algebra
polynomial
symmetry
modular arithmetic
number theory proposed
number theory

We have $c(c^2-c+1)=abX$, so $a=x_1y_1,\,b=x_2y_2$, where $x_1$ and $x_2$ are divisors of $c$; $y_1$ and $y_2$ are divisors of $c^2-c+1$. Let $x_2=c/x_3,\,y_2=(c^2-c+1)/y_3$. Then $a+b=x_1y_1+\frac{c(c^2-c+1)}{x_3y_3}$ is divisible by $c^2+1$, hence $x_1y_1+\frac{1}{x_3y_3}$ is divisible by $M:=c^2+1$ (because $c(c^2-c+1)\equiv 1$ modulo $M$). So, $x_1x_3y_1y_3+1=kM$. Note that $x_1x_3y_1y_3$ is a divisor of $c^2(c^2-c+1)^2$, so we may write $c^2(c^2-c+1)^2=(kM-1)x$. Considering both sides modulo $M$ we have $x\equiv -1$, so $x=lM-1$. Note that $k_0=1,\,l_0=c^2-2c+2$ or viceversa satisfy this equation, and it corresponds to $\{a,b\}=\{c,c^2-c+1\}$. Then, $(kM-1)(lM-1)=(M-1)(l_0M-1)$, from here $klM-(k+l)=l_0M-(l_0+1)$. We see that $k+l\equiv l_0+1$ modulo $M$. If $k+l=l_0+1$, then $kl=l_0$ and $\{k,l\}=\{1,l_0\}$. If not, $k+l\ge l_0+1+M$. We have $klM\ge (k+l-1)M$, so $l_0M-(l_0+1)=klM-(k+l)\ge (k+l-1)M-(k+l)=(k+l)(M-1)-M\ge (l_0+1+M)(M-1)-M$, it is a contradiction.

a and b are the roots of X^2 - (a+b)X + ab
replace a and b by the relations given and chek that the discriminant must be positive because it's the same polynomial.This leads immediately to the result.

senouy: I'm afraid that it donesn't trivialy comes from the problem. You have no equations, but only divisibilities there....

Since the plynomial X^2 + (a+b)X + ab has two roots a and b
Check that the numbers c and c^2 - c +1 LIES btween a and b (plug them in the polynomial and verify that we got two negative numbers by using the divisibility relation.And this is an impossible thing if one of the a or b i sdifferent from one of the c's.

Can you please post your complete solution  ?
The hints you are giving are clear enough

In fact c and c^2 - c + 1 lies betwwen a and b but i am wrong in saying that this  is impossible unless a and b equal th c's.

What do you  mean by sets of coincidence?

By symmetry, we can assume that  $a \leq b$. We will analyze two situations:

$i) \: a \leq c \: \Longrightarrow$

Since  $c^2+1 \mid a+b \; , \; a+b \geq c^2+1$ and $b \geq c^2-c+1$

Let $c-a=x \; , \; b-(c^2-c+1)=y$. Since $a>0$ , then  $x<c$ and $x,y \geq 0$

$c^2+1 \mid y-x$

Let $y=x+k(c^2+1)$ where $k \in \mathbb{Z}$

$b \mid c(c^2-c+1) \; \Longrightarrow \; c^2-c+1+y \mid cy$

$y \neq 0 \; \Longrightarrow \; cy \geq c^2-c+1+y \; \Longleftrightarrow \; c \neq 1\; , \; y \geq c+ \frac{1}{c-1}>c$

$\Longrightarrow \; y>x$ because $x \geq y \; \Longrightarrow \; c>x \geq y >c$

$y > x \; \Longrightarrow \; k>0$

$c^2-c+1+y \mid cy \; \Longrightarrow \; \frac{c[x+k(c^2+1)]}{(k+1)c^2-c+k+1+x} \in \mathbb{Z}$

$\Longrightarrow \; \frac{kc^2+xc}{(k+1)c^2-c+k+x+1} \in \mathbb{Z}$

$y \neq 0 \; \Longrightarrow \; k>0$

$\Longrightarrow \; kc^2+xc \geq (k+1)c^2-c+k+x+1 \; \Longrightarrow \; x(c-1) \geq c^2-c+k+1$

$\Longrightarrow \; c \neq 1 \; , \; x \geq c+\frac{k+1}{c-1}>c$. It is a contradiction since $c>x$.

$y=0 \; \Longrightarrow \; b=c^2-c+1 \; , \; a \mid c \; , \;  c^2+1 \mid c-a$

It is easy to show that the last one holds if and only if $c=a$.

$ii) \; a>c \; \Longrightarrow$

$b<c^2-c+1$. Let $a=c+x \; , \; b=c^2-c+1-y$

$\Longrightarrow \; x,y >0 \; , \; c^2+1 \mid x-y$

$y>x \; \Longrightarrow \; y>c^2+1=b+c+y$ which is a contradiction.

$x \geq y \; \Longrightarrow \; a \geq c+y \; , \; (c+y)(c^2-c+1-y)>c(c^2-c+1)$

$\Longleftrightarrow \; (c-1)^2>y \; \Longleftrightarrow \; b>c $

$\Longrightarrow \; ab>c(c^2-c+1)$ because

$b \leq c \; \Longrightarrow \; a \leq b \leq c$ which is a contradiction.

$ab>c(c^2-c+1)$ is also a contradiction.

Solution is done. :wink:

[quote="erdos"]Let $a,b$ and $c$ be positive integers such that $ab$ divides $c(c^{2}-c+1)$ and $a+b$ is divisible by $c^{2}+1$.

Prove that the sets $\{a,b\}$ and $\{c,c^{2}-c+1\}$ coincide.[/quote]
here is hint to another process:
here $gcd(c,c^2-c+1)=1$
so $ab|c^2-c+1$ or $ab|c$,but $c^2+1|a+b$ so $a+b\ge c^2+1$

[quote="mathmdmb"]
here $gcd(c,c^2-c+1)=1$
so $ab|c^2-c+1$ [u][b]or[/b] [/u]$ab|c$
[/quote]
Sorry if I`m wrong but just put $a=c=3 , b=7$ !!! We have $ab \nmid c^2-c+1$ and $ab \nmid c$ .

[quote="mathmdmb"][quote="mahanmath"][quote="mathmdmb"]$a=c$ is to be proved and here is a step to show that  $a|c$ and $b|c^2-c+1$[/quote]
Sorry , Maybe I was really bad in describe the wrong point .

 You said if $gcd(A,B)=1$ and $XY \mid AB$ then $XY \mid A$ or $XY \mid B$ . ($A,B,X,Y$ are natural.)
And it`s not true.[/quote]
that`s i pointed out,sorry(me too) for bad language,
here we have wlog, $a|c \implies c\ge a$ and $b|c^2-c+1$,so let $c^2-c+1=bk,c^2+1=bk+c|a+b$
Then $a+b\ge bk+c\implies a-c\ge b(k-1)$ but $c\ge a$,so $a=c$ and then $k=1,b=c^2-c+1$[/quote]

I think you still doesn't understand. You said that "wlog  $a|c$", but it isn't necessary to be true. Actually we prove that wlog $a=c$ and hence $a \mid c$, but this proof is the main problem. :wink:

[quote="mathmdmb"]can you explain why we are wrong here?Since $gcd(c,c^2-c+1)=1$,$gcd(a,b)=1$ because otherwise $c$ and $c^2-c+1$ would share a common factor and so I assumed $a|c$,not $a=c$,and I took it for an obvious conclusion.We must have this.[/quote]

Yes, I understand your idea, but, unfortunately your claim is not true. As mahanmath said, we can easily find $a,b,c$ which don't satisfy your claim. Let us take

$c=10, a=35, b=26$

As you said, $gcd(a,b)=gcd(c,c^2-c+1)=1$ and

$ a \nmid c \; , \; a\nmid c^2-c+1 \; , \; b \nmid c \; , \; b \nmid c^2-c+1$, but

$ab=c(c^2-c+1)$ and hence $ab \mid c(c^2-c+1)$

Hopefully this is right... I think this is basically what [b]senouy[/b] suggested above.

Let $c(c^2-c+1)=mab$ and $a+b=n(c^2+1)$. Then
\begin{align*}
(a-b)^2
=n^2(c^2+1)^2-\frac{4}{m}c(c^2-c+1)
&=n^2c^4-\frac{4}{m}c^3+\left(2n^2+\frac{4}{m}\right)c^2-\frac{4}{m}c+n^2\\
&\ge n^2c^4-\frac{4}{m}c^3+\left(2n^2+\frac{4}{m^2n^2}\right)c^2-\frac{4}{m}c+n^2\\
&=\left(nc^2-\frac{2}{mn}c+n\right)^2.
\end{align*}WLOG set $a\ge b$. Then
\[a-b\ge nc^2-\frac{2}{mn}c+n\implies b\le\frac{1}{mn}c.\]But
\[c^2+1|a+b\implies -mb^2\equiv mab=c(c^2-c+1)\equiv-c^2\pmod{c^2+1},\]so for some $k\ge0$,
\[c^2+k(c^2+1)=mb^2\le m\left(\frac{1}{mn}c\right)^2=\frac{1}{mn^2}c^2.\]Clearly $k=0$ and $m=n=1$, so we're done.

[hide="Solution"]WLOG assume $a\ge b$. Suppose $\frac {c(c^2-c+1)}{ab}=r$. We claim $r=1$. If not, then $r\ge 2$. Note that we must have $c\ge b$. Otherwise $c^2-c+1\ge \frac {c(c^2-c+1)}{b}\ge 2a\ge a+b\ge c^2+1$ which is not possible. Set $s=c-b$.

If $s=0$, we'll obviously have $b=c$ and from the given relations, it is not hard to prove $a=c^2-c+1\implies r=1$. Therefore assume $s$ is positive. So $a\ge c^2+1-b>c^2-c+1\implies \frac {c^2-c+1}{a}<1\quad (1)$
So $\frac {c(c^2-c+1)}{ab}=\frac {c^2-c+1}{a}+\frac{s(c^2-c+1)}{ab}\ge 2\quad (2)$

Note that we must have $s\ge b$, because otherwise (2) is less than 2. Now we claim, for $n\ge 1$, $s\ge nb\implies s> ((n+1)^2-2)b$
[hide="Proof"]$c=b+s\ge (n+1)s\implies c^2+1\ge (n+1)^2b^2+1$.
Notice that $0\equiv c(c^2+1)-(c^2+1)\equiv rab-1\equiv r(-b)b-1\equiv rb^2+1\pmod{c^2+1}$

So $rb^2+1\ge c^2+1\ge (n+1)^2b^2+1\implies r\ge (n+1)^2$

Hence $\frac {c(c^2-c+1)}{ab}=\frac {c^2-c+1}{a}+\frac{s(c^2-c+1)}{ab}=r\ge (n+1)^2>(n+1)^2-1\quad (3)$

From (1), $\frac {c^2-c+1}{a}<1$. So if we have $\frac{s(c^2-c+1)}{ab}\leq (n+1)^2-2$, it will be a contradiction to (3).

Hence $\frac s b >\frac{s(c^2-c+1)}{ab}>(n+1)^2-2$

So our claim is true. [/hide]
So $s$ has no upper bound. But for fixed pair of $a,b$, $s$ must have upper bound. Because $2a\ge a+b\ge (b+s)^2+1$ A contradiction. So we must have $r=1$

Now we claim $\frac {a+b}{c^2+1}=1$. Because otherwise $2a\ge a+b\ge 2(c^2+1)\implies \frac {c^2-c+1}{a}<1$
Also $c^2+1|rb^2+1\implies c\leq b$. If $c<b$, then \[1=\frac {c(c^2-c+1)}{ab}<\frac {b(c^2-c+1)}{ab}<1\]
A contradiction. So $a+b=c^2+1$. Now we have $ab=cd$ and $a+b=c+d$ for four positive integers. So we must have $\{a,b\}=\{c,c^2-c+1\}$[/hide]

Wlog assume that $a \le b$. There exists integer $k$ with $abk=c(c^2-c+1)$. $c^2+1|a+\frac{c(c^2-c+1)}{ak} \implies c^2+1|a^2k+c(c^2-c+1) \implies c^2+1|a^2k+1 \implies k \ge \frac{c^2}{a^2} \implies c(c^2-c+1) \ge \frac{bc^2}{a} \implies \frac{c^2-c+1}{c} \ge \frac{b}{a}$. Note that $c^2+1 \le a+b$. Thus $\frac{a+b-c}{c} \ge  \frac{c^2-c+1}{c} \ge \frac{b}{a} \implies a \ge c \implies b \le c^2-c+1$ because of the first condition. Let $a=c+m, b=c^2-c+1-n$. First condition yields $m(c^2-c+1) \ge mn+cn ... (1)$, second condition yields $c^2+1|m-n ... (2)$, $a \le b \implies c^2-2c+1\ge m+n ... (3)$. If $m=n\neq 0$, we get $m \ge c^2-2c+1$ from $(1)$ and this contradicts with $(3)$. If $m \neq n$, we get $c^2-2c+1 \overset{(3)} \ge m+n \ge |m-n| \overset{(2)} \ge c^2+1$, contradiction. Thus the only possibility is $m=n=0$, sets $\{a,b\}$ and $\{c,c^{2}-c+1\}$ coincide

Let $a+b=k(c^2+1)$ and $c(c^2-c+1)=\ell ab$ for some positive integers $k,\ell$. Then, working $\pmod {c^2+1}$ we obtain

$1 \equiv c(c^2-c+1) \equiv \ell ab \equiv -\ell a^2 \pmod {c^2+1},$

and so $(c^2+1) \mid (\ell a^2+1)$. Similarly, $(c^2+1) \mid (\ell b^2+1)$. Hence, we may let $\ell a^2=(c^2+1)m+c^2$ and $\ell b^2=(c^2+1)n+c^2$, for some $m,n \geq 0$. Then,

$c^2(c^2-c+1)^2=(ab\ell)^2=((c^2+1)m+c^2)((c^2+1)n+c^2),$

or equivalently

$(c^2+1)mn+c^2m+c^2n=c^2(c-1)^2$.

Working $\pmod c^2$, we find that $c^2 \mid mn$. However, if $mn \neq 0$ this implies that $mn \geq c^2,$ and so

$c^2(c-1)^2 \geq (c^2+1)mn \geq c^2(c^2+1) >c^2(c^2-2c+1)=c^2(c-1)^2,$

a contradiction. Thus, $mn=0$. WLOG assume that $m=0$, and so $n=(c-1)^2$. Note that $m=0$ implies that $a^2\ell=c^2$. However, this implies that

$c(c^2-c+1)=ab\ell=ab \cdot \dfrac{c^2}{a^2},$

that is

$\dfrac{b}{a}=\dfrac{c^2-c+1}{c}$.

Therefore, $(c^2+1)k=b+a=\dfrac{a(c^2+1)}{c},$ which implies that $kc=a$. This, combined with $a^2\ell=c^2$ implies that

$a=kc \geq c=a\sqrt{\ell}  \geq a,$

so equality must hold i.e. $k=\ell=1$, and now it's easy to conclude that $a=c$ and $b=c^2-c+1$, as desired.


Let $ n$ and $ k$ be positive integers such that $ 1 \leq k \leq n\minus{}1$.
When is the binomial coefficient $ C^{k}_{n}$ prime ?

details please.
 :P

Does anybody can give solutions?
I have thought for a couple of times.

