algebra
polynomial
induction
function
algebra proposed

Find all polynomials $P(x) \in \mathbb{R}[x]$ such that $P(x^2+1)=P(x)^2+1$ holds for all $x \in \mathbb{R}$.

I remember  a problem from our olympiad 2 years ago (Iran 3rd round 2003) that:
Find all polynomials $p(x)\in \mathbb{C}[x]$ that $p(x)^2+c=p(x^2+c)$. that: $c\in \mathbb{C}$ with conditions that:

a)$|c|>2$
b)$c$ is rational and non integer.
c)$c$ is not algebraic (i.e. not root of any polynomial in $\mathbb{Z}[x]$)
d)$c\in \mathbb{R}$ and $c$ is outside $ [-2,\frac14]$

For all of these conditions we could prove all of polynomials are $f_0(x),f_0(f_0(x)),...$ that $f_0(x)=x^2+c$.
The proof was easy you proved that: if $deg(p)>1$ then there exist polynomial $p_1$ that $p_1^2+c=p$.

This solves your problem

Also you see for $c=-2$ all solution are those $p$ that there exist $n$ that $\forall x\in \mathbb{R}$ we have $p(x+\frac 1x)=x^n+\frac1{x^n}$ and for $c=-1$ the only polynomial with odd degree is $p(x)=x$. I think the problem in general situation has some relations with Mandelbrat (I'm not sure the spelling) set in complex plane.

here is my solution:
let $x_0=1, x_(n+1)=x_n^2+1$, then by induction,we obtain $p(x_n)=x_n$, by polynomial identity theorem, we obtain $p(x)=x$.

[quote="Omid Hatami"]
... we could prove all of polynomials are $f_0(x),f_0(f_0(x)),...$ that $f_0(x)=x^2+c$.
[/quote]

like he said, there is unique polynomial of every even degree, and there are no polynomials of odd degree grater than 1

First I proved that there exist a unique polynomial of degree $2^n$ for $c$ with the conditions I said. Also your solution is wrong xiaozg. How do you say that $p(1)=1$ ? You see that $x^2+1$ is also a solution.

And another thing you can prove that for each degree $n$ there exist at most one polynomial with the condition $p(x^2+c)=p(x)^2+c$

[quote="Omid Hatami"]

For all of these conditions we could prove all of polynomials are $f_0(x),f_0(f_0(x)),...$ that $f_0(x)=x^2+c$.
The proof was easy you proved that: if $deg(p)>1$ then there exist polynomial $p_1$ that $p_1^2+c=p$.

This solves your problem
[/quote]

i dont understand how THAT solves the problem...can anybody tell me??

:) I put together all your ideas, added some others and finally got this:

Let A be the set of all polynomials P in the statement.
For each P in A we have $P^2(-x)=P^2(x) \Rightarrow P(-x)=\pm P(x) \forall x\in R$, so at least one of the equalities $P(-x)=P(x)$ and $P(-x)=-P(x)$ holds for an infinit number of x.

In the first case, by polynomial identification we obtain
$(1) P(x)=\sum_{k=0}^{n}a_kx^{2k}$,
while in the second
$P(x)=x \sum_{k=0}^{n}a_kx^{2k}$
(half of the coefficients equal 0 in both cases).
It follows that one and only one of the two equalities holds for all $x \in R$.

I). If $P(-x)=-P(x), \forall x\in R$, we have P(0)=0.
Define $x_0=0, x_{n+1}=x_n^2+1$ and we obtain by induction $P(x_n)=x_n, \forall n \in N$, so $P(x)=x$ (polynomial identification; the sequence is strictly increasing, so it contains an infinite number of distinct real values, which legitimates the identification).

II). If $P(-x)=P(x), \forall x\in R$, let t be a number in $[1,\infty)$. For $x=\sqrt{t-1}$ we have $x^2+1=t$, hence $P(t)=P^2(\sqrt{t-1})+1, \forall t \in [1, \infty)$. From (1) we infer that $P(\sqrt{t-1})$ is a polynomial function, say $Q(t)$ (no sqrts remain after replacing in (1)).

Consequently, for each P in A we have  either
   $P(x)=x$
or
$(2) P(x)=Q^2(x)+1$ for some polynomial Q.
In the second case we can assume $Q \in A$. Indeed: from (2) we get $P(x^2+1)=Q^2(x^2+1)+1$, thus $P^2(x)+1=Q^2(x^2+1)+1$, so
$P(x)=\pm Q(x^2+1)$
We obtain
$Q(x^2+1)=Q^2(x)+1$
or
$-Q(x^2+1)=Q^2(x)+1$.
In first case $Q \in A$.
In the second $-Q \in A$. If we denote $S=-Q$, relation (2) becomes
$P(x)=S^2(x)+1$, where $S \in A$.
Cosequently, each polynomial P in A which satisfies $P(-x)=P(x)$ can be written $P(x)=Q^2(x)+1$ for some polynomial Q [b]in A[/b].
Denote $P_0(x)=x$ and $f(x)=x^2+1$. We have either $P=P_0$ or $P=f \circ Q$ for some polynomial Q in A.
We can now easily prove  by induction that each polynomial P in A of degree > 1 can be written $P=f\circ f\circ...\circ f$.

So $A=\{P_0\}\cup\{P/P=f\circ f\circ...\circ f , \mbox {where \"}\circ \mbox{\" is applied n times}, n\in N\}$ (there is no constant polynomial in A).

A generalization:
Find all polynomials $P\in R[X]$ such that
(1)          $P(x^2+c)=P^2(x)+c, \forall x\in R$,
where
(2)          $c>\frac{-1}{4}$.

We need condition (2) to prove that the sequence $x_0=0, x_{n+1}=x_n^2+c$ is strictly increasing, so its terms take an infinite number of distinct real values (otherwise we wouldn't have been allowed to infer $P(x)=x$ from $P(x_n)=x_n$; see solution #7 above).

[quote="joshua_mex"]and what about c=1, how do i find that there is only one solution... (Poland NMO 2000 problem 6)[/quote]
You can't find such a thing, because it's not true! :)
Just check these two functions:
$f_1(x)=x, \forall x\in \mathbb{R}$;
$f_2(x)=x^2+1, \forall x\in \mathbb{R}$.
Actually, there are infinitely many suitable functions (see the complete solution for c=1 above at #7).

[quote=joshua_mex]and what about c=1, how do i find that there is only one solution... (Poland NMO 2000 problem 6)[/quote]

$c=-1$ in Poland 2000 problem 6.!!!


Find all non-negative integral solutions (n_1,n_2,\cdots,n_14) to
\[n_1^4+n_2^4+\cdots+n_{14}^4=1599\]

Well $7^4>1599$ and if you have $6^4$ in expression you can have only one $4^4$ etc. There is not much to think, just compute "law of brut force"  :D  :D

let me restate the problem:
find all $(n_1,n_2,...,n_{14})$, in which:

$\displaystyle n_1^4+n_2^4+\cdots+n_{14}^4=1599$

my solution:
there are no such forth powers.

take a look at equation mod $16$. all the $n_i^4$ are either $0$ or $1$ mod $16$. so the $\sum_{i=1}^{14}n_i^4 \equiv 0\rightarrow 14 (\mod 16)$, but $1599\equiv -1 (\mod 16)$ which is impossible.

BTW, chess 64, i like your signiture.

I tried mod 4, it did not work.
Can you tell me how did you figure out that mod 16 will solve the problem?

Did you just guess or did you actually have a technique of doing these?

well i knew that forth powers are either 0 or 1 mod 16.
also if you look at the forth powers less than 1599, yousee that the only ones are 1, 16, 81, 256, 625, 1296 , which are either 0 or 1 mod 16.

Nice solution.... For that one who asked how to guess module, well you see that every fourth power of even number is divisible with 16 and even powers of odd numbers gives 1 like residue when divides with $2^l$

[quote="amirhtlusa"]BTW, chess 64, i like your signiture.[/quote]

:D lol thanks :)


I've heard about "Go" all over but I don't know what it is. Can someone please explain what it is and, more importantly, how to play?

Thanks

Yep. :)
Here for example : http://en.wikipedia.org/wiki/Go_(board_game)


wouldn't there be multiple answers...? unless the blanks are the same number...?

[quote="grizzland"]3x^2 + 5x^2 + _____ = 3(x+____)^2

thanks for the help. it really helps me figure out how to do the problem once i have the answer[/quote]

Don't you mean 3x^2+5x+____ on the left hand side? (It should be, I believe)

Then
[hide]
\[3x^{2}+5x+\square = 3 \left(x+\frac{5}{6}\right)^{2}\]
which becomes
\[3x^{2}+5x+\frac{25}{12}= 3 \left( x+\frac56 \right)^{2}. \]
[/hide]


If $a,b,c$ are positive real numbers such that $a+b+c+2=abc$,prove that the following inequality occurs:\[(a-1)(b-1)(c-1)\leq 1\]


Prove that for positive x, y, z
x^2*y/z+y^2*z/x+z^2*x/y>= x^2+y^2+z^2.

Have you tried using Cauchy for (x,y,z), (y/z;z/x;x/y) considering them as triplets of the same monotony? :D

hmmm Cauchy doesn't involve monotony, but rather Cebasev does. What did you mean?

I was silly ... I meant Cabasev.. sorry for that!

Well ordering both triplets, (x, y, z) and (x/y,etc) looks really nasty to me.  The usual method of finding a, b, c such that a+b+c=1 and                     \sum (a*(x^2*y)/z) >= x^2 and the others (eventually multiply by xyz)doesn't work because you find a negative b and you can't apply AM-GM. I don't know how to apply Murhead or Schur in this particular case.

i have an half solution.
Suppose x>=y>=z>0.

since a+1/a>=2 wa have:
(x^2y/z+y^2z/x+z^2x/y)+(x^2z/y+y^2x/z+z^2y/x)>=2(x^2+y^2+z^2).
Then if we prove that (x^2y/z+y^2z/x+z^2x/y)>=(x^2z/y+y^2x/z+z^2y/x),
we are done.
we have
(x^2y/z+y^2z/x+z^2x/y)>=(x^2z/y+y^2x/z+z^2y/x)
<=>
x(y^2-z^2)+yz(y-z)>=0  (after simple computation)
which is true as x>=y>=z>0

then the inequality is true if x>=y>=z>0.
but it remain the case where 0<=x<=y<=y.

bye.

WLOG,suppose z>=x>=y
and A=xy ,B=yz , C= zx then C>=B>=A and the inequality becomes
A  2 /B + B 2 /C + C 2 /A >=
AB/C + BC/A + CA/B.After clearing the denominators,
A^3*C + B^3*A + C^3*B >=
(AB) 2   + (BC) 2 + (CA) 2
After some calculations
A^4 + B^4 + C^4 + 2A^3C + 2B^3 A + 2C^3B >= (A 2  + B 2  + C 2 ) 2
BY CAUCHY WE HAVE
(A+B+C)(A^3 + B^3 +C^3) >= (A 2  + B 2  + C 2 ) 2
AFTER EXPANDING WE SEE THAT IT IS ENUF TO PROVE
A^3C + B^3A +C^3B >= A^3B + B^3C + C^3A
WHIC IS EQUIVALENT TO
(C-B)(B-A)(C-A)(A+B+C)>= O WHICH IS TRUE BY ASSUMPTION
I hopw this is correct :D
cheers

Arne observed the following:

" compare this post:
http://www.mathlinks.ro/phpBB/viewtopic.php?t=293
to problem 6 on this page:
http://www.kalva.demon.co.uk/vietnam/viet91.html
Strange, isn't it? "

10x Arne! :)

cheers! :D

The official text is :

Let x,y,z positive real numbers with x \geq y \geq z. Prove

 x 2 y/z + y 2 z/x + z 2x/y \geq x 2 + y 2 + z 2

A delightful solution is by the substitution x=a   y=a+b   z=a+b+c

We get (a+b+c)^3(ab+b 2 )+a^3c(a+b+c) \geq (a+b)^3(ab+ac) which is true by simple inspection

I guess that this solves the problem (both the corect text and the solution)! :D

Manlio thank you for the corection and solution! :)

cheers!!!

Ok, I'll send the correct version to John Scholes.

nevermind I already did that. I also told him to keep an eye here. Maybe register? :)

try to convince him of that. :D I tried. seems that I have failed.

I'll keep saying it, each time I send a correction or so :)
He's a member of the yahoo group.

maybe we will see him around here at some point. I certainly hope so.


[quote="LordoftheMorons"]Just wondering what type of movies AoPSers like  :D

My favorite three movies:

1. The Shawshank Redemption
2. Lord of the Rings: Return of the King
3. Monty Python and the Holy Grail[/quote]

1."9 reinas" (argentinian movie)
2."Una mente brillante" (a beautiful mind)
3."Truman Show".

I like JIm Carrey and BEn Stiller comdies.

in no order
Bananas-Wood Allen
Seven Samurai and Magnificent Seven
Blazing Saddles, Spaceballs, Young Frankenstien-Mel Brooks
Great Escape
Lord of the Rings

good recent movies (not sure if these will be classic favorites of mine or not probably will)
V for Vendetta
the Matrix (original then they got worse and worse but still not   bad)
Serenity

lotr
holy grail
meaning of life
life of brian
HOSTEL (jk?)

probably will like sentinel.
JACK BAUER IS REAL, KIEFER SUTHERLAND IS FICTIONAL (obviously)

The Bicycle Thief
A Roman Holiday
The Bridge on the River Kwai
Psycho
Ben Hur
Pather Panchali
Sonar Kella
Goopi Gyne Bagha Byne
Singin' in the Rain


factorial
floor function
ceiling function
puzzles

Express all the integers between 1 and 10 inclusive using three 3's and any combination of ordinary operations (ex. adding, subtracting, multiplying, dividing, exponentiation...)

