geometry
3D geometry
algebra unsolved
algebra

The seventh grade class at a certain middle school decided to have an ice cream party to celebrate their outstanding grades. They decided to make ice cream cones. But they were all very independent thinkers, and they each wanted to have a different cone. Unfortunately the school was only willing to provide them with two ice cream flavors, vanilla and chocolate. They knew that if they used only single scoop cones, they could make only two different cones.

They figured out that if they used double scoop cones they could make four different cones.

There are 256 students in this outstanding class. How many scoops of ice cream would have to be used in each cone in order to make a different cone for each person.

If there are n scoops, there are 2^n choices.
So 8 scoops should be used

Well. In fact it isn't really true because for example double scoop cones are symetrical, that is : the cone chocolate-vanilla and the cone vanilla-chocolate isn't very different.

Oh, I see.
So there are n+1 choices when there are n scoops(the no. of vanilla can be 0 to n)
255 scoops should be used in this case
(It seems quite impossible!? :P )

No, because fo example for $7$ scoops you can put $6$ of them in an hexagonal pattern with one in the center, and then two cones with exactly one vanilla scoop, one in which it is the center one, and one in which it is a border one, are different.
And for certain numbers $n$ of scoops you can position the scoops in several unequivalent ways, for example with six scoops, either you place the six in a hexagon, or you place one in the middle and five around.
 :?

So only one scoop is needed (put it on top of the cone, or push it to the bottom, or ...)

jdawg please clarify your question :D


inequalities
calculus
inequalities unsolved

Can you help me to find the largest value of k so that this inequality is right :
$a^3+b^3+c^3+kabc \geq (1+\frac{k}{3})(a^2b+b^2c+c^2a) \forall a,b,c>0.$

I find the answer is $9\sqrt[3]{\frac{1}{4}}-3$
and also I think i can prove it :)  :D

I'm waiting for your solution,please post it as soon as possible for me to acknowledge.Thank you so much.

you last post just ask to find the largest k,so I didn't post it.
now let me post it.
in fact,it is nothing sepiecal. :)
WLOG $c=min{a,b,c}$
just $f(a,b,c)=a^3+b^3+c^3+kabc - (1+\frac{k}{3})(a^2b+b^2c+c^2a)$
let $g(x)=f(a+x,b+x,c+x)$
prove $g'(x) \geq 0$,when $c+x \geq 0$
then we have $f(a,b,c) \geq f(a-c,b-c,0)$
(or you can prove   $f(a,b,c) \geq f(a-c,b-c,0)$ directly)
then  is to find the largest k such that
$x^3+y^3 \geq (1+\frac{k}{3})x^2y$
where $x=a-c,y=b-c$
not hard.
have you understood?

Nice problem and nice solution , Zhaobin! Here , I don't use mixing variable to solve this problem because SOS looks nicer .
   First , put $c=0$ , we get $3(\frac{a^3+b^3}{a^2b}-1) \geq k$
  We'll find the minimum of $\frac{a^3+b^3}{a^2b}$
 Suppose $\frac{a^3+b^3}{a^2b} \geq m$ . Using calculus , we find $m =\frac{3}{\sqrt[3]4}$ . So $k \leq 3(\frac{3}{\sqrt[3]4}-1)$
 Change the ineq into : $\sum A(b-c)^2 \geq 0$ , where $A=(12+k)b+(6-k)c-3ka$
  The rest is easy .
  Apparently , this problem is Mr Namdung's ?


geometry
circumcircle
trigonometry
algebra
polynomial
geometry proposed

Let $ABC$ be a right triangle, $AC < BC < AB$. Consider the circle passing through $A$ and touching $BC$ at $B$, the circle passing through $B$ and touching $CA$ at $C$, and the circle passing through $C$ touching $AB$ at $A$. Show that these circles have a common point $P$ and show that $11/16 < PA^{2}+PB^{2}+PC^{2}< 1$.

P is one of the Brocard points of $\triangle ABC,$ regardless of $\angle C = 90^\circ.$ Let the circle $(O_{A})$ passing through A and tangent to BC at B meet again at P the circle $(O_{B})$ passing through B and tangent to CA at C. Since $(O_{A})$ is tangent to BC at B, $\angle PAB = \angle PBC.$ Since $(O_{B})$ is tangent to CA at C, $\angle PBC = \angle PCA.$ Since $\angle PAB = \angle PCA,$ the circumcircle $(O_{C})$ of $\triangle PAC$ paccec through C is tangent to AB at A.

Assume now $\angle C = 90^\circ$ is right. Denote $\omega = \angle PAB = \angle PBC = \angle PCA$ the Brocard angle. Since $\angle PCB = \angle C-\angle PCA = 90^\circ-\omega$ and $\angle PBC = \omega,$ it follows that $\angle BPC = 90^\circ$ is right and $PB^{2}+PC^{2}= BC^{2}= a^{2}.$ Let $D \in AB$ be the foot of the C-altitude. The right triangles $\triangle ACD \sim \triangle ABC$ are similar. Since $\angle PAD \equiv \angle PAB = \angle PCA = \omega,$ P is also Brocard point of $\triangle ACD,$ which implies $\angle PDC = \omega$ and $\angle APD = 90^\circ$ is also right. Thus $PA^{2}= AD \cos \omega = \frac{b^{2}}{c}\cos \omega$ and

$PA^{2}+PB^{2}+PC^{2}= a^{2}+\frac{b^{4}}{c^{2}}\cos^{2}\omega$

For any $\triangle ABC,$ $\cot \omega = \cot A+\cot B+\cot C$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=273]Brocard angle satisfies cot w = cot A + cot B + cot C[/url]). For example, if AP meets the circle $(O_{B})$ again at Q, then $\angle PQC = \angle PBC = \angle PAB,$ which means $CQ \parallel AB.$ Since $(O_{B})$ is tangent to CA at C and $CQ \parallel AB$, $\angle QBC = \angle A.$ If R is the foot of a normal from Q to AB, then $QR = AD = h_{c}$ is the C-altitude and $\angle QBR = 180^\circ-\angle B-\angle A = \angle C,$ so that

$h_{c}\cot \omega = AR = AD+DB+BR = h_{c}\cot A+h_{c}\cot B+h_{c}\cot C$

If $\angle C = 90^\circ,$ then $\cot A = \frac{b}{a},\ \cot B = \frac{a}{b},\ \cot C = 0,$ so that $\cot \omega = \frac{b}{a}+\frac{a}{b}= \frac{c^{2}}{ab}$ and

$\cos^{2}\omega = \frac{\cot^{2}\omega}{\cot^{2}\omega+1}= \frac{c^{4}}{c^{4}+a^{2}b^{2}}$

$PA^{2}+PB^{2}+PC^{2}= a^{2}+\frac{b^{4}}{c^{2}}\cdot \frac{c^{4}}{c^{4}+a^{2}b^{2}}= \frac{c^{4}a^{2}+a^{4}b^{2}+b^{4}c^{2}}{c^{4}+a^{2}b^{2}}=$

$= \frac{c^{6}-c^{4}b^{2}+a^{4}b^{2}+b^{4}c^{2}}{c^{4}+a^{2}b^{2}}= \frac{c^{6}-a^{2}b^{2}c^{2}+a^{4}b^{2}}{c^{4}+a^{2}b^{2}}= \frac{c^{6}-a^{2}b^{4}}{c^{4}+a^{2}b^{2}}$

From this, it is clear that $\frac{PA^{2}+PB^{2}+PC^{2}}{c^{2}}< 1$ (typo in the problem or condition c = AB = 1 missing). On the other hand,

$\frac{11}{16}< \frac{PA^{2}+PB^{2}+PC^{2}}{c^{2}}\ \Longleftrightarrow\ 11c^{6}+11a^{2}b^{2}c^{2}< 16c^{6}-16a^{2}b^{4}\ \Longleftrightarrow$

$11a^{4}b^{2}+11a^{2}b^{4}+16a^{2}b^{4}< 5(a^{2}+b^{2})^{3}\ \Longleftrightarrow\ 12a^{2}b^{4}-4a^{4}b^{2}< 5(a^{6}+b^{6})\ \Longleftrightarrow$

$P(x) = 5x^{3}+4x^{2}-12x+5 > 0$ for any $x = \frac{a^{2}}{b^{2}}> 0.$ This cubic polynomial has its local maximum and minimum at

$P'(x) = 15x^{2}+8x-12 = 0,\ \ x = \frac{-8 \pm 28}{30}=-\frac{6}{5}$ and $+\frac{2}{3}.$ The value of P(x) at the local minimum is

$P(2/3) = \frac{40+48-216+135}{27}= \frac{7}{27}> 0$

This means that P(x) has a single negative real root $r <-\frac{6}{5},$ so that P(x) > 0 for any x > 0.


Let ABCD be an convex quadrilateral and put m,M be the minimum and maximum of the edges.Prove that $2(AC^2+BD^2)\geq (7m^2+M^2)$

This is probably geometry...
This is probably inequalities...
This is probably algebra...
But it's for sure not combinatorics!

I put it in Combinatorics because its idea is from an old Problem: Prove that $\frac{\sqrt{AC^2+BD^2}}{2}\geq m$.

This Problem can be solved in 2 ways. The firmilar way is on combinatorics.I solved it in other way,so I proposed the first.

no one can solve it?I've just completed its solution.I think it's interesting

First ,consider ifAC or BD$\geq 2m$we take the inequality hold the case that the are 3 vertices colinear.Otherwise we take it hold the case that the quadrilateral has 4 edge equal


[quote="X-man"]Let $\{ a_n \}_{n \ge 1}$ be a sequnce of positive integers such that there is a constant $a>0$ and $a_n \le an$ for every positive integer $n$. Prove
 1. If $a<5$ then the sequence contain infinitely many numbers such that $5$ doesn't divide there digital sum
 2. If $a=5$ then 1. is not correct.[/quote]
I think $a_n$ sequence of positive distinct integers, else 1 and 2 are wrong.
 Exactly one of numbers from (5x,5x+1,5x+2,5x+3,5x+4) had sum of digits s(n), suth that 5|s(n).
1.Therefore at least $t(n)=n-1-[\frac{an}{5}]$ terms $a_i,i\le n$ satisfyed: $5\not |s(a_i),i\le n$. It proved 1, because $t(n)\to \infty$.
2. Let $a_1=1,a_2=5,a_{2i+1}=10i+d(i),a_{2i+2}=a_{2i+1}+5$, were d(i)=s(i)(mod 5).


articles
LaTeX
linear algebra
matrix
Vieta

Well, I got the first formula to fit by uncentering it, but now the second one is messed up. sigh...

The same problem occurs on the PIE article in the 5-set example.  I found it in the forums, and thought it would be a nice addition to show an extended use of PIE.  However, it runs off the page.  I fixed it by breaking the LaTeX into chunks, but is there a nicer way?

[quote="perfect_radio"]What's to be done when formulas don't fit the width of the screen?

Look [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas]here[/url]. I have my screen's resolution set to $1024 \times 768$ and I get a big gap between:

[quote]Expanding out the right hand side gives us[/quote]

and

[quote]$\left. a_{n}x^{n}-a_{n}(r_{1}+r_{2}+\cdots+r_{n})x^{n-1}+a_{n}(r_{1}r_{2}+r_{1}r_{3}+\cdots+r_{n-1}r_{n})x^{n-2}+\cdots+(-1)^{n}a_{n}r_{1}r_{2}\cdots r_{n}\right.$.[/quote][/quote]

Hmmm... I was going to suggest maybe a size manipulation command, like \small or \footnotesize, but before I posted, I tried them out on the Wiki (with the preview button), and this version of $\LaTeX$ seems to not recognize them for some reason...

Yeah, the TeX in the wiki only recognizes the basic AMS Math stuff.  So you'll notice there are a few things that don't work.  Like those size commands, \binom, you can't put characters under the brace in \underbrace the usual way (though you can use a matrix to do this), etc.

Ok, I fixed both the PIE and Vieta's pages.  For PIE I just broke the formula up onto multiple lines by interrupting the <math> ... </math> and then putting the rest of the formula in other math tags.

For [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas]Vieta's[/url] I took the spaces out between the plus signs and the \cdots.  It's not perfect, but it works.


If $A$, $B$, and $C$ are integers from $0$ to $9$, and \[ \frac{.\overline{ABC}+.\overline{ACB}+.\overline{BAC}+.\overline{BCA}+.\overline{CAB}+.\overline{CBA}}{.\overline{A}+.\overline{B}+.\overline{C}} \] is integer, find the number of possible values of $(A,B,C)$.

Something smells here:

1) If overline means "periodic decimal number" then let $M=999$ and $N=9$
2) If overline means "this is only a numeric representation and not a product", then let $M=1000$ and $N=10$

In any case, we'll have $\frac{\frac{222(A+B+C)}{M}}{\frac{A+B+C}{N}}=\frac{222N}{M}$, which is [u]always[/u] an integer in the first case (equals to $2$) and [u]never[/u] an integer in the second (equals to $2.22$), and that [i]doesn't[/i] depend on $A,B,C.$

Since the digits repeat they are divided by $999$ and $9$, and so we simplify out to $\frac{222(A+B+C)}{99} \cdot \frac{9}{A+B+C}=2$

So all the possibilities work except $(A,B,C)=(0,0,0)$ or in other words $\boxed{999}$ possibilites.

[quote="jhcreinhardt"]Converting into fractions and simplifying yields 2 for all a, b, and c..I think...[/quote]

yes, which makes it simple because we are simply finding the number of solutions, so in other words, there is ONLY one case that can't work $(0,0,0)$

Yes, I forgot that A,B,C are single digits

Hey, chess, what's the source of this problem?


