what is the solution for $ x^{\frac 23} \equal{} 9$?

Is it $ x \equal{} 27$ or $ x \equal{} \pm 27$?

I guess I do not know the definition of fractional exponents well. The (not so great) school math textbook says its only $ x \equal{} 27$, but I feel that that is wrong. $ \minus{} 27$ works no matter what I think. Either way of interpreting the fractional exponent works for $ x \equal{} \minus{} 27$

If you interpret $ ( \minus{} 27)^{\frac 23}$ as $ \sqrt [3]{( \minus{} 27)^2} \equal{} \sqrt [3]{729} \equal{} 9$,

If you interpret $ ( \minus{} 27)^{\frac 23}$ as $ (\sqrt [3]{ \minus{} 27})^2 \equal{} ( \minus{} 3)^2$

However, solution of $ x \equal{} \minus{} 27$ implies that $ 9^{\frac 32} \equal{} \minus{} 27$ (or does it imply this??). That is my real question, I guess. Is that true?

I guess my question simplifies to whether $ x^{\frac 12} \equal{} \sqrt x$ or $ x^{\frac 12} \equal{} \pm \sqrt x$.

EDIT:oops, the textbook said $ x\equal{}27$, not $ x\equal{}9$.

your final question idk

the first one:
you cube both sides to get
x^2=729
x can be negative or positive 27.
because
x^2-729=0
(x-27)(x+27)=0
and there are no restrictions on the value.


This is a Mendelian inheritance problem.

Capital letters are dominant alleles and lower case are recessive.

If we cross the genotypes AaBbCc with AAbbcc what is the probability of the offspring exhibiting two recessive phenotypic traits?

Could anyone tell me if this is correct :

chance of aa = none; can discount this
chance of both bb and cc = (chance of bb, 1/2) x (chance of cc, 1/2) = 1/4 ?


1)let a,b,c are real numbers .prove:
 a(a+b)(a^2+b^2)+b(b+c)(b^2+c^2)+c(c+a)c^2+a^2)>=0
2) let x,y,z >0 and x+y+z=k(k>0);
find the great value of expression:
k(xy+yz+zx)-2xyz

[quote="andrew-mathmaster"]
2) let x,y,z >0 and x+y+z=k(k>0);
find the great value of expression:
k(xy+yz+zx)-2xyz[/quote]

This problem posted in lectures of Hojoo Lee, and it lie his website  :wink:


In Morse code, each symbol is represented by a sequence of dashes and
dots. How many distinct symbols can be represented using sequences
of 1, 2, 3, or 4 dots and/or dashes?

In each symbol, each character can be a dot or a dash (two possibilities).
So for example, there are two possibilities for each character of a three-character symbol, leading to $ 2^3\equal{}8$ different symbols.
Overall, there are $ 2^1\plus{}2^2\plus{}2^3\plus{}2^4\equal{}2\plus{}4\plus{}8\plus{}16\equal{}\boxed{30}$ different 1,2,3, or 4-letter symbols.


[quote="white_horse_king88"]Prove that $A - B = - (B - A)$ geometrically.[/quote]

[hide]What do you mean by geometrically?
I have another proof, but am not sure it is geometrical:
Let $A$ be $[a_1,a_2]$ and $B$ be $[b_1,b_2]$

$A - B = [a_1-b_1,a_2-b_2]$

$B - A = [b_1-a_1,b_2-a_2]$

$-(B - A) = -[b_1-a_1,b_2-a_2] = [a_1-b_1,a_2-b_2]$

QED[/hide]

I think maybe he means by congruent triangles... perhaps...

Well, I mean using the properties of vector subtraction. Start by drawing vectors A and B...

One of the thick arrows is $A-B$ and the other one is $B-A$.
Since they have opposite directions, we conclude $A-B=-(B-A)$.


[quote]Huge Meat-Eating Dinosaur Discovered

Updated 10:01 PM ET April 17, 2006

Move over, Tyrannosaurus Rex. There's a new dinosaur in town, and he's bigger and badder than you ever were.

A ferocious meat-eater named Mapusaurus roseae and weighing almost eight tons was recently discovered in Argentina. Because the bones of several individuals were found in one place, scientists believe that the carnivores lived together in social units and hunted with deadly efficiency in well-organized packs.

Their prey? Mapusaurus probably dined on the flesh of the largest dinosaur that ever lived, Argentinosaurus, a massive 125 foot-long, 100-ton giant.

"They would have been quite formidable," said Philip Currie, paleontologist at the University of Alberta who worked on the Mapusaurus site. "These were definitely the top dogs in the Southern Hemisphere."

The Family that Preys Together
The Mapusaurus bones were found in a bone bed on a hill in the Patagonia region of southern Argentina. "Bone beds are a good area to work for giving info on more than one animal," said Currie, who made the Mapusaurus discovery with Rodolfo Coria of the Museo Carmen Funes in Plaza Huincul, Argentina.

Carnivorous dinosaurs like T. Rex were once believed to be solitary hunters, but new evidence suggests otherwise. Currie speculates that the Mapusaurus he and Coria discovered were all living together in a social group until their death in some catastrophic event about 100 million years ago.

"They all died together, so you can reason that they all lived together," Currie said. "Carnivores, when you find them all together like this, were probably packing animals with coordinated hunting efforts."

The group found included several younger animals about 18 feet long, who were no less dangerous than the adults. "The juveniles were faster than the adults -- they were longer-legged and more lightly built," Currie said.

When Size Matters
Based on the remains found at the site, an adult Mapusaurus would have measured about 40 feet in length. This puts the carnivore ahead of Tyrannosaurus Rex and on par with another meat-eater, Giganotosaurus.

"Over the last decade, people have become increasingly aware of a group of gigantic meat-eating dinosaurs called carcharodontosaurids," explained Currie. This group included Mapusaurus, Giganotosaurus and other large carnivores.

It's difficult to determine the exact size of Mapusaurus, Currie explained, because the complete skeleton of a single adult has never been found. But much information can be gleaned from the skull and other bones that were uncovered at the site.

"This new animal we have is slightly larger than T. Rex, and it may be larger than Giganotosaurus," said Currie. "No matter how you cut it, this is a big animal." [/quote]

Stunning news no doubt, but it could well end up as one of those dinosaur frauds the last century has seen so many of.


Let $ x,y$ and $ z$  be a real numbers greater that $ 1$ such that :

$ \begin{cases} xy^2\minus{}y^2\plus{}4xy\plus{}4x\minus{}4y\equal{}4004 \\ xz^2\minus{}z^2\plus{}6xz\plus{}9x\minus{}6z\equal{}1009 \end{cases}$

Find all possibles values of the expression :

$ xyz\plus{}3xy\plus{}2xz\minus{}yz\plus{}6x\minus{}3y\minus{}2z$

The first equation is equivalent to
\[ (x\minus{}1)(y\plus{}2)^2\equal{}4000,\]and the second to
\[ (x\minus{}1)(z\plus{}3)^2\equal{}1000.\]
Also note that
\[ xyz\plus{}3xy\plus{}2xz\minus{}yz\plus{}6x\minus{}3y\minus{}2z\equal{}(x\minus{}1)(y\plus{}2)(z\plus{}3)\plus{}6.\]
So from our first two equations,
\[ (x\minus{}1)(y\plus{}2)(z\plus{}3)\plus{}6\equal{}6\pm2000\equal{}\{\minus{}1994,2006\}.\]


Prove that for all positive intergers n:
$ S_n \equal{} C^0_{2n \plus{} 1}2^{2n} \plus{} C^2_{2n \plus{} 1}.2^{2n \minus{} 2}.3 \plus{} ... \plus{} C^{2n}_{2n \plus{} 1}.3^n$ is always sum of two uninterrupted squares .
I mean two uninterrupted squares are two squares like:$ a^2,(a\plus{}1)^2$
and C stand for combination ? :(

[quote="No Reason"]Prove that for all positive intergers n:
$ S_n \equal{} C^0_{2n \plus{} 1}2^{2n} \plus{} C^2_{2n \plus{} 1}.2^{2n \minus{} 2}.3 \plus{} ... \plus{} C^{2n}_{2n \plus{} 1}.3^n$ is always sum of two uninterrupted squares .[/quote]

Two questions:
1) What is "uninterrupted squares"?
2) What does $ C^a_{b}$ stand for?

I will never understand how a thread with such a title can be posted in the unsolved problem section... :wink:

Using binomial expansion theorem, it is easy to verify that
$ S_n \equal{} \frac 1 4 ( (\frac 1 q)^{2n\plus{}1} \plus{} q^{2n\plus{}1} )$, where $ q\equal{}2\plus{}\sqrt {3}$ (so that $ \frac 1 q \equal{}2\minus{}\sqrt {3}$).

Thus $ S_n \equal{} a_n^2\plus{}(a_n \plus{}1)^2$ where $ a_n \equal{} \frac {1\plus{}\sqrt 3}{4} q^n \plus{}\frac {1\minus{}\sqrt 3}{4} (\frac 1 q)^n \minus{}\frac 1 2$.
It is easy to verify that $ a_0\equal{}0,a_1\equal{}2$ and $ a_{n\plus{}1} \equal{} 4a_n\minus{}a_{n\minus{}1}\plus{}1$ which ensures that $ a_n$ is an integer.

Pierre.


algebra
polynomial
number theory
composite number
prime numbers
IMO Shortlist

All the coefficients of $f(x)=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)=$ $Sx^2+(ab+bc+ca-de-ef-fd)x+(abc+def)$ are multiples of $S$. Evaluating $f$ at $d$ we get that $f(d)=(a+d)(b+d)(c+d)$ is a multiple of $S$.
So this implies that $S$ is composite, since $a+d,b+d,c+d$ are all strictly less than $S$.

davron

My solution (some kind of minimality idea).

Suppose $S$ is prime.
We observe that if $(a, b, c, d, e ,f)$ is a solution then $(a-1, b-1, c-1, d+1, e+1, f+1)$ is also a solution. Note that $S$ does not change. Denote $X=\min\{a,b,c\}$. We can claim that $(a-X, b-X, c-X, d+X, e+X, f+X)$ also satisfies the conditions.
So we get $S \mid (d+X)(e+X)(f+X)$. But this is impossible, since $d+X,e+X, f+X$ are all $<S$. :)

[hide="Hidden due to length"]
For the sake of contradiction assume $S$ to be prime.

It's easy to prove $S|(a+b)(b+c)(c+a)+(d+e)(e+f)(f+d)\quad (1)$

Also $S|S(b+c)(c+a)=(a+b)(b+c)(c+a)+(c+d+e+f)(b+c)(c+a)\quad (2)$

Subtracting (1) from (2) gives $S|(c+d+e+f)(b+c)(c+a)-(d+e)(e+f)(f+d)\quad (3)$

Simplifying (3), we derive $S|(c+f)[(b+c)(c+a)+(d+e)(c-f)]$

Since $S$ is a prime and $S>c+f$, hence $S|(b+c)(c+a)+(d+e)(c-f)$

$\Longrightarrow S|c^2+cd+ce+(ab+bc+ca-ef-fd)$

$\Longrightarrow S|c^2+cd+ce+de$

$\Longrightarrow S|(c+d)(c+e)$

But $c+d<S,c+e<S$, So a contradiction, hence $S$ can't be a prime.[/hide]

Well, note $(a+d)(b+d)(c+d)=A+dB+d^2S$ now so we get $S|(a+d)(b+d)(c+d)$ , suppose $S$ is prime then it must be less than one of $a+d,b+d,c+d$ but as $S=a+b+c+d$ so absurd and done.

Assume $S$ is a prime. Then $(x-a)(x-b)(x-c)$ and $(x+d)(x+e)(x+f)$ are the same polynomial, mod $S$, so by Lagrange's theorem $\{a,b,c\} = \{-d,-e,-f\}$. Thus $a+b+c+d+e+f \ge 3S > S$.

[hide=Solution]
Let $P(x)=\frac{(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)}{S}$. Since $S|abc+def$ and $S|ab+bc+ca-de-ef-fd$, $P(x) \in \mathbb{Z}$ for all integers $x$.

However, $P(d)=\frac{(a+d)(b+d)(c+d)}{S}$. Suppose $S$ is a prime. Then $S|a+d$ or $S|b+d$ or $S|c+d$ which means one of $\{b+c+e+f, a+c+e+f, a+b+e+f\}$ are nonpositive, contradiction.
[/hide]

For the sake of contradiction, let $S$ be prime. Note $S$ divides $(x-a)(x-b)(x-c) - (x+d)(x+e)(x+f)$. Substitute $x= a,b,c,-d,-e,-f,$ we get $S$ divides $(d+a)(d+b)(d+c) , (e+a)(e+b)(e+c), (f+a)(f+b)(f+c)$ and

$S|(a+d)(a+e)(a+f)$
$S|(b+d)(b+e)(b+f)$
$S|(c+d)(c+e)(c+f)$

From the first set of 3 conditions, WLOG, we can let $S|(a+d)$, $S|(b+e)$ and $S|(c+f)$. However, each of $(a+d),(b+e), (c+f)$ are smaller than $S$, a contradiction.

Let $p$ be a fixed prime. We say a $6$-tuple $(a,b,c,d,e,f)$ of integers is [i]good[/i] if $a+b+c+d+e+f = p$ and $p \mid abc+def$ and $p \mid ab+bc+ca-de-ef-fd$.

[b][color=red]Claim:[/color][/b]  	If $(a,b,c,d,e,f)$ is good then so is 	$(a+1, b+1, c+1, d-1, e-1, f-1)$.

[i]Proof.[/i]  	Direct computation. $\blacksquare$

We claim there exists no good $6$-tuple of [i]positive[/i] integers. If not, WLOG $f$ is the smallest of the six numbers. Then by repeating the claim, the tuple $(a+f, b+f, c+f, d-f, e-f, 0)$ is good. But now $p \mid (a+f)(b+f)(c+f)$ and $p > \max(a+f, b+f, c+f)$, contradiction.

We see that
\begin{align*}
a+b+c&\equiv -(d+e+f)\pmod{S} \\
ab+bc+ca&\equiv -(de+ef+fd)\pmod{S} \\
abc &\equiv -def\pmod{S}.
\end{align*}
Now, assuming that $S$ is prime, we see that
\[(X-a)(X-b)(X-c)=(X+d)(X+e)(X+f)\]
over $\mathbb{F}_S[X]$, so because of unique factorization, we have WLOG that $a\equiv -d\pmod{S}$, $b\equiv -e\pmod{S}$, and $c\equiv -f\pmod{S}$. In particular, this means that $S\mid a+d$. However, $S>a+d$, so this isn't possible. Therefore, $S$ couldn't have been prime to start with. $\blacksquare$

[hide=Solution]Yay I got an ISL!! I thought it said any integers not just positive, so I was confused for a while.

Let $P(x)=(x-a)(x-b)(x-c)-(x+d)(x+e)(x+f)=-(a+b+c+d+e+f)x^2+(ab+bc+ca-de-ef-fd)x-(abc+def)
$. Note that $S|P(\alpha)\forall\alpha\in\mathbb{Z}$. Thus $S|P(a)=-(a+d)(a+e)(a+f)$, but $S$ doesn't divide any of its factors due to size reasons (otherwise the sum of the other 4 variables would be zero, which is impossible given they're positive integers), so $S$ cannot be prime by Euclid's Lemma.[/hide]

AFSOC $S = p$ for some prime $p$. Define \begin{align*}
    P(x) &= (x + a)(x + b)(x + c) = x^3 + x^2(a + b + c) + x(ab + bc + ca) + abc \\
    Q(x) &= (x - d)(x - e)(x - f) = x^3 - x^2(d + e + f) + x(de + ef + fd) - def,
\end{align*} and consider \begin{align*}
    R(x) &= P(x) - Q(x) \\
    &= x^2 \cdot p + x(ab + bc + ca - de - ef - fd) + abc + def.
\end{align*} Taking mod $p$, we find that $P(x) \equiv Q(x) \pmod p$.