It can be seen that if and only if k=1 or n-1
and n is a prime will the conditions goes right.


geometry
3D geometry
search
function
algebra
partial fractions

[quote="mathmanman"]Well :
$\zeta(3) = 1.2020569...$, so...  :P
Just kidding, I'll search how to prove it though. (for the problem, not for the value of zeta(3) ;))[/quote]
Of course, I post this problem, which is not for applying R-zeta function :) . Try to solve it! :D

btw, what is the general formula to calculate $\zeta(n)$?

There isn't one...
I mean, for $\zeta(2k)$, then, yes, there's a formula with the [url=http://mathworld.wolfram.com/BernoulliNumber.html]Bernoulli numbers[/url].
But, for $\zeta(2k+1)$, good luck! ;)

[quote="Lovasz"]btw, what is the general formula to calculate $\zeta(p)$?[/quote]
$\zeta(3)\leq\frac{7-4\gamma}{2(3-2\gamma)}=1.2709192066782574508177\cdots.$


geometry
circumcircle
trapezoid
trigonometry
geometry proposed

Let $ K,L,M$ and $ N$ be the midpoints of sides $ AB,$ $ BC,$ $ CD$ and $ AD$ of the convex quadrangle $ ABCD.$ Is it possible that points $ A,B,L,M,D$ lie on the same circle and points $ K,B,C,D,N$ lie on the same circle?

$ (O_a)$ is circumcircle of the $ \triangle ABD,$ $ (O_c)$ is circumcircle of the $ \triangle BCD.$ $ LM \parallel BD$ is midline of the $ \triangle BCD,$ if $ L, M \in (O_a),$ then $ BMLD$ is isosceles trapezoid and the $ \triangle BCD$ is isosceles. Similarly, $ NK \parallel BD$ is midline of the $ \triangle ABD,$ if $ N, K \in (O_b),$ then $ BDNK$ is isosceles trapezoid and $ \triangle ABD$ is isosceles. Therefore, $ ABCD$ is at least a kite, let $ E$ be its diagonal intersection. Assuming $ ABLMD, CDNKB$ are both cyclic, $ LM \equal{} \frac{BD}{2} \equal{} NK$ implies $ \angle DAB \equal{} \angle BCD$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=204352[/url]) and  $ ABCD$ is a rhombus. Power of $ A$ to $ (O_c)$ is

$ AK \cdot AB \equal{} \frac {AB^2}{2} \equal{} (AE \plus{} EO_c)^2 \minus{} R_c^2$

Substituting $ AB \equal{} \frac {BD}{2 \sin \frac {A}{2}},$ $ AE \equal{} \frac {BD}{\tan \frac {A}{2}},$ $ EO_c \equal{} R_c \cos A,$ $ R_c \equal{} \frac {BD}{2 \sin A}$ and reducing by $ \frac {BD^2}{4 \sin^2 A},$ we get the following equation for the angle $ \angle DAB \equiv \angle A$:

$ 2 \cos^2 \frac {A}{2} \plus{} 1 \equal{} \left(2\cos^2 \frac {A}{2} \plus{} \cos A\right)^2$

$ \cos A \plus{} 2 \equal{} (2 \cos A \plus{} 1)^2$

with roots $ \cos A \equal{} \frac {_1}{^4}$ and $ \minus{} 1,$ only the 1st one acceptable.


if possible... put only the equation on how to solve the volume of a ring.... thanks... ^^

hello, what kind of ring do you mean,a torus?
Sonnhard.

hello, this will help you
http://whistleralley.com/torus/torus.htm
Sonnhard.

Sorry, never mind, I thought he meant rings as in College Algebra....


ARML
geometry
3D geometry
octahedron
probability
circumcircle
AMC

[please read everything]

[note this is in the AMC forum; but this gets more traffic; if any of the moderators have a problem with this, could you kindly move this topic to the ARML forum]

[color=green][size=167]Mock ARML Individual for 2007 [/size][/color]

Written by Alex Anderson (Altheman)

[Thanks to one of my ARML coaches, Mark Fritz, for helping me check problems and revise the test.]

Attached is the .pdf with the problems. The rules are written on the cover sheet. Everybody pretty much knows the rules...no calcluators, etc. [b]The time limit is 11 minutes per round.[/b] I suppose that we are going by the honor system.

I will be accepting answers for the next week [[b]everything must be sent to me by sunday night at 11:59 Central time[/b]].

Submit answers to me, Altheman, by sending me a private message.

Please don't make it harder on me to compile results, [b][color=red]Put messages in this form[/color][/b]:

[Aops Name]

1.
answer
2.
answer
...

1.
comments
2.
comments
...

By comments, I mean a short sentence as to how you solved the problem. [b]A short description of how you did the problem is required.[/b] I want to be able to gauge the types of solutions that people are giving; that helps me understand how to better write the tests. Furthermore, I like to discuss problems with other people.

[i]Please refrain from discussing the problems with other people for the next week or so. If you just want to talk about any of the problems, just send me a PM. If you take the individual, it would be nice if you would send me a pm to say what you thought about the test. [/i]

[b]Assume that the problems are correct before you send a complaint. While I have been quite careful to make sure that everything is correct, mistakes can happen. Send all complaints to me by using an PM. Don't post them in the thread; that causes confusion. [/b]

[Problems attached]

My thoughts: I tried my best to make the problems fair for the time limit. Nothing is meant to be misleading. All problems have an elementary solution. Lastly, good luck and have fun!

--Alex

Looks great! I'll [size=25]probably[/size] do it tonight

You probably need to download Adobe Reader, which is free. Google it.

Rules:
1. No calculator, books, etc.
2. Standard arml standards for simplifying answers.
3. 11 minutes per pair of questions.
4. Write your answer in the box.

[hide="Part 1"]1. If $\sqrt[3]{41\sqrt{5}+54\sqrt{3}}=a\sqrt{5}+b\sqrt{3}$ where $a$ and $b$ are integers, then determine the ordered pair, $(a,b)$.

$\boxed{1. \qquad \qquad \qquad \qquad \qquad \qquad }$

2. Find the smallest positive integer, $N$, such that: $11$ divides $N$ and the sum of the digits of $N$ is equal to $27$.

$\boxed{2. \qquad \qquad \qquad \qquad \qquad \qquad }$[/hide]

[hide="Part 2"]3. Three verticies of a regular octahedron are: $(3,-1,-2)$, $(8,9,8)$, and $(-11,1,3)$. Provided that the other $3$ vertices are $(x_{i},y_{i}, z_{i})$ for $i=1,2,3$, compute $x_{1}+x_{2}+x_{3}$.

$\boxed{3. \qquad \qquad \qquad \qquad \qquad \qquad }$

4. Alex and Sam are playing a game with an unfair coin where the probability of tossing a head is $h>0$. The first person to toss a head wins. Alex starts and they alternate turns. Alex flips the coin once on his turn and Sam flips the coin twice on his turn. Given that the game is fair, find $h$.

$\boxed{4. \qquad \qquad \qquad \qquad \qquad \qquad }$[/hide]

[hide="Part 3"]5.  Let $p(x)=x^{5}-1$. Suppose that $a$, $b$, $c$, $d$, and $e$ are the roots of $p(x)$. Compute $t=(2+b^{2}+c^{2}+d^{2}+e^{2})(a^{2}+2+c^{2}+d^{2}+e^{2})(a^{2}+b^{2}+2+d^{2}+e^{2})(a^{2}+b^{2}+c^{2}+2+e^{2})(a^{2}+b^{2}+c^{2}+d^{2}+2)$.

$\boxed{5. \qquad \qquad \qquad \qquad \qquad \qquad }$

6.  Let $a_{n}=2^{3}+3^{3}+...+n^{3}$ for $n=2,3,...$. If $2<k\le 2007+2$, then determine all six values of $k$ for which $a_{k}$ is divisible by $2007$.

$\boxed{6. \qquad \qquad \qquad \qquad \qquad \qquad }$[/hide]

[hide="Part 4"]7. Provided that $x$ and $y$ are real numbers, find the minimum value of
$(x-y)^{2}+(x-20)^{2}+(y-21)^{2}$.

$\boxed{7. \qquad \qquad \qquad \qquad \qquad \qquad }$

8. Triangle $ABC$ has a circumcircle $O$. Construct circle $O'$ such that $O'$ is internally tangent to $O$ at $A$ and $O'$ is tangent to $BC$. Determine the radius of $O'$ given that $AB=6$, $BC=10$, $AC=8$. Express your answer as an improper fraction reduced to lowest terms.

$\boxed{8. \qquad \qquad \qquad \qquad \qquad \qquad }$[/hide]

Well, real ARMLs are 10 minutes, but Altheman was nice and gave us an extra minute. I'm OK with it. :D

Shhhhh. I'd love to talk about #7, but the competition's still going on.

I don't think Altheman would be very happy if we all jumped out and started discussing the trivial-ness or non-trivial-ness of #7.

Just wanted to make sure I was reading it right.
Didn't imply anything with my comment.
 :wink:

#7's are usually relatively easy, giving you time for 8 (or something to do if you can't do 8).

These are great problems. Thanks Altheman.

BTW: Is it just me or does this topic belong in the ARML Forum.

EDIT: Wow, I suck at reading.

[quote="Altheman, at the top of the first post"]

[note this is in the AMC forum; but this gets more traffic; if any of the moderators have a problem with this, could you kindly move this topic to the ARML forum]
[/quote]

Some of these problems are excellent teaching/coaching problems. I've already borrowed #5  :)

[quote="aufha"]Does the form we submit our answers with say to alternate between pairs of answers and comments or does the ellipsis after number 2's answer mean to post the answers consecutively down to problem 8?[/quote]

either way is fine i guess

@Altheman: It doesn't matter if this is in AMC forum although I think you will get more traffic if the thread was in ARML forum though (since it's approaching ARML). AMC forum is not active this time of year so it's fine.  :)

Solutions thread [in the ARML forum]:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151467[/url]

[b]Contest is officially closed.[/b]

all problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3733598_2007_mock_arml_i_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


1. if $\frac{1 + \frac{b}{a} + \frac{b}{c}}{1 + \frac{a}{b} + \frac{a}{c}} = 11$,

and if $b + a - 1 = 35$, what is $b$?

2. find the sum $\frac{1}{8} + \frac{1}{24} + \frac{1}{48} + \frac{1}{80} + \frac{1}{120}$

3. how many three digit numbers $abc$ has such properties that $a + c = b^2$?

[quote="zmli"]For problem 3, can a and c be zero?[/quote] if its a 3-digit number, a shouldn't be 0. i would say there's no reason why c cannot be 0.

[quote="zmli"]For problem 3, can a and c be zero?[/quote]

a shouldn't be zero and for $110$, we have $1^2+0^2=1^2$

[hide="hint for 1"]
well.. find the common denominator on both top and bottom and add.. then the problem is clear.
[/hide]

[hide="hint for 2"]
separate fractions.. if you know what i mean
[/hide]

[hide="hint for 3"]
there are only certain number which can be b...
[/hide]

Here's #1:

1.

[hide]
multiplying both sides by a/b,

(a + b + ab/c) / (a + b + ab/c) = 11a/b

Hence 11a/b = 1 and 11a = b.

But we're given a + b = 36, so    a + 11a = 36, 12a = 36 and a = 3. It follows that b = 33.[/hide]

How does that thing with the telescoping series work?

[hide="The way I did it was"]
to take a $\frac{1}{120}$ off of both sides to get:

$\frac{1}{120}(15+5+\frac{5}{2}+\frac{3}{2}+1)$ and it was easy from there to get $\frac{5}{24}$
[/hide]

[quote="shobber"][quote="zmli"]For problem 3, can a and c be zero?[/quote]

a shouldn't be zero and for [hide=""Hide me""]$110$, we have $1^2+0^2=1^2$[/hide][/quote]
Hi shobber, can you at least hide some parts of your answers even if you are making a point on an argument?

[quote="sky9073"][quote="xxreddevilzxx"]3. how many three digit numbers $abc$ has such properties that $a + c = b^2$?[/quote]

There has got to be a more elegant solution than this..

[hide]b is an integer < 10; b^2 can be either 1, 4, 9 or 16.
(b^2 can't be 25 because a + c can't add up to it)

casework from here:

when b = 1, abc = 110 only. 1 number.
when b = 2, abc = 420, 321, 222, 123 only. 4 numbers.
when b = 3, abc = 930, 831, 732, 633, 534, 435, 336, 237, 138. 9 numbers.
when b = 4, abc = 747, 848, 947 only. 3 numbers.

Then add... 1 + 4 + 9 + 3 = [b]17[/b].[/hide][/quote]

A little comment for those who took this answer.

[hide]I think in your b = 4, abc should = 749 not 747. Must be a typo, cause you got 947. Nonetheless, that's a good way of solving it. I can't think of a better way. [/hide]

1.
[hide]
$\frac{\frac{ac + bc + ba}{ac}}{\frac{bc+ac+ab}{bc}} = 11$
Everything will cancel and we will be left with b/a = 11 so a = b/11
So b + b/11 = 36 and b = 33
[/hide]

3.
[hide]
We know that $b^2$ must be no greater than 18 since that is the highest sum of two digits.
The possible values of b are then
0, 1, 2, 3, 4
We can eliminate 0 since no two a and c will add up to 0.
For 1 (1^2 = 1)
a  +  c
1  +  0

For 4 (2^2 = 4)
a  +  c
4  +  0
1  +  3
3  +  1
2  +  2

For 9 (3^2 = 9)
a  +  c
9  +  0
1  +  8
8  +  1
2  +  7
7  +  2
3  +  6
6  +  3
4  +  5
5  +  4

For 16 (4^2 = 16)
a  +  c
9  +  7
7  +  9
8  +  8

So 17
[/hide]


algebra
polynomial
integration
vector
real analysis
real analysis unsolved

This problem seems interesting but I can't find a solution:
Does there exist a constant C such that for all polynomials of degree less than or equal 2005 we have
\[|P(0)| \leq C\int_{-1}^{1} |P(x)| dx\]

Thank you

The set of polynomials of degree at most $n$ restricted to $[-1,1]$ is a finite-dimensional real vector space $\mathcal V$. The sup norm, $\|P\|_{\infty}=\sup_{[-1,1]}|P|$ and the $L^1$ norm $\|P\|_1=\int^1_{-1}|P|\mbox{d}t$ are equivalent, since this is true for all normed finite-dimensional vector spaces, which means that there are constants $C_1,C_2>0$ s.t. for all polynoiamls $P\in\mathcal V$ we have $C_1\|P\|_1\le\|P\|_{\infty}\le C_2\|P\|_1$.

So the answer is "yes" :).


The product of two positive integers minus their sum is 39. The integers are relatively prime, and each is less than 20. What is the sum of the two integers?


The ratio of the length to the width of a rectangle is 12 to 5 and the area of the rectangle is 540 square units. What is the number of units in the length of the rectangle?

540/60=9
sqrt9=3

Thus the ratio of sides to a 12x5 rectangle is 3:1 so its 36x15. Thus 36.

