function
inequalities
algebra
system of equations
algebra proposed

Consider the equation
y^3=9(x^2-3x+3)
Discriminant = 9-12=-3. No real roots. The given function {(x^2-3x+3)}
is always greater than 0.
=> x,y,z > 0.
Now,
(y-3)(y^2+3y+9)=9x(x-3)
(z-3)(z^2+3z+9)=9y(y-3)
(x-3)(x^2+3x+9)=9z(z-3)

Multiply these 3 eqs to get
((y^2+3y+9)(z^2+3z+9)(x^2+3x+9)-729xyz)(x-3)(y-3)(z-3)=0
Case(i):
=>(3+y+9/y)(3+z+9/z)(3+x+9/x)=729
As x,y,z > 0. AM GM inequality is applicable.
=> y+9/y >= 6
So, (3+y+9/y)(3+z+9/z)(3+x+9/x) >= (3+6)(3+6)(3+6)=729
Equality holds when y+9/y=6 => (y-3)^2 = 0 => y=3
Case(ii):
x=3
=> 9z(z-3)=0 => z=0 or 3.
If z=0 => y^2-3y+3=0 has no real solutions as discriminant is less than zero
=> z=3 => y=3
Same is the procedure for y=3 and z=3 cases.
The only solution to the given system of equations
is x=y=z=3.

[quote="zaya_yc"]$ x,y,z$ real number solve:

$ y^{3} \minus{} 9x^{2} \plus{} 27x \minus{} 27 \equal{} 0$
$ z^{3} \minus{} 9y^{2} \plus{} 27y \minus{} 27 \equal{} 0$
$ x^{3} \minus{} 9z^{2} \plus{} 27z \minus{} 27 \equal{} 0$[/quote]

the systeme <=>^$ (y^2\plus{}3y\plus{}9)(z^2\plus{}3z\plus{}9)(x^2\plus{}3x\plus{}9)\equal{}9^3xyz$

and we know tha $ (y\minus{}3)^2>0$ :rotfl:  so $ y^2\plus{}3y\plus{}9>9y$ the same thing for $ (x.y)$

so the systeme equivalent to $ (y\minus{}3)^2 \equal{}0$

so $ S\equal{}(3.3.3)$


$ \frac{dN}{dt}\equal{}k_1N\minus{}k_2N^2\minus{}r$ where $ N(t_0)\equal{}N_0$

If k1, k2, and r are constants, then you are to integrate

$ \frac {dN}{k_2N^2 \minus{} k_1N \plus{} r} \equal{} \minus{} dt$.

Just by curiosity, what is the context of this differential equation?

Complete the square, and make a linear substitution. The solutions to $ y'\equal{}a^2\minus{}y^2$ are hyperbolic tangents, and the solutions to $ y'\equal{}\minus{}a^2\minus{}y^2$ are cotangents.


I've read the chapter so many times it's unreal, but I still can't seem to grasp this concept.

Here are the instructions:

Rewrite each expression as a sum or a difference of multiples of logarithms:

log (x^2y)
ln (16)

I have the answers for these but I don't understand the steps or HOW to arrive at these answers - please help :)

Log rules tell us that $\log a^b=b\log a$

So $\log x^{2y}=2y\log x$

$\ln 16=\ln 2^4=4\ln 2$  ;)

[quote="aidan"]By the way, what does x^2y refer to? It could be $x^{2y}$ or ${x^2}{y}$  ;)[/quote]

well then , since we also have the rules $\log ab=\log a+\log b$ so

$\log x^2y=\log x^2+\log y=2\log x+\log y$  :)

Thanks for esplaining...how simple that ended up being.

BTW it was the second option :)

Okay two more...hopefully I get a reply soon so i'll know how to do these problems on the test:

x*ln(3)-x=ln(7)

ln(x+2)-ln(x-10)=ln(2)

Okay the second one is pretty easy figured it out based on the previous explanation - 22, but still have no idea how to do the first guess ill just miss that one.

If $x\ln 3-x=\ln 7$ then $x(\ln 3-1)=\ln 7$ and so $x=\frac{\ln 7}{\ln 3-1}$. If $\ln(x+2)-\ln(x-10)=\ln(2)$ note that x+2>0 and x-10>0 i.e. x>10. Now $\ln \frac{x+2}{x-10}=\ln 2$ so $\frac{x+2}{x-10}=2 \Leftrightarrow x=22$. Now $22>10$ so that $x=22$ is indeed a solution. I used only basic logarithmic formulas and that $\ln x=\ln y\Leftrightarrow x=y$.


A regular 1997-gon is decomposed into triangles using non-intersecting diagonals. How many of these triangles are acute?

Cute!

Inscribe the polygon into a circle, with center $O$. That a triangle $T$ is acute is equivalent to the fact that $O$ is an interior point of $T$. Since we have a triangulation of the polygon, the interior of the triangles are pairwise disjoints. Therefore, there is at most one acute triangle.
In another hand, if there is no acute triangle, it means that $O$ belongs to the side of a triangle, so that there are two vertices which form a diameter. But this is impossible since $1997$ is odd. Thus, there is at least one acute triangle.

Then, there is exactly one acute triangle.

This is clearly true for any odd $n$ instead of $1997$. The first part of the reasoning remains true for even $n$. If I've no mistake, if $n=4,6$ there is no acute triangle. If $n \geq 8$ there can be one acute triangle or not, according to chosen triangulation of the polygon.

Pierre.


trigonometry
geometry
trig identities
Law of Sines
geometry unsolved

Let $ ABC$ be a triangle with $ (O)$ is circumcircle.Let $ P$ be intersection of tangent at $ B,C$ of $ (O)$.Let $ D$ be intersection of $ AP$ and $ BC$.Let $ E,F$ be points on $ AC$,$ AB$ respectively such that $ DE$ paralell to $ AB$,$ DF$ paralell to $ AC$.Prove that $ B,C,E,F$ lie on a circle with centre $ A_1$. Definding similar with $ B_1,C_1$.Prove that $ AA_1,BB_1,CC_1$ are concurent.

It is well-known that ABQC is a harmonious quadrangle
So we have ∠BQD=∠MQC=∠ACB   where M is the midpoint of BC
CK∥EF  =>  H is the midpoint of CK  =>  MH∥AB
=> H,M,Q,C are concyclic  =>  ∠MHC+∠MQC=∠BKC+∠MQC=∠BFE+∠ACB=180
=> E,F,B,C are concyclic

[quote="plane geometry"]I am still working on part2 but I don't have much time now , do you have ideas,dear mathlinkers?[/quote]
I'm thinking about Trig Ceva.

If $ A_1$ is the center of the circle and $ R$ is the radius, then let $ \angle{BAA_1} \equal{} \alpha$, $ \angle{CAA_1} \equal{} \beta$, by the Law of Sines, we have:

$ \frac {\sin{\alpha}}{R} \equal{} \frac {\sin{\angle{A_1BF}}}{EA_1}$ and $ \frac {\sin{\beta}}{R} \equal{} \frac {\sin{\angle{A_1CE}}}{EA_1}$

Then: $ \frac {\sin{\alpha}}{\sin{\beta}} \equal{} \frac {\sin{\angle{A_1BF}}}{\sin{\angle{A_1CE}}} \equal{} \frac {\cos{\angle{FCB}}}{\cos{\angle{EBC}}}$.

Then I got lost when I try to prove, in the following figure, that $ \angle{FCB} \equal{} \angle{ICA}$.

[quote="nsato"]See
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98972[/url][/quote]
Oh, I posted this Chinese problem 3 years ago and I totally forgot it......


What is the remainder when $ 1999^{2000}$ is divided by 5?

I don't know if there is a better way to do this, but I looked for a pattern:
1999^1 has a ones digit of 9, so the remainder is 4.
1999^2 has a ones digit of 1, so the remainder is 1.
1999^3 has a ones digit of 9, so the remainder is 4.
... the pattern is 4,1,4,1...
Since 2000 is an even power, the answer is 1.

We know that the remainder will be based solely on what the units digit is, since any 2+ digit number can be written as the units digit + $ 10^1 *$ tens digit + $ 10^2 *$ hundreds digit, etc, and $ 10^n$ will always be divisible by 5 as long as $ n$ is an integer greater than $ 0$.

Now, for any number $ x^a$, the units digit will be equal to:  $ ($units digit of x$ )^{a\%4}$, where % is the modulo operator.  I don't know how to prove this, but just try it for the ten different digits, and you'll see it's true.

So instead of $ 1999^{2000}$ you just do $ 9^0 \equal{} 1$.  The remainder when 1 is divided by 5 is 1, so the answer is 1.

[hide="Modular Arithmetic"]
We know that $1999\equiv 4\mod 5\equiv (-1)\mod 5$. Therefore, \[1999^{2000}\equiv (-1)^{2000}\mod 5\implies\boxed{1}.\][/hide]


Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that

$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\geq\frac{9}{8}\plus{}\frac{5}{2(a\plus{}b\plus{}c)}$

[quote="Ligouras"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {9}{8} \plus{} \frac {5}{2(a \plus{} b \plus{} c)}$[/quote]
We only need to check your inequality for $ c\rightarrow0^\plus{}$ and $ b\equal{}c.$ :wink:

[quote="arqady"][quote="Ligouras"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {9}{8} \plus{} \frac {5}{2(a \plus{} b \plus{} c)}$[/quote]
We only need to check your inequality for $ c\rightarrow0^ \plus{}$ and $ b \equal{} c.$ :wink:[/quote]

Thank you Arkady :(

[quote="pifi"][quote="arqady"][quote="Ligouras"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {9}{8} \plus{} \frac {5}{2(a \plus{} b \plus{} c)}$[/quote]
We only need to check your inequality for $ c\rightarrow0^ \plus{}$ and $ b \equal{} c.$ :wink:[/quote]

 :blush:   Can you share us your proof, Arkady? Thanks.[/quote]
Let $ a,$ $ b$ and $ c$ are non-negatives, $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2,$ where $ v>0$ and $ abc\equal{}w^3.$
Hence, $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {9}{8} \plus{} \frac {5}{2(a \plus{} b \plus{} c)}\Leftrightarrow$
$ \Leftrightarrow\frac{9u^2\plus{}3v^2}{9uv^2\minus{}w^3}\geq\frac{3\sqrt3}{8v}\plus{}\frac{5}{6u}\Leftrightarrow w^3\geq f(u,v).$
But $ w^3$ gets a minimal value when $ w^3\equal{}0$ or when two numbers from $ \{a,b,c\}$ are equal.


What is the ninth digit from the right in 101^20?

[hide]
Expand by Binomial Theorem,
$ 101^{20} \equal{} (10^2 \plus{} 1)^{20} \equal{} \sum_{k \equal{} 0}^{20}\binom{20}{k}10^{2k}$
We want to find the coefficient of $ 10^8$, so all that matters is
$ \binom{20}{4}10^8 \plus{} \binom{20}{3}10^6$
With some calculation we find the factorial coefficient of first term has a unit digit $ 5$, and the second term equals to $ 11.4\times 10^8$. Therefore, the 9th digit from right is $ 5 \plus{} 1 \equal{} \boxed{6}$.
[/hide]


algebra
polynomial
calculus
derivative
absolute value
calculus computations

I once heard that one way to prove the existance of transcedental numbers was to prove a lower bound on the error of rational approximations to algebraic numbers (in terms of the denominator). How do you do this?

If $ r\equal{}\frac{p}{q}$ is not a root of a polynomial $ f$ of degree $ n$ with integer coefficients, $ |f(r)|\ge \frac1{q^n}$. Now, we can find $ C$ such that the derivative of $ f$ near a root $ x$ is at most $ C$ in absolute value, and thus $ |r\minus{}x|\ge \frac1{Cq^n}$.


LaTeX
number theory
greatest common divisor
relatively prime

I'll do #2 later but...

[hide="1"]
Hint: Use proof by contradiction. If a+x is rational, what happens?
[hide="The proof"]
Assume the contrary that $a+x$ is actually rational. Since $a$ is rational, then $(a+x)-a$ is also rational, because the rationals are closed under addition and subtraction. Thus $x$ would be rational, which contradicts our original assumption. Therefore $a+x$ is irrational. Q.E.D.
[/hide]
[/hide]

1. Try with a proof by contradiction, i.e. assume the premise ($a \in{\bf Q}, \: x \not \in{\bf Q}$) is true [b]and[/b] the conclusion ($a+x \not \in{\bf Q}$) is false, and this leads to a contradiction.

2. Use the fact that every non-square natural number $n$ can be expressed as a product of a square ($k^{2}$) and a squarefree (not divisible by any square > 1) number $m>1$. Hence $\sqrt{n}= \sqrt{k^{2}\cdot m}= k\sqrt{m}$. Thus $\sqrt{n}\in{\bf Q}$ iff $\sqrt{m}\in{\bf Q}$, i.e. there are two relatively prime natural numbers $a$ and $b$ such that $\sqrt{m}= a/b$. Squaring we obtain $a^{2}= m \cdot b^{2}.$ Then $m$ has a prime divisor $p$, hence $p|a^{2}$. Therefore $p|a$, so $p^{2}|mb^{2}$. Now $b$ is not divisible by $p$ since $p|a$ and GCD($a,b$)=1. This fact implies that $p^{2}|m$, a contradiction which completes the proof.

[quote="b-flat"]In $\LaTeX$, use \sqrt{n} for $\sqrt{n}$.

Also, I do not know if these are getting started level (not sure if proofs should be allowed...), but someone else might.[/quote]Please move this topic to Intermediate Topics. If the moderators there think it should be moved up, they will move it.

#1 if you have the sum $\frac{a}{b}+\frac{c}{d}$ you just ad them $\frac{ac+bd}{bd}$ I think

Hey, thanks guys. I came up with this last night:

$a \in{\bf Q}, \: x \not \in{\bf Q}$
$a+x \in{\bf Q}$
$x \in{\bf Q-a}$
$x \in{\bf Q}$ (because a rational minus a rational is rational) This contradicts the original statement.