If you could find the integers between 11 and 20, that would be nice.

This is not homework. I have the answers already.

[hide="hint"]
There is no hint! :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol:
[/hide]

Is factorial allowed?
[hide]
1) $3 - \frac{3!}{3}$ Sorry. I couldn't think of one without factorial.

2) $\frac{3+3}{3}$

3) $\frac{3\cdot3}{3}$

4) $3 + \frac 33$

5) $3 + \frac{3!}{3}$ again, I couldn't think of one without factorial

6) $3\cdot3-3$

7) $3! + \frac 33$ again, factorial.

8) $3! + \frac {3!}{3}$

9) $\frac {3^3}{3}$

10) Well....I can't think of one for 10...unless $3 \cdot 3 + \lfloor\sqrt 3 \rfloor$ is allowed...
[/hide]

[hide="11"]$3\cdot 3+\lceil\sqrt{3}\rceil$[/hide]

[hide="12"]$3\cdot 3+3$[/hide]

[hide="13"]$3!+3!+\lfloor\sqrt{3}\rfloor$[/hide]

[hide="14"]$3!+3!+\lceil\sqrt{3}\rceil$[/hide]

[hide="15"]$3!+3!+3$[/hide]

[hide="16"]$\lceil(3\cdot \sqrt{3^3})\rceil$[/hide]

I can't believe I got something for 16... I don't know if it's possible to go any higher.

[b]EDIT:[/b] My dad found a way to do $17$, and I was just thinking of going consecutively. So:

[hide="17"]$3!\cdot 3-\lfloor\sqrt{3}\rfloor$[/hide]

[hide="18"]$3\cdot (3+3)$[/hide]

[hide="19"]$3!\cdot 3+\lfloor\sqrt{3}\rfloor$[/hide]

[hide="20"]$3!\cdot 3+\lceil\sqrt{3}\rceil$[/hide]

[hide="21"]$3!\cdot 3+3$[/hide]

And I can find multiple ways of getting $24$...

[hide="Easy Way"]$3^3-3$[/hide]

[hide="Fun Way"]$3!\cdot (3+\lfloor\sqrt{3}\rfloor)$[/hide]

Yes, factorial is allowed.

Wow, you people are creative.

One thing, though: I don't think you can get higher than 9 (with consecutive integers) without using floor/ceiling functions. ;)

Can anyone get 22/23?

[quote="JesusFreak197"]Can anyone get 22/23?[/quote]

yep, by floor/ceiling abuse.

[hide="22"]
$\lceil\sqrt{(3!)!}\rceil-\lceil\sqrt{3}\rceil-3$
[/hide]

[hide="23"]
$\lceil\sqrt{(3!)!}\rceil-\lfloor\sqrt{3}\rfloor-3$
[/hide]

might as well continue :P

[hide="25"]
$\lceil\sqrt{(3!)!}\rceil+\lfloor\sqrt{3}\rfloor-3$
[/hide]

[hide="26"]
$\lceil\sqrt{(3!)!}\rceil+\lceil\sqrt{3}\rceil-3$
[/hide]

[hide="27"]
$\lceil\sqrt{(3!)!}\rceil*\frac 33$
[/hide]

[hide="28"]
$\lceil\sqrt{(3!)!}\rceil+\frac 33$
[/hide]

[hide="29"]
$\lfloor\sqrt{(3!)!}\rfloor+3!-3$
[/hide]

[hide="30"]
$\lceil\sqrt{(3!)!}\rceil+3!-3$
[/hide]

With enough factorials, floors/ceilings, and roots, wouldn't you be able to get any number using only one 3?

Probably, but it would look ridiculously complicated. For example, $\displaystyle\left\lfloor\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3!!!}}}}}}\right\rfloor=1278$. :roll:


A circle with center [i]I[/i] is inscribed in $ \triangle{ABC}$. Extend $ AI$ to intersect side $ BC$ at point $ D$, and let $ P$ be the intersection of $ IC$ with the angle bisector of $ \angle{IAC}$. If $ \angle{A}=40^\circ$ and $ \angle{B}=110^\circ$, find the measure of $ \angle{APD}$ in degrees.

[quote="gotztahbeazn"][hide]

Since $ AP$ is the angle bisector of $ \angle IAC$ and $ IC$, which is the angle bisector of $ \angle C$ also intersects this angle bisector at $ P$, we know $ P$ is the incenter of $ \triangle ADC$. Now, drawing $ DP$, we know that $ DP$ is the angle bisector of $ \angle ADC$ as $ DP$ clearly passes through the incenter

knowing this the rest is just easy angle chasing and our final answer is $ \angle APD = 105^{\circ}$

[/hide][/quote]
Yeah, I realized those facts, but I just messed up on the angle-chasing or something, because I just couldn't get the answer. Could you show the angle-chasing?


counting
geometry
combinatorial geometry
combinatorics
IMO Shortlist

Let $ n \geq 4$ be a fixed positive integer. Given a set $ S \equal{} \{P_1, P_2, \ldots, P_n\}$ of $ n$ points in the plane such that no three are collinear and no four concyclic, let $ a_t,$ $ 1 \leq t \leq n,$ be the number of circles $ P_iP_jP_k$ that contain $ P_t$ in their interior, and let \[m(S)=a_1+a_2+\cdots + a_n.\] Prove that there exists a positive integer $ f(n),$ depending only on $ n,$ such that the points of $ S$ are the vertices of a convex polygon if and only if $ m(S) = f(n).$

It is strange to see such an easy and evident problem in IMO SL.

We claim that the function $f(n)=2 \binom{n}{4}$ satisfies the requirements.

Let the [i]score[/i] $s(a,b,c,d)$ of the four points $P_a$, $P_b$, $P_c$ and $P_d$ be the number of points $P_i$ where $i \in \{a,b,c,d \}$ such that $P_i$ is properly contained in the circle passing through the remaining three points. Observe that

\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d)\]

First we prove the following lemma.

[b]Lemma 1.[/b] The score of a convex quadrilateral $2$.

[i]Proof.[/i] Let the vertices of the convex quadrilateral be denoted as $A, B, C$ and $D$. We have that point $D$ lies within the circumcircle of $\triangle{ABC}$ if and only if

\[\angle{ABC} > 180 - \angle{ADC} \quad \Leftrightarrow \quad \angle{ABC} + \angle{ADC} > 180\]

Therefore if $D$ lies within the circumcircle of $\triangle{ABC}$, it also follows by symmetry that $B$ lies within the circumcircle of $\triangle{ADC}$. If not, then since the sum of the interior angles of $ABCD$ is $360$,

\[\angle{ABC} + \angle{ADC} < 180 \quad \Rightarrow \quad \angle{BAD} + \angle{ACD} > 180\]

Therefore $A$ lies within the circumcircle of $\triangle{BCD}$ and $C$ lies within the circumcircle of $\triangle{ABD}$. In both cases, the score of $ABCD$ is equal to $2$.

[b]Lemma 2.[/b] The score of a concave quadrilateral is $1$.

[i]Proof.[/i] Let $ABCD$ denote the concave quadrilateral. Without the loss of generality, let $A$ be such that interior angle $\angle{BAC} > 180$. It follows that $A$ lies in the interior of triangle $\triangle{BCD}$. Therefore, the circumcircle of $\triangle{BCD}$ contains $A$. However, none of the remaining three circles passing through three of the points $A, B, C$ and $D$ contain the remaining point. Hence the score of $ABCD$ is equal to $1$.

[b]If Direction.[/b] If the vertices of $S$ form a convex $n$-gon, then each quadrilateral formed by four distinct points in $S$ is a convex quadrilateral and therefore

\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d) = 2 \binom{n}{4}\]

[b]Only If Direction.[/b] If the vertices of $S$ form a concave $n$-gon, then at least one of the quadrilaterals formed by four distinct points in $S$ is a concave quadrilateral and therefore

\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d) < 2 \binom{n}{4}\]

The function $f(n)=2 \binom{n}{4}$ therefore satisfies that $m(S)=f(n)$ if and only if the points in $S$ are the vertices of a convex polygon.

[b]Lemma[/b] Every quadrilateral $S$ satisfies $m(S) \le 2$ with equality if and only if it is convex,

Proof:If a quadrilateral $ABCD$ is convex then one out of $\angle{ABC}+\angle{ADC}$ and $\angle{DAB}+\angle{DCB}$ is greater than $\pi$ and the other is less than $\pi$.This implies $m(S)=2$.One the other hand if it is concave then clearly $m(S)=1$

Back to our main problem.If S forms a convex polygon then every quadrilateral $P_{i}P_{j}P_{k}P_{l}$ contributes $1$ to two of $a_{i},a_{j},a_{k}$ and $a_{l}$ (from the lemma) and thus contributes two to $m(S)$.This immediately implies $f(n)=2\binom{n}{4}$

Now let $m(S)=f(n)=2\binom{n}{4}$.Each quadrilateral   $P_{i}P_{j}P_{k}P_{l}$  contributes atmost $2$ to $f(n)$ and thus $m(S) \le 2\binom{n}{4}$.As there is equality,we have that each quadrilateral contributes exactly $2$ to $f(n)$,hence from the lemma every quadrilateral is convex,implying that $S$ forms a convex polygon.

Note that given any 4 points, they add 2 to the count if their convex hull is a quadrilateral, and 1 otherwise. So, $f(n)=2\binom{n}{2}$, with equality achieved iff all quadruplets of points form convex quadrilaterals. However, if there exists a point $P_i$ inside the convex hull $Q_1,Q_2,\ldots,Q_k$, then it must be in one of the triangles $Q_1Q_2Q_3$, $Q_1Q_3Q_4,\ldots Q_1Q_{k-1}Q_k$, which cover the convex hull. So, there exist 4 points whose convex hull is a triangle. Therefore, the only way equality is reached is if all points are on the convex hull, as desired.

The key idea is to consider four-tuples. \\

[i] Lemma: [/i] If $ABCD$ is a non-cyclic convex quadrilateral, $m(ABCD) = 2$. \\

[i] Proof: [/i] Suppose $ABCD$ is convex and non-cyclic. Note that $\angle ABC + \angle ADC \neq \angle BCD + \angle BAD \neq 180^\circ$. Hence, exactly one of these two sums is greater than $180^\circ$. WLOG, suppose $\angle ABC + \angle ADC > 180^\circ$ and $\angle BCD + \angle BAD < 180^\circ$. Since $\angle ABC > 180^\circ - \angle ADC$ it follows that $B$ lies within $(ADC)$. Similarly, $D$ lies within $(ABC)$. Meanwhile, since $\angle BCD < 180^\circ - \angle BAD$, it follows that $A$ does not lie within $(BCD)$, and similarly, $C$ does not lie within $(ABD)$. This implies that $m(ABCD) = 2$, proving the lemma. $\blacksquare$

[i] Lemma: [/i] If $ABCD$ is a concave quadrilateral, $m(ABCD) = 1$. \\

[i] Proof: [/i] Say $\angle BAD > 180^\circ$. Since $A$ and $C$ both lie on the same side of $\overline{BD}$, and $\angle BCD < \angle BAD$, it follows that $C$ does not lie within $(ABD)$, while $A$ lies within $(BCD)$. Meanwhile, since $\angle ABC + \angle ADC < 180^\circ$, by the same argument as in the previous lemma we find that $B$ does not lie within $(ACD)$ and that $D$ does not lie within $(ABC)$. Hence, $m(ABCD) = 1$. $\blacksquare$ \\

We now return to the given problem. We claim that $m(S) \leq 2\binom{n}{4}$, and that equality holds iff the points of $S$ form a convex quadrilateral. Indeed, as no four points in $S$ are concyclic, we have
\begin{align*}
	m(S) &= \sum_{1 \leq w < x < y < z \leq n} m(P_wP_xP_yP_z) \\
	&\leq \sum_{1 \leq w < x < y < z \leq n} 2 \\
	&\leq 2\binom{n}{4}.
\end{align*}
Equality holds here iff each of the quadrilaterals $P_wP_xP_yP_z$ is convex, which occurs iff the points of $S$ form a cyclic quadrilateral. This completes the proof. $\Box$

Rather easy for a C3.

All angles are in degrees.

[color=#f00][b]Claim:[/b][/color] Suppose $Q$ consists of four points $A,B,C,D$. Then $m(Q)$ is $2$ if $A,B,C,D$ form a convex quadrilateral and $1$ otherwise.
[i]Proof.[/i] Notice that if $A,B,C,D$ form a convex quadrilateral, then $D$ appears inside the circumcircle of $ABC$, $\angle D$ must be greater than $180-\angle B$, which is equivalent to $\angle B + \angle D >180$. Since exactly one opposite pair of angles sum to greater than $180$ degrees, there will be exactly two points that are contained in the circumcircle of the other three.

If $A,B,C,D$ are not convex, WLOG let $A,B,C$ be the convex hull. Then it is clear that the only point that is contained in the circumcircle of the other three is $D$. $\blacksquare$

The key step is to notice that $$m(S)=\sum_{1\le a<b<c<d\le n} m\left(\{P_a,P_b,P_c,P_d\}\right)$$ because both quantities count the amount of ordered pairs containing a single point of $S$ and a set of three points in $S$, with all four points distinct, such that the single point is inside the circumcircle of the three points.