A little bit  more explained:
length: $ a\equal{}12x$, width: $ b\equal{}5x$
$ Area\equal{}ab\equal{}60x^2\equal{}540$
So $ x^2\equal{}9$, so $ x\equal{}3$
Length: $ a\equal{}36$


limit
number theory
group theory
superior algebra
superior algebra unsolved

Prove or disprove:
A cyclic group with only 1 generator can have at most 2 elements.

Can anyone help?Thx :)

Do you know the formula for the number of generators of a cyclic group with $ n$ elements?

If $ \zeta$ is a generator of $ Z_n$, a little number theory will tell you that $ (k,n)\equal{}1$ if and only if $ \zeta^k$ is a generator as well.

Rephrased, the question becomes: for which integers $ n\geq$ does $ n$ have a common factor with every single $ 2\leq k\leq n$? :)  I can think of three ways to proceed: Bertrand's postulate, the Chinese Remainder Theorem, or the classic formula for Euler's phi function.  Pick your poison.  :lol:

i just did this question recently, in Fraleigh's abstract algebra book..

by the definition of a cyclic group $ \{a^n|n \in Z\}$, $ a^{\minus{}1}$ must be a generator of the same group as $ a$ generates. we then must have $ a\equal{}a^{\minus{}1}$ so $ a^{2k}\equal{}a^k a^k\equal{}a^k a^{\minus{}k}\equal{}e$ ($ e$ is identity) and $ a^{2k\plus{}1}\equal{}a$

Whoops, yes, that's a *much* simpler approach.  :blush: $ (n\minus{}1,n)\equal{}1$ of course...

What about a cyclic group with only two generators? What can be said about such a group?

If your group has exactly $ k$ generators, then it can have $ n$ elements for any $ n$ with $ \varphi(n) \equal{} k$. As $ \lim_{n \to \infty} \varphi(n) \equal{} \infty$, there will always be just finitely many.

The group of integers under addition has infinitely many elements and only 2 generators. So that would only work if $ k\neq 2$.


could you help me by giving me the link of this inequality because it already posted

$ \sum_{cyc} \frac{\sqrt{b\plus{}c\minus{}a}}{\sqrt{b}\plus{}\sqrt{c}\minus{}\sqrt{a}} \le 3$

http://www.mathlinks.ro/Forum/viewtopic.php?p=741368#p741368
 here is it :)


It is a mechanics discussion regarding a problem of electrostatics.

Couple of months ago I dealt with a problem:Given two electric dipoles, separated at a distance r and are perpendicular to each other.We are to find the torque exerted on each other.
I took p1 as vertical and p2,right to p1,at a distance r,facing towards right.
I used co-ordinate free form of E_dip which gave the torques...

The torques were in the same direction and were not cancelling one another.It resulted from the fact that there were also dipole-forces exerted by one on the other---which I did not need to consider for the purpose of the problem.If one finds the torques due to both the factors---one arising from pxE and the other from (p.grad)E,the total torque on the system P1+P2 can be shown equal to zero.It is OK as the angular momentum of the system does not change.I did not use a frame to avoid two different frames involved in the same problem.

Everything is clear.What I want to know is that when I used co-ordinate free form,how can I be sure that I am working from an inertial system?

Is my qustion clear?I am not referring to any particular co-ordinate system...but the end result is consistent with that as viewed from an inertial system...
So,I think using a co-ordinate free form somehow makes it possible.

can you try out something to make it clear?

What I got is that in STRthe co-ordinate free forms have meaning only in inertial co-ordinate systems...


linear algebra
matrix
linear algebra unsolved

how many non-zero eigenvalues  has  $ A$ where $ A\in M_n(R)$ and $ rank(A) \equal{} k$,
is it $ k$ the answer?if yes how can i prove it? :D

If $ A$ is a diagonalizable matrix, then we can easily find the eigenvalues and the diagonal matrix $ D$ is such that each of its diagonal entries are the eigenvalues and all other entries are 0.  Then since we are saying it has rank $ k$, it has $ k$ linearly independent rows/columns and hence it must have $ k$ non-zero eigenvalues.

What about if $ A$ is not diagonalizable?  You try.

[quote="JRav"]
What about if $ A$ is not diagonalizable?  You try.[/quote]
i guess that for $ A$ not diagonalizable this is not true.:whistling:
$ A\equal{}\begin{bmatrix}0&1\\0&0\end{bmatrix}.$ ..eigenvalues are 0 and 0, but rank is 1.


let a,b,c>0 . prove that :
\[ (ab\plus{}bc\plus{}ac)(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2})\geq \frac{9}{4}\]

Crux No.4(1940)     Problem 1940
[b]Geometric background[/b]
In $\triangle ABC$,have $$\frac{r^2_a}{a^2}+\frac{r^2_b}{b^2}+\frac{r^2_c}{c^2}\geq\frac{9}{4}$$.
[url=http://www.artofproblemsolving.com/community/c6h411853p2311707]here[/url]

[quote=vo thanh van]Please see here http://www.mathlinks.ro/viewtopic.php?t=230372
and http://www.mathlinks.ro/viewtopic.php?t=239240[/quote]

[quote=BaBaK Ghalebi]it is also posted [url=http://www.mathlinks.ro/viewtopic.php?t=222268]here[/url] in persian with a different solution by M.A...[/quote]

These pages you are looking for have been deleted.. But you can see here
 [url]http://ac.upg-ploiesti.ro/vcirtoaje/vcirtoaje.php[/url]
in
 [b]Mathematical Inequalities, Vol. 2 - Symmetric Rational and Nonrational Inequalities[/b]
the following generalization (P 1.71):
$$\sum\frac{ab+(k-1)bc+ca}{b^2+kbc+c^2}\ge \frac {3(k+1)}{k+2}, \ \ \ \ k>-2.$$


geometry
incenter
circumcircle
geometric transformation
reflection
geometry unsolved

Dear mathlinkers,
let ABC be a triangle, I the incenter of ABC, UVW the cevian triangle of I wrt ABC,
O the center of the circumcircle of ABC,
X the meetpoint of the parallel to AI with BC,
Y, Z symmetrically…
U' the symmetric of U wrt X,
V', Z' symmetrically…
Prove that the parallels to AI, BI, CI passing through U', V', W' are concurrent.
Sincerely
Jean-Louis

Dear Mathlinkers,
a synthetic proof of this result originated has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


floor function
number theory unsolved
number theory

Let $ m$ be a positive integer and define $ f(m)$ to be the number of factors of $ 2$ in $ m!$ (that is, the greatest positive integer $ k$ such that $ 2^k|m!$). Prove that there are infinitely many positive integers $ m$ such that $ m \minus{} f(m) \equal{} 1989.$

[hide]
I claim that $ m\minus{}f(m)$ equals the number of ones in the binary representation of $ m$. From there, it will follow that there exist infinitely many such $ m$ (for instance, just take $ m\equal{}2^x(2^{1989}\minus{}1)$ for any nonnegative integer $ x$).

Suppose $ m\equal{}2^{d_1}\plus{}2^{d_2}\plus{}...\plus{}2^{d_n}$ is the binary representation of $ m$. Of course, $ n$ is the number of ones in the binary representation. Now it's well known that

$ f(m)\equal{}\sum_{k\equal{}1}^{\infty}\left\lfloor\frac{m}{2^k}\right\rfloor\equal{}\sum_{k\equal{}1}^{\infty}\lfloor 2^{d_1 \minus{} k}\plus{}2^{d_2 \minus{} k}\plus{}\cdots\plus{}2^{d_n \minus{}k}\rfloor\equal{}\sum_{k\equal{}1}^{\infty}\sum_{1\le i\le n,d_i\ge k}2^{d_i\minus{}k}$

Rearranging this is just

$ \sum_{i\equal{}1}^{n}(1\plus{}2\plus{}\cdots\plus{}2^{d_i\minus{}1})\equal{}\sum_{i\equal{}1}^{n}(2^{d_i}\minus{}1)\equal{}\left(\sum_{i\equal{}1}^{n}2^{d_i}\right)\minus{}n\equal{}m\minus{}n$.

Hence $ f(m)\equal{}m\minus{}n$, so $ m\minus{}f(m)\equal{}n$, as desired.
[/hide]

By legendre's formula,
\[f(m)=\frac{m-S(m)}{2-1}=m-S(m)\Longrightarrow m-f(m)=S(m)\]
where $S(m)$ is the sum of digits of $m$ in base $2$.
Now, we can choose
\[m=2^k(2^{1989}-1)\text{  or  }m=2^{1989}(2^k+1)-1\]


[color=darkred]Construct outside of the triangle $ABC$ the right triangles $\{\begin{array}{c}\triangle ABP\ ,\ BP\perp BA\\\ \triangle ACQ\ ,\ CQ\perp CA\end{array}$.

Denote $\{\begin{array}{c}D\in BC\\\ AD\perp BC\end{array}$. Prove that $AD\cap BQ\cap CP\ne\emptyset$ $\Longleftrightarrow$  $\widehat{PAB}\equiv\widehat{QAC}$ ([u][b]A. Zanoschi - Moldova[/b][/u]).[/color]

[color=darkblue][b]Remark.[/b] You can use the my method from http://www.mathlinks.ro/Forum/viewtopic.php?t=154396 [/color]

[quote][color=darkred][b]An equivalent enunciation.[/b] Let $ABCD$ be a cyclic convex quadrilateral for which $BA\perp BC$.
Let $M,N$ be two mobile points so that $B\in (CM)$, $D\in (CN)$ and $\widehat{BAM}\equiv\widehat{DAN}$.
Ascertain the geometrical locus of the intersection $L\in BN\cap DM$.[/color][/quote]

Through vertices $B,$ $C$ of $\bigtriangleup ABC,$ we draw two lines $BX,$ $CZ$ respectively, perpendicular to $BC.$

So, the segment lines$BP,$ $BX,$ are isogonal conjugates with respect to the angle $\angle B$ and similarly the segment lines $CQ,$ $CZ,$ are isogonal conjugates wrt $\angle C.$

Hence, because of $\angle PAB = \angle QAC,$ based on [b][size=100]Jacobi’s theorem[/size][/b] $($ considering that the parallel lines $AD,$ $BX,$ $CZ,$ are concurrent at infinity $),$ we conclude that the segment lines $AD,$ $BQ,$ $CP,$ are concurrent at one point and the proof is completed.

Kostas vittas.


Let's say you included a form in a webpage, and then now u've got a text box saying enter your name.
Now HOW would you extract each letter of the name?
(For example user inputs Tim, if the first letter is T, I'm supposed to display 20)
In java it's charAt(), so is there any such corresponding function in Javascript?

Typing 'javascript string characters' into google and looking at result #2 tells you that charAt works in javascript.

Hmm...i'll try again but it says 'object not found' whenever I do it.
Also, how do u display multiple images?
(For eg. if input=apple, display apple.jpg. If input=all, display apple.jpg, orange.jpg, grapes.jpg, etc)

Are you sure you initiated the string before trying charAt?

Also, make sure the string name is right before the function. i.e. (stringnamemustberighthere).charAt(blah);

I did...but there still seems to be a problem...will post the exact details a little later...it's the part about storing form.text1.val in a string..it says 'Object Cannot Be Found'

-Spelling?
-Post the entire example here, it could be a corrupt object or something.


limit
algebra
polynomial
number theory unsolved
number theory

For natural $ n$ we define $ s(n)$ as the sum of digits of $ n$ (in base ten)
Does there exist a positive real constant $ c$ such that for all natural $ n$ we have

$ \frac{s(n)}{s(n^2)} \le c$ ?

No, the expression is unbounded. I managed to produce the following counterexample after a lot of computer work. I have no idea how one could get such an idea without a computer in a couplle of hours time :wink: :blush:

[b]lemma[/b]
if $ a,b,c,d$ are natural numbers than we can conclude $ \max(a,b) \equal{} \max(c,d)$ and $ \min(a,b) \equal{} \min(c,d)$ from $ 2^a \plus{} 2^b \equal{} 2^c \plus{} 2^d$. this lemma will be used later on. it's proof is trivial.

[b]proof[/b]
consider numbers of the form $ a_n \equal{} 5\cdot 10^{2^n \minus{} 1} \minus{} (10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}} \plus{} 1)$.

It's a number with $ 2^n$ digits that are mostly $ 9$'s. We can calculat the exact value of $ s(a_n)$ the following way:
we have the number equal to $ 499....99 \minus{} (10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}})$ where each of $ n \minus{} 1$ "summands" of the form $ \minus{} 10^k$ turns one $ 9$ in the decimal representation into an $ 8$.
therefore $ s(a_n) \equal{} 1\cdot 4 \plus{} (n \minus{} 1)\cdot 8 \plus{} (2^n \minus{} 1 \minus{} (n \minus{} 1))\cdot 9 \equal{} 9\cdot 2^n \minus{} 4 \minus{} n$

Now we need to calculate $ a_n^2$. We have:
$ a_n^2 \equal{} 25\cdot 10^{2^{n \plus{} 1} \minus{} 2} \minus{} 2\cdot 5\cdot 10^{2^n \minus{} 1}(10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}} \plus{} 1) \plus{} (10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}} \plus{} 1)^2 \equal{} 25\cdot 10^{2^{n \plus{} 1} \minus{} 2} \minus{} 10^{2^n}(10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}} \plus{} 1) \plus{} (10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}} \plus{} 1)^2$
On this point note that $ 10^{2^{n \plus{} 1} \minus{} 2} \equal{} 10^{2^n}\cdot 10^{2^n \minus{} 2}$ and that for each $ i$, $ 1\leq i\leq n$ we have $ 10^{2^n}\cdot 10^{2^n \minus{} 2^i} \equal{} (10^{2^n \minus{} 2^{i \minus{} 1}})^2$ therefore we can simplify $ a_n^2$ to:
\[ 24\cdot 10^{2^{n \plus{} 1} \minus{} 2} \plus{} 1 \plus{} 2\cdot \prod_{1\leq j < k\leq n}(10^{2^n \minus{} 2^j}\cdot 10^{2^n \minus{} 2^k})\]
Now due to the lemma we now that $ 10^{2^n \minus{} 2^a}\cdot 10^{2^n \minus{} 2^b}\neq 10^{2^n \minus{} 2^c}\cdot 10^{2^n \minus{} 2^d}$ or otherwise that would contradict the lemma. Therefore $ a_n^2$ in a decimal representation can be seen as $ 24...1$ where between $ 4$ and $ 1$ there are some $ 2$s and all other zeroes. The number of $ 2$s is exactly the number of choices of $ 2$ elements out of $ n$.

Therefore $ s(a_n^2) \equal{} 2 \plus{} 4 \plus{} 1 \plus{} \binom{n}{2}\cdot 2 \equal{} n^2 \minus{} n \plus{} 7$ meaning that:
\[ \frac {s(a_n)}{s(a_n^2)} \equal{} \frac {9\cdot 2^n \minus{} 4 \minus{} n}{n^2 \minus{} n \plus{} 7}\]
so $ \lim_{n \to \infty}\frac {s(a_n)}{s(a_n^2)} \equal{} \infty$ and the expression is unbounded

[quote="Jure the frEEEk"]No, the expression is unbounded. I managed to produce the following counterexample after a lot of computer work. I have no idea how one could get such an idea without a computer in a couplle of hours time :wink: :blush:

$ a_n \equal{} 5\cdot 10^{2^n \minus{} 1} \minus{} (10^{2^n \minus{} 2} \plus{} 10^{2^n \minus{} 4} \plus{} \cdots \plus{} 10^{2^n \minus{} 2^{n \minus{} 1}} \plus{} 1)$. \[ \frac {s(a_n)}{s(a_n^2)} \equal{} \frac {9\cdot 2^n \minus{} 4 \minus{} n}{n^2 \minus{} n \plus{} 7}\]
so $ \lim_{n \to \infty}\frac {s(a_n)}{s(a_n^2)} \equal{} \infty$ and the expression is unbounded[/quote]

Very very nice demo! (and I carefully checked it)

I tried some numbers in this direction but did not see any solution.

