label, its color and its size are different from those of [b]MP[/b]. Here is an example:
[color=red]
path P=D((0,0)..(1,2)..(3,1)..(4,6)..(5,7),MidArrow);
MC(50,"\mbox{beginning}",16,P,0.1,NW,red);
MC(62,"\mbox{end}",7,P,0.9,E,black);
[/color]
results in
[asy]path P=D((0,0)..(1,2)..(3,1)..(4,6)..(5,7),MidArrow);
MC(50,"\mbox{beginning}",16,P,0.1,NW,red);
MC(62,"\mbox{end}",7,P,0.9,E,black);[/asy]

6) The [b]MA()[/b] or [b]MarkAngle()[/b] command is a competitor of the olympiad [b]anglemark()[/b]. The usage is
[b]MA([color=green]a,s,f,p[/color],[color=red]B,A,C,r[/color],[color=green]m,S,q[/color])[/b]
(again, the necessary arguments are in red and the optional ones in green). Here

[b]a[/b] is the angle by which you want to rotate the label in degrees (0 by default);
[b]s[/b] is the string labelling the angle (empty string by default); the dollar (or now rather \(...\)) delimeters will be added automatically as in other cse5 labelling commands;
[b]f[/b] is the fontsize for your label (currently 9 by default; can be also reset by the [b]anglefontsize=...[/b] command);
[b]p[/b] is the pen you wish to use for the label (red by default; can be also reset by the [b]anglefontpen=...[/b] command);
[b]B,A,C[/b] are the pairs determining the angle $ \angle BAC$ with the vertex $ A$;
[b]r[/b] the radius of the largest arc in the anglemark;
[b]m[/b] the number of arcs in the anglemark (1 by default). Can be set to $ 0$, in which case no arcs will be drawn, $ 1$ through $ 7$, in which case $ m$ arcs will be drawn, and anything $ >7$ in which case the sector will be filled with color;
[b]S[/b] is almost useless; it allows you to position the label manually. The automatic positioning is where you want it to be most: on the bisector of the angle right beyond the largest arc in the anglemark. I added it just in case but I doubt it will ever be used.
[b]q[/b] is the pen used for drawing the arcs or filling the sector (green by default; can be also reset by the [b]anglepen=...[/b] command).

[b]MA()[/b] returns the vertex of the angle.
Example of usage:
[color=red]
size(200);
pair A=dir(-40), B=dir(0);
for(int k=0;k<9;++k)
{
pair C=dir(40*(k+1));
D(A--D(MA(k*40+90,"\beta_{"+string(k)+"}",12,black,C,B,A,0.1,k,orange))--C);
A=B;B=C;
}
[/color]
results in
[asy]size(200);
pair A=dir(-40), B=dir(0);
for(int k=0;k<9;++k)
{
pair C=dir(40*(k+1));
D(A--D(MA(k*40+90,"\beta_{"+string(k)+"}",12,black,C,B,A,0.1,k,orange))--C);
A=B;B=C;
}[/asy]

7) The [b]CC()[/b] or [b]CopyClean()[/b] command just returns a copy of currentpicture and cleans currentpicture. So, if you want to concatenate 2 or more pictures for which you already wrote the codes, you can do something like
[color=red]
picture p1,p2;
[i]<code for the first picture>[/i]
p1=CC();
[i]<code for the second picture>[/i]
p2=CC();
add(shift(...)*p1); add(shift(...)*p2);
[/color]
with an obvious result. Let's try:
[color=red]
size(150);
picture p1,p2;
D(unitcircle);
D((0,0));
p1=CC();
D(unitsquare);
D((0,0));
p2=CC();
add(shift((-1,0))*p1); add(shift((1,0))*p2);[/color]
results in
[asy]size(150);
picture p1,p2;
D(unitcircle);
D((0,0));
p1=CC();
D(unitsquare);
D((0,0));
p2=CC();
add(shift((-1,0))*p1); add(shift((1,0))*p2);[/asy]

Those are basically all the important changes. I also fixed a few bugs but since nobody noticed them anyway, I'll not tell what they were :P.

If you notice some problems with the new version or have ideas about what else you'd like to see, post your comments in this thread :)

Is the default pen now black?

[asy]unitsize(2cm);draw(D((0,1))--D((0,sqrt(5))));[/asy]

I suggest having a default black pen + black dot (as in the dot command) and a simple option (global value) to change to cse5 old behaviour (blue lines + red dots).

No, it is blue for paths, but you should use D() instead of draw() everywhere to get that:
[asy]unitsize(2cm);D(D((0,1))--D((1,sqrt(5))));[/asy]
The options for resetting defaults are there already: there are 6 pens

pathpen (blue) for drawing paths
pathpointpen (magenta) for marking curves
pointpen (red) for dotting points
pointfontpen (black) for marking points
anglepen (green) for drawing anglemarks
anglefontpen (red) for marking angles

Each of them can be reset to anything you want at any place in the code. If you want, I can introduce AB()/AllBlack() command resetting all pens to black. Just confirm that it is, indeed, what you wish and I'll send you the new code. I just happen to like bright colors more: compare
[asy]pair A=D("A",(0,0)), B=D("B",(3,1)), C=D("C",(-3,2));
MC(20,"\mbox{angle}",D(B--MA("\theta",B,A,C,0.5,3)--C),0.3);[/asy]
which is the current default, with
[asy]pathpen=black;
pathfontpen=black;
pointpen=black;
pointfontpen=black;
anglepen=black;
anglefontpen=black;
pair A=D("A",(0,0)), B=D("B",(3,1)), C=D("C",(-3,2));
MC(20,"\mbox{angle}",D(B--MA("\theta",B,A,C,0.5,3)--C),0.3);[/asy]
which is the black-and-white style.

I do wish there was a FD() command for filldraw...

Are we allowed to edit the code and add our own functions to the package?


Hello everyone.

I am looking for a free PHP web hosting with MySQL support to put up my forum and test my computer skills.  Can anyone give me a suggestion as to which free hosting (preferably no ads) is satisfactory at least?

[url=http://www.trashthesystem.co.uk]Trash the System[/url] has free MySQL support (and 75 MB storage, FTP access, etc).  Beyond that, I don't know; I have a TTS account but haven't used the MySQL feature before.

Hmm, thanks for the website!
I've been trying to find a free hosting for a while, and it's really hard to find one that supports PHP and MySQL and without ads.
I will just sign up and see what happens.

Thanks again!


trigonometry
inequalities
function
inequalities proposed

[quote="doicu"]hi,
Prove that for every $ y \geq 1$ we have
\[ (1 \plus{} 2\sin^{2}{(x)})^{y} \plus{} (1 \plus{} 2\cos^{2}{(x)})^{y} \geq 2^{y \plus{} 1},
\]
$ \forall x \in \mathbb{R}$.[/quote]
Use the power-mean inequality: $ \frac {a^y \plus{} b^y}{2} \ge \left ( \frac {a \plus{} b}{2} \right ) ^ y$. So:
$ (1 \plus{} 2\sin^{2}{(x)})^{y} \plus{} (1 \plus{} 2\cos^{2}{(x)})^{y} \ge 2 \cdot \left ( \frac {1 \plus{} 2\sin^{2}{(x)} \plus{} 1 \plus{} 2\cos^{2}{(x)}}{2} \right ) ^ y \equal{} 2^{y \plus{} 1}$

hi,
yes its a nice solution .
i constructed a function in x and showed that the min. is greater than 2^(y+1).
can this problem be generalized ?


My three-digit code is 023. Reckha can't choose a code that is the same as mine in two or more of the three digit-positions, nor that is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine). Reckha can otherwise choose any three-digit code where each digit is in the set $ \{0, 1, 2, ..., 9\}$. How many codes are available for Reckha?

We use complementary counting.

Case 1: Two or more digits are the same.
$ \overline{0a3}$, where $ 0\leq a\leq9$, which has $ 10$ cases.
$ \overline{02a}$, where $ 0\leq a\leq9$, which has $ 10$ cases.
$ \overline{a23}$, where $ 0\leq a\leq9$, which has $ 10$ cases.
However, in each of these cases, we have counted $ 023$, so for these three cases, there are only $ 10\plus{}10\plus{}10\minus{}2\equal{}28$ cases total.