However, plug in $x = d$. We know $Q$ has a root at $d$, so \begin{align*}
    P(d) = (d + a)(d + b)(d + c) \equiv 0\pmod p,
\end{align*} but $p > d + a, d + b, d + c$ and so we are done.

$S\geq6$, so assume for the sake of contradiction that $S$ is prime. Let $P(x)\coloneqq (x-a)(x-b)(x-c)$ and $Q(x)\coloneqq (x+a)(x+b)(x+c)$. Multiplying out, we get that $P(x)\equiv Q(x)\;(\text{mod}\,S)$ for all integers $x$. In particular, we get that for an integer $x$, $S$ divides $P(x)$ if and only if $S$ divides $Q(x)$, which, because $S$ is assumed to be a prime, means that $\{a, b, c\}$ is just some permutation of $\{-d, -e, -f\}$ modulo $S$. WLOG, assume that $a\equiv -d$, $b\equiv -e$ and $c\equiv -f$ modulo $S$. But this means that $a+d$, $b+e$ and $c+f$ are all multiples of $S$, which is impossible, because they are positive integers which sum up to $S$.

Note that $(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)$ has all of its coefficients divisible by $S$. This implies $S|(d+a)(d+b)(d+c)$. FTSOC assume $S$ is prime. Then it must divide either $d+a,d+b,d+c$, all of which are smaller than $S$, a contradiction, so we are done.

Assume for the sake of contradiction that $S$ is prime. Notice that $$S \mid (a+k)(b+k)(c+k)+(d-k)(e-k)(f-k)$$
for all $k \in \mathbb{Z}$. \\
Then taking $k$ to be $-a,-b,-c$ we have that $\{d,e,f\} = \{-a,-b,-c\}$ in $\mathbb{F}_S$ using that $S$ is prime, yet, adding any such values of $a,b,c,d,e,f$, we have that $$S = a+b+c+d+e+f \geq 3S$$ which is a contradiction. $\blacksquare$ \\

Suppose that $S$ is a prime and let $x=-d,y=-e,z=-f$,so that $abc \equiv xyz \pmod S$
Now consider the polynomial $ (a+k)(b+k)(c+k)+(d-k)(e-k)(f-k)$ and clearly inputting $x,y,z$ implies $S|(a+d)(a+e)(a+f)$,which implies that $S$ divides one of $(a+d),(a+e),(a+f)$,a contradiction.

Note that the polynomial $f(x) = (x-a)(x-b)(x-c)-(x+d)(x+e)(x+f) = (ab+ac+bc-de-ef-df)x-(abc+def)$ is divisible by $S$ for all integers $x$. Thus,
\[S\mid f(a) = -(a+d)(a+e)(a+f)\]
To finish, $S$ cannot be prime because $a+d,a+e,a+f<a+b+c+d+e+f = S$, so we're done. $\blacksquare$.

[hide=Sol][quote=ISL 2005 N3]Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ be positive integers and let $ S = a+b+c+d+e+f$.
Suppose that the number $ S$ divides $ abc+def$ and  $ ab+bc+ca-de-ef-df$. Prove that $ S$ is composite.[/quote]
Define $P(x)=(x+a)(x+b)(x+c),Q(x)=(x-d)(x-e)(x-f)$
Note that
$S|P(x)-Q(x) \implies S|P(d)-Q(d)=(d+a)(d+b)(d+c)$.
Since $(d+a),(d+b),(d+c)<S$,
$S$ can't be prime and we are done$\blacksquare$

First N3 and surely sillies it cuz I was like staring at $(d+b),(d+c),(d+a)$, before coming to my wits

[/hide]

Notice \begin{align*}S\mid f(x)&=Sx^2+(ab+bc+ca-de-df-fd)x+(abc+def)\\&=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)\end{align*} Letting $x=d,$ we have $S\mid (d+a)(d+b)(d+c),$ which is absurd if $S$ is prime as $S>d+a,d+b,d+c.$ $\square$

Assume for the sake of contradiction $S$ is prime, and take polynomials $P(x)=(x-a)(x-b)(x-c)$ and $Q(x)=(x+d)(x+e)(x+f).$ Note that \[P(x)\equiv x^3 - (a+b+c)x^2 + (ab+bc+ca)x-abc \pmod S\]
Note that on the other hand, \[Q(x)\equiv x^3 + (d+e+f)x^2 + (de+ef+fd)x+def \pmod S\]
Now, $Q(x)-P(x)=Sx^2+(de+ef+fd-ab-bc-ca)x+abc+def.$ Note that this is divisible by $S.$ In particular, $Q(a)-P(a)=(a+d)(a+e)(a+f)$ is divisible by $S.$ Note that each of those factors are less than $S$ but $S$ is prime, which is a contradiction.

Observe that we have the system of congruences
\begin{align*}
a+b+c &\equiv -d-e-f \pmod S \\
ab+bc+ca &\equiv de+ef+fd \pmod S \\
abc &\equiv -def \pmod S.
\end{align*}
This means that $$(x-a)(x-b)(x-c) \equiv (x+d)(x+e)(x+f) \pmod S$$ for all $x \in \mathbb Z$ as the resulting polynomials are equivalent under modulo $S$. Letting $x=a$, $$S \mid (a+d)(a+e)(a+f),$$ but $a+d, a+e, a+f < S$, and thus $S$ must be composite.

Suppose otherwise that $S$ is prime $p$. Then we have the following equations.
\begin{align*}
a+b+c \equiv -(d+e+f) \pmod p \\
ab+bc+ca \equiv de+ef+df \pmod p \\
abc \equiv -def \pmod p
\end{align*}
We will construct numbers $a_1,b_1, c_1, d_1, e_1, f_1$ such that:
\begin{align*}
a_1+b_1+c_1 \equiv -(d_1+e_1+f_1) \pmod p \\
a_1b_1+b_1c_1+c_1a_1 \equiv d_1e_1+e_1f_1+d_1f_1 \pmod p \\
a_1b_1c_1 \equiv -d_1e_1f_1 \pmod p
\end{align*}
In fact, this is not so hard to do. It is just enough to take $a_1=a+1, b_1=b+1, c_1 =c_1$ and $d_1 =d-1, e_1 =e-1, f_1 =f-1$. It is easy to see that they satisfy the first congruence. Note that:
\begin{align*}
(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1) \equiv (d-1)(e-1)+(e-1)(f-1)+(f-1)(d-1) \pmod p\\
ab+bc+ca +2(a+b+c)+3 \equiv de+ef+fd -2(d+e+f)+3 \pmod p
\end{align*}
We see that the second desired congruence is also satisfied by given congruences. Also, observe that:
\begin{align*}
(a+1)(b+1)(c+1) \equiv -(d-1)(e-1)(f-1)  \pmod p \\
a+b+c+(ab+bc+ca)+abc+1 \equiv -((d+e+f)-(de+ef+fd)+def-1) \pmod p \\
(a+b+c+de+f) +(ab+bc+ca-de-ef-fd)+(abc+def) \equiv 0 \pmod p
\end{align*}
We see that the third desired congruence is also satisfied by given congruences.

Since $a,b,c, d,e,f$ are positive integers and their sum it $p$, then they all are less than $p$. WLOG $a$ is the largest among $a,b,c$. With each construction we increase $a,b,c$ by $1$ and decrease $d,e,f$ by $1$. We can keep doing it until $a$ becomes $p$. Suppose that we needed for that $k$ constructions, in other words $a+k =p=a_k$. Consider congruence:
$$ a_kb_kc_k \equiv d_ke_kf_k \equiv 0 \pmod p $$
We see that it implies that $d_ke_kf_k \equiv 0 \pmod p$. WLOG assume that $d_k = d-k$ is the one which is divisible by $p$. Then $0 \equiv a_k + d_k = a+k+d-k = a+ d \pmod p$, which is contradiction, since it forces $b,c,e,f$ to be $0$.

Our initial assumption was wrong, therefore $S$ composite as desired.

I'm not smart as others to see that you can really form a polynomial... Anyways here is a solution by induction which kind of makes the polynomial in a different way which I feel is intuitive too...

[i]Solution.[/i] Assume on the contrary that $S$ is prime. We will make use of these handy identities in the solution.
    \begin{align}
        (a+1)(b+1)(c+1) &= abc + (ab+bc+ca) + (a+b+c) + 1 \\
        (d-k)(e-k)(f-k) &= def - (de+ef+df) + (d+e+f) - 1
    \end{align}
The following claim is the crux of the problem.

[b][color=red]Claim:[/b][/color] $S \mid (a+x)(b+x)(c+x) + (d-x)(e-x)(f-x)$ for all $x \in \mathbb{Z}_{\ge 0}$.

[i]Proof:[/i] The proof is by induction with the base case being trivial. As an induction hypothesis, assume that $S \mid (a+k)(b+k)(c+k) + (d-k)(e-k)(f-k)$. For ease of notation label $a+k = A$, $d - k = D$ and so on. Now finally for the inductive step, we will just work backward.
        \begin{align*}
            S \mid (A+1)(B+1)(C+1) &+ (D-1)(E-1)(F-1) \\
            \iff S \mid (ABC + DEF) + (AB+BC+CA) &- (DE+EF+FA) + (A+B+C+\ldots)
        \end{align*}
By the induction hypothesis, we will cancel of $ABC + DEF$ term. It is not hard to see that $S \mid (A+B+C+D+E+F)$ as well.
        \begin{align*}
            \iff S \mid AB + BC + CA - DE - EF - FA
        \end{align*}
It isn't hard to expand this out and realize its a tautology. So, the induction is complete.$\square$
------------
Finally choose $x = d$. This would give
    \[S = a + b + c + d + e + f \mid (a + d)(b + d)(c+d).\]
Since $S$ is a prime, it must divide one of the factors. But since all the factors are positive and less than $S$, its the desired contradiction. $\blacksquare$

Let's allow $0$ as value of $d,e,f$.
FTSOC, let's assume that $S = p$ is a prime.
Here, \begin{align*}
S &= a+b+c+d+e+f\\ &= (a+1) + (b+1)+(c+1) + (d-1)+(e-1)+(f-1)
\end{align*}

Now, $p\mid a+b+c+d+e+f$, $p\mid abc+def$ and $p\mid ab+bc+ca-de-ef-df$. Summing these up gives, $$p\mid (a+1)(b+1)(c+1) + (d-1)(e-1)(f-1)$$
Also, \begin{align*}
&\sum_{cyc}(a+1)(b+1) - \sum_{cyc}(d-1)(e-1)\\
=& ab+bc+ca-de-ef-df +2(a+b+c+d+e+f)
\end{align*}

Therefore, $\displaystyle p\mid \sum_{cyc}(a+1)(b+1) - \sum_{cyc}(d-1)(e-1)$.
So, if $(a,b,c,d,e,f)$ is a solution then, $(a+1, b+1, c+1, d-1,e-1,f-1)$ is also a solution with $S= p$.

Now we continue this process until one of the elements becomes $0$. WLOG that be $f$.
Then we will get, $p\mid abc$. But therefore $p$ must divide one of $a,b,c$ which isn't possible as, $p = a+b+c+d+e$. This gives a contradiction. $\blacksquare$

Suppose that $S$ was prime. Then the polynomials $(x+a)(x+b)(x+c)$ and $(x-d)(x-e)(x-f)$ have equivalent expansions modulo $p$, hence we must have $\{-a,-b,-c\} \equiv \{d,e,f\} \pmod{S} \implies \{S-a,S-b,S-c\}=\{d,e,f\}$. But then $d+e+f=3S-a-b-c \implies S=a+b+c+d+e+f=3S$: contradiction. $\blacksquare$

Consider $P(x)=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)$\\

$P(x)=Sx^2+x(ab+bc+ca-ed-ef-df)+(abc+def) \implies S|P(x)$ $\forall x \in \mathbb{Z}$\\

$P(d)=(a+d)(b+d)(c+d)$ hence as $S|P(d)$. Now , $\mathrm{FTSOC}$ assume $S$ is prime, then as we have $S|(a+d)(b+d)(c+d)$ we have at least one of $(a+d), (b+d) , (c+d)> S$ but since $S=a+b+c+d+e+f$ this is absurd for $a,b,c,d,e, f \in \mathbb{Z}^{+}$ , hence our assumption false and hence $S$ is composite $\blacksquare$.

Solved with [b]xor, xook, and xonk[/b], also known as [b]fuzimiao2013, sixoneeight, popop614[/b].

We can rewrite the condition as
\[
(t-a)(t-b)(t-c) \equiv (t+d)(t+e)(t+f) \pmod{S}
\]
for all integer $t$. Suppose for the sake of contradiction that $S$ was prime. Take $t=a$. Then, $S$ divides one of $a+d$, $a+e$, $a+f$. However, since $a,b,c,d,e,f$ are positive integers, that would mean that one of $a+d, a+e, a+f$ is equal to $S$. This is a contradiction as the rest of the variables could not be positive, so $S$ must be composite.

FTSOC let $S$ be a prime
$a+b+c \equiv -(d+e+f)$
$abc \equiv (-d)(-e)(-f)$
$ab+bc+ca \equiv (-d)(-e)+(-e)(-f)+(-d)(-f)$
In $\mathbb{F}_S$
$(x-a)(x-b)(x-c) \equiv (x+d)(x+e)(x+f)$
therefore $\{a,b,c\} \equiv \{-d,-e,-f\}$ in $\mathbb{F}_S$
$\therefore$  we're done.


geometry
incenter
vector
geometric transformation
reflection
analytic geometry
geometry unsolved

Let $ABC$ be triangle, and $T$ its centroid (i. e. the point of intersection of the three medians of triangle $ABC$). Let $p$ be a line passing through the point $T$.

The line $p$ intersects the line $BC$ at a point $P$;
the line $p$ intersects the line $CA$ at a point $Q$;
the line $p$ intersects the line $AB$ at a point $R$,
such that the points $Q$ and $R$ lie on one side of the point $T$.

Prove that $\frac{1}{TR}+\frac{1}{TQ}=\frac{1}{TP}$.

[b]Theorem of the transversal.[/b]
\[\{A,B,C\}\subset d,\ V\not \in d;\ M\in VA,\ N\in VB,\ P\in VC,\ P\in MN\Longrightarrow \]

\[\overline{\underline{\left| \frac{\overline{MA}}{\overline{MV}}\cdot \overrightarrow{BC}+\frac{\overline{NB}}{\overline{NV}}\cdot \overrightarrow{CA}+\frac{\overline{PC}}{\overline{PV}}\cdot \overrightarrow{AB}=\overrightarrow{0}\right| }}. \]
[u]A particular cases.[/u] For the centroid $N$ of the triangle $VAC$ we obtain the following relation: $\frac{\overline{MA}}{\overline{MV}}+\frac{\overline{PC}}{\overline{PV}}+1=0$, i.e. $\frac{\overline{AM}}{\overline{MV}}+\frac{\overline{CP}}{\overline{PV}}=1$. If $M\in (VA)$, $P\in (VC)$ then $\frac{MA}{MV}+\frac{PC}{PV}=1.$ For the incenter $N$ of the triangle $VAC$ what becomes the relation $(*)\ ?$ a.s.o.
[b]Remark.[/b] The standard notations: the [u]line[/u] $AB$ or the [u]length[/u] $AB$ of the [u]segment[/u] $[AB]$; the [u]rays[/u] $[AB$ or $(AB$; the [u]segments[/u] $[AB]$, $(AB)$, $[AB)$, $(AB]$; the [u]vector[/u] $\overrightarrow{AB}$; the [u]algebraic measure[/u] $\overline{AB}$ of the [u]ordered pair[/u] $(A,B)$ on the line $d$, i.e. $A(a)$, $B(b)\Longrightarrow \overline{AB}=b-a\in R$.