Using our claim, $m(S)$ must be equal to twice the number of four-point subsets of $S$ that consist of points forming a convex quadrilateral plus the four-point subsets of $S$ that do not. Moreover, every subset of four points form a convex quadrilateral iff $S$ does, so $m(S)=2\binom{n}{4}$ iff $S$ forms a convex quadrilateral. We are done.

Let $m(\{a,b,c,d\})$ be the number $m(S)$ for the quadrilateral $P_aP_bP_cP_d.$ It is easy to see that $m(S)$ is the sum of $m(\{a,b,c,d\})$ for all choices of subset $\{a,b,c,d\} \subseteq S$. Note that $(P_aP_bP_c)$ contains $P_d$ if only $P_d$ lies inside of the angle $P_aP_bP_c$ and $\angle P_b+\angle P_d> 180^\circ.$ Clearly, when $P_aP_bP_cP_d$ then $m(a,b,c,d)=2$ and when it is nonconvex $m(a,b,c,d)=1$. Therefore, $m(S)=2\tbinom{n}4$ if and only if $S$ is convex.


1. What is the smallest positive integer which when divided by 10, 9, 8, 7, 6 leaves the remainder 9, 8, 7, 6, 5 respectively?

Can this be easily solved by applying the modulus (Modulo) properties? I'm not too familar with mod, but I know they are quite useful in solving questions that involve remainders.

Here's the jist of the problem.

[hide]Lets do a smaller problem: Find the smallest number that if divided by 3 results in a remainder of 2 and if divided by 5 gives the remainder of 3.

Express it like this:  N=3a+2=5b+3

3a+2=5b+3 so 3a=5b+1; express it in mod

3a=1 mod 5 --> 3a=6 mod 5 --> a=2 mod 5

Now!  a=5c+2 and remember that N=3a+2....

Plug in a to get N=3(5c+2)+2=15c+6=6 mod 15.

If you have more cases, you use this result (6 mod 15) and combine it with the others.

Maybe you can solve it from here.[/hide]

Actually, there's a shortcut...

[hide]The number must be congruent to -1 modulo each of 6,7,8,9, and 10, so it is -1 mod LCM(6,7,8,9,10). The smallest number is LCM(6,7,8,9,10)-1.[/hide]


function
inequalities unsolved
inequalities

Let $ x,y\in (0;1)$. Find the minimum of
$ A\equal{}\frac{x^2}{1\minus{}y}\plus{}\frac{y^2}{1\minus{}x}\plus{}x\plus{}y\plus{}\frac{1}{x\plus{}y}$

you mean : ]0 , 1[   :wink:
using simple ineqs , i could prove that , f(x,y) (denoting f(x,y) the proposed expression to minimize) is >= 62/27 , i hope i made no mistake but i just can't find the point where it's reached  :oops_sign:

sorry it's not a solution just some possible minimizations :
x + y + 1/(x+y) >= 2
a look at the variations of the function t ---> t sqrt(1-t) gives after cmputations : 1-x < 2 / (3 * sqrt(3) * x² ) , doing the same with x and y and multiplying respectively by y² and x² , i found that : x² / (1-y) + y² / (1-x) >=  4/27 * (x²/y² + y²/x²) which is obviously >= 8 / 27
this gives a possible minimum which is : 2 + 8/27 = 62/27
sorry for delay


probability
Stanford
college
probability and stats

Assume that $ A,B,C,D$ are four people in a competition, with a corresponding powers $ a,b,c,d$. The probability for $ X$ to defeat  $ Y$ is $ \frac{x}{x\plus{}y}$ (which $ x,y$ are the powers of $ X,Y$ respectively).

Knowing that $ A$ and $ B$ will compete first, then  $ C,D$ will compete in the second match. The winner of each math will play one more game to find the Winner of the whole competition.

Find the probability for $ A$ to be the winner of the whole competition.

That seems to be very basic, but I am not sure 100% about the answer, so please help me.

Just do a step by step analysis: A needs to win the game with B. The probability of this is $ P_1\equal{}?$ Now, conditioned on that event, the probability that he'll have to play against $ C$ in the final round is $ P_2\equal{}?$ and the probability that he'll play against $ D$ is $ P_3\equal{}1\minus{}P_2\equal{}?$. In the case of playing against $ C$, the probability of winning is $ P_4\equal{}?$ and in the case of playing against $ D$, it is $ P_5\equal{}?$. The final answer is, thus
\[ P_1\cdot(P_2P_4\plus{}P_3P_5).\]

I do the same way, but still I feel there is something not absolutely accurate.  :(

Well, let's see. Of course we have to make the assumption that the outcomes of the games are independent. Without it, all our computations aren't worth half-penny. Other than that, there is no flaw: you just use the two standard laws of probability at each step: the probabilities of the independent events multiply and the probabilities of mutually exclusive events add. If it makes you feel better, you can consider all 8 possible scenarios, find the probability of each scenario by multiplying the probabilities of the corresponding events and add the two that correspond to A being the winner. The only little problem is that you'll get exactly the same answer :).

An interesting question to answer based on this: Suppose the "powers" of teams A,B,C,D are $ 4,3,2,1$ respectively. In other words, A is the strongest team and D is the weakest team. (A playing D is something like USC playing Stanford in football ... oops!)

Case one: we "seed" the tournament, as seeding is generally understood. In the first round, A plays D and B plays C, with the winners of the first round matches meeting for the championship. What probability does each team have of winning the tournament?

Case two: we do what "seeding" theory suggests is the worst thing we can do. In the first round, A plays B and C plays D. Ask the same question as in the first case.

Case three: try the possibility that is left. In the first round, A plays C and B plays D. Repeat the question.

For each team, which scenario is best for their chances and which scenario is worst?

Interestingly, what Kent suggests is what I had to study before posting this question (and my question here is to make sure that I am not wrong at the first step). Thank Kent and thank you very much if you can let me [i]alone[/i] to think about it further (without give me any of your further suggestion).  :blush:


induction
LaTeX
linear algebra
matrix
algebra
polynomial

Let $ A_1,A_2,\dots,A_n$ be idempotent matrices with real entries. Prove that:
\[ \mbox{N}(A_1)\plus{}\mbox{N}(A_2)\plus{}\dots\plus{}\mbox{N}(A_n)\geq \mbox{rank}(I\minus{}A_1A_2\dots A_n)\]
$ \mbox{N}(A)$ is $ \mbox{dim}(\mbox{ker(A)})$

This is quite trivial, isn't it? For $ n\equal{}1$ this is an equality since the rank of an idempotent is the trace and $ A$ is idempotent if and only if $ I\minus{}A$ is. For greater $ n$ it follows by induction using that $ I\minus{}A_1...A_n\equal{}(I\minus{}A_1...A_{n\minus{}1})A_n\plus{}I\minus{}A_n$, that the rank is sub-additive and does not increase when multiplying by matrices. Of course, one uses also the case $ n\equal{}1$.

Yeah, it's really easy once you have the $ n\equal{}1$ case. You don't even need the nice fact about the trace; it's easy to see that if Av = 0, then (I - A)v = v, so $ \ker(A) \subset Im(I\minus{}A)$. Furthermore, $ A(I\minus{}A) \equal{} A\minus{}A^2 \equal{} 0$, so $ Im(I\minus{}A) \subset \ker(A)$.

Can someone please elaborate on the case n=1, I don't understand why N(A1) = rank(I-A1)

Also, can you please recommend some online article on idempotent matrices, I only menaged to find a brief introduction.

Thanks

[quote="Makaveli"]Can someone please elaborate on the case n=1, I don't understand why N(A1) = rank(I-A1)[/quote]

If $ A_1$ is idempotent then by definition $ A_1^2 \equal{} A_1$, so the eigenvalues of $ A_1$ are all either 0 or 1.  $ N(A_1)$ is the number of 0 eigenvalues.  $ \textrm{rank} (I \minus{} A_1)$ is the number of non-0 eigenvalues of $ I \minus{} A_1$, and this is the number of non-1 eigenvalues of $ A_1$, and this is the number of 0 eigenvalues of $ A_1$.

Sorry, I am new to this:

Is the number of non-zero eigenvalues equal to the number of pivots? (otherwise why would you say (I-A1) is the number of non-zero eigenvalues of I-A1)

Also, how do you know that it is the number of non-1 eigenvalues of A1?
By the way, what does Im stand for?

Thanks

[quote="Makaveli"]Is the number of non-zero eigenvalues equal to the number of pivots? (otherwise why would you say (I-A1) is the number of non-zero eigenvalues of I-A1)[/quote]
I've never used the "pivot" terminology, so I used terminology that I thought was clear.  However, I made a LaTeX error that omitted the word "rank" from that sentence -- I hope now it's clear.

[quote]Also, how do you know that it is the number of non-1 eigenvalues of A1?[/quote]  The rank of a matrix is the number of non-0 eigenvalues.  If the eigenvalues of $ M$ are $ \lambda_1, \ldots, \lambda_n$ then the eigenvalues of $ I \minus{} M$ are $ 1 \minus{} \lambda_1, \ldots, 1 \minus{} \lambda_n$.  Thus, the rank of $ I \minus{} M$ is the number of non-1 eigenvalues of $ M$.

Thanks!!!

Many things are clearer to me now, except, how do you conclude that if M has eigenvalues l1,l2,... I-M has eigenvalues 1-l1,1-l2,..? Does it simply follow from the form of the polynomial whose roots are the eigenvalues, and the conclusion is made when looking at the traces of M and I-M, or there is a neater way? What is the most general form of that rule?

And, what does the Im notation stand for?

If $ M v \equal{} \lambda v$ for some nonzero vector $ v$ then $ (I\minus{}M)v \equal{} v \minus{} \lambda v \equal{} (1\minus{}\lambda) v$.

I don't know about "the most general form", but for instance $ p(\lambda)$ is an eigenvalue of $ p(M)$ for any polynomial $ p(x)$ over the relevant field, for the same obvious reason.

This problem was used at oral examination ENS École Normale Supèrieure 2019


Let a,b>0, and a+b=1. Then
 \sqrta +  \sqrtb + 1/\sqrta + 1/\sqrtb  \geq  \sqrt(5 + 10\sqrt(ab) + 4/\sqrt(ab)) \geq 3 \sqrt2,
equality holds for a = b = 1/2.

It's an easy problem
We have
1. A = \sqrta + \sqrtb + 1/\sqrta + 1/\sqrtb = ( \sqrta + \sqrtb)(1+ 1/\sqrtab),
=> A^2 = (1+2\sqrtab)(1+2/\sqrtab + 1/ab) = 5 + 1/ab + 4/\sqrtab + 2\sqrtab.
It suffices to show 1/ab >= 8\sqrtab. It's trivial b'se  ab <= 1/4.
2. The right one inequality equivalent to
        5 + 10\sqrtab + 4/\sqrtab >= 18 (1)
Apply AM-GM, we have 10\sqrtab + 5/(2.\sqrtab) >= 10
Also  3/(2.sqrtab) >= 3.
From these we have (1) and problem is solved.


Prove that for every real number x, there exists two sequences of integers an and bn such that an+bn*sqrt2 is convergent to x.

Z+ \sqrt 2.Z is dense in R, every x real there exist a_n, b_n sequences
of integers such that a_n + b_n \sqrt 2 converges to x

do we have for any irrational number q that Z+qZ is dense in R????

Yes, because {nq} is dense in (0,1) ({} is the fractional part).


[quote=10maths_tp0609][quote="NguyenDungTN"]Let $ a,b,c >0$.Prove that:
$ (a^{2}+b^{2}+c^{2})(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}) \ge 4(a^{2}+b^{2}+c^{2})-(ab+bc+ca)$

Hello IMO 2007! :welcomeani:
Welcome to Vietnam!  :welcome:[/quote]

The inequality is equivalent to
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\ge 3+k$ for $ k=1$
This is the solution for $ k=2$:
Rewrite the inequality as the following form:
$ f(a,b,c)=M(a-b)^{2}+N(a-c)(b-c) \ge 0$
In which:
$ M=\frac{1}{ab}-\frac{2}{a^{2}+b^{2}+c^{2}}\ge 0$
$ N=\frac{1}{ac}-\frac{2}{a^{2}+b^{2}+c^{2}}\ge 0$

The best constant $ k$ for the given inequality is $ k \approx 2.76...$ but I only can prove for $ k \le 2.55$[/quote]
k=2
Let $ a,b,c >0$.Prove that:
$$ (a^{2}+b^{2}+c^{2})(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}) \ge 5(a^{2}+b^{2}+c^{2})-2(ab+bc+ca)$$

[quote=sqing]k=2
Let $ a,b,c >0$.Prove that:
$$ (a^{2}+b^{2}+c^{2})(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}) \ge 5(a^{2}+b^{2}+c^{2})-2(ab+bc+ca)$$[/quote]

See: http://diendantoanhoc.net/topic/152779-fracxy-fracyz-fraczx-frac2xy-yz-zxx2-y2-z2-geq-5/


In how many ways can we change the sign $ *$ with $ \plus{}$ or $ \minus{}$, such that the following equation is true?
\[ 1 *2*3*4*5*6*7*8*9*10\equal{}29\]


In triangle $ ABC$,point L lies on the side $ BC$ .Extend segment $ AB$ through $ B$ to $ M$ such that $ \hat{ALC}\equal{}2 \hat{AMC}$.Extend segment $ AC$ through $ C$ to $ N$ such that $ \hat{ALB}\equal{}2\hat{ANB}$ .Let $ O$ be the circumcenter of triangle $ AMN$.Prove that $ OL$ perpendicular to $ BC$.