A bag contains blue and green marbles. If  $6$  green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be  $\frac{5}{13}$  . If instead  $7$  blue marbles are added to the bag, the probability of drawing a blue marble would be  $\frac{15}{26}$ . What was the number of blue marbles in the bag before any changes were made?

[hide]You can tell that there are 19 marbles in the bag.

So by subtraction, you get 8 blue marbles in the bag originally.[/hide]


Does someone here play "Antichess"?

I really like it; I got introduced to it in a math competition, and I know well how to play...

http://en.wikipedia.org/wiki/Antichess

See here.

I love that game. For some reason I'm only decent at regular chess yet I win 'Loser's Chess' everytime.

@mathking, look at this link: http://lmgtfy.com/?q=antichess   :D

Fun.  Loser's chess and giveaway are completely different, though.  I'm not too good, but on ICC (chessclub.com), in the Leauge of Losers (LOL), there are some really great loser's players

The position posted in the diagram at the link below begs the question: how does white win after either 5. ... a1K(ing) or 5. ... a1N?

http://en.wikipedia.org/wiki/Antichess

a1N, you promote to bishop, and move to a1 (the night must move)
a1K, promote to a rook and corner the king; you make sure that you stay away from the king so that it can't get captured by you, and you stay 2 squares away diagonally. He has to move toward a corner, and once trapped there, he loses.


function
calculus
derivative
limit
LaTeX
real analysis
real analysis unsolved

Suppose that $ f$ and $ g$ are two differentiable functions which satisfy $ f g' \equal{} f' g$.  Prove that if $ a$ and $ b$ are adjacent zeros of $ f$, and $ g(a)$ and $ g(b)$ are not both $ 0$, then $ g(x) \equal{} 0$ for some $ x$ between $ a$ and $ b$.

I think maybe the mean value theorem would be used or something but I can't find how.

adjacent zeros:  I mean $ a < b$ and $ f(a) \equal{} f(b) \equal{} 0$ but for any $ a < x < b \ \  f(x) \neq 0$.

What do you mean by "adjacent zeros"? If $ f\equal{}0$ then it satisfies the hypotheses for practically any choice of $ g$...

[quote="CommandoGuard"]What do you mean by "adjacent zeros"? If $ f \equal{} 0$ then it satisfies the hypotheses for practically any choice of $ g$...[/quote]

original post updated to answer this

Then this is very simple. Note that $ (fg)^{'} \equal{} fg^{'} \minus{} f^{'}g \equal{} 0$ so $ fg$ is constant. Since $ f$ has a zero, $ fg\equal{}0$. In particular $ f(x)g(x)\equal{}0$ for all $ x \in (a,b)$ but since $ f(x) \ne 0$ for such $ x$, we must have $ g(x) \equal{}0 \: \forall x \in (a,b)$.

$ (f g)' \equal{} fg'\minus{}f'g$? I thought $ (f g)' \equal{} fg' \plus{} f'g$.  What am I missing?

Right, sorry. We'll do it this way then: $ \left(\frac{f}{g}\right)^{'}\equal{}\frac{f^{'}g\minus{}fg^{'}}{g^{2}}\equal{}0$ in points at which $ g(x) \ne 0$. If $ g(x) \ne 0 \quad \forall a<x<b$ then $ f\equal{}cg$ on the interval $ [a,b]$, where $ c \in \mathbb{R}$ Since $ f(a)\equal{}0$ and $ g(a) \ne 0$ we have $ c\equal{}0$, but then $ f\equal{}0$ which is a contradiction.

[quote="amacfie"]Suppose that $ f$ and $ g$ are two differentiable functions which satisfy $ f g' \equal{} f' g$.  Prove that if $ a$ and $ b$ are adjacent zeros of $ f$, and $ g(a)$ and $ g(b)$ are not both $ 0$[/quote]  This situation is impossible: $ f$ is differentiable and nonzero on $ (a, b)$.  Then we can show in any of several ways that we must have $ g \equal{} cf$ for some constant $ c$ and so $ g(a) \equal{} \lim_{x \to a^\plus{}} g(x) \equal{} \lim_{x \to a^\plus{}} c f(x) \equal{} 0$ and likewise $ g(b) \equal{} 0$.

By the way, CommandoGuard, when you use a ', LaTeX knows where to put it -- adding an additional carrot makes it harder to read.


geometry
rectangle
trigonometry
geometric transformation
rotation
homothety
analytic geometry

Let $ ABCD$ be a rectangle with center $ O$, $ AB\neq BC$. The perpendicular from $ O$ to $ BD$ cuts the lines $ AB$ and $ BC$ in $ E$ and $ F$ respectively. Let $ M,N$ be the midpoints of the segments $ CD,AD$ respectively. Prove that $ FM \perp EN$.

[hide="solution :huh: "]
Call $ AB\equal{}a$, $ BC\equal{}b$ and $ P: NE \cap MF$, w.l.o.g. $ a>b$, so we have $ EB\equal{}\frac{a^2\plus{}b^2}{2a}$ and  $ FB\equal{}\frac{a^2\plus{}b^2}{2b}$ and then $ AE \equal{} \frac{a^2\minus{}b^2}{2a}$ and $ FC \equal{} \frac{a^2\minus{}b^2}{2b}$, then $ \tan ANE \equal{} \frac{a^2\minus{}b^2}{ab}\equal{}\tan CMF$ and so MDPN is cyclic and $ \angle MPN \equal{} \frac{\pi}{2}$[/hide]

[quote="Valentin Vornicu"] [color=darkred]Let $ ABCD$ be a rectangle with center $ O$ and  $ AB\neq BC$.

Denote the points $ E\in AB$ , $ F\in BC$ so that $ O\in EF$ and $ EF\perp BD$ .

Let $ M,N$ be the midpoints of the segments $ [CD]$ , $ [AD]$ respectively. Prove that $ FM \perp EN$.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ L\in FM\cap EN$ . Observe that the quadrilaterals $ OAED$ , $ OFCD$ are cyclically

(diameters $ [DE]$ . $ [DF]$ respectively). Thus, $ \left\|\begin{array}{c} \widehat {DEF}\equiv\widehat {DEO}\equiv\widehat {DAO}\equiv\widehat {DAC} \\
 \\
\widehat {DFE}\equiv\widehat {DFO}\equiv\widehat {DCO}\equiv\widehat {DCA}\end{array}\right\|$ $ \implies$

$ \boxed {\ \triangle DEF\ \sim\triangle DAC\ }$ $ \implies$ $ DE\perp DF\ \implies\ BEDF$ is cyclically (diameter $ [EF]$ ) $ \implies$

$ \widehat {CFD}\equiv\widehat {AED}$ $ \implies$ $ \boxed {\ \triangle CFD\ \sim\ \triangle AED\ }$ $ \implies$ $ \widehat {CMF}\equiv\widehat {ANE}$ $ \implies$ $ \widehat {CML}\equiv\widehat {DNL}$ $ \implies$

The quadrilateral $ DMLN$ is cyclically (diameter $ [MN]$ )  $ \implies$ $ LM\perp LN$ $ \implies$ $ FM\perp EN$ .

[b][u]Remark I.[/u][/b] Otherwise, if denote the intersection $ P\in AD\cap EF$ , then $ BP\parallel FD$ and

$ P$ is the orthocenter of the triangle $ BDE$ , i.e. $ BP\perp DE$ . Therefore, $ DE\perp DF$ a.s.o.

[b][u]Remark II.[/u][/b] Generally, for $ N\in (AD)$ , $ M\in (CD)$ so that $ \frac {NA}{ND} \equal{} \frac {MC}{MD}$ $ \implies$ $ FM\perp EN$ .

$ \triangle DAE$ results from $ \triangle DCF$ by rotation $ \left(D\ ,\ 90^{\circ}\right)$ and homothety $ \left(D\ ,\ \frac {DA}{DC}\right)\ !$[/color]

[b]Here is my solution using Complex numbers.[/b]

Treat O as the origin and the lines through O parallel to AB and BC as the X and Y axes.
Denote the coordinate of each uppercase letter by a lowercase letter. The coordinates of A,B,C,D are a,b,-a,-b respectively.
Let <DBC=x
Then (d-c)/(b-c)=itanx or equivalently
[b]itanx=(a-b)/(a+b)  ......(1)[/b]

We also have f/b=itanx or
[b]f=(a-b)b/(a+b)  .............(2)[/b]        (using (1))

Also b/e=itanx=(a-b)/(a+b) or
[b]e=b(a+b)/(a-b) ...............(3)[/b]

It is easy to note that m=(c+d)/2=-(a+b)/2 and n=(a+d)/2=(a-b)/2

From these we get
[b] (f-m)/(e-n)=(a-b)/(a+b)=itanx.[/b]
So FM is perpendicular to EN as desired.

We need to prove $\frac{CM}{CF}=\frac{AN}{AE}$ or $\frac{2CM}{CF}=\frac{2AN}{AE}\iff\frac{CD}{CF}=\frac{AD}{AE}\ (\ *\ )$. Since $BEDF$ is a kite, we have $DF\bot DE$, $\triangle ADE\sim\triangle CDF$ ( as [b]Virgil[/b] mentioned ) and, indeed, $(*)$ is true.

Best regards,
sunken rock


given an  equation $ { 8t^3\minus{}4t^2\minus{}4t\plus{}1\equal{}0}$ have a  root x on $ {(0,\frac { 180}{13})}$ , then find x.

[quote="CCMath1"]given an  equation $ { 8t^3 \minus{} 4t^2 \minus{} 4t \plus{} 1 \equal{} 0}$ have a  root x on $ {(0,\frac { 180}{13})}$ , then find x.[/quote]

The three roots of this equation are $ \cos(\frac{\pi}7)$,  $ \cos(\frac{3\pi}7)$ and $ \cos(\frac{5\pi}7)$.

Both values $ \cos(\frac{\pi}7)$ and  $ \cos(\frac{3\pi}7)$ belong to $ (0,\frac{180}{13})$ and so :

either $ x\equal{}\cos(\frac{\pi}7)$ $ \equal{}0,900968868...$

either $ x\equal{}\cos(\frac{3\pi}7)$ $ \equal{}0,222520934...$


Afraid Im not the best at wrapping my head around proofs, I can do the math itself, I just never know where to start. Came here hoping sombody could point me in the right direction- or even guide me to the end of this if you would be so kind...  ;)

The question is worded as follows: Suppose the natural number [i]N[/i] is ([i]a[size=75]n-1[/size]...a[size=75]0[/size][/i])[size=75]b[/size] (in base [i]b[/i]). Prove that [i]n[/i] - 1 = [log[i][size=75]b[/size] N[/i]] (and hence that [i]N[/i] has 1 + [log[i][size=75]b[/size] N[/i]] digits in base [i]b[/i]).

Any help would be appreciated. :)

Just look what's going on in base 10.

If a number $N$ has $n$ digits,  $10^{n-1} \leq N < 10^{n}$ then "take log".

Same thing in any base.


function
number theory proposed
number theory

Let $ n\in N$
Find
$ min_{k\in N }(k\plus{}[{\frac{n}{k}}]$

[hide="Hint"]Analyze the function $ k\plus{}\frac{n}{k}$ with its minimum at $ n^{1/2}$ and increasing at both sides [/hide]


Problem 1: Solve for $ x$ in positive real numbers,
\[ x^{x^5}\equal{}5\]

Problem 2: Let $ n$ be a natural number,
\[ x^{x^{\cdots^{x^n}}}\equal{}n\]

Find a solution for $ x$ in positive real numbers, in terms of $ n$.

I think this works.... I might be missing some solutions though...

#1 [hide]
$ f(n) = x{}^{x{}^n}$
Holding x constant we know:
$ 5 = f(f(5))$
$ f(f(5)) = 5 = f(f(f(f(5))))$
Combining we get:
$ 5 = f(f(5)) = f(f(f(f(5))))$
Going on like this we can find out that any number of combined $ f(f(f(\cdots f(5))))$ works. Continuing this onto infinity, we get one of those old power tower equations where:
$ x{}^x{}^x^\cdots = 5$
$ x^5 = 5$
$ x = \sqrt [5]{5}$
[/hide]
#2 [hide]A solution by similar substitution above is $ \sqrt [n]{n}$[/hide]


trigonometry AMC AIME USA(J)MO USAMO geometry

Hello everybody,

The first admissions test for this summer is running Jan. 15th to Feb. 5th.

You can obtain the test here:

http://awesomemath.org/wp-content/uploads/2010/01/Test_A.pdf

Good Luck!

Hello!
    My friend stated that the Test A is only for people applying for level 1 courses. Is that true? (i really hope not, because I spent considerable time today and yesterday working on the problems already :( )

And secondly, is it true that we won't need trigonometry to solve the admission test questions?

First question: no, your friend is wrong.

Second question: That is correct, but trigonometry is very useful and not too complicated, you might want to learn the basics anyways.

All the admissions tests are the same level, just to clarify what AIME15 said.

Just out of curiosity, how difficult are the questions intended to be (although they are proofs are they AMC, AIME or USAMO or somewhere between one of these three levels)? Also, is there a reason why there is no geometry on Test A?

I'd say easy AIME level, others might disagree.

Thanks. Would you say that some would qualify as medium-difficult AIME?

The cover of the book says that the questions use only the most basic concepts, although they are be used in creative methods. All the questions up to the point that I have solved only invoked basic concepts, although they are creatively used.

Yeah definitely agree. I think though they are competition-like problems, the AMP qualifying test seems to be different from AIME, and thus is kind of difficult to compare. This is based on what I have solved.