Case 2: Two digits are switched, but nothing else.
The $ \dbinom32\equal{}3$ codes $ 032,203,320$ can't be made.

Since Case 1 and Case 2 don't overlap and are all-encompassing, our final answer is $ 10^3\minus{}28\minus{}3\equal{}\boxed{969}$.

[quote=cool972]probably but please don't bump topics up it had not been posted for 8 years[/quote]

Actually, if you have a question, you can ask it.

[quote=NoDealsHere][quote=lexd]Don't you have to account for different order in Case 1?[/quote]

That's 8 years after the previous post

[quote=mathleticguyyy]lol 10choose3 is not number of possible 3 digit codes[/quote]

That's about 2 years after the post before[/quote]

That was unnecessary. All the previous posts were contributing to the thread, unlike your post.

Oh god...I wrote 9969 instead of 969.  :blush:

Edit: I got it again today, and did it correctly :D


articles geometry rectangle perimeter SFFT videos blogs

I just keep forgetting what it is!! Can you please give me examples and how the trick would help? Thanks!!!

I can't think of any examples, but here's the basic idea:

$xy+x+y=c$

$x(y+1)+y=c$

$x(y+1)+y+1=c+1$

$(x+1)(y+1)=c+1$

$(x+1)(y+1)-1=c$

$\therefore xy+x+y=(x+1)(y+1)-1$

Main wiki article:
http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick

Two examples:
http://www.mathlinks.ro/Forum/viewtopic.php?p=454191
http://www.artofproblemsolving.com/Wiki/index.php/2000_AMC_12/Problem_6

Wow, there have only been about a billion topics on this...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116843[/url]

A more recent one. HEHEHE!

And there was a problem about it on the latest mathleague

Here is an example. A nonsquare rectangle has integer side lengths and a perimeter numerically equal to its area. Find the area (or the perimeter).

It has far too exalted a name.  It's just FOIL in reverse.

[hide]
Factoring like the first example shows:
$-xy+2x+2y=0$
$-x(y-2)+2y=0$
Subtract four from each side (because we want  $(y-2)$ factors):
$-x(y-2)+2y-4=-4$
$-x(y-2)+2(y-2)=-4$
$(2-x)(y-2) =-4$

This is how you would use the SFFT
In the case of this problem, all you have to do is check various integer side lengths.

In this case, equality is achieved when X and Y are both integers not equal to each other.
[/hide]

[quote="Karth"][quote="jeez123"]The trick is basically about adding number in if it helps with factoring the given "expansion"...?

The area(or perimeter) problem:

Length: x
Width: y
Area: xy
Perimeter: 2(x+y)

$2(x+y)=xy$
$2x+2y=xy$
$-xy+2x+2y = 0$

Oh no... I'm stuck!![/quote]

[hide="OH NO A BEEHIVE"]
USE FFFT HERE:
$-(x-2)(y-2) =-4$
$(x-2)(y-2) = 4$
WOLOG, $x-2 = 1, y-1 = 4$
$x = 3, y = 5$

So the area is 15. The perimeter is then 16.[/hide][/quote]
Dude, the area is supposed to be the perimeter. I think you made a typo or something and then accepted the typo to be true.

Or, if you want a more "visual" experience, try watching the videos, staring Richard Ruscyzk  :)  :)

[url=http://www.artofproblemsolving.com/Movies/simon1.php]click here for part 1[/url]

[url=http://www.artofproblemsolving.com/Movies/simon2.php]click here for part 2[/url]
I don't remember if the second one was ever truly released, but I found out about it from PenguinIntegral's blog this summer.

Yeah, I remembered gaping at the videos :D  It slipped my mind that SFFT was being discussed there.

Thank you for all your help!


It is clear that for |a_i|<=1 we have |a_1x+a_2x^3+...+a_nx^2n-1|<=|x|+|x|^3+...+|x|^2n-1. Thus, it is enough to prove that for t>0 we have t+t^3+...+t^2n-1+1/2<1+t^2+...+t^2n. This can be written (1+t^2n+1)/(1+t)>1/2, or 1+2t^2n+1>t. If t>1 then 1+2t^2n+1>t^2n+1>t and if t<=1 it is trivial.


Is there some place you can find or buy the questions and solutions to previous AP exams as a student?

I can't find them or figure out how to buy them.. :(
http://www.collegeboard.com/student/testing/ap/prep_free.html#compsciab

How about [url=http://apcentral.collegeboard.com/apc/members/exam/exam_questions/1999.html]this[/url] link instead?
While the link you gave just gives the questions, the link I give also includes scoring guidelines (I don't think you can do better than the scoring guidelines for knowing the solutions), sample responses with a score commentary at the bottom of it, and more.

Or if you want to go for different subjects, just click on the preferred subject [url=http://apcentral.collegeboard.com/apc/public/exam/exam_questions/index.html]here[/url].


A few minutes ago (Sept. 19) the site wasn't responding, although it's back now. And for a while yesterday evening (Sept. 18), I couldn't log in.

Planned or unplanned?


function
calculus
trigonometry
limit
conics
parabola
real analysis

Find all functions $f : \mathbb R \to \mathbb R$ s.t. $\exists F, \, F^\prime = f$, and \[\left( x f(x)-2 F(x) \right) \left( F(x)-x^{2}\right) = 0, \, \forall x \in \mathbb R .\]

For $x \neq 0$ we divide by $x^{5}$ and take :

$\\ \frac{xf(x)-2F(x)}{x^{3}}\left( \frac{F(x)}{x^{2}}-1 \right) =0 \Rightarrow \\ \\ \\ \left( \frac{F(x)}{x^{2}}-1 \right)' \left( \frac{F(x)}{x^{2}}-1 \right) =0 \Rightarrow \\ \\ \\ \left[ \left( \frac{F(x)}{x^{2}}-1 \right)^{2}\right] ' =0$ .

This gives

$\\ \left( \frac{F(x)}{x^{2}}-1 \right)^{2}= c^{2}\Rightarrow F(x)= a x^{2}, \,\, a \in \mathbb{R}$

hence $f(x) = b x , \, b \in \mathbb{R}$ .

[quote="perfect_radio"]Find all functions $f : \mathbb R \to \mathbb R$ s.t. $\exists F, \, F^\prime = f$, and $( x f(x)-2 F(x) ) ( F(x)-x^{2}) = 0\ \ \ (*)\ ,\ \forall x \in \mathbb R\ .$[/quote][color=darkblue][b]Proof[/b] (similarly to Ramanujan). [b][u]Suppose for the time being[/u][/b] $x\ne 0\ .$
The relation $(*)$ is equivalently with the relation $[xF'(x)-2F(x)]\cdot[F(x)-x^{2}]=0$ $\Longleftrightarrow$
$[x^{2}F'(x)-2xF(x)][\frac{F(x)}{x^{2}}-1]=0$ $\Longleftrightarrow$ $2\bullet[\frac{F(x)}{x^{2}}-1]\bullet[\frac{F(x)}{x^{2}}-1]'=0$ $\Longleftrightarrow$ $[(\frac{F(x)}{x^{2}}-1)^{2}]'=0$ $\Longleftrightarrow$
$\frac{F(x)}{x^{2}}=1\pm\sqrt m\ ,$ where $m\ge 0\ .$ Therefore, $F(x)=\{\begin{array}{c}(1+\sqrt m)x^{2}\mathrm{\ \ for\ \ }x\in D\\\\ (1-\sqrt m)x^{2}\mathrm{\ \ for\ \ }x\in \mathbb R-D\end{array}$

But the function $F$ is derivatively on $\mathbb R$, in particularly in the point $0$ a.s.o.

[b]Similar exercises.[/b] Ascertain the function $f: \mathbb R\rightarrow\mathbb R$ for which there is a function $F: \mathbb R\rightarrow\mathbb R$ so that $F'(x)=f(x)$
for any $x\in\mathbb R$ and which answers for the following relation :

$1.\blacktriangleright\ f(x)\cdot F(1-x)=1\ ,\ x\in\mathbb R\ .$

$2.\blacktriangleright\ f(0)=0\ ,\ F(x)+f(x)=\sin x\ ,\ x\in\mathbb R\ .$

$3.\blacktriangleright\ f(x)\cdot F(1-x)=1+x^{2}\ ,\ x\in\mathbb R\ .$

$4.\blacktriangleright\ 2F(x)=x[f(x)-x^{2}-x^{3}]\ ,\ x\in\mathbb R\ .$

$5.\blacktriangleright\ F(x)\cdot f(x)>x\ ,\ x\in\mathbb R\Longrightarrow$  the function $F$ isn't bounded.

$6.\blacktriangleright\ F(x)f(x)\le\frac{\sin 2x}{1+\sin^{2}x}\ ,\ x\in\mathbb R$ $\Longrightarrow$ the limit $\lim_{x\to\infty}F(x)$ doesn't exist.

$7.\blacktriangleright\ f(x)\le\frac{|x|}{1+|x|}\ ,\ x\in\mathbb R\Longrightarrow$ the function $F$ has a fixed point and this point is uniquely.[/color]

[quote="Kent Merryfield"]Hey, guys - the difficulty level of the "Calculus Computations and Tutorials" section keeps creeping upward. As a general principle, contest problems - even from a local contest - should go in the "Unsolved and Proposed Problems," not the "Computations and Tutorials."[/quote]
It depends a lot on the specific problem. Since we're talking about a local contest, some of them really belong to the "Computations" area.
As for this problem, I think you're right. It's of another type. Plus, I couldn't get nowhere with it.

PS: According to the solution given in the book, Ramanujan's and Virgil Nicula's answers are [i]wrong[/i]. There is a pretty subtle [i]catch[/i].

[color=red]EDIT:[/color] [color=blue]JBL's right.[/color]

[quote="perfect_radio"]PS: According to the solution given in the book, Ramanujan's and Virgil Nicula's answers are [i]wrong[/i]. There is a pretty subtle [i]catch[/i].[/quote]

We can piece together two different parabola-halves at $x = 0$, right?


MATHCOUNTS
AMC
AMC 8
AMC 10
AMC 12
logarithms
trigonometry

Would it be beneficial to use amc 8's to study for mathcounts? Obviously I would white out the answer choices.

[color=blue]It could be beneficial but it depends which MATHCOUNTS you are studying for. AMC8 would be best for studying for chapter because it is not as difficult as the other levels of MATHCOUNTS.[/color]

AMC8 is actually easier. As flyagaric stated, it is Chapter level. AMC8 does not require calculators even though MC sometimes does for radicals and other calculator bashing formulas. AMC8 questions can be solved without a calculator. AMC8 is mainly like MC but easier.

[quote="Thunder365"]
Really? It wont be beneficial even for states? I mean I know the first 10 or so are chapter, but wouldn't the last 15 AMC 8's be easy to hard state level? And definitely the last 5 would be hard state level, right?

And Im not worried about chapters at all. Im asking this because so far I only have 13 state tests, and Im worried that might run out soon. I heard that some people studied 8 hours a day *caughmewtwo55555*, but obviously at 8 hours a day, the tests would have run out, right?

So how about the AMC 10 questions #1-15. Would those be good practice for states?

[/quote]

Who said that everyone only studies Math competition materials? There are Curriculum based textbooks and lots of other things. Lots. They will run out but hey. There's the internet where you can order stuff.
AMC10 questions are actually about State level. I'm not sure if the questions progress in difficulty as the questions go on. I don't know. But AMC12 is about Nationals I think.

You cannot accurately compare AMC 12 and MATHCOUNTS, as the AMC 12 involves topics such as logarithms and trigonometry, which do not appear in MATHCOUNTS.

It's, in general, not the best idea to compare MATHCOUNTS with AMC at all, as they are two completely different style competitions.

Also, I'd say the problems in AMC 10 are more difficult than State MATHCOUNTS...

By the way, how are 2003 AMC 8 Problems 21, 23, and 25 not State MATHCOUNTS level?  Problem 21 could easily fit around 18 in the Sprint Round, 23 would also be around there, and 25 would be in the early 20's.

Finally, in response to the OP: Taking AMC 8's would probably be helpful, however, you should take them in less time (such as 25 minutes, or 30 minutes).  Also, don't white out answer choices - some problems depend on them (see 2003 AMC 8 Problem 24, and similar ones).  Just don't look at them unless the problem asks for it.

The best thing to study for Mathcounts is Mathcounts.  I have very rarely used anything other than old Mathcounts tests and old Mathcounts questions.  It's not important that the problems are old.  What is important are the techniques that are used on those problems.  Those same techniques are going to show up again only in different situations.

The first year we won nationals I used old high school math league questions.  I through out all the trig problems and then looked for problems that only 15 percent of the HS kids got right and used them to amke up some extra team tests.

But all you need to do is to study Mathcounts materials.  All of the top kids in my program have done every school, chapter, state and national test at least three times.  If you concentrate on methods you can still learn something the 2nd or 3rd time through the test (once per year).

[quote="athunder"]Dont worry about practice sets. They will not make you smart. I suggest only doing practice sets when you are at the point where you have to make a difference from 44,45,46. Other then that you have to know everything. Praactice sets are for practice if you dont know how to do something stick with the books.[/quote]

Not necessarily. Books can cover topics in general but cannot cover every strategy or technique. You will have to also do practice tests if you want to improve your speed on the test. It will be useful to read books as well as doing practice tests.

To do list:
Every MathCounts test EVER.
AoPS Volume 1
Introductory set
* Every Purple Comet test EVER.
* Every Rocket City Math League test EVER.

* means that it might not be necessary.

I study 6.5 hours a day, 8hours on weekends

You can study that much if you have nothing else to do, other than AoPS and math, like me.

Using USAMO problems might even be helpful, considering how I did on nationals and what my preparation consisted of...

Doing USAMO problems gives you a really, really, good intuition, so like you can just look at any USAMO #1/#4 or easier problems and just be like "Oh you obviously solve this by doing this." (Assuming you know the right theorems...)

pythag011.... -.- not everyone can do usamo problems in middle school....

i suggest you do aops books/get a good basis /grasp on basic topics. then do as many mathcounts problems for mathcounts as possible.
 agreed with JBoyd.

If you are able to do every problem on every old Mathcounts, and are able to recognize pretty quickly what techniques to do on a problem, you are fine. As mentioned earlier, if you can make nats, AMC 8's are pretty useless in terms of prep.

Good luck with Olympiad Geometry without algebra fundamentals...

[quote="batteredbutnotdefeated"]My to do list for math competitions:
finish intro to geo course
finish intro to prob course
finish difficult problems in geo
finish olympiad geo course
finish SOME practice AIMEs and USAMOs

I think that ought to help! :lol:

P.S. Topofmath, how do you study 6.5 hours on school days?! I get home at 4:00 and study till 8:00...[/quote]

Yeah, this is a pretty useless study method for MATHCOUNTS, which is the sole purpose of this thread. Olympiad Geo, contrary to how trivial pythag011 says it is, will NOT show up even on a MATHCOUNTS contest. I knew all the topics from intro to geo, and that was enough to do every single nationals geometry problem. Really, at this point, everything referring to the productivity of AMC 8 in MC studying (very little past chapter/state) has been discussed, so a mod should probably lock this or merge it with the tips thread.

@Mew: I just like calling things trivial, Olympiad Geo isn't actually trivial....

I wouldn't recommend you take olympiad classes without having a good foundation in every subject.

Alright, so you guys are saying that even amc 10 problem #'s 1-15 would be too hard and not beneficial for using as practice mathcounts tests?

[quote="shelly32494"]no. amc & mathcounts are different styles of problems... amc is multiple choice. making it quite a bit easier.[/quote]

But like you said, white out the answers so it is a little harder. The last few on amc8 are about state difficultyish.

You can't compare AMC and MC in a sense. There are so many differences between the two.

Every problem related to finding maximums/minimums could be resolved by SV (or MV) calculus, in both AMC8/10/12 and Mathcounts. However, since they expect MC contestants not to know anything about calculus... would one lose credit if calculus (i.e. Lagrange multipliers and other such things) was used to find a correct answer?

I thought it would be like USAMO, where any math is authorized in solutions.

P.S.: I'd most likely [i]expect[/i] USAMO contestants from any academic year to know at least some calculus. Even 8th-graders who actually made it to USAMO.

Solutions aren't required in MATHCOUNTS - the whole test is short answer.

However, it's probably faster to stay with elementary techniques on a MATHCOUNTS test.

I distinctly remember a state test where I managed to forget the formula for the volume of a sphere and rederived it using calculus...

Also, you could have missed 7 more sprints on the test and still have made nationals. :wink:


inequalities
linear algebra
matrix
linear algebra unsolved

$A,B \in M_{n}(R)$ suppose $^{t}A=A$ ; $^{t}B=B$

denote $a_{k}$ the eigenvalues of A
$b_{k}$ the eigenvalues of B

Prove $det(A+B)\leq max_{q\in S_{n}}\prod_{k=1}^{n}(a_{k}+b_{q(k)})$

Here $S_{n}$ group of permutation of the set {1,2,...,n}

What is strange is that this inequality was the subject of Mines Ponts contest this year, even if the solution for it is very simple if we use the fact that there exists a matrix $C$ such that $C^[*] AC$ and $C^[*] BC$ are simultenously diagonal.


Integers $a,b,c,d,e$ are such that $n|a+b+c+d+e$ and $n|a^{2}+b^{2}+c^{2}+d^{2}+e^{2}$ for the odd integer $n$. Prove that $n|a^{5}+b^{5}+c^{5}+d^{5}+e^{5}-5abcde$.

Using the two given divisibilities, we can find that n|(a_1a+b_1b+c_1c+d_1d+e_1e)^2+(e_1a+a_1b+b_1c+c_1d+d_1e)^2+...
so n fits into a lot of stuff... then you just manipulate more to get the last expression...


As the title says, who's going to the CTY summer programs at LA?

I hope I'm not going to be the only AoPSer there...  :(

...nah. If I were to go to a summer program I'd go to Mathcamp or something of the sort. But seeing as I'm a lazy bum...  :P

i went for the last two years
but this year i'm going to mathpath :)
you should try to apply, its probably gonna be a lot better than CTY
there's not much time left though... :ninja:
check out http://www.mathpath.org

You wish... it's supposed to be $3500, but last year they only had 72 people that made it so they increased it to$4500.

(It's worth it anyway!)

[quote="CircleSquared"]I would have applied for mathpath this year if the tuition weren't $3000 or so.[/quote]
Same.  I might go to CTY things this summer.
By the way,  CircleSquared, you made SET... right?  If so have you gotten your confirmation letter?


[quote="236factorial"]Is there a forum specifically for sports? I haven't seen one....[/quote]

That sounds like a Fun and Games tppic.

[quote="MCrawford"]That sounds like a Fun and Games tppic.[/quote]Indeed, there are many sports topics in the [url=http://www.mathlinks.ro/Forum/index.php?f=139]Games and Fun Factory[/url] forums.


Prove that:

\[ x(x\plus{}1)(x\plus{}2)(x\plus{}3) \equal{} y^{2}\]

has no Solutions for $ x,y\in\mathbb{N}$

\[ x(x\plus{}1)(x\plus{}2)(x\plus{}3) \equal{}\left[(x\plus{}1)(x\plus{}2)\minus{}2\right]\left[(x\plus{}1)(x\plus{}2)\right]\]
\[ \equal{}\left[(x\plus{}1)(x\plus{}2)\minus{}1\right]^{2}\minus{}1\]
No two positive integer squares differ by $ 1$.


quadratics
modular arithmetic
number theory unsolved
number theory

Find at least one quadruple naturals $ (k,l,m,n)$ such that:

a) $ 2l^2$ is quadratic nonresidue $ mod(k^2 \minus{} 2)$
b) $ 4mnl^2$ is quadratic nonresidue $ mod(k^2 \minus{} 4mn)$

That's impossible.

Note that for any positive integer $ k$,
\[ 2\equiv k^2\pmod{k^2 \minus{} 2}
\]
and furthermore for any integer $ l$,
\[ 2l^2\equiv (kl)^2\pmod{k^2 \minus{} 2}.
\]

Therefore, $ 2l^2$ is always a quadratic residue modulo $ k^2 \minus{} 2$.


geometry trapezoid parallelogram circumcircle BritishMathematicalOlympiad

Let $ ABPC$ be a parallelogram such that $ ABC$ is an acute-angled triangle. The circumcircle of triangle $ ABC$ meets the line $ CP$ again at $ Q$. Prove that $ PQ \equal{} AC$ if, and only if, $ BAC \equal{} 60^{\circ}$ . [i]The circumcircle of a triangle is the circle which passes through its vertices.[/i]

if $ <CAB\equal{}60$ then $ <BQP\equal{}60$ and we have also $ <QPB\equal{}60$ so $ PQ\equal{}PC\equal{}AC$.

now if $ PQ\equal{}AC\equal{}PB$ let $ x\equal{}<PQB\equal{}<QBP$,we have $ <ABP\equal{}x\plus{}<QBA\equal{}x\plus{}180\minus{}<ACQ\equal{}x\plus{}180\minus{}<ABP$ so $ 2<ABP\equal{}180\plus{}x$.

and $ <ABP\plus{}<QPB\equal{}180$ so $ 90\plus{}x/2\plus{}<QPB\equal{}180$ and we know that $ 2x\plus{}<QPB\equal{}180$ therefore $ 180\minus{}2x\equal{}90\minus{}x/2$ so $ x\equal{}60$

so $ <QPB\equal{}<CAB\equal{}60$.

Clearly $AC=BP$. Also, $\angle BAC=\angle BQP=\angle BPQ$ so triangle $QBP$ is isosceles. But then $BP=PQ$ if and only if $\triangle BQP$ is equilateral, that is, when $\angle BAC=180^\circ $.


abstract algebra
modular arithmetic
group theory
superior algebra
superior algebra unsolved

When are the monoids $(m\mathbb Z+1, \cdot)$ and $(n\mathbb Z+1, \cdot)$ isomorphic?

Aren't they both $m\mathbb Z+1$ and $n\mathbb Z+1$ free abelian monoids on a countable infinity of generators, namely the numbers of the form $km+1$ and $kn+1$ respectively which have no divisors of those respective forms other than $1$ and themselves? :?

I did not understand many things from these advanced words, but I have two remarks: $3Z+1$ and $4Z+1$ seem to be isomorphic (and if I'm not wrong, it's even a problem proposed by ROmania for IMO). There was also a problem of Laurentiu Panaitopol given in Balkan olympiad  on this theme. I really don't think the problem is that easy, but who knows?

What I wanted to say is that I think those two monoids are isomorphic for any pair $(m,n),\ m,n\ge 2$.

I don't know what on earth I was thinking about :). Sorry for that..

Edit:

It shouldn't be too hard to prove that the two monoids are isomorphic whenever the multiplicative groups of invertible residues modulo $m$ and $n$ are isomorphic, so $3\mathbb N+1$ and $4\mathbb N+1$ are indeed isomorphic. We can prove this by considering an isomorphism $\varphi: \mathbb Z_m^{\#}\to\mathbb Z_n^{\#}$ (for an integer $k,\ \mathbb Z_k^{\#}$ denotes the mentioned multiplicative group of invertible residues modulo $k$), mapping the primes of the form $t\pmod m,\ t\in\mathbb Z_m^{\#}$ to those of the form $\varphi(t)\pmod n$ bijectively for all $t\in\mathbb Z_m^{\#}$, and then extending this map multiplicatively. The restriction of this map to $m\mathbb N+1$ is the isomorphism we want.

This gives a first positive result, if I'm not making any stupid mistakes (again :)).

Yes, I did have Dirichlet's Theorem in mind. Right now we don't even know if we can solve this problem at all, so our concern isn't solving it in an elementary way :).

We have $2^2\equiv 1\pmod 3$, so $2$ is mapped onto a prime $p$ which satisfies $p^2\equiv 1\pmod 4$, and $4$ is mapped onto $p^2$. This is what I meant by "the map is extended multiplicatively".

P.S.

Where exactly did you encounter it anyway? :)

I think the converse might be true. I haven't finish it completely, but maybe the following approach is helpful.
Suppose p is a prime not dividing m, k is its order mod m. Suppose f( $p^k$ ) =qr, where q is a prime power and r is coprime with q.
Now f( $p^{kN}$ ) = $(qr)^N$. Suppose r>1. For large enough N, we can choose l>1 so that $q r^l$ = 1 mod n. But this implies
$p^{kN}$ = AB, where A=1 mod m and is not a power of $p^k$ (by taking inverse images), which is impossible, since k is the order of p. Thus the image of a prime power in mZ+1 must be a prime power in nZ+1. Since f is bijective, the inverse image of a prime power is also a prime power. Now suppose f( $p^k$ ) = $q^{ls}$, where l is the order of q mod n. But then the inverse image of $q^l$ must be of form $p^u$, where 0< u $\leq$ k, so u=k, since k is the order of p mod m. This implies s=1. Thus for each prime p that doesnot divides m, we can assign g(p)=q, which is a prime that doesnot divide n. This is a bijection from the primes coprime with m to the primes coprime with n. Now if a= $p_1 ^ {b_1} ... p_l ^{b_l}$ is in mZ+1, take N so that N is divisible by all the orders of $p_i$. This shows the prime factors of $f(a) ^ N$ and hence of f(a) are precisely the $g(p_i)$s.
Now let t be a prime and an inverse of p mod m. So t and p have the same order k mod m.
$f( tp) ^k$=f($(tp)^k$)=f( $t^k$) f( $p^k$ ) would then imply the order of g(p) mod n is divisible by k, the order of p mod m. Since p and g(p) have the same role, this implies the order of g(p) mod n is the order of p mod m. In particular, that shows that if we extend g(x) multiplicatively from the primes not dividing m, the restriction of g on mZ+1 is precisely f by looking at the above prime factorization. Note that then, if p, p' are primes congruent mod m, then by taking a prime t that is the inverse of p mod m, we have g(p)g(t)=g(p')g(t) mod n, thus g(p)=g(p') mod n. This would also show in particular that the unit group of Z/mZ and Z/nZ must have the same order, and the number of elements mod m of order k is the same as the number of elements of order k mod n . I think we can prove that they are actually isomorphic if we look at it more carefully , but I will think more about this later.
Hope that there are no stupid mistakes

Hello cimabue,
If t,p are distinct primes that are inverses of each other mod m, then they have the same order mod p, namely k. Then
tp, $t^k$ and $p^k$ are in mZ+1. So we get the indentity. But we have shown that $f( p^k )$ = $g(p) ^l$, where l is the order of g(p) mod n. Also $f(t^k)$ is a prime power and coprime with g(p), so by looking at prime factoriztion of $f( tp )^k$, we see that k divides l. Since the role of p and g(p) is reversible, the opposite is also true and k=l.
I hope that cleared your first question.
For your second question: I think the nunber of elements of the same order does not characterise an abelian group, since I vaguely remember someone showing me a counter-example, but I might be wrong. But anyway, we don't actually need that. We can prove that if x=x' mod m are units in (Z/mZ), then g(x)=g(x') mod n, simply by taking a prime that is a common inverse of x,x' mod m. Thus g induces an isomorphism from (Z/mZ)* to (Z/nZ)*, and we are done.


[quote="shaaam"]Given that:
$ a$ and $ b$ are [b]real[/b] numbers that satisfies :

$ 3a^2 \plus{} 9b^2 \minus{} 2a \minus{} 13 \geq 0$

Prove that:

$ a^2 \plus{} b^2 \plus{} 10ab \geq 0$

 :lol:[/quote]
Try $ a \equal{} \minus{}1$ and $ b \equal{} 1.$

[quote="shaaam"][quote="arqady"][quote="shaaam"]Given that:
$ a$ and $ b$ are [b]real[/b] numbers that satisfies :

$ 3a^2 \plus{} 9b^2 \minus{} 2a \minus{} 13 \geq 0$

Prove that:

$ a^2 \plus{} b^2 \plus{} 10ab \geq 0$

 :lol:[/quote]
Try $ a \equal{} \minus{} 1$ and $ b \equal{} 1.$[/quote]

Sorry  :blush:
Thank you  for pointing out my error... Okay, how about this:

$ a$ and $ b$ are [b]real[/b] numbers that satisfies :

$ 8a^2 \plus{} 8b^2 \minus{} 10b \plus{} 13 \geq 0$

Prove that

$ 3a^2 \plus{} b^2 \plus{} 18a \minus{} 24 \geq 0$[/quote]
$ a\equal{}b\equal{}1$, my friend. ;)

okay...........

 :blush:  :blush:  :blush:

I give up.. sorry for this unnecessary post..

Moderator, please delete this topic.. I did bad calculations....


Prove that the equation $ z^3\plus{}z^2\plus{}z\minus{}6\equal{}0$ doesn't have roots $ x$ with $ |x|\equal{}2$.

Denote $ f(x)\equal{}x^3\plus{}x^2\plus{}x\minus{}6$. Then $ f'(x)>0$ for $ x\in\mathbb R$, $ f(1)<0<f(1.5)$, therefore, there're $ x_0\in(1;1.5)$ and $ z_0\in\mathbb C\setminus\mathbb R$ such that $ f(z)\equal{}(z\minus{}x_0)(z\minus{}z_0)(z\minus{}\Bar z_0)$. Then $ |x_0|\equal{}x_0<2$, $ |z_0|\equal{}|\Bar z_0|\equal{}\sqrt{6/x_0}>2$.


In triangle ABC(AB>AC)  AD is median and AF is bisector. FN is parallel to AC (N is on AB) and it intersects AD at L. Prove that CL is perpendicular to AF.

Let $ K \equiv AF \cap CL$ and $ E \equiv AC \cap DK.$ Since $ LF \parallel AC,$ it follows that the cevians $ AF$ and $ CL$ of $\triangle ADC$ cut on its D-median $ \Longrightarrow$ $ E$ is the midpoint of $ AC$ $ \Longrightarrow$ $ DE \parallel BA.$ Consequently, $ \angle AKF \equal{} \angle DKF \equal{} \angle BAF \equal{} \angle KAE$ $ \Longrightarrow$ $ EA \equal{} EK \equal{} EC$ $\Longrightarrow$ $ \angle AKC \equal{} 90^{\circ}$ as desired.


Given the polynomial F(x)=x^n+a_(n-1)x^(n-1)+...+a_1x+a_0 with integral coefficients a_0, a_1, ..., a_(n-1), and given theat there exist four distinct integers a, b, c, d such that F(a)=F(b)=F(c)=F(d)=5, show that there is no integer k such that F(k)=8.

More generally ?

[hide]

F(a)=F(b)=F(c)=F(d)=A

that there is no integer k such that F(k)=B with |A-B| prime ?[/hide]

[color=cyan]i/3, is that true or are you guessing?[/color]

JBL, no, it's easy to show.

[hide]Assume that F(n_1)=F(n_2)=F(n_3)=F(n_4)=a and that for some n F(n) = b. Then F(x) - a = (x - n_1)(x - n_2)(x - n_3)(x - n_4)f(x) for some polynomial f, and we get F(n) - a = b - a = (n - n_1)(n - n_2)(n - n_3)(n - n_4)f(n) which is impossible, it's easy to generalize this even further.[/hide]

Oh, I see.  But it wouldn't work for only 3, because [hide]you could have 1, -1, and -p as the three parts of the product.  Right?[/hide]  Very cool.

The generalization is nasty, though, isn't it?  I mean, you can talk in terms of numbers of prime factors and things, but it isn't very nice, is it?


geometry
ratio
parallelogram
analytic geometry

Q1-   The point which divides segment joining points (4, -2) and (8, 6) in the ratio of 7:5 externally is:

Q2- The exterminities of the diagonal of a parallelogram are the points (3, -4) and (-6, 5) if the third vertex is the point (-2, 1) then the fourth coordinate of the fourth vertex are :

no.... only the fourth vertex is required. Its coordinates are required only in question 2.

...the three points are collinear.  :maybe:

thanks its  it (-1,0)
but kindly can you guide me about how to solve such problems ?

I got another question :

Q-The points (0,0), (1,0) and (0,2) are the vertices of which triangle ?

i- Right
ii- Isosceles
iii-Oblique
iv-Equilateral
?????????????

n how to make it sure which one is it ?
how to determine the answer of such question ?

Well, it's of course a right triangle, you can draw it and see that in the origin you have the right angle...
As a general method, you can calculate the equations of the lines on which the sides of the triangle lie... Then you compare the angular coefficients, to see if something is perpendicular to something else...
To see if a triangle is isosceles or equilateral it suffices to calculate the lenghts of the sides...  :wink:

Thanks                                                         :blush:


real analysis
advanced fields
advanced fields unsolved

I read that if $A \in R^{2}$ with $\mathcal{H}^{1}(A) > 0$ and $f$ is a Lipschitz map from $A$ to $R$ it is not necessarily true that $\lambda(A) > 0$ (where $\lambda$ is the Lebesgue measure on $R$). Some Russian dudes constructed a counterexample in the 60s.

1) I don't understand precisely what is implicitly assumed that makes this non-trivial. Are they assuming that the Lipschitz constant (that is, the smallest $L$ such that $|f(x)-f(y)| < L|x-y|$ for all $x, y \in A$) is greater than 0? Or that the map is invertible? Otherwise why not just send everything to a point?

2) Anyone want to try to construct their own counterexample? I don't think it's too hard (in fact if I understand the problem properly I think I might even have a counterexample myself, but I'm not confident).

You misunderstood the problem. The question was: given a set $A\subset \mathbb R^{2}$ with $\mathcal H^{1}(A)>0$, does there exist a Lipschitz map $f\colon A\to\mathbb R$ such that $\lambda(f(A))>0$? Vitushkin-Ivanov-Melnikov answered it in the negative by constructing a set $A$ for which no such $f$ exists.


Does anyone know of an article or a website about Gaussian integers?

Have you tried googling? :)

http://mathworld.wolfram.com/GaussianInteger.html
http://planetmath.org/encyclopedia/GaussianIntegers.html
http://en.wikipedia.org/wiki/Gaussian_integer

These are great, but I was looking for a good-sized article, something at least 5 pages.

Oh, I see :) :
http://www.maths.ox.ac.uk/current-students/undergraduates/lecture-material/A/rings/pdf/RandALect9.pdf
http://www.emis.de/journals/INTEGERS/papers/e16/e16.pdf
http://home.wlu.edu/~dresdeng/papers/factorrings.pdf


Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0,$ given that there are no multiple roots.

[hide]The $x^{2001}$ coefficient disappears after expanding.

The $x^{2000}$ coefficient is $\frac{2001}{2}$.

The $x^{1999}$ coefficient is $-\frac{\binom{2001}{2}}{4}=-\frac{2001}{2}\times 500\Rightarrow \boxed{500}$.[/hide]

[hide=Clever Solution]Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$  We make the substitution $y=x-\frac{1}{4}$ so $$(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.$$ Expanding gives $$2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots$$ so by Vieta, the sum of the roots of $y$ is 0.  Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is $$2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.$$[/hide]

Clearly, the degree of the polynomial is $2000$. Also note that the coefficient of $x^{2000}$ is, by the binomial theorem, $\frac{2001}{2}$. The coefficient of $x^{1999}$ is $-\frac{1}{4} \cdot \binom{2001}{2} = \frac{2001 \cdot 2000}{2} \cdot \frac{-1}{4} = -500 \cdot \frac{2001}{2}$. Therefore, the sum of the roots, is by Vieta’s formulas: $-\frac{-500 \cdot \frac{2001}{2}}{\frac{2001}{2}} = \boxed{500}$.

Since $2001$ is odd, the $x^{2001}$ term cancels out.

Hence we just need the coefficient of the $x^{2000}$ and $x^{1999}$.

By the binomial theorem the coefficient of $x^{2000}$ will be:
$$2001\cdot\frac{1}{2} = \frac{2001}{2}$$

By the binomial theorem, again, the coefficient of $x^{1999}$, is:
$$-\frac{\binom{2001}{2}}{4}$$

By Vieta's our answer is just:
$$\frac{\frac{\binom{2001}{2}}{4}}{\frac{2001}{2}} = \boxed{500}$$

[hide=Solution]Since the $x^{2001}$ term cancels, this is a $2000$ degree polynomial. However, the roots come in pairs that add up to $\frac{1}{2}$, so we have $1000$ pairs that add up to $\frac{1}{2}$ for a sum of $\boxed{500}$ in total.[/hide]

This solution has a somewhat weak motivation for the transformation I proposed - other than the fact that I looked at the answer and reverse engineered another solution.

[hide="Hindsight solution"]Note that there are $2000$ roots, as the $x^{2001}$ terms cancel.

Let $y = x - \frac{1}{4}$, or $x = y + \frac{1}{4}$.  We have:

$(y + \frac{1}{4})^{2001} + (\frac{1}{2} - (y + \frac{1}{4}))^{2001} = 0$

$(y + \frac{1}{4})^{2001} - (y - \frac{1}{4})^{2001} = 0$

$(y + \frac{1}{4})^{2001} = (y - \frac{1}{4})^{2001}$

As before, the $y^{2001}$ term cancels.  But that's not the only term that cancels - all odd exponents cancel, because they're equal on both sides.  That means we have an even polynomial.  The sum of the roots will thus be $0$.

Since we transformed by subtracting $\frac{1}{4}$ from every root, we have to add $\frac{1}{4}$ back to every root, which gives $\frac{2000}{4} = \boxed{500}$ as the sum of the roots when expressed in $x$ instead of in $y$.[/hide]

If $x$ is a root, $\frac{1}{2}-x$ is also a root, so each pair of roots add up to $\frac{1}{2}$, and since there are no multiple roots the total amount of distinct roots is $1000$, for an answer of $1000 \cdot \frac{1}{2} = \boxed{500}$.


Hi  :)
Find the value of $ \sin\frac{\pi}{10}$ and $ \cos\frac{\pi}{10}$
Without using the calculator  :!:

Calculate $ cos\frac{\pi }{5}\equal{}\frac{1\plus{}\sqrt 5 }{4}$, then $ sinx\equal{}\sqrt{(1\minus{}cos2x)/2},cosx\equal{}\sqrt{(1\plus{}cos2x)/2}.$

I don't think that it's good answer because we dont have $ \cos\frac{\pi}{5}$ or $ \ sin\frac{\pi}{5}$.
I have another sollution  :)

My sollution:
[hide]we have : $ \cos2a = 1 - 2\sin^2a$                       and                   $ \sin3a = 3\sinh - 4{sina}^3$

Denote :     $ a = \frac {\pi}{10}$  (nonegative)

and :     $ \sin(a) = x$

$ \cos2a = \sin3a$

$ \longleftrightarrow \,\, - 4x^3 + 2x^2 + 3x - 1 = 0$

$ \longleftrightarrow \,\, (x - 1)(4x^2 + 2x - 1) = 0$

$ \longleftrightarrow \,\, (x - 1)(x - \frac {\sqrt {5} - 1}{4})(x - \frac { - \sqrt {5} - 1}{4}$

$ \longrightarrow \,\, \sin\frac {\pi}{10} = \frac {\sqrt {5} - 1}{4}$[/hide]


inequalities
function
calculus
derivative
inequalities unsolved

If $a,b,c$ are positive numbers, then

$\displaystyle \frac{1}{7a^2+2bc}+\frac{1}{7b^2+2ca}+\frac{1}{7c^2+2ab} \geq \frac{3}{(a+b+c)^2}$.

This is a nicer variant of my inequality (unsolved on this forum yet)
$\displaystyle \frac{1}{22a^2+5bc}+\frac{1}{22b^2+5ca}+\frac{1}{22c^2+5ab} \geq \frac{1}{(a+b+c)^2}$.

I think this Sysmetric kind can be solve by classical way.

Vasc,I guess we have a best constant $k$ st the inequality satisfy for all $a,b,c\ge 0$ :

$\frac{k+1}{ka^2+bc}+\frac{k+1}{kb^2+ac}+\frac{k+1}{kc^2+ab} \ge \frac{9}{(a+b+c)^2}$

edit:  all this stuff is wrong because the fn isnt actually convex.

[hide]

Normalize abc=1.  Consider the function .  It is obvious this is convex.

Now by jensen, .

Remains to show that .

or,

which is true by AMGM (remember, abc = 1.)

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

To solve variant, we apply same principle.  In the key Jensen step, we have .

It remains to show

This simplifies again to .

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

We now prove hungkhtn's generalization.  He wants to find the least constant k, such that:

Again, normalize abc = 1.  Consider the function . It is concave (**im 99% sure, but i didn't check.)

Once again, .

Put .  We just need to show .

This is the same thing as .  This is always true.

So the least constant k is 0.  We prove this now.

Normalize abc = 1. Then we have equivalently, .  But , so it is true, with no equality.

Because the inequality is true for any k, infact, any constant k suffices.

[/hide]

You remember jensen inequality wrongly.
If they can be so easy, everyone will be very happy..

and in fact, even $\frac{1}{7x^2+\frac{2}{x}}$ is not concave. It is sometimes concave sometimes convex.

[quote="Singular"]how is jensen used wrong??

[b]Jensen Inequality (nonweighted)[/b]

For a convex function:

$\sum f(x_i) \geq nf(\frac{\sum x_i}{n})$[/quote]

why is the function concave/convex?
If you are using RCF or LCF, please clarify it..

[quote="Singular"]Sorry, there was a typo, I meant convex, but wrote concave.

Yes, $\frac{1}{7x^2 + 2/x}$ is convex.  Look at x close to zero, then the whole thing is close to infinity.

Look at x close to infinity, then the whole thing is close to zero.

I didn't bother to check, but just thinking about it, I am 99% sure that this function is convex.[/quote]

i differentiated the function and it's not always convex.

do you wanna share your differentiation?  maybe mine is wrong.

edit:  If x^3 <= 1/7, then I think its wonky,.. not sure yet though.

[quote="Vasc"]If $a,b,c$ are positive numbers, then
$\displaystyle \frac{1}{22a^2+5bc}+\frac{1}{22b^2+5ca}+\frac{1}{22c^2+5ab} \geq \frac{1}{(a+b+c)^2}$.[/quote]
$\sum_{cyc}\frac{1}{22a^2+5bc}=\sum_{cyc}\frac{(b+c)^2}{(b+c)^2(22a^2+5bc)}\geq\frac{4(a+b+c)^2}{\sum_{cyc}(b+c)^2(22a^2+5bc)}.$
Remain to prove that  $\frac{4(a+b+c)^2}{\sum_{cyc}(b+c)^2(22a^2+5bc)}\geq\frac{1}{(a+b+c)^2}.$
But $\frac{4(a+b+c)^2}{\sum_{cyc}(b+c)^2(22a^2+5bc)}\geq\frac{1}{(a+b+c)^2}\Leftrightarrow$
$\Leftrightarrow4\cdot\sum_{cyc}(a^4+4a^3b+4a^3c+6a^2b^2+12a^2bc)\geq$
$\geq\sum_{cyc}(22a^2b^2+22a^2c^2+10b^2c^2+44a^2bc+5a^3b+5a^3c)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(4a^4+11a^3b+11a^3c-30a^2b^2+4a^2bc)\geq0\Leftrightarrow$
$\Leftrightarrow4\cdot\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+15\cdot\sum_{cyc}(a^3b+a^3c-2a^2b^2)\geq0.$
The last inequality is true by Schur and Muirhead. :)

Again a beautiful solution in a Vasc's problem congratulations Arqady.
But how did you invent this trick which is in fact <<all the money>> and all the solution because after it all things are easier
[quote="arqady"]
$\sum_{cyc}\frac{1}{22a^2+5bc}=\sum_{cyc}\frac{(b+c)^2}{(b+c)^2(22a^2+5bc)}$[/quote]

[quote="silouan"]
But how did you invent this trick which is in fact <<all the money>> and all the solution because after it all things are easier
[/quote]
I used the vasc's hint to other problem. :lol:

[quote="Vasc"]If $a,b,c$ are positive numbers, then

$\displaystyle \frac{1}{7a^2+2bc}+\frac{1}{7b^2+2ca}+\frac{1}{7c^2+2ab} \geq \frac{3}{(a+b+c)^2}$.

[/quote]
Dear sir, how can you come up to create this lovely problem??

The above inequality is equivalent to
$\sum \frac{a^2+b^2+28ab}{(7a^2+2bc)(7b^2+2ac)}(b-a)^2\\+\sum \frac{42(a^2+ac+c^2)}{(7a^2+2bc)(7b^2+2ca)(7c^2+2ab)}(a-c)^2(b-a)^2\geq 0$

I hope i didn't make any mistake

It is not very hard to create it. See
http://www.mathlinks.ro/Forum/viewtopic.php?p=196538#p196538.
But, how did you find this solution?

[quote="arqady"]
Remain to prove that  $\frac{4(a+b+c)^2}{\sum_{cyc}(b+c)^2(22a^2+5bc)}\geq\frac{1}{(a+b+c)^2}.$[/quote]

I found that
$4(a+b+c)^4-\sum_{cycl}(b+c)^2(22a^2+5bc)=4\sum_{cycl}a^2(a-b)(a-c)+15\sum_{cycl}ab(a-b)^2$
Also
\[ 4(k+1)(a+b+c)^4-27\sum(b+c)^2(ka^2+bc)= 4(k+1)\sum_{cycl} a^2(a-b)(a-c)+ (20k-7)\sum_{cycl} bc(b-c)^2 + (5k-22)\sum_{cycl} a^2(b-c)^2 \]
Hence the above approach can be used to prove the one hungkhtn mentioned at least for $k\ge 22/5$.

[quote="Vasc"][quote="barra"]The above inequality is equivalent to
$\sum \frac{a^2+b^2+28ab}{(7a^2+2bc)(7b^2+2ac)}(b-a)^2\\+\sum \frac{42(a^2+ac+c^2)}{(7a^2+2bc)(7b^2+2ca)(7c^2+2ab)}(a-c)^2(b-a)^2\geq 0$
[/quote]
But, how did you find this solution?[/quote]

barra, could you please detail how you reach this equivalent form.

Actually, the result of Barra is not true, because the second sum is not symmetric.


A standing wave is maintained in a homogenous string of cross-sectional area $ a$ and density$ \rho$  . It is formed by the superposition of two waves travelling in opposite directions given by the equations

$ y_1 \equal{} Asin(wt \minus{} kx)$
$ y_2 \equal{} 2Asin(wt \plus{} kx)$

Find the total mechanical energy confined between the section corresponding to the adjacent antinodes.

[quote="Conjurer"]...is equality of amplitude also a must?[/quote]

I think amplitudes must be equal but I'm not fully sure....

See this for more:
http://en.wikipedia.org/wiki/Standing_wave

No, I do not think that the amplitudes need to be equal. You can break the second equation(2A =A+A) and then trigonometrically add y1 and one of y2, and leave the other part of y2 as it is.

I dont think amplitude must be equal because addition of these two waves also gives nodes and antinodes.

Yes, I agree with you. But I was trying to say that you will get only a [b]partially [/b]standing wave, with properties quite different from a [b]complete[/b] standing wave.

What kind of properties are you saying would be different. I actually think it will almost be the same.

[quote="Conjurer"]OK atleast tell me : can these waves make a standing wave or is equality of amplitude also a must?[/quote]

A [b]standing wave[/b] results whenever there are two waves of the same frequency moving in opposite directions in the same medium.

If said waves have equal amplitudes, then perfect interference occurs at the [b]nodes[/b] (locations of minimum amplitude). Where perfect interference occurs the combined amplitude is zero.

If said waves have different amplitudes then the combined amplitude at the nodes is a minimum, but not equal to zero.

[b]Anti-nodes[/b] are locations of maximum amplitude for a standing wave.

The ratio of maximum amplitude to minimum amplitude is called the [i]standing wave ratio[/i] (SWR).


Euler
logarithms
calculus
derivative
inequalities
real analysis
real analysis solved

$\gamma_m(p) = \sum_{k=1}^m\frac{(ln\;k)^p}{k}- \frac{(ln\;m)^{p+1}}{p+1}$
When $p=0$ we the above sequence converge to Euler constant.
Prove that when $p\geq1$, $\gamma_m(p)$ converge when $m$ tend to infinity.

There's a much more general setting for this problem. I think I can remember seeing a problem from G.M. which generalized this sort of result as follows:

Let $f:[1,\infty),\ f(x)=\frac{(\ln x)^{p+1}}{p+1}$. We have $f'(x)=\frac{(\ln x)^p}x$. $f$ is increasing (this isn't very important), and $f'$ is decreasing for large enough $x$ (you can check this for yourself by taking the derivative of $f'$). Moreover, $f'(x)\to 0$ as $x\to\infty$. Let's show that in this case $x_n=f'(1)+f'(2)+\ldots+f'(n)-f(n)$ is convergent.

Since $x_{n+1}-x_n=f'(n+1)-(f(n+1)-f(n))=f'(n+1)-f'(c_n),\ c_n\in(n,n+1)$ and $f'$ is decreasing on a neighbourhood of $\infty$, it means that $x_n$ becomes decreasing and $f'(n+1)-f'(n)\le x_{n+1}-x_n\le 0$ for large $n$. The first inequality shows that $x_n$ is bounded (by summation after $n$), and since it's monotonic for large $n$, we're done: $x_n$ is convergent.


rotation
inequalities
algebra
polynomial
complex numbers
geometry solved
geometry

several points are given on a unit circle so that the product of the distances from any point on the circle to the given points does not exceed 2. prove that the given points are the vertices of a regular polygon.

i believe (if my memory doesnt betray me) that this has been solved here before but i dont remember the proof and cant find it either :) so some1 please help

Hello, galois!

I think it is pure algebraic fact.

I have following idea.
Let A1A2...An is n-gon. Suppose arguments (in complex-number sense) of vertices of n-gon are a1, a2, ..., an (0 \leq ai < 2\pi ).
Consider arbitrary point X on circle, let x be its argument. Then AiX= \pm 2sin((ai-x)/2). Thus product of AiX is \pm 2n.sin((a1-x)/2)...sin((an-x)/2)=P.
Using identities sin(u)=2q/(q2+1) and cos(u)=(1-q2)/(q2+1), where q=tan(u/2), we obtain P=\pm 2n(u12t+v1(1-t2))...(u1n2t+vn(1-t2))/(1+t2)n,
where ui=cos(ai/2), vi=sin(ai/2), t=tan(x/4).

Since |P|\leq 2, we conclude (1+t2)n \geq Q(t)=2n-1(u12t+v1(1-t2))...(u1n2t+vn(1-t2)) \geq -(1+t2)n.

Let us vary x from -2\pi to 2\pi . It follows that t varies from -\infty to +\infty.

We have 1 \geq Q(0)=2n-1v1...vn \geq 0 (as vi \geq 0).
First trick: For regular n-gons similar product varies in the same bounds, so we can find regular n-gon for which v'1...v'n=v1...vn, where v'i are corresponding sin-s. Let Q'(t) be corresponding polynom for this regular n-gon.
(*) Now Q(t) and Q'(t) are two polynom degree of 2n with equal leading coefficients.

Second trick: (**) It is easy to see for regular n-gon Q'(t) attains bounds (1+t2)n and -(1+t2)n by turns 2n times (!!!). It follows that equation Q(t)=Q'(t) has at least 2n roots (taking into account order of roots)==> Q(t)=Q'(t) (due to (*)).

That is all. Any comments?

P.S. (**) I have in mind that for regular n-gon P can be represented as xn-e (in some meaning), where x and e are unit complex numbers, x runs along unit circle, i.e. x is complex representation of X.

Hey! It is for one year here! I know it is not a brilliant proof.
Maybe someone knows something better?

we can rotate the complex numbers and supposse their product is 1.

let $P(x)=(x-a_1)(xa_2)....(x-a_n)$ we evaluate $P(a^t)$ for $a$  a primitive root of unity and $t=0,1,2,3...,n-1$, summing all this expressions we get an inequality that must be an equality, and therefore, we get our result from the equality condition!

In fact:  $P(a^0)+P(a^1)...+P(a^{n-1})=2n$  So we must have $P(a^t)=2$

Well, I hope that the below approach is little bit more brilliant. Let $z_1,\,z_2,\,\dots,\,z_n$ be complex numbers, which correspond to our points. Then we have for $f(z)=(z-z_1)(z-z_2)\dots (z-z_n)$ an estimate $|f(z)|\le 2$ for $|z|=1$. Without loss of generality, $f(z)=z^n+a_{n-1}z^{n-1}+\dots+a_1z+1$ (elsewhere rotate our points). Then $\sum_{w^n=1} f(w)=2n$, since the sum of $k$-th powers of roots of unity equals $n$ if $n$ divides $k$ and 0 otherwise. On the other hand, $|\sum_{w^n=1} f(w)|\le \sum_{w^n=1} |f(w)|\le 2n$. So, the last inequality is equality at fact, and $|f(w)|=2$ for any $w$ such that $w^n=1$. Moreover, $f(w)=2$ for each such $w$ (elsewhere the real part of $\sum_{w^n=1} f(w)$ is strictly less than $2n$). So, $f(z)-2$ is a unitary polynomial of degree $n$, for which every root of unity of degree $n$ is a root. So, $f(z)-2=z^n-1$, and $f(z)=z^n+1$, and our points are roots of $-1$.


At the 2000 Olympics in Sydney, Australia, 10,200 athletes
participated in 28 different sports over a period of 16 days.
What was the average number of athletes per sport? Round
your answer to the nearest whole number.

In this problem, they gave a lot of extra information. All you need to know is that there are 10200 athletes and 28 sports. $ 10200 \div 28 \approx \boxed{364}$.


inequalities
LaTeX
inequalities unsolved

Prove the following inequality:

\frac{a^3}{b^2 - bc + c^2} + \frac{b^3}{c^2 - ca + a^2} +\frac{c^3}{a^2 - ab + b^2} \geq a + b+ c

hi.could you tell me please how do i use latex when writing on the forum ?


Prove that each year has at least one Friday 13th :)

There are 2 types of years to consider: 366-days and 365-days (dunno their english names)

356:
say the 13th day is x in January (mod 7 of course)
then x+3 in feb, x+3 in march, x+6 in april, x+8 in may, etc

oh well go on and you'll see it, got no time to write it out now :D


inequalities
IMO Shortlist
maximum value
maximization

Determine the maximum value of the sum
\[ \sum_{i < j} x_ix_j (x_i \plus{} x_j)
\]
over all $ n \minus{}$tuples $ (x_1, \ldots, x_n),$ satisfying $ x_i \geq 0$ and $ \sum^n_{i \equal{} 1} x_i \equal{} 1.$

Write $f(x_1,x_2,\dots,x_n)=\sum_{i<j}{x_ix_j(x_i+x_j)}$ where the sum runs over all $n$ tuples.Wlog let $x_1 \ge x_2 \ge\dots\ge x_k > x_{k+1}=\dots=x_n=0$We sall now solve the problem by using variable mixing.It is not hard to show that $f(x_1,x_2,\dots,x_{k-1}+x_k,0,0,\dots) > f(x_1,x_2,\dots,x_k0,0,\dots)$.So using this over and over again,we see that it suffices to check the maximum value of $f(x_1,x_2,0,0\dots,0)$ with $x_1+x_2=1$ which is clearly $\frac{1}{4}$,So this is the required answer with equality when $x_1=x_2=\frac{1}{2},x_3=\dots=x_n=0$.

Notice that $\sum_{i < j} x_i x_j (x_i + x_j) = \sum_{i \neq j} x_i^2 x_j = \sum_{i = 1}^n x_i^2 \sum_{j \neq i} x_j$. Using the fact $\sum_{i = 1}^n x_i = 1$, this expression can be rewritten as $\sum_{i = 1}^n x_i^2(1 - x_i)$. Finally, manipulating the inequality $x_i (x_i - 1/2)^2 \geq 0$ yields $x_i^2(1 - x_i) \leq \frac{1}{4} x_i$, thus yielding the result with equality case possible.

[quote=sqing]Determine the [b]minimum [/b]value of the sum
\[ \sum_{ i < j} x_ix_j (x_i \plus{} x_j)
\]
over all $ n \minus{}$tuples $ (x_1, \ldots, x_n),$ satisfying $ x_i \geq 0$ and $ \sum^n_{i \equal{} 1} x_i \equal{} 1.$[/quote]

Let $P$ be $\sum_{i<j}x_ix_j(x_i+x_j)$.
Note that $P\ge0$.
When $x_1=1$ and $x_2=x_3=\cdots=x_n=0$, $P=0$.
Thus the minimum value of $P$ is $0$.