Could someone kindly email some previous competition MathCounts questions to Whitestation583@yahoo.com. Thank you for your help!

1. There's already a topic open for this.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]here[/url].

2. Edit your email address, unless you want spam-bots to send you advertisements for fake Viagra...

[quote="AIME15"]
2. Edit your email address, unless you want spam-bots to send you advertisements for fake Viagra...[/quote]

Maybe that's why my junk box is filled with some [i]special[/i] types of pills.


Find all the $ (x,y,z)\in \mathbb{N}\times \mathbb{N}\times \mathbb{N}$ such that :

$ \sqrt{\frac{2005}{x\plus{}y}}\plus{}\sqrt{\frac{2005}{y\plus{}z}}\plus{}\sqrt{\frac{2005}{z\plus{}x}}\in \mathbb{Z}$

[hide]No perfect square except $ 1$ divides $ 2005\equal{}5\cdot 401$, so the only cases where all three of the terms are rational is when $ x\plus{}y\equal{}y\plus{}z\equal{}z\plus{}x\equal{}2005$ so that $ x\equal{}y\equal{}z\equal{}2005/2$ which is not an integer, or $ x\plus{}y\equal{}y\plus{}z\equal{}z\plus{}x\equal{}9\cdot 2005$ so that $ x\equal{}y\equal{}z\equal{}9\cdot 2005/2$, which is also not an integer. $ \sqrt{a}\plus{}\sqrt{b}\equal{}k\Leftrightarrow \sqrt{ab}\equal{}k^2\minus{}a\minus{}b$, so it's not possible for exactly one of the terms to be rational. $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\equal{}k\Leftrightarrow \sum \sqrt{ab} \equal{}(k^2 \minus{} a\minus{}b\minus{}c)/2$; squaring again, $ 2\left(\sum b\sqrt{ac}\right)\equal{}(k^2 \minus{} a\minus{}b\minus{}c)/2\minus{}ab\minus{}bc\minus{}ca$, so that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$ is rational only if $ \sum b^n\sqrt{ac}$ is, which is clearly false, so there are no solutions.[/hide]

Temperal: That solution is wrong.
This problem was Bulgaria MO 2005/Problem 1
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=233034#233034
An extremely useful lemma for this problem:
[hide]If $ \sqrt{a_1}\plus{}\sqrt{a_2}\plus{}\cdots \plus{} \sqrt{a^n}$ is rational, and all the $ a_i$ are rational, then each $ a_i$ must be the perfect square of a rational; that is, $ \sqrt{a_i}$ is rational for each $ a_i$.
Of course, you only need prove this for $ n\equal{}3$.[/hide]

Actually, yes, that is what I proved, but for some reason I mistakenly assumed the sum of three nonintegers to be a noninteger; the same mistake as the second poster in the thread you linked to.  :oops:


$ f(x) \equal{} x^4 \minus{} x^3 \minus{} 1$ prove that the sum of the real roots ($ S \geq \frac{1}{2}$) and the propagation of the real roots is ($ \minus{}1 \geq P$)

[quote="KapioPulsar"][quote="pco"][quote="KapioPulsar"]$ f(x) \equal{} x^4 \minus{} x^3 \minus{} 1$ prove that the sum of the real roots ($ S \geq \frac {1}{2}$) and the propagation of the real roots is ($ \minus{} 1 \geq P$)[/quote]

What is the meaning of "propagation" ?

And, btw, this equation has no real roots. So I wonder how we can show that the sum of the real roots ($ S \geq \frac {1}{2}$) ...   :huh:[/quote]  ... this equation has real roots......  do the grafic in mathematika (if x1,x2,x3,x4 the real roots P=x1*x2*x3*x4)[/quote]

Sorry. I considered $ x^4 \minus{} x^3 \plus{} 1$

But, once again, what is the meaning of "propagation", please ... ?

[quote="pco"][quote="KapioPulsar"]$ f(x) \equal{} x^4 \minus{} x^3 \minus{} 1$ prove that the sum of the real roots ($ S \geq \frac {1}{2}$) and the propagation of the real roots is ($ \minus{} 1 \geq P$)[/quote]

What is the meaning of "propagation" ?

And, btw, this equation has no real roots. So I wonder how we can show that the sum of the real roots ($ S \geq \frac {1}{2}$) ...   :huh:[/quote]
But $ f(\minus{}1)\equal{}1$, $ f(0)\equal{}\minus{}1$, $ f(2)\equal{}7$ so there are at least two real roots.

[quote="MeKnowsNothing"][quote="pco"][quote="KapioPulsar"]$ f(x) \equal{} x^4 \minus{} x^3 \minus{} 1$ prove that the sum of the real roots ($ S \geq \frac {1}{2}$) and the propagation of the real roots is ($ \minus{} 1 \geq P$)[/quote]

What is the meaning of "propagation" ?

And, btw, this equation has no real roots. So I wonder how we can show that the sum of the real roots ($ S \geq \frac {1}{2}$) ...   :huh:[/quote]
But $ f( \minus{} 1) \equal{} 1$, $ f(0) \equal{} \minus{} 1$, $ f(2) \equal{} 7$ so there are at least two real roots.[/quote]

Yes. As I said, I considered a wrong equation $ f(x) \equal{} x^4 \minus{} x^3 \plus{} 1$. I'm sorry for that.

The original equation $ f(x) \equal{} x^4 \minus{} x^3 \minus{} 1$ indeed has exactly two real roots. So $ S$ is their sum.

But I dont understand what is the propagation $ P$.  Maybe their product (but KapioPulsar seemed to say that $ P$ is the product of all roots x1*x2*x3*x4, so I dont understand).

[quote="KapioPulsar"]$ f(x) \equal{} x^4 \minus{} x^3 \minus{} 1$ prove that the sum of the real roots ($ S \geq \frac {1}{2}$) and the propagation of the real roots is ($ \minus{} 1 \geq P$)[/quote]

$ f'(x) \equal{} 4x^3 \minus{} 3x^2 \equal{} x^2(4x \minus{} 3)$ and so $ f(x)$ is decreasing on $ ( \minus{} \infty,\frac 34)$ and increasing on $ (\frac 34, \plus{} \infty)$ and $ f(0) < 0$

Hence only two real roots $ u < 0 < \frac 34 < v$

It's easy to check that $ f( \minus{} \frac 5{6}) > 0$ while $ f( \minus{} \frac 34) < 0$ and so $ \minus{} \frac 5{6} < u < \minus{} \frac 34$

It's easy to check that $ f(\frac 43) < 0$  and so $ \frac 43 < v$

So $ u \plus{} v > \frac 43 \minus{} \frac 56 \equal{} \frac 12$

And $ uv < \minus{} \frac 43\frac 34 \equal{} \minus{}1$


\[ \frac {{dy}} {{dx}} \equal{} \frac {{y\varphi '(x) \minus{} y^2 }} {{\varphi (x)}}\]

Rearranging, we have

$ \frac{y\varphi \prime (x)\minus{}\varphi (x)\frac{dy}{dx}}{y^{2}}\equal{}1$

$ y\equal{}\frac{\varphi (x)}{x\plus{}C}$


logarithms
calculus
calculus computations

If $e^\pi>\pi ^e,$ then $\ln e^\pi>\ln \pi ^e\Longleftrightarrow \frac{\ln e}{e}>\frac{\ln \pi}{\pi}.$

Now we are to use the graph of $y=\frac{\ln x}{x}.$

[quote="drunner2007"]Find, with proof, the larger number: $e^{\pi}$ or $\pi^{e}$[/quote]

What I have done is:

$ln (e^{\pi}) = \pi$

$ln (\pi^{e}) < \pi$

if you compute the second equation, you can, by observation, notice the proof.

I hope this helped

This question was already asked here. ;)


calculus
integration
real analysis
real analysis solved

$C=\{1;1/2;1/3;...;1/n;...\}$

$F:[0,1]\rightarrow R$ define by

$F(x)=1/x$ if $x\in C$
$F(x)=x$ if $x$ not in $C$

1/ Prove that $F$ is integrable on any interval $[\epsilon,1]$ with $0<\epsilon<1$

2/ Does the generalise integral $\int_{0}^{1}F(x)dx$ converge ?

If yes, compute its value

Since all of the points of $C$ are isolated, $F$ restricted to $[\epsilon,1]$ is bounded and continuous at all but finitely many points. Bracketing those points with tiny intervals, it is clear that $F$ is Riemann integrable on this set and $\int_\epsilon^1F(x)\, dx=\int_\epsilon^1x\, dx=\frac12-\frac{\epsilon^2}{2}.$

This clearly makes the improper integral $\int_0^1F(x)\, dx$ converge with value $\frac12.$


[quote="kx526627"]anyone knows the questions for#7 and #14?[/quote]
7 was the perfect squares, 14 was the circle with radius 9


geometry
geometric transformation
dilation

A monitor displays the graph y = 3 sin 5x. What will be the amplitude after a dilation of 2?

(1) 5     (3) 7
(2) 6     (4) 10

[hide="um"]
The amplitude now is 3. A dilation multiplies the amplitude by the dilation factor. Therefore the amplitude will be 6.
[/hide]

Correct me if I'm wrong.

[hide]
For y=1sin x, the amplitude is 1.
For y=3 sin 5x, the amplitude is 3.
After the dialation, it is 3*2, or 6.
[/hide]


Let $\cal{H}$be a family of finite subsets of an infinite set $\cal{X}$can be represented as the union of two disjoint sets from $\cal{H}$.prove that for everypositive integer $k$ there is a subset of $\cal{X}$ that can be represented in at least$ k$ diffrent ways as the union of two disjoint sets from $\cal{H}$.

Hmm... Maybe I am missing something, but I think, there is a typo in the first sentence.
Do you mean that each finite subsets of $\cal{X}$ can be represented as a union of two disjoint subsets from $\cal{H}$?

You are right Myth.
And this problem comes from Schweitzer 1978.

Pierre.


AMC AIME probability trigonometry USA(J)MO USAMO calculus

Darn it.  You went ahead and posted without my verification of #14.  Tsk tsk... one day this lack of hubris is going to backfire on you probability1.01 with a probability of 1.01 (pretty tricky huh)

Well whatev... see my PM your answer is correct anyway (UBERLY DARN IT) and whatev.

And you're welcome for verifying, editing, advertising, and iCertifying.  Yup, everyone, this AIME has been stamped by THE BUBALA and the number of questions does indeed equal the number of answers.  Trust me, probability1.01 will easily gain back the extra four corn husks (what i charged him since people where he lives don't have paper money) given that the cost of this test is exactly zero dollars and everyone knows that something*zero=corn because corn is nothing anyway.

If you doubt for even one second that the iCertification process works, just remember that you are taking the test, so it must have been what lured you into it.  If you read this post and don't take the test, you're giving probability exactly zero, which as I proved above, is exactly what he wants.  If you don't read this post.... well you should because it is written by the BUBALA who now has exactly six hundred posts plus more, which is MORE than zero.

Nice file name :P. It's not a regular AIME... its a ZHAIME!

[quote="bubala"]

If you doubt for even one second that the iCertification process works, just remember that you are taking the test, so it must have been what lured you into it.  If you read this post and don't take the test, you're giving probability exactly zero, which as I proved above, is exactly what he wants.  If you don't read this post.... well you should because it is written by the BUBALA who now has exactly six hundred posts plus more, which is MORE than zero.[/quote]

Hey hey hey hey!!!! I offer FCertification for only 1.99!!!

BTW bubula can you iCertify my mock AIME 2?? :lol: Maybe if you do that the participation will go from 7 to 30, 000 in the last two days!!!

Edit: Alex how much does your ZCertification cost?

woo I verified all of one problem and I pretty much let zhai work out all the numbers by himself anyway

I AM A TEAM PLAYER

[quote="joml88"][quote="probability1.01"]Well actually I think you meant excess of hubris as opposed to lack. Oh well.[/quote]

I was just about to ask about that :P

Btw, I'm sorry to inform you, but you have not been $j\mathbb{C}\scriptstyle\mathsf{ERTIFIED}$ on account of the lack of complex numbers problems on the test :([/quote]

It's hard to write good complex number problems  :(

I'm going to sticky this topic so more people can compete.

[quote="joml88"][quote="probability1.01"]Well actually I think you meant excess of hubris as opposed to lack. Oh well.[/quote]

I was just about to ask about that :P

Btw, I'm sorry to inform you, but you have not been $j\mathbb{C}\scriptstyle\mathsf{ERTIFIED}$ on account of the lack of complex numbers problems on the test :([/quote]

Lol, you haven't been $_A\textsf{C} \texttt{ERTIFIED}$ either :P for the same reason

[quote="bubala"][quote="probability1.01"]Well actually I think you meant excess of hubris as opposed to lack. Oh well.[/quote]

Yup.  Darn it.  I'll correct that.  :P
And I'll have you know, probability made a conscious decision not to put any trig/complex numbers on the ZHAIME, saying "Trig is like whatev."  So trig-lovers, assail him now![/quote]

Well, well... I'm just glad I didn't make an [i]unconscious[/i] decision. :-) Now I've written a solutions manual, which I will post in a week or two. Meanwhile, if you're stumped on anything or question my answers (HOW DARE YOU?? :D), I can send it to you privately too.