Let $ a_0,a_1$,... be positive integers such that $ a_0 \equal{} 1, a_1 > 1$ and

$ a_{n \plus{} 1} \equal{} \frac {a_1\cdot a_2\cdot ...\cdot a_n}{a_{\lfloor{\frac {n}{2}}\rfloor}}$

for all $ n \equal{} 1,2,....$

Show that $ \sum_{n \equal{} 1}^{\infty}\frac {1}{a_{n \plus{} 1}a_{\lfloor{{\frac {n}{2}}\rfloor}}}$ is a rational number.

[quote="feliz"]It is not.

[hide="Hint to show the number´s irracionality"]Racionality is periodicity in decimal base. But 10 is not a good base in this case.[/hide][/quote]

I don't think you've proved the sum is irrational...

Let $ a_1\equal{}a>1$, then $ a_n\equal{}a^{f(n)}$ for any n, $ a_1\equal{}a_2\equal{}a_3\equal{}a,a_4\equal{}a^2,a_5\equal{}a^4,...$
Therefore $ S\equal{}\sum_n a^{\minus{}g(n)}, \ g(n)\equal{}f(n\plus{}1)\plus{}f([n/2])$. If S rational, then exist k,m suth that $ g(n\plus{}k)\equal{}g(n)\plus{}k \ \forall n>m$.
In these case $ g(n)$ increase too fast and $ g(n\plus{}k)>g(n)\plus{}k$ for big n.

Can you explain some more? I don't get a thing...


You meant
$ \frac{{a^3 }}{{a^2  \plus{} ab \plus{} b^2 }} \plus{} \frac{{b^3 }}{{b^2  \plus{} bc \plus{} c^2 }} \plus{} \frac{{c^3 }}{{c^2  \plus{} ac \plus{} a^2 }} \ge \frac{{a \plus{} b \plus{} c}}{3}(a,b,c > 0)$
Well the easiest proof that I have known so far is conducted by following steps
Firstly, we prove the identity which is obvious and simple:
$ \frac{{a^3\minus{}b^3}}{{a^2  \plus{} ab \plus{} b^2 }} \plus{} \frac{{b^3 \minus{}c^3}}{{b^2  \plus{} bc \plus{} c^2 }} \plus{} \frac{{c^3\minus{}a^3 }}{{c^2  \plus{} ac \plus{} a^2 }}\equal{}0$
Meanwhile, the equivalent inequality is
$ \frac{{a^3\plus{}b^3}}{{a^2  \plus{} ab \plus{} b^2 }} \plus{} \frac{{b^3\plus{}c^3 }}{{b^2  \plus{} bc \plus{} c^2 }} \plus{} \frac{{c^3\plus{}a^3 }}{{c^2  \plus{} ac \plus{} a^2 }}\geq2\frac{{a \plus{} b \plus{} c}}{3}(a,b,c > 0)$
and finally, applying Cauchy- Schwartz, we get the immediate result! :)
$ \frac{{a^3\plus{}b^3}}{{a^2\plus{} ab \plus{} b^2 }}\equal{} \frac{(a\plus{}b)(a^2\minus{}ab\plus{}b^2)}{a^2  \plus{} ab \plus{} b^2 } \geq \frac{a\plus{}b}{3}$

by Cauchy Schwarz, we have
\[ LHS \equal{} \sum \frac {a^2}{a \plus{} b \plus{} \frac {b^2}{a}} \ge \frac {(a \plus{} b \plus{} c)^2}{2a \plus{} 2b \plus{} 2c \plus{} \frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c}} \equal{} t.\]

We need to prove that
\[ \frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c} \le a \plus{} b \plus{} c.\]
which is trivial by rearrangement.

This concludes \[ t \ge \frac {(a \plus{} b \plus{} c)^2}{3(a \plus{} b \plus{} c)} \equal{} \frac {a \plus{} b \plus{} c}{3}.\]

So it's clear now that \[ \frac {a^3}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^3}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^3}{c^2 \plus{} ca \plus{} a^2} \ge \frac {a \plus{} b \plus{} c}{3}.\]

QED.

[quote="great math"][quote="Raja Oktovin"]by Cauchy Schwarz, we have

We need to prove that
\[ \frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c} \le a \plus{} b \plus{} c
\]
which is trivial by rearrangement.
[/quote]
I think your inequality above is totally wrong.
The exact form is
$ \frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c} \ge a \plus{} b \plus{} c$[/quote]

Today, I solve the problem. I think it is interesting
\[ \begin{array}{l}
 \frac{{a^3 }}{{a^2  \plus{} ab \plus{} b^2 }} \equal{} a \minus{} \frac{{ab \plus{} b^2 }}{{a^2  \plus{} ab \plus{} b^2 }} \equal{} a \minus{} \frac{{ab(a \plus{} b)}}{{a^2  \plus{} ab \plus{} b^2 }} \\
 a^2  \plus{} ab \plus{} b^2  \ge 3ab \\
  \Rightarrow  \minus{} \frac{{ab(a \plus{} b)}}{{a^2  \plus{} ab \plus{} b^2 }} \ge  \minus{} \frac{{a \plus{} b}}{3} \\
  \Rightarrow a \minus{} \frac{{ab(a \plus{} b)}}{{a^2  \plus{} ab \plus{} b^2 }} \ge a \minus{} \frac{{a \plus{} b}}{3} \\
  \Rightarrow \frac{{a^3 }}{{a^2  \plus{} ab \plus{} b^2 }} \ge \frac{{2a \minus{} b}}{3} \\
 .............................. \\
 \end{array}
\]


Isosceles triangle ABC has sides AB and AC congruent. Square EFGH is the largest possible square that can be inscribed inside ABC. Given that B, E, and F are collinear, BE = 10, BE < BF, and the altitude from point A to side BC has length 75, find the area of triangle AGH.

It's definitely doable, but this problem might be a little difficult - it's getting late and I pulled this problem off the top of my head.

Can you please post your solution?

I don't think this problem is hard, and I'm not extremely smart or anything...

[hide="Hint"]Similar triangles[/hide]
[hide="Solution"]
[asy]pair B,A,C,H,G,I,F,E,J; B=(0,0); A=(25,75); C=50*dir(0); H=(10,30); G=(40,30); I=midpoint(B--C); F=(40,0); E=(10,0); J=(25,30); draw(A--B--C--cycle); draw(E--F--G--H--cycle); draw(A--I); dot(A); label("A",A,N); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(E); label("E",E,S); dot(F); label("F",F,S); dot(G); label("G",G,NE); dot(H); label("H",H,NW); dot(I); label("I",I,S); dot(J); label("J",J,NE);[/asy]

There's the picture.
Note that $ \triangle BHE \sim \triangle HAJ$. Let the side length of the square be $ s$. Then by CPSTS (congruent parts of similar triangles are similar),
$ \frac{75-s}{s/2}= \frac{s}{10}$

My TI-89 tells me that the positive solution for $ s$ is $ s=30$.

Note that
$ [AGH] = AJ\cdot HJ = \frac{(75-s)s}{2}$.

Plug in $ s=30$:
$ [AGH] = \boxed{675}$.[/hide]

[hide]I also got $ \boxed{675}$ by similarity... I don't think it's much harder than normal Mathcounts, i think there was a similar problem this year in the National Mathcounts team round, but I'm not sure.[/hide]


Let $a,b,c$ be positive numbers. Prove that
   $\frac{(a+b+c)(a^2+b^2+c^2)}9 \geq  \sqrt{\frac{(a^2+bc)(b^2+ca)(c^2+ab)}8}.$

[quote="dduclam"]Let $ a,b,c$ be positive numbers. Prove that
   $ \frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{9} \geq\sqrt{\frac{ (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}{8}}.$[/quote]
$ \frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{9} \geq\sqrt{\frac{ (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}{8}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(8a^6\plus{}16a^5b\plus{}16a^5c\plus{}24a^4b^2\plus{}16a^4c^2\minus{}65a^4bc\minus{}49a^3b^3\plus{}$
$ \plus{}32a^3b^2c\plus{}32a^3c^2b\minus{}38a^2b^2c^2)\geq0\Leftrightarrow19(a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}$
$ \plus{}\sum_{cyc}(8a^6\plus{}16a^5b\plus{}16a^5c\plus{}5a^5b^2\plus{}5a^4c^2\minus{}27a^4bc\minus{}11a^3b^3\minus{}$
$ \minus{}6a^3b^2c\minus{}6a^3c^2b)\geq0,$ which true by Muirhead.

[quote="arqady"]
 which true by Muirhead.[/quote]
 Can you speak more louder?

Now,we let's check the following similar inequality
 $ 8(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab) \leq (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2$   with $ a,b,c \geq 0$

[quote="dduclam"]Now,we let's check the following similar inequality
 $ 8(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \leq (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2$   with $ a,b,c \geq 0$[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="dduclam"]Thank you arqady :D. But I hope the following inequality is true
$ (a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \leq (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2$   with $ a,b,c \geq 0$[/quote]
We have:
$ (a\plus{}b)(a\plus{}c) \geq a^2\plus{}bc$
$ (b\plus{}c)(b\plus{}a) \geq b^2\plus{}ca$
$ (c\plus{}a)(c\plus{}b) \geq c^2\plus{}ab$
So $ (a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \leq (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2$
Done!

The following inequality is true too:
For non-negative $ a,$ $ b$ and $ c$ prove that:

\[ 4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \leq (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\]

[quote="arqady"]The following inequality is true too:
For non-negative $ a,$ $ b$ and $ c$ prove that:
\[ 4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \leq (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2
\]
[/quote]

The inequality is clearly true if we can prove that:
\[ (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2 \ge 4(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)\plus{}32a^2b^2c^2\]
Assume that $ c\equal{}min(a,b,c)$, we have:
\[ LHS\minus{}RHS\equal{}(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}8abc^2(a\minus{}b)^2\plus{}4abc(a\plus{}b)(a\minus{}c)(b\minus{}c) \ge 0\]
Therefore, we have the desired result.

Given $ a,b,c >o$ and $ abc\equal{}1$.Provethat
 $ \sqrt{(\frac{a}{b}\plus{} \frac{b}{c}\plus{}\frac{c}{a})(\frac{a}{c}\plus{}\frac{c}{b}\plus{}\frac{b}{a})} \geq 1\plus{} \sqrt[3]{(a^3\plus{}1)(b^3\plus{}1)(c^3\plus{}1)}$

[quote="dduclam"]Given $ a,b,c > o$ and $ abc \equal{} 1$.Provethat
 $ \sqrt {(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})(\frac {a}{c} \plus{} \frac {c}{b} \plus{} \frac {b}{a})} \geq 1 \plus{} \sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)}$[/quote]
We have  $ \sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)}\equal{}\sqrt[3]{(\frac{a^2}{bc}\plus{}1)(\frac{b^2}{ac}\plus{}1)(\frac{c^2}{ba}\plus{}1)}\equal{}t\ge 2$
Ineq <=> $ \sqrt{t^3\plus{}1}\ge 1\plus{}t$
       <=> $ t(t\minus{}2)(t\plus{}1)\ge 0$
So ineq was proved  :D

[quote="Mather"][quote="dduclam"]Given $ a,b,c > o$ and $ abc \equal{} 1$.Provethat
 $ \sqrt {(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})(\frac {a}{c} \plus{} \frac {c}{b} \plus{} \frac {b}{a})} \geq 1 \plus{} \sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)}$[/quote]
We have  $ \sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)} \equal{} \sqrt [3]{(\frac {a^2}{bc} \plus{} 1)(\frac {b^2}{ac} \plus{} 1)(\frac {c^2}{ba} \plus{} 1)} \equal{} t\ge 2$
Ineq <=> $ \sqrt {t^3 \plus{} 1}\ge 1 \plus{} t$
       <=> $ t(t \minus{} 2)(t \plus{} 1)\ge 0$
So ineq was proved  :D[/quote]
Thank you Mather,nice solution :D

[quote="dduclam"]Let $ a,b,c$ be positive numbers. Prove that
 \[ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{9} \geq\sqrt {\frac { (a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}{8}}.\][/quote]
Setting $ x\equal{}a^2\plus{}b^2\plus{}c^2,y\equal{}ab\plus{}bc\plus{}ca$, then $ x \ge y$.
By the AM-GM Inequality, we get
\[ (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab) \le \frac{1}{27}(x\plus{}y)^3\]
It suffices to prove that
\[ (a\plus{}b\plus{}c)^2(a^2\plus{}b^2\plus{}c^2)^2 \ge \frac{3}{8}(x\plus{}y)^3\]
\[ \Leftrightarrow 8x^2(x\plus{}2y) \ge 3(x\plus{}y)^3\]
Since
\[ 2x \ge x\plus{}y\]
It suffices to prove that
\[ 4x(x\plus{}2y) \ge 3(x\plus{}y)^2\]
\[ \Leftrightarrow x^2 \plus{}2xy\minus{}3y^2 \ge 0\]
\[ \Leftrightarrow (x\minus{}y)(x\plus{}3y) \ge 0\]
which is true.
We have done. :)

This problem 's easy. My solution using normalization technique and AM-GM,same yours,Can  :)

However,we have a following stronger result:

$ \bigstar$ For $ a,b,c\ge 0$ than $ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)^2\ge 3(2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab)$

And a hard problem:

$ \bigstar$ Find the best constant for inequality: $ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{9} \geq\sqrt {\frac { (ka^2 \plus{} bc)(kb^2 \plus{} ca)(kc^2 \plus{} ab)}{(k \plus{} 1)^3}}$  where $ a,b,c$ be nonnegative numbers.


algebra
polynomial
real analysis
real analysis unsolved

A have a problem:
Let $ P(x)$ be a polynomial with real coefficients and with degree $ n$.When this polynomial has a root with multiple $ m$?($ m\in N$)

You mean "complex root of multiplicity $ m$"? Then it happens if and only if $ P,P',P'',\dots,P^{(m\minus{}1)}$ have a common factor (which can be easily checked using the Euclidean algorithm).