Great congrats Jure!  :)

Thanks :)

Has anyone taught about how many generalisations this problem has. Just some ideas (not sure all are true:D)

$ \frac{s(n^2)}{s(n)}$ being unbounded

changing the base from $ 10$ to something else (i think the same argument work for bases $ 2k$ where $ k>1$). Is it true for base $ 2$?

$ \frac{s(n)}{s(n^3)}$ unbounded?

$ \frac{s(n)}{s(p(n))}$ unbounded? for any non-constant polinomial $ p$

[quote="Jure the frEEEk"]$ \frac {s(n)}{s(p(n))}$ unbounded? for any non-constant polinomial $ p$[/quote]

For this one, it's not true for all non-constant $ P(n)$ : Choose $ P(n)\equal{}10n$, for example  :wink:

I've done for over 8 hours but still can't solve it.
I can just give some examples:\
$49^2=2401,149^2=22201,14499^2=210221001,...$
it's hopeless,I think_

if one requires a similar problem tested earlier,see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222814&sid=3e12baf39d7ba34dcb3b3eb4565fc3ea#p1222814

Can't something close to $ n \approx 10^k \sqrt{10}$ be done? (I'm not sure)

[quote=WizardMath]
It is a known result that a non zero digit appears infinitely many times in the decimal expansion of $\sqrt{10}$ so we are done. (For eg see Sierpinski's paper).[/quote]

I think it's generally not true. Can you fully explain it?

[quote=WizardMath]Let $n=[10^k\sqrt{10}]$
where$[.]$ is the floor function.
It is a known result that a non zero digit appears infinitely many times in the decimal expansion of $\sqrt{10}$ so we are done. (For eg see Sierpinski's paper).[/quote]

No your proof is wrong. taking that $ n^2<10^{2k+1} $ and $ s(n^2) $ is really big. probably upper will be more suitable
and we don't even know how $ n^2 $s decimal expression will be. Whatever you have in mind, I think it might be wrong.
(Because if it was right, the proof would be at least 5 times longer than that. )

There does not exist such $c$.

Select $n = 10^{4k} + 2 \cdot 10^{3k} - 10^{2k} + 2 \cdot 10^k + 1$ for some large positive integer $k$. Then it is easy to see that we may make $s(n)$ as large as we wish while
\[s(n^2) = s(10^{8k} + 4 \cdot 10^{7k} + 2 \cdot 10^{6k} + 11 \cdot 10^{4k} + 2 \cdot 10^{2k} + 4 \cdot 10^k + 1) \le 16\]
when $k$ is large. This implies the conclusion.


calculus
integration
function
calculus computations

[quote="cmad"]Calculate $ I = \int_{0}^{2}( \: \sqrt{x^{3}+1}+\sqrt[3]{x^{2}+2x}\: ) dx$[/quote]
[color=darkblue][b]Lemma.[/b] Let $ f: [a,b]\rightarrow [c,d]$ be a continue, increasing and bijective function. Then $ \int_{a}^{b}f(x)\ dx+\int_{c}^{d}f^{-1}(x)\ dx=bd-ac$.

[b]Proof of the proposed exercise.[/b] Observe that the function $ \{\begin{array}{c}f: [0,2]\rightarrow [1,3]\\\\ f(x)=\sqrt{x^{3}+1}\end{array}$  is continue, increasing, bijective.

The its inverse function is $ \{\begin{array}{c}f^{-1}: [1,3]\rightarrow [0,2]\\\\ f^{-1}(x)=\sqrt [3]{x^{2}-1}\end{array}$ for which $ f^{-1}(x+1)=\sqrt [3]{x^{2}+2x}$.

Therefore, $ I=\int_{0}^{2}f(x)\ dx+\int_{0}^{2}f^{-1}(x+1)\ dx=$  $ \int_{0}^{2}f(x)\ dx+\int_{1}^{3}f^{-1}(x)\ dx=6$.

[b]Remark (generally).[/b] Function $ f: [a,b]\rightarrow [a+\delta, b+\delta]$ is continue, increasing and bijective $ \implies$ $ \boxed{\int_{a}^{b}[f(x)+f^{-1}(x+\delta )]\ dx=(b-a)(b+a+\delta )}$.[/color]


What is an example of an equidistributed sequence in [0,1]?

How do you prove that it is an equidistributed sequence?


A jet of water from a nozzle 2.5 cm in diameter has a velocity of 30m/s and strikes a flat plate held normal to the jet and moving away from the jet at 6m/s.
a)Compute the force exerted by the jet on the moving plate
b) What is the force if the plate is moving toward the jet at 6m/s?Water has a density of 10^3kg/m^3

Solution.

It helps enormously to change our frame of reference to the rest frame of the moving plate. So by Galilean relativity, questions (a) and (b) are in fact the same problem - a stationary plate, with water jet coming at $ 24 ms^{ \minus{} 1}$ and $ 36 ms^{ \minus{} 1}$ respectively. So we can tackle the problem with a general water jet velocity $ v$.

Now consider an element of volume $ dV$ of water arriving at the plate. Let the corresponding elements of mass and length of this volume be $ dm$ and $ dl$ respectively. Further let the time taken for this volume element to hit the plate (from when the front hits the plate, until when the back hits the plate) be $ dt$.

$ dm \equal{} \rho dV \equal{} 10^3dV \equal{} 10^3 \pi r^2 dh \equal{} 10^3 \pi (1.25 \times 10^{ \minus{} 2})^2 dh \equal{} 0.49087vdt$

Now if we let $ dV$ tend towards zero, we can treat this element as a "rigid" body.

$ p_{before} \equal{} vdm \equal{} 0.49087 v^2 dt$
$ p_{after} \equal{} 0$

$ I \equal{} \Delta p \equal{} 0.49087 v^2 dt$

$ F \equal{} \frac {I}{dt} \equal{} 0.49087 v^2$

Substitution of $ v \equal{} 24ms^{ \minus{} 1}$ yields: $ F \equal{} 2.8 \times 10^2 N$
Substitution of $ v \equal{} 36ms^{ \minus{} 1}$ yields: $ F \equal{} 6.4 \times 10^2 N$

Very nice solution sludgethrower! Your solution is exactly the same as my friend's. I see a strange phenomenon:Most people can't understand what this problem tries to say, including me :blush: . I'm curious why the way physics problems and its phenomenons are expressed in English is so difficult to understand, at least to me?

Apart from very basic German, I do not know how to speak any other language. So, sadly, I am probably not the best person to answer your question. However, I do believe that problems should be stated rigourously, and without ambiguity, like this one.


now find $\displaystyle\sum_{k=n}^{m} {k\choose n}$.
i think it's quite a nice thing :)

[quote="sheepwarrior"][quote="liyi"]Use Pascal's Identity
${n\choose m}+{n\choose m+1} = {n+1\choose m+1}$[/quote]

can u explain how this relates?? sry  :blush:[/quote]
add them togather the lots of them cancel out.

$\displaystyle {2\choose2} = {3\choose3}$ :D
so you have that $\displaystyle {2\choose2} + {3\choose2} + \cdots + {n\choose 2} = {4\choose3} + {4\choose2} + \cdots {n\choose2}$. now you can use induction ;)
same trick for the more general question i've posed...

so is it ${n+1\choose3}$?

and for the general that ma_go posted is it
${m+1\choose{n+1}}?$

for my original problem, if you observe the Pascal Triangle, you will see it is actualy to find:
\[1+3+6+10+15+21......\]
The same trick still works for ma_go's general question, i think.


The exponential function $ f(x)$ is defined by $ f(x) \equal{} a\cdot 2^{bx}$ and contains the points (0,3) and (1,24). Find the sum of $ a$ and $ b$.

Since anything multiplied by $ 0$ is $ 0$, and anything to the $ 0$ power is $ 1$, $ 2^{bx} \equal{} 1$ if $ x \equal{} 0$. Thus, if $ f(x) \equal{} 3$, then $ a$ must be $ \frac {3}{1}$, or simply $ 3$. Then, we plug in the second point, $ (1,24)$ into the function and solve for $ b$.

$ f(x) \equal{} 3\cdot2^{bx}$
$ 24 \equal{} 3\cdot2^{b}$ (substitute: $ 24 \implies f(x)$, $ 1 \implies x$)
$ 8 \equal{} 2^b$
$ b \equal{} 3$

Thus, $ a$ and $ b$ are both $ 3$, and the sum of $ a$ and $ b$ would be $ 3 \plus{} 3 \equal{} 2(3) \equal{} \fbox{6}$.


I read a while ago about a mathematical proof showing that democracy is a bad form of government.

Was this just a dream or what? could someone link me? thanks.

[quote="PenguinIntegral"]Move this to the round table.[/quote]

i believe this should go in fun in games...lol, these types of math proofs for real world things are just jokes

It can't logically happen. A lot of times there is a subtle fallacy in the proof that allows you to make new and untrue statements. For example, I  once saw a proof showing that Winston Churchhill is a carrot. Its really stupid but I guess its funny. It went like this:

Let $a=b=1$. Then $b^2=ab$. Also $a^2=a^2$ so $a^2-b^2=a^2-ab$ by subtracting the first equation from the second. Factor and divide by $a-b$ to get $a+b=a$. Thus $b=1=0$.

We know winston churchhill has $0$ leafy tops. But $0=1$ so we can say winston churchhill has $1$ leafy top. Multiply $0=1$ by $2$ to get $0=2$. Churchill has 2 legs, and therefore no legs. Multiply $0=1$ by churchhill's waist size so that (churchhill's waistsize)=$0$. So churchhill tapers to a point.

Take a photon that is hitting winston churchhill. Multiply $0=1$ by the photon's wavelength. So (churchhill's photon's wavelength)=$0$. Multiply $0=1$ by $640$ nanometers, the wavelength of orange light. So $640=0$. Then (churchhill's photon's wavelength)=$640$. So winston churchhill is orange.

So winston churchill has no legs or head, tapers to a point, is orange, and has a leafy top: winston churchhill is a carrot.

Do you see the fallacy in the proof?

[quote="drunner2007"]Do you see the fallacy in the proof?[/quote]

[hide]  You cannot divide by $a-b=0$.[/hide]


Can you prove the following theorem?
The number of factors of p in n! (p prime, n positive int) is $(n-s_n)/(p-1)$ where $s_n$ is [i]the sum of the digits of n when expressed in base p[/i].

This is due to Legendre.
Let $n = n_0 + n_1p + \cdots + n_kp^k$ be the expansion of $n$ in base $p$.

It is well-known that the valuation of $n!$ modulo $p$, say $\nu_p (n!)$ (that is the exponent of $p$ in the prime decomposition of $n!$) is : $\dsp \nu_p(n!) = \sum_{i=1}^{+ \infty} [ \frac n {p^i} ]$,
where $[.]$ denotes the integer part.

It easily follows that
$\nu_p (n!) = (n_1 + n_2p \cdots + n_kp^{k-1}) + (n_2 \cdots + n_kp^{k-2})  + \cdots + (n_k) $
$= n_1 + n_2(1+p) + n_3(1+p+p^2) + \cdots + n_k(1+p+ \cdots p^{k-1})$
$ = \sum_{i=1}^k n_i \frac {p^i - 1} {p-1} $
$= \sum_{i=0}^k n_i \frac {p^i - 1} {p-1}$
$ = \frac {n-s_p(n)} {p-1}$,
as desired.

Pierre.


Happy Mole Day everyone!

How did you celebrate? I got to go out to IHOP at 6:02 am with my school's chemistry classes, and then my AP chemistry class had a party. Our class got to "mole" other classes by bursting in the door, throwing moles, and shouting "$ 6.02\times10^{23}$!"

Oh, haha, I wasn't even aware there was a mole day.

What is mole day?Why is it celebrated?Is it a worldwide event?

I thought it was Oct 23. I noticed you posted on Oct 24. I have heard that Mole lovers cheer at 6:02 AM/PM. Do people really wake up early on mole day to celebrate at 6:02 AM?  :lol:


In triangle $ABC$ we have $BC$ = 4 and the medians from $A$ and $B$ are perpendicular to each other. Find the biggest possible value of the height from $A$ to $BC$.

"medians from A and B perp. to each other" strongly suggests vectors.

Put B as the origin, 4 as vector BC and a as vector BA.  Then $\frac{a+4}{2}$ is the median from B, and $\frac{-a + (4-a)}{2}$ is the median from A.  Their dot product is 0, so:  (where $a_x$ is the real part of a, and $a_y$ the imag. part of a)

$(4+a_x)(4-2a_x) + (a_y)(-2a_y) = 0$

We want to maximize $a_y$.  Then $a_y^2 = (4+a_x)(2-a_x)$ has a maximum when $a_x = -1$, at $a_y^2 = 9$.  So $a_y = 3$ is the biggest height.


number theory
combinatorics unsolved
combinatorics

Let p is prime.Prove that $p|C_n^{p}-[\dfrac{n}{p}]$

What does that mean...for me it has no sense...can you explain it

First, note that if $p>n$ then $C_n^p = [\frac n p ] = 0$ so that the result is obvious.
Let $n \geq p$ and $n=pq+r$ where $r \in \{0,1, \cdots , p-1 \}.$
Note that $q = [\frac n p ].$

Then $C_n^p = \dfrac{n(n-1) \cdots (n-p+1)} {p \times (p-1)!}.$ Among the $p$ consecutive integer which form the numerator exactly one is divisible by $p$, namely $n-r = pq$. Thus :
$C_n^p = q \dfrac{n(n-1) \cdots (n-r+1)(n-r-1) \cdots (n-p+1)} { (p-1)!}.$
Now, using Wilson's theorem, we deduce that the numerator and the denominator are both equal to $-1 \mod [p].$
Therefore $C_n^p = q \mod [p]$ and we are done.

Pierre.


inequalities
induction
number theory unsolved
number theory

let d(n) be the number of the multiples of n for every integer

find all integers like n such that there exists no perfect square in the following numbers

n , d(n), d(d(n)), d(d(d(n))), ...

I assume d(n) is the number of [i]divisors[/i] of n.

If $n=\Pi_i p_i^{a_i}$, then $d(n)=\Pi_i (a_i+1)$, and I claim that for $n>1$, $d(n)\le n$ with equality if and only if n=2. This follows from the simple fact that $p_i^{a_i}\ge 2^{a_i}\ge a_i+1$ (where the second inequality is easily proved with induction), with equality only when $p_i=2$ and $a_i=1$. Thus, for $n>1$, the sequence $n,d(n),d(d(n)),\ldots$ steadily decreases until it reaches 2, at which point it becomes fixed. We now consider [i]when[/i] it reaches 2. (If $n=1$ then there is a square in the list, namely 1.)

If $n=2$, then the sequence is fixed from the beginning, and there are no squares in the list.