A remark:

  I wonder what kind of name is Transversal?

  Shouldn't the correct spelling be transVersailles?

  Or maybe it would be more appropriate to call it

  the theorem  of TransSilvan?  It's there where it originated,

  wasn't it?

  Thank you.

  M.T.

[i]Arm_pish[/i], thanks for the your refined and clever words ! The Transilvania is a beautiful country (a paradise on earth), like to Pennsylvania, if you was there. There I was born long ago, my dear !
With friendship, [i]V.N.[/i]

[[b]Moderator edit:[/b] This post is merged from a different thread, and it uses different notations. In fact, the following notations are used:

[quote="dule_00"]Let $ABC$ be triangle, and $M$ its centroid (i. e. the point of intersection of the three medians of triangle $ABC$). Let $p$ be a line passing through the point $M$.

The line $p$ intersects the line $BC$ at a point $R$;
the line $p$ intersects the line $CA$ at a point $S$;
the line $p$ intersects the line $AB$ at a point $T$,
such that the points $R$ and $S$ lie on one side of the point $M$.

Prove that $\frac{1}{MR}+\frac{1}{MS}=\frac{1}{MT}$.[/quote]

Darij]

name the intersection of $AM,BC$ ,   $F$.

its clear that $AM=2MF$.from $A$ draw a paralell line to $BC$,and let it intersect $MT$ at $K$.

triangles $AKM$ and $MFR$ are simmilar to each other, then we should have: $MK=2MR$

it is known that: $(KMTS)=-1$    (why? if u dont know, u can ask me later)

then we should this fact about $K,M,T,S$:

$SK.MT=KT.MS$

then $SK.MT=(MK-MT).MS$

then $SK.MT=MK.MS-MS.MT$

then $(MS+MK).MT+MS.MT=MK.MS$

then $\frac{2MS+MK}{MK.MS}=\frac{1}{MT}$

and then: $\frac{2}{MK}+\frac{1}{MS}=\frac{1}{MT}$

and because $MK=2MR$, we should have:

$\frac{1}{MR}+\frac{1}{MS}=\frac{1}{MT}$

and every thing is ok...................................

i start to solve some kinds of problems by a very very very incredible tool.may be more incredible than $INVERSION$.

its name is..............................................$PROJECTION$ :D

ok first i should say an important theorem.

theorem:

suppose $A,B,C$ are three non colinear point in plane $K$. and $M,N,P$, are also three non colinear point in plane$K'$

we can fix $K,K'$ in space, inwhich, if a paralell projection from $K$ to $K'$, changes $A,B,C$ to $A',B',C'$ ,then

triangle $A'B'C'$ be simmilar to triangle $MNP$.

(parallel projection, is like sun and shade)

theorem:

suppose $AB$ is a segment on plane$K$, and $A'B'$ is its projection form on plane$K'$.

for each point $C$ on segment $AB$, we have:

$\frac{AC}{A'C'}=\frac{BC}{B'C'}$

hmmmmmmmmmmm,,, dears im sorry because i dont have time to post the proofs of these two theorem. ;)

ok, according to theorems it is enough ,if i prove the problem for an equaliteral triangle.(why?)

then the problem

[quote="dule_00"]Let $ABC$ be triangle, and $M$ its centroid (i. e. the point of intersection of the three medians of triangle $ABC$). Let $p$ be a line passing through the point $M$.

The line $p$ intersects the line $BC$ at a point $R$;
the line $p$ intersects the line $CA$ at a point $S$;
the line $p$ intersects the line $AB$ at a point $T$,
such that the points $R$ and $S$ lie on one side of the point $M$.

Prove that $\frac{1}{MR}+\frac{1}{MS}=\frac{1}{MT}$.[/quote]

changes to this one:

$ABC$ is an equaliteral triangle. and $M$ is its centroid. an arbitrary line goes through $M$ and intersects the sides

at $R,S,T,$, inwhich $R,S$ are in the same side.

prove that: $\frac{1}{MR}+\frac{1}{MS}=\frac{1}{MT}$.

This problem is the first particular case from the generalization posted to  http://www.mathlinks.ro/Forum/viewtopic.php?t=86470

[quote="Virgil Nicula"][b]Theorem of the transversal.[/b]
\[ \{A,B,C\}\subset d,\ V\not \in d;\ M\in VA,\ N\in VB,\ P\in VC,\ P\in MN\Longrightarrow
\]

\[ \overline{\underline{\left| \frac {\overline{MA}}{\overline{MV}}\cdot \overrightarrow{BC} \plus{} \frac {\overline{NB}}{\overline{NV}}\cdot \overrightarrow{CA} \plus{} \frac {\overline{PC}}{\overline{PV}}\cdot \overrightarrow{AB} \equal{} \overrightarrow{0}\right| }}.
\]
[u]A particular cases.[/u] For the centroid $ N$ of the triangle $ VAC$ we obtain the following relation: $ \frac {\overline{MA}}{\overline{MV}} \plus{} \frac {\overline{PC}}{\overline{PV}} \plus{} 1 \equal{} 0$, i.e. $ \frac {\overline{AM}}{\overline{MV}} \plus{} \frac {\overline{CP}}{\overline{PV}} \equal{} 1$. If $ M\in (VA)$, $ P\in (VC)$ then $ \frac {MA}{MV} \plus{} \frac {PC}{PV} \equal{} 1.$[/quote]

Stupid question to ask after 2 years, but in how far does this solve the problem? It rather looks like a solution to http://www.mathlinks.ro/viewtopic.php?t=4828 ...

By the way, see [url=http://www.cip.ifi.lmu.de/~grinberg/S64.pdf]Solution of problem S64 from Mathematical Reflections[/url] for another solution of the original problem.

  darij

[quote="dule_00"]Let $ ABC$ be triangle, and $ T$ its centroid (i. e. the point of intersection of the three medians of triangle $ ABC$). Let $ p$ be a line passing through the point $ T$.

The line $ p$ intersects the line $ BC$ at a point $ P$;
the line $ p$ intersects the line $ CA$ at a point $ Q$;
the line $ p$ intersects the line $ AB$ at a point $ R$,
such that the points $ Q$ and $ R$ lie on one side of the point $ T$.

Prove that $ \frac {1}{TR} \plus{} \frac {1}{TQ} \equal{} \frac {1}{TP}$.[/quote]
It is not hard to prove that $ \frac{TP}{TR}\plus{}\frac{TP}{TQ}\equal{}1$ using Affine coordinates


geometry
circumcircle
ratio
trigonometry
cyclic quadrilateral
angle bisector
perpendicular bisector

Let $ABCD$ be a cyclic quadrilateral, and let $L$ and $N$ be the midpoints of its diagonals $AC$ and $BD$, respectively. Suppose that the line $BD$ bisects the angle $ANC$. Prove that the line $AC$ bisects the angle $BLD$.

You must prove two affirmations are equivalent.
In fact one of them is equivalent to following: the tangents at A,C intersect on BD;
the other with: the tangents at B,D intersect on AC.
These two affirmation are both equivalent to AB*CD=AD*BC
This problem was invented in Kvant (by Sharyghin I think) about 20-25 years ago

Here is my proof :
         Let R, S be the intersections of OL and BD, ON and AC. Hence RANC is cyclic.  Thus $RL.RO=RI.RN=RA^2$ (by showing that the $ RAI ~ RNA$). This shows that RA is the tangent of (O) at A. Similarly, we have RC is the tangent of (O). Therefore, ABCD is a harmonic quadilateral. We obtain SB and SC are tangents of (O).
         Now we have $SN.SO=SB^2=SI.SL$ .Thus $\measuredangle {SBC}=\measuredangle {BLS}$. In the same manner, we finally obtain LS is the bisector of $\measuredangle {BLD}$. :cool:

I remember posting this some time ago, but it got locked because it appeared in a not-so-old issue of Komal :). My solution was basicly the same as treegoner's.

[quote="treegoner"]Here is my proof :
         Let R, S be the intersections of OL and BD, ON and AC. Hence RANC is cyclic.  Thus $ RL.RO \equal{} RI.RN \equal{} RA^2$ (by showing that the $ RAI \sim  RNA$). This shows that RA is the tangent of (O) at A. Similarly, we have RC is the tangent of (O). Therefore, ABCD is a harmonic quadilateral. We obtain SB and SC are tangents of (O).
         Now we have $ SN.SO \equal{} SB^2 \equal{} SI.SL$ .Thus $ \measuredangle {SBC} \equal{} \measuredangle {BLS}$. In the same manner, we finally obtain LS is the bisector of $ \measuredangle {BLD}$.[/quote]

I was asked by a friend to explain this solution in a more detailed way. I'm doing this publicly, since other people may also be interested (in fact, the above solution was written up very summarily, even I had many troubles with understanding it, moreover the point I was not properly introduced).

We will need three lemmas (you probably know some of them):

[b]Lemma 1.[/b] For any triangle ABC, the point of intersection of the angle bisector of the angle CAB with the perpendicular bisector of the segment BC lies on the circumcircle of triangle ABC.

[i]Proof.[/i] The angle bisector of the angle CAB meets the circumcircle of triangle ABC at two points: One of them is obviously A; the other one will be denoted by X. Then, < BAX = < XAC. Hence, the chords BX and CX in the circumcircle of triangle ABC have equal chordal angles; this yields that they have also equal lengths, i. e. we have BX = CX. Consequently, the point X lies on the perpendicular bisector of the segment BC. In other words, the point X is the point of intersection of the angle bisector of the angle CAB with the perpendicular bisector of the segment BC. But since our point X lies on the circumcircle of triangle ABC, we thus have proven that the point of intersection of the angle bisector of the angle CAB with the perpendicular bisector of the segment BC lies on the circumcircle of triangle ABC. Lemma 1 is proven.

[b]Lemma 2.[/b] Let the tangent to the circumcircle of a triangle ABC at the vertex A meet the sideline BC at the point D. Then,

  $ \displaystyle \frac {BD}{CD} \equal{} \frac {AB^{2}}{CA^{2}}$.

[i]Proof.[/i] As an angle between a tangent and a chord, the angle < DAB is equal to the chordal angle < ACB of the chord AB. So we have < DAB = < ACB = < DCA. (Our angles are not directed, so we don't care for the orientation.) Furthermore, < ADB = < CDA. Thus, the triangles DAB and DCA have two equal pairs of angles, and therefore are similar. Consequently,

  $ \displaystyle \frac {BD}{AD} \equal{} \frac {AD}{CD} \equal{} \frac {AB}{CA}$.

Hence,

  $ \displaystyle \frac {BD}{CD} \equal{} \frac {BD}{AD}\cdot \frac {AD}{CD} \equal{} \frac {AB}{CA}\cdot \frac {AB}{CA} \equal{} \frac {AB^{2}}{CA^{2}}$,

and Lemma 2 is proven.

[b]Lemma 3.[/b] Let ABCD be a cyclic quadrilateral with the circumcircle k. Then we have the following equivalence: The tangents to k at the vertices A and C intersect on the diagonal BD if and only if the tangents to k at the vertices B and D intersect on the diagonal AC.

[i]Proof.[/i] Let the tangent to k at the vertex A meet the diagonal BD at a point A', and let the tangent to k at the vertex C meet the diagonal BD at a point C'. Then, the tangents to k at the vertices A and C intersect on the diagonal BD if and only if A' = C'. This is obvious. But now we are interested in seeing when A' = C' occurs. Applying Lemma 2 to the triangles DAB and BCD, we find the ratios in which the points A' and C' divide the diagonal BD:

  $ \displaystyle \frac {BA^{\prime }}{DA^{\prime }} \equal{} \frac {AB^{2}}{DA^{2}}$;
  $ \displaystyle \frac {BC^{\prime }}{DC^{\prime }} \equal{} \frac {CB^{2}}{DC^{2}}$.

Hence, we have A' = C' if and only if these two ratios are equal, what is equivalent to $ \displaystyle \frac {AB^{2}}{DA^{2}} \equal{} \frac {CB^{2}}{DC^{2}}$, i. e. to $ AB^{2}\cdot DC^{2} \equal{} DA^{2}\cdot CB^{2}$. This condition can be rewritten as $ AB^{2}\cdot CD^{2} \equal{} DA^{2}\cdot BC^{2}$, or as $ AB\cdot CD \equal{} DA\cdot BC$. Hence, we have proven that the tangents to k at the vertices A and C intersect on the diagonal BD if and only if $ AB\cdot CD \equal{} DA\cdot BC$. Similarly, we can show that the tangents to k at the vertices B and D intersect on the diagonal AC if and only if $ AB\cdot CD \equal{} DA\cdot BC$ (the condition is exactly the same, since it is symmetric!). Hence, the tangents to k at the vertices A and C intersect on the diagonal BD if and only if the tangents to k at the vertices B and D intersect on the diagonal AC. This completes the proof of Lemma 3.

Now to the solution of our problem:

Let the diagonals AC and BD of our cyclic quadrilateral ABCD intersect at the point I.

If O is the circumcenter of our cyclic quadrilateral ABCD, then the line ON is the perpendicular bisector of the segment BD. This is because the point O is the circumcenter of the quadrilateral ABCD, i. e. the center of a circle through the points A, B, C, D, so that it lies on the perpendicular bisector of the segment BD, and the midpoint N of this segment BD obviously lies on its perpendicular bisector, too. Similarly, the line OL is the perpendicular bisector of the segment AC. Let the lines OL and BD meet at R.

The point R, lying both on the line OL, i. e. the perpendicular bisector of the segment AC, and on the line BD, i. e. the angle bisector of the angle ANC (according to the problem!), can be considered as the point of intersection of the angle bisector of the angle ANC with the perpendicular bisector of the segment AC. Hence, after Lemma 1, this point R lies on the circumcircle of triangle ANC. In other words, the points A, N, C, R lie on one circle. Therefore, < RAC = < RNC. But < RNC = < ANR (since the line BD bisects the angle ANC). Hence, < RAC = < ANR. We can rewrite this as < RAI = < RNA. Also, obviously, < ARI = < NRA. Hence, the triangles RAI and RNA are similar, and it follows that $ \displaystyle \frac {RA}{RI} \equal{} \frac {RN}{RA}$, so that $ RA^{2} \equal{} RN\cdot RI$.