Would this work as a proof by contradiction? I'm a little new to proof writing. The only thing is I don't know how to prove a rational minus a rational is rational. I know it is logically because of the way a number line works, but I'm not sure I can prove it mathematically. mathnerd314's proof is also hinged on this, too.

Thanks again.

[quote="skibum14"]The only thing is I don't know how to prove a rational minus a rational is rational.[/quote]

Good point...

Lemma: A rational minus a rational is another rational.
Proof:
Let $x=\frac{a}{b}$ and $y=\frac{c}{d}$
So x and y are both rational.
Then $x-y=\frac{ad-bc}{bd}$.
Letting $ad-bc=p$ and $bd=q$, $x-y=\frac{p}{q}$ which is also rational.
Q.E.D.

[quote="skibum14"]$a \in{\bf Q}, \: x \not \in{\bf Q}$
$a+x \in{\bf Q}$
$x \in{\bf Q-a}$
$x \in{\bf Q}$ [/quote]

I think you probably want to reconsider that third line, unless that's a notation you are using in your class (or wherever these questions came from).

Mathnerd gives a good proof that the difference of two rationals is rational, provided that you already "know" (=have proved with the appropriate degree of rigor) that products and differences of integers are integers.  I suppose you also need to be able to show that $x-y=\frac{ad-bc}{bd}$ is really true, based on whatever your definitions of rationals, addition, subtraction and multiplication might be (again with the appropriate degree of rigor).

[hide="1"]Since $a$ is rational, let $a=\frac{p_{1}}{q_{1}}$.  Suppose that $a+x$ is rational, so that $\frac{p_{1}}{q_{1}}+x=\frac{p_{2}}{q_{2}}$.  Then $x=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}$, which implies that $x$ is rational, a contradiction.[/hide]

[quote="skibum14"]These should be fairly simple proofs, but I don't know where to begin.

1. If $a$ is rational and $x$ is irrational, prove that $a+x$ is irrational.

2. Prove that the following is not rational: $\sqrt{n}$ where the integer $n$ is not a perfect square

There are a few more of these types problems I'm working that don't make sense to me, but hopefully if these are explained, I'll understand how to do those.

Thanks in advance. I appreciate it.[/quote]

I don't think that all simple ideas are not simply proved.
Prove that there is one and only one line that passes through two distinct points A and B.

[quote="nutz_for2.718281828"]I don't think that all simple ideas are not simply proved.
Prove that there is one and only one line that passes through two distinct points A and B.[/quote]

You [b]don't[/b] think that all simple ideas are [b]not[/b] simply proved? So you do think that all simple ideas are simply proved. Prove it.

I meant that I don't think that the are simply proved.


In a $ n*n$ array each of the numbers $ 1,2,3,...,n$ appear exactly $ n$ times .show that there is a ropw or a clumn in the array with at least $ n^{1/2}$ distinct numbers.


1.    (ab-b+1)(ac-a+1)(bc-c+1)<abc  if abc=1 and a,b,c are positive numbers
2.    2/b(a+b) + 2/c(b+c) + 2/a(a+c)>27/(a+b+c)^2 for positive reals a,b,c
3.    a^2+b^2+c^2=3 then prove 1/(1+2ab) + 1/(1+2ac) + 1/(1+2bc)>1 for positive reals a,b,c
4.    (2a+b+c)(2b+c+a)(2c+a+b)>8(a+b)(a+c)(b+c) for positive reals a,b,c
 Good luck and look when equality holds in each inequality  :)

[/img]

for the first inequality
$ (ab\minus{}b\plus{}1)(ac\minus{}a\plus{}1)(bc\minus{}c\plus{}1) \leq abc$  <=> $ \left(a\minus{}1\plus{}\frac{1}{b} \right)\left(b\minus{}1\plus{}\frac{1}{c} \right)\left(b\minus{}1\plus{}\frac{1}{c} \right) \leq 1$  (and it is offered problem has IMO on 2000)
set  $ a\equal{}\frac{x}{y}$ ; $ b\equal{}\frac{y}{z}$ and $ c\equal{}\frac{z}{x}$
and inequality becomes
$ (x\plus{}y\minus{}z)(x\minus{}y\plus{}z)(\minus{}x\plus{}y\plus{}z) \leq xyz$.  and this is right

4.
$ L.H.S \equal{} ((a \plus{} b) \plus{} (a \plus{} c))((b \plus{} c) \plus{} (b \plus{} a))((c \plus{} a) \plus{} (c \plus{} b))\geq 8(a \plus{} b)(b \plus{} c)(c \plus{} a)$
Let
a+b=x
b+c=y
c+a=z
the problem turn to
$ (x \plus{} y)(y \plus{} z)(z \plus{} x)\geq 8xyz$
which is clearly true by AM-GM
the equality held when a=b=c
3.
From CS we have
$ \sum_{cyc} \frac {1}{1 \plus{} 2ab} \geq \frac {9}{3 \plus{} 2\sum_{cyc} ab} \equal{} \frac {3\sum_{cyc} a^2}{(\sum_{cyc} a^2 \plus{} 2\sum_{cyc}ab)}$
since we have \sum_{cyc} a^2 \geq \sum_{cyc} ab
$ \frac {3\sum_{cyc} a^2}{(\sum_{cyc} a^2 \plus{} 2\sum_{cyc}ab)}\geq 1$ is true already

a simple resolution for the third inequatily:
$ S=\sum_{cyc}\frac{1}{2ab}\geq\sum_{cyc}\frac{1}{1+a^2+b^2}$
and we have with cauchy:
$ \left h(\sum_{cyc}\frac{1}{1+a^2+b^2} ) (3+2(a^2+b^2+c^2)) \geq 9$
<=>$ S\geq1$

[quote="peine"]for the second inequality. it is not easy. :wink: but I am going to search a resolution  :)[/quote]
basically for the second one , AM-Gm directly helps :D
$ LHS \geq \frac{6}{\sqrt[3]{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}} \geq \frac{6}{\sqrt[3]{(\frac{a\plus{}b\plus{}c}{3})^3.(\frac{2a\plus{}2b\plus{}2c}{3})^3}} \equal{} \frac{6}{\frac{(a\plus{}b\plus{}c)}{3}.\frac{2a\plus{}2b\plus{}2c}{3}}\equal{} \frac{27}{(a\plus{}b\plus{}c)^2}\equal{} RHS$


parameterization
function
limit
quadratics
algebra
functional equation
algebra unsolved

We are given real numbers $a$ and $b$ which satisfy $ 0<a,b< \frac{1}{2}$. Find all continuous functions $ \mathbb R \to \mathbb R$ such that $ f(f(x))\equal{}af(x)\plus{}bx$ for all real $x$.

[quote="Dumel"]reals a,b satisfy $ 0 < a,b < \frac {1}{2}$. find all continuous functions $ R \to R$ such that $ f(f(x)) \equal{} af(x) \plus{} bx$[/quote]

Since $ b\ne 0$, $ f(u)\equal{}f(v)$ implies $ f(f(u))\minus{}af(u)\equal{}f(f(v))\minus{}af(v)$ and so $ bu\equal{}bv$ and so $ f(x)$ is injective, and so monotonous (since continuous).
If $ \lim_{x\to\infty}f(x)\equal{}l$, then the equality $ f(f(x))\minus{}ax\equal{}bx$ is wrong when $ x\to\infty$ since $ LHS\to f(l)\minus{}l$ while $ RHS\to\infty$. So $ f(x)$ is surjective and so is a bijection from $ \mathbb R\to\mathbb R$

Let $ h(x)\equal{}f(x)\minus{}x$ : $ h(f(x))\plus{}(1\minus{}a)h(x)\equal{}(a\plus{}b\minus{}1)x$ and, since $ a\plus{}b\minus{}1\ne 0$ and $ 1\minus{}a>0$, $ h(x)$ cant have a constant sign over $ \mathbb R$ (since then LHS would have a constant sign while RHS would not). So $ \exists u$ such that $ h(u)\equal{}0$ and so $ f(u)\equal{}u$ and so $ f(f(u))\equal{}af(u)\plus{}u$ implies $ (a\plus{}b\minus{}1)u\equal{}0$ and so $ u\equal{}0$ since $ a\plus{}b<1$ and so $ f(0)\equal{}0$

Then $ g(x)\equal{}f^{[\minus{}1]}(x)$ exists, is monotonous and respects the equation $ x\equal{}ag(x)\plus{}bg(g(x))$

Let then $ c\in\mathbb R$ and $ f(c)\equal{}d$ and consider the two sequences :
$ u_0\equal{}c$, $ u_1\equal{}d$ and $ u_{n\plus{}2}\equal{}au_{n\plus{}1}\plus{}bu_n$ such that $ u_{n\plus{}1}\equal{}f(u_n)$
$ v_0\equal{}d$, $ v_1\equal{}c$ and $ v_{n\plus{}2}\equal{}\minus{}\frac abv_{n\plus{}1}\plus{}\frac 1bu_n$ such that $ v_{n\plus{}1}\equal{}g(v_n)$

Let the quadratic $ x^2\minus{}ax\minus{}b\equal{}0$. It has two real roots $ r_1<0<\minus{}r_1<r_2$ (since $ r_1\plus{}r_2\equal{}a>0$ and $ r_1r_2\equal{}\minus{}b<0$

We get $ u_n\equal{}\minus{}\frac{d\minus{}r_2c}{r_2\minus{}r_1}r_1^n\plus{}\frac{d\minus{}r_1c}{r_2\minus{}r_1}r_2^n$ and $ v_n\equal{}\minus{}\frac{d\minus{}r_2c}{r_2\minus{}r_1}r_1^{1\minus{}n}\plus{}\frac{d\minus{}r_1c}{r_2\minus{}r_1}r_2^{1\minus{}n}$

Consider then $ r_n\equal{}\frac{u_{n\plus{}1}}{u_n}\equal{}\frac{f(u_{n})}{u_n}$ and $ s_n\equal{}\frac{v_{n\plus{}1}}{v_n}\equal{}\frac{g(v_{n})}{v_n}$

Since $ f(x)$ is monotonous and $ f(0)\equal{}0$, we get that $ r_n$ must always have the same sign $ \forall n$
Since $ g(x)$ is monotonous and $ g(0)\equal{}0$, we get that $ s_n$ must always have the same sign $ \forall n$

$ r_n\equal{}\frac{\minus{}(d\minus{}r_2c)r_1^{n\plus{}1}\plus{}(d\minus{}r_1c)r_2^{n\plus{}1}}{\minus{}(d\minus{}r_2c)r_1^{n}\plus{}(d\minus{}r_1c)r_2^{n}}$ $ \equal{}\frac{\minus{}(d\minus{}r_2c)r_1\plus{}(d\minus{}r_1c)r_2(\frac{r_2}{r_1})^{n}}{\minus{}(d\minus{}r_2c)\plus{}(d\minus{}r_1c)(\frac{r_2}{r_1})^{n}}$

So, if $ d\ne r_1c$, $ \lim_{n\to\plus{}\infty}r_n\equal{}r_2>0$ and $ f(x)$ is increasing and so the only decreasing solution would be such that $ d\equal{}r_1c$ and so $ f(x)\equal{}r_1x$

$ s_n\equal{}\frac{\minus{}(d\minus{}r_2c)r_1^{\minus{}n}\plus{}(d\minus{}r_1c)r_2^{\minus{}n}}{\minus{}(d\minus{}r_2c)r_1^{1\minus{}n}\plus{}(d\minus{}r_1c)r_2^{1\minus{}n}}$ $ \equal{}\frac{\minus{}(d\minus{}r_2c)(\frac{r_2}{r_1})^n\plus{}(d\minus{}r_1c)}{\minus{}(d\minus{}r_2c)r_1(\frac{r_2}{r_1})^n\plus{}(d\minus{}r_1c)}$

So, if $ d\ne r_2c$, $ \lim_{n\to\plus{}\infty}s_n\equal{}\frac 1{r_1}<0$ and $ g(x)$ is decreasing so $ f(x)$ is decreasing and so the only increasing solution would be such that $ d\equal{}r_2c$ and so $ f(x)\equal{}r_2x$

It's then easy to check that $ r_1x$ and $ r_2x$ indeed are solutions.

And so the only two solutions of this functional equation are $ f(x)\equal{}r_1x$ and $ f(x)\equal{}r_2x$


Prove that, given a positive integrer $n$, there exists a positive integrer $k_n$ with the following property: Given any $k_n$ points in the space, $4$ by $4$ non-coplanar, and associated integrer numbers between $1$ and $n$ to each sharp edge that meets $2$ of this points, there's necessairly a triangle determined by $3$ of them, whose sharp edges have associated the same number.

This is just a particular case of Ramsey's Theorem. Anyway, I think it belongs to the Combinatorics section :).


percent geometry parallelogram rhombus trapezoid

What percent of the quadrilaterals in the diagram below are parallelograms?

[asy]pair A,B,C,D,E,F;
A=(1,1.7); B=(2,0); C=(1,-1.7); D=(-1,-1.7); E=(-2,0); F=(-1,1.7);
draw(A--B--C--D--E--F--cycle);
draw(A--D);
draw(B--E);
draw(C--F);[/asy]

We can form a rhombus with any two adjacent oppositely-facing triangles, of which there are 6.  These are obviously parallelograms.  We can also form trapezoids with any 3 consecutive triangles, and we can form 6 of those.  Unless there are some additional quadrilaterals which I can't find, [b]half[/b] are parallelograms.


trigonometry
geometry
trig identities
Law of Sines

Given this diagram:
http://img108.imageshack.us/img108/5936/diagramau2.jpg

Solve for c and d (two seperate equations) in terms of alpha, beta, gamma, a, and b.

Took me a really long time to figure this one out.  :wink:

[hide]Note that $ \angle CBD \equal{} 180 \minus{} \beta \minus{} \delta_2$. Apply the law of sines to triangle CBD to find c.
Do likewise in triangle ABD to find d.