If $d(n)=2$, then $\Pi_i^k (a_i+1)=2$ (where k is the number of prime factors), which implies that $k=1$, $a_1=1$, and so n is equal to a prime p. So the sequence goes $p,2,2,\ldots$, and again there are no squares.

Now assume that the sequence $n,d(n),d(d(n)),\ldots$ first reaches 2 at the $m^{th}$ term with $m>2$. (Notice that this happens if and only if $n$ is composit.) Call the the $(m-2)^{th}$ term $z$, so that $d(d(z))=2$ and $d(z)\neq2$. As before, since $d(d(z))=2$, we must have $d(z)$ equal some prime p, and since $d(z)\neq2$, p is an odd prime. Since $d(z)$ is odd, $z$ is a perfect square.

So the sequence $n,d(n),d(d(n)),\ldots$ has no perfect squares if and only if $n$ is prime.


induction
search
modular arithmetic
linear algebra
matrix
number theory
Diophantine equation

A positive integer $ n$ is said to be [i]cool[/i] if there exists a positive integer $ N$ such that $ N^2$ can be written as the sum of the squares of $ n$ consecutive positive integers. Is it true that there are infinitely many positive integers which are cool? Justify your answer. (As an example, the number $ n\equal{}2$ is cool since $ 3^2 \plus{} 4^2 \equal{} 5^2$)

[hide="Hidden by mod"]Lets start with the fact that for an infinite number of integers $ n$, there exists a Pythagorean triples $ 0 < x < n$ such that $ x^2 \plus{} (n \minus{} 1)^2 \equal{} n^2$ (Proof: rearranging, $ x^2 \equal{} 2n \minus{} 1$.  Since the square of any odd number is an odd number, and there are an infinite number of odds, there are an infinite number of odd squares and thus an infinite number of $ (x,n)$ pairs, one for every odd $ x$).

Now let $ a \equal{} n \minus{} 1$ and $ b \equal{} n$.  Hence we have from above that $ x^2 \equal{} b^2 \minus{} a^2$ for an infinite number of numbers.  Assume $ x^2 \equal{} \sum_{i \equal{} 1}^k y_i^2$.   The initial case, $ k \equal{} 2$, is equivalent to the fact that there are an infinite number of Pythagorean triples (one particular set of Pythagorean triples is shown to be infinite above).  Assume that the square of some odd number $ x$ can be written as the sum of $ j$ perfect squares.  Then, for every odd $ x$ we can write $ \sum_{i \equal{} 1}^j y_i^2 \plus{} a^2 \equal{} b^2$ for consecutive numbers $ a$ and $ b$; hence $ x^2 \equal{} \sum_{i \equal{} 1}^j y_i^2 \implies b^2 \equal{} \sum_{i \equal{} 1}^{j \plus{} 1} z_i^2$ for odd $ x$; Thus, by induction, all positive integers greater than or equal to $ 2$ are cool.

For instance, one possible series
$ 3^2 \plus{} 4^2 \equal{} 5^2$
$ 3^2 \plus{} 4^2 \plus{} 12^2 \equal{} 13^2$
$ 3^2 \plus{} 4^2 \plus{} 12^2 \plus{} 84^2 \equal{} 85^2$
$ 3^2 \plus{} 4^2 \plus{} 12^2 \plus{} 84^2 \plus{} 3612^2 \equal{} 3613^2$
$ 3^2 \plus{} 4^2 \plus{} 12^2 \plus{} 84^2 \plus{} 3612^2 \plus{} 6526884^2 \equal{} 6526885^2$... et cetera[/hide]

[quote="aidan"]A positive integer $ n$ is said to be [i]cool[/i] if there exists a positive integer $ N$ such that $ N^2$ can be written as the sum of the squares of $ n$ [b]consecutive[/b] positive integers. Is it true that there are infinitely many positive integers which are cool? Justify your answer. (As an example, the number $ n \equal{} 2$ is cool since $ 3^2 \plus{} 4^2 \equal{} 5^2$)[/quote]

In that vein, it's easy to show that $ 4$ is not cool- the sum of four consecutive perfect squares is congruent to $ 2$ mod $ 4$, and can't be a square.

oops I forgot about that word... :-(

So lets try again... let $ a$ be the smallest of the $ n$ consecutive integers, so we have the Diophantine equation $ a^2\plus{}(a\plus{}1)^2\plus{}(a\plus{}2)^2 \plus{} \cdots \plus{} (a\plus{}n\minus{}1)^2 \equal{} b^2$ which simplifies $ na^2 \plus{} 2a\sum_{i\equal{}0}^{n\minus{}1}i \plus{} \sum_{i\equal{}0}^{n\minus{}1}i^2 \equal{} b^2 \iff na^2 \plus{}n(n\minus{}1)a \plus{} (n\minus{}1)n(2n\minus{}1)/6 \equal{} b^2$.

Now, completing the square: $ n(a\plus{}\frac{n\minus{}1}{2})^2 \plus{} \frac{(n\minus{}1)n(2n\minus{}1)}{6} \minus{} \frac{n(n\minus{}1)^2}{4} \equal{} b^2$

Just wondering... is there any theorem that the Diophantine equation $ a^2\minus{}xb^2\equal{}y$ only has solutions for certain $ y$ values?  ($ y\equal{}\pm 1$ gives Pell's equation, but are others solvable?)

anyone is continuing the proof?i dun know the answer but i want to know it. :)

[quote="facis"]is there any theorem that the Diophantine equation $ a^2 \minus{} xb^2 \equal{} y$ only has solutions for certain $ y$ values?  ($ y \equal{} \pm 1$ gives Pell's equation, but are others solvable?)[/quote]

Asked recently [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=916458739&t=213317]here[/url].  The full answer is quite complicated.

So lets rewrite $ \frac {(n \minus{} 1)n(2n \minus{} 1)}{6} \minus{} \frac {n(n \minus{} 1)^2}{4} \equal{} b^2 \minus{} n(a \plus{} \frac {n \minus{} 1}{2})^2$.  So we can rewrite it as $ b^2 \minus{} n/4(2a \plus{} n \minus{} 1)^2 \equal{} \frac {(n \minus{} 1)n(n \plus{} 1)}{12}$.

So now we may have a pell-like equation which only has integer coefficients when $ n \equiv 0 \pmod 4$.  But again manipulating it:

$ (2b)^2 \minus{} n(2a \plus{} n \minus{} 1)^2 \equal{} \frac{(n \minus{} 1)n(n \plus{} 1)}{3}$

So now we are interested in solutions to the equation $ b^2\minus{}xa^2 \equal{} c$ with $ b$ even and $ a$ and $ n$ with the opposite parity.

All integers of the form $ f^2=(24k+11)^2$ for $ k \in \mathbb N$ are cool. Really,
$ \sum_{j=\frac{f^3-11f^2-11f+25}{24}}^{\frac{f^3-11f^2-11f+25}{24}+f^2-1} j^2 = (\frac{f(f^3+f^2+13f-11)}{24})^2$
and limits of the sum are natural.

Maybe, the solution seems like black magic. :) So let's explain, how to figure out these formulas.
Let's seek for the natural numbers $ m,n,N$, which satisfy the equation: $ (m-(n-1)/2)^2+...+(m+(n-1)/2)^2 = N^2$.
As it is known, $ 1^2+2^2+...+k^2=k(k+1)(2k+1)/6$, so our equation transforms to: $ n(m^2+(n^2-1)/12)=N^2$.
Let's search for $ n$ in the form $ n=f^2$. Then we must get:
$ f^2(m^2+(f^4-1)/12)=N^2$, or
$ 12(N/f+m)(N/f-m)=f^4-1$.
We may try to find such solutions, for which:
$  \{ \begin{matrix} 12(N/f+m)=f^3+f^2+f+1 \\ (N/f-m)=f-1. \end{matrix}$
We may try some other decompositions, but not each gives an acceptable result.

Solving the mentioned system relatively to $ m$ and $ N$, we get:
$ m=(f^3+f^2-11f+13)/24$,
$ N=f(f^3+f^2+13f-11)/24$.
For $ f=24k+11$, where $ k \in \mathbb N$, the numbers $ m,n,N$ are natural. Even more, we have $ m-(n-1)/2>0$, which is required for our task. Hence $ n=f^2=(24k+11)^2$ is cool.

[hide="Solution"]We suspect that there are infinitely many cool integers, so we will try to find an infinite family of them.  Suppose $ n$ is cool.  First assume $ n$ is odd, and let $ n\equal{}2m\plus{}1$.  Then we can write

$ (a\minus{}m)^{2}\plus{}(a\minus{}m\plus{}1)^{2}\plus{}\cdots \plus{}(a\plus{}m)^{2}\equal{}k^{2}$

Expanding gives $ (2m\plus{}1)a^{2}\plus{}2(1^{2}\plus{}2^{2}\plus{}\cdots \plus{}m^{2})\equal{}k^{2}$.  Using the sum of squares formula, we get $ (2m\plus{}1)a^{2}\plus{}\frac{m(m\plus{}1)(2m\plus{}1)}{3}\equal{}k^{2}$.  We want to factor out $ 2m\plus{}1$, so assume $ m$ is divisible by 3 so that the second term on the left side of the equation is divisible by $ 2m\plus{}1$.  Factoring, we get

$ (2m\plus{}1)\left(a^{2}\plus{}\frac{m(m\plus{}1)}{3}\right)\equal{}k^{2}$

We already ensured that $ \frac{m(m\plus{}1)}{3}$ is an integer.  Assume $ 2m\plus{}1$ is a perfect square.  Then it is sufficient for $ a^{2}\plus{}\frac{m(m\plus{}1)}{3}$ to be a perfect square.  We can find $ a$ for which is is true as long as $ \frac{m(m\plus{}1)}{3}$ is not 2 modulo 4, and then we'll have a good integer $ n\equal{}2m\plus{}1$.  To make sure $ \frac{m(m\plus{}1)}{3}$ is not 2 modulo 4, assume $ m$ is divisible by 4.

Now we just have to make sure all our assumptions can be true simultaneously and we'll get our infinite family.  We want $ m$ to be divisible by 3 and 4, and we want $ 2m\plus{}1$ to be a perfect square.  In other words we want $ n$ to be 1 modulo 24 and also a perfect square.  Our infinite family is $ n\equal{}(24j\plus{}1)^{2}$ for all integers $ j>1$.  You can show that all these integers are cool basically by reversing the process above.[/hide]


A wheel has a diameter of $ \frac9{\pi}$ feet. How many revolutions does the wheel make when rolled for 60 yards?

Well its circumference is 9 ft or 3 yds. 60/3=20

That's not correct:
[hide="My Solution"]
The circumference is $\frac{9}{\pi}\cdot\pi=9$, but notice that it is 60 [b]yards[/b]. Therefore the answer is

$180\div9=\boxed{20\text{ revolutions}}$
Thank you.
[/hide]

Both of your solutions are correct, Mewto converted the circumference and you converted the distance rolled.


If a and b are the solutions to the equation $x^2 - 5x + 9 = 0$, what is the value of
(a-1)(b-1)?

[quote="236factorial"]Where did it say that ab = 0?[/quote]

Sorry, I meant 9 :blush:

[quote="kunny"]We can write $x^2-5x+9=(x-a)(x-b)$.Plugg $x=1$ into this equality we have $(1-a)(1-b)=1^2-5\cdot 1+9=5$[/quote]

Multiplying by $(-1)^2$, we have $-1(1-a) \cdot -1(1-b)=5 \Rightarrow (a-1)(b-1)=5.$

[hide="Answer"]The roots will be in the form $\frac{5\pm\sqrt{x}}{2}$, but we do not need to find x, as will be shown later. $(a-1)(b-1)=ab-a-b+1$, so we have $9-\frac{5+\sqrt{x}}{2}-\frac{5-\sqrt{x}}{2}+1=5$.[/hide]

[quote="chess64"][quote="kunny"]We can write $x^2-5x+9=(x-a)(x-b)$.Plugg $x=1$ into this equality we have $(1-a)(1-b)=1^2-5\cdot 1+9=5$[/quote]

Multiplying by $(-1)^2$, we have  :? $-1(1-a)-1(1-b)=5 \Rightarrow $(a-1)(b-1)=5.$[/quote]

Hmm, about 3 different approaches. I think mine is easiest :D

Actually, our approaches are the same, but I explained more than you did. The explanation isn't necessary when solving the problem in a competition, but I thought it would be useful for some here who don't know how to do it.

nice job kunny, that is the intended approach and if you see that now then problems from higher level math will be easier!

another solution is:

(a-1)(b-1) = ab -(a+b) + 1= 9-5+1= 5.

Thank you, Barnacle.

In my opinion, I don't think the relation between root and coefficient (vieta's formula) is not always need .

Because it is important that you can expand rapidly the following equatiuon.

1. $(x-a)(x-b)$

2. $(x-a)(x-b)(x-c)$

3. $(x-a)(x-b)(x-c)(x-d)$

kunny

[quote="kunny"][quote="chess64"][quote="Barnacle"](a-1)(b-1) = ab -(a+b) + 1= 9-5+1= 5.[/quote]

 :roll: Me and JF197 already said that! :rotfl:

Kunny, why do you need to know how to expand (3)? :?[/quote]

Let's solve the following problem.

Let $a,b,c,d$ be the roots of the quartic equation $3x^4-2x^3-6x+5=0$.

Find the value of $\left(a-\frac{2}{3}\right)\left(b-\frac{2}{3}\right)\left(c-\frac{2}{3}\right)\left(d-\frac{2}{3}\right)$

kunny[/quote]

Well, I don't think that I will need to know that. It's easy but complex. Mathcounts won't make you do that.

[quote="236factorial"][/quote][quote="kunny"][quote="chess64"][quote="Barnacle"](a-1)(b-1) = ab -(a+b) + 1= 9-5+1= 5.[/quote]

 :roll: Me and JF197 already said that! :rotfl:

Kunny, why do you need to know how to expand (3)? :?[/quote]

Let's solve the following problem.

Let $a,b,c,d$ be the roots of the quartic equation $3x^4-2x^3-6x+5=0$.

Find the value of $\left(a-\frac{2}{3}\right)\left(b-\frac{2}{3}\right)\left(c-\frac{2}{3}\right)\left(d-\frac{2}{3}\right)$

kunny[/quote][quote="236factorial"]

Well, I don't think that I will need to know that. It's easy but complex. Mathcounts won't make you do that.[/quote]

You're not going to be in MATHCOUNTS forever! Aren't you in grade 8, anyway?

[quote="236factorial"][/quote][quote="kunny"][quote="chess64"][quote="Barnacle"](a-1)(b-1) = ab -(a+b) + 1= 9-5+1= 5.[/quote]

 :roll: Me and JF197 already said that! :rotfl:

Kunny, why do you need to know how to expand (3)? :?[/quote]

Let's solve the following problem.

Let $a,b,c,d$ be the roots of the quartic equation $3x^4-2x^3-6x+5=0$.