On the other hand, the triangles RLI and RNO are similar (since < RLI = 90° and < RNO = 90°, so that < RLI = < RNO, and evidently < LRI = < NRO). Hence, $ \displaystyle \frac {RL}{RI} \equal{} \frac {RN}{RO}$, so that $ RL\cdot RO \equal{} RN\cdot RI$. Comparing with $ RA^{2} \equal{} RN\cdot RI$, we get $ RA^{2} \equal{} RL\cdot RO$, so $ \displaystyle \frac {RA}{RL} \equal{} \frac {RO}{RA}$. Together with < ARL = < ORA, this entails that the triangles ARL and ORA are similar. Hence, < RAO = < RLA. But < RLA = 90°; hence, < RAO = 90°, and the line RA is perpendicular to the line AO. This means that the line RA is tangent to the circumcircle k of the quadrilateral ABCD; in other words, the line RA is the tangent to k at the vertex A. Similarly, we can prove that the line RC is the tangent to k at the vertex C. But since the point R lies on the diagonal BD, we see that the tangents to k at the vertices A and C intersect on the diagonal BD.

Now we can apply Lemma 3, which tells us that the tangents to k at the vertices B and D intersect on the diagonal AC. Let S be the point of intersection of these tangents. Then S lies on the diagonal AC. Also, since SB = SD (the two tangents from a point to a circle have equal length), the point S lies on the perpendicular bisector of the segment BD, i. e. on the line ON.

Since the point S lies on the tangent to k at the vertex B, the line SB is perpendicular to the line BO. This means that < SBO = 90°. We also have < SNB = 90°. Therefore, < SBO = < SNB; furthermore, we trivially have < BSO = < NSB. Therefore, the triangles SBO and SNB are similar, and we get $ \displaystyle \frac {SB}{SO} \equal{} \frac {SN}{SB}$, so that $ SB^{2} \equal{} SN\cdot SO$. But since < SLO = < SNI (this is because < SLO = 90° and < SNI = 90°) and < LSO = < NSI, the triangles SLO and SNI are also similar to each other, and hence we have $ \displaystyle \frac {SL}{SO} \equal{} \frac {SN}{SI}$, and thus, $ SL\cdot SI \equal{} SN\cdot SO$. Comparing this with $ SB^{2} \equal{} SN\cdot SO$, we get $ SB^{2} \equal{} SL\cdot SI$, or, in other words, $ \displaystyle \frac {SB}{SL} \equal{} \frac {SI}{SB}$. Together with < BSL = < ISB, this yields that the triangles BSL and ISB are similar.

Therefore, < BLS = < IBS. In other words, < BLS = < DBS. Similarly, < DLS = < BDS. But since the triangle BSD is isosceles (in fact, we already have noted that SB = SD), we have < DBS = < BDS, and therefore, < BLS = < DLS. This shows that the line AC bisects the angle BLD.

  Darij

[quote="dreammath"][color=darkred]Let $ ABCD$ be a cyclic quadrilatedral and let $ L$ , $ N$ be the midpoints of its diagonals $ AC$ and $ BD$ respectively.
Prove that $ \widehat{BNA}\equiv\widehat{BNC}\Longleftrightarrow$ $ AB\cdot CD=AD\cdot BC$ $ \Longleftrightarrow$ $ \widehat{CLB}\equiv\widehat{CLD}$.
[/color][/quote]
[color=darkblue][b]Proof I (similarly to [u]Iura[/u]'s)[/b]. Observe that $ \{\begin{array}{c}ON\perp BD\\ \ OL\perp AC\end{array}$, where $ O$ is the circumcenter of $ ABCD$. Denote $ \{\begin{array}{c}S\in AC\cap BD\\ \ X\in AC\cap ON\\ \ Y\in BD\cap OL\end{array}$.

$ 1\blacktriangleright$ Suppose that $ \widehat{ANB}\equiv\widehat{CNB}$. Since the rays $ [NS$, $ [NX$ are the bisectors of the angles between the lines $ NA$, $ NC$,
then the points $ X$, $ S$ are conjugate w.r.t. the points $ A$, $ C$. Since $ SN\perp OX$, then the line $ BD$ is the polar of the point $ X$ and $ X\in BB\cap DD$.
From an well-known property obtain that the quadrilateral $ ABCD$ is harmonically, i.e. $ AB\cdot CD = AD\cdot BC$.
Therefore, $ \boxed{\widehat{ANB}\equiv\widehat{CNB}\Longleftrightarrow AB\cdot CD = AD\cdot BC}$.

$ 2\blacktriangleright$ Suppose that $ \widehat{BLC}\equiv\widehat{DLC}$. Since the rays $ [LS$, $ [LY$ are the bisectors of the angles between the lines $ LB$, $ LD$,
then the points $ Y$, $ S$ are conjugate w.r.t. the points $ B$, $ D$. Since $ SL\perp OY$, then the line $ AC$ is the polar of the point $ Y$ and $ Y\in AA\cap CC$.
From an well-known property obtain that the quadrilateral $ ABCD$ is harmonically, i.e. $ AB\cdot CD = AD\cdot BC$.
Therefore, $ \boxed{\widehat{BLC}\equiv\widehat{DLC}\Longleftrightarrow AB\cdot CD = AD\cdot BC}$.

[b]Proof II (metrical - [u]ugly !?[/u])[/b]. Denote $ \{\begin{array}{c}AB = a\ ,\ BC = b\ ,\ CD = c\\ \ DA = d\ ,\ AC = e\ ,\ BD = f\end{array}$. For the point $ S\in AC\cap BD$ prove easily that
$ \frac{SA}{da}=\frac{SB}{ab}=\frac{SC}{bc}=\frac{SD}{cd}=\frac{e}{ad+bc}=\frac{f}{ab+cd}$. From the [b]Ptolemeu's relations [/b]$ \{\begin{array}{c}ef = ac+bd\\ \\ \frac{e}{f}=\frac{ad+bc}{ab+cd}\end{array}$ obtain $ \{\begin{array}{c}e^{2}=\frac{(ac+bd)(ad+bc)}{ab+cd}\ \ (1)\\ \\ f^{2}=\frac{(ac+bd)(ab+cd)}{ad+bc}\ \ (2)\end{array}$
$ 1\blacktriangleright$ the ray $ [NB$ is the bisector of the angle $ \widehat{ANC}$ $ \Longleftrightarrow$ $ \frac{NA}{NC}=\frac{SA}{SC}$ $ \Longleftrightarrow$ $ \frac{4\cdot NA^{2}}{4\cdot NC^{2}}=\frac{a^{2}d^{2}}{b^{2}c^{2}}$ $ \Longleftrightarrow$ $ \frac{2(a^{2}+d^{2})-f^{2}}{2(b^{2}+c^{2})-f^{2}}=\frac{a^{2}d^{2}}{b^{2}c^{2}}$ $ \Longleftrightarrow$
$ f^{2}=\frac{2b^{2}c^{2}(a^{2}+d^{2})-2a^{2}d^{2}(b^{2}+c^{2})}{b^{2}c^{2}-a^{2}d^{2}}$ $ \Longleftrightarrow$ $ 2b^{2}c^{2}(a^{2}+d^{2})-2a^{2}d^{2}(b^{2}+c^{2}) = (ac+bd)(ab+cd)(bc-ad)$ $ \Longleftrightarrow$ $ (ab-cd)(bc+ad)(bd-ac) = 0$ $ \Longleftrightarrow$ $ ab = cd\ \ \vee\ \ ac = bd$ $ \Longleftrightarrow$ the line $ AC$ is the bisector of the segment $ [BD]$ [b]OR[/b] the quadrilateral $ ABCD$ is harmonically.

$ 2\blacktriangleright$ the ray $ [LC$ is the bisector of the angle $ \widehat{BLD}$ $ \Longleftrightarrow$ $ \frac{LB}{LD}=\frac{SB}{SD}$ $ \Longleftrightarrow$ $ \frac{4\cdot LB^{2}}{4\cdot LD^{2}}=\frac{a^{2}b^{2}}{c^{2}d^{2}}$ $ \Longleftrightarrow$ $ \frac{2(a^{2}+b^{2})-e^{2}}{2(c^{2}+d^{2})-e^{2}}=\frac{a^{2}b^{2}}{c^{2}d^{2}}$ $ \Longleftrightarrow$
$ e^{2}=\frac{2c^{2}d^{2}(a^{2}+b^{2})-2a^{2}b^{2}(c^{2}+d^{2})}{c^{2}d^{2}-a^{2}b^{2}}$ $ \Longleftrightarrow$ $ 2c^{2}d^{2}(a^{2}+b^{2})-2a^{2}b^{2}(c^{2}+d^{2}) = (ac+bd)(ad+cb)(dc-ab)$ $ \Longleftrightarrow$ $ (ad-bc)(cd+ab)(bd-ac) = 0$ $ \Longleftrightarrow$ $ ad = bc\ \ \vee\ \ ac = bd$ $ \Longleftrightarrow$ the line $ BD$ is the bisector of the segment $ [AC]$ [b]OR[/b] the quadrilateral $ ABCD$ is harmonically.

[b][u]In conclusion[/u][/b] : if a diagonal is the bisector of the another, then is evidently that $ \widehat{BNA}\equiv\widehat{BNC}\Longleftrightarrow$ $ ac=bd$ $ \Longleftrightarrow$  $ \widehat{CLB}\equiv\widehat{CLD}$ ;

in contrary case, $ \widehat{BNA}\equiv\widehat{BNC}\Longleftrightarrow$ $ ac = bd$ $ \Longleftrightarrow$ $ \widehat{CLB}\equiv\widehat{CLD}$.
[/color]

an easier solution:
let O be center of the circle and G be intersection point of diagonals.
let ON intersect AC at E and OL intersect BD at F. cause NB and ON are in/ex-ternal angle bisector of ANC hence $ (EG,AC) \equal{}\minus{}1$ and cuase $ ON\perp BD$ hence BD is polar of E .
F lies on BD hence its polar pass through E and cause $ EC\perp OF$ hence its polar is AC and $ (FG,BD) \equal{}\minus{}1$ therefore LA is angle bisector of BLD.

Let $ M,\ K$ be, the intersection points of the circucircle $ (O)$ of the given quadrilateral $ ABCD,$ from the line segments $ ON,\ AN$ respectively and it is easy to show that $ \angle CNM \equal{} \angle MNK$ $ ,(1)$ from $ \angle ANB \equal{} \angle BNC$ and $ BN\perp OM.$

So, we have that the line segment $ AC$ is the $ A$-symmedian of the triangle $ \bigtriangleup ABD.$

Hence, based on the below [b][size=100]lemma[/size][/b], we conclude that $ \angle BLC \equal{} \angle CLD$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ A'$ be, the intersection point of its circumcircle $ (O),$ from the $ A$-symmedian. Prove that $ \angle BMA' \equal{} \angle CMA',$ where $ M$ is the midpoint of the segment $ AA'.$[/color][/size][/b]

Kostas Vittas.

PS. I will post her later the easy proof of the above [b][size=100]lemma[/size][/b], I have in mind.

[quote="vittasko"]... [b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ A'$ be, the intersection point of its circumcircle $ (O),$ from the $ A$-symmedian. Prove that $ \angle BMA' \equal{} \angle CMA',$ where $ M$ is the midpoint of the segment $ AA'.$[/color][/size][/b]

Kostas Vittas.
[/quote]
If I'm not mistaken, then $ \angle BMA'$ and $ CMA'$ are all equal to $ \angle BAC$. Here is my complicated way (???):

Let $ A_1$, $ B_1$, $ C_1$ be the middle points of the sides $ BC$, $ CA$, $ AB$. Then we have $ \triangle BAA'\sim\triangle A_1AC$. It implies that $ \angle BMA'\equal{}\angle A_1B_1C\equal{}\angle BAC$. Similarly, $ \angle CMA'\equal{}\angle BAC$. Hence we conclude that $ \angle BMA' \equal{} \angle CMA'$.

Thank you dear [b][size=100]Son[/size][/b] for a nice solution of the lemma I mentioned, which is simpler than the one I have in mind. So, I like better your approach.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ A'$ be, the intersection point of its circumcircle $ (O),$ from the $ A$-symmedian. Prove that $ \angle BMA' \equal{} \angle CMA',$ where $ M$ is the midpoint of the segment $ AA'.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] – Let $ B',\ C'$ be, the orthogonal projections of $ B,\ C,$ on $ AB,\ AC,$ respectively.

Because of $ AA'$ is the $ A$-symmedian of $ \bigtriangleup ABC,$ we have that $ \frac {AB}{AC} \equal{} \frac {A'B'}{A'C'}$ $ ,(1)$

From the similar right triangles $ \bigtriangleup B'A'B,\ \bigtriangleup C'A'C,$ we have $ \frac {A'B'}{A'C'} \equal{} \frac {A'B}{A'C}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {A'B}{A'C} \equal{} \frac {AB}{AC}$ $ ,(3)$ $ \Longrightarrow$ $ \frac {A'B}{AB} \equal{} \frac {A'C}{AC}$ $ ,(4)$

So, from $ (4),$ based on the argument of $ (3),$ we conclude that the line segment $ BC$ is the $ B$-symmedian of the triangle $ \bigtriangleup BAA'$ and then we have $ \angle A'BC \equal{} \angle ABM$ $ ,(5)$

By the same way from $ (4),$ we conclude also that the line segment $ BC$ is the $ C$-symmedian of the triangle $ \bigtriangleup CAA'$  and then, we have $ \angle A'CB \equal{} \angle ACM$ $ ,(6)$

From the triangle $ \bigtriangleup MAB$ $ \Longrightarrow$ $ \angle BMA' \equal{} \angle A'AB \plus{} \angle ABM \equal{} \angle A'CB \plus{} \angle A'BC$ $ ,(7)$

From the triangle $ \bigtriangleup MAC$ $ \Longrightarrow$ $ \angle CMA' \equal{} \angle A'AC \plus{} \angle ACM \equal{} \angle A'BC \plus{} \angle A'CB$ $ ,(8)$

From $ (7),$ $ (8)$ $ \Longrightarrow$ $ \angle BMA' \equal{} \angle CMA'$ $ ,(9)$ and the proof is comcpleted.

[b][size=100]NOTES.[/size][/b]

[b][size=100](1)[/size][/b] – It is easy to show that $ \angle BMA' \equal{} \angle CMA' \equal{} \angle A.$

[b][size=100](2)[/size][/b] – Also, from $ \angle ABM \equal{} \angle A'BC \equal{} \angle A'AC$ and $ \angle ACM \equal{} \angle A'CB \equal{} \angle A'AB,$ we conclude that the circles $ (O_{1}),\ (O_{2}),$ taken as their cords the side-segments $ AB,\ AC$ of $ \bigtriangleup ABC$ and tangent to $ AC,\ AB$ respectively, pass through the midpoint $ M$ of the segment $ AA'$

Kostas Vittas.

cyshine was too lazy to post his solution, so I'll post for him. It's similar to Amir's solution, but obtained independently (and very quickly).

    Let $ C_0$, $ C_1$ and $ C_2$ be the circumcircles of $ ABCD$, $ ANC$ and $ BLD$, respectively, and $ P$ the intersection of diagonals.
    Let $ E$,$ F$ the intersection of $ C_1$ and $ BD$, $ C_2$ and $ AC$.

    By Lemma 1, posted by darij, $ BD$ bisects $ \angle ANC \Leftrightarrow EL \bot AC$, so we need just prove $ FN \bot BD$ and we'll have $ AC$ bisects $ \angle BLD$.
    It's easy to see that P is the radical center of $ C_0$, $ C_1$ and $ C_2$, so $ EP\cdot PN \equal{} FP\cdot PL$. But this means $ EFNL$ cyclic, which implies $ \angle ENF \equal{} \angle ELF \equal{} 90^o$, ie, $ FN \bot BD$.