[hide]You should arrive at:
$ c \equal{} \frac{b \sin {(\beta \plus{} \delta_2)}}{\sin {\beta}}$

$ d \equal{} \frac{a \sin {(\alpha \plus{} \delta_1)}}{\sin {\alpha}}$[/hide][/hide]

I probably didn't make this clear, but you have to put it in terms of the GIVEN variables, not just any.
Angle BCD is not given, so you can not use it in the final answer.

Ah -- I see! That makes the problem much more interesting.

Anyone going to give this a try?  It is very fun...  :P


1. -2 -(-2)^3 = what?

2. A number which is a multiple of 15, but not a multiple of 18 is
a.)180
b.)320
c.)369
d.)420
e.)540

3. A mathematics test consists of ten questions.  Ten points are given for each correct answer and 3 points are deducted for each incorrect answer.  If John did all questions and scored 61 what would his number of correct answers be?

4. If all two-digit natural numbers were written down in a column, then what would the sum of the digits in the tens place be?

5. If x/4 + y/5 = 19/20, where x and y are positive integers, then x + y equals what?

Chinaboy gets #2.
MysticTerminator gets #3.

on the last post i made, it was to question 5, not 4

Mystic Terminator gets #4 & 5.
Next set posted in 2 minutes.

I bet 10 on mystic, 7 on theone, and 5 on MathFiend (I'm just going to observe this one :))

I'm going to quickly look over the thread to make sure everything was assigned correctly

Leaderboard after 1 round:
MysticTerminator-4
ChinaBoy-2
MathFiend-1

Next set in 2 minutes...


function
advanced fields
advanced fields unsolved

Consider the Dirichlet boundary problem ( in PDE  ) on the hemisphere in $\mathbb{R}^3$. How can we find the Green function for this problem?? :(

 i.e : Find the function $g(P,Q)$ such that : $g(P,Q)$ is harmonic in the interior of the hemisphere and $g(P,Q)$ when restrict to the boundary of the hemisphere is $d(P,Q)$

I am a little bit confused about the question: when saying "hemisphere", do you actually mean half-ball, i.e., the 3-dimensional region whose boundary consists of a hemisphere and a flat disk?

Yes. This is a boundary problem. The boundary here is the hemisphere and the flatdisk ( that is the boundary of the half-ball). Sorry :lol:

Then it's easy. Let $G_y(x)$ be the Green function for the ball, i.e., $G_y=0$ on the boundary of the ball and the Laplacian of $G_y$ is the delta-measure at a given point $y$. Then your Green function for the half-ball is given by $G_y(x)-G_y(S(x))$ where $S(x)$ is symmetric to $x$ with respect to the plane containing the flat part of the boundary.


[quote="NeverOddOrEven"]if you qualified through the amc 10b, which aime do you take?[/quote]

You take whatever AIME your school chooses to take. The AMC you took is irrelevent.

[quote="NeverOddOrEven"]hmm even if i qualified through the amc b when my school only offered the amc a?[/quote]

If you qualified for the AMC from a different school than your own... you may have to contact the AMC about that. I believe you can still take the AIME at your own school.


geometry
3D geometry
quadratics
calculus
trigonometry
algebra
polynomial

if a and b are both irrational numbers with no common irrational or rational factors (except 1), can it be proven that a/b is irrational? :huh:

Hmm your question sounds like :
"If a,b are irrational and such that a/b si irrational, can we say that a/b is irrational?"
It doesn't have so much sense speaking of "factors" of an irrational number without any other precisation.

Let me restate.

a and b are both irrational numbers.

a and b have no common irrational factors such that the irrationality of both factors is cancelled when a is divided by b.

Can it be proven that a/b is irrational?

[quote="trancestorm"]a and b have no common irrational factors such that the irrationality of both factors is cancelled when a is divided by b.[/quote]

Well, if the irrationality is not canceled when a is divided by b, then it's still irrational.

Perhaps you're asking if there's a way to prove if a number such as $\frac{\pi}{e}$ is irrational. I think they're still working on that.

I'm assuming that when you say an irrational factor of a, you mean a is a rational mutiple of that irrational...

So, in other words, you're saying that a and b have no factors such that one is a rational multiple of the other?  Because then it's trivial.  If a(rational number)/b(rational number) is irrational (ie we have the quotient of the two factors is irrational), then a/b is a rational multiple of that, and is thus irrational...

Unless you mean something else?  The idea of an "irrational factor" has to be better defined.

At first define your "irrational factors" in a formal way, not by words (especially what "cancelling out" means). Otherwise your statement is either "if the quotient is irrational, the quotient is irrational" or "if you can't show it is rational by a trivial way, it is irrational" (both nonsense).

[quote="trancestorm"]the "irrational factors" part is that when a is divided by b, irrationality cannot simply be cancelled out to make a rational number[/quote]

Your definition remains unclear. What is "simply"? If you write $\sqrt{7}-2 | 3$, I guess it's not "simply cancelled out". But if you write $\sqrt{7}-2 | \left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right) = 3$, it is... Really, there isn't something like an "irrational factor".

[quote="trancestorm"]What I mean by the "irrational factors" part is that when a is divided by b, irrationality cannot simply be cancelled out  to make a rational number, as is the case with (root 2)/(root 2).

pi/e and root(2)/e are good extensions of my question, but I am looking for any general rules.[/quote]

So, for example, $\sqrt{2}$ is irrational, and $\sqrt{3}$ is irrational. Their quotient, $\frac{\sqrt{2}}{\sqrt{3}}= \sqrt{\frac{2}{3}}$ is also irrational.

What you're asking doesn't really make sense. You are saying that irrationality is not "canceled out" when a is divided by b. That means it's still irrational. If a is irrational, b is irrational, and a/b is not irrational, then a and b obviously had some sort of common irrational factor which dividing canceled out.

[quote="trancestorm"]with no common irrational or rational factors [/quote]

This is meaningless.  An irrational number is simply an irrational number.  There is no way to prefer a particular "factorization" of an irrational over another since there is no concept of primality in the irrationals.

I see where I am erring.  Thanks for all the comments.

Let me try and ask more relevant questions:

A and B are both irrational.  If A*B is irrational, is A/B always irrational?

is pi/e irrational?

Thanks

[quote="trancestorm"]A and B are both irrational.  If A*B is irrational, is A/B always irrational?[/quote]
No, take $A = B = \sqrt[3]{2}$

[quote="Mathworld"]It is not known if $\pi+e$ or $\frac{\pi}{e}$ is irrational. [/quote]
http://mathworld.wolfram.com/e.html

One really cool theorem though is $e^{i\pi}=-1$  :P

[quote="Gyan"][quote="Mathworld"]It is not known if $\pi+e$ or $\frac{\pi}{e}$ is irrational. [/quote]
Good april fools day  theorems: (In days of slide rules or log tables where you calculate 3-4 sig figures)
$e^{\pi}-\pi = 20$

Of course the classic one (Scientific Americans famous April fool issue) is:

$e^{\pi \sqrt{163}}= 262537412640768744$[/quote]
didn't martin gardner write that one?
then people didn't realize it was april fools? (the rest of that issue was pretty wierd too, him writing about the science of spirits and stuff...)
but it was based off of a calculation by ramanujan

Oh is that one of the fake equations that seem true on any calculator?

I've never been able to get a straight answer from anyone as to whether that is true or not. I've tried proving it before but to no avail. Does anyone have any idea?

[quote="me@home"]well everyone knows it can't be true... just put 6 into the power series of e... then you get a polynomial kind of thing back... which can't be pi^4 + pi^5... because it would be a lot more complicated...[/quote]
Well... by that reasoning, $e^{x}$ could never be some given number except very special cases ;)

Haven't there been power series-like represenations of pi before? I seem to recall something adding up to pi/4 or something.

And as much as logic would indicate that such a relationship would not exist, I would really like to see it happen that pi and e get some kind of relationship.

hey do you have a better "proof"?
okay I'll get into SAGE in a couple minutes I'll post back with the exact decimals to prove it...

okay here we are:
e^6=403.428793492735
pi^4+pi^5=403.42877581928383

and they are different... at least according to sage (see here: http://modular.math.washington.edu/sage/)

[quote="me@home"]well everyone knows it can't be true... just put 6 into the power series of e... then you get a polynomial kind of thing back... which can't be pi^4 + pi^5... because it would be a lot more complicated...[/quote]
It might be better to say "Well everyone knows it can't be true... if it were true, everyone would already know that formula!"

[quote="xsquaredster"] Haven't there been power series-like represenations of pi before? I seem to recall something adding up to pi/4 or something. [/quote]
Wallis' product?
$\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdots = \frac{\pi}{2}$
Liebniz' formula?
$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... = \frac{\pi}{4}$

[quote="xsquaredster"]And as much as logic would indicate that such a relationship would not exist, I would really like to see it happen that pi and e get some kind of relationship. [/quote]
$e^{\pi i}+1 = 0$
 :rotfl:

[quote="Xevarion"][quote="xsquaredster"]And as much as logic would indicate that such a relationship would not exist, I would really like to see it happen that pi and e get some kind of relationship. [/quote]
$e^{\pi i}+1 = 0$
 :rotfl:[/quote]

Yeah, but that's just the identity $e^{i\theta}= \mbox{cis}(\theta)$, and it's imaginary, so it really doesn't count for me. I want a real relationship.

You know how "realm" starts with "real"? Well, my realm ends there, too. :wink:

[quote="ZetaX"]What do you have against imaginaries¿ (real numbers are dumb  :P )
And you should define what you want, otherwise I give you $\left(e^\pi\right)^{2}= e^{2 \pi}$. ;)[/quote]

He means something involving only reals which links the two transcendentals together computationally.  In $e^{\pi i}=-1$, is true that the 2 are linked because of a mixture of calculus and trigonometry, but are the two numbers linked in any other way (if one is being raised to the other's power, and there's an i thrown in there, it's not so computational any more).

Yeah, I want something along the lines of $e^{6}= \pi^{4}+\pi^{5}$, if only that were true...

And real numbers are the backbone of society.

[quote="K81o7"] In $e^{\pi i}=-1$, is true that the 2 are linked because of a mixture of calculus and trigonometry, but are the two numbers linked in any other way (if one is being raised to the other's power, and there's an i thrown in there, it's not so computational any more).[/quote]

$e^{\pi i}=-1$ is true because of the axioms of arithmetic and the definitions of $e, \pi, i$. Calculus and trigonometry are just names to keep things organized in our heads and have no other meaning. Why should the labels associated with that formula when you learned it matter so much? Indeed I also first learned that expression in a calculus context, but now I think of it as more related to algebra!

Observe that $a, b, c, n \in \mathbb{Z}, n > 2, a^{n}+b^{n}= c^{n}\Rightarrow 0 \in \{a, b, c\}$ describes a computational relation between integers (much simpler than transcendental numbers!). But look what it takes to prove that!

Also, observe Xevarion's Identity:

$\pi = 4\sum_{n=0}^\infty (-1)^{n}e^{\left(\sum_{k=1}^\infty (-1)^{k+1}\frac{1}{k}\left(\frac{-2n}{2n+1}\right)^{k}\right)}$

[hide=":D :D"] pwned! [/hide]

Exactly.  Calculus and trigonometry ARE names of things.  But the things they represent have meaning.  And I remember those only because I understand hwo to prove the identity, and I realize that it uses both of those, and so it is not so apparent at first sight.  Maybe use something WITHOUT imaginary numbers, as has been asked?

I'm assuming you made up that identity.  It's interesting, but still, it's a limit of an infinite series, so even if it is true, you have $e$ raised to an exponent which is an infinite series, the nature of which is not so obvious...basically we want you to show a relationship between $e$ and $\pi$ without introducing more transcendentals...

Okay, some fun stuff..From my old books/notes  about Ramanujan. Ramanujan  found a fun number, of course it is [color=darkblue] [b][size=150]744 [/size][/b][/color]

Now check this out,

  $744+e^{\sqrt{20}\pi }\approx 2^{3}(25+13\sqrt 5)^{3}$
  $744+e^{\sqrt{52}\pi }\approx 30^{3}(231+9\sqrt{13})^{3}$
 $744+e^{\sqrt{148}\pi }\approx 60^{3}(2837+468\sqrt{37})^{3}$

There are MANY such examples!! Something very special about 744.

Of course, the most famous ones are: (I mentioned it above)
$e^{\sqrt{163}\pi}-744 \approx 640320^{3}$  (the diff is about   ($7.5 \cdot 10^{-13}$   or accurate  about 30 significant figures!)

And the Best of all:
$e^{\sqrt{427}\pi}-744 \approx 5280^{3}(236674+30303\sqrt{61})^{3}$
The diff is about $10^{-23}$ (Accuracy - more than 50 significant figures)

(Special thing about $744$ is there are Many numbers $n$  such that
${e^{\pi \sqrt n}\pm 744}$ is very close to cube of a  solution of quadratic equation!

(One develops a special respect for Ramanujan to find such interesting relationships when there were no Mathematica/Maple/TI-89's)


trigonometry
geometry
AMC
AIME
geometric transformation
rotation
MATHCOUNTS

I spent like 30 minutes on #12 just to simplify this huge gradical mess... still ended up getting it wrong because I miscopied a sqrt3 as a sqrt2(so i ended up being off by 1 on the coefficients of sqrt2 and sqrt3)

simplifying 20*(20/sin45)*sin75*sin60/2 - (20 - 5/sin15)^2 tan75/2 is not fun!

[quote="Altheman"]there are some cool cyclic quadrilaterals...they just might be useful  :wink:[/quote]

I've notice that too, but do they lead to a solution with decent amount of calculation?

i drew the diagram wrong on the test... :| after i did all the rationalizing...like a page for what i did...i looked back to reread just before i was going to put my answer down...and i was like - that...so i ono...i get something that seems pretty quick...you get two triangles that can be found with the sine area formula...i was also thinking of something with the carpet formulas...but i did not explore that at all

I did the two triangles thing, but uh, looked easy at first. Then... a page and a half of working with radicals and multiplying gigantic radical expressions. Yuck.

anyone?