Find the value of $\left(a-\frac{2}{3}\right)\left(b-\frac{2}{3}\right)\left(c-\frac{2}{3}\right)\left(d-\frac{2}{3}\right)$

kunny[/quote][quote="236factorial"]

Well, I don't think that I will need to know that. It's easy but complex. Mathcounts won't make you do that.[/quote]

How can you solve this? :)

[quote="chess64"][quote="236factorial"]

Well, I don't think that I will need to know that. It's easy but complex. Mathcounts won't make you do that.[/quote]

You're not going to be in MATHCOUNTS forever! Aren't you in grade 8, anyway?[/quote]

No, I'm in grade 7, and I do know how to expand that, except I'm not interested right now.  :P

Aagh! Stop quoting everything! You can delete stuff from inside the quote, you know... try just quoting the most recent thing the last person said.


is there any way to get some harder problems?
or maybe some probles of different topics?

You have to pass the subjects you're currently working on.

No, the problem is that there are only so many problems in Alcumus, and hard problems are by far harder to make and find.

It depends on what's a hard problem for you. The problems I'm getting right now seem hard to me, but probably are easy for a lot of people.

All current problems in Alcumus are Introduction to Counting and Probability level, thus most appropriate for those in the 6th to 9th grade range or to those who have not mastered those topics (which is most kids and adults in America.)

We have no estimated time frame for additional problems. Next on the list will likely be Introduction to Algebra level, so if you find the current problems too easy, you'd probably also find those too easy. For more difficult problems, I suggest our intermediate series textbooks available in our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]Bookstore[/url].


[color=darkblue]The sequence $ (a_n)_{n \in \mathbb{N}}$ is defined as follows:
\[ a_n \equal{} \dfrac{2}{3 \plus{} 1} \plus{} \dfrac{2^2}{3^2 \plus{} 1} \plus{} \dfrac{2^3}{3^4 \plus{} 1} \plus{} \ldots \plus{} \dfrac{2^{n \plus{} 1}}{3^{2^n} \plus{} 1}
\]
Prove that $ a_n < 1$ for any $ n \in \mathbb{N}$[/color]


geometry
geometric transformation
reflection
incenter
trigonometry
power of a point
trig identities

In triangle ABC, $ \angle A\equal{}90$, $ \angle B\equal{}30$, and $ AB\equal{}6$. D is on AB such that a semicircle with diameter AD and center E is tangent to BC at F. Let CE and AF intersect at G. Let DG intersect the semicircle at H. Find GH.

[hide]
Let $ C', F'$ be reflection of $ C, F$ respectively across $ \overline{AB}$; then $ E$ is the incenter of equilateral $ \triangle BCC'$. Then $ E$ is also the centroid, so $ AE \equal{} EF \equal{} 2$, and $ AG \equal{} GF \equal{} 6 / \sqrt {3} / 2 \equal{} \sqrt {3}$. $ \triangle AFF'$ is the medial triangle so $ \angle BAF \equal{} 30^{\circ}$, and if $ G'$ is the projection of $ G$ onto $ \overline{AB}$, we have $ GG' \equal{} \sqrt {3}/2,\ AG' \equal{} 3/2,\ G'D \equal{} 4 \minus{} AG' \equal{} 5/2,\ GD \equal{} \sqrt {G'D^2 \plus{} G'G^2} \equal{} \sqrt {7}$. Then Power of a Point on $ G$ gives $ GH \equal{} \frac {AG \cdot FG}{DG} \equal{} \frac {3}{\sqrt {7}}$. [/hide]

By the way, I've been doing all your questions (though my bashings are too ugly too post), and they are nice.  :)

Nice solution, more elegant than mine  :)
[hide="My Solution"]
Let DH intersect AC at I. By 30-60-90 triangles the radius of the semicircle is 2, $ AC \equal{} 2 \sqrt {3}$, $ AG \equal{} GF \equal{} \sqrt {3}$, $ CG \equal{} 3$, $ GE \equal{} 1$. Now by Menelaus on $ \triangle AID$,
$ \frac {ED}{AD} \cdot \frac {AI}{IC} \cdot \frac {CG}{GE} \equal{} 1$.
Since $ CI \plus{} IA \equal{} 2 \sqrt {3}$, it follows that $ AI \equal{} \frac {4\sqrt {3}}{5}$ and $ IC \equal{} \frac {6\sqrt {3}}{5}$. So by Pythag on $ \triangle AID$ we have $ ID \equal{} \frac {8\sqrt {7}}{5}$. By power of a point, $ IA^2 \equal{} IH \cdot ID$, so $ IH \equal{} \frac {6\sqrt {7}}{35}$, so $ HD \equal{} ID \minus{} IH \equal{} \frac {10\sqrt {7}}{7}$. But $ \angle GED \equal{} 120$, so by the law of cosines, $ GD \equal{} \sqrt {7}$, hence $ HG \equal{} HD \minus{} GD \equal{} \frac {3\sqrt {7}}{7}$. [/hide]


Find all $p>2$ primes such that  $(p,n^n-3n+1)=1$ for all $n>2$

 ;)

Umm. This prime should not divide this expression for all $n$ or for one certain $n$?

Oh, you've corrected the statement :).

Ok, suppose that $p \equiv 2 \mod{3}$.
Then, take: $n=\frac{(p-1)(p-2)}{3}$ to obtain that:
$n^n - 3n + 1 \equiv (\frac{(p-1)(p-2)}{3})^{\frac{(p-1)(p-2)}{3}} - (p-1)(p-2) + 1 \equiv 1 - 2 + 1 \equiv 0 \mod{p}$
So, in case this case there are no solutions (I am using Fermat's Little Theorem here).
Now, suppose that $p \equiv 1 \mod{3}$ and take: $n=\frac{(p-1)(2p-2)}{3}$. We see that:
$n^n - 3n + 1 \equiv (\frac{(p-1)(2p-2)}{3})^{\frac{(p-1)(2p-2)}{3}} - (p-1)(2p-2) + 1 \equiv 1 - 2 + 1 \equiv 0 \mod{p}$
Just like before there are no solutions.
We are left with the case $p=3$. It's very easy to verify that $3|5^5 - 3*5 + 1$ so $p=3$ also doesn't work.

[quote="elchuy"]Is for all $n>2$

[quote="mathmanman"]Indeed, this is unclear, since, if we take $p=n=3$, we're done...

Maybe he meant "divides" instead of "doesn't divide"?  :?[/quote]
$gcd(3, 3^3-3(3)+1)=gcd(3, 19)=1$[/quote]
The previous statement was wrong.
Now he has corrected it, and [b]TomciO[/b] has solved it. ;)

Ok,

$n^n - 3n + 1 \equiv \left(\frac{(p-1)(p-2)}{3}\right)^{\frac{(p-1)(p-2)}{3}} - (p-1)(p-2) + 1 \equiv 0 \mod{p}$

How you applied the Fermat's Little Theorem here?

 :oops:

$(\frac{(p-1)(p-2)}{3})^{\frac{(p-1)(p-2)}{3}} \equiv 1 \mod{p}$
because the exponent is divisible by $p-1$.


The perfect squares 1, 4, 9, 16, 25, $ \ldots$ are put into a string of digits 1491625$ \ldots$. What is the $ 40^{\mathrm{th}}$ digit in the string?

hello, the $ 40^{th}$ number is $ 3$.
Sonnhard.


function
trigonometry
limit
integration
real analysis
real analysis unsolved

Let $ F(x) \equal{} x^2 \cdot \sin(\frac {1}{x^2})$, with $ 0 \equal{} F(0)$.  Show that $ F'$ is not integrable on $ [ \minus{} 1,1]$.  This reduces to showing that $ \cos(\frac {1}{x^2})/x$ is not (absolutely) integrable on $ [ \minus{} 1,1]$.

Make the change of variables $ t\equal{}\frac{1}{x^2}$.

I don't think that works.  $ F'$ not absolutely integrable, yet $ \lim_{e \rightarrow 0} F(e) \rightarrow 0$.  I don't think finding the antiderivative is the right way to go, in this case.

Define $ x_k$ by $ \frac1{x_k^2}\equal{}\left(k\plus{}\frac12\right)\pi.$

Then $ \int_{x_{k\plus{}1}}^{x_k}|F'(x)|\,dx\ge\left|\int_{x_{k\plus{}1}}^{x_k}F'(x)\,dx\right|\equal{} |F(x_k)\minus{}F(x_{k\plus{}1})|.$

Since $ F(x_k)$ and $ F(x_{k\plus{}1})$ have opposite signs, we have $ |F(x_k)\minus{}F(x_{k\plus{}1})|\ge|F(x_k)|\equal{}x_k^2.$

Now we need to estimate $ x_k.$ We see that $ x_k\approxeq \frac{c}{\sqrt{k}}$ so that $ x_k^2$ is comparable to $ \frac1k.$

Hence $ \int_0^{x_1}|F'(x)|\,dx\equal{}\sum_{k\equal{}1}^{\infty}\int_{x_{k\plus{}1}}^{x_k}|F'(x)|\,dx\ge \sum_{k\equal{}1}^{\infty}x_k^2$

and that sum diverges by comparison to the harmonic series.

The same proof also directly shows that $ F$ is not of bounded variation. One technical note: the points $ x_k$ are not exactly the maxima and minima of $ F,$ but there's close enough to that to accomplish what we want.

I don't see your problem. $ \int_0^1 \left| \frac {\cos(\frac {1}{x^2})}{x}\right|dx \equal{} \left<{t \equal{} \frac {1}{x^2}}\right> \frac 12\int_0^{\infty}\left| \frac {\cos(t)}{t}\right|dt$.


induction
calculus
inequalities proposed
inequalities

First, \[
\begin{gathered}
  a_1^3  + a_2^3  \leqslant \frac{1}
{n}\left( {a_1  + a_2 } \right)^2  \hfill \\
   \Leftrightarrow a_1^2  - a_1 a_2  + a_2^2  \leqslant \frac{1}
{n}\left( {a_1  + a_2 } \right) \hfill \\
   \Leftarrow a_1^2  \leqslant \frac{1}
{n}a_1 {\text{ }}and{\text{ }}a_2 \left( {a_2  - a_1 } \right) \leqslant \frac{1}
{n}a_2  \hfill \\
\end{gathered}
\]
which is true.
Then,
\[
a_1^3  + a_2^3  + ... + a_n^3  \leqslant a_1^3  + \left( {a_2  + ... + a_n } \right)^3  \leqslant \frac{1}
{n}\left( {a_1  + a_2  + ... + a_n } \right)^2
\]

[quote="siuhochung"]\[a_1^3  + \left( {a_2  + ... + a_n } \right)^3  \leqslant \frac{1}
{n}\left( {a_1  + a_2  + ... + a_n } \right)^2
\][/quote]

This is not true. Consider n=3, $a_{i}= \frac{i}{n}$. Then LHS=126/27>36/27=RHS.

yehy.my solution is like $E. Leung$,but use the induction directly. :)
and I filmize the inquality  from an calculous problem. :)


Is it a necessity to have Ph.D in Mathematics in order to obtain a job as an olympiad coach, or IMO team leader?

I don't think so. As long as your country's math society selects you to be a leader/trainer/coach, you can be one. In fact, many leaders (especially deputy leaders) are current university students.

But I think it's more common for leaders to be professors.


Given that $ABCD$ is a square, $CF$ bisects angle $ACD$.  $BPQ$ is perpendicular to $CF$. Prove that $DQ=2PE$
This problem came from challenging problems in geometry. I think its weird because I don't think that $DQ=2PE$
[hide=" my solution..."]
$CQ=CP$ by A.S.A. If you set the side length of the square to $x$, then diagonal $AC$ will have length $\sqrt2x$.
Then AE will be half of that or $\sqrt2x$/2. If you let $CQ=CP=y$, then $DQ=x-y$ and $PE= (\sqrt2x$/2)-y. From this, it is obvious that 2PE does not equal DQ. My method seems right...Is it a misprint or am I doing something wrong?[/hide]

Some questions...

First, what do you mean by "$BPQ$ is perpendicular to $CF$"?
Second, where is $E$?

E is the point of intersection of diagonals AC and BD.  BPQ is the line that meets CF at P and  AB and Q.


Two different numbers are randomly selected from the set
$ \{1, 2, 3, 4\}$ and they are multiplied. What is the probability
that the product is even? Express your answer as a common
fraction.

[hide="SOLUTION"]The total number of products is $ {4} \choose {2}$ $ \equal{} 6$. There is only one possible product that is odd, $ 3 \times 1$. The answer is thus $ \boxed{\frac {5}{6}}$.[/hide]

The number of ways to select $ 2$ numbers is $ \binom{4}{2}\equal{}6$. The only way to get an odd product is $ 1\times{3}\equal{}3$. Thus, the probability is $ 1\minus{}\frac{1}{6}\equal{}\boxed{\frac56}$.


Ring Theory
calculus
integration
greatest common divisor
number theory unsolved
number theory

Assume $x^2+dy^2=p$ is satisfied where $p$ is a prime number.Show that if $d\ge2$
then there is unique $(x,y)$ satisfying the condition.
Show that if $d=1$ then there are exactly two pairs satisfying the condition.
[hide]Actually I have solved  the second part.But the first part is a little difficult. [/hide]

I think you are wrong because if $p=a^2+b^2$ then this presenting is unique.(Without $a\rightarrow b$)

If there are $(x_1;y_1)\neq(x_2;y_2)$ then we can write $\left(x_1+y_1\sqrt{-d}\right)\left(x_1-y_1\sqrt{-d}\right)=\left(x_2+y_2\sqrt{-d}\right)\left(x_2-y_2\sqrt{-d}\right)=p$ and the numbers $\ x_i\pm y_i\sqrt{-d}$ are prime in $Z[\sqrt{-d}]$ because of their norm is prime.

Well, this interchange $(a,b) \leftrightarrow (b,a)$ is what gives the two solutions ;)

Lets give it a try for $d \geq 2$:
Assume that $p = s^2+dt^2 = x^2+dy^2$.
We have:
$p^2 = (s^2+dt^2)(x^2+dy^2) = (sx-dty)^2 + d (sy+tx)^2 = (sx+dty)^2 + d (sy-tx)^2$
$(sx+dty)(sy+tx) = xy(s^2+dt^2) + st(x^2+dy^2) = p(xy+st)$

1. Case: $p|sx+dty > 0$.
Then $sy-tx=0$ because $p^2 = (sx+dty)^2 + d (sy-tx)^2 = p^2 + d (sy-tx)^2$.
Thus $\frac{s}{t} = \frac{x}{y}$, which gives $\frac{p}{y^2} = \frac{x^2}{y^2}+d = \frac{s^2}{t^2}+d = \frac{p}{t^2}$, clearly implying the result.

2. Case: $p|sy+tx > 0$, so $sy+tx = kp$ with $k \geq 1$.
Then we get $(sx-dty)^2 = p^2 - d (sy+tx)^2 = p^2 (1-dk^2) < 0$, contradiction.
(for $d=1$ we would get the second solution here)

@bilarev: I think you need some more argument than just being prime.

I think that bilarev solution doesn't work because we don't know is the $\mathbb{Z}[\sqrt{-d}]$ an UFD?

What he needs is that the prime can be written as product of two new primes then.
For that to show, unique ideal factorisation suffices (and we have that ofter taking the integral closure), but that's not that trivial than just being prime.
But probably he knows some nicer way (I somehow think I think to complicated).