From a cursory glance over the test, it looks pretty good; thoroughly easier than Beta's. :thumbup:

Just to let you know, after 30 minutes of working on these problems, this is a really good set of problems  :D  Not too easy, not too hard.

Well, that's definitely good to hear! :) And if any of the few of you who have participated so far want some solutions, I plan on posting the solutions manual shortly before Christmas (I'll be away right before then). If you're really anxious, I can also give it to you individually ;).

[quote="JesusFreak197"]When should we have done the problems by? I haven't had a chance to really work on them yet, and may not for a few days.[/quote]

Ehhh as long as its before 2006, and before Christmas if you don't want to have the solutions available.

woohoo time to remind people that the solutions shall be posted tomorrow in the afternoon at some random time, assuming that my computer will be nice to me.

Meanwhile, since I have had so much success with this mock AIME -- over 30,000 participants -- I'm going to be also making a mock USAMO, which will crown the national champion, who will receive a free ipod.* Well it will probably meaning definitely be easier than a real USAMO, but it might meaning will be good practice for proofs and stuff. So yeah I only have 6 shortlisted problems right now, but I might say something later when I'm close to being finished.

* An "ipod" is defined to be any private message whose subject line is "ipod."

STOP BRAGGING STEPHEN MY COMPETITION (WHICH DID NOT TECHNICALLY EXIST) HAD OVER 3 BILLION MICE, UM I MEAN PARTICIPANTS

[quote="probability1.01"]woohoo time to remind people that the solutions shall be posted tomorrow in the afternoon at some random time, assuming that my computer will be nice to me.

Meanwhile, since I have had so much success with this mock AIME -- over 30,000 participants -- I'm going to be also making a mock USAMO, which will crown the national champion, who will receive a free ipod.* Well it will probably meaning definitely be easier than a real USAMO, but it might meaning will be good practice for proofs and stuff. So yeah I only have 6 shortlisted problems right now, but I might say something later when I'm closed to being finished.

* An "ipod" is defined to be any private message whose subject line is "ipod."[/quote]

WOOT!! Alex stop bragging about your 30,000 participants, I had 30,001 participants!!! I sent the distance formula by AIM to all 146 people on my buddylist and I told them to also send the distance formula to everyone on their buddylist... yeah! Well you might want to consider to buy my JCertification for 999 dollars!

WOOT first mock USAMO on AoPS! As you see, it will be definitely the premier competition that challenges only the best of the best of the best of the best of the best of the best in United States of America, next time it will challenge  the best of the best of the best of the best of the best of the best of the best in the world! Winners will receive the national champion title(formal international recognition)?

[quote="beta"]WOOT!! Alex stop bragging about your 30,000 participants, I had 30,001 participants!!! [/quote]

HA YOU STILL CAN'T MATCH MY MICE UM I MEAN PARTICIPANTS ARRGH WHY DO I KEEP DOING THAT

[quote="probability1.01"]Here's the PDF! The rules are the same as an official AIME. Submit your answers at any time you feel like before 2006. I'll make a list of people who score above 8. Actually I'll probably be too lazy. Hmmm oh well.

A couple notes:
- There is no complex numbers/trig in this AIME. Sorry!
- The order of difficulty may be somewhat off.
- The overall difficulty should be slightly harder than the 2005 AIME I.

Thanks to bubula and mysticterminator for some help verifying solutions.
"Thanks" to bubula for advertising for me.

When you're done, please tell me what you did/didn't like about this AIME or any specific problems.

So far:

1. scorpius119 -- 10 (one trivial mistake)
2. sirorange -- 10
3. SamE -- 9 (two trivial mistakes)
4. isaacchao -- 9[/quote]

WHOA I THOUGHT 14 USED CALCULUS BUT GO FIGURE ... NO CALCULUS AT ALL

AND HOW DID I MISS THE ONE PROBLEM THAT I ACTUALLY CHECKED
THAT'S PRETTY STUPID

I guess it was a pretty good test. I wish there had been complex numbers on there, as that's one of my favorite subjects and I probably would have gotten a few more right...

Anyway, I answered 9, and I'm expecting to miss around 3, but I only spent about an hour and fifteen minutes, so I guess that's pretty good. Spending about fifteen minutes of that time looking at the other six, I really have no idea how to approach any of them, though, and that includes #2. :huh: Either I'm just not very intuitive with geometry (which is definitely possible), or else it required some kind of theorem that isn't commonly enough known to be in the AIME. ;) I think maybe the questions weren't always placed correctly for the difficulty, but other than that, I suppose it was just on the hard side for an AIME.

[quote="JesusFreak197"]...and that includes #2. :huh: Either I'm just not very intuitive with geometry (which is definitely possible), or else it required some kind of theorem that isn't commonly enough known to be in the AIME. ;)[/quote]

#2 was one of the few questions that I did get right, and it was a geometry problem, ... yeah you know how badly I stink at geometry :roll: so it should have been pretty easy :huh:

[quote="MysticTerminator"]

WHOA I THOUGHT 14 USED CALCULUS BUT GO FIGURE ... NO CALCULUS AT ALL

AND HOW DID I MISS THE ONE PROBLEM THAT I ACTUALLY CHECKED
THAT'S PRETTY STUPID[/quote]

Well I did 14 for verification purposes using integrals, which made the problem quite simple, but indeed, there is a non-calulus solution.

Really?? Well I did not come up with the non-calculus solution, but Alex assured me that there was one...

[quote="probability1.01"][quote="MysticTerminator"]btw (if we want to be *serious* and all that), no there isn't[/quote]

Well actually I thought we decided there was :-). But it requires a lot of definitions and stuff.[/quote]

sorry I forgot about that one

NEVER MIND I WAS JUST LYING

[quote="darktreb"]Haha seeing this thread makes me wish I could go back to MOP this year.[/quote]

you could if you tried hard enough

[quote="probability1.01"]OK, this mock AIME can be considered over now, I guess. So thanks Silverfalcon for putting it up as a sticky thingy. You can take it down now though :).[/quote]

Wish granted. By the way, this was a really nice Mock AIME. We should AoPS Mock AMC/AIME Convention in 2006.  :D

[quote="SamE"][quote="probability1.01"][quote="MysticTerminator"]btw (if we want to be *serious* and all that), no there isn't[/quote]

Well actually I thought we decided there was :-). But it requires a lot of definitions and stuff.[/quote] No, you just need to draw a picture. :)[/quote]

REALLY, I DOUBT IT?

TURKEYS SMELL INORDINATELY BAD I AM GETTING A TURKEYWHIFF

WOOHOO I CAN USE THE CAPS-LOCK KEY TOO!!!!!

[quote="SamE"]Okay, we're talking about using calculus on #14, right? I just graphed each of the probability regions and compared the graphs. That's what I meant when I said you can draw a picture.[/quote]

WHY IS THE ANSWER NOT 105

AND YES FEIQI I AGREE THE CAPSLOCK KEY IS AWESOME I USE IT FREQUENTLY AS WELL AS OMITTING PUNCTUATION

[quote="chess64"][quote="JesusFreak197"][quote="chess64"]#2 was one of the few questions that I did get right, and it was a geometry problem, ... yeah you know how badly I stink at geometry :roll: so it should have been pretty easy :huh:[/quote]

Well, I'm probably just more horrible at geometry than you are. :roll:[/quote]

I don't think that's even possible :D[/quote]

#2 was pretty good problem, just draw a triangle inside the circle (let the radius to the point of tangency be a hpy. of a right triangle)

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

[url=https://artofproblemsolving.com/community/c3744787_2005_alex_zhai_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS.1. Official solutions pdf is [url=https://artofproblemsolving.com/community/c5h67284p396580]here[/url].
PS.2. [b]probability1.01[/b] is [b]Alex Zhai [/b].


self explanatory. the first person posts the first 3 words of a sentence, then the next person posts the next 3 words, ect. when you want to end a sentence, please put a period. and make sure you review the topic to see what the sentence is about. on this other forum, there was like a 75 word sentece. ill start.

there once was

country. This made

(good to see this effort again - its a long time since mathlinks was at its creative best  :) )


[quote="Makaveli"]a, b, c are positive reals.Show that
((a-b)^2)/ab+((b-c)^2)/bc+((c-a)^2)/ac>=
3/((a^2)+(b^2)+(c^2))*((a-b)^2+(b-c)^2+(c-a)^2)[/quote]

Wrong for a = 2, b = 2, c = 3.

The weaker inequality

$\frac{\left(b-c\right)^2}{bc}+\frac{\left(c-a\right)^2}{ca}+\frac{\left(a-b\right)^2}{ab}\geq 2\frac{\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2}{a^2+b^2+c^2}$

holds and is pretty trivial.

  darij


conics
parabola
calculus
calculus computations

Find all common tangent lines of the parabola $ y \equal{} \frac {1}{2}x^2 \plus{} x$ and the circle $ (x \minus{} 1)^2 \plus{} (y \plus{} 1)^2 \equal{} 2$.


geometry
3D geometry
sphere
trigonometry

This is a interesting problem I found on Fermat.

[quote="amcavoy"][hide="Solution"]Take the equilateral triangle formed by the tops of the cones.  The center of this triangle, which is above the center of the sphere, is $\frac{2}{3}\sqrt{100^{2}-50^{2}}=\frac{100\sqrt{3}}{3}$ from each vertex.  When you draw this out you can make a few right triangles and you end up with

$r\sin\left(\tan^{-1}\frac{12}{5}\right)=\frac{100\sqrt{3}}{3}-50+x$

$\frac{12}{5}=\frac{120-r-r\cos\left(\tan^{-1}\frac{12}{5}\right)}{x}$

Solving this system gives $r=\frac{200\sqrt{3}}{9}\approx 38.5\implies\boxed{D}$.[/hide]
Nice problem :).[/quote]
nice solution. :lol:
that's the right answer.

Opps, I didn't read that the top of the sphere was on the same plane as the tops of the cones.  :blush:


geometry
area of a triangle
Heron\u0027s formula

What is the area in square units of the convex quadrilateral with vertices $ (\minus{}1, 0)$, $ (0, 1)$, $ (2, 0)$ and $ (0, \minus{}3)$?

If we draw this, we get:

[asy]draw((-1,0)--(0,1)--(2,0)--(0,-3)--cycle);[/asy]

We can break this figure into two triangles.

[asy]draw((-1,0)--(0,1)--(2,0)--(0,-3)--cycle);
draw((-1,0)--(2,0),red);[/asy]

Using the distance formula, we find the side lengths of each side.

[asy]draw((-1,0)--(0,1)--(2,0)--(0,-3)--cycle);
draw((-1,0)--(2,0),red);
label("$\sqrt{2}$",(-0.5,0.5),NW);
label("$\sqrt{5}$",(1,0.5),NE);
label("$\sqrt{13}$",(1,-1.5),SE);
label("$\sqrt{10}$",(-0.5,-1.5),SW);
label("$3$",(0.5,0),N);[/asy]

If we use Heron's Formula, we find that the top triangle has an area of $ \frac {3}{2}$ and the bottom triangle has an area of $ \frac {9}{2}$.

$ \frac {9}{2} \plus{} \frac {3}{2} \equal{} \boxed{6}$

You do not have to do that. In a quadrilateral with perpendicular diagonals, the area is half the product of the lengths of the diagonals.


Hi, I'm considering applying to MathZoom and i was just wondering ...would it be possible for me to take AoPS classes while at the camp? I'd like to take the AIME class which runs on Tuesday 4:30 - 6:00 (PT). Also, is there any transportation to and from the airport? And how does laundry work? Thanks in advance.

Internet access is available in the dorm, and the AOPS class time doesn't really conflict with the Math Zoom camp schedule, so it would be OK.  Just make sure you don't miss dinner (which is available from 5:30 to 7:00).

Math Zoom staff will arrange the transportation to and from the LAX airport.  Once you book the airline, send us the flight information.

You will do your own laundry.  Laundry room is in the same dorm building.  Detergent will be provided to you.


Derive the sum of squares formula without using induction.

Sum of squares formula?

[hide="ummm...."]$a^2+b^2=(a+b)^2-2ab=(a+b+\sqrt{2ab})(a+b-\sqrt{2ab})$. [/hide]

By "the sum of the squares", (I think) he meant $\sum_{k=1}^n k^2 =\frac{n(n+1)(2n+1)}{6}$.

This topic has already been posted several times in the AoPS :roll:


LaTeX
factorial
floor function
symmetry
number theory
prime factorization

Is the binomial coefficient
 $(\frac{500}{212})$
 500
(	)
  212
even or odd?

In general, how can you discover this without explicitly calulating the value?

see http://mathworld.wolfram.com/LucasCorrespondenceTheorem.html
 :wink:

The latex is
{a \choose b} or \binom{c}{d} for ${a \choose b}$ or $\binom{c}{d}$.

Also, you can make your parentheses larger by using \Big or by using the paired commands \left and \right.

So, \left( \frac{17^{18}}{i_{j_k}} \right) gives $\left( \frac{17^{18}}{i_{j_{k}}}\right)$, which looks a lot nicer than $( \frac{17^{18}}{i_{j_{k}}})$

As for this question, you can always do it by counting the number of 2s in the factorials:

${500 \choose 212}= \frac{500!}{212!\cdot 288!}$.
There are $\sum_{i = 1}^{\infty}\left\lfloor \frac{n}{2^{i}}\right\rfloor$ powers of 2 in the prime factorization of $n!$, so in the numerator we have $250+125+62+31+15+7+3+1 = 494$ while in the denominator we have $(106+53+26+13+6+3+1)+(144+72+36+18+9+4+2+1) = 208+286 = 494$.  So in fact that binomial coefficient is odd.