If the AIME committee does not have a solution with a moderate amount of computation, this problem could potentially be THE WORST problem in the AIME history.

[quote="Altheman"]let B', C' be the images of B, C, respectively

let B'C' intersect AC and BC at C'' and A'', and AB' intersect BC at B''

we want [AA''B'']+[A''C''A]

note that CC'AA'' and A''B'BA are cyclic

hence a similarity,  [AA''B'']/[BB'B'']={AB''/B''B}^2=tan^2 75

[A''C''A]/[CC'C'']={AC''/C'C''}^2=sin^2 30/ sin^2 15

it should be clear how to find [BB'B'']=[AB'B]-[AB''B], etc[/quote]'

the calculations is still quite long...

(20tan15*20/2 - 20/sin15 * 20/cos15 *1/2)* tan^2 75 is already a pain, and [CC'C''] is even harder to find.

The total amount of calculation appears to be more than

20*(20/sin45)*sin75*sin60/2 - (20 - 5/sin15)^2 tan75/2, which requires a multiplication of two trinomials and one binomial --> leaves tons of rooms for computational mistakes.

[quote="K81o7"]People, suck it up.  Don't be afraid of a little computation, and don't blame the people who made the problems for your careless mistakes.  I'm sorry if I sound harsh, but everyone has to learn this sometime...you have to understand that being able to do the problems doesn't completely dictate your score: you can't take anything for granted.

Here's what I did...the common area is the area of the triangle's rotation, minus the section of the triangle adjacent to the y-axis and sitting on the original triangle, minus the small right triangle formed near the bottom right vertex of the rotated triangle.  Using a little of the absinC/2 formula,

Area=$100(2+\sqrt{3})(1-\frac{\sin{15^{\circ}}}{\sin{45^{\circ}}}-2(1-\cos{15^{\circ}})^{2})$
That isn't so bad at all.[/quote]

we are not afraid of computation; it is just that a lot of compution makes the problem bad in comparison to an elegant problem with little computation

[quote="kyyuanmathcount"]I devoted a few pages of brute force to this one, and it was worth it, because by some stroke of luck, it turned out right.

Without trig:

Basically...um, just solve for all your points and lines and brute force to get intersections then use shoestring.   :oops:  :blush:  :lol:

Yep, it was pretty bad.[/quote]

typical mathcounts
I tried doing that, although I used some trig, but I would have gotten it right had I realized B' was outside ABC.

[quote="beta"][quote="K81o7"]People, suck it up.  Don't be afraid of a little computation, and don't blame the people who made the problems for your careless mistakes.  I'm sorry if I sound harsh, but everyone has to learn this sometime...you have to understand that being able to do the problems doesn't completely dictate your score: you can't take anything for granted.

Here's what I did...the common area is the area of the triangle's rotation, minus the section of the triangle adjacent to the y-axis and sitting on the original triangle, minus the small right triangle formed near the bottom right vertex of the rotated triangle.  Using a little of the absinC/2 formula,

Area=$100(2+\sqrt{3})(1-\frac{\sin{15^{\circ}}}{\sin{45^{\circ}}}-2(1-\cos{15^{\circ}})^{2})$
That isn't so bad at all.[/quote]

Usually AIME problems only needs at most half a page to a page of calculation(with my normal hand writing), this one took like more than 2 pages, which suggests that I probably did not find the most efficient way to do it.[/quote]

Often there's a problem like that on the AIMe...and I don't think the above method takes 2 pages...

[quote="zhou_qiang"]Notice that sin 15= sin(60-45). Then use compound angle.[/quote]

Before I started any computation, I just wrote down, using the half-angle formula,

$\sin{15^{\circ}}=\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$
$\cos{15^{\circ}}=\frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

at the top of the page.  Once I found the method I demonstrated before, it took me just what was remaining on the page of another problem.

[quote="kyyuanmathcount"]I devoted a few pages of brute force to this one, and it was worth it, because by some stroke of luck, it turned out right.

Without trig:

Basically...um, just solve for all your points and lines and brute force to get intersections then use shoestring.   :oops:  :blush:  :lol:

Yep, it was pretty bad.[/quote]

Just wondering.....did your solution actually involve anything with angles? I tried that but it was too long

[quote]DPatrick (19:45:53)
Yes it was. There doesn't seem to be any way of avoiding long calculations. [/quote]

There's a solution in the mathjam transcript, but it's not as "nice" as you'd like I'm betting.

[quote="sapphyre571"]No one has a complete solution, do they?[/quote]

AoPSWiki has $3\frac{1}{2}$ full solutions [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_12]here[/url]. But all of them still involve a lot of multiplication of trinomials, etc. (and took a long time to type up)  :( .

[quote="sapphyre571"]I just typed up a solution to post here, but I accidentally closed the tab, so I'm not going to type it up again, but it took me less than half a page of computation, so it wasn't that bad (other than in involved finding $(\tan75^\circ)(1-\sin75^\circ)^{2}$, but I know sin75 and tan75 off the top of my head)[/quote]

That’s the area of the little $\triangle$ in the bottom right corner, perhaps that was being subtracted from the $45-60-75 \triangle$?

If so, its solution 1 on the link in my previous post.


[quote="yanglei1979"]i want to study at here.

how  to input Mathematical formula?
     [code]\frac{\pi}{4}[/code][/quote]

um add the dollar signs and take out the code part
like this $ \sin{\frac{\pi}{4}}\equal{}\frac{\sqrt{2}}{2}$

Why take out the code part? They're actually helpful in giving a nicer looking expression. Certainly $ \sin$ looks better than $ sin$.

To be more specific and clear, we just add $ \$$ signs on both ends of the code we want to put in.
Example: given your code
[code]\frac{\pi}{4}[/code] just add $ \$$ signs at both ends:
[code]$\frac{\pi}{4}$[/code]
and you get this:
$ \frac {\pi}{4}$

You can also check [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]here[/url] to learn more codes, especially for characters like $ \alpha$, $ \sum$, $ \prod$, and much more.

The rest is simple to learn independently.  :) (although I'm struggling to learn independently how to create asymptote graphs on these forums  :( )

[quote="Patterns_34"](although I'm struggling to learn independently how to create asymptote graphs on these forums  :( )[/quote]

Same here, practice helps though  :) .

I've finally figured out how to draw lines!  :D

I have several questions: how do you draw curves or circles (and is there a list of commands for these scenarios similar to [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]this[/url] for latex?)?

[b][Edit][/b] I apologize for going off topic for a moment.

This might be deviating off topic so try to keep Asymptote questions in the LaTeX and Asymptote forum.

As for codes, I find that [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]this[/url] helps.

Remember, as with LaTeX, the codes are often experimentable (if that's even a word), so Asymptote is all about trying out codes you can quote from other users.


While eating out, Mike and Joe each tipped their server $ 2$ dollars. Mike tipped $ 10\%$ of his bill and Joe tipped $ 20\%$ of his bill. What was the difference, in dollars between their bills?

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 20$

[hide]
Mike's bill:

$ 2\equal{}.1m$
$ 20\equal{}m$

Joe's bill:
$ 2\equal{}.2j$
$ 10\equal{}j$

$ 20\minus{}10\equal{}\boxed{10, (D)}$

[/hide]


Do any of you know when the Garfield Annual Mathematics Extravaganza is going to take place this year?  When does registration end?  How much does registration cost?  And any other general information.  Thanks

P.S. are there any links giving information about the GAME

i think its on next saturday like the 8th. I still haven't decided on whether to participate or not.

April 8th.
Registration ends at the competition: please send rough numbers and tell us whether you're coming before that, but you can register until a few minutes before it starts.
$\$$40/team.
 There are 4 rounds: purple (short answer), white (mc), team (short answer), and bulldog (team vs. team oral).
Our goal is to have hard enough problems that even the top students are seperated by several points (to make arithmetic mistakes matter less), although we've added a few easier/more typical problems this year.
There's no website, sorry.
Incidentally, middle schoolers could participate too, if you're insane enough.

-Adam Hesterberg

How many people are in a team? And are there different divisions for people in different math levels like Mu Alpha Theta?

usually 4 per team

no there arent any divisions, just 11/12 or 9/10...

its a fun competition :)

im pretty sure i'm going

[quote]Incidentally, middle schoolers could participate too, if you're insane enough.[/quote]

Hesterberg says that after making third place team as an 8th grader.

P.S. Washington's team got 4th this year.


An enormous party has an infinite number of people. Each two people either know or dont know each other. Given a positive integer n, prove there are n people in the party such that either they all know each other, or nobody knows each other (so the first possibility means that if A and
B are any two of the n people, then A knows B, whereas the second possibility means
that if A and B are any two of the n people, then A does not know B.

This is not very interesting -- it follows immediately from Ramsey's Theorem, for example.  Slightly more interesting is to do one of the following:

1) Show that for every such party, there is an [i]infinite[/i] clique or anti-clique, or
2) Give a coloring in which there are arbitrarily large finite cliques or anti-cliques but no infinite cliques or anti-cliques.

(A clique is a set of people who pairwise know each other and an anti-clique is a set of people who pairwise don't know each other.)

Also, it's extremely likely that this has appeared on the forum before.  Try a search for "Ramsey."


group theory
abstract algebra
linear algebra
matrix
geometry
3D geometry
superior algebra

Let $ G$ be the special linear group $ SL_2(5)$ of $ 2 \times 2$ matrices of determinant $ 1$ over the field $ \mathbb{F}_5$ of integers modulo $ 5$, so that the arithmetic in $ G$ is modulo $ 5$. Show that $ G$ is a group of order $ 120$. [b](I've done this by considering $ ad \minus{} bc \equal{} 1$ and the multiplication table $ \mod \,5$)[/b]

Find a subgroup of $ G$ of order $ 8$ isomorphic to $ Q_8$, and an element of order $ 3$ normalizing it in $ G$. Deduce that $ G$ has a subgroup of index $ 5$, and obtain a homomorphism from $ G$ to $ S_5$. Deduce that $ SL_2/\{\pm I\}$ is isomorphic to the alternating group $ A_5$. Show finally that $ SL_2(5)$ has no subgroup isomorphic to $ A_5$.

For the second part, I can only do the first line: I have the subgroup of $ G$ generated by $ I, \begin{pmatrix} 0 & 1 \\ 4 & 0\end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 3\end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 2 & 0\end{pmatrix}$, but I can't find the element or order $ 3$ normalizing this subgroup. Hints for this and subsequent parts are appreciated :)

I'll use your generators, but re-ordered a little bit.

Let i=2 mod 5 be a square root of negative one.  Then set I = [i,0;0,-i] and J = [0,1;-1,0].  Then over any ring containing i and not of characteristic 2, < I, J > is a quaternion group of order 8.

Set X = [-1-i,-1+i;1+i,-1+i]/2.  Then X^3 = 1, I^X = J, and J^X = -I*J, so X is an element of order 3 in the normalizer of <I,J>.  <X,I,J> is isomorphic to SL(2,3).

A subgroup isomorphic to A5 would have index 2, and so give a non-identity abelian quotient.  However SL(2,5) is perfect.

[quote="JackSchmidt"]Set X = [-1-i,-1+i;1+i,-1+i]/2.  Then X^3 = 1, I^X = J, and J^X = -I*J, so X is an element of order 3 in the normalizer of <I,J>.  <X,I,J> is isomorphic to SL(2,3).[/quote]

How would you think of this set $ X$? o.0 Also, what does I^X mean - a matrix to the power of a matrix?

[quote="JackSchmidt"]Set X = [-1-i,-1+i;1+i,-1+i]/2.  Then X^3 = 1, I^X = J, and J^X = -I*J, so X is an element of order 3 in the normalizer of <I,J>.  <X,I,J> is isomorphic to SL(2,3).[/quote]

[quote="aidan"]How would you think of this set $ X$?[/quote]

X is a 2x2 matrix, not a set.  I don't understand your question.  I just looked for an automorphism of Q8, then looked for a matrix in GL(2,C) inducing it.

[quote="aidan"]Also, what does I^X mean - a matrix to the power of a matrix?[/quote]

A^B = B^-1 * A * B is conjugation of A by B.

One wants I*X = X*J, which gives four equations, two of which are superfluous.  Setting X=[a,b;c,d] one gets:  0=b+ai=bi-a=d-ci=-c-di.

One also wants J*X = -X*I*J, so 0=ib+c=ia+d.  In particular, a=bi=-c=di, and X is now determined up to a scalar.  One chooses the scalar so the resulting matrix has order 3, that is, one chooses a to be some cube root of 1/(2+2i), like (-1-i)/2.

[quote="JackSchmidt"][quote="aidan"]But the group <I, J> has 8 elements. Why must it be that I*X = X*J and J*X = -X*I*J?[/quote]

You could be asking several different questions.  I'll restate the possible questions and answer each one.  Feel free to ignore the ones you did not intend to ask.

Where did the equation come from?

The equations say that I^X = X^-1*I*X = J and J^X = X^-1*J*X = -I*J.

Ok, where did these equations come from?

These equations say that g → g^X is the automorphism of Q8 that takes i to j to -k, (i,j,-k)(-i,-j,k).  This is clearly an automorphism of order 3 of Q8.

Why did you choose that particular automorphism?

Random choice amongst the possibilities.  It seems pretty enough to me.

Why did you choose randomly?  Why did you think that would work?

The real answer is: why bother doing anything more formal when this just works?  Maybe I would have done something more clever if it failed, but it worked.  However some people like to make up reasons for things:

The group <I,J> is the quaternion group of order 8.  It is not a mystery; it is a tiny concrete object which can be understood very thoroughly.  Its automorphism group is isomorphic to the symmetric group on 4 points: Aut(Q8) = S4.  Its outer automorphism group is isomorphic to the non-abelian group of order 6, Out(Q8) = Aut(Q8)/Inn(Q8) = S4/V4 = S3, Inn(Q8) = Q8/Z(Q8) = V4.  This group S3 has only one subgroup H of order 3, so if Q=<I,J> is contained as a normal subgroup in G, then either G has no element acting as an automorphism of order 3 on Q or it has the full pre-image of that unique subgroup H worth of automorphisms of order 3.

Either my equations for X can be solved, or there is no element acting as an automorphism of order 3 on Q8.

Granted your original question could have allowed X to centralize Q8, but that is a very dull situation that only happens for this matrix representation when the field contains a primitive 3rd root of unity (which F5 does not).[/quote]

I actually meant to ask the last 2 questions; this pretty much clears everything up  :lol: Thanks a lot!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let H be a subgroup of G, and let a be in G. Show that aHa^-1 is a subgroup of G that is isomorphic to H

You can see easily that $f: H\longrightarrow aHa^{-1}$ whcih sends $h$ to $aha^{-1}$ is an isomorphism.


I need a hint on these two problems:
1. Show that $[(2+\sqrt3)^n]$ is odd for all n>0
2. Prove that $(\frac{n}{3})^n<n!<(\frac{n}{2})^n$ for all n>5

Edit: Problem 1 should be the floor of it.

Something is definitely lacking in number 1 -- that thing isn't integral for n > 0, so it definitely isn't odd.  Do you perhaps mean the floor of it?

For the second one, that's just like Sterling's approximation, which gives $n! \approx \left( \frac{n}{e} \right) ^n$, but obviously that isn't sufficient to solve it.  For the upper bound, [hide]try splitting n! as $\left( 1\cdot n \right) \cdot \left( 2 \cdot (n-1) \right) \cdots$ and see what you can do from there.[/hide]

For (1)   consider  the expression  (2+ :sqrt: 3)^n +  (2- :sqrt: 3)^n
For (2) Use induction. ,,,, Remember ratio of  (n+1)! /n! is just (n+1)

[hide]You notice   (and can easily prove) that it is (The expression (2+:sqrt:3)^n +(2-:sqrt:3)^n ))

a)   A positive integer:
b)    the integer is even.

[hide]Think  of Binomial terms,   of  (2+ :sqrt: 3)^n    all  "odd" terms cancel out with (2- :sqrt: 3)^n  and what is left all all integers counted twice) [/hide]

Now   the BIG HINT:   (2- :sqrt: 3)^n  is less than 1
[/hide]
QED

For the   second easiest to use Induction:   [hide]Again Hint:   All you have to prove that (1+1/n)^n  is >2 and less than 3.
(Assume it is true for n! and try to prove it for n+1 by multiplying both sides by n+1[/hide]

[quote="Gyan"]For the   second easiest to use Induction:
Again Hint:   All you have to prove that (1+1/n)^n  is >2 and less than 3.
(Assume it is true for n! and try to prove it for n+1 by multiplying both sides by n+1[/quote]

I can prove that (the limit is e) but I don't see how it helps.
I'm stuck trying to prove that $\(frac{n+1}{3})^n+1<(n+1)(n/3)^n$ and vice versa on the other side.

I got the first one.  It also shows why, if you switch sqrt(3) with sqrt(5) it alternates odd, even, odd, even.  And why the decimal part approaces 0.

[b](1+1/n)^n  is greater than 2 and less than e (and e is less than 3) .
[/b]
Easiest is to expand it:

1+ n*(1/n) +   (1/2!) n(n-1)/(n^2)+ ....
=   1 + 1  + (1/2) (1)((1-1/n) + 1/(3!) (1-1/n)(1-2/n) + ...
[b]
Obviously it is greater than 2 [/b]
  writing,  noticig  3! > 2^2, 4!> 2^3 etc  (That is  .((1- a/n) < 1  and n! > 2^(n-1)))

S < 1 + 1 + 1/2 + 1/4 + 1/8  + 1/16 + .... [b] < 3.[/b]

Here are the  details,  if you want to see it,  for the  Induction proof.
[hide]Now if we assume  for n it is true that

[b] n^n/(2^n)  >   n!   > n^n /3^n  [/b]

(Multiply left side by (1+1/n)^n / 2  which is greater than 1, and right side by (1+1/n)^n/3 which is less than 1) and we will still have:

((1+1/n)^n * n^n) /( 2^(n+1)) > n!  > ((1+1/n)^n * n^n /(3^(n+1)
or  (n+1)^n /(2^(n+1)) > n! > (n+1)^n / (3^(n+1))

Mulitply both side by (n+1) and notice  (n+1)! = (n+1)*n!

we get  [b](n+1)^(n+1) /(2^(n+1)) > (n+1)! > (n+1)^(n+1) / (3^(n+1))[/b]

So it is true for n+1 too, but a  quick check shows that  it is true for n=5  so it is true for all n>5  too.

Hope this helps.

[/hide]

Yeah, thanks, I got it now.  I read your thing about a step at a time until I figured out where I was stuck.  I was stuck at here: $(1+1/n)^n*n^n=(n+1)^n$.  It's not complicated, really, but I guess I just wasn't connecting it.

[quote="Gyan"]Glad,  it was helpful.

[quote]Isn't (1 + 1/n) ^ n
n -> oo e?  [/quote]
Yes, the  limit is e. For any given n   (finite)  the value is less than e.[/quote]
*Sigh* Finally, took me a while to find this link
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=12913&highlight=#12913[/url]
It's an old link. It may help a bit, I think.

[quote="Ragingg"]Gyan... How do you prove that (2- :rt3: )^n is large enough to change [(2+:rt3:)^n+(2-:rt3:)^n] from an even to an odd?[/quote]

(2- :sqrt: 3) is less than 1, (and yet greater than zero) . Any power of this would be less than 1 (yet greater than zero)   Now   If  take an even integer  say K (Notice: [(2+:rt3:)^n+(2-:rt3:)^n]  is  EXACTLY an even integer]  and subtract a quantiity less than 1, the quantity would be something like [(K-1) + x]  where 0<x<1   hence K-1 is odd.

Ohh... Right right.. I forgot that it was exactly an even integer.  Thanks!


For how many two-digit prime numbers is the sum of its digits 8?

Such a number would have to be some combination of 0-8, but 0 can't be the first number.

Beginning with 1, we have 17, which is prime.
With 2, it would have to be 26 but that's not prime.
With 3, we would have 35 but that's no good.
With 4, 44, but not prime.
Continuing, 53 is prime, but 62 isn't, and 71 is.

So we have 17, 53, and 71.

ya, but you stopped one step too early

it asks how many, and you gave us a list of numbers
the answer to the question is 3

(but you could probably figure that out, just wanna put it in black and white)

Yes, I'm assuming that the people on this site can draw the connection that the set {17, 53, 71} has three elements.


induction
pigeonhole principle
analytic geometry
number theory unsolved
number theory

Let $n$ be a positive integer. A $corner$ is a finite set $S$ of ordered $n$-tuples of positive integers such that if $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ are positive integers with $a_k \geq b_k$ for $k = 1, 2, \ldots, n$ and $(a_1, a_2, \ldots, a_n) \in S$, then $(b_1, b_2, \ldots, b_n) \in S$. Prove that among any infinite collection of corners, there exist two corners, one of which is a subset of the other one.

it is in my book . if noone ..., i will post solution? :ninja:

A point $(a_{1}, a_{2}, \ldots a_{n})$ is said to pwn a point $(b_{1}, b_{2}, \ldots b_{n})$ if $a_{i}\ge b_{i}$ for all $i$.

Suppose that there is an infinite set $S$ of corners such that no corner is the subset of another. We call such a set evil; any infinite subset of an evil set is clearly evil. We will construct an infinite sequence $T_{k}$ of points such that $T_{i}$ pwns $T_{j}$ only if $i \le j$. To aid our construction, we will also define a sequence $S_{k}$ of evil sets and a sequence $C_{k}\in S_{k}$ of corners.

We will define $S_{k}$ inductively, with $S_{1}= S$. Suppose that we have already defined $S_{k}$ as well as $T_{i}$ for $1 \le i \le k-1$. Further suppose that for each $i \le k-1$, there is no element in a corner in $S_{k}$ which pwns $T_{i}$. This is vacuously true in the base case when $k = 1$.

We choose $C_{k}$ to be an arbitrary element of $S_{k}$. Let $T_{k}$ be an element of $C_{k}$ which is not contained in infinitely many other corners in $S_{k}$. Such a $T_{k}$ exists since $S_{k}$ is evil. By the induction hypothesis, $T_{k}$ does not pwn $T_{i}$ for any $i \le k-1$.

We now define $S_{k+1}$ to be the set of all corners $C \in S_{k}$ such that $T_{k}\not\in C$. We know that $S_{k+1}$ is evil since it is infinite and $S_{k+1}\subset S_{k}$. Obviously, no element in a corner $C \in S_{k+1}$ pwns $T_{k}$, or else $T_{k}$ would be an element of $C$. Also, since $S_{k+1}$ is a subset of $S_{k}$, no element in a corner in $S_{k+1}$ pwns $T_{i}$ for $i \le k-1$.

This completes our inductive definitions, so that we have in fact constructed a sequence $T_{k}$ such that $T_{i}$ pwns $T_{j}$ only if $i \le j$.

But now we will show that the existence of this sequence is contradictory. Indeed, we construct an infinite subsequence $X_{k}$ of $T_{k}$ such that for no $i \ne j$ does $X_{i}$ pwn $X_{j}$. This can be done, since if we have picked the first $a$ points in $X_{k}$, where $X_{a}= T_{b}$, then note that the points $X_{1}, X_{2}, \ldots X_{a}$ collectively pwn only a finite number of points, so that there must be some $c > b$ such that $X_{i}$ does not pwn $T_{c}$ for any $i \le a$. Then choose $X_{a+1}$ to be $T_{c}$, and continue ad infinitum in a similar manner.

However, it is not possible to have an infinite set of lattice points no two of which pwn each other. This can be proven by induction on number of dimensions. It's trivial in dimension 1. Suppose it holds for $n-1$ dimensions. Assume that there is an infinite set $X$ of points no two of which pwn each other in $n$ dimensions.

Let $(a_{1}, a_{2}, \ldots a_{n})$ be a point in $X$. Then every other element $(x_{1}, x_{2}, \ldots x_{n})$ in $X$ must satisfy $x_{i}\le a_{i}$ for at least one $i$. Then by pigeonhole, there is some $i$ such that infinitely many elements of $X$ have $i$th coordinate at most $a_{i}$. Then by pigeonhole, there is some $a \le a_{i}$ such that infinitely many elements of $X$ have $i$th coordinate equal to $a$. However, this is not possible, since it implies that there exists an infinite set of points in $n-1$ dimensions, no two of which pwn each other, which there doesn't.

Thus we have reached a contradiction, and no evil set exists.

We can extend the result: in any infinite set of corners, there exist infinitely many corners, each of which is a subset of infinitely many other corners.  We can similarly induct on the number of dimensions: if each corner is denoted by a foremost point (ie the point which pwns all other poitnts in the corner, by probability's jargon), then we either have that,

in each of the n places (the components, the dimensions), we hit infinitely many values as we vary over the foremost points, in which case we can create an elimination scheme to make, for any point in the list, an infinite list of successively pwning points, or

we have some place where there are infinitely many points which have the same value in that dimension, and then the problem reduces to one of $n-1$ dimensions.

This is just an outline of the solution, but, reading it over, I realize it makes little sense if you haven't thought about it...so i can post the whole thing if needed.


On a triangle $ABC$ with the usual side lengths $a,b,c$, we draw circles centered at each of $A$, $B$, and $C$, with radii $r_{1}$, $r_{2}$, and $r_{3}$ respectively. Given the restrictions:
\[a \geq r_{2}+r_{3}, \ \ b \geq r_{1}+r_{3}, \ \ c \geq r_{2}+r_{3}\]
find the radius $r$ of the largest circle that is tangent to all three circles, and whose center is contained within $\triangle ABC$.

[quote="Altheman"]kissing circles problem[/quote]
But that's when they're all tangent to one another. In this problem, the three initial circles might not be tangent.

yea i kno i just had the wrong name...it is on wolfram though

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


hello ,I am a newbie to group theory and I just tried some random very basic problems and in many cases I am doubtful whether my solutions are right ,so I just want to see generally how you all solve them .
Let $ G$ be a nonempty set closed under an associative product ,which in addition satisfies
(a) There exists an $ e \in G$ such that $ a.e\equal{}a$ for all $ a \in G$
 (b) Give $ a \in G$ there exists an element $ y(a) \in G$ such that $ a.y(a)\equal{}e$.
prove that $ G$ must be a group under this product .

That's the definition of a group.  A group is a set with a binary operation $ (G,\circ)$ such that $ G$ is closed under a binary operation, $ G$ is associative under $ \circ$, there is an identity element $ e$ such that $ g\circ e\equal{}e\circ g\equal{}g$ and every element has an inverse $ g^{\minus{}1}$ such that $ g\circ g^{\minus{}1}\equal{}g^{\minus{}1}\circ g\equal{}e$.

It is not the definition of the group. The question is how to prove that $ ea\equal{}a$ and $ y(a)a\equal{}e$, i.e. the commutativity of the identity and inverse element.

$ ay(a)y(y(a))$ can be seen in 2 different ways. Grouping the last 2 terms, we have it equal to $ a$, and grouping the first 2 terms, we have it equal to $ y(y(a))$. Therefore, $ a\equal{}y(y(a))$.

From (b), $ y(a)y(y(a))\equal{}e$, which according to the above gives $ y(a)a\equal{}e$. Therefore, $ y(a)$ commutes with $ a$.

$ ea\equal{}ay(a)a\equal{}ae$, so the identity commutes as well.

Cheers,

Rofler

this problem is from IN Herstein, Topics in Algebra, sec 2.3, q12. there is another part to this problem, which is:
Prove that the conclusion of this problem is false if we assume
(b') Given $a\in G$, there exists $y(a)\in G$ such that $y(a)\cdot a=e$
instead of condition (b).


number theory
prime numbers
number theory unsolved

Let $p_{1},p_{2}$ be any prime numbers, for any $n>1$, define $p_{n+1}$ as the maximum prime divisor of $p_{n}+p_{n-1}+2000$. Prove that the sequence $\{p_{n}\}$ is bounded.

Let $C=max\{p_{1}, p_{2}, 2003\}!+1$
We will prove inductively that $p_{n}\leq C$ and the proof will be finished.
The first step is obvious. So, suppose that $p_{n}\leq C$, $p_{n+1}\leq C$ and both of $p_{n}, p_{n+1}$ are odd. Then $p_{n+1}+p_{n}+2000$ is even and it's obvious that:
$p_{n+2}\leq \frac{1}{2}(p_{n+1}+p_{n}+2000) \leq \frac{1}{2}(C+C+2000) = C+1000$
But we can't have $C+1 \leq p_{n+2}\leq C+1000$ since the numbers from $C+1$ to $C+1000$ are all composite (as one can easily verify). So $p_{n+2}\leq C$.
Now, suppose that for example $p_{n}=2$ and $p_{n+1}$ is odd (other case is obvious). Then of course:
$p_{n+2}\leq p_{n+1}+2+2000 \leq C+2002$
Analogously we conclude that $p_{n+2}\leq C$ and the proof is completed.


trigonometry
inequalities proposed
inequalities

[quote="luke999"]if 1<=x_1,x_2,...,x_n<=-1 than proove that
    \[ x_1 \sqrt{1-x_2^2}+x_2 \sqrt{1-x_3^2}+...+x_n \sqrt{1-x_1^2} \leq \frac{n}{2}  \][/quote]
  I think you've done a mistake.It must be 1>$\ x_i$ ;)

In this case we can use trigonometry, taking $\ x_i$=$\ cos^2$ $\ A_1$...

[quote="Lovasz"][quote="luke999"]if $x_i \in [-1,1],\ i=\overline{1,n}$ then prove that:
    \[ x_1 \sqrt{1-x_2^2}+x_2 \sqrt{1-x_3^2}+...+x_n \sqrt{1-x_1^2} \leq \frac{n}{2}  \][/quote]
This trivial, AM-GM only.[/quote]
  how? $x_i$ can be negative?

Of course,
\[ ab \le \frac{a^2+b^2}{2},\ \ \forall a,b \in \mathbb{R}  \]

well we use that x_k=sin y_k
so let's say

\[ x_k\sqrt{1-x_{k+1}^2}= sin x_k cos x_{k+1} \]
....

That's my idea! :D We use $\ sin^2 a$+$\ cos^2 a$=1.

well...it's my problem and my solution  :D

Sorin have u been to the NMO???And what class are u??

I'm seventh grade, but unfortunately I haven't been to NMO because my mother didn't let me to go to Iasi.I'm from Slatina and Iasi is too far.I took the second place in my district.But mom... :( Don't remind me about NMO!It's too paithfull! :stretcher:


linear algebra
matrix
graph theory
\/closed

Can I get a program of Hamiltonian Cycle Problem somewhere..??

What do you mean, the "Hamiltonian cycle problem"? Do you want a program that takes an adjacency matrix (or other representation of a graph) and outputs whether it has a Hamiltonian cycle?


Find the positive numbers $ x,y,z$ such that

$ x\plus{}y\plus{}z\equal{}1$ and $ \displaystyle\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\equal{}\frac{x\plus{}y}{y\plus{}z}\plus{}\frac{y\plus{}z}{x\plus{}y}\plus{}1$

If $ x\equal{}y\equal{}z$ then $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$.

If $ y\equal{}z$ then $ x\equal{}1\minus{}2y$ and substituting, I get $ 10y^{3}\minus{}2y^{2}\minus{}6y\plus{}3\minus{}0$ which seems to have one non-integral real root.

Hope this is of some help.

hello, i think $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$ is the only positive real solution.
Sonnhard.

maybe this will help;

Notice that the fractions of the $ LHS$ are the mediants of the fractions on the $ RHS$. We write

$ \left(\frac{x}{y}\plus{}\frac{y}{z} \minus{} \frac{x\plus{}y}{y\plus{}z}\right)\plus{}\left(\frac{z}{x}\plus{}\frac{y}{y} \minus{} \frac{y\plus{}z}{x\plus{}y}\right)\equal{}2$

$ \frac{z}{y\plus{}z}\cdot\left(\frac{x}{y}\right) \plus{} \frac{y}{y\plus{}z}\cdot\left(\frac{y}{z}\right) \plus{} \frac{y}{y\plus{}x}\cdot \left( \frac{z}{x} \right) \plus{} \frac{x}{y\plus{}x}\equal{}2$

This looks nicer, like some weighted AM-GM perhaps. (but it's late, so i'll come back to this tomorrow)

Okay, here is the full solution

[hide="solution"]
Notice that the fractions on the $ RHS$ are the mediants of the fractions on the $ LHS$, we write

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \equal{} \frac {x \plus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{x \plus{} y} \plus{} 1$

$ \therefore \left(\frac {x}{y} \plus{} \frac {y}{z} \minus{} \frac {x \plus{} y}{y \plus{} z}\right) \plus{} \left(\frac {z}{x} \plus{} \frac {y}{y} \minus{} \frac {y \plus{} z}{x \plus{} y}\right) \equal{} 2$

$ \therefore \frac {z}{y \plus{} z}\cdot\left(\frac {x}{y}\right) \plus{} \frac {y}{y \plus{} z}\cdot\left(\frac {y}{z}\right) \plus{} \frac {y}{y \plus{} x}\cdot\left(\frac {z}{x}\right) \plus{} \frac {x}{y \plus{} x} \equal{} 2$

Now we substitute $ a \equal{} \frac {x}{y}, b \equal{} \frac {y}{z}, c \equal{} \frac {z}{x}$ and we have $ abc \equal{} 1$

$ \therefore \frac {a}{b \plus{} 1} \plus{} \frac {b^2}{b \plus{} 1} \plus{} \frac {c}{a \plus{} 1} \plus{} \frac {a}{a \plus{} 1} \equal{} 2$

Sub in $ c \equal{} \frac {1}{ab}$ and then multiply though $ (a \plus{} 1)(b \plus{} 1)$

$ (a \plus{} b^2)(a \plus{} 1) \plus{} \frac {(a^2b \plus{} 1)(b \plus{} 1)}{ab} \equal{} 2(a \plus{} 1)(b \plus{} 1)$

$ a^2 \plus{} b^2 \plus{} ab^2 \plus{} a \plus{} ab \plus{} a \plus{} \frac {1}{a} \plus{} \frac {1}{ab} \equal{} 2(ab \plus{} a \plus{} b \plus{} 1)$

$ \therefore (a \minus{} b)^2 \plus{} (ab \minus{} \frac {1}{ab})^2 \plus{} \frac {1}{a}(ab \minus{} 1)^2 \equal{} 0$

Equality occurs when $ a \equal{} b$ and $ ab \equal{} 1$ therefore $ a \equal{} b \equal{} 1 \longrightarrow c \equal{} 1$

Therefore we have $ x \equal{} y \equal{} z$, and from our condition $ x \plus{} y \plus{} z \equal{} 1$ we get $ (x,y,z) \equal{} \left(\frac {1}{3},\frac {1}{3},\frac {1}{3}\right)$ as the only solution  :) [/hide]

I think it should be $ (a\minus{}b)^2\plus{}(\sqrt{ab}\minus{}\frac1{\sqrt{ab}})^2\plus{}(\sqrt{ab^2}\minus{}\sqrt{1/a})^2\equal{}0$...
The square roots are fine here since the problem asks for positive real numbers, but would further analysis be required if the problems asks for any real numbers (so $ a,b,c$ could be negative)


Please explain how to get the answer for this question. THanks.

If a + b = 1 and a^2 + b^2 = 2, then what is a^4  + b^4?

can this method be applied to $x^{n}+y^{n}$?


modular arithmetic
number theory
least common multiple

Find the last three digits of $ 8^{3^{2008}}$.


Let $P$, $Q$, $R$ be points on the sides $AB$, $BC$, $CA$ of a triangle $ABC$ such that $AP=CQ$ and the quadrilateral $RPBQ$ is cyclic. The tangents to the circumcircle of triangle $ABC$ at the points $C$ and $A$ intersect the lines $RQ$ and $RP$ at the points $X$ and $Y$, respectively. Prove that $RX=RY$.

The triangles $APY,CQR$ have pairs of equal angles: $\angle PAY=\angle QCR,\ \angle APY=\angle CQR$, hence they are similar. Since we also have $AP=CQ$ they must be congruent, so $PY=RQ\Rightarrow RY=RP+PY=RP+RQ$. In the same way we show that $RX$ is also equal to $RP+RQ$.


conics parabola algebra function domain inequalities quadratics

Let $ a,{\rm{b,c}} \in \left[ {{\rm{0;1}}} \right].$
Find Max value of
$ P\equal{}(a\minus{}b)^4\plus{}(b\minus{}c)^4\plus{}(c\minus{}a)^4$

Let $ a \minus{} b \equal{} x,\ b \minus{} c \equal{} y$, we have $ c \minus{} a \equal{} \minus{} (x \plus{} y)$.
$ \therefore P \equal{} x^4 \plus{} y^4 \plus{} (x \plus{} y)^4 \equal{} (x^2 \plus{} y^2)^2 \minus{} 2x^2y^2 \plus{} (x \plus{} y)^4$.
Let $ x \plus{} y \equal{} s,\ xy \equal{} t$,
$ P \equal{} (s^2 \minus{} 2t)^2 \minus{} 2t^2 \plus{} s^4 \equal{} 2s^4 \minus{} 4s^2t \plus{} 2t^2 \equal{} 2t^2 \minus{} 4s^2t \plus{} 2s^4 \equal{} 2(t \minus{} s^2)^2$.
Since $ a,\ b,\ c\in[0,\ 1]\Longrightarrow |x|\leq 1,\ |y|\leq 1,\ |x \plus{} y|\leq 1$, yielding $ D \equal{} \{(s,\ t)|\ |s|\leq 1, |t|\leq 1,\ s^2\geq 4t\}$, which is described by shaded region including boarder line.
Suppose that for $ s,\ t\in D,\ t \minus{} s^2$ would take some value $ k$, considering the condition for which the parabola $ t \equal{} s^2 \plus{} k$ have more than one intersecion point with the domain $ D$ on $ s \minus{} t$ plane gives $ \minus{} 2\leq k\leq 0$, yielding $ 0\leq k^2\leq 4$. Therefore $ 0\leq P\leq 8$, that is to say, $ P_{max} \equal{} 8$ when $ k \equal{} \minus{} 2\Longleftrightarrow (s,\ t) \equal{} (\pm 1, \minus{} 1)$ by the attached graphs.$ (a \minus{} b,\ b \minus{} c) \equal{} \left(\frac {1 \plus{} \sqrt {5}}{2},\ \frac {1 \minus{} \sqrt {5}}{2}\right),\ \left(\frac {1 \minus{} \sqrt {5}}{2},\ \frac {1 \plus{} \sqrt {5}}{2}\right), \left(\frac { \minus{} 1 \plus{} \sqrt {5}}{2},\ \frac { \minus{} 1 \minus{} \sqrt {5}}{2}\right), \left(\frac { \minus{} 1 \minus{} \sqrt {5}}{2},\ \frac { \minus{} 1 \plus{} \sqrt {5}}{2}\right)$.

U said $ x\equal{}a\minus{}b \le 1$
But $ \frac{1\plus{}\sqrt{5}}{2} >1$ !!!.
Sorry, I'm not good about English :(.

The expression $ P$ is symmetric in $ a,b,c$, so we can WLOG assume that $ a\leqslant b\leqslant c$. Let $ x \equal{} b \minus{} a$, $ y \equal{} c \minus{} b$. Then $ x,y\geqslant 0$, $ x \plus{} y\leqslant 1$ and
$ P \equal{} x^{4} \plus{} y^{4} \plus{} \left(x \plus{} y\right)^{4}$.
By the Karamata inequality applied for the convex function $ f\left(u\right) \equal{} u^{4}$,
$ x^{4} \plus{} y^{4}\leqslant \left(x \plus{} y\right)^{4}$
with equality for $ x \equal{} 0$ or $ y \equal{} 0$, since $ \left(x,y\right)\prec\left(x \plus{} y,0\right)$. Thus
$ P\leqslant 2\left(x \plus{} y\right)^{4}\leqslant 2$
with equality for $ x \plus{} y \equal{} 1$. Therefore $ \max P \equal{} 2$ and the maximum is reached if one of $ a,b,c$ is $ 0$ and the other two are $ 1$ or one is $ 1$ and the other two are $ 0$.

Yes, the answer is 2, Rzeszut. :)

Here is my edited solution. I can't edit the solution posted in #2.

Let $ a \minus{} b \equal{} x,\ b \minus{} c \equal{} y$, we have $ c \minus{} a \equal{} \minus{} (x \plus{} y)$, we have
$ \therefore P \equal{} x^4 \plus{} y^4 \plus{} (x \plus{} y)^4 \equal{} (x^2 \plus{} y^2)^2 \minus{} 2x^2y^2 \plus{} (x \plus{} y)^4$.
Let $ x \plus{} y \equal{} s,\ xy \equal{} t$,
$ P \equal{} (s^2 \minus{} 2t)^2 \minus{} 2t^2 \plus{} s^4 \equal{} 2s^4 \minus{} 4s^2t \plus{} 2t^2 \equal{} 2t^2 \minus{} 4s^2t \plus{} 2s^4 \equal{} 2(t \minus{} s^2)^2$.
Since $ a,\ b,\ c\in[0,\ 1]\Longrightarrow |x|\leq 1,\ |y|\leq 1,\ |x \plus{} y|\leq 1$.
Consider the quadratic function of $ u$ ; $ \displaystyle f(u) \equal{} u^2 \minus{} su \plus{} t \equal{} \left(u \minus{} \dfrac{s}{2}\right)^2 \minus{} \frac {1}{4}s^2 \plus{} t$,
$ x,\ y$ are the roots of the quadratic equation : $ f(u) \equal{} 0$. Thus
$ \left\{\begin{array}{ll} f( \minus{} 1)\leq 0 & \quad \\
f(1)\leq 0 & \quad \\
Discriminant\geq 0 & \quad \\
the\ axis : \displaystyle \minus{} 1\leq \frac {s}{2}\leq 1 , & \quad \end{array} \right.$ Combining the condition : $ |x \plus{} y|\leq 1\Longleftrightarrow |s|\leq 1$, the condition for which $ s,\ t$ should satisfy is as follows.
$ |s| \minus{} 1\leq t\leq \frac {1}{4}s^2$ and $ |s|\leq 1$.Let $ D \equal{} \{(s,\ t)| |s| \minus{} 1\leq t\leq \frac {1}{4}s^2$ and $ |s|\leq 1 \}$, which is described by
shaded region including boarder line.Especially the line $ t \equal{} |s| \minus{} 1$ is the tangent lines at $ (s,\ t) \equal{} (\pm 2,\ 1)$ on the parabola $ \displaystyle t \equal{} \frac {1}{4}s^2$.
Suppose that for $ s,\ t\in D,\ t \minus{} s^2$ would take some value $ k$, considering the condition for which the parabola $ t \equal{} s^2 \plus{} k$ have more than
one intersecion point with the domain $ D$ on $ s \minus{} t$ plane gives $ \minus{} 1\leq k\leq 0$, yielding $ 0\leq k^2\leq 1$. Therefore $ 0\leq P \equal{} 2k^2\leq 2$.
$ \therefore P_{max} \equal{} 2$ when $ k \equal{} \minus{} 1\Longleftrightarrow (s,\ t) \equal{} (0, \minus{} 1)$ by the attached graphs.That is to say, $ (s,\ t) \equal{} (0, \minus{} 1)\Longleftrightarrow x \plus{} y \equal{} 0$ and $ xy \equal{} \minus{} 1$
$ \Longleftrightarrow (x,\ y) \equal{} ( \minus{} 1,1),\ (1, \minus{} 1)$, yielding $ (a \minus{} b,\ b \minus{} c,\ c \minus{} a) \equal{} ( \minus{} 1,\ 1,\ 0),\ (1, \minus{} 1,0)$.

[quote="bbbbbbbbbb"]Let $ a,{\rm{b,c}} \in \left[ {{\rm{0;1}}} \right].$
Find Max value of
$ P \equal{} (a \minus{} b)^4 \plus{} (b \minus{} c)^4 \plus{} (c \minus{} a)^4$[/quote]
This is my solution .[u]WLOG[/u] assum $ 1\geq a\geq b\geq c\geq 0$
we have $ P\equal{}(a\minus{}b)^4\plus{}(b\minus{}c)^4\plus{}(c\minus{}a)^4\equal{}2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^2$,so we prove that:
$ A\equal{}a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\leq 1$
from  $ 1\geq a\geq b\geq c\geq 0$ we have $ (a\minus{}b)(1\minus{}b)\leq 0\Rightarrow b^2\minus{}ab\leq b\minus{}a$
and $ c^2\minus{}bc\leq 0$,so it's remain to prove that
$ a^2\minus{}ac\plus{}b\minus{}a\leq 1$ which trivial because $ a^2\minus{}a\minus{}ac\leq 0,b\leq 1$
SO $ max_P\equal{}2$  :lol:


geometry
circumcircle
trigonometry
national olympiad

1. Let $n$ be a positive integer with $n \geq 2$, and $0<a_{1}, a_{2},...,a_{n}< 1$. Find the maximum value of the sum
$\sum_{i=1}^{n}(a_{i}(1-a_{i+1}))^{\frac{1}{6}}$
where $a_{n+1}=a_{1}$

2. Find the smallest positive real $k$ satisfying the following condition: for any given four DIFFERENT real numbers $a,b,c,d$, which are not less than $k$, there exists a permutation $(p,q,r,s)$ of $(a,b,c,d)$, such that the equation $(x^{2}+px+q)(x^{2}+rx+s)=0$ has four different real roots.

3. In $\triangle PBC$, $\angle PBC=60^{o}$, the tangent at point $P$ to the circumcircle$g$ of $\triangle PBC$ intersects with line $CB$ at $A$. Points $D$ and $E$ lie on the line segment $PA$ and $g$ respectively, satisfying $\angle DBE=90^{o}$, $PD=PE$. $BE$ and $PC$ meet at $F$. It is known that lines $AF,BP,CD$ are concurrent.
 a) Prove that $BF$ bisect $\angle PBC$
 b) Find $\tan \angle PCB$

4. Assuming that the positive integer $a$ is not a perfect square, prove that for any positive integer n, the sum ${S_{n}=\sum_{i=1}^{n}\{a^{\frac{1}{2}}\}^{i}}$ is irrational.

5. Let $S=\{n|n-1,n,n+1$ can be expressed as the sum of the square of two positive integers.$\}$. Prove that if $n$ in $S$, $n^{2}$ is also in $S$.

6. $AB$ is a diameter of the circle $O$, the point $C$ lies on the line $AB$ produced. A line passing though $C$ intersects with the circle $O$ at the point $D$ and $E$. $OF$ is a diameter of circumcircle $O_{1}$ of $\triangle BOD$. Join $CF$ and produce, cutting the circle $O_{1}$ at $G$. Prove that points $O,A,E,G$ are concyclic.

7. Let $k$ be a positive integer not less than 3 and $x$ a real number. Prove that if $\cos (k-1)x$ and $\cos kx$ are rational, then there exists a positive integer $n>k$, such that both $\cos (n-1)x$ and $\cos nx$ are rational.

8. Give a positive integer $n\geq 2$, let $B_{1},B_{2},...,B_{n}$ denote n subsets of set $X$, which each $B_{i}$ contains exactly two elements. Find the minimum value of $|X|$ such that for any choose of $B_{1},B_{2},...,B_{n}$, there exists a subset $Y$ of $X$ such that:
 (1) $|Y|=n$
 (2) $|Y \cap B_{i}|\leq 1$ for $n=1,2,...,n$

I am lucky that I get a Gold with 87/120

Great!  :lol:

I've posted the problems. Here are the links:

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=118084
Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=118085
Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=118088
Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?t=118086
Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?t=118090
Problem 6: http://www.mathlinks.ro/Forum/viewtopic.php?t=118089
Problem 7: http://www.mathlinks.ro/Forum/viewtopic.php?t=118087
Problem 8: http://www.mathlinks.ro/Forum/viewtopic.php?t=118091

Please kindly put this into the "Resources" part to make it a full set of all the CWMO questions yayayay~~ thx

Edit: this is the China WMO 2006

[quote="brianchung11"]The result of the official best question award is very unusual, so why don't we vote for the best question of this CWMO paper in mathlinks?[/quote]

Which is the official best problem?


What is the difference between escape velocity and rotational velocity?  Can someone provide me with an equation for both?

i'm assuming by "rotational" velocity you mean orbital velocity.  normally rotational velocity refers to angular velocity.  the escape velocity of an object in orbit is the square root of 2 times its orbital velocity.

Do you have sample equations for both velocities?

Escape velocity is

$v_e=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2g}{R}}$

where M is the mass and R the radius of Earth, g gravitational acceleration and G is the gravitational constant. Orbital velocity is

$v_e=\sqrt{\frac{GM}{h}}$

where h is the radius of the orbit.

Exactly what I was looking for.  Thanks.


[color=blue][b]One cable has $ d\equal{}3.5mm$. The fiber has its own $ d\equal{}70micro m$. How many fibers inside the cable????????????=))[/b] :roll:[/color]

I believe it has more than one possible. :wink:

Is it asking for the maximum amount of fibers?

$ \left(\frac {3.5\cdot 10^{ \minus{} 3}}{70\cdot 10^{ \minus{} 6}}\right)^2\cdot\frac {\pi\sqrt{3}}{6}\approx 2267.$
[url=http://mathworld.wolfram.com/CirclePacking.html]Circle Packing[/url]

[b][img]http://i42.photobucket.com/albums/e316/wlp01/circle.jpg[/img]

Big circle diameter:$ D = 3,5 mm = 3,5.10^{ - 3} m, R = 1,75.10^{ - 3} m$
Small circle diameter: $ d = 70 micro m = 70.10^{ - 6} m, r = 35.10^{ - 6} m$

1) Assuming there aren’t any spaces between small circles, then you can just divide the areas and rounding down the result to estimate the number of small circles that can be fit inside the big circle.

2) Another idea is to assume that small circles fit orderly inside the big circle, there will be $ 25$ rounds or levels of small circles stacking up inside the big circle.

$ \frac {R}{d} = 25$

To find out how many small circles fit inside the outer level, you would calculate the circumference of the level circle and divide by the diameter of the small circle.

Circumference of outer level $ = \Pi.d_1$ (where $ d_1$ is $ R - r$)

# of small circle fit inside the outer level would be $ = \frac {\Pi.d_1}{d}$

Continue with the other $ 24$ levels of small circle and add up the answers.

3) You can find an estimate of the total space inside the big circle after fitting all the small circles by doing step 2 and figure out the total area of the all small circles, and then subtract that from the area of the big circle.

4)  Area of the space between 3 circles:

[img]http://i42.photobucket.com/albums/e316/wlp01/circle1.jpg[/img]

A = Area of Equilateral Triangle $ = \frac {s^2\sqrt {3}}{4}$, where s is the side of the e-triangle.
B = Area of Circle Sector $ = \frac{\del}{360\pir^2}$

Area of space between there circles = A – 3B

In this case: $ s = d = 70 micro m = 70.10^{ - 6} m$
----------------------------------------------------------------

I hope this is helpful[/b]


I installed once the TeXnicCenter , but i had problems with the computer and i had to install it again. Since i installed it long time ago when i tried to install it again i think i made some mistakes (i downloaded first the Miktex,but i think i forgot to install it and then i installed TeXnicCenter) and it doesn't have any output profiles so the build function doesn't work. I added some output profiles that i remembered but it still doesn't work. Now i deleted all the folders,files, etc. about TeXnicCenter and Miktex and i tried to download and install them in the correct order as in the guide from [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]here[/url], but it has the same problem. I had some projects saved in the computer, but i dunno how this may influence the TexnicCenter. Should i delete these projects too and try again? Or what shall i do?

Sorry if i ask a stupid question and sorry for my bad english. Thanks for your help.

Ensure that MiKTeX is installed. Then tell TexnicCenter where to find MiKTeX.
From [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Help#I.27ve_upgraded_MiKTeX_and_now_TeXnicCenter_won.27t_work]here[/url]
[quote]You need to tell TeXnicCenter where MiKTeX has moved to. If you are using MiKTeX 2.7 it will probably be C:\Program Files\MiKTeX 2.7\miktex\bin (this folder contains lots of exe files including latex.exe). You do this by clicking on Build, Define Output Profiles and clicking on the Wizard button in the bottom left corner. [url=http://www.youtube.com/watch?v=-h8Oi1oCTQg]This video[/url] will show you what to do.[/quote]


Is it possible to arrange 1,2,3,...,41 in a line such that for every two consecutive integers in the line, the sum of them is a prime?

I hate these kind of problems where you have to find examples.....

Consider the following arrangement: $ 1,40,3,38,5,36,7,34,...,39,2,41$.

The sum of every two consecutive terms is either $ 41$ or $ 43$, which are both primes.


analytic geometry
geometry
geometry proposed

Could anybody prove this?

Given a triangle $\triangle ABC$ and three points $P_{1}, P_{2}, P_{3}$ on its plane. Let $P_{1}=\left(x_{1},y_{1},z_{1}\right)$, $P_{2}=\left(x_{2},y_{2},z_{2}\right)$, $P_{3}=\left(x_{3},y_{3},z_{3}\right)$ be their normalized barycentric coordinates. Then

$\left[P_{1}P_{2}P_{3}\right]=\left[ABC\right]\cdot\left| \begin{array}{ccc}x_{1}& y_{1}& z_{1}\\ x_{2}& y_{2}& z_{2}\\ x_{3}& y_{3}& z_{3}\\ \end{array}\right|$.

See the end of [url=http://mathworld.wolfram.com/BarycentricCoordinates.html]this article[/url].


Consider the set of all equations $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$, where $ a_2$, $ a_1$, $ a_0$ are real constants and $ |a_i| < 2$ for $ i \equal{} 0,1,2$. Let $ r$ be the largest positive real number which satisfies at least one of these equations. Then

$ \textbf{(A)}\ 1 < r < \frac{3}{2}\qquad
\textbf{(B)}\ \frac{3}{2} < r < 2\qquad
\textbf{(C)}\ 2 < r < \frac{5}{2}\qquad
\textbf{(D)}\ \frac{5}{2} < r < 3\qquad \\
\textbf{(E)}\ 3 < r < \frac{7}{2}$

$ \minus{} 2 < a_0 < 2$
$ \minus{} 2 < a_1 < 2$
$ \minus{} 2 < a_2 < 2$

Suppose that $ 3 \le r$ then there are $ a_i$ for $ 27 \plus{} 9a_2 \plus{} 3a_1 \plus{} a_0 \equal{} 0$ however $ \minus{} 26 \equal{} \minus{} 18 \minus{} 6 \minus{} 2 < 9a_2 \plus{} 3a_1 \plus{} a_0 < 18 \plus{} 6 \plus{} 2 \equal{} 26$
then $ r < 3$
Take for example $ r \equal{} \frac {8}{3}$ then
$ \frac {8^3}{3^3} \plus{} \frac {8^2}{3^2} a_2 \plus{} \frac {8}{3}a_1 \plus{} a_0 \equal{} 0$

$ 512 \plus{} 192 a_2 \plus{} 72 a_1 \plus{} 27 a_0 \equal{} 0$
with solution for $ a_2 \equal{} a_1 \equal{} \minus{} 1,9$, getting $ a_0 \equal{} \frac { \minus{} 512 \plus{} 364,8 \plus{} 136,8}{27} \equal{} \minus{} \frac {10,4}{27}$
therefore $ \frac {8}{3}\le r$ the answer is $ (D)$

Letting $a_0 = a_1 = a_2 \approx -2$ will let you put the largest values (relative to the outputs) into the polynomial. So trying 3 gives 1 and trying $\frac{5}{2}$ gives a negative, so the largest value must be in this range

so D


la ce clasa e propusa problema ?
solutie :
[hide]notam $\sum x_{i}^{5}=\sum x_{i}^{6}=s$.
din inegalitatea mediilor cu puteri avem
$\frac{\sum x_{i}}{10}\leq\sqrt[5]{\frac{s}{10}}\leq\sqrt[6]{\frac{s}{10}}$
dar prima inegalitate ne spune ca $s\geq 10$, iar a doua ne spune ca $s\leq 10$, deci $s=10$, si cum avem egalitate, rezulta ca toate numerele sunt egale, deci $x_{1}=x_{2}=\ldots=x_{10}=1$.[/hide]

Eu n-am stiut sa o fac. :( Dar dl Marius Ghergu a propus-o pentru clasa a VII-a.

pai d`asta am intrebat
solutia mea nu e chiar la nivelul clasei a VII`a, si sunt convins ca exista si o alta solutie

[quote="stancioiu sorin"]Determinati numerele $\ x_i,i=1,10$ stiind ca
$\sum x_i=10$
$\sum x_i^5=\sum x_i^6$.[/quote]
[hide="Hint (pentru o solutie la nivelul clasei a VII-a)"]Adunam 10 la ambii membri ai egalitatii $\sum x_i^5=\sum x_i^6$. [/hide]

Uitati care e soluita dansului:
$\sum\ (x_i^6-x_i^5-x_i-1)=0$, deci $\sum\ (x_i-1)^2(x_i^4+x_i^3+x_i^2+x_i+1)=0$ Pe de alta parte avem
$\ x_i^4+x_i^3+x_i^2+x_i+1=\left(x_i^2+\frac{x_i}{2}\right)^2+$
$\ +\left(\frac{x_i\sqrt 3}{2}+\frac{x_i}{\sqrt 3}\right)^2+\frac{2}{3}>0$, deci $\ x_1=x_2=...=x_{10}=1$
   O solutie extraordinar de frumoasa!Seamana putin cu hintul dat de dl M.Damian.


geometry
circumcircle
geometric transformation
homothety
incenter
geometry proposed

[quote="sunken rock"]Let ABC be a triangle and O its circumcircle. A circle O' is drawn inside the circle O, tangent at A' to O and tangent to AB and AC.
Similarly we get B' and C'.
Then AA', BB' and CC' have a common point.

Best regards,
Sunken rock

A blind man sees better the details[/quote]
Obviously their intersection point is the point of homothety that turns incircle to circumcirlce.

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=201342]here[/url] for the costruction.

Let the tangential triangle of the mid-arc triangle, the homothetic center of this and ABC is the point where concur AA', BB' and CC' and it stay on the line OI.

Regarding the construction of circle O', I have a different idea:
1) It is a well known fact by now that the perpendicular through I at AI cuts AB, AC at the points of contact of this circle inscribed into the triangle made by AB, AC and the arc BC of the circumcircle, so this circle is "constructible", we know 2 points on it and 2 tangent lines.
2) Further more, the line joining the middle of the arc BC containing A of the circumcircle and the incenter of ABC cuts the circumcircle in the tangency point between circles O and O' ( see http://www.mathlinks.ro/viewtopic.php?t=202329)

Best regards,

sunken rock

For more developpement, you can see
http://perso.orange.fr/jl.ayme   vol. 5  A new mixtilinear incircles adventure I
Sincerely
Jean-Louis

[quote="luisgeometra"]$A$ is the exsimilicenter of the incircle $(I)$ of $\triangle ABC$ and the A-mixtilinear incircle. $A'$ is the exsimilicenter of $(O)$ and $(O_a)$. Hence, by Monge & d'Alembert theorem, it follows that $AA'$ cuts $IO$ at the exsimilicenter $X_{56}$ of $(I),(O)$. Similarly, lines $BB',CC'$ go through $X_{56}.$[/quote]


If the following three numbers are placed in ascending order, which number would be second? $ 5^{56}$, $ 7^{49}$, $ 13^{35}$

Yes, there is a non-caculator solution, but I forgot what it was.  All I can remember now is that you divide one by the other, then somehow simplify it.  It actually isn't too difficult.

[quote="AIME15"]Reduce these to $ 5^8, 7^7, 13^5$.  It's a workout problem, so calculators are allowed.

$ 5^8\equal{}390625$, $ 7^7\equal{}823543$, $ 13^5\equal{}371293$, so the answer is $ \boxed{5^8}$.

Interested if there's a non-calculator solution.[/quote]

No, the correct answer is $5^56.


modular arithmetic
abstract algebra
number theory unsolved
number theory

Prove that:
g is a generator of Fp* if and only if g^(p-1) = 1 (mod p) and gq ≠ 1 (mod p) for all prime divisors q of (p – 1).

Request help, thanks.

There are two mistakes in your post.

1) If you meant $ gq \neq 1 \pmod{p}$, the statement is false: example $ \mathbb{F}_7$ with $ g \equal{} 2$.
We have $ 2^6 \equal{} 1 \pmod{7}$, $ 2\cdot 2 \neq 1 \pmod{7}$ and $ 2\cdot 3 \neq 1 \pmod{7}$, but $ 2^3 \equal{} 1 \pmod{7}$, so $ 2$ is not a generator of $ \mathbb{F}_7^*$.
This must have been a typo, so we'll correct it to $ g^q \neq 1 \pmod{p}$.

2) The second mistake is that it must happen for all proper (different from $ p \minus{} 1$ itself) divisors $ q$ of $ p \minus{} 1$, not just the prime divisors $ q$. An example is $ \mathbb{F}_{17}$; take a generator $ g$, then $ (g^2)^{16} \equal{} 1 \pmod{17}$ and $ (g^2)^2 \neq 1 \pmod{17}$, with $ 2$ being the only prime divisor of $ 17 \minus{} 1 \equal{} 16$, but $ g^2$ is not a generator, since $ (g^2)^8 \equal{} 1 \pmod{17}$ (a simpler example is to take $ g \equal{} 1$, and you are not allowed to use $ q \equal{} 1$, since $ 1$ is not a prime).

So the correct equivalent statement is "$ g^{p \minus{} 1} \equal{} 1 \pmod{p}$ and $ g^q \neq 1 \pmod{p}$ for any proper divisor $ q$ of $ p \minus{} 1$"
($ g^{p \minus{} 1} \equal{} 1 \pmod{p}$ is only needed in fact in order to remove the possibility $ g \equal{} 0$, since any non-zero element obeys it).
If $ g$ is a generator, then the statement above is trivial, by the definition of the generator. Conversely, if $ g$ obeys the conditions of the statement above, since its order must be a divisor of $ p \minus{} 1$ by Lagrange's theorem, it follows it can only be $ p \minus{} 1$, hence $ g$ is a generator.

You have to be careful with both avoiding typos, and using the proper terms - mathematics is much less tolerant than common spoken language.

Well...you are right about the typo.
Did you mean that if q is only a prime divisor of (p-1), it cannot imply that g is a generator?

YES. Just take a closer look at point 2) above. In all generality, if $ r^3 \mid p \minus{} 1$ for some prime $ r$, let's consider $ a$ a generator of $ \mathbb{Z}_p^*$ and check $ g \equal{} a^r$. Clearly $ g^q \neq 1 \pmod{p}$ for any prime $ q \mid p \minus{} 1$ (including $ q \equal{} r$), but $ g$ is not a generator, since $ g^{(p \minus{} 1)/r} \equal{} 1 \pmod{p}$.


vector
linear algebra
linear algebra unsolved

basis of a vector space V is defined as a linearly independent subset of V, which spans V. But, till now, I have never read a text which addresses the question of existence of such a set. A possible way to prove it is this:

Let a1 be a non zero element of V. Let L1 be the linear span of a1. If there is no element of V, not in L1, then, a1 is a basis. If not, let a2 be an element of V, not spanned by a1. The set (a1,a2) is obviously linearly independent.
Let L2 be the linear span of (a1,a2). If there is no element of V, not spanned by (a1,a2), then, (a1,a2) is a basis. Else, we find an element a3, not spanned by a1 & a2, and consider the set (a1,a2,a3) and so on......

Now, my problem is, this process will surely give a basis if it terminates, but, what if it does not terminate?

It is clear that the process terminates if V has finite dimensions. But for an infinite dimensional space, it doesn't terminate. But, 'dimension' is defined after a basis. So, We cannot suppose that the process terminates at all.

That procedure won't work in infinite dimensions. For infinite-dimensional spaces, you need Zorn's lemma or an equivalent; consider the family $ S$ of all independent sets, partially ordered by inclusion. The union of a chain is an independent set, so $ S$ has a maximal element, which must be a basis.


trigonometry
inequalities
triangle inequality
number theory unsolved
number theory

[size=100]if $a,b,c$ are integer sides of a triangle with angle $C = 60 ^o$ and $a \geq b$. Prove that if $a$ is a prime then the triangle is equilateral.[/size]

Please use topic names that say what the thread deals with; "very hard problem" is subjective (yes my good old troll "pleasread"/"A cute angle", I partially agree ;) ) and doesn't tell other's the theme of it.
Also not an open problem.

Well by the cosine theorem we get $c^2=a^2+b^2-ab=a^2+b(b-a) \leq a^2$, so $c \leq a$.
Rearrange the equation to $a(a-b) = (c+b)(c-b)$. Since $a$ is prime, it divides one of the factors.
1. case: $a \mid b-c$. But because of $0 < b,c \leq a$ this gives $b=c$ and the rest is easy.
2. case: $a \mid b+c$.
With $b+c \leq 2a$ and the triangle inequality $a <b+c$ we get $b+c=2a$. Setting this into the cosine equation $a(a-b) = (c+b)(c-b)$ gives $a-b=2(c-b) \iff a+b = 2c$.
Now solve this two linear equations to get $a=b=c$.


geometry
circumcircle
inradius
analytic geometry
incenter

What is the distance between the centers of the incircle and circumcircle of a triangle with sides 3, 4, and 5?

[hide="My Solution"]Consider $ \triangle ABC$ in the Cartesian plane such that $ C \equal{} (0,0)$, $ B \equal{} (3,0)$, $ A \equal{} (0,4)$. If $ r$ is the inradius of $ \triangle ABC$, the coordinates of the incenter are $ (r,r)$. We have $ r \equal{} \frac{3 \plus{} 4 \minus{} 5}{2} \equal{} 1$, so let $ I \equal{} (1,1)$. The circumcenter of a right triangle is the midpoint of the hypotenuse, so we have $ O \equal{} \left(\frac{3}{2}, 2\right)$. Therefore, $ IO \equal{} \sqrt{\left(\frac{3}{2} \minus{} 1\right)^2 \plus{} (2 \minus{} 1)^2} \equal{} \boxed{\frac{\sqrt{5}}{2}}$.[/hide]

[hide="Alternatively"] We know that the area of the triangle is $ \frac{3(4)}{2}=6$. Thus, we have that $ rs=6 \implies 6r=6 \implies r=1$. In the same way, we have that $ \frac{abc}{4R}=6 \implies 60=24R \implies R=\frac52$. The distance from the incenter to tthe circumcenter in a triangle is given by

$ d=\sqrt{R(R-2r)}$.

Hence,

${ d=\sqrt{\frac52(\frac52-2)} \implies d=\sqrt{\frac54}=\boxed{\frac{\sqrt{5}}{2}}}$.[/hide]


Find a point on the y-axis that is equidistant from the point (5,-5) and (1,1).

I used the midpoint formula, but then realized that that point isn't on the y-axis.  Please tell me how to set up this problem.

[hide="Hint"]
If a point is equidistant from $(5,5)$ and $(1,1)$, then the point has to be...[hide]on the perpendicular bisector of the line segment whose endpoints are $(5,5)$ and $(1,1)$ :) [/hide] [/hide]

[hide="Hint"]Use distance formula. The point you are looking for is $(0,y)$. Equate distances and solve for $y$. :)[/hide]


linear algebra
matrix
superior algebra
superior algebra solved

it seems that the first one is n+1...

Denote the determinant of order n by D_n.
Expand the first row, we find that

D_n = 2D_{n-1} - D_{n-2}

Solve this recurrence relation using characteristic roots.
The only root of x^2-2x+1 = 0 is 1. hence,
D_n = \alpha_1 1^n + \alpha_2 n 1^n = \alpha_1 + \alpha_2 n.
for some constants \alpha_1 and \alpha_2. Then solve them using some initial conditions.

An interesting problem that is somehow related to the 1st problem Moubinool posted is:
COmpute the det:
|x  1   0  0 0 ...   0   0|
|1 2x  1  0 0 ...   0   0|
|0  1  2x 0 0 ...   0   0|
|..............................|
|0  0   0  0 0 ... 2x   1|
|0  0   0  0 0 ...  1  2x|
It should give you cos(narccos(x)).
cheers! :D :D

The nice determinant that Lagrangia posted is an occasion to post
an easy version, because there are indications.
the left top corner should be 2x and not x

x be real An(x) the matrix \in Mn(R)

[2x 1 0  0 ..................0]
[1  2x 1 0 ..................0]
[..................................]
[..................................]
[0 0  0  0 ................1 2x]

so the main diagonal of An(x) is (2x. 2x, ....,2x) with 2x written n times
the diagonal just above and under the main diagonal are
(1,1,....,1)  with 1 written n-1 times

Un(x)=det A_n(x)
1) Show that U_n(x)=2x.U_(n-1)(x)-U_(n-2)(x)
Give an expression of U_n(x) for |x|>1
Find U_n(1), U_n(-1)
2)For |x|<1 we introduce \alpha = Arccos x
Prove that U_n(cos \alpha) = sin((n+1)\alpha))/sin(\alpha)
Find the roots of the equation U_n(x)=0

how to pronouce oubinool
Mou bee (like be ) nool (like cool)