This which I wrote isn't proof...it's an idea for the proof.
This works for the UFD fields but in the other cases I don't know.
Actually for d=1 the proof is simple using $Z[\sqrt{-1}]$  ;)

For all cases the non-elementary way:
Assume that the rational prime $p$ can be written as $x^2+dy^2$ in two essentially different ways, so there are two factorisations $p= \mathrm{p}_1 \mathrm{p}_2 = \mathrm{p}_3 \mathrm{p}_4$.
Then in the integral closure of $\mathbb{Z}[\sqrt{-d}]$, this gives two different factorisations of the ideal generated by $p$:
$(p) = \mathfrak{p}_1 \mathfrak{p}_2 = \mathfrak{p}_3 \mathfrak{p}_4$ with $\mathfrak{p}_i = ( \mathrm{p}_i)$.
But that contradicts unique prime ideal factorisation in the Dedekind ring $R$.

Thank you very much,ZetaX. :)
I could also solve, but it didn't came to my mind to write
$(sx+dty)(sy+tx) = xy(s^2+dt^2) + st(x^2+dy^2) = p(xy+st)$ :(

Could you please explain me, or give some link(when you have time) about
for example $gcd(x-i,x+i)=1$,$gcd(x-\sqrt{z},x+\sqrt{z})$.What does these mean?
I'm now training for imo and when I see  these I really feel upset. :( ,because I don't understand but I see that they are very useful.
Looking forward for your help!! :lol:

[quote="TomciO"]You should precise your question. $gcd(a, b)$ means, as always, the Greatest Common Divisor of $a, b$ where $a, b$ are integers in some field.[/quote]
:D I know gcd(a,b) for a,b integers.My question is that what does it mean $gcd(x-i,x+i)=1$,$gcd(x-\sqrt{z},x+\sqrt{z})$.
i-imaginary number.

I think the problem is that I don't know about  "in some field". :!: Thank you for your help.

I posted a short explaination in http://www.mathlinks.ro/Forum/viewtopic.php?t=84491 (but don't know any stuff online  :( )

$gcd(x+i,x-i)=1$ hold's everytime ? Thanks for useful posts ZetaX .

No, for $x$ even only. For $x$ odd it gives $1+i$ as $\gcd$.

Could you please refer me to some online texts and books relating this subject? Thanks in advance.

Masoud Zargar

Thanks to my teacher: very-very helpful online material
[url=http://www.math.uiuc.edu/~r-ash/]www.math.uiuc.edu/~r-ash/[/url]


Find the zeros of the linear function f(x) = x + 3


can someone help me with this?

4. A dumbbell of length 2L is initially at rest on a horizontal surface. The
accompanying diagram shows a view looking down at the surface. Both
cubes at the ends of the dumbbell have mass m. The connecting rod is
massless. An object with mass M and velocity v perpendicular to the rod
slides across the surface and strikes the dumbbell a distance αL from its
center. The collision is perfectly elastic. There is no friction between any of
the objects and the surface. Assume the masses M and m can be treated as
point masses. After the collision, mass M is at rest.
(15) a. Find M in terms of some or all of the quantities α, L, v, and m.
(5) b. What is M if α = 1, i.e., M strikes m at the end of the rod?
(5) c. What is M if α = 0, i.e., M strikes at the center of the rod?

Hmm...possibly one can do something like this:

No external torque acts on the system, so angular momentum must be conserved: $M \alpha L v = 2mL^2 \omega \Rightarrow \omega = \frac{\alpha Mv}{2mL}$. In fact, there aren't really any external elements at all, so linear momentum should also be conserved: $Mv = 2mv' \Rightarrow v' = \frac{Mv}{2m}$. As the collision is elastic, total kinetic energy is also conserved and $Mv^2 = I \omega^2 + 2mv'^2 = (2mL^2) \left(\frac{\alpha Mv}{2mL}\right)^2 + 2m\left(\frac{Mv}{2m}\right)^2 = \frac{(\alpha Mv)^2}{2m} + \frac{(Mv)^2}{2m} = (Mv)^2 \frac{\alpha^2 + 1}{2m} \Rightarrow M = \frac{2m}{\alpha^2 + 1}$.


The number of inches in the perimeter of a square is equal to
the number of square inches in its area. What is the number of
inches in the side length of the square?

The perimeter of a square is $ 4s$, and the area is $ s^2$. Thus, the only way for the perimeter to be equal to the square is if $ s$ is equal to $ \boxed{4}$.

SImple algebra:

$ x^2 \equal{} 4x$

We can divide out x because $ x\neq{0}$

$ x \equal{} 4$


calculus
integration
trigonometry
logarithms
calculus computations

$ \int \frac{1}{1\plus{} \cot x} \ dx \equal{} \int \frac{\tan x}{1 \plus{} \tan x} \ dx \equal{} \int \frac{1}{2} \left[ 1 \minus{} \frac{1 \minus{}\tan x}{1\plus{}\tan x} \right] \ dx$

$ \equal{} \frac{x}{2} \minus{} \int \tan \left(\frac{\pi}{4}\minus{}x \right) \ dx \equal{} \frac{x}{2} \plus{} \sec \left(\frac{\pi}{4}\minus{}x \right)\tan \left(\frac{\pi}{4}\minus{}x \right) \plus{} C$

[hide]Considering $ \tan x \equal{} u$,

$ P \equal{} \int \frac{u}{(1\plus{}u)(1\plus{}u^2)}\,du \equal{} \frac12 \int \left(\frac{1}{1\plus{}u^2} \plus{} \frac12 \frac{2u}{1\plus{}u^2} \minus{} \frac{1}{1\plus{}u}\right)\,du$

$ \equal{} \frac12 \arctan u \plus{} \frac14 \ln(1\plus{}u^2) \minus{} \frac12 \ln|1\plus{}u| \plus{} C$

$ \equal{} \frac12 x \minus{} \frac12 \ln|\cos x| \minus{} \frac12 \ln|1 \plus{} \tan x| \plus{} C$

$ \equal{} \frac12 x \minus{} \frac12 \ln|\sin x \plus{} \cos x| \plus{} C$

$ \equal{} \frac12 \left[x \minus{} \ln \left|\sin \left(x \plus{} \frac{\pi}{4}\right)\right|\right] \plus{} C'$.[/hide]

[url=http://www.goiit.com/posts/list/integral-calculus-please-solve-its-urgent-982652.htm]one more way[/url]
 [hide="@hsbhatt sir"]r u sure about the last step[/hide]

Let $ I\equal{}\frac{\sin x}{\sin x\plus{}\cos x}\ dx,\ J\equal{}\int \frac{\cos x}{\sin x\plus{}\cos x}\ dx$,

$ I\plus{}J\equal{}x\plus{}A$

$ J\minus{}I\equal{}\ln |\sin x\plus{}\cos x|\plus{}B$

$ \therefore I\equal{}\int \frac{1}{1\plus{}\cot x}\ dx\equal{}\frac 12 x\minus{}\frac 12|\sin x\plus{}\cos x|\plus{}C$.


When writing mechanisms for reactions of carbonyl compounds, in many instances an acid is involved (which liberates a proton). My question is that during a nucleophilic addition, does the carbonyl oxygen get protonated and then attacked by a nucleophile or does the attack occur before protonation? I have seen both but not sure which one is right. In fact my book is sometimes internally inconsistent. When describing the formation of imines both mechanisms occur in two different sections of the book.

In reactions of carbonyl compounds in the presence of an acid the first step is always the protonation of the carbonyl oxygen, which activates (further more) the carbonyl carbon to nucleophilic attack.


Let $\mu_{n}(m)$ be the number of positive integers smaller than $m$ and co-prime to $n$. Prove that:

$\lim_{m\rightarrow\infty}\frac{\mu_{n}(m)}{m}=\frac{\phi(n)}{n}$

http://www.mathlinks.ro/Forum/weblog_entry.php?t=152294

What is that title supposed to say?
And moved to Pre-Olympiad.


Coin $ A$ is flipped three times and coin $ B$ is flipped four
times. What is the probability that the number of heads obtained from
flipping the two fair coins is the same?

I did essentially the same thing. Although, first I found the number of ways for you to achieve that, and then put it all over $ 2^7$, or $ 128$. You don't need to simplify the values at all.

Hi everyone! Sorry for reviving!

I have no idea what's wrong with [hide=this]Cases:

0 heads: $\left (\frac 12\right) ^7=\frac{1}{128}$

1 head: $\left (\frac 12\right) \left (\frac 12\right) =\frac{1}{4}$

2 heads: $\left (\frac 12\right) ^2\left (\frac 12\right) ^2=\frac{1}{16}$

3 heads: $(1/128)$, same as case 1
[/hide]. :(

It all adds up to [hide]41/128. and then i didn't considerate case of 0 heads and that, 5/16, was also wrong. :([/hide]

are you considering ordering? Also, could you explain where you got (1/2)(1/2) from? I think you are only just counting the probability of getting a head independently without considering the other 2 and 3 rolls.

I remembered the answer cause I'm going through all of the AMC 10 tests :)

I don't understand the solution to this question, actually. How is the probability of heads 3/8 and 4/16? Thank you!


Prove: If G and H are regular graphs then so is
a.) G + H
b.) G x H
c.) G[H]
please help me...

I would recommend posting this in the "Computer Science and Informatics" forum, under "Arts and Sciences". This will most likely get you more responses.


If you know the double angle formula then this problem becomes trivial. But since $15^o$ is very special, it is half of $30^o$, so we can construct a right triangle $ABC$ so that $\angle{A}=30^o$, $\angle{C}=30^o$. Extend $CA$ to $D$ so that $AD=AB$, then $\angle{ADB}=\angle{ABD}=15^o$. Now we have two right triangles: $\triangle{ABC}$ and $\triangle{DBC}$. So you can use $a^2+b^2=c^2$ and the definition of sine and cosine to find the answer.

[quote="236factorial"]If angle A is 30 and angle C is 30, then angle B must be 120, and that's not a right triangle...[/quote]
Didn't you notice that it was only a typo? C=90.

[quote="shyong"]do you know double angle theorem , 236factorial ?

[hide="double angle theorem"]

$\sin 2A=2\sin A\cos A$

[/hide][/quote]

Yeah I do, given that i just looked it up in my book  :P

[quote="236factorial"]Is there a way to show that $\sin15\cos15=0.25$?[/quote]
[hide]we know that $sin(2\alpha)=2 \times sin(\alpha) \times cos(\alpha) \rightarrow sin30=2 \times sin15 \times cos15 \rightarrow sin15 \times cos15=\dfrac{1}{4}=0.25$[/hide]

[quote="eryaman"]Another way (longer, but sometimes useful) is to use the difference and sum formulas:

$\sin(x+y)=\sin x\cos y+\sin y\cos x$
and
$\cos(x+y)=\cos x\cos y-\sin x\sin y$

You can derive the double angle formulas by setting $x=y$.

To find $\sin 15$ and $\cos 15$, use $\sin 15=\sin(45-30)$, and do the same with $\cos$, and multiply the results.[/quote]
your solution is nice but in this problem it is so long,but i think it is really nice. :D

Apply the double angle forumla. If you are not sure where this formula is from, here it is.

$\sin (a+b)=\sin a\cos b+\sin b\cos a$
$\sin 2a= \sin (a+a) =2 \sin a\cos a$

Now, $.5 = \sin 30= \sin (15+15) = 2 \sin15 \cos 15$
$2 \sin 15 \cos 15 = .5$
$\sin 15 \cos 15 = .25$

Done!!  :D

You can also use the product-to-sum identities, but I never remember what they are, so... :D


!این جا برای مساله های نظریه اعداد حالا دیگه لطف کنید سوالهاتون رو با همون سیستم بفرستید دیگه

khob sharmandeam. yeki az soalaatam hal nashod; majboor shodam eftettah konam :blush:
tamame $a,b,c \in N$ raa biabid ke $b^2+1$ va $c^2+1$ aval baashand va $a^2+1=(b^2+1)(c^2+1)$

na na bikhiyal post man sho!!!! :blush:  :blush:

[quote="fatehi"]na na bikhiyal post man sho!!!! :blush:[/quote]
khob deletesh kon :lol:

let b^2 + 1 =p Therefore a = b (modp) or a = -b mod p.
Similarly if c^2 + 1 =q we get that a = +/- c mod q
now i shall assume that b>c ( b is not c ) so that we can say that:
a = nbc + r
Then a^2 + 1 = (nbc+r)^2 + 1 = n^2 b^2 c^2 + 2nbc + 1 +r^2 = b^2 c^2 + b^2 + c^2 +1
Now thhis impies that pretty much (!!!) n=1.
nemikham dige tamame step ha ro neshonn bedam...

pas: a = bc +r . and 2bc + r^2 = b^2 + c^2
or that r = b-c.

hence (b-c)^2 + 1 is divisible by p and q.
so b-c = +/- b mod p that is c=0 mod p or c = 2b mod p
similarly b-c = +/- c mod q so that b=0 mod q or b = 2c mod q.

Anyways i guess the rest is not tough....
shayad inha albatte aslan mofid nabashe chonke ziyad aslan tozih nadadam.

-Ali

[quote="ali"]
Now thhis impies that pretty much (!!!) n=1.
nemikham dige tamame step ha ro neshonn bedam...
-Ali[/quote]
Hamoon tor ke hads zadi aslan mofid nabood; chon man hichi nafahmidam!! :roll:

oon gesmate rahe hal benazaram kheily vazehe ke n=1.
daghigan chisho nemifahmi?

az bozorgan mikham ke age momkene biyan va ba ma mennat bezaran va ye kam az ide hayi ke daran dar nazarie adaad benevisan.(zakate elm amookhtane an ast!)

in soale avale poland 2004 ast.to resources mathlinks hallesh hast,

khob hala man do masale ziba midam.albatte khodam hal kardam vali daram baraye shoma minevisam. :)
$*$tamame javab haye tabiei moadale zir ra biyabid:
$k!+48=48(k+1)^n$

adade tabiei $n>1$ mafrooz ast.neshan dahid tedad tarkib haye $r$ az $n$ ke baghimande anha bar $3$ barabare $1$ ast az an tedad ke baghimandeshan bar $3$ barbar $2$ ast bishtar ast. :)


[quote="nmtvn"]Let $ x$, $ y$, $ z$ be three positive numbers. Show that
$ \sum\limits_{\text{cyc}} \frac {x\left(y \plus{} z\right)^2}{2x \plus{} y \plus{} z} \geq \sqrt {3\left(x \plus{} y \plus{} z\right)xyz}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167965

arqady is the inequality
$ \sum_{cyc} \frac{a^{2}(s \minus{} a)}{b \plus{} c} \geq \sqrt {3} \Delta$
solved over there???

So, can anybody give the solution to this?

$ \sum\limits_{\text{cyc}} \frac {x\left(y \plus{} z\right)^2}{2x \plus{} y \plus{} z} \geq \sqrt {3\left(x \plus{} y \plus{} z\right)xyz}\Leftrightarrow$
$ xy\plus{}xz\plus{}yz\minus{}\sqrt{3(x\plus{}y\plus{}z)xyz}\geq\sum_{_cyc}\left(yz\minus{}\frac{x(y\plus{}z)^2}{2x\plus{}y\plus{}z}\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{z^2(x\minus{}y)^2}{2(xy\plus{}xz\plus{}yz\plus{}\sqrt{3(x\plus{}y\plus{}z)xyz})}\geq\sum_{cyc}\frac{y^2z\plus{}z^2y\minus{}y^2x\minus{}z^2x}{2x\plus{}y\plus{}z}.$
But $ \sum_{cyc}\frac{y^2z\plus{}z^2y\minus{}y^2x\minus{}z^2x}{2x\plus{}y\plus{}z}\equal{}\sum_{cyc}\left(\frac{y^2(z\minus{}x)}{2x\plus{}y\plus{}z}\minus{}\frac{z^2(x\minus{}y)}{2x\plus{}y\plus{}z}\right)\equal{}$
$ \equal{}\sum_{cyc}\left(\frac{z^2(x\minus{}y)}{2y\plus{}x\plus{}z}\minus{}\frac{z^2(x\minus{}y)}{2x\plus{}y\plus{}z}\right)\equal{}\sum_{cyc}\frac{z^2(x\minus{}y)^2}{(2x\plus{}y\plus{}z)(2y\plus{}x\plus{}z)}.$
Id est, it remains to prove that
$ (2x\plus{}y\plus{}z)(2y\plus{}x\plus{}z)\geq2\left(xy\plus{}xz\plus{}yz\plus{}\sqrt{3(x\plus{}y\plus{}z)xyz}\right).$
But $ (2x\plus{}y\plus{}z)(2y\plus{}x\plus{}z)\geq2\left(xy\plus{}xz\plus{}yz\plus{}\sqrt{3(x\plus{}y\plus{}z)xyz}\right)\Leftrightarrow$
$ \Leftrightarrow2x^2\plus{}2y^2\plus{}z^2\plus{}3xy\plus{}xz\plus{}yz\geq2\sqrt{3(x\plus{}y\plus{}z)xyz},$ which true because $ xy\plus{}xz\plus{}yz\geq\sqrt{3(x\plus{}y\plus{}z)xyz}$ and $ x^2\plus{}y^2\plus{}z^2\geq\sqrt{3(x\plus{}y\plus{}z)xyz}.$

from titu lemma and ago
$ \sum\limits_{\text{cyc}}\frac{\left(y\plus{}z\right)^{2}}{2\plus{}\frac{y}{x}\plus{}\frac{z}{x}}\geq \frac{(2(x\plus{}y\plus{}z))^2}{(x\plus{}y\plus{}z)(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})\plus{}3}$ and  $ xy\plus{}yz\plus{}xz\geq \sqrt{3\left(x\plus{}y\plus{}z\right)xyz}$

we prove $ \frac{(2(x\plus{}y\plus{}z))^2}{(x\plus{}y\plus{}z)(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})\plus{}3}\geq xy\plus{}yz\plus{}xz$

after multiplying and simplifying we have $ [2,0,0] \geq [2,1,\minus{}1]$ is true for muirhead

[quote="mestav"]from titu lemma and ago
$ \sum\limits_{\text{cyc}}\frac {\left(y \plus{} z\right)^{2}}{2 \plus{} \frac {y}{x} \plus{} \frac {z}{x}}\geq \frac {(2(x \plus{} y \plus{} z))^2}{(x \plus{} y \plus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}) \plus{} 3}$ and  $ xy \plus{} yz \plus{} xz\geq \sqrt {3\left(x \plus{} y \plus{} z\right)xyz}$

we prove $ \frac {(2(x \plus{} y \plus{} z))^2}{(x \plus{} y \plus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}) \plus{} 3}\geq xy \plus{} yz \plus{} xz$

after multiplying and simplifying we have $ [2,0,0] \geq [2,1, \minus{} 1]$ is true for muirhead[/quote]
your solution is wrong because $ [2,1, \minus{} 1]\succ[2,0,0]$, just the contrary  :(

Theorem 4.3 in
[url=http://www.cip.ifi.lmu.de/~grinberg/Yugoslavia1998.pdf]Darij Grinberg, [i]An inequality involving $ 2n$ numbers[/i] (version 22 August 2007)[/url],
applied to $ a\equal{}y\plus{}z$, $ b\equal{}z\plus{}x$, $ c\equal{}x\plus{}y$, $ u\equal{}yz$, $ v\equal{}zx$ and $ w\equal{}xy$.

That theorem is not new actually, I remember seeing it on MathLinks.

  darij


Let $ X_1,X_2 \cdots X_n$ be independently exponentially distributed random variables with rates $ \lambda_1,\lambda_2, \cdots \lambda_n$, respectively.

Let $ Z_k \equal{}$ min of the $ \binom{n}{k}$ possible product of $ k$ $ X_i$'s.
What is $ E(Z_k)$?

Example:
Say we have $ X_1,X_2,X_3$.
Then:
$ Z_1 \equal{} \min{(X_1,X_2,X_3)}$
$ Z_2 \equal{} \min{(X_1X_2,X_1X_3,X_2X_3)}$
$ Z_3 \equal{} X_1X_2X_3$

is it an exercice or a question of your own. Even the density of the product of $ 2$ independant exponential random variables has an ugly density (but maybe a good closed form formula for the expectation). Just wondering ..


[quote="civilmath"]ΡΕ ΠΑΙΔΙΑ ΚΑΛΗΣΠΕΡΑ.
ΜΠΟΡΕΙΤΕ ΝΑ ΜΕ ΒΟΗΘΗΣΕΤΕ ΜΗΠΩΣ ΝΑ ΒΡΩ ΠΡΟΣΕΓΓΙΣΤΙΚΑ ΤΟ 1.0002 ΣΤΗΝ 50 Κ ΤΟ 3 Η ΡΙΖΑ ΤΟΥ 1.009.ΕΥΧΑΡΙΣΤΩ.
ΜΠΟΡΕΙΤΕ ΝΑ ΜΟΥ ΠΕΙΤΕ Κ ΜΕ ΠΟΙΟ ΠΡΟΓΡΑΜΜΑ ΓΡΑΦΩ ΜΑΘΗΜΑΤΙΚΑ.??[/quote]

First off, welcome  :welcomeani:

$ 1.0002^{50} \sim 1 \plus{} 50*0.0002 \equal{} 1.01$ σε προσέγγιση πρώτης τάξης, για καλύτερη προσέγγιση χρειάζεσαι περισσότερους όρους στο ανάπτυγμα Taylor της $ (1 \plus{} x)^{50} \equal{} 1 \plus{} 50x \plus{} \binom{50}{2}x^2 \plus{} \cdots$

$ \sqrt [3]{1.009} \sim 1 \plus{} \frac {1}{3}0.009 \equal{} 1.003$ σε πρώτη τάξη. Ομοίως, χρειάζεσαι περισσότερους όρους του αναπτύγματος για καλύτερη προσέγγιση.

Μαθηματικά με LaTeX, υπάρχει ένα stickie για οδηγίες χρήσης. Takes some getting used to, but it's *really* worth it :)

Cheerio,

Durandal 1707

οριζεις την συναρτηση $ f(x,y) \equal{} x^y$ με $ x_0 \equal{} 1$ και $ y_0 \equal{} 50$, $ \Delta x \equal{} x \minus{} x_0 \equal{} 1.0002 \minus{} 1 \equal{} 0.0002$

και εφαρμοζεις τον τυπο $ f(x_0 \plus{} \Delta x, y_0 \plus{} \Delta y) \approx f(x_0, y_0) \plus{} df(x_0,y_0) \equal{} 1 \plus{} y_0\cdot x_0^{y_0 \minus{} 1}dx$

Θα βρεις $ 0,0002^{50} \approx 1 \plus{} 50\cdot 0,0002 \equal{} 1,01$

Το ιδιο και για την ριζα.

ΥΓ. δηλαδη αυτο που εκανε ο Durandal  :roll:


geometry
geometric transformation
dilation
parallelogram
circumcircle
Gauss
similar triangles

Let ABCD be a cyclic quadrilateral. Let M and N be the midpoints of AC and BD, respectively. Define E as the intersection point of AB and CD, and let F be the intersection point of AD and BC. Prove that

$\frac{2MN}{EF}= \left | \frac{AC}{BD}-\frac{BD}{AC}\right |$

I would appreciate if anyone could provide a synthetic solution for this problem.

Again some collaborative work by darij and me. The first part of the solution was straightforward for me, the second was straightforward for darij.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=440[/img]

Let $P$ be the point such that $ABCP$ is a parallelogram. Let $Q$ be the point of intersection of $CP$ with $AD$. Then we have the following two results:

Lemma 1. $DP = 2MN$.

Proof. This is clear, as a dilation with factor 2 and center $B$ sends $M$ to $P$(diagonals in a parallelogram intersect at their common midpoint) and $N$ to $D$.

Lemma 2. $DQ = \left|\frac{BD}{AC}-\frac{AC}{BD}\right| FC$.

Proof. This can be split into the two equalities $FD = \frac{BD}{AC} FC$ and $FQ = \frac{AC}{BD} FC$. The first identity follows from the well-known similarity of $FBD$ and $FAC$. For the second identity, note that $FQC$ and $FAB$ are similar (since CQ || AB by construction). Therefore $\frac{FQ}{FC} = \frac{FA}{FB}$. But $\frac{FA}{FB} = \frac{AC}{BD}$ follows again from the similarity of $FBD$ and $FAC$.

Looking at Lemma 1 and Lemma 2, we see that it is enough to prove that the triangles $DPQ$ and $FEC$ are (inversely) similar. The interpretation of the crazy equality, especially the term $\frac{AC}{BD}-\frac{BD}{AC}$, is done - maybe this was the hard part.
The similarity again follows from the following two lemmata:

Lemma 3. $\sphericalangle PQD = \sphericalangle FCE$.

Proof. This is a simple angle chase. We take oriented angles mod 180 degree. Then $\sphericalangle PQD = \sphericalangle BAD = \sphericalangle BCD = \sphericalangle FCE$.

Lemma 4. $\sphericalangle DPQ = \sphericalangle CEF$.

Proof. This is the part found by darij. First, let us simplify a bit. $\sphericalangle DPQ = (DP; AB) = (MN; AB)$, where $(g; h)$ denotes the angle (oriented angle mod 180 degree) between two lines. On the other hand, we clearly have $\sphericalangle CEF = (CE; EF) = (CD; EF)$. So we are supposed to show $(MN; AB) = (CD; EF)$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=441[/img]

Let $O$ be the circumcenter of $ABCD$. Let $K$ be the midpoint of $AB$ and let $G$ be the intersection of $AC$ and $BD$. Then we know that $OMGN$ and $OKBN$ are cyclic quadrilaterals.

It is well-known that $EF$ is the polar of $G$ with respect to the circumcircle of $ABCD$. Therefore $(OG; EF) = 90^{\circ}$. But then $(CD; EF) = (CD; OG) + (OG; EF) = (CD; OG) + 90^{\circ}$.
On the other hand, $(MN; AB) = (MN; OK) + (OK; AB) = (MN; OK) + 90^{\circ}$. It is left to show that $(CD; OG) = (MN; OK)$. But this is easy:

$(MN; OK) = (MN; ON) + (ON; OK) = (MG; OG) + (BN; BK)$ $= (AG; OG) + (BD; BA) = (AC; OG) + (CD; AC) = (CD; OG)$.

This completes the proof.

[Moderator edit: I have made the images, Peter, but alas they seem to be too wide for being normally displayed on ML, so one has to click on them in order to view them in full size. Darij]

This problem appeared on the 1997 Bulgarian olympiad.  For another synthetic solution, see pages 17-18 of this document:
[url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf]www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf[/url]

Here is my solution(?):

  We have to show some respect to Gauss,Bodenmiller,Steiner and JPE.

  Draw Gauss line thru diagonals midpoints and construct a circle thru M,N,E

  orthogonal to circle with diameter EF.  Midpoint of EF is P.  We have

                                     EP*EP = PM*PN

                          (PN + NM) / EP = EP / PN

                            MN / EP = EP / PN  - PN / EP

  In similar triangles   EPN and MPE we have  ME / NE = EP / NP

  And ME / NE = AC / BD

   Thank you.

   M.T.


What is the greatest number of points of intersection for 8 distinct lines in a plane?

For extra credit, give at least 1 formula for the number of points of intersection of n lines.

Billy

[hide="problem"]List and find pattern:
1:0
2:1
3:3
4:6
5:10
...
Triangular #s!!!!
6:15
7:21
8:28
>>28<<[/hide]
[hide="formula1"]l=# of lines drawn
l(l-1)/2
(the l-1th triangular number)[/hide]
[hide="reasoning"]Every line you draw can intersect every other line once. ie,
first line starts it off. (+0)
second can cross first (+1)
third can cross 1, 2 (+2)
...
and so on and so forth
[/hide]


logarithms
limit
real analysis
real analysis unsolved

[quote="kalazealot"]Why $\sum \frac{n^{c}}{2^{n}}<\infty$? $c>0$ constant.[/quote]
when will be :
$n>\frac{(c+2)\ln{n}}{\ln{2}}$ then
$\frac{n^{c}}{2^{n}}<\frac{1}{n^{2}}$
and for last terms the sum converge

By the D'Alembert test:

$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}= \lim_{n\to\infty}\frac{\frac{(n+1)^{c}}{2^{n+1}}}{\frac{n^{c}}{2^{n}}}= \lim_{n\to\infty}\frac{1}{2}\left(1+\frac{1}{n}\right)^{c}= \frac{1}{2}< 1$

and so the series converges $\forall c \in R$.

As $\lim_{n\to\infty}\sqrt[n]{a_{n}}= \lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}= \frac{1}{2}< 1$

the Root test also allow us to conclude the convergence of the series.


For a and b positive  , prove that $ \frac{x}{x^{4}\plus{}y^{2}}\plus{}\frac{y}{y^{4}\plus{}x^{2}}\le \frac{1}{xy}$

Please give me some techiniques or strategies for solving inequalities .  Thank you

[hide="my solution"]
$ x^4\plus{}y^2\ge2x^2y$
$ \frac{x}{x^4\plus{}y^2}\le\frac1{2xy}$(1)

$ y^4\plus{}x^2\ge2xy^2$
$ \frac{y}{y^4\plus{}x^2}\le\frac1{2xy}$ (2)

Adding (1) & (2), we get the desired inequality

[/hide]

This is the problem of Russia 1995.

Here is my solution:

$ \frac {x}{x^4 \plus{} y^2} \equal{} \frac {x}{(x^2 \minus{} y)^2 \plus{} 2x^2y}\leq \frac {x}{2x^2y} \equal{} \frac {1}{2xy}$, similarly we have $ \frac {y}{y^4 \plus{} x^2}\leq \frac {1}{2xy}$,

Adding those inequalities, yielding $ \frac {x}{x^4 \plus{} y^2} \plus{} \frac {y}{y^4 \plus{} x^2}\leq \frac {1}{xy}$.

Alternative Solution:

I prove that $ \frac {x^2y}{x^4 \plus{} y^2} \plus{} \frac {xy^2}{y^4 \plus{} x^2}\leq 1$.

Regard $ x$ as the constant,  $ \frac {x^2y}{x^4 \plus{} y^2} \equal{} \frac {x^2}{y \plus{} \frac {x^4}{y}}\leq \frac {x^2}{2\sqrt {y\cdot \frac {x^4}{y}}} \equal{} \frac {1}{2}$,

the symmetry of $ x, y$, we have $ \frac {xy^2}{y^4 \plus{} x^2}\leq \frac {1}{2}$, adding those gives the proof.

Let $ R, E$ be given positive constants and $ r$ be postive variable. Find the maximum value of $ P \equal{} \frac {E^2R}{(R \plus{} r)^2}$.