Yes.But i mean coefficients of this polynomial satisfies which condition?Can you help me?
(If degP=2 then it has root with multiple 2 iff D=0)

For roots of multiplicity 2, the [url=http://mathworld.wolfram.com/Resultant.html]resultant[/url] of $ P$ and $ P'$ must be $ 0$. Beyond that I don't know a simple condition. Of course, vanishing of all pairwise resultants is necessary but I do not see why it would be sufficient.


The download site for $\LaTeX$ TexicCenter doesn't work now. (http://texniccenter.org/front_content.php?idcat=26) Does anyone know an alternative site :o


Does anyone when the results for the 2007 apmo will be released?

Mexico got its results from the Apmo 2007 , we got six bronze(i got one) ,one silver and 2 honorable mentions. we got 9 out of 22 countries.  :D

The results are now available!

[url]http://www.kms.or.kr/competitions/apmo/[/url]

BanishedTraitor//oh, thanks!, problems in APMO i have been searching on Internet... finally they're found!
marko avlia//one more q. please. period of APMO. Is that annual contest, or biennial contest, and so on.?


abstract algebra
LaTeX
number theory
relatively prime
superior algebra
superior algebra unsolved

Suppose G is a group and a,b are two relatively prime integers such that all powers of exponent a commute and so do the powers of exponent b. Prove that G is abelian. My solution is very ugly, but I wonder if there might be something nice about this problem.

Hi there,
Nice & Easy:Since (a,b)=1,there exists elements r,s s.t ra+sb=1 (Bezout).I claim that for every x,y in G
x^a* y^b=y^b*x^a (ie all powers of exponent a commute with all powers of exponent b) then
xy=x^(ra+sb)* y^(ra+sb)=x^ra*x^sb*y^ra*y^sb=...=y^(ra+sb)*x^(ra+sb)=yx.
Proof:(x^a*y^b)^ra=x^a*(y^b*x^a)^ra*(y^b*x^a)^(-1)*y^b=
         (y^b*x^a)^ra*x^a*(y^b*x^a)^(-1)*y^b=(y^b*x^a)^ra (similarly for sb)  thus
         x^a*y^b=(x^a*y^b)^(ra+sb)=(x^a*y^b)^ra*(x^a*y^b)^sb=(y^b*x^a)^ra*(y^b*x^a)^sb=
         (y^b*x^a)^(ra+sb)=y^b*x^a .This completes the proof.

[quote="The Captain"]
proof:(x^a*y^b)^ra=x^a*(y^b*x^a)^ra*(y^b*x^a)^(-1)*y^b
          [/quote]
can you explain this step in more details?

it is easy to see $[x^{a},y^{b}]^{b}=1=[x^{a},y^{b}]^{a}$ so it yields that $[x^{a},y^{b}]=1$ and it is also straightforward to see that a-th power and b-th power generate $G$.

@-oo- Look at the time of my post. :wink: One more point: Can't edit it any more.


Determine the largest possible value of $ z$, if $ x,y,z$ are real numbers satisfying $ x\plus{}y\plus{}z\equal{}5$ and $ xy\plus{}yz\plus{}zx\equal{}3$.

sorry, i put x instead of z in last line

[quote="sax"]x^2 + y^2 +z^2 +2(xy + yz + xz )=25
19=x^2 + y^2 +z^2>=z^2+2yx=z^2 +3 - 2z (y+x)= z^2 +3+2z^2 - 10z=3z^2-10z +3
so 3z^2-10x-16>=0[/quote]
I think you are wrong sax...
[hide="solution"]
$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 19$
But:
$ 19\ge z^2 \plus{} 2xy \Longleftrightarrow19\ge z^2 \plus{} 2[3 \minus{} z(5 \minus{} z)]\Longleftrightarrow 19\ge z^2 \plus{} 6 \minus{} 10z \plus{} 2z^2 \Longleftrightarrow 3z^2 \minus{} 10z \minus{} 13\le 0$
From here we can easily determine the largest value of $ z$
With your way sax i think $ z\rightarrow \plus{} \infty$......  [/hide]

thanks but in last line you put z^2 instead of [b]3z^2[/b]


Anyone have pictures from MathCounts Nationals that you are willing to share?
Post them in the "Math Contests" category in the album...

Thanks!

just take pictures of the tv when the countdown's on national tv.
wait, they didn't show it yet did they??

I uploaded some things....
I might upload actual pictures later

[quote="Zmastr"][quote="joml88"][quote="Zmastr"]Can anyone say when ESPN 2 is broadcasting it? I remember seeing somewhere it would be around the start of June.  :huh:[/quote]

I don't think they are this year.[/quote]

Oh, I remember hearing that they were having it, but there's a chance you may be right.  I think it'll take a while for them to show the comp because they need to edit it and work out legal stuff.  It should probably be in 2-3 weeks.[/quote]

such as andrew on the die problem

[quote="Treething"]XD ignite's countdown summary-ish sheet.

Somebody will submit it soon.[/quote]

If anybody is looking for my summary sheets
http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=858
http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=859

I took 3 pictures of the 2nd page of the top 57 too but they were all really crappy pictures as I kept getting glares or something and lots of people in the way.

[quote="Klebian"]Yeah, here's the link to the msnbc video. (Note: You need IE to watch it)
http://www.msnbc.msn.com/id/12772703

Andrew... curses... after missing the die problem. :rotfl:[/quote]

Lol i didn't see that even though i was right next to him. My thoughts after his answer were

"Hmm, the diagram is extraneous. Wait...am i sure it's extraneous? Hmmm....I still have 20 seconds to determine whether or not it's extraneous. Ok, it's extraneous. Really? Why is it there then? .... *buzz* five?"

There's also a picture of the VA team at the whitehouse [url=http://www.whitehouse.gov/news/releases/2006/05/images/20060515-1_p051506pm-0142-515h.html]at the whitehouse website[/url].

omg... every single one of the first 25 or so countdown questions were so blatantly obvious... then the last 25 or so were like "I DIDN'T SEE THE FINE PRINT, THIS IS CHEAP".

Kind of explains why Neal and I missed like five problems each...


okay, so this is DICE GAME II:

Death
In
Crazy
Escapes

After escaping from THE LAB WHERE CRAZY EXPERIMENTS HAPPEN, 8 survivors make their way through the LAND OF CRAZY LABS to find THE PIE.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=269167]Click here[/url]

Above is a link to the first dice game.

The point is:

BE AS ACTIVE AS POSSIBLE!

In this version of the game, you have 2 options:

Run or hide.

If you run, you escape the LAB MEN.

if you hide, you have a 1/2 chance to survive. I will roll a die and if its even, you escape, if its odd, you are found and CAUGHT!!!!

If you are very active but you must go on a trip, please notify me.

Have fun! AND DONT GET CAUGHT BY LAB MEN!!!!

However, if you run more than 3 times, you die of no energy.

The last escaper (one who survived the whole game) is the winner.

You also have an option of being a LAB MAN a.k.a. Game mod, in case I'm gone for a while. BUT ONLY ONE SPOT YOU HAVE!

Signups:
1.
2.
3.
4.
5.
6.
7.
8.

Mod:

Okay, I guess I join. Will the turns go in order of signups, like the first game?

signups:

1.PowerOfPi
2.spaceguy524
3.bluecarneal
4.ytrewq
5.mathtyro
6.abcak

spaces left:
#7, #8, and THE MOD


Given $8$ distinguashable rings, find the number of possible five-ring arragement on $4$ fingers.

[hide]Number of ways to select, in order, the $5$ of the $8$ rings:
$P^{8}_{5}=8\times 7\times 6\times 5\times 4$
Number of ways to place $3$ separators (to make $4$ groups) in $5+3$ places:
$C^{8}_{3}=\binom{8}{3}$
(maybe there are no rings in a finger, maybe one, maybe up to five, so we have to assume big fingers)
So, my first solution would be
$P^{8}_{5}C^{8}_{3}$
(this counts even the order of the rings in each finger)[/hide]

Maybe you mean $5!C^{8}_{5}C^{8}_{3}$, which is equivalent to my solution $P^{8}_{5}C^{8}_{3}$.
I think this topic was discussed a little more in a posterior double-post.


modular arithmetic
number theory
relatively prime

[hide]The last two digits of $x^y$ is a cyclic pattern. The last two digits will repeat on only factors of 100. The only way this can happen if the pattern ends in $01$. Which leads us to conclude that $x^y$ has the same last two digits of $x^{y+100}$.[/hide]

[quote]...The last two digits of  is a cyclic pattern. The last two digits will repeat on only factors of 100....[/quote]

Yeah, that's what you have to prove...

I think proofs are above most kids in this forum

[quote="rcv"][quote="chess64"]Prove that the last two digits of $x^y$ are the same as the last two digits of $x^{100+y}$. ($x^{100+y} \equiv x^y \pmod {100}$. For example, $2^{103}$ ends in $08$.)[/quote]
Generally, proofs are above the level of MATHCOUNTS.

This specific proof is especially difficult, because it is false.

$2^{101}=2535301200456458802993406410752$.

$05^1 = 05$
$05^{101} = \hbox{xxxxx}25$[/quote]

Lol. How did you calculate 2^101?

[hide]
Assume for now that x is relatively prime to 100.
By Fermat's Little Theorem, we have
\[
x^{40} \equiv 1 \mod {100}
\]
By squaring or taking square roots, we have
\[
x^{80} \equiv 1 \mod {100}
\]
and
\[
x^{20} \equiv \pm 1 \mod {100}
\]
Multiplying these, we have
\[
x^{100} \equiv \pm 1 \mod {100}
\]
\[
x^{100 + y} \equiv \pm x^{y} \mod {100}
\]
In order  for x^100 to be -1 mod 100, x^100 must end in nine.  3^100 = 9^50 = 81^25 ends in one.  5^100 ends in 5.  7^100 = 49^50 = (something that ends in 1)^25 ends in 1. Finallly, 9^100 ends in 1.Thus x^100 can't be -1 mod 100 and so it must be 1.  That proves it for all x not divisible by 2 or 5.
Suppose x = 5n for some n.  5^100 ends in 25.  If n is odd, x^y = 00, 25 or 75 mod 100 and x^(100+y) = 25*00, 25*25 or 25*75 =00, 25 or 75 mod 100.
We can similarly prove it for x=2n by noting that 2^100 ends in 76, and for p>1, 76 * 2^p ends in whatever 2^p ends in.[/hide]

[quote="nr1337"]
By Fermat's Little Theorem, we have
\[
x^{40} \equiv 1 \mod {100}
\]
[/quote]

I thought that Fermat's little theorem was that $A^{p-1} \equiv 1 \pmod {p}$ where $p$ is prime ...?

He means Euler's Theorem:

If $(a,n)=1$, then $a^{\phi(n)} \equiv 1 \mod{n}$ where $\phi(n)$ is Euler's Totient Function.

[quote="paladin8"][quote="xxreddevilzxx"]How do you find out $\phi{n}$? is it just the regular way, where it is the number of all the positive integers less than $n$ that is relatively prime to $n$?[/quote]

$\phi(n) = n\prod_{p|n}(1-\frac{1}{p})$ is the explicit formula.[/quote]

could you give me an example, maybe? I still dont understand... not quite.

oic. I thought the $p$ in the formula was merely a divisor of $n$, but not prime. That's where i got confused. Thanks.

Didn't you only prove it for $1 \leq x \leq 9$?

[quote="blahblahblah"]You can't take square roots with $\mod$'s.[/quote]

Can someone confirm this? This is important to the proof!


man how come the locked threads are always threads I've posted in

I SIMPLY DO NOT GET IT

which reminds me dude at what point exactly did you start copying me

[quote="Altheman"][quote="bpms"]Solving the problem isn't that hard, but does anyone have a closed form for the nth n-gon that has a perfect square number of diagonals?[/quote]

use pell equations[/quote]
But like what would a closed form be?
I'm looking for an explicit closed form.
Does one even exist?


probability
algebra
polynomial
algorithm
geometry
summer program
MathPath

Prove that if the remainder of (x^2+17) when divided by 12 is 6, then x is a prime number.

Does anyone know of any other fast and easy ways of determining if a number is prime?

Um, is this result true?  What if you try x = 25 say?

[quote="ComplexZeta"]Unfortunately, there isn't anyone who knows of a fast and easy way of determining whether a number is prime. Many of the greatest mathematicians have worked on this problem, but none of them made any real progress.[/quote]

Fermat's Little Theorem, when applied with multiple bases, gives a strong indication that a number is prime.  [But there are some pseudoprimes that get through.]

Another technique, called the Miller-Rabin test is about as computationally complex as Fermat's Little Theorem.  You can determine whether a number is prime to any finite degree of probability that you need.  (E.g., is one false prime in a googol sufficient?)

But I think you are right that a fast *proof* of the general case is elusive.

Recently, some researchers in India found a (deterministic) polynomial-time algorithm to determine whether a number is prime.  Here is a link:
[url]http://www.cse.iitk.ac.in/news/primality.html[/url]

I don't know how fast their algorithm is in practice, but it solved a major theoretical problem.

[quote="tlhoo"]Prove that if the remainder of (x^2+17) when divided by 12 is 6, then x is a prime number.
[/quote]

I think you've got this problem around the wrong way. If x is prime, then the remainder of (x^2+17) when divided by 12 is 6, (apart from in the trivial cases of x=2, 3). But definitely not around the other way - if that was true, then you'd be famous :)

Ah, yes, that was the starting point for their algorithm.  It turned to be a fairly simple algorithm in spite of the many years the problem was open.