I am surprised that such an easy problem is on such an olympiad (when I say easy, I mean that if you know some theory, the problem is immediate?).

Anyway, let the perpendicular through $ N$ to $ BD$ intersect $ AC$ at $ E'$. Then $ E'B\equal{}E'D$ by triangle congruency. By a common lemma, $ E'AEC$ is a harmonic quadruple. It follows that $ E$ lies on the polar of $ E'$ wrt the circumscribed circle. But the polar of $ E'$ is determined by taking a line perpendicular to $ OE'$ through $ E$. But this is precisely $ BD$. So $ BE',BD'$ are tangents. Thus $ ABCD$ is harmonic. Let $ F$ be the intersection of the tangents from $ A$ and $ C$. Then $ F$ is the polar of $ AC$ so $ FBED$ is harmonic. By congruent triangles, $ \angle ALF\equal{}90$. By the common lemma, $ \angle ALB\equal{}\angle ALD$.

Well, I will use trigonometry, it is far easier for me :D

I will show an interesting result:
Let $ ABCD$ be a cyclic quadrilatedral, and let $ L$ and $ N$ be the midpoints of its diagonals $ AC$ and $ BD$, respectively.

Show that $ BN$ is angle bisector of $ \angle ANC$ if and only if $ AB\cdot CD\equal{}AD\cdot CB$.

Let $ \angle BAC\equal{}\angle BDC\equal{}\alpha$, $ \angle CBD\equal{}\angle CAD\equal{}\beta$, $ \angle ADB\equal{}\angle ACB\equal{}\delta$, $ \angle ABD \equal{} \angle ACD\equal{}\gamma$ with $ \alpha\plus{}\beta\plus{}\gamma\plus{}\delta\equal{}\pi$.
I will suppose the circumradius is 1, so that $ AB\equal{}2\cdot\sin(\delta)$,$ BC\equal{}2\cdot\sin(\alpha)$,$ CD\equal{}2\cdot\sin(\beta)$,$ DA\equal{}2\cdot\sin(\alpha)$ and $ AC\equal{}2\cdot\sin(\alpha\plus{}\delta)\equal{}2\cdot\sin(\beta\plus{}\gamma)$,$ DB\equal{}2\cdot\sin(\gamma\plus{}\delta)\equal{}2\cdot\sin(\beta\plus{}\alpha)$

Let it $ \angle ANB\equal{}x_1$, $ \angle ANC \equal{} x_2$.

Sines, triangle $ BAN$:
$ \dfrac{BN}{BA}\equal{}{\sin(\gamma\plus{}x_1)}{\sin(x_1)}$
$ \dfrac{\sin(\gamma\plus{}\delta)}{2\sin(\delta)}\equal{}{\sin(\gamma\plus{}x_1)}{\sin(x_1)}$
$ \sin(\gamma)\cot(\delta)\plus{}\cos(\gamma)\equal{}2\sin(\gamma)\cot(x_1)\plus{}2\cos(\gamma)$

$ 2*\cot(x_1)\equal{}\cot(\delta)\minus{}\cot(\gamma)$

Analogously,
$ 2*\cot(x_2)\equal{}\cot(\alpha)\minus{}\cot(\beta)$

Both are equal if and only if

$ \cot(\alpha)\minus{}\cot(\beta)\equal{}\cot(\delta)\minus{}\cot(\gamma)$
$ \dfrac{\cos(\alpha)}{\sin(\alpha)}\plus{}\dfrac{\cos(\gamma)}{\sin(\gamma)}\equal{}\dfrac{\cos(\beta)}{\sin(\beta)}\plus{}\dfrac{\cos(\delta)}{\sin(\delta)}$
$ \dfrac{\sin(\alpha\plus{}\gamma}){\sin(\alpha)\sin(gamma)}\equal{}\dfrac{\sin(\beta\plus{}\delta}){\sin(\beta)\sin(delta)}$
$ \sin(\alpha)\sin(\gamma)\equal{}\sin(\alpha)\sin(\delta)$
$ AB \cdot DC \equal{} AD \cdot BC$

Well, that is all, folks! This can be demonstrated analogously at the other diagonal.

[quote="Altheman"]I am surprised that such an easy problem is on such an olympiad (when I say easy, I mean that if you know some theory, the problem is immediate?).
[/quote]

Well, I disagree to you. It is not a "very easy" problem. Eg, there are IMO problems very easy if you apply some approach and very hard if you do not apply the same approach (with trigonometry this problem was easy to me, but I do not think nothing using just synthetic constructions...).

Indeeed, this projective solution is beautiful! And easy to me to understand (I know some of it, but I have few opportunities to apply...), but very few geometry problems use extensively projective principles.

[color=#00f]Lemma[/color]: [color=#ff9a00]sinx/siny=sinx'/siny' and x-y=x'-y' (<180) then x=x', y=y'[/color]
[b][color=#00f]Proof:[/color][/b] Construct two similar triangle ABC and A'B'C' with two equivalent angle and apply law of sin we have our proof
[color=#0B5394]Back to problem[/color]: The appolinus circle of DMB cuts (BD) at G then GMF=90, M midpoint so GA=GC and note that GD/GB=FD/FB=AD/AB.CD/CB. In another way, GD/GB=GD/GC.GA/GB =sinGCD/sinBDC.sinABD/sinBAG=sinGCD/sinBAG.AD/BC. We will prove ABCD is harmonic <-> sinGCD/sinBAG=CD/AB<->sin(GCA+ACD)/sin(GCA-BDC) =sinDAC/sinADF=sin(ADC+ACD)/sin(ADC-BDC)
[color=#0f0]Apply the lemma[/color], we claim that GCA=ADC hence GA=GC and GC tangent to (O) and ABCD is harmonic. Now we prove NB is bisector. It is so easy now because AN is symmedian hence AND=ABC and CND=CBA so BD bisect ANC (q.e.d)


linear algebra
matrix
algebra
polynomial

Let X be a matrix with integer elements of order (k-1,k-1) with det(X+e*I_(k-1))=0 where k>=3, a prime number and e=cos(2*pi/k)+i*sin(2*pi/k). Prove that:
a) the matrices X and (X-I_(k-1)) are invertible;
b) X^(-1) and (X-I_(k-1))^(-1) have only integer elements.

thank you very much!
bye bye!!

I know this from a Romanian contest. The main idea is that, because k is prime, the minimal polynomial of cos2Pi/k+i*sin2Pi/k is 1+x+x^2+..+x^(k-1), so this must also be the polynomial
P(x)=det(X+x*I_(k-1)). It's not that hard from here.


Let $a,b,m>0$ and $P=\sqrt{\frac a{a+mb}}+\sqrt{\frac b{ma+b}}, Q=\frac2{\sqrt{1+m}}$
Prove that $P\geq Q$ when $m\geq3$, and $P\leq Q$ when $0<m\leq2$.

[quote]Is it too difficult or easy?[/quote]
I don't think this one is too difficult.But can you tell me if m is an integer, since you question suggests this point.

$\ P=\sqrt{\frac a{a+mb}}+\sqrt{\frac b{b+ma}}, Q=\frac2{\sqrt{1+m}}$
Assume $\ A = \sqrt{\frac a{a+mb}}, B =\sqrt{\frac b{b+ma}}$
$\ P=A+B, AB = \sqrt{\frac{1}{(1+\frac{am}{b})(1+\frac{bm}{a})}} = \sqrt{\frac{1}{m^2+m(\frac{a}{b}+\frac{b}{a})+1}} \leq \sqrt{ \frac{1}{1+2m+m^2}} = \frac{1}{1+m}$
$\ P^2-Q^2= \frac a{a+mb} +\frac b{b+ma} + 2\sqrt{\frac{1}{m^2+m(\frac{a}{b}+\frac{b}{a})+1}} -(\frac2{\sqrt{1+m}})^2$
If $\ m \in (0,2]$ then $\ \frac{2+(\frac{a}{b}+\frac{b}{a})m}{1+(\frac{a}{b}+\frac{b}{a})m+m^2}-\frac{2}{1+m} \geq \frac{2}{1+m} - 2\sqrt{\frac{1}{m^2+m(\frac{a}{b}+\frac{b}{a})+1}}\geq 0$
Elseif $\ m \in [3,+\infty)$ then  $\ 0 \leq \frac{2+(\frac{a}{b}+\frac{b}{a})m}{1+(\frac{a}{b}+\frac{b}{a})m+m^2}-\frac{2}{1+m} \leq \frac{2}{1+m} - 2\sqrt{\frac{1}{m^2+m(\frac{a}{b}+\frac{b}{a})+1}}$

[quote="Kalimdor"]
I don't think this one is too difficult.But can you tell me if m is an integer, since you question suggests this point.
...[/quote]
$m\in\mathbb R$


Multiply all inequalities of the form (1+x_1)^2<=(1+x_2)(1+(x_1)^2/x_2).


inequalities
function
inequalities unsolved

For triangle $ABC$, prove that
\[\cot A + \cot B + \cot C \ge 3 \cot \sqrt[3]{ABC}.\]

Do you know the Taylor's formula for the function $f(x)=cotx$?
Then the problem would be solved by proving $f(x)+f(y)\ge 2f(\sqrt{xy})$.


What is the integer value of $ x$ that minimizes the absolute value of $ (6x^2 \minus{} 2x)$?

The answers to these are $ \minus{}\frac{b}{2a}$.

[quote="emotionlessone"]The answers to these are $ \minus{} \frac {b}{2a}$.[/quote]

according to emotionlessone, the answer should be $ \minus{}\frac{\minus{}2}{12}$, which is $ \frac{1}{6}$.

But the actual answer is 0...

I don't get this problem at all...

$ 6x^2\minus{}2x$ is $ 2x(3x\minus{}1)$, so as you increase the value of x, the absolute of value of $ 6x^2\minus{}2x$ also increases...

This is really weird...how can the value be maximized, when the graph open upward?

There is only a minimum, so I would have to agree with emotionless1.

Wait...unless it's the minimum absolute value value possible. Which could be 0.

It asked for minimum absolute value the answer could be o OR 1/3 so it still doesn't makes sense...

[quote="FantasyLover"]yeah the answer should be $ \infty$...[/quote]
But then if $x$ is $ \infty $, then $ (6x^{2}-2x) $ would be zero, because $6\cdot \infty^2=2\cdot \infty$.
The question asks for the integer value of $x$ that minimizes $|6x^2-2x|$. Therefore we need $6x^2-2x=0$. Clearly $x=0$ satisfies this.

But $\infty$ can't be treated as a real number.

hello, but $\infty$ is not an integer value.
Sonnhard.

Maybe they meant minimum instead of maximum.  Indeed the minimum is $x = 0$ resulting in the quantity being $0$.

Yea, the question asks for the minimum. I think Gamebot typed it wrong

Yep! It's definitely "minimizes", which is how it currently reads in the live version of the Alcumus problem. I've updated the wording in GameBot's post to reflect this.


Given is a rhombus $ABCD$ of side 1. On the sides $BC$ and $CD$ we are given the points $M$ and $N$ respectively, such that $MC+CN+MN=2$ and $2\angle MAN = \angle BAD$. Find the measures of the angles of the rhombus.

[i]Cristinel Mortici[/i]

Let the rhombus $ADKL \equiv ABCD$ and let $P \in (DK)$ such that $BM=DP.$
[img]http://i19.photobucket.com/albums/b200/mariusdamian/ProblemaC.png[/img]
Obviously, $\Delta ABM \equiv \Delta ADP.$
Result $AM=AP, \angle BAM \equiv \angle DAP.$
Follows $\Delta MAN \equiv \Delta PAN,$ so $MN=PN.$
From $MC+CN+MN=2$ and $BM=DP$ we obtain $MN=BM+DN=DP+DN.$
But $MN=PN$
So $PN=DP+DN.$
Result that points $N, D, P$ are collinear, so $ABCD$ is a square.

[b]Remark.[/b] Let $w$ be the exincircle w.r.t. the vertex $C$ for the triangle $CMN$. From the relation $CM+MN+NC=2$ results $B,D\in w$ and $MN=BM+ND$, a.s.o.


function
limit
inequalities
real analysis
triangle inequality
real analysis unsolved

Let $\{f_{n}\}$ be sequence of measurable functions on $[a,b]$ and $f$ be a real-valued function on $[a,b]$. If for any $\epsilon>0$ it holds that
$\lim_{n\to\infty}m^\ast(\{x: [a,b]: |f_{n}(x)-f(x)|>\epsilon\})=0$
do we have that $f$ is measurable?

The triangle inequality gives that $(f_{n})_{n}$ is Cauchy in measure, whence converging in measure to some measurable function $h$. But the given inequality yields $m^{*}(|f-h|>\epsilon)=0$ for every $\epsilon>0$. This means that $f=h$ a.e, which guarantees the measurability of $f$ since the measure is complete (assuming we are working with the Lebesgue measure). You need to check it out, I'm rather sleepy now ... - yeah, I'm in China also  :lol:  -


[quote="daermon"]solve the system of equations
$\frac{x+2y}{xy}=\frac{11}{12}\\ \frac{3x-3y}{xy}=\frac{2}{3}$[/quote]
[hide]Let's break this up
$\frac{1}{y}+\frac{2}{x}=\frac{11}{12}$
Multiplying by twelve gives and subbing $a=\frac{1}{y}$ and $b=\frac{1}{x}$
gives you $12a+24b=11$
Repeating with the second we get $3a-3b=\frac{2}{3}$
Which is the same as $12a-12b=\frac{8}{3}$. Subtracting from the first we get
$36b=\frac{25}{3}$
Meaning $b=\frac{25}{108}$
meaning $x=\frac{108}{25}$
Plugging back in we find that $y=\frac{108}{49}$
This method doesn't take as long as it seems because I had to show all the steps...[/hide]


Let z be a root of a polynomial p(z) such that |z| = 1.  Must z be an nth root of unity for some positive integer n?

Certainly not, if you do not have any condition on the polynomial p.

Pierre.

p(z)=(z-exp(i \pi  \sqrt 2))(z-exp(-i \pi  \sqrt 2))
=z^2-2cos( \pi  \sqrt 2).z+1
the modulus of the roots are 1
suppose there exist n s.t (exp( i \pi  \sqrt 2))^n=1=>
n \pi  \sqrt 2=2k \pi =>  \sqrt 2 is rationnal wich is not

Yes Moubi, or just p(z) = z - exp(i \pi  \sqrt 2).

Pierre.

Duh, I wasn't thinking.  What if the coefficients of p are integers?

--J.H.

No again.

- If exp(i@) is a n-root of the unity for some integer n then @ = q \pi  for some rational q.

- In another hand, it is well-known that if @ = q \pi  with q in Q and such that cos(@) is a rational number then @ = 0, \pi/6, \pi/4, \pi/3 mod[\pi/2] (the usual proof is via Tchebychev's polynomials).

- Thus, if cos(@) = 1/3 then @ is not of the form q\pi for some rational q, and then cannot be a root of the unity.
But p(z) = 3z 2 + 2z + 3 = 3(z 2 + 2cos(@)z + 1) is a polynomial with integer coefficients, whose roots are exp(i@) and exp(-i@). Then exp(i@) is an algebric number with modulus 1 but it is not a root of the unity.

Pierre.


[quote="Totally Zealous"]A $3\times3\times3$ cube is formed by gluing together 27 standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the $3\times3\times3$ cube is
(A) 60
(B) 72
(C) 84
(D) 90
(E) 96[/quote]
[hide]
Corners:1+2+3=6, 8 corners. 48
Edges:1+2=3, 12 edges. 36
centers: 6.
48+36+6=$D$[/hide]

[quote="Totally Zealous"]A $3\times3\times3$ cube is formed by gluing together 27 standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the $3\times3\times3$ cube is
(A) 60
(B) 72
(C) 84
(D) 90
(E) 96[/quote]

[hide]It is 6*8+1*6+3*12=90

$90$

D[/hide]

[hide]the minimum is $6\times1$ for centers, $12\times 3$ for edges, and $8\times 6$ for corners.  Adding these together gives $90$,so $D$[/hide]

This problem could be made harder by making the cube bigger. This would make more edges and centers to account for, but if you knew how to do the problem it would not make it much more difficult.

[quote="lotrgreengrapes7926"]Extension:
An $n\times n\times n$ cube is formed by gluing together $n^{3}$ standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) What is the smallest possible sum of all the numbers showing on the surface of the $n\times n\times n$ cube?[/quote]
[hide]
$12(n-2)$ edges. 1 and 2 . So then $36(n-2)$ for the edges.

8 corners. 1,2,3. SO then 48 for the corners.
$(n-2)^{2}$ center pieces. 1 for those. so $n^{2}-4n-4$ for the center pieces.

Total is $n^{2}+32n-28$[/hide]

[hide="ans to extension"]
$6n(n+2)$
[/hide]

riddler, how do you find your answer? It is different from ch1n353ch3s54a1l's answer, which appears to be right.

Also, here's another extension:

What is the largest possible sum of all the numbers showing on the surface of the n*n*n cube?

doesnt my solution work?
6n(n+2) for 3*3*3 is 90 which is correct.
This is how I got to my solution...
6(n-2)(n-2)+48+3*12*(n-2)

Ooooooooooh. I see. Yours is right. ch1n353ch3s54a1l forgot that there are 6 faces for the center...

[quote="laughinghead505"]
Also, here's another extension:
What is the largest possible sum of all the numbers showing on the surface of the n*n*n cube?[/quote]
[hide]
same as before except u have 6's in the centre areas and in the corners will be 4+5+6=15
and then the edges.....
[/hide]

Doesn't the question of an n*n*n cube depend on parity (i.e. whether or not n is even)?


The average newborn sleeps $ \frac{5}{8}$ of a full day, and the average adult sleeps $ \frac{1}{3}$ of a full day. How many more hours per day does the average newborn sleep than the average adult?

there is no dividing by 24: then the newborn would sleep $8\frac{7}{24}$ hours instead of 15 hours.

@andyzhou: it's asking for how many [i]hours[/i], not how many days.


trigonometry
ratio
logarithms
geometry
geometric sequence
AoPS Books

The problems are here: http://www.math.sc.edu/contest/1991/exam/PDF/exam1991.pdf

Remark: this is the contest one of our SC math team coaches won. :)

[hide="16"]We can factor the expression as such:

$5!(3(10 \cdot 9 \cdot 8 \cdot 7 \cdot 6)+12+4(7 \cdot 6))$
$5! \cdot 12(10 \cdot 9 \cdot 7 \cdot 6 \cdot 2+1+7 \cdot 2)$
$5! \cdot 12(7575)$

Thus, the maximum value of $k=3$.
[/hide]
[hide="17"]After simplifying the discriminant, we have that $\sqrt{t}$ must be a perfect square. Thus, the values of $t$ for the given interval are $1,4,9,16,25,36,49,64,81,100$.[/hide]

[hide="18"]For $\arcsin{x}$, we can create a right triangle with legs $x$ and $\sqrt{1-x^{2}}$, and hypotenuse $1$. Using the double angle formula for $\tan$ with this right triangle and rationalizing the denominator, we get the value as $\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}}$.[/hide]

[hide="19"]This simplifies to $\sin{20}(\frac{\sin^{2}{20}+\cos^{2}{10}}{\sin{10}\cos{10}}=\frac{\sin{20}}{\sin{20}/2}=2$[/hide]

[hide="20"]Create two right triangles with the radii of the two circles as one of the legs of the respective triangles and their hypotenuses collinear. These two triangles are similar, and using similar triangle ratios, we have that the smaller circle’s radius is $\frac{3-2\sqrt{2}}{2}$.
[/hide]
[hide="21"]The sum is $\log{2/1}+\log{3/2}+...\log{100/99}\Longrightarrow \log{2}-\log{1}+\log{3}-\log{2}+...+\log{100}-\log{99}\Longrightarrow \log{100}-\log{1}=2$
[/hide]
[hide="23"]$z$ must be in the form $a+bi$ as well, and $a$ and $b$ can change values as well as $\frac{b}{a}$ is constant. Thus we form a line of possible $z$s.[/hide]

[hide="24"]Using $x=0,1,2,3,4,5,6$, we find that the values of $x$ that work are $1$ and $5$ $\mod{7}$. Listing the numbers out for the interval, there are $29$ values of $x$ that work.[/hide]

[hide="26"]After simplifying the expression to $2^{101}(2^{202}+1)(2^{101}+1)+1$ and using long division with $2^{100}+1$ repeatedly. The remainder is $(5)(2)+1=11$.[/hide]

[hide="27"]The sum of the $x$ values are $\frac{15}{3}=5$.[/hide]

[hide="28"] For $(i)$, if $x \in S$ and we let $y=0$, then we assume that $0 \not \in S$, but since $0+x=x \in S$, then we have a contradiction, therefore, $0 \in S$. For $(ii)$, we assume the contrary once again, that $x \in S$ and $-x \not \in S$. But $x+(-x)=0$, which is $\in S$. Once again a contradiction, therefore $(ii)$ is also true. For $(iii)$, we already established that $0 \in S$, so if we let $y=0$, clearly $(iii)$ is valid. For $(iv)$, based on $(iii)$, is true. Therefore, all four statements are true.[/hide]

[hide="29"]Seeing that the circle is in the form $(x-0)^{2}+(y-r)^{2}=r^{2}$ and using the fact that $y=x^{2}$, we have that:

$x^{2}+(y-r)^{2}\geq r^{2}$ for all $x$
$x^{2}+(x^{2}-r)^{2}-r^{2}\geq 0$ for all $x$.
$x^{2}+1-2r\geq 0$ for all $x$
$1-2r\geq 0$
$1/2\geq r$

Thus, the maximum diameter is $1$.[/hide]

[hide="30"]Let $x_{1},x_{2},...,x_{40}=1,2,...,40$ and $y_{1},y_{2},...,y_{40}=40,39,...,1$. The sum therefore simplifies to $1+2+3+...+40+41(40)=\frac{41(40)}{2}+41(40)=2460$.[/hide]

[hide="22"]
As a fraction, $(.333\ldots)_{b}$ equals $\frac{3}{b-1}$.  Thus \[\frac 3{b-1}=\frac{1}{5}.\] Solving, we get $\boxed{b=16}$.

In this solution, we used a trick to convert repeating decimals to fractions in base $b$.  In base 10, for example, if we want to find $.565656\ldots$ we make it $\frac{56}{99}$.  Likewise if it's $.124124\ldots$ we make it $\frac{124}{999}$.  This method generalizes and in base $b$ we the repeating part is the string $a_{1}a_{2}\cdots a_{n}$ then our number is $\frac{a_{1}a_{2}\cdots a_{n}}{b^{n}-1}$.[/hide]

[hide="25"]Assume that the segment were not missing. Then there are 12 choose 6 ways. Now we subtract those ways going along the desired segment. There are 7 choose 3 ways to get to its base, and 4 ways to travel from its top to the upper right corner. This makes a total of 4*35=140 ways going through the segment, subtracted from 924 total ways. Hence our answer is 784, and we have answer choice $\boxed{f=typo}$.[/hide]

I took this contest today for practice, it was a relatively easy year.  I have an alternate solution for #20 that doesn't use geometry, since I'm incrediby bad at it.  I admit that your way is much easier, but my way took no time at all to see.

[hide="23"]
I won't do the math again, but basically, i made an infinite geometric sequence to find the ratio of the diameter of the small circle to the large circle.  Using the infinite sum $\frac{a}{1-r}=S$, I set $\frac{1}{1-r}=\frac{1+\sqrt{2}}{2}$ and solved.  It only took like ten seconds, and it was easier for my algebra-minded self.
[/hide]

[quote="joml88"]
In this solution, we used a trick to convert repeating decimals to fractions in base $b$.  In base 10, for example, if we want to find $.565656\ldots$ we make it $\frac{56}{99}$.  Likewise if it's $.124124\ldots$ we make it $\frac{124}{999}$.  This method generalizes and in base $b$ we the repeating part is the string $a_{1}a_{2}\cdots a_{n}$ then our number is $\frac{a_{1}a_{2}\cdots a_{n}}{b^{n}-1}$.[/quote]

Nice.  But what if we have the decimal $.11212121...$.  Normally, we would convert to a fraction by doing (nonrepeating and repeating) - (nonrepeating) over (a 9 for every repeating, a 0 for every nonrepeating).  This simplifies to $\frac{112-1}{990}=\frac{111}{990}$.  How do we convert that to base x?

I'm guessing that we just express the denominator $990 = 10^{3}-10$, so in base x, it would be $x^{3}-10$.  Or would it be $x^{3}-x$?

If we have  $.1\overline{12}$, we can evaluate this as \[\frac 1{10}+\frac 1{10}.\overline{12}= \frac 1{10}+\frac{12}{990}= \frac{99+12}{990}.\] Let's go a bit further and look at $.x\overline{ab}$.  This is \[\frac x{10}+\frac{ab}{990}= \frac{99x+ab}{990}.\] Your method gives $\frac{xab-x}{990}.$  These are the same since $xab-x = 100x+ab-x=99x+ab.$

The method I used above is what I use.  So as another example, suppose we had $.15\overline{62}_{7}$.  I would evaluate this as $\frac{15}{100}+\frac{62}{6600}$ and simplify that to whatever it comes out to (gotta remember we're in base 7 so multiplication rules and stuff are different).

That's really interesting.  I've never seen a good explanation of all this base stuff.  Is it in the AoPS books somewhere?  I saw Billy's paper on fractional bases, but it seems like an extraordinarily long read.  Where can I find more about this stuff.

Also, how can we go ahead and simplify the above quantity? ($\frac{62}{6600}$) in base 7?

So we have \begin{eqnarray*}.15\overline{62}_{7}&=& \left(\frac{15}{100}+\frac 1{100}\cdot\frac{62}{66}\right)_{7}\\ &=& \left(\frac{12}{49}+\frac 1{49}\cdot \frac{44}{48}\right)_{10}\\ &=& \frac{155_{10}}{588_{10}}\\ &=& \frac{311_{7}}{1500_{7}}\end{eqnarray*} So basically I converted to base 10 to do the arithmetic, and then converted back afterwards.

If you want to do this without converting to base 10 and then back...well, let's give that a try (oh boy...):

We start with $\frac{15}{100}+\frac 1{100}\cdot\frac{62}{66}$ in base 7.

Combining fractions we get \[\frac{66\cdot 15+62}{6600}.\] Now we need to compute $66\cdot 15$.  We multiply like usual but we have to carry 7's.  This is hard to write out, so I'll just explain it and hopefully you'll follow.  So set up a multiplication with 66 on top and 15 on bottom.  First we multiply the 6 and the 5 to get 30 but this is in base 10, so when converted to base 7 we get 42.  So we write down a 2 and carry the 4.  We multiply the other 6 with the 5 and we get 42.  Add in the 4 that we carried and we get 46.  The bottom row of the multiplication is just 660 (multiplying by 1 is easy).

So our multiplication has 462 on top with 660 on the bottom.  We add these to get 1452 (base 7 arithmetic...).

So then our numerator is $1452+62=1544$.  Thus we have $\frac{1544}{6600}$.  Now this thing is very hard to simplify.  We need to find prime factorizations of 1544 and 6600 but to do that we need to know basic base 7 multiplication/division! (If you don't understand why we can't just factor it like usual, consider this quickly: $12_{7}$...you might think this is divisible by 2, right?  But is it?  $12_{7}=9_{10}$ so you see it's quite a bit more complicated).

Ok, so to factor in base 7 we need to create a base 7 "times table" (you have this memorized for base 10 so well that you never even think about it): \begin{tabular}{c|c|c|c|c|c|c}& 1 & 2 & 3 & 4 & 5 & 6\\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 2 & 2 & 4 & 6 & 11 & 13 & 15 \\ \hline 3 & 3 & 6 & 12 & 15 & 21 & 24\\ \hline 4 & 4 & 11 & 15 & 22 & 26 & 33\\ \hline 5 & 5 & 13 & 21 & 26 & 34 & 42\\ \hline 6 & 6 & 15 & 24 & 33 & 42 & 51\\ \end{tabular} So now we're ready to do factor 1544 and 6600.  Let's start with 6600.  We start with $66\cdot10^{2}$.  10 is prime.  Now 66.  We check for divisibility using long division using that times table.  So divisibility by 2 yields 33.  And then from the table $33=6\cdot 4 =2^{3}\cdot3$.  So $6600=2^{4}\cdot3\cdot10^{2}$.

Now for 1544.  Check for divisibility by 2: $1544=2\cdot 622$.  Again: $2\cdot 622=2^{2}\cdot 311$.  Just keep going...$2^{2}\cdot 311=2^{2}\cdot 5\cdot 43$.  43 is prime.

Thus $\frac{1544}{6600}=\frac{2^{2}\cdot5\cdot 43}{2^{4}\cdot 3\cdot10^{2}}=\frac{311}{1500}.$

Woohoo...I'd actually never done that much arithmetic in base 7.  It was [i]almost[/i] as horrible as I'd expected it to be.


A calculator has a key which replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let $y$ be the final result if one starts with an entry $x \neq 0$ and alternately squares and reciprocates $n$ times each. Assuming the calculator is completely accurate (e.g., no roundoff or overflow), then $y$ equals

A. $x^{((-2)^n)}$
B. $x^{2n}$
C. $x^{-2n}$
D. $x^{-(2^n)}$
E. $x^{((-1)^n 2n)}$

[hide]writing out the first few terms gives: $x^{-2}$, $x^4$, $x^{-8}$, $x^{16}$... so the only answer choice that fits this is $\boxed{A}$[/hide]

[hide="Answer"]We have $x$. First it is squared, reciprocaled (yay, I verbed a word!), then squared again, reciprocaled again, etc. Obviously, the power that $x$ is raised to will always be equal to $2^n$, since it will still increase the power whether $x$ is itself or the reciprocal. We will have the reciprocal if $n$ is negative and standard if $x$ is positive, so our answer is $x^{(-2)^n}\Rightarrow \boxed{A}$.[/hide]

[hide]

let $f(x) = x^{-2}$, that is clearly what happens after 1 time of pressing those buttons
so we want to find $f^n(x)$ which denotes the nth compounding of the fuction
clearly we have $x^{((-2)^n)}$ (if one wanted to prove this, they could use induction, but that is unnecessary for the AMC)

[/hide]

personally mathematical induction is helpful just for the sake of confidence(obviously i dont have 4000 posts either) since i'm just starting to practice on these problems. Reasoning also works pretty darn well on this problem as b, c, and d can be eliminated immediately since they remain "large" the entire time or "small" the entire time. Looking at A and E all one has to do is plug in x=3. this keeps the values relatively small and allows the individual to avoid putting in 2 which corresponds in the equations 2x and x^2.

Please do not revive old threads!  Especially when they are several years old!

Thanks,
[color=red]b-flat[/color]

Locked.


its friday, and i dont have school! yay! :D
do any of you not have labor day weekend?

We all do! It's the last holiday till school starts  :(

i wish we didnt have labor day off

cant wait till school starts

[quote="aznmonkey1992"]my school started on the 29th... :mad:[/quote]
SO did mine! :)

yay 4 day weekend!

I REALLY needed this weekend..high school has been rough on me

my school hasn't started yet.  It starts wednesday.

So glad we have labor day off!

Today ends the 4th week of school.

My school starts this Tuesday and I can't wait for it to start!  :lol:

Yes Labor Day weekend is awesome! 3-day weekend galore!

wow. i didn't know so many schools hadn't started yet. i've been in school for a month now. labor day does provide a nice break.

I'm happy about it - being a 65-year old retiree is a hard life.

Our school started this Wednesday and now we have a 3 day weekend! :D yay! I couldn't wait for school to be over today.


Let a, b, c be positive real numbers such that ab + bc + ca =1. Prove that 1/(a+b) + 1/(b+c) + 1/(a+c) >= sqrt(3) + ab/(a+b) + ac/(a+c) + bc(b+c).

Same as  \sum 1/(a+b) \geq sqrt(3) + \sum ab/(a+b).

1/(a+b) = (ab + bc + ca)/(a+b), so the inequality is equivalent to

\sum (bc + ca)/(a+b) \geq sqrt(3)

or

\sum a \geq sqrt(3).

Square to obtain

\sum a^2 + 2 \sum ab \geq 3 = 3 \sum ab,

which is equivalent to

\sum a^2 \geq \sum ab, which is trivial.

x put a=tgx,b=tgy,c=tgz =>x+y+z= \pi /2
      \sum  1/(tgx+tgy)  \geq   \sqrt 3+  \sum tgxtgy/(tgx+tgy)
=> \sum 4(cosxcosy) 2 -4 \sqrt 3cosxcosycpsz- \sum sin2xsin2y \geq 0
    =>2 [( \sum cos  2 (x+y))+3-2 \sqrt 3cosxcosycosz-
                      2 \sum sin 2 ]
   we prove  (\sum cos 2 (x+y))+3/4  \geq  2 \sum sin 2 x  => cos(x+y)cos(x+z)cos(z+y) \leq 2sinxsinysinz-1/8
 for examal y<45    => we prove
   :)


if n - natural number than prove that
 2n{n(7)^(1/2)}>1

Is it
 $ 2n.\{n.\sqrt{7}\}>\sqrt{\frac{9}{7}}$ for all natural $ n$?

Let $ m\equal{}[n\sqrt 7 ]$ and $ x\equal{}\{n\sqrt 7 \}\equal{}n\sqrt 7 \minus{}m>0$, then $ x\equal{}\frac{7n^2\minus{}m^2}{n\sqrt 7 \plus{}m}>\frac{7n^2\minus{}m^2}{2n\sqrt 7}$. Obviosly $ 7n^2\minus{}m^2\equal{}3,5,6,7\mod 7$ or $ 7n^2\minus{}m^2\ge 3$. It give $ 2n\{n\sqrt 7\}\equal{}2nx>\sqrt{\frac 97 }.$


calculus
integration
calculus computations

Can someone find the integral of ((x^2 -1)/(x-1))  upper bound 2a and lower bound 2?

Thanks a lot.

Simplify the fraction using the "difference of two squares"

[quote="iamhe"]So, that would be x+1. Fundamental Theo. of Calc ... antiderivative =(x^2)/2 +x  right?  So, plug in 2a and 2 and then subtract?

So the answer would be 2a^2 + 2a -4?

Is that right, tonyftw?[/quote]

Yes, thats right, but I get the impression that he wants this done for his homework, hence why I only hinted


Find four positive whole numbers, the sum of every two of which is a perfect fourth power.

The question is to find such $a_1,a_2,a_3,a_4 \in \mathbb{N}$ that $\forall _{i,j} a_i+a_j=k^4$, $k \in \mathbb{N}$, yes [I'm askin cuz i dont know if i understand the problem correctly] ? And shouldnt there be another condition that $a_1,a_2,a_3,a_4$ are different ?

[quote="Megus"]And shouldnt there be another condition that $a_1,a_2,a_3,a_4$ are different ?[/quote]

I think it is implicit in all the problems of that kind proposed by Orlando.

Pierre.


Find twelve consecutive integers that are divisible by $ 3$, $ 5$, $ 7$, $ 9$, $ 11$, $ 13$, $ 15$, $ 17$, $ 19$, $ 21$, $ 23$ and $ 25$ respectively.

Note that some conditions are unnecessary, for example, if the 12th integer is divisible by 25, then the 2nd and 7th will also be divisible by 5.
So let $ n$ be the first of the 12 integers, and our conditions can be simplified into
$ n\equiv \minus{} 2\mod 7$
$ n\equiv \minus{} 3\mod 9$
$ n\equiv \minus{} 4\mod 11$
$ n\equiv \minus{} 5\mod 13$
$ n\equiv \minus{} 7\mod 17$
$ n\equiv \minus{} 8\mod 19$
$ n\equiv \minus{} 10\mod 23$
$ n\equiv \minus{} 11\mod 25$
Let $ a_i$ represent the product of every number listed above except the $ i$th number, ex. $ a_1 \equal{} 9\times11\times13\times...\times25$. Let $ a^{ \minus{} 1}_{i}$ denote the inverse of $ a_i$ modulo the $ i$th number. Then the sum
$ \minus{} 2a_1a^{ \minus{} 1}_1 \minus{} 3a_2a^{ \minus{} 1}_2 \minus{} 4a_3a^{ \minus{} 1}_3 \minus{} ... \minus{} 11a_8a^{ \minus{} 1}_8$
will satisfy the above conditions.

In particular, the smallest $ n$ satisfying the above congruences is $ 836598264$. (I wrote a program a while back to solve linear congruences)


If $ n > 3$ then prove
$ n^{n \plus{} 1} > (n \plus{} 1)^{n}$

[quote="enndb0x"]If $ n > 3$ then prove
$ n^{n + 1} > (n + 1)^{n}$[/quote]
Induction
For $ n = 3$ is true
So we have that
$ n^{n + 1} > (n + 1)^n$ which is
$ n > (1 + \frac {1}{n})^n$
For $ n + 1$ we have to prove that
$ n + 1 > (1 + \frac {1}{n + 1})^{n + 1} \rightarrow n + 1 > (1 + \frac {1}{n + 1})^n\dot(1 + \frac {1}{n + 1})$
But $ {\frac {1}{n} > \frac {1}{n + 1}}$ so
$ (1 + \frac {1}{n}) > (1 + \frac {1}{n + 1})$
But we know that
$ n > (1 + \frac {1}{n + 1})^n$ So the only we have to prove is
$ n(1 + \frac {1}{n + 1}) < n + 1$, which is true... :)
Another method is:
$ n^{n + 1} > (n + 1)^n$ is
$ n > (1 + \frac {1}{n})^n$
But $ (1 + \frac {1}{n})^n$ is the number $ e$, but $ 3 > e$  :wink:

That is almost right, but it says if $ n>3$. That means that $ n$ is greater than but not equal to 3.

Forgive me for being nit-picky, but $ \left(1 \plus{} \frac {1}{n}\right)^n < e$, and $ \lim_{n\to\infty}\left(1 \plus{} \frac {1}{n}\right)^n \equal{} e$ (it doesn't interfere with your proof, just for correctness sake).
BTW, try to prove $ \left(1 \plus{} \frac {1}{n}\right)^n < 3$, it's a nice exercise.

[quote]That is almost right, but it says if $ n>3$. That means that $ n$ is greater than but not equal to $ 3$.[/quote]

Well, essentially, Dimitris proved a sharper claim, and there's no problem about that. The problem should have said '$ n\ge 3$.

We use $ y\equal{}\frac{\ln x}{x}$.

$ e<p<q \Rightarrow \frac{\ln p}{p}>\frac{\ln q}{q} \iff p^{1/p}>q^{1/q} \iff p^q>q^p$.


[url]http://www.mathlinks.ro/viewtopic.php?t=172148[/url]
Στο link βρίσκεται μια άσκηση. Να γενικευτεί η λύση του Α αν έχουν μείνει m επιβάτες, n στάσεις
και στην κάθε στάση $ i$ μπορούν να κατέβουν επιβάτες $ x_i$ μεταξύ των αριθμών $ a_i$ και $ b_n$.

Η λύση είναι[b]ΦΑΝΤΑΣΤΙΚΗ[/b].

[hide]
Το πρόβλημα είναι στην ουσία η εύρεση του πλήθους των ριζών της διοφαντικής εξίσωσης:
$ x_1 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_n \equal{} m$ με $ a_i <\equal{} x_i <\equal{} b_i$
(υπάρχει συνδυαστικός τύπος για το πλήθος αυτό, το θέμα ήταν να μεταφραστεί το πρόβλημα σε σύμβολα)
[/hide]

Όντως ωραία ασκηση


Find all $ f:\mathbb{N}\rightarrow\mathbb{R}$ that for a given $ n\in\mathbb{N}$,
\[f(m+k)=f(mk-n)\]
 holds for all positive integers $k$ and $m$ with $mk>n$.

[quote="Amir.S"]find all $ f:\mathbb{N}\rightarrow\mathbb{R}$ that for a given $ n\in\mathbb{N}$ we have:
$ f(m\plus{}k) \equal{} f(mk\minus{}n)\ ,\ m,k\in\mathbb{N}\ ,\ mk > n$[/quote]

$ m\equal{}x\minus{}1$ and $ n\equal{}1$ $ \implies$ $ f(x)\equal{}f(x\minus{}(n\plus{}1))$ $ \forall x>n\plus{}1$ and so $ f(x)\equal{}f(x\plus{}n\plus{}1)$ $ \forall x>0$
$ k\equal{}n\plus{}1$ implies then $ f(m\plus{}n\plus{}1)\equal{}f((m\minus{}1)(n\plus{}1)\plus{}1)$ $ \forall m>0,n>0$
But, since $ f(x)\equal{}f(x\plus{}n\plus{}1)$ $ \forall x>0$, this last equation may be written $ f(m)\equal{}f(m\plus{}n\plus{}1)\equal{}f((m\minus{}1)(n\plus{}1)\plus{}1)\equal{}f(1)$

And so the unique solution $ f(m)\equal{}c$ $ c\in\mathbb{R}$


Let $ a,b \in R$
Solve system equations follow:
\[ \left \{ \begin{array}{ll}
                          |b\minus{}2a\minus{}ab|\equal{}\sqrt{3(a^2\plus{}b^2)} \\

                          |ab|\equal{}20
                                    \end{array} \right.  \]

can you solve the above problem for me?
many thanks to you!
Best, Luan.

Thats going to be a hard problem. I don't have any answers for you, but I do have some insight:

Because of the absolute values I think we really have 4 equations:

$ b\minus{}2a\minus{}ab\equal{}\sqrt{3(a^2\plus{}b^2)}$
$ b\minus{}2a\minus{}ab\equal{}\minus{}\sqrt{3(a^2\plus{}b^2)}$
$ ab\equal{}20$
$ ab\equal{}\minus{}20$

As for solving, maybe use substitution. Theres probaly something Im missing.


A student uses an audio oscillator of adjustable frequency to measure the depth of a water well. The student hears two successive resonances at 51.5 Hz and 60.0 Hz. How deep is the well? The speed of sound is 343 m/s.

[hide="hint"]
the well is same as a tube with one end closed[/hide]

$ \frac {(n + 1)v}{4l} = 60$
$ \frac {nv}{4l} = 51.5$

therfore $ \frac {v}{4l} = 8.5$
on substituiting $ v = 343$m/s
$ l = 10.088$m

[size=150][color=blue]MOD EDIT:I $ \text{\LaTeX}$ ified it[/color][/size]
[size=150][color=blue]Learn $ \text{\LaTeX}$ [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url][/color][/size]

[quote="valeriummaximum"]$ \frac {(n \plus{} 1)v}{4l} \equal{} 60$
$ \frac {nv}{4l} \equal{} 51.5$

therfore $ \frac {v}{4l} \equal{} 8.5$
on substituiting $ v \equal{} 343$m/s
$ l \equal{} 10.088$m
[/quote]

valerium you're wrong. If its similar to a tube closed at one end then 2 succesive  resonances implies thats it should be

$ \frac {(2(n\plus{}1) \plus{} 1)v}{4l} \equal{} 60$
and

$ \frac {(2n\plus{}1)v}{4l} \equal{} 51.5$

solve these 2 equations and get $ l$;

you are right hash_include coz for resonance

f=(2*n+1)*v/4*l

SO the answer is 20.176 m...


Hi all.
Prove that if m and n are positive integers, then 5^m + 5^n can be written as the sum of two perfect squares if and only if
m - n is even.
Thanks.

If $m,\;n$ are both even we're obviously done.
Suppose $m,\;n$ are both odd: $m = 2m' + 1,\;n = 2n' + 1$ then \[5^m + 5^n = (1^2 + 2^2)(5^{2m'} + 5^{2n'}) = (5^{m'} + 2 \cdot 5^{n'})^2 + (2 \cdot 5^{m'} - 5^{n'})^2.\]
If $m$ and $n$ have opposite parities then it can be checked that \[5^m + 5^n = 6\;(\textrm{mod}\;8).\]
But all perfect squares are 0, 1 or 4 (mod 8). Hence result.


Find all the Real solutions:

$ \begin{cases} 2x\plus{}x^2y\equal{}y\\2y\plus{}y^2z\equal{}z\\2z\plus{}z^2x\equal{}x\end{cases}$

By analysis, a solution is $ (x,y,z)\equal{}(0,0,0)$.

[hide]
Re-arranging gives $ y \equal{} \frac{2x}{1\minus{}x^2}$ and so forth. Let $ x \equal{} \tan \theta$; then we have $ y \equal{} \tan 2\theta, z \equal{} \tan 4\theta$, so $ \tan \theta \equal{} x \equal{} \tan 8\theta$. Then $ 8\theta \equal{} \theta \plus{} \pi n \Longrightarrow \theta \equal{} \frac{n\pi}{7}$. Thus, our answer is $ (x,y,z) \equal{} \left(\tan \frac{k\pi}{7}, \tan \frac{2k\pi}{7}, \tan \frac{4k\pi}{7}\right)$ for $ k \equal{} 0,1,2,3,4,5,6$. [/hide]

Slick indeed, thanks for the solution azjps.


abstract algebra
superior algebra
superior algebra unsolved

Prove that $ \mathbb{Z}[i]/(i \minus{} 1)$ es isomorphic to $ \mathbb{Z}_2$

From the isomorphism theorem it suffices to find $ \phi : \mathbb{Z}[i] \rightarrow \mathbb{Z}_{2}$ such that $ \phi$ is "onto" and $ ker \phi \equal{} (i\minus{}1)$. We can define $ \phi$ on the generators of $ \mathbb{Z}[i]$: $ \phi(1)\equal{}\phi(i)\equal{}1$ (in other words, if $ z\equal{}a\plus{}bi$, then $ \phi(z)\equal{}a\plus{}b \ mod \ 2$). Now every element $ z \in (i\minus{}1)$ is equal to $ (a\plus{}b)(i\minus{}1)\equal{}\minus{}(a\plus{}b)\plus{}(a\minus{}b)i$, and you can easily check that indeed $ (i\minus{}1) \equal{} ker\phi$ (to show that $ ker\phi \subseteq (i\minus{}1)$, take any $ z\equal{}a\plus{}bi \in ker\phi$, $ a\equal{}b \ mod \ 2$, and show that you can find such $ x, y$ that $ a\equal{}\minus{}(x\plus{}y), b\equal{}x\minus{}y$).