Without calculating the value... hmm....

[hide="solution"]$\binom{500}{212}=\frac{500*499*498*\cdots*290*289}{212*211*\cdots*3*2*1}$

We calculate the number of 2's between 501 and 288:

Number of 2's in 1-500:

$\lfloor{\frac{500}{2}\rfloor}+\lfloor{\frac{500}{4}\rfloor}+\lfloor{\frac{500}{8}\rfloor}+\lfloor{\frac{500}{16}\rfloor}+\lfloor{\frac{500}{32}\rfloor}+\lfloor{\frac{500}{64}\rfloor}+\lfloor{\frac{500}{128}\rfloor}+\lfloor{\frac{500}{256}\rfloor}=250+125+62+31+15+7+3+1=494$

Number of 2's in 1-288:

$\lfloor{\frac{288}{2}\rfloor}+\lfloor{\frac{288}{4}\rfloor}+\lfloor{\frac{288}{8}\rfloor}+\lfloor{\frac{288}{16}\rfloor}+\lfloor{\frac{288}{32}\rfloor}+\lfloor{\frac{288}{64}\rfloor}+\lfloor{\frac{288}{128}\rfloor}+\lfloor{\frac{288}{256}\rfloor}=144+72+36+18+9+4+2+1=286$

494-286=208

So there are 208 2's in 500-289.

Number of 2's in 1-212:

$\lfloor{\frac{212}{2}\rfloor}+\lfloor{\frac{212}{4}\rfloor}+\lfloor{\frac{212}{8}\rfloor}+\lfloor{\frac{212}{16}\rfloor}+\lfloor{\frac{212}{32}\rfloor}+\lfloor{\frac{212}{64}\rfloor}+\lfloor{\frac{212}{128}\rfloor}=106+53+26+13+6+3+1=208$

So there are 208 2's in 1-212.

Since there are an equal number of 2's, $\binom{500 }{212}$ is odd.[/hide]

@JBL: I have a differet definition for x choose y. It's essentially the same:

$\frac{x*(x-1)*\cdots*(x-y+1)}{y*(y-1)*\cdots*3*2*1}=\binom{x}{y}$

[quote="nutz_for2.718281828"]@JBL: I have a differet definition for x choose y. It's essentially the same:

$\frac{x*(x-1)*\cdots*(x-y+1)}{y*(y-1)*\cdots*3*2*1}=\binom{x}{y}$[/quote]

One would hope they are the same: they count the same thing.  Do you know how to translate between our two expressions?  Mine has the advantage of making the symmetry ${x \choose y}={x \choose x-y}$ abundantly obvious.  Your expression will look prettier if you use \cdot instead of * for multiplication.


We have already discussed this problem. It is a russian problem and I remember to have posted it. Unfortunately, I don't remember who gave a solution.

snif.... I did....

http://www.mathlinks.ro/viewtopic.php?t=589

Pierre.

further more, Dapet wanted to make a slight particularisation of this. It's a sticky :)


In triangle $ ABC$ there are two altitudes $ BD$ and $ CE$ where $ D$ is on $ AC$ and $ E$ is on $ AB$.  $ BC=50$ and $ DE=48$.  If $ M$ is the midpoint of $ BC$ and $ N$ is the midpoint of $ DE$, then what is the length$ MN$

Which MATHCOUNTS competition was this problem from?

Hmm there probably is a way but I think that it would be way beyond the mathcounts level.

[hide]Let $ w$ be the circle withe diameter $ BC$.

Then of course $ M$ is the center of $ w$ and $ MB$ is its radius

Also, $ D,E \in w$. So $ MD=ME=MB=25$

$ \triangle MDE$ is isosceles, so $ MN\perp DE$

From the right triangle $ MND$ we get $ MN^{2}= MD^{2}-DN^{2}= 25^{2}-24^{2}= 49\Rightarrow $

$ \boxed{MN=7}$[/hide]

[quote="pontios"][hide]Let $ w$ be the circle withe diameter $ BC$.

Then of course $ M$ is the center of $ w$ and $ MB$ is its radius

Also, $ D,E \in w$. So $ MD=ME=MB=25$

$ \triangle MDE$ is isosceles, so $ MN\perp DE$

From the right triangle $ MND$ we get $ MN^{2}= MD^{2}-DN^{2}= 25^{2}-24^{2}= 49\Rightarrow$

$ \boxed{MN=7}$[/hide][/quote]

DO you have to draw the circle to realize this?

It seems the solution is actually quite simple (computational wise) but very hard to see (like how did you know to draw circle

[quote="archimedes1"]I really don't think cyclic quadrilaterals ( quads that can be inscribed in a circle have special properties) are MATHCOUNTS level...
spell it out 11, the reason it seems very hard to see is that MATHCOUNTS never uses problems with cyclic quadrilaterals, so if you've been doing only MATHCOUNTS, then you don't have any practice with them.  There are different ways to recognize cyclic quadrilaterals but you'll never need them in MATHCOUNTS.[/quote]

well rather than looking at it as a cycli quad loook at it as two right $ \triangle{BCE}$ and $ \triangle{BDC}$ now if we draw a semicircle that inscribes $ \triangle{BCE}$ it must inscribe $ \triangle{BDC}$ because $ \angle{BDC}=90$ and i know that MATHCOUNTS does have problems where you need to know that an inscribed triangle with the diameter as a side is a right triangle.

[quote="pontios"][hide]Let $ w$ be the circle withe diameter $ BC$.

Then of course $ M$ is the center of $ w$ and $ MB$ is its radius

Also, $ D,E \in w$. So $ MD=ME=MB=25$

$ \triangle MDE$ is isosceles, so $ MN\perp DE$

From the right triangle $ MND$ we get $ MN^{2}= MD^{2}-DN^{2}= 25^{2}-24^{2}= 49\Rightarrow$

$ \boxed{MN=7}$[/hide][/quote]
wait, how do you know what MD is?

The quadrilateral is cyclic, and MD is a radius.  All radii are the same.


Let $ S$ be a set, and let $ *$ be a binary operation on $ S$ satisfying the laws:

$ x*(x*y)\equal{}y$ for all $ x,y$ in $ S$,
$ (y*x)*x\equal{}y$ for all $ x,y$ in $ S$.

Show that $ x*y\equal{}y*x$ for all $ x,y$ in $ S$.

[hide]
$ x*(x*y) = y$ for all $ x,y$ in $ S$ implies $ y*(y*x)=x$.
$ (y*x)*x = y$ for all $ x,y$ in $ S$ implies $ (x*y)*y=x$.
So, we have \begin{eqnarray}x*(x*y)&=&y\\
y*(y*x)&=&x\\
(y*x)*x &=& y\\
(x*y)*y&=&x\end{eqnarray} Plugging in $ (x*y)$ for $ x$ in the first equation yields \[ (x*y)*((x*y)*y)=y.\] By equation four, this is equivalent to \[ (x*y)*x=y.\] Similarly, plugging in $ (y*x)$ for $ x$ in the third equation yields \[ (y*(y*x))*(y*x)=y\] and by equation two, this is equivalent to \[ x*(y*x)=y.\] Therefore, \[ (x*y)*x=x*(y*x).\] Now, \begin{eqnarray*}x*y&=&x*((y*x)*x)\\
&=&(x*(y*x))*x\\
&=&y*x,\end{eqnarray*} where the second line comes from plugging in $ y*x$ for $ y$ in the above identity.[/hide]

[quote="azn6021023"]Why can't we just start with the transitive property, then divide out x?[/quote]Because your mind is thinking that the binary operation is [u]multiplication[/u], there is no indication from the given conditions that there exists an inverse element property. Therefore, the words 'divide out' would not make sense here. towersfreak2006 did an excellent demonstration that is very instructive on how to apply the 'algebraic' laws.


Im learning chi-square test in statistic now but I dont quite understand what is meant by " level of significant " ?

Like if  given data , after you work out chi-square test at 5% level of significant(with certain degree of freedom) and found out the deviation is very small (almost $0$ ) . We will mostly likely accept the null-hypotesis saying that the difference between the observed data and expected data is not significant . So does this actually means that your observed data can be trusted  :huh: ?

Also , another example which Im confused also . Say you throw 3 dices at once for 1000 times and record the outcome of "six" in each throw . Say if you get $10$ times for outcome of zero "six" , one "six" , two "six"  and $970$ times for three "six" . Now if we want to test whether it follows binomial distribution of where $p=\dfrac{1}{6}$ , most probably we get a very big number for chi-square calculation which lead to rejection hypotesis (if we test at 5% lvl of significant) . Hence conclude that it doesnt follows Binomial distribution .
But then if we calculate the $p$ our self and get a very high $p$ (almost $1$) and use that $p$ to calculate our chi-square calculation , we are likely to accept it at 5% significant (degree of freedom is of course $1$ less than d.o.f the above where $p$ is given) .So what does this mean ? Accept the hypotesis for this (which is the data follows binomial distribution) and reject the hypotesis for above ? How come that happen ? Thats really confusing me  :mad:

Can someone please clear my doubt  :?

Since deviation is small, and your chi-square value is less than that of the 5% mark, then you would accept the null hypothesis.  That means your data is very likely to be correct since it's close to the expected values.

For the second part, since the observed number of "sixes" were much higher than expected, it obviously would be far off from the expected distribution.  So, null hypothesis is rejected.  I don't understand what you mean by calculating the probability and then running a chi-square test.

Thanks for reply dakyru .

[quote]
That means your data is very likely to be correct since it's close to the expected values.
[/quote]

What do you mean by the word " correct " ? It somehow doesnt explain the situation clearly .

As for the second part , what i mean is re-calculate the probability for the binomial distribution again ( like making the table and find the mean from the table then set the mean = np to find $p$ . )

By correct, I mean accurate because the comparison is based on established values.

As for the second part, aren't you already calculating $p$ for the chi-square test?
The expected distribution (0,1,2,3 sixes) would be $\dfrac{125}{216}$, $\dfrac{75}{216}$, $\dfrac{15}{216}$, and $\dfrac{1}{216}$, respectively.  Then you just plug in the observed and expected numbers to compute the chi square value.  I don't see another way you could do it.  Could you explain with an example?

[quote="dakyru"]By correct, I mean accurate because the comparison is based on established values.[/quote]

Well , i try to look up for the meaning for " level of significant " and I found out the explaination to be " chances to make type I error " . So Im wondering , if the deviation is small , you will get a big " level of significant " (from chi-square table) , and does this mean the "chances of making type I error " increases ?

[quote="dakyru"]
As for the second part, aren't you already calculating $p$ for the chi-square test?
The expected distribution (0,1,2,3 sixes) would be $\dfrac{125}{216}$, $\dfrac{75}{216}$, $\dfrac{15}{216}$, and $\dfrac{1}{216}$, respectively.  Then you just plug in the observed and expected numbers to compute the chi square value.  I don't see another way you could do it.  Could you explain with an example?[/quote]

Ok , the probability you listed out is so-called " given " probability where we already assume the probability to get a $6$ is $\dfrac{1}{6}$ . The second test is , we assume we do not know the probability is $\dfrac{1}{6}$ (or you can say that we havent known the distribution is binomial) . So we will calculate the $p$ based on our observed data to obtain a new $p$ (obviously it would be higher than $\dfrac{1}{6}$ since the dice is somehow biased ) . And we will use " this $p$ " to carry out chi-square " to see whether it fits Binomial .


Let $x,y,z$ be three positive real numbers.

Prove that:

\[ \sqrt{\frac{xyz(x+y+z)}{3}}\leq \frac3{8} \cdot \frac{(x+y)(y+z)(z+x)}{x+y+z}  \]

When does the inequality take place?

[b][EDIT][/b] Factorized the RHS
[b][EDIT 2][/b] Improved the inequality

[quote="cezar lupu"]Can you please write it corectly? :)[/quote]

How correctly do you want it? :huh:

[quote="Ezbakhe Yassin"]Let $a,b,c$ be three positive real numbers.

Prove that:

\[ \sqrt{\frac{xyz(x+y+z)}{12}}\leq x^2(y + z) + x(y + z)^2 + yz(y + z)  \][/quote]

 Are you sure that it's $yz(y + z)$ ?:?

I think that you have missed some condition. For example put $x=y=z=\alpha$, where $\alpha<\frac{1}{16}$

\[ \sqrt{\frac{xyz(x+y+z)}{3}}\leq \frac3{8} \cdot \frac{(x+y)(y+z)(z+x)}{x+y+z} \]

is equivalent to

\[ xyz(x+y+z) \leq \frac{27}{64} \cdot \frac{((x+y)(y+z)(z+x))^2}{(x+y+z)^2} \]

thus to

\[ xyz\left(\frac{x+y+z}{3}\right)^3 \leq \left(\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{z+x}{2}\right)^2 \]

Which is trivial with popoviciu's inequality with $f(x)=-ln(x)$.