Ah yes. Whatever happened with AKS? It really hasn't been implemented into any "commercial" prime program, as far as I know. ECPP is of more practical value than it is, or so it seems.

if p>3 is prime, then p^2+17 is divisible by 6

proof:

[hide]

all the integers can be represented by:

6k

6k+1

6k+2

6k+3

6k+4

6k+5

Obviously 6k, 6k+2, 6k+3, 6k+4 is divisible by at least one of 2,3

by our original hypothesis, 6k, 6k+2, 6k+3, 6k+4 cannot be a prime that is greater than 3.  Hence all the primes greater than 3 are in the form of 6k+1 or 6k+5

we make this substitution

(6k+1)^2+17=36k^2+12k+18=6(6k^2+2k+3)

(6k+5)^2+17=36k^2+60k+42=6(6k^2+10k+7)

Obviously both numbers are divisible by 6.

[/hide]

Maybe I'm just really confused, but why is the Manhattan Mathematical Olympiad at Kansas State University?

[quote="rcv"]
My daughter attended the Grade 6 2003 MMO last year.  Check out Grade 9-12, Problem 4, from the same year:  [url=http://www.math.ksu.edu/~soibel/olympiad.html]Professor Jan Soibelman's Manhattan Mathematical Olympiad[/url].[/quote]

Okay, that explains a lot.

*feels silly*

[quote="JBL"][url=http://www.mapquest.com/maps/map.adp?country=US&address=&city=Manhattan&state=KS]Manhattan, Kansas.[/url]

[url=http://www.ci.manhattan.ks.us/]Manhattan, Kansas.[/url] <-- this one may not work[/quote]

Several questions have been raised concerning practical efficiency, and impact on cryptography.  I think these questions (and many others) are well-answered at the following link.  [Look especially at Q8, Q9, and Q13.]

[url]http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.html[/url]

The proof that "PRIMES is in P" had eluded some of the best mathematicians for years.  In retrospect, the proof is so basic, it was within reach of the very best high school students.

It makes me wonder what *other* unsolved problems are within the reach of high school students.

For those who care, Manhattan, Kansas is "The Little Apple", and is home to Kansas State University.

[quote="rcv"][quote="Ravi B"]Recently, some researchers in India found a (deterministic) polynomial-time algorithm to determine whether a number is prime.[/quote]

Yes, this was a major theoretical step!

My daughter attended the Grade 6 2003 MMO last year.  Check out Grade 9-12, Problem 4, from the same year:  [url=http://www.math.ksu.edu/~soibel/olympiad.html]Professor Yan Soibelman's Manhattan Mathematical Olympiad[/url].[/quote]

That doesn't make sense. Colorado is your location and your daughter's location is in Kansas.

Your comment doesn't make sense. Ever heard of separated parents?

[quote="churchilljrhigh"]That doesn't make sense. Colorado is your location and your daughter's location is in Kansas.[/quote]

Already answered via PM, but since speculation has ensued...

Don't you folks ever attend math competitions outside your immediate area?

AoPS has 1300 registered users.  Perhaps some of them won't qualify for USAMO.

Where else can a 5th or 6th grader attend an Olympiad-style competition with questions at their level?  The Manhattan Mathematical Olympiad requires no prequalification and no preregistration.  The founder, Yan Soibelman, has been a lecturer at Mathcamp.  Last year, they drew contestants from as far away as Colorado and Alabama.  MMO was held last weekend.

Aside from Mathpath (\[$) or Mathcamp (\]$), where else can a student hear a lecture on a current problem in mathematics and meet authors and judges of Olympiad-style competitions?  The Colorado Mathematical Olympiad requires no prequalification, but preregistration is requested.  It is the largest "Olympiad" held in the United States.  The founder, Alexander Soifer has been a USAMO judge.  The competition is this coming Friday.  They expect contestants from as far away as the Phillipines.

Expand your horizons.  Attend a regional math competition.


Greetings all,

I have been studying this problem for a while and I cannot figure out a way to solve it. Any help would be appreciated.

Show that: For every n in Z+, n can be written as the power of 2 and an odd integer.

Thank you all,

Try splitting it into two cases, $n$ odd and $n$ even. $n$ even -- there's only one thing you can possibly do. Why? What are the powers of 2? For $n$ odd, there are many possible ways you can write the sum. Why? Again, what are the powers of 2? What happens if you subtract a power of 2 from $n$?

Edit: The above is assuming you meant adding. If you meant factoring $n$ into a power of 2 and an odd number, try induction. Suppose $n$ is even. Then $n = 2 * m$. Repeat. Why can't it continue forever?

Also, note that this factorization is unique on $\mathbb{N}$.
Indeed, let's assume that :
$n = 2^{m}(2r+1) = 2^{m'}(2r'+1)$.
If $m \geq m'$, $2^{m-m'}(2r+1) = 2r'+1$.
Since the RHS is odd, so has to be the LHS. Hence : $m=m'$.
So now it remains : $2r+1 = 2r'+1$. Hence $r=r'$.

Besides, it shows a bijection from $\mathbb{N}$ to $\mathbb{N}_{0}^{2}$.


[quote="JRav"]You could use Cramer's Rule.  I don't know how you feel about evaluating 2010 determinants of 2009x2009 matrices... :D I'm joking of course.[/quote]

The equations aren't linear.

We can ask the general question of when there exists a solution to the system $ f(x_i) \equal{} x_{i\plus{}1}$ (for $ x_1, x_2, ... x_n$ where the indices are interpreted cyclically) such that $ \exists i \neq j : x_i \neq x_j$.  There is a great thread where a partial answer was given to this more general question, but I can't find it.


http://www.freerice.com/index.php

My mom is doing this- I just set up a macro

here it is:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171712[/url]

dang it, you beat me
oh well, ill post it again  :D

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171712

its still active, as a matter of fact. hint: use search

Search was down when I made this thread >>


Does anyone have the answers to the AMC 12 from 1983, or could someone point me in the direction where I'd find the answers? thanks!

[url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]
tests and solutions from 1980-2001

That is a cool website. I never knew about that one. :D


vector
rotation
geometry
geometric transformation
reflection
geometry proposed

Find the locus of pt M inside an equilateral triangle ABC s.t. MA^2=MB^2+MC^2.

MB^2+MC^2-MA^2 is a moment of p. M relatively mass-points (A,-1), (B,1), (C,1) ==> due to Steiner's theorem
MB^2+MC^2-MA^2=(1+1-1)*MZ^2+(ZB^2+ZC^2-ZA^2) where Z is a centre of mass of (A,-1), (B,1), (C,1), it will be symmetrical point to p.A relatively BC ==> MZ=AB ==> locus is arc of circle with R=AB which connect B and C.

hey myth
wot is steiner's theorem (i 've heard abt it in sumthinng related to physics :D ,moment of inertia)

you can look at this site:

http://mathworld.wolfram.com/SteinersTheorem.html

cheers!

This is the Steiner thm I know (I guess Myth used a generalization): let G be the centroid of ABC and M a point. Then MA^2+MB^2+MC^2=-3MG^2+GA^2+GB^2+GC^2. So I guess the generalization is obtained by replacing 1 (the coefficients of MA^2 etc.) with the masses attributed to pts A, B, C.

Let (A1,m1),...,(An,mn) are mass-points (i don't know exact english term).

If M is a point then value IM=\sum mi*MAi2 is called moment of inertia.

Let Z be a center of mass of (A1,m1),...,(An,mn) (it means vector \sum mi*ZAi equals zero-vector).

Steiner's Theorem
IM=IZ+(m1+...+mn)ZM2

Rotate $\triangle ACM$ $60^\circ$ about $A$, so that $C$ goes to $B$; $M$ will go to $M'$ so that $\triangle AMM'$ is equilateral and $\triangle MM'B$ is made by the segments $|MB|, |M'B|=|CM|, |MM'|=|MA|$, hence it is right-angled at $B$.
Consequently, $\angle ABM+\angle ACM=\angle ABM+\angle ABM'=\angle MBM'=90^\circ$, and from $\angle ABC+\angle ACB=120^\circ$ it yields $\angle MCB+\angle MBC=30^\circ$, so $M$ belongs to the circle $(A', A'B)$, $A'$ being the reflection of $A$ in $BC$.

Best regards,
sunken rock


Let $f(n)$ denote the sum of the digits of $n$. Let $N = 4444^{4444}$. Find $f(f(f(N)))$.

Classic problem!

Posted earlier here http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=4444&t=31409


inequalities function logarithms calculus derivative

Let $a,$ $b$ and $c$ are positive numbers such that $abc=1.$ Prove that
$a^{2}+b^{2}+c^{2}+10(ab+ac+bc)\geq11(a+b+c).$

Here is a non detaited solution

[hide]Consider the function $f(x)=x^{2}+\frac{10}{x}-11x+19\ln x$
and notice that $f(1)<f(\frac{5}{2})$[/hide]

I did first as yours, because $f'(x)=\frac{(x-1)(x-2)(2x-5)}{x^{2}}$ for $x>0$.

Can we have the solution without using differentiation?

Wow, silouan! Very nice proof. :lol:
I proved it by SOS method. :D

I know that some modern techniques are invented for inequalities(something like sos).
But i dont know them :(
From where can i learn these methods?

Moon

[quote="arqady"][quote="kunny"]Oh,really. Didn't you make the problem behind silouan's solution? I was assured that[/quote]
No, kunny. I have missed it. :(

Moonmathpi496, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127
To explain on an example of this problem?[/quote]

Sounds strange! I know the tool for solving inequality such as $A.M.-G.M.-H.M. , Cauchy-Schwarz$ rather than that's all what I can use.
Muirhead, S.O.S. is beyond my ability. As the final method, I will use differentiation. Anyway proving inequality is enjoyable! Oh,so I add to the list, Jensen.
I wonder, Japanese IMO contestant use Muirhead, S.O.S. Anyway you are an expert of inequality, which everyone recognize.

arquady :

We know that a^2 + b^2 + c^2 >= ab + ac + bc, then a^2 + b^2 + c^2 + 10(ab + ac + bc) >= 11(ab + ac + bc)

We have to show now that 11(ab + ac + bc) >= 11(a + b + c) <=> ab + bc + ac >= a + b + c ...

you are wrong zweig,take $a=100,b=\frac{1}{10}.c=\frac{1}{10}$ then it becomes:
$20+\frac{1}{100}\geq 100+\frac{1}{5}$ contradiction...

But, if X >= Y and Y >= Z, then X >= Z, no ?  :huh:

if you want to know whats wrong with your solution:
the step that you used the inequality $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ in this step you made the $RHS$ too weak,so the inequality became wrong...

Excuse me, I odn't understand what you mean ... But a^2 + b^2 + c^2 >= ab + bc + ac is true ....

If $ab+bc+ca\ge a+b+c$ then it would imply the inequality given in the problem. However, because $ab+bc+ca\ge a+b+c$ is not true (see BaBaK Ghalebi's counter example), you have to prove the problem's inequality a different way. If P implies Q, it doesn't necessarily mean Q implies P. While this inequality would imply the problem's inequality, the problem's inequality does not imply the inequality $ab+bc+ca\ge a+b+c$. Though $a^{2}+b^{2}+c^{2}>ab+bc+ca$ is true, it is too weak.

let me give you an example:
assume that we want to prove that $10>7$ for this one we know that $10>6$ and $6>4$ so we have to prove that $4>7$ which is wrong...
your solution is like the solution above which is wrong...

It's also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197.
I have forgotten about it! I am sorry.


superior algebra
superior algebra unsolved

I have a quite dumb question. I understand this theory poorly.

I did an exercise to show that every finite extension field $ E$ of $ \mathbb{R}$ is either $ \mathbb{R}$ itself or is isomorphic to $ \mathbb{C}$.

This is what I did:
[hide="Solution"]
Let $ \overline{E}$ be an algebraic closure of $ E$. Now, since $ E$ is finite, it's in particular algebraic over $ \mathbb{R}$, and since $ \overline{E}$ is also algebraic over $ E$ it follows that $ \overline{E}$ is algebraic over $ \mathbb{R}$ (so indeed, $ \overline{E}$ is an algebraic closure of $ \mathbb{R}$). Choose $ \alpha \in \overline{E}$ with $ \alpha^2 \plus{} 1 \equal{} 0$. Clearly $ \text{irr}(\alpha, \mathbb{R}, x) \equal{} x^2 \plus{} 1$, so  $ \mathbb{R}(\alpha) \equal{} \{a \plus{} b\alpha \, : \, a,b \in \mathbb{R}\} \cong \mathbb{C}$. So $ \mathbb{R}(\alpha)$  is algebraically closed, and hence has no strict algebraic extension, and hence $ \mathbb{R}(\alpha) \equal{} \overline{E}$. Since $ 2 \equal{} [\overline{E} : \mathbb{R}] \equal{} [\overline{E} : E] [E : \mathbb{R}]$ it follows that either $ E \equal{} \overline{E} \cong \mathbb{C}$ or $ E \equal{} \mathbb{R}$
[/hide]

I have two questions:
1) Is this a good proof? Can one avoid referring to the existence of an algebraic closure, which in my opinion has a very esoteric proof that I cannot check if it's correct.
2) This is what I really don't understand: what goes wrong when $ E$ is not a finite extension? Seems counter-intuitive that there are other extensions

1. it's ok. the proof for the existence of a algebraic closure is very clear using infinite tensor products. you just repeat the method of kronecker infinitely many times.
2. consider pure transcendental extensions.

How may I construct a transcendental extension?

When $ x$ is trancendental over $ \mathbb{R}$, we can have other extensions e.g. $ \mathbb{R}(x)$.

@-oo-: can you elaborate? what are infinite tensor products?

I mean the coproduct in the category of algebras, see eisenbud or mac lane.

And how do you use this for the existence of algebraic closure?

It seems like you can avoid talking about algebraic closures by using the Primitive Element Theorem and the Fundamental Theorem of Algebra to show that the extension has degree at most 2.


Congrats to Alex and Cynthia for making HM in USAMO!

They're only grade 11...2 years of IMO for them ^^

IMO would be a piece of cake for them. I bet they're dreaming of getting gold medals already :D

Lol, I'm so proud of you two! Alex and Cynthia, Congrats!!!
You two so deserve it!!!! Enjoy your time in Vietnam!!!!

so they're automatically on the canadian IMO team now???

[color=red][edited by mod] [/color]..., we are all eagerly waiting to know who is the 6 new faces of IMO team this year!!! And congrats to all who made it!!!! You guys will make Canadian proud!!!

Every year, when the Canadian IMO team members are informed of their eligibility at the beginning of May, they are told not to disclose this information until the CMS does so officially near the end of May.

In other words, please do not discuss who is on the team until the official announcment has been made by the CMS.


MY 100th post!!! If I have 100 posts and I post 2 times a day, except on February 29, when I post 3. when did I post my 100th post if I started posting on December 31, 2003?

[hide] :D  :D  :D  :D  :D  :D  :D  :D[/hide]
[i]Click to see to many smileys - A+MATH[/i]

I edited down your number of emoticons to about a hundredth of what there were.  Please don't fill posts with emoticons.  Just loading them will stall browsers on slower computers.  Thanks.

Well.. smilies or not - assuming you posted 2 on december 31st...

[hide]January has 31 days, and you already have 2 posts, so $2 * 31 +2=64$ and you need 100 more and $100-64=36$ - so you need 36 more posts. $36/2=18$ thus you would have 100 posts on February 18th.. not even up to february 29th. Also, a day after Michael Jordan's bday  :lol: [/hide]

[hide]
How many posts does he make in 2003? Only 2.
In january, he makes $31 \cdot 2 = 62$ posts. So far, he has $64$ posts, so he needs $36$ more. IT will take him $18$ days to make 36 posts, so he will post his $100^{\text {th}}$ post on February 18th.
[/hide] :coolspeak:


algebra
functional equation
algebra proposed

Find all solutions to the functional equation: \[ 2f(x)\plus{}f\left(1\minus{}x\right)\equal{}x^2\qquad \forall x\in\mathbb{R}\]

[quote="April"]Find all solutions to the functional equation:
\[ 2f(x) \plus{} f\left(1 \minus{} x\right) \equal{} x^2\qquad \forall x\in\mathbb{R}
\]
[/quote]

take instead of $ x\rightarrow 1 \minus{} x$

so you get:

$ 2f(1 \minus{} x) \plus{} f(x) \equal{} (1 \minus{} x)^2$

multiply the initial equation by $ 2$ an subtract one from another:

$ 3f(x) \equal{}x^2\plus{} 2x \minus{} 1$

So $ f(x) \equal{} \dfrac{x^2\plus{}2x \minus{} 1}3$

[quote="sandu2508"][quote="April"]Find all solutions to the functional equation:
\[ 2f(x) \plus{} f\left(1 \minus{} x\right) \equal{} x^2\qquad \forall x\in\mathbb{R}
\]
[/quote]

take instead of $ x\rightarrow 1 \minus{} x$

so you get:

$ 2f(1 \minus{} x) \plus{} f(x) \equal{} (1 \minus{} x)^2$

multiply the initial equation by $ 2$ an subtract one from another:

$ 3f(x) \equal{} 2x \minus{} 1$

So $ f(x) \equal{} \dfrac{2x \minus{} 1}3$[/quote]

The method was the correct but your answer is wrong. Please, check your answer.

The answer is $ f(x)\equal{} \dfrac{x^2\plus{}2x\minus{}1}3$

[quote="BlAcK_CaT"][quote="sandu2508"][quote="April"]Find all solutions to the functional equation:
\[ 2f(x) \plus{} f\left(1 \minus{} x\right) \equal{} x^2\qquad \forall x\in\mathbb{R}
\]
[/quote]

take instead of $ x\rightarrow 1 \minus{} x$

so you get:

$ 2f(1 \minus{} x) \plus{} f(x) \equal{} (1 \minus{} x)^2$

multiply the initial equation by $ 2$ an subtract one from another:

$ 3f(x) \equal{} 2x \minus{} 1$

So $ f(x) \equal{} \dfrac{2x \minus{} 1}3$[/quote]

The method was the correct but your answer is wrong. Please, check your answer.

The answer is $ f(x) \equal{} \dfrac{x^2 \plus{} 2x \minus{} 1}3$[/quote]

Yes, you're right. I forgot to multyply by 2 be right part of the equality. I edited it.


Hi, I need some tips about solving problems.

Math was one of my strongest subjects ever since I started school. (It is still even in High School). However, I find that I have the ability, but not the approach. What i mean is, for example, I am in the most advanced math class in my school, I can solve some problems, but I find that sometimes I'm off route. I would be trying to solve a problem in a specific approaches. Needless to say, I try another one if it doesn't work. In the end, I usually give up after trying a couple of methods. I open up the solution booklet and read the solution, and i feel like jumping off the building. It always turns out to be a simple problem if approached from another angle.

I know everyone tells me "IT COMES WITH PRACTICE" but I was just wondering if you guys would give me some tips on how to determine an approach. Like are there hints in a problem? or a certain approach for certain types of problems?

thx

A couple of methods? That isn't nearly enough. On one of this year's Putnam problems, I looked at more than five different ideas before finding one that worked, and the method that worked was one I had already rejected.

For contest problems, you just have to try something- and then something else if that doesn't work, and keep coming back until you find something that works. With experience, you learn more methods and get faster. It helps to have some sense of what might work; if you can reject a method in fifteen seconds, you have more time to try other ideas.
"Brute force" methods are also an essential part of the toolbox. These are the methods that will probably be ugly, but are almost guaranteed to work. There may be a nicer finesse solution, but all correct solutions get the credit.
If a problem does give away a hint on how to solve itself, that hint probably leads to a brute force method. A sum over permutations, caring about the number of fixed points? Group permutations by their sets of fixed points, use inclusion/exclusion. That was another hard Putnam problem, and that idea led to a routine solution.

There's no silver bullet here- I can get problems the first time most of the time, but I've had many years of practice, and I know an enormous library of tricks. If your first few tricks fail, look at something else and come back later with new ideas. Sooner or later, something will probably work. If you've spent lots of time and most of your tricks, then it's time to look at the solution.

Start from the most basic problems and try to gather the ways of solving this kind of problems.


Is there a formula for the sum of all factors of a number?

For example, I have the number 72
that is 3^2*2^3

I know there are 12 factors, but is there a way to caculate their sums?

Yes there is.
Say you have a number A.
Prime factorize it to $a^{n_{1}}_{1}\cdot a^{n_{1}}_{2}\cdots$
Then the formula is $(1+a_{1}+a^{2}_{1}+\cdots+a^{n_{1}}_{1})\cdot(1+a_{2}+a^{2}_{2}+\cdots+a^{n_{2}}_{2})\cdots$
For $72=2^{3}\cdot3^{2}$, it's
$(1+2+4+8)\cdot(1+3+9)$
which turns out to be 195.

I think that's it.

[quote="CircleSquared"]Prime factorize it to $a^{n_{1}}_{1}\cdot a^{n_{1}}_{2}\cdots$
[/quote]

The second term should be $a_{2}^{n_{2}}$.
Otherwise the formula is correct.  :)


function
calculus
derivative
calculus computations

I have to find the absolute max and min of the function

f(x) = -4x+3 on the interval (-2,4], (-2,4)

I took the derivative of the function, which would be f'(x) = -4 and then set that equal to 0.  Seeing as how -4!=0, does that mean that it has no absolute max or min on those intervals?  Thanks for your help.

Absolute max and min are found by an elimination process. At the first step all [i]interior[/i] points $x$ at which $f'(x)\ne 0$ are eliminated from consideration. All the second step you inspect the remaining points, which are: $x=4$ (for the interval $(-2,4]$) or nothing at all (for the interval $(-2,4)$).

It may well happen that there is no point within a given interval at which $f$ attains its absolute maximum (or minimum).

[quote]$f(x)=-4x+3$[/quote]
Graph of the function you wrote is a line...
(HINT: no need to derivate the function, just sketch it at the given intervals and see from picture)


What's the way in general to get rid of the outer radical in expressions like

$ \sqrt{2\plus{}\sqrt{2}}.\quad\sqrt{2\minus{}\sqrt{2}}$,
$ \sqrt{2\plus{}\sqrt{3}},\quad\sqrt{4\plus{}\sqrt{7}}$

Also are there cases when it's impossible?

By getting rid, I mean, say, $ \sqrt{2\plus{}\sqrt{3}}\equal{}(\sqrt{2}\plus{}\sqrt6)/2$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1487073#1487073]see here[/url]
for my particular derivation (took me a while to realize it, frankly... lucky it's just a practice; if on the actual test I'm gonna pull my hair out!!! :D  :P )

You can let
$ x\plus{}\sqrt{y} \equal{}a^2\plus{}2ab\plus{}b^2$

But I don't use it unless I can see the answer right away

$ \quad\sqrt{4\plus{}\sqrt{7}}$
= $ \frac{\sqrt{2}}{2}\plus{}\sqrt{\frac{7}{2}}$

Sometimes "squaring" it is the trick, but rarely:

[hide="Example for squaring"]
2009 MATHCOUNTS Nationals #28: Evaluate $ \sqrt {15 \plus{} 6\sqrt6} \plus{} \sqrt {15 \minus{} 6\sqrt6}$
[/hide]

As for your radicals, perhaps an example would help?

$ \sqrt {3 \plus{} 2\sqrt2}$

This will be able to be expressed as $ a \plus{} b\sqrt2$ (can you see why?)

Thus, $ (a \plus{} b\sqrt2)^2 \equal{} 3 \plus{} 2\sqrt2$

$ a^2 \plus{} 2b^2 \plus{} 2ab\sqrt {2} \equal{} 3 \plus{} 2\sqrt {2}$

$ a^2 \plus{} 2b^2 \equal{} 3$ and $ 2ab \equal{} 2 \implies ab \equal{} 1$

From here, we use trial and error to deduce $ a \equal{} b \equal{} 1$, so $ \sqrt {3 \plus{} 2\sqrt2} \equal{} 1 \plus{} \sqrt2$

[quote="Mewto55555"]
This will be able to be expressed as $ a \plus{} b\sqrt2$ (can you see why?)[/quote]

No, unfortunately this is not always true. For example, $ \sqrt {2 \plus{} \sqrt 3} \equal{} \frac 12(\sqrt 6 \plus{} \sqrt 2)$.

Also,
[quote="2009 National MATHCOUNTS Competition"]
Evaluate $ \sqrt {15 \plus{} 6\sqrt 6} \plus{} \sqrt {15 \minus{} 6\sqrt 6}$.[/quote]
[b]THAT[/b] was the third last problem on the Nationals competition???? Even last year's #28 was like 10000x better...

What's the level of national mathcounts?
for that particular mathcounts problem it's not difficult to realize $ 6\sqrt6\equal{}2\cdot3\sqrt6$ and $ 15\equal{}6\plus{}9$...

I think that example is confusing. $ \sqrt{2\plus{}\sqrt{3}}\equal{}\frac{\sqrt{2}}{2}\cdot(1\plus{}\sqrt{3})$.

I feel like there is something going on here that can be stated clearly but I'm not sure how to prove it. First of all, we'd have to define what we are asking. What expressions are allowed?

In general, I'd say look at $ \sqrt{u\plus{}v\sqrt{d}}\equal{}\sqrt{c} (a\plus{}b\sqrt{d})$ where we are given $ d$, $ u$ and $ v$.

$ u\plus{}v\sqrt{d}\equal{}c(a^2\plus{}b^2d)\plus{}2abc\sqrt{d}$

Then $ u\equal{}c(a^2\plus{}b^2d)$ and $ v\equal{}c(2ab)$

Then $ uv\equal{}u(2ab)\equal{}(a^2\plus{}b^2d)v$ and search that for integer solutions...

Loosely speaking, what is going on here is that you take the $ \sqrt{3}$ part and the $ 1$ part and they are separable. The issue is that I'm not sure how to characterize the coefficients of $ 1$ and $ \sqrt{3}$. Usually they are integers, but sometimes they are an integer times $ \sqrt{c}$ where $ c$ is a rational number.


A traveller starts from a certain place on a certain day and travels 1km on the 1st day and on subsequent days, he travels 2km more than the previous day. After 3 days, a second traveller sets out from the same place and on his first day he travels 12km and on subsequent days he travels 1km more than the previous day. $On how many days$ would the 2nd traveller be ahead of the 1st?

I'm not sure abot the meaning of "ahead" and also I don't have a dictionary now...

[hide]The first distance is $\frac{(2 + 2(N - 1))N}{2}$, while the second distance is $\frac{(20 + N)(N - 3)}{2}$.
$(20 + N)(N - 3) \geq 2N^2$
$N^2 - 17N + 60 \leq 0$. Hence $5 \leq N \leq 12$. So the 2nd is ahead for $6$ days.[/hide]

For Andreas, "ahead" in this problem means to have travelled more distance.  Also, there are 8 solutions, not 6 for $5\leq N\leq12$.
[hide="My Answer"]When traveller 2 starts, it is the 4th day.  Traveller 1 has already travelled 5 miles.  Therefore, 2 will be ahead.  Then, find the point where they will have travelled the same distance.
$1+2n-2=12+n-3$
$n=10$
Therefore, after 10 days they will have gone the same distance.  From this, you get the second traveller will be ahead from day 4 to day 9 (he is not actually ahead after the 10th day).  There are $\boxed{6}$ days he is ahead.[/hide]


Let S be the sum of all real coeffients of the expansion of $ (1\plus{}ix)^{2009}$ What is $ log_2{S}?$

[quote="bigman"]Let S be the sum of all real coeffients of the expansion of $ (1 \plus{} ix)^{2009}$ What is $ log_2{S}?$[/quote]

$ S\equal{}R((1\plus{}i)^{2009})\equal{}R((\sqrt{2}e^{i\frac{\pi}{4}})^{2009})\equal{}R(2^{\frac{2009}{2}}e^{i\frac{\pi}{4}})\equal{}2^{\frac{2009}{2}} \times \frac{1}{\sqrt{2}}\equal{}2^{1004}$

So $ log_2{S}\equal{}1004$

stephenchang, where did you get the 1/rt(2)?


Let ABC be a triangle inscribed in a circle of radius R

 and a, b, c  be respectly sides of the triangle ABC.

  Prove that:

      1/ab + 1/bc + 1/ca >= 1/R^2.

This one is straight forward...
LHS is ( :Sigma: a)/( :prod: a) = ( :Sigma: sin A)/4R2 ( :prod: sin A)  :ge: 3/4R2 ( :prod: sin A)2/3 using AM-GM inequality as all SinA , SinB, SinC are +ve and  :prod: sin A reaches its maximum when A=B=C=60 :^o: (this can as well be derived from AM-GM ineq). hence giving what is required.


UKRAINIAN TEAM
Shyshatskyy Yuriy (bronze at IMO 2007)
Semikina Yuliya
Mishchenko Pavlo (gold at IMO 2007)
Sobolyev Dmytro
Serdyuk Nazar
Senin Vitaliy.
Leader - Sergiy Torba
Deputy Leader - Nataliia Prokopenko.

Also, what date will the olympiad (solving problems, 1st and 2nd days) take place on?

[quote="BlackMax"]Also, what date will the olympiad (solving problems, 1st and 2nd days) take place on?[/quote]

Wednesday, July 16th and Thursday, July 17th, 2008: http://www.imo-2008.es/programme.html


There are so many smart people in this forum and I was wondering if anyone had taken any formal IQ tests.

For those who have taken those tests, how old were you when you took it, and what is your score if you don't mind me asking?And do you think IQ scores should play a role in young children's education?
I know someone who's very smart and he was able to count from z to a very naturally as a young child right after he'd learned to count from a to z (he's not a native speaker of English and I'm uncertain whether it makes any difference). Later he turned out to be very gifted in Maths and scored a high score(top 0.003%) in an IQ test. Do you also know people who are intelligent and can do difficult things naturally and easily beside traditional problem solving? I can't count from z to a.

[quote="Bugi"]Well the men who test you usually don't say your results to you(at least where I am). Also, IQ doesn't mean anything. There are many 'average' people who do well in life and many 'geniuses' who never do much in life. So, don't mind very much on your IQ.[/quote]
Yea, I've realised that.
I actually won't mind scoring low in an IQ test.
But I figure it will still be kinda nice to be able to solve puzzles and have a better spatial ability and stuff.
And what I'm more interested in is the correlation between IQ scores and other stuff like counting from z to a.
(I know it's too hasty to say there's any correlation at all between the two :P  )

Well, I took a formal IQ test when I was 5 years old, and my IQ is 129, which I think is high average.

I did a test for a job placement service this summer. I scored over 130, which is high average or low genius I guess, I study math at university and it didn't stop them from sending me to work in a warehouse, so take it for what it's worth, not much. The IQ test is an old test that doesn't say much about anything. It will tell an analphabet from a genius for sure but not much in between.

hmm just out of the blues

when i took it when i was 6 years old, i got 152.

now i took it again few months ago (i am now 14) i got 183.

I feel bad and happy (happy that it increased, and bad that it only increased 4 per year)

[quote="chrischris"][color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding yours.[/color]
I've heard it's illegal to require people to take IQ tests before hiring them. Why did they make you do the test?[/quote]

it's a job placement company, and I had no idea it was illegal

Quoting the green line is particularly ironic.

I don't think IQ measures anything particularly useful. A one-dimensional measure of "intelligence" doesn't say very much, especially at the extremes the tests aren't designed for.

I think IQ isnt directly linked with intellegence, as intellegence is something that cannot be measured by a simple number. It is more of a matter of opinion.

I took the IQ test when I was (I think) 7 and scored 145 (I believe).
Yeah I'm a low genius too.  :blush:

big brain @above

this thread was made before I was born

I didn't take the IQ test

Your IQ can change.
Scientists say that you have to retake the test every 10 months if you want a reliable score.
It doesn't work well if you have a learning divergency or don't regularly take timed tests.
Plus people are smart in many different ways, but an IQ test doesn't measure all of them.

/request lock.
this has turned into people flexing their IQ test scores.

hmm
considering mensa accpeted 2 year olds idk if they are very accurate

[quote name="Mathboyisepic123" url="/community/p25514592"]
I suppose work ethic might be the most important factor for success.
[/quote]

Or one of them. Opportunity and startup capital are also very important, as are leadership skills, integrity, confidence, and the support of others (including a spouse).

These were pulled from a list of responses to his question by CEOs/founders of the Fortune 500. Seems like they have some clout here lol.

Richard Feynman said according to the only IQ test he took his IQ was 125, yet he's one of the most respected physicists in the world. I don't think one needs an IQ of 160+ to be at the top of his/her field.

I got 145 on one IQ test and 92 on another...

[quote=fuzimiao2013]I took the 16-17 test and got 135.  I'm 14, so if this is even somewhat accurate, then my actual IQ (measured by mental age/actual age) is about 16.5 * 1.35 / 14 * 100 = 159.

Thanks, but no thanks, Mensa.

You tried.[/quote]
[quote=mahaler]then my actual iq would be: 16.5*1.42/13.8*100 = 170...yeah there's no way that I'm smarter than albert einstein lol[/quote]
No. You are using the ratio IQ formula on Correlation IQ. Please learn the difference.

[quote=s12d34][quote=peelybonehead]Blaming his or her IQ and genes for not becoming successful is an excuse people make a lot. Personally, I have never taken an IQ test before in my life but I believe that hard work is the true key to success.[/quote]

Yeah, pretty much. There's no real point in taking an IQ test, it's just another psychological metric like MBTI. It might be fun to do it once, though. Also, fun fact: this thread was originally posted 5 days after my birthday, August 25, 2008.[/quote]

Well, MBTI is more of spectrum criteria than a metric. Although both of them are theories that have not been proven(solid proof) yet.

[quote=solasky][quote=Lionking212]I took it when I was 9 and i got 178.[/quote]

[quote=larkwhisker]I have

Some freeiqtest. net thing or so

173[/quote]

Almost every single online IQ test is absolute BS (are acronyms like these allowed on AoPS?) Just saying, going by standard deviation of IQ, approximately only 35 people in the US have an iq of 178 or above. If you'd like to make the claim that you're one of the 35, go ahead :) [hide=.]And IQ itself is severely outdated and never used seriously. No single metric can accurately measure anything about a person. For example, I simply solved a math equation for every single spacial recognition question on an IQ test and it gave me an IQ score of 210. Does this mean I have an IQ of 210? That's ridiculous.[/hide][/quote]

what iq test

I've got an IQ of 110. IQ does not limit your intelligence at all. I am the literal highest person in academics in school. The only thing that actually matters is whether you're willing to put in the hard work.

[quote=QuickSort1][quote=nsd08]i got 148 on the WISC-V as a kid
i have never ever made the AIME or done anything remarkable
i've always had very bad organization which led to a 76.5 on the AMC 10 despite lots of prep
when i improved my organization i started doing well in school
i'll graduate college with a BA in math specializing in pure math when I'm 20, and then head for my PhD or something[/quote]

you make yourself sound like you're an adult[/quote]
I'm 14. Definitely not an adult, but still :)
in all honesty that's what i probably will end up doing as an adult if i can't find anything else to do - my high school allows me to take courses at an actual t75 college and then get automatically accepted into that college. because the courses I've taken at that college will count as credit to the degree, so I can get a degree in 2 years from that college if i take the pure math track.

Just wondering, where can you take an official IQ test?


geometry
perimeter
search
inequalities
geometry solved

In a unit square one hundred segments are drawn from the centre to the sides dividing the square into one hundred parts (triangles and possibly quadrilaterals). If all the parts have equal perimeter p show that 1.4<p<1.5.

... i'm getting a very small lower bound..no where near 1.4, 1.5

can anyone help me out.. here's my approach ...
sum SQRT(s(s-a)(s-b)(s-c))=1 (a,b,c are sides of all triangles and s=2P)
also s(s-a)(s-b)(s-c)<[(s+s-a+s-b+s-c)/4]^4 (=s/2)^4 (AM/GM)
am not getting any where with this....

The upper bound is easy to obtain. Let $OA_i$ be the segments starting from the center $O$. The the condition is equivalent to $\sum_{i=1}^{100}OA_i<73$ (this is obtained by adding the perimeters of all the polygons, subtracting $4$, which is the sum of the segments on the sides of the square, and then dividing by $2$, because each $OA_i$ is added twice). However, $OA_i\le \frac{\sqrt 2}2$, so $\sum_{i=1}^{100}OA_i<50\sqrt 2<71<73$.

I'll think about the lower bound. The same procedure shows that we should prove $\sum_{i=1}^{100}OA_i>68$.

thanks for the reply grobber. any thoughts on the lower bound?

i thought i would dig this one up from the archives...

I think it has been posted somewhere else and solved by Myth. Try looking for it (use teh search button). I can't remember exactly where I saw it, but you could try the Inequalities section first.

Actually, grobber, both bounds are easy to obtain, but I won't say anything more because both Arne and I didn't see it the first time we saw the question either :blush: .

http://www.mathlinks.ro/Forum/viewtopic.php?p=121976


function
integration
calculus
inequalities
analytic geometry
geometry

This is a good problem.

$k$  is the ground. This does not move

$s$ is a bar. This bar have mass $m_{3}$ and it is homogeneous

$M_{1}$ and $M_{2}$  are 2  blocks with mass
$m_{1}$ and $m_{2}$

this system is in Equilibrium

Determine $m_{1}$ in function of the others constantes given

[hide="tips"]
The center of mass of $s$ is the half of the bar and it can be in 3 different positions, analise these 3 possitions
[/hide]

I soon realized how much of a hand cramp this problem would be so i'm just gonna setup the problem for if the center of mass is between points a and b.

first i determined a point p to which the sum of torques on either side of p will be equal, and p is between points b and c.  if these two contidions are met then the system will remain in static equillibrium, i say this because p can be anywhere between b and c, so there isnt a single answer, the answer will be a range.

here's my first equation: (i dont know how to make a line over two letters; each set of uppercase letters is a line)
$m_{1}* AP+m_{3}* (AD / 2-DP)+\int_{0}^{PB}\frac{m_{1}+m_{2}+m_{3}}{k}g*x dx = m_{2}* PD+\int_{0}^{PC}\frac{m_{1}+m_{2}+m_{3}}{k}g*x dx$

the integrals represent my sumations of all the torques due to the normal forces between the points b and c.

this can be made without calculus

using inequalities 12 or more in total

but, i dont understant what is this $k$ on your solution

in my problem $k$ is the ground,   $k$ is not the mass, length or nothing, is just the name of the ground

it does not need to appear in solution

can you explain what is your $k$ ?

and thaks for solution

another thing

in this system, suppose this is in equilibrium

i can put a mass $m_{6}$  in on side of this that it will stay in equilibrium

have a minimum mass in each side for equilibrium and a maximum mass in each side

in your equation how can i solve this?

the minimu/maximum mass that it will stay in equilibrium

I was rushing at the time, i meant to write the lenght of k, BC, m1 plus m2 plus m3 divided by BC is force per area.  The integral is linear so algebra can be used to solve it, but i wanted to show that i was summing the torgues due to every normal force on top of k.  ushually the normal force can be replaced with an equivelent single force, but here i have to take each normal force by itself.

i'll try to post a minimum and maximum mass for each of the three cases on monday.

hmm........ k is BC

i  did not realise you type pressure

hum

i understand what you say now

i just dont know how  to find the maximum/and minimum mass for each case using what you say

i will try later use a cartesian coordinates graph

and find the torques

force per area

you said this

is force per Length

because k is the length  BC

just correcting

The equation i have above is overly complicated.  what i was doing was i was summing the torques about a variant point p, and solve for m1 then set p to either the right or left side to get my min and max values, unnecessarily complicated.

a much better way which i will put here is i'll sum the torques about point b and solve for m1 then sum the torques about c and solve to m1 again.  Here i'm assuming AB > AD/2 (AD/2 being the length from one end to the center of mass)

Sum of Torques about point B Counter Clock Wise:
$\sum \tau_{B ccw}= 0 = m_{1}gAB+m_{3}g(AB-\frac{AD}{2})+\frac{BC}{2}F_{n}-m_{2}gBD$
$F_{n}= [m_{2}gBD-m_{1}gAB-m_{3}g(AB-\frac{AD}{2})]\frac{2}{BC}$

now for the system to remian in static equillibrium not only do the torques have to balance, as do the x and y forces, which may look strange but its true:
$\sum F_{y}= 0 = F_{n}-m_{1}g-m_{2}g-m_{3}g$
$F_{n}= m_{1}g+m_{2}g+m_{3}g$
$m_{1}g+m_{2}g+m_{3}g = [m_{2}gBD-m_{1}gAB-m_{3}g(AB-\frac{AD}{2})]\frac{2}{BC}$
assuming i didnt screw up solving for m1 the maximum value for m1 given the center of mass of s is between points A and B is:
$m_{1}= \frac{BC+2AB}{BC}[[m_{2}BD-m_{3}(AB-\frac{AD}{2})]\frac{2}{BC}-m_{2}-m_{3}]$

the five other cases are all the same idea, so it would be a lot of writing and no creativity so i'll just descibe how i would do it.    To find the minimum value for m1 where it AD/2 is less than AB then you would use the same first equation except the arm lenghts of the torques would have to be adjusted to about point c and the sign on the torque of the normal force would be flipped.
if AD/2 is less than CD then its the same idea just the center of mass torque would be going the other direction and different arm lenghts for the torques.   basically you just have to setup the first equation differently each time then repeat.