I think it's more natural to look at
$ \mathbb{Z}\to \mathbb{Z}[i]/(i\minus{}1), a\mapsto a\plus{}(i\minus{}1)$ onto with kernel $ 2\mathbb{Z}$.

Here is a method which often works: $ \mathbb Z[i]/(i\minus{}1)\stackrel{X\mapsto i}\equal{}\mathbb Z[X]/(X^2\plus{}1,X\minus{}1)\stackrel{X\mapsto 1}\equal{}\mathbb Z/(1^2\plus{}1)\equal{}\mathbb F_2$. owk


[hide="Answer"]
$40^\circ$[/hide]

[hide="Hint"]
Prove that $\triangle DEF$ is isosceles[/hide]

I already have solutions i want your different ways.And also it is true the answer is 40.

Question;
[hide][img]http://img154.imageshack.us/img154/9675/fggrl0.png[/img][/hide]
First solution;
Choose F point circumcenter of ABD.Than there will be equilateral trangle AFD.Draw EF line ABE isosceles & ABF isosceles so EF is bisector of ABE. And also AE is bisector of AFD so AEFD is a deltoid and AE is also bisector of FED than we find BDE 40.
[hide][img]http://img184.imageshack.us/img184/5045/dfdfdfdt6.png[/img][/hide]

Second solution;
intersect extension of BK and EK such that the EK perpendicular to AC.Than draw AK and DE.There will be equilateral trangle AEK.And DEK will be 20 degree because of simetry than we find BDE 40.

[hide][img]http://img130.imageshack.us/img130/3289/nhntrn0.png[/img][/hide]


[quote="Makaveli"]What is the right time to join the contest and when new problems will be given?[/quote]
I think you will have one week to solve the 3 problems.


the critical factor driving yeast to use fermentation to metabolize sugar is lack of oxigen????? is this true?

Yes, there is no oxygen to accept the electrons from the electron transport chain so the yeast must ferment to free the energy "holding" molecules of their H's.  (If i remember correctly)


http://s126.photobucket.com/albums/p108/ckck07/?action=view&current=Circleotop.jpg

Assume circle O is tangent to Circle P.
What is the ratio of the area of circle P to circle O?

[quote="ckck"]i_like_pie, how did you find the radius of circle P? That's what I always have trouble with.[/quote]
I messed up... :blush:

[hide]Lets say the radius of O is r and the radius of P is y. First you draw an isoscles right triangle with side lenghts r+y, r-y and r-y.
Then you get that $r\sqrt{2}-y\sqrt{2}= r+y$. If you simplify this then you get $\frac{y}{r}= \frac{\sqrt{2}-1}{\sqrt{2}+1}$

Then you multiply the bottom by ${\sqrt{2}-1}$ and you get the ratios to be $\frac{2-2\sqrt{2}+1}{1}$ Now since you want the area you square each side and get

$\frac{17-12\sqrt{2}}{1}$ .[/hide]

[quote="JavaMan"][quote="i_like_pie"][hide]Let the side length of the square be $x.$

The radius of $O$ is $\frac{x}2,$ so it's area is $\frac{\pi x^{2}}4.$

The radius of $P$ is $\frac{x\sqrt{2}-x}2,$ so it's area is $\frac{\pi\left(3x^{2}-2x^{2}\sqrt{2}\right)}4.$

The ratio is $\frac{\pi\left(3x^{2}-2x^{2}\sqrt{2}\right)}4: \frac{\pi x^{2}}4,$ or $\boxed{3-2\sqrt{2}: 1}.$[/hide][/quote]

Are you sure radius of $P$ is $\frac{x\sqrt{2}-x}2$?

[hide]
Let $r$ be the radius of $O$ and $R$ be the radius of $P$.
Dropping perpendiculars from their intersection and center of $P$.
By similarity,
$\frac{(\sqrt{2}-1)r}{\frac{(\sqrt{2}-1)r}{\sqrt{2}}}= \sqrt{2}= \frac{(\sqrt{2}-1)r-R}{R}\implies (\frac{r}{R})^{2}= 12\sqrt{2}+17$
[/hide][/quote]

Are you sure about the similarity part because I can't understand how you came up with the equation.

[quote="JavaMan"]Yeah I'm sure, alternatively you can see that $(\sqrt{2}-1)r$ is the length of C to the intersection of the circles, which is also $(\sqrt{2}+1)R$[/quote]

Then from there don't you set up an equation like this:

$(\sqrt{2}+1)R=(\sqrt{2}-1)r$  because they both equal the same thing?

[quote="JavaMan"]That's correct. Ratio of areas becomes $(\frac{r}{R})^{2}$ which is just what you have, the reciprocal of my answer (misread P:O in the beginning).[/quote]

Oh, okay I see.


Let $ a$, $ b$, and $ c$ be positive real numbers such that $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 6$. Prove that
\[ \frac {a \plus{} 1}{(a \minus{} 1)^2} \plus{} \frac {b \plus{} 1}{(b \minus{} 1)^2} \plus{} \frac {c \plus{} 1}{(c \minus{} 1)^2} \ge \frac {5}{3}.
\]

Edit: $ a,b,c>1$

i find this ineq in a beautiful way... :)

Edit: OMG.. we need this one $ a,b,c>1$. :( sorry for my mistake..

Nice inequality, Raja. :)
[hide="You need patience in reading this solution"]
$ (\sum\sqrt {a})^2\le 3\sum a$ from Cauchy.
$ a\plus{}b\plus{}c\ge 12$

Now, observe that from CS, we have
$ \sum\frac {(a \minus{} 1)^2}{a \plus{} 1}\cdot\sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge 9$................(1)

Thus it suffices to prove
$ \sum \frac {(a \minus{} 1)^2}{a \plus{} 1}\ge \frac {27}{5}$

I prove this in a long way.
$ LHS \equal{} \sum\frac {(a \plus{} 1)^2 \minus{} 4a}{a \plus{} 1} \equal{} a \plus{} b \plus{} c \plus{} 3 \minus{} 4\sum\frac {a}{a \plus{} 1}$
$ \equal{} a \plus{} b \plus{} c \plus{} 3 \minus{} \sum\frac {a \plus{} 1 \minus{} 1}{a \plus{} 1} \equal{} a \plus{} b \plus{} c \plus{} 3 \minus{} 4\times 3 \plus{} 4\sum\frac {1}{a \plus{} 1}$
$ \ge 12 \plus{} 3 \minus{} 12 \plus{} 4\sum\frac {1}{a \plus{} 1} \equal{} 3 \plus{} 4\sum\frac {1}{a \plus{} 1}$........(2)

Now, we have, from CS(again!?) $ (a \plus{} b \plus{} c \plus{} 3)\sum\frac {1}{a \plus{} 1}\ge 9$
or, $ \sum\frac {1}{a \plus{} 1}\ge\frac {9}{15} \equal{} \frac {3}{5}$

Using this, we get, from (2) that
$ LHS\ge 3 \plus{} \frac {12}{5} \equal{} \frac {27}{5}$.
Or, we have obtained that $ \sum \frac {(a \minus{} 1)^2}{a \plus{} 1}\ge\frac {27}{5}$

Hence (1) gives,
$ \frac {27}{5}\cdot\sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge 9$
OR, $ \sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge \frac {9\times 5}{27} \equal{} \frac {5}{3}$

$ QED$!!!!  :blush: [[/hide]

Hope this is correct(I think it's correct.) :)

[quote="Potla"]

$ (\sum\sqrt {a})^2\le 3\sum a$ from Cauchy.
$ a \plus{} b \plus{} c\ge 12$

Now, observe that from CS, we have
$ \sum\frac {(a \minus{} 1)^2}{a \plus{} 1}\cdot\sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge 9$................(1)

Thus it suffices to prove
$ \sum \frac {(a \minus{} 1)^2}{a \plus{} 1}\ge \frac {27}{5}$

...

Hence (1) gives,
$ \frac {27}{5}\cdot\sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge 9$
OR, $ \sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge \frac {9\times 5}{27} \equal{} \frac {5}{3}$

[/quote]

sorry, my friend, but your proof is not valid...

$ \sum\frac {(a \minus{} 1)^2}{a \plus{} 1}\cdot\sum\frac {a \plus{} 1}{(a \minus{} 1)^2}\ge 9$

and

$ \sum \frac {(a \minus{} 1)^2}{a \plus{} 1}\ge \frac {27}{5}$

does not imply the desired inequality...

$ \sum \frac {a \plus{} 1}{(a \minus{} 1)^2}$ can be smaller...

a ya...
my inequality is actually like this..

Let $ a$, $ b$, and $ c$ be positive real numbers satisfying $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {a^2 \plus{} 1}{(a^2 \minus{} 1)^2} \plus{} \frac {b^2 \plus{} 1}{(b^2 \minus{} 1)^2} \plus{} \frac {c^2 \plus{} 1}{(c^2 \minus{} 1)^2} \ge \frac {135}{32}.
\]
But i think this one has bad constant so i change it into the nice one..

but this one will be easier to see.. :)


inequalities
Cauchy Inequality
algebra open
algebra

Solve the equation [tex]\sqrt{x} + \sqrt[3]{x+7}=\sqrt[4]{x+80}[/tex]

i dont know but something tells me it is just $1$.(if you want i can tell you , what is that SOMETHING)

Any hints please??? I'm not getting anything except an idea for substitution :spider:

Here is my Idea:

from the question
$\sqrt{x} + \sqrt[3]{x+7}=\sqrt[4]{x+80}$

we change it in this form:
$ \sqrt[12]{x^{6}} + \sqrt[12]{(x+7)^{4}}=\sqrt[12]{(x+80)^{3}}$

from Cauchy Inequality
$\sqrt[12]{x^{6}} + \sqrt[12]{(x+7)^{4}} \geq 2\sqrt[12]{x^{3}(x+7)^{2}}$

Hence, we have
$\sqrt[12]{(x+80)^{3}} \geq 2\sqrt[12]{x^{3}(x+7)^{2}}$

I think that it does not hold in a large area.(we can find it by comparing the degree of the polynomial. Of course, I have not finished the proof)

then the remain should be solved with complex procedure.

This is just my idea..  ( In Fact, this question could be converted to the problem of inequality, Indeed, I don't want to use Cauchy-Ineq  because it is so weak  ;) )

I think 1 should be  the unique answer in this equation.

I think this is a solution:

it is obvious that x is non-negative, then consider 2 possibilities when x>1 and x<1;
for x>1;=>  \sqrt[3]{x+7}>2; =>

         \sqrt{x}+2<sqrt[4]{x+80} => let's take 4th power both sides
        =>
         64>{x}^{2}+23*x+8*x*(4+\sqrt{x})
        but right side is >64,because x>1.Contradiction! For x<1 analogously.
        So only solution is when x=1. :)

[quote="coffeechao"]Here is my Idea:

from the question
$\sqrt{x} + \sqrt[3]{x+7}=\sqrt[4]{x+80}$

we change it in this form:
$\sqrt[12]{x^{6}} + \sqrt[12]{(x+7)^{4}}=\sqrt[12]{(x+80)^{3}}$

from Cauchy Inequality
$\sqrt[12]{x^{6}} + \sqrt[12]{(x+7)^{4}} \geq 2\sqrt[12]{x^{3}(x+7)^{2}}$

Hence, we have
$\sqrt[12]{(x+80)^{3}} \geq 2\sqrt[12]{x^{3}(x+7)^{2}}$

I think that it does not hold in a large area.(we can find it by comparing the degree of the polynomial. Of course, I have not finished the proof)

then the remain should be solved with complex procedure.

This is just my idea..  ( In Fact, this question could be converted to the problem of inequality, Indeed, I don't want to use Cauchy-Ineq  because it is so weak  ;) )

I think 1 should be  the unique answer in this equation.[/quote]
By computer, the solution to $\sqrt[12]{(x+80)^{3}} \geq 2\sqrt[12]{x^{3}(x+7)^{2}}$ is $x\leq 1.24424$, so you are right.


An equilateral triangle is divided into $9000000$ congruent equilateral triangles by lines parallel to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that there exist three points of the same colour being the vertices of a triangle with its sides parallel to the lines of the original triangle.

I solved it when I was on my way from the home to the school.

Suppose a contrary.
Consider the bottom side of the triangle. It contains 3001 vertices. So there are at least 1001 points of the same color (color 1). Consider all possible triangles, whose two 2 vertices coincide with two of these points. There are at least $1001\cdot 1000/2=500500$ such triangles. Their third vertices of color 2 or 3, and they are placed on 3000 horizontal lines, so there is a horizontal line, which contains at least 167 of these vertices. At least 84 have the same color (color 2). If we consider a triangle, whose two vertices coincide with two of these 84 vertices, then third vertex is colored (!) in color 3. There is at least $84\cdot 83/2=3486$ such triangles, so there is two vertices of color 3 on the same horizontal line, but then third vertex for the corresponding  triangle cannot be colored in color 1, 2 or 3. Contradiction.


algebra
system of equations
algebra unsolved

Let $x = p, y = q, z = r, w = s$ be the unique solution of the system of linear equations \[ x + a_i \cdot y + a^2_i \cdot z + a^3_i \cdot w = a^4_i, i = 1,2,3,4. \] Express the solutions of the following system in terms of $p,q,r$ and $s:$ \[ x + a^2_i \cdot y + a^4_i \cdot z + a^6_i \cdot w = a^8_i, i = 1,2,3,4. \] Assume the uniquness of the solution.


quadratics
number theory unsolved
number theory

Evaluate $ (\frac{p\minus{}1}{2})!$ modulo $ p$, for each prime number $ p$.

Let $ A\equal{}(\frac{p\minus{}1}{2})!\mod p, \ B\equal{}\prod_{1\le x\le (p\minus{}1)/2} (p\minus{}x)\equal{}\theta A\mod p, \theta \equal{}(\minus{}1)^{(p\minus{}1)/2}$.
Therefore $ \theta A^2\equal{}\minus{}1.$
0. For $ p\equal{}2, A\equal{}1$
1. For $ p\equal{}1\mod 4$ $ \theta \equal{}1$, therefore $ A^2\equal{}\minus{}1$
2. For $ p\equal{}3\mod 4$ $ A^2\equal{}1$ and $ (\frac{A}{p})\equal{}(\minus{}1)^m$, were m is number of non quadratic residius mod p les than p/2.

Not really understand your solution, Rust. Please write more clearly. I asked you to evaluate $ A$, not $ A^2$.

1. We had 2 solutions $ A^2\equal{}\minus{}1$ for $ p\equal{}1\mod 4$. Let $ m\equal{}\frac{p\minus{}1}{4}$.
Let g is primitive rote and k is number of $ \{i|g^i>p/2,i\equal{}1,..,2m\minus{}1 \}$. Then $ A\equal{}(\minus{}1)^{m\plus{}k}g^m$.
I don't know simple expression for it.
2. $ A\equal{}(\minus{}1)^k$, were k is non quadratic residius in $ 1,1,..,(p\minus{}1)/2$.