The problem becomes easy after applying the ineq : $9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$  :)

Evaluate  the  sum

   Σ 1/(2n-1)(3n-1)  from  1 to infinity

Why apply Popoviciu to a such an easy inequality? Anyway this solution with Popoviciu is well-known. Like nttu said we have:

 [color=blue][b]Problem.[/b]

    Let $x, y, z>0$. Prove that the following inequality holds:

                            \[ \sqrt{\frac{xyz(x+y+z)}{3}}\leq\frac{3}{8}\cdot\frac{(x+y)(y+z)(z+x)}{x+y+z}.  \]
  [/color]

   [i] Solution.[/i]

   Like nttu said let us apply the following inequality:(1) \[ (x+y)(y+z)(z+x)\geq\frac{8}{9}(x+y+z)(xy+yz+zx).  \]
   To prove this nice and useful inequality we just need to use this well-known identity:
     \[ (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz.  \]
    Well, (1) becomes \[ (x+y+z)(xy+yz+zx)-xyz\geq\frac{8}{9}(x+y+z)(xy+yz+zx)\Leftrightarrow  \]
     \[ (x+y+z)(xy+yz+zx)\geq 9xyz.  \]
    This last inequality follows from AM-GM. Let's get back to our problem. By (1), we only need to prove that
    \[ xy+yz+zx\geq\sqrt{3xyz(x+y+z)},  \] which follows immediately from $(a+b+c)^{2}\geq 3(ab+bc+ca)$ by
     putting $a=yz, b=zx, c=xy$. So the conclusion follows. :)  ;)


inequalities
function
inequalities unsolved

[quote="Detective"]Let $a,b,c$ be positive real numbers. Prove that

$\frac{(a+b+c)^{3}}{27abc}\geq \frac{3(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}$[/quote]
The mine  inequality $(a+b+c)^{6}\geq27(a^{2}+b^{2}+c^{2})(ab+ac+bc)^{2}$ is  stronger.

Nice...

Do you have another proof for my original inequality without using arqady's inequality?

Arqady, can apply the your technique (I like it very much) to the identities $(a+b+c)^{3}=\sum_{\mathrm{cyc}}a\left[a^{2}+3(b+c)a+2bc\right]=\sum_{\mathrm{cyc}}\left[a^{3}+(a+b)(b+c)(c+a)\right]\ .$ Maybe you obtain a nice inequality !

Let $a,$ $b$ and $c$ are non-negative numbers such that $(a+b)(a+c)(b+c)=8.$
Prove that $\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^{3}+b^{3}+c^{3}}{3}}.$
Very nice, Virgil! :lol:

[quote="Detective"]Let $ a,b,c$ be positive real numbers. Prove that

$ \frac {(a \plus{} b \plus{} c)^{3}}{27abc}\geq \frac {3(a^{2} \plus{} b^{2} \plus{} c^{2})}{(a \plus{} b \plus{} c)^{2}}$[/quote]
Suppose $ c\equal{}min\{a;b;c\}$ then we have:
$ (a\minus{}b)^2[\frac{a\plus{}b\plus{}7c}{27abc}\minus{}\frac{2}{(a\plus{}b\plus{}c)^2} ]\plus{}(a\minus{}c)(b\minus{}c)[\frac{4a\plus{}4b\plus{}c}{27abc}\minus{}\frac{2}{(a\plus{}b\plus{}c)^2}] \geq 0$

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 8.$
Prove that $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [27]{\frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{3}}.$
Very nice, Virgil! :lol:[/quote]

$ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [27]{\frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{3}}\Longleftrightarrow \frac{(a\plus{}b\plus{}c)^{27}}{a^3\plus{}b^3\plus{}c^3}\geq 3^{26}$.

Let $ x\equal{}(a\plus{}b\plus{}c)^3\ (x>24)$, the left side of the above inequality $ f(x): \equal{}\frac{x^9}{x\minus{}24}\Longrightarrow f'(x)\equal{}\frac{8x^8(x\minus{}27)}{(x\minus{}24)^2}$,

we have $ f(x)\geq f(27)\equal{}3^{26}$.

Here is my proof:
$ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}24\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3\plus{}...\plus{}3\geq9\sqrt[9]{(a^3\plus{}b^3\plus{}c^3)\cdot3^8}.$

[quote="arqady"]Here is my proof:
$ (a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 24 \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3 \plus{} ... \plus{} 3\geq9\sqrt [9]{(a^3 \plus{} b^3 \plus{} c^3)\cdot3^8}.$[/quote]

Wonderful! :)

Indeed. Yours is stronger. How do you prove your inequality arqady?

[url=http://www.math123xyz.com]Interactive Math Resources[/url]

[quote="kally"]How do you prove your inequality arqady?
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=945097#945097

[quote="arqady"]
The mine  inequality $ (a \plus{} b \plus{} c)^{6}\geq27(a^{2} \plus{} b^{2} \plus{} c^{2})(ab \plus{} ac \plus{} bc)^{2}$ is  stronger.[/quote]

There exists $ x,y\ge 0$ such that $ \frac{a\plus{}b\plus{}c}{3}\equal{}\frac{x\plus{}y}{2}$ and $ \frac{ab\plus{}bc\plus{}ca}{3}\equal{}xy$. Then it is a two variable inequality that can be factored.

As a new user recently registered to this forum I browsed especially the inequality-section.
Thereby I spotted quite a few inequalities of the type

$f(a, b, c) \geq g(a, b, c)$

where a, b and c are positive real numbers and functions f and g are symmetric.

Honestly, [i]I admire all the artificial methods invented to prove them and also the time spent to develop these ideas.[/i]

But to my understanding very many of the inequalities can be done (with the aid of a CAS) simply by the substitution
b = a + s and c = a + s + t, s, $t \geq 0$
leading to an inequality like

$h(a, s, t) \geq 0$

with h(a, s, t) being a sum of nonnegative expressions.
(See also the attached solution.)

[i]What does this mean for proofs of such inequalities in the future?[/i]  :roll:

Your method will work, but often it involves a bit too much calculation. Also, for those inequalities involving many roots and/or rational functions, it would require an incredible amount of multiplying out. For example, this famous problem:

For all positive reals $ a,b,c$,

\[ \sqrt[3]{\frac{2a}{b\plus{}c}} \plus{} \sqrt[3]{\frac{2b}{c\plus{}a}} \plus{} \sqrt[3]{\frac{2c}{a\plus{}b}} \ge 3\]

would be extremely difficult (computation-wise) to solve with such a method (unless you use a computer, but then most such inequalities are rather pointless :wink: )

[quote="not_trig"]
For all positive reals $ a,b,c$,
\[ \sqrt [3]{\frac {2a}{b \plus{} c}} \plus{} \sqrt [3]{\frac {2b}{c \plus{} a}} \plus{} \sqrt [3]{\frac {2c}{a \plus{} b}} \ge 3\]
[/quote]
Maybe are you mean:
For all positive reals $ a,b,c$,
\[ \sqrt [3]{\left(\frac {2a}{b \plus{} c}\right)^2} \plus{} \sqrt [3]{\left(\frac {2b}{c \plus{} a}\right)^2} \plus{} \sqrt [3]{\left(\frac {2c}{a \plus{} b}\right)^2} \ge 3\]
or
For all positive reals $ a,b,c$,
\[ \sqrt [5]{\left(\frac {2a}{b \plus{} c}\right)^3} \plus{} \sqrt [5]{\left(\frac {2b}{c \plus{} a}\right)^3} \plus{} \sqrt [5]{\left(\frac {2c}{a \plus{} b}\right)^3} \ge 3\]
? :wink:
[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 8.$
Prove that $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [27]{\frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{3}}.$
[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 8.$ Prove that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [53]{\frac {a^{4} \plus{} b^{4} \plus{} c^{4}}{3}}\]

[quote="not_trig"]
For all positive reals $ a,b,c$,
\[ \sqrt [3]{\frac {2a}{b \plus{} c}} \plus{} \sqrt [3]{\frac {2b}{c \plus{} a}} \plus{} \sqrt [3]{\frac {2c}{a \plus{} b}} \ge 3\]
[/quote]

I think arqady is right ,this problem is not true.

By Am-Gm inequality we obtain :

$ 3\sqrt[3]{\frac{4(b\plus{}c)}{2a}} \leq \frac{4(b\plus{}c)}{2a} \plus{}1 \plus{}1 \equal{}\frac{4(a\plus{}b\plus{}c)}{2a}\implies \sqrt[3]{\frac{2a}{b\plus{}c}}\geq  \frac{6\sqrt[3]{4}a}{4(a\plus{}b\plus{}c)}$

We have that  $ \sum_{cyc}{ \sqrt[3]{\frac{2a}{b\plus{}c}}} \geq \frac{9\sqrt[3]{4}}{2}$

[quote="arqady"]
Maybe are you mean:
For all positive reals $ a,b,c$,
\[ \sqrt [3]{\left(\frac {2a}{b \plus{} c}\right)^2} \plus{} \sqrt [3]{\left(\frac {2b}{c \plus{} a}\right)^2} \plus{} \sqrt [3]{\left(\frac {2c}{a \plus{} b}\right)^2} \ge 3\]
[/quote]

In similar form we solve problem proposed by arqady.

$ 3\sqrt [3]{\left(\frac {b\plus{}c}{2a}\right)^2} \leq \frac{b\plus{}c}{2a} \plus{}\frac{b\plus{}c}{2a} \plus{}1\equal{}\frac{a\plus{}b\plus{}c}{a} \implies \sqrt [3]{\left(\frac {2a}{b \plus{} c}\right)^2} \geq \frac{3a}{a\plus{}b\plus{}c}$

We have that $ \sum_{cyc}{\sqrt [3]{\left(\frac {2a}{b \plus{} c}\right)^2}} \geq 3$

[quote="arqady"][quote="enndb0x"]

We have that  $ \sum_{cyc}{\sqrt [3]{\frac {2a}{b \plus{} c}}} \geq \frac {9\sqrt [3]{4}}{2}$
[/quote]

Maybe $ \sum_{cyc}{\sqrt [3]{\frac {2a}{b \plus{} c}}} \geq \frac {3\sqrt [3]{4}}{2}$ $ ?$ :wink:
By the way, $ \sum_{cyc}{\sqrt [3]{\frac {2a}{b \plus{} c}}} \geq 2\sqrt [3]2$ is  stronger.[/quote]

Yeah ,I really meant that ...

 $ \sum_{cyc}{\sqrt [3]{\frac {2a}{b \plus{} c}}} \geq 2\sqrt [3]2$

$ 3\sqrt[3]{\frac{4(b\plus{}c)}{2a}} \leq \frac{4(b\plus{}c)}{2a}\plus{}1\plus{}1 \implies \sqrt[3]{\frac{4(b\plus{}c)}{2a}}\leq \frac{4(a\plus{}b\plus{}c)}{6a}\implies \sqrt[3]{\frac{2a}{4(b\plus{}c)}} \geq \frac{6a}{4(a\plus{}b\plus{}c)}$

We have that  $ \sum_{cyc}{\sqrt [3]{\frac {2a}{b \plus{} c}}} \geq \frac{3}{2} \cdot \sqrt[3]{4}$

arqady,can you check again the stronger inequality  ?

Or perhaps the old inequality
$ \sum_{cyc} \sqrt{\frac{2a}{a\plus{}b}}\geq 3$  :blush:
Nice proofs, endd0box :)

[quote="enndb0x"][quote="arqady"]

By the way, $ \sum_{cyc}{\sqrt [3]{\frac {2a}{b \plus{} c}}} \geq 2\sqrt [3]2$ is  stronger.[/quote]

arqady,can you check again the stronger inequality  ?[/quote]
We obtain the infimum when $ a\equal{}b$ and $ c\rightarrow0^\plus{}.$ :wink:

[quote="Detective"]Let $ a,b,c$ be positive real numbers. Prove that

$ \frac {(a \plus{} b \plus{} c)^{3}}{27abc}\geq \frac {3(a^{2} \plus{} b^{2} \plus{} c^{2})}{(a \plus{} b \plus{} c)^{2}}$[/quote]

Dear Detective, which is your book?

[quote="enndb0x"][quote="arqady"]

For all positive reals $ a,b,c$,
\[ \sqrt [5]{\left(\frac {2a}{b \plus{} c}\right)^3} \plus{} \sqrt [5]{\left(\frac {2b}{c \plus{} a}\right)^3} \plus{} \sqrt [5]{\left(\frac {2c}{a \plus{} b}\right)^3} \ge 3\]
[/quote]

Arqady ,Do you think this inequality can be proved just by AM-GM as I posted above ? If yes , can you give me a hint ?[/quote]
Maybe it's possible. I don't see it. I have found another proof.


function
parameterization
calculus
calculus computations

err.. beta function need 2 parameters, not only one...

can you make more clear the second parameter?

I was wondering the same thing... However, in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=155825]this topic[/url], I deduced that $ \beta (1) = \frac{\pi}4$.

[quote="JBL"]The beta function in question is most likely [url=http://mathworld.wolfram.com/DirichletBetaFunction.html]this one[/url].  (It is given by a zeta-function-like series.)[/quote]

that is correct.  :)

i know, i could've simply said $ \frac{\pi}{4}$ instead of saying $ \beta{(1)}$,
but the text i took the problem out of stated it using $ \beta{(1)}$.

also, if i am [u]not[/u] mistaken, back when i studied analysis
from whittaker & watson's text, we studied the beta function
(which yagaron stated above), and i remember it was represented
using $ \textbf B(m,n)$ ([i]capital[/i] 'beta') ...... maybe today's textbooks
use $ \beta (m,n)$ instead, that i don't know.... :|

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

not today's notation. the central beta function use it. and this function is not new :)

or another old fashioned http://mathworld.wolfram.com/BetaExponentialFunction.html


Any updates on when the new classroom will be implemented?