See what i thougth

In this problem we need to make some equations

$P$ is the point you described. But $P$ is not a point!!! $P$ can be all points in between $BC$
SO, the minimum torque of the weigth on $A$ is when $P = B$, and the maximum torque of the weigth $A$ is when $P = C$
this ideia is intuitive. because the length

Lets the equations we have. (This is not a system!! its just some possible equations to determine the minimum and maximum weigth)
First for the center of mass of the bar between $AB$

${\text{AB}g m_{1}+\frac{1}{2}\text{AD}g m_{3}=m_{2}g \text{BD}}$
${\text{AB}g m_{1}+\frac{1}{2}\text{AD}g m_{3}=m_{2}g \text{CD}}$

${\text{AC}g m_{1}+\frac{1}{2}\text{AD}g m_{3}=m_{2}g \text{BD}}$
${\text{AC}g m_{1}+\frac{1}{2}\text{AD}g m_{3}=m_{2}g \text{CD}}$

now for the center of mass of the bar between $CD$

${m_{1}g \text{AB}=\text{BD}g m_{2}+\frac{1}{2}\text{AD}g m_{3}}$
${m_{1}g \text{AB}=\text{CD}g m_{2}+\frac{1}{2}\text{AD}g m_{3}}$

${m_{1}g \text{AC}=\text{BD}g m_{2}+\frac{1}{2}\text{AD}g m_{3}}$
${m_{1}g \text{AC}=\text{CD}g m_{2}+\frac{1}{2}\text{AD}g m_{3}}$

now for the center of mass of the bar between $BC$

${m_{1}g \text{AB}=m_{2}g \text{BD}}$
${m_{1}g \text{AB}=m_{2}g \text{CD}}$

${m_{1}g \text{AC}=m_{2}g \text{BD}}$
${m_{1}g \text{AC}=m_{2}g \text{CD}}$

This is why i said we have $12$ equations

With a analisis we can determine the minimum and the maximum in each case.
In each case we can elimine $2$ equations, so we have $6$ equations

But in each case i need to solve for $m_{1}$ or $m_{2}$ to analise. If is a minimum or maximum.
at the last we put $Min(m_{1}) \leq m_{1}\leq Max(m_{1})$
analogous for $\(m_{2}\)$

Because of this i want a easy solution!!!!
We can simplfy the equations i made to $6$ equations, but i need to solve to $\(m_{1}\)$ and $\(m_{2}\)$. So i have to analise $12$ equations. better than $24$.
but... its too hand-writing... does not make sense this. Need to have a easy way!!!

In your last solution, what is $\frac{BC}{2}F_{n}$ ?

analising what you put, is the normal,

but why the normal generate a torque!?!?

i never see this,

i think this is incorrect, or im wrong?!

thanks for helping

EDIT:  i made a mistake and put only $\frac{AD}{2}$

the rigth is $\frac{AD}{2}-BD$ or similaries...

but i dont want to edit, because you understand me like that, sorry :(

i fogert

in each 3 possibilities for the center of mass of the bar we have 8 equations

not 4

beucase  AD/2-BD or AD/2-CD

so we get more 4 equations and we cannot alimine nothing

we have 24 equations!!!

in the end we have 48 equations, because we need to solve for the masses

$F_{n}$ indicates the normal force, wikipedia briefly describes it here:
[url]http://en.wikipedia.org/wiki/Normal_force[/url]

[img]http://www.geocities.com/rcory44/fig1.jpg[/img]

Just say you have a bar with a mass on the end of it, m1.  And the mass is resting on a table.  Also the end of the bar on the left side is fixed, and the distance from the point where the bar is fixed to the center of mass of m1 is d.  also the bar has a mass of m2 and lenth of l.  and finally the entire system is at rest.  Just say you were to sum the torques on the bar about the point where the bar is fixed, what would you do?

would you write:
$\sum \tau = m_{2}g \frac{l}{2}+m_{1}gd$
Now the system is in static equillibrium and when a system is in static equillibrium the torques about any point are zero therefore:
$0=m_{2}g \frac{l}{2}+m_{1}gd$
this may look kinda funny, and it should b/c there's no such thing as a negative mass, g 9.8m/s^2 and l and d are always positive, so the expression cannot equal zero.

that means we are missing a force, the normal force.  the normal force as wikipedia says acts perpendicular to the surface meaning it is vertical.  so the correct equation is:
$\sum \tau = m_{2}g \frac{l}{2}+m_{1}gd-F_{n}d = 0$

now there are an infinite number of normal forces here, a normal force for every point of contact, however we can mathematically replace the infinite normal forces with a single force which will be equivalent.

As for the 24 equations i dont know...
There are three cases:
1.  AB > AD/2
2.  CD > AD/2
3.  AB <= AD/2 and CD <= AD/2

For each case you have to find a minimum and maximum value of m1.  And to find a minimum or maximum you need one equation plus the sum of forces in the y direction.

so you need 7 equations.

more on the normal force, in the figure for the problem m1g pull down and m2g pulls down, so if there were no normal force wouldnt the system accelerate downwards?

i make a error

m_3 -> m_1 + m_2 + m_3

and what i said, is

why the normal force make a torque?!

i understand your point

because without it does not make sense

but, i never read a physics book that say that the normal force creat a torque

because you are saying this

wikipedia does not say this and google not too

i just think this is strange, i never read this

sorry


Let $f:R\rightarrow R$ satisfy
$f(x^2+y^2+2f(xy)) = (f(x+y))^2$
Is it necessary that $f(0)\geq 0$?

$f(x^2+y^2+2f(xy)) = (f(x+y))^2$
so the right side has to be $(f(x+y))^2 \ge 0$ by the trivial inequality (assuming that we are dealing with real numbers) so
$f(x^2+y^2+2f(xy)) \ge 0$
we need to have $f(0) = ?$
so let $0 = x^2+y^2+2f(xy)) $ for some $x$ and $y$ (we do this because we want to find $f(0)$ so
$f(0) \ge 0$
i might be guilty of circular reasoning, but this seems correct

[quote="AntonioMainenti"][quote="Altheman"]
so let $0 = x^2+y^2+2f(xy)) $ for some $x$ and $y$[/quote]The question really is whether such $x$ and $y$ necessarily exist. I haven't gotten it yet though. :?[/quote]

OK, then you have to prove they exist


Prove: For integers n>1,
 $n^{4}+4$
 is not a prime.

[quote="Potato Theory"]It's equivalent to $n^{4}+4*1^{4}$. This factors out using Sophie-Germain's identity into $(n^{2}+2+2n)(n^{2}+2-2n)$.

Isn't this a little easy for pre-olympiad? :maybe:[/quote]

Your proof is incomplete; you forgot to show that one of the factors could never equal $1$.
$n^{4}+4$
$=n^{4}+4n^{2}+4-4n^{2}$
$=(n^{2}+2)^{2}-(2n)^{2}$
$=(n^{2}+2n+2)(n^{2}-2n+2)$

When $n>1$,
$n^{2}-2n+2>1$
$n^{2}+2n+2>5$

Thus, when $n>1$, $(n^{2}+2n+2)(n^{2}-2n+2) = n^{4}+4$ is not a prime.

What is the Sophie-Germain's identity, as used by Potato Theory?

[quote="vishalarul"]
What is the Sophie-Germain's identity, as used by Potato Theory?[/quote]
It's just what you did in the first part.
$a^{4}+4b^{2}=a^{4}+4a^{2}b^{2}+4b^{2}-4a^{2}b^{2}=(a^{2}+2b^{2})^{2}-2^{2}a^{2}b^{2}$
$=(a^{2}+2ab+2b^{2})(a^{2}-2ab+2b^{2})$
This problem is a special case when $b=1$

[quote="vishalarul"]
Your proof is incomplete; you forgot to show that one of the factors could never equal $1$.
[/quote]

Yeah, realized that after walking away.  :blush:


function
limit
real analysis
real analysis unsolved

For a real function $ f$ call a point $ x_0 \in \mathbb{R}$, a discontinuity point of type 1 for $ f$ if \[ \lim_{x\to x_0,x<x_0}f(x) \neq \lim_{x\to x_0,x>x_0}f(x)\] or \[ \lim_{x\to x_0,x<x_0}f(x) \equal{} \lim_{x\to x_0,x>x_0}f(x) \neq f(x_0).\]
Otherwise, call $ x_0$ a discontinuity point of type 2.

a) Prove that the number of discontinuities of type 1 of a real function is at most countable.
b) Construct a function with uncountable discontinuities of type 2.

Let $ f\equal{}\chi_{\mathbb{Q}}$, which is discontinuous everywhere.

[quote="pwahs"]Alex just told us, that we further have to assume that f is monotonic for the a)-problem, since we found a counter example, based on the Cantorset.

Cheers ;)[/quote]

I would like to see this example.  :)  It seems to me that the original statement of a) is correct. To prove this statement one needs to assume contrary that the set $ A$ of type-1 discontinuities is uncountable and to proceed with the following three steps.

[b]Step 1.[/b] Demonstrate that there are a $ t > 0$ and an uncountable $ B \subset A$ such that the "size" of discontinuity in every $ x \in B$ is bigger that $ t$,  i.e. that for every neighbourhood $ U$ of every $ x \in B$
\[ \sup_U f \minus{} \inf_U f > t.\]
[b]Step 2.[/b]  Demonstrate that there is a point $ y \in B$ which is a cluster point for $ B$.

[b]Step 3.[/b] Find contradiction with the fact that in this $ y$ our function has both one-side limits -- from the right and from the left.

In my example, every point in B is a cluster point of B, thats not a contradiction:
I define a function $ f$ from $ [0,1]$ to $ [0,1]$:
Start with the undefined interval $ [0,1]$, and take out the middle third $ (\frac {1}{3},\frac {2}{3})$ (open interval), define the function to be $ \frac {1}{2}$ in this interval. You are left with two undefine intervals $ [0,\frac {1}{3}]$ and $ [\frac {2}{3},1]$.
In the nth step of defining, take all undefined intervals (there will be $ 2^n$) $ [a,b]$ and take out the middle third $ (\frac {2a \plus{} b}{3},\frac {a \plus{} 2b}{3})$ and define f to be $ \frac {1}{2^{n \plus{} 1}}$ in this intervall. Take the limit of this process, and define the remaining points (points of the Cantor set) to be $ 0$.
Now all points of the Cantor set are borders of some constant interval of $ f$, thus from this direction the limit obviously exists. From the other side, the limit also exists and is $ 0$. The Cantor set is uncountable, so we have what we want.

Cheers,
pwahs

[quote]Now all points of the Cantor set are borders of some constant interval of f, thus from this direction the limit obviously exists. From the other side, the limit also exists and is 0. The Cantor set is uncountable, so we have what we want.[/quote]

Well, there is only countable collection of the constant intervals of $ f$: one is constructed on the first step, two more on the second, etc. So only a countable number of the Cantor set elements consists of these intervals borders. The remaining points of the Cantor set are points of continuity of $ f$.

Ah, you are right. At least, I convinced the TA that there is a mistake in the problem ^^

As for question a) I have found the attached reference.

[quote="The Bogy Man"]
a) Prove that the number of discontinuities of type 1 of a real function is at most countable.
[/quote]


Show that for all positive reals a,b,c:
$ \ \sqrt[3]{4a^3\plus{}4b^3}\plus{}\sqrt[3]{4b^3\plus{}4c^3}\plus{}\sqrt[3]{4c^3\plus{}4a^3} \le \frac{4a^2}{a\plus{}b}\plus{}\frac{4b^2}{b\plus{}c}\plus{}\frac{4c^2}{c\plus{}a}$

It's my old problem  :wink:
And it's not hard because Am-Gm solved it easily  :)

[quote="duongptnkvn"]Show that for all positive reals a,b,c:
$ \ \sqrt [3]{4a^3 \plus{} 4b^3} \plus{} \sqrt [3]{4b^3 \plus{} 4c^3} \plus{} \sqrt [3]{4c^3 \plus{} 4a^3} \le \frac {4a^2}{a \plus{} b} \plus{} \frac {4b^2}{b \plus{} c} \plus{} \frac {4c^2}{c \plus{} a}$[/quote]
$ \sqrt[3]{\frac{a^3\plus{}b^3}{2}}\leq\frac{a^2\plus{}b^2}{a\plus{}b}.$ :wink:

[quote="arqady"][quote="duongptnkvn"]Show that for all positive reals a,b,c:
$ \ \sqrt [3]{4a^3 \plus{} 4b^3} \plus{} \sqrt [3]{4b^3 \plus{} 4c^3} \plus{} \sqrt [3]{4c^3 \plus{} 4a^3} \le \frac {4a^2}{a \plus{} b} \plus{} \frac {4b^2}{b \plus{} c} \plus{} \frac {4c^2}{c \plus{} a}$[/quote]
$ \sqrt [3]{\frac {a^3 \plus{} b^3}{2}}\leq\frac {a^2 \plus{} b^2}{a \plus{} b}.$ :wink:[/quote]
Yes, that's it. It's obviously trues by Am-Gm for 3 numbers  :)


Jorge bought a TV at a $ 15\%$ discount.  Later he found that there should have been a $ 20\%$ discount, so the store gave him the $ \$17$ he was owed.  What was the price of the TV before any discount?

5% of the answer is 17, so the answer is 17*20=340.