Trust me this is the place where you will find them when the time comes ;)


Let $ a,b,c\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ \dfrac{a}{b\plus{}2}\plus{}\dfrac{b}{c\plus{}2}\plus{}\dfrac{c}{a\plus{}2}\leq 1$

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \dfrac{a}{b \plus{} 2} \plus{} \dfrac{b}{c \plus{} 2} \plus{} \dfrac{c}{a \plus{} 2}\leq 1$[/quote]
$ \iff ab^2 \plus{} bc^2 \plus{} ca^2 \leq 2 \plus{} abc$
assume , $ b \leq max(a,c)$ and $ b \geq min(a,c)$
$ \equal{} > a(b \minus{} a)(b \minus{} c) \leq 0 \iff a^2b \plus{} abc \geq ab^2 \plus{} ca^2 (1)$
We have :
$ (b \minus{} 1)^2(b \plus{} 2) \geq 0 \iff 2 \geq a^2b \plus{} bc^2 (2)$ (because $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$)
$ (1) \plus{} (2) \equal{} > ab^2 \plus{} bc^2 \plus{} ca^2 \leq 2 \plus{} abc$


What's the best way to express a restriction map? \[f\mid_{E}\] restriction of a function f on the subset E of the domain of f.
There is a space between f and the \mid part.

You could use a negative space f\hspace{-0.3em}\mid_{E} which gives $f\hspace{-0.3em}\mid_{E}$ - you can adjust the spacing to your taste.

Incidentally, amsmath has \restriction which looks very similar to \mid but has a small tick at the top right $f\restriction_{E}$. I haven't seen it used for function restriction - has anyone seen it used for that or for anything else?

No, never seen stevem !:|
For redkimchi, You can use the stevem's trick to define a new command.
For example [b]restr[/b]. Put this in the premble:

[code]\newcommand{\restr}[1]{\hspace{-0.3em}\mid_{#1}[/code]
To use it, f\restr{G} will give you $f\hspace{-0.3em}\mid_{G}$


trigonometry
inequalities
inequalities proposed

Prove that $ \alpha\plus{}\beta\plus{}\gamma\geq \alpha\cdot\dfrac{\sin\beta}{\sin\alpha}\plus{}\beta\cdot\dfrac{\sin\gamma}{\sin\beta}\plus{}\gamma\cdot\dfrac{\sin\alpha}{\sin\gamma}$   for any numbers $ \alpha, \beta, \gamma \in (0,\pi/2)$.

Note that on this interval $ \sin x$ is an increasing function.  What's less obvious is that $ \frac{x}{\sin x}$ is also an increasing function.  (We can write $ \frac{\sin x}{x}$ as an infinite product of cosines, all of which are decreasing on this interval.)  It follows that $ \left[ \frac{\alpha}{\sin \alpha}, \frac{\beta}{\sin \beta}, \frac{\gamma}{\sin \gamma} \right]$ is similarly sorted to $ [ \sin \alpha, \sin \beta, \sin \gamma ]$, so the result follows by Rearrangement.

yes, that's my solution too... we can prove also that $ \dfrac{x}{\sin x}$ is increasing, because $ \left(\dfrac{\sin x}{x}\right)' \equal{} \dfrac{(x \minus{} \tan x)\cos x}{x^2} < 0$...


Find $ x,y,z\in R$ such that
$ log(x^3\minus{}x^2)\equal{}log_5 y^2$
$ log(y^3\minus{}y^2)\equal{}log_5 z^2$
$ log(z^3\minus{}z^2)\equal{}log_5 x^2$

We have $ x;y;z \geq 1$
Suppose that $ x\equal{}max\{x;y;z\}$
then $ log_5 x^2 \geq log_5 y^2$
=>$ log(y^3\minus{}y^2) \geq log(x^3\minus{}x^2)$
=>$ y \geq x$ =>$ x\equal{}y$
Similar we have $ x\equal{}z$ then $ x\equal{}y\equal{}z$
=>$ log(x^3\minus{}x^2)\equal{}log_5 x^2$

hello, yes, we have only one real solution.
Sonnhard.


inequalities
vector
linear algebra
matrix
inequalities unsolved

Real numbers $ a_{i}, b_{i}$ satisfy $ \sum_{i\equal{}1}^{n}{a_{i}^2}\equal{}\sum_{i\equal{}1}^{n}{b_{i}^2}\equal{}1$ and $ \sum_{i\equal{}1}^{n}{a_{i}b_{i}}\equal{}0$. Prove that $ \left (\sum_{i\equal{}1}^{n}{a_{i}} \right)^2\plus{} \left (\sum_{i\equal{}1}^{n}{b_{i}^2} \right)^2 \leq n$.

you probably mean
[quote="Nak"]Real numbers $ a_{i}, b_{i}$ satisfy $ \sum_{i \equal{} 1}^{n}{a_{i}^2} \equal{} \sum_{i \equal{} 1}^{n}{b_{i}^2} \equal{} 1$ and $ \sum_{i \equal{} 1}^{n}{a_{i}b_{i}} \equal{} 0$. Prove that $ \left (\sum_{i \equal{} 1}^{n}{a_{i}} \right)^2 \plus{} \left (\sum_{i \equal{} 1}^{n}{b_{i}} \right)^2 \leq n$.[/quote]

By trying to generalise (or rather sharpen) it, I have found an easy solution very naturally:
Take $ n \minus{} 2$ vectors from $ \mathbb R^n$ that form an orthonormal basis together with $ a \equal{} (a_1,\dots,a_n)$ and $ b \equal{} (b_1,\dots,b_n)$. If $ C \equal{} (c_{ij})$ is the corresponding orthogonal $ n$x$ n$ matrix and $ J$ denotes the all-1-matrix, we have $ (CJ)^T(CJ) \equal{} JC^TCJ \equal{} J^2 \equal{} nJ$, so all the entries of this matrix are equal to
\[ n \equal{} \sum_{j \equal{} 1}^n\left(\sum_{i \equal{} 1}^nc_{ij}\right)^2\ge \sum_{j \equal{} 1}^2\left(\sum_{i \equal{} 1}^nc_{ij}\right)^2 \equal{} \left(\sum a_i\right)^2 \plus{} \left(\sum b_i\right)^2
\]
:)

I wonder if there is a short solution without matrices. :?:

Let $ A\equal{}\sum_{i \equal{} 1}^{n}{a_{i}}$, $ B\equal{}\sum_{i \equal{} 1}^{n}{b_{i}}$ then $ A^2\plus{}B^2 \leq n \Longleftrightarrow \sum_{i \equal{} 1}^{n}{(1\minus{}Aa_{i}\minus{}Bb_{i})^2} \geq 0$.


Prove that $ \triangle\ ABC$ is right $ \Longleftrightarrow\ \cos\frac A2\cos\frac B2\cos\frac C2\minus{}\sin\frac A2\sin\frac B2\sin\frac C2\equal{}\frac 12$ .

$ \cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2} \equal{} \frac {1}{4} (\sin A \plus{} \sin B \plus{} \sin C)$

$ \sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2} \equal{} \frac {1}{4} (\cos A \plus{} \cos B \plus{} \cos C) \minus{} \frac {1}{4}$

$ \therefore \cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2} \minus{} \sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2} \equal{} \frac 12$

$ \Longleftrightarrow \sin A \plus{} \sin B \plus{} \sin C \minus{}(\cos A \plus{} \cos B \plus{} \cos C) \equal{} 1.$

[quote="kunny"].... $ \Longleftrightarrow \sin A \plus{} \sin B \plus{} \sin C \minus{} (\cos A \plus{} \cos B \plus{} \cos C) \equal{} 1$ .[/quote]

I"ll continue (finish) the [b]Kunny's proof.[/b]

Using the well-known relations  $ \left|\begin{array}{c} \tan\frac A2 \equal{} \frac {r_a}{p} \equal{} \frac {r}{p \minus{} a} \\
 \\
r_a \plus{} r_b \plus{} r_c \equal{} 4R \plus{} r\end{array}\right|$ obtain easily $ \left|\begin{array}{c} \tan\frac A2 \plus{} \tan\frac B2 \plus{} \tan\frac C2 \equal{} \frac {4R \plus{} r}{p} \\
 \\
\tan\frac A2\tan\frac B2 \plus{} \tan\frac B2\tan\frac C2 \plus{} \tan\frac C2\tan\frac A2 \equal{} 1 \\
 \\
\tan\frac A2\tan\frac B2\tan\frac C2 \equal{} \frac rp\end{array}\right|$ .

With other words, the equation $ f(x)\equiv p\cdot x^3 \minus{} (4R \plus{} r)\cdot x^2 \plus{} p\cdot x \minus{} r \equal{} 0$ has the roots $ \left\{\tan\frac A2,\tan\frac B2,\tan\frac C2\right\}$ .
Observe that $ f(1) \equal{} 2[p \minus{} (2R \plus{} r)]$ . From the well-known relations $ \left|\begin{array}{c} \sin A \plus{} \sin B \plus{} \sin C \equal{} \frac pR \\
 \\
\cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} \frac rR\end{array}\right|$ obtain

$ \sum\sin A \minus{} \sum\cos A \equal{} 1\Longleftrightarrow p \equal{} 2R \plus{} r\Longleftrightarrow f(1) \equal{} 0$ $ \Longleftrightarrow$ $ 1\in \left\{\tan\frac A2,\tan\frac B2,\tan\frac C2\right\}\Longleftrightarrow 90^{\circ}\in \{A,B,C\}$ .

[b][u]Remark[/u].[/b] $ \left|\begin{array}{c} \prod\sin\frac A2 \equal{} \sqrt {\frac {(p \minus{} b)(p \minus{} c)}{bc}\cdot\frac {(p \minus{} c)(p \minus{} a)}{ca}\cdot\frac {(p \minus{} a)(p \minus{} b)}{ab}} \equal{} \frac {(p \minus{} a)(p \minus{} b)(p \minus{} c)}{abc} \equal{} \frac {pr^2}{4Rpr} \equal{} \frac {r}{4R} \\
 \\
\prod\cos\frac A2 \equal{} \sqrt {\frac {p(p \minus{} a)}{bc}\cdot\frac {p(p \minus{} b)}{ca}\cdot\frac {p(p \minus{} c)}{ab}} \equal{} \frac {pS}{abc} \equal{} \frac {pS}{4RS} \equal{} \frac {p}{4R}\end{array}\right|$ . Therefore,

$ \prod\cos\frac A2 \minus{} \prod\sin\frac A2 \equal{} \frac 12\ \Longleftrightarrow\ \frac {p \minus{} r}{4R} \equal{} \frac 12\ \Longleftrightarrow\ p\minus{}r\equal{}2R\ \Longleftrightarrow\ p \equal{} 2R \plus{} r$ .

[quote="Virgil Nicula"]Prove that $ \triangle ABC$ is right $ \Longleftrightarrow$ ...[/quote]
[quote="kunny"].... $ \Longleftrightarrow \sin A \plus{} \sin B \plus{} \sin C \minus{} (\cos A \plus{} \cos B \plus{} \cos C) \equal{} 1$.[/quote]
is problem [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=134218[/url].


Let $ M \subset {1,2,...15}$ such that if $ a,b,c$ are three distinct  elements then $ abc$ is a perfect square
Find the maximum of$ /M/$ (the maximum of number of elements)

Answer: 3.
Example: $ M\equal{}\{1,4,9\}$.

Assume $ |M| \geq 4$. Assume to the contrary that $ M$ contains at least 4 elements. Consider 4 elements $ a, b, c, d \in M$. Numbers $ abc$, $ abd$, and $ acd$ are perfect squares. Therefore, $ (abc)(abd)/(acd) \equal{} ab^2$ is a square of a rational number. Thus $ a$ is a perfect square. Similarly, $ b$, $ c$, and $ d$ are perfect squares. But there are only 3 distinct perfect squares between 1 and 15: 1, 4, and 9.

[quote="Yury"]Answer: 3.
Example: $ M \equal{} \{1,4,9\}$.

Assume $ |M| \geq 4$. Assume to the contrary that $ M$ contains at least 4 elements. Consider 4 elements $ a, b, c, d \in M$. Numbers $ abc$, $ abd$, and $ acd$ are perfect squares. Therefore, $ (abc)(abd)/(acd) \equal{} {ab}^2$ is a square of a rational number. Thus $ a$ is a perfect square. Similarly, $ b$, $ c$, and $ d$ are perfect squares. But there are only 3 distinct perfect squares between 1 and 15: 1, 4, and 9.[/quote]
Thanks,but can you detail this line
 :blush:
 [quote=Yury]Therefore, $ (abc)(abd)/(acd) \equal{} {ab}^2$ is a square of a rational number. Thus $ a$ is a perfect square[/quote]
 I can't understand it :oops: ,sorry

Let $ abc\equal{}x^2$, $ abd\equal{}y^2$, and $ acd\equal{}z^2$. Then
\[ a\equal{} \frac{x^2 y^2}{z^2 b^2}\equal{}\left(\frac{xy}{zb}\right)^2.\]
Now if the square of a rational number is a natural number, then this rational number is a natural number. Therefore, $ \frac{xy}{zb}$ is a natural number. So $ a$ is equal to the square of a natural number.

Oh sorry,it iseems very clearly,and i had a stupid question Sorry :blush: