calculus
integration
trigonometry
geometry
geometric transformation
reflection
real analysis

Define $ K(k)\equal{}\int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1\minus{}k^2\sin^2 x}}$ .

Can you show $ K\left(\frac{\sqrt{6}\minus{}\sqrt{2}}{4}\right)\equal{}\frac{2^{\minus{}7/3}\, 3^{1/4}\, \Gamma^3\left(\frac13\right)}{\pi}$

and $ K\left(\frac{\sqrt{6}\plus{}\sqrt{2}}{4}\right)\equal{}\frac{2^{\minus{}7/3}\, 3^{3/4}\, \Gamma^3\left(\frac13\right)}{\pi}$ ?

Hey Doe John,

here's a passing thought:
\[ K(k)\equal{}\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sqrt{1\minus{}k^{2}\sin^{2}x}} \equal{} \int_{0}^{\frac{\pi}{2}} \sum_{n\equal{}0}^{\infty} (k \sin(x))^{2n} \ {n\minus{}1/2 \choose n} \ dx\]
\[ \implies 2K \equal{}  \sum_{n\equal{}0}^{\infty}   \ {n\minus{}1/2 \choose n} \ k^{2n} \frac{\Gamma(\frac{1}{2}) \Gamma(n \plus{} \frac{1}{2})}{\Gamma(n\plus{}1)} \equal{} \sum_{n\equal{}0}^{\infty}  {n\minus{}\frac{1}{2} \choose n}^2 \pi \ k^{2n}\]

This holds for $ |k| \leq 1$.

I have not put much thought into this last sum, but it seems reminiscent of ones that you have been discussing lately. Do you have an expression for it?

I found a way to compute the values with several substitutions.

For $ k\equal{}\sin\left(\frac{\pi}{12}\right)\equal{}\frac{\sqrt{6}\minus{}\sqrt{2}}{4}$ , thus $ k^2\equal{}\frac{2\minus{}\sqrt{3}}{4}$ ,

we want to compute $ 2\, K(k)\equal{}\int_0^\pi \frac{dx}{\sqrt{1\minus{}k^2 \sin^2 x}}$ .

[i]1.Substitution[/i] $ x\mapsto 2\arctan x$ :

$ 2\,K(k)\equal{}\int_0^\infty \frac{\frac{2\, dx}{1\plus{}x^2}}{\sqrt{1\minus{}\frac{2\minus{}\sqrt{3}}{4}} \left(\frac{2x}{1\plus{}x^2}\right)^2}$

$ \Rightarrow K(k)\equal{}\int_0^\infty \frac{dx}{\sqrt{(1\plus{}x^2)^2\minus{}(2\minus{}\sqrt{3}\,)\, x^2}}\equal{}\int_0^\infty \frac{dx}{\sqrt{x^4\plus{}\sqrt{3} \, x^2\plus{}1}}$

[i]2.Substitution[/i] $ x\mapsto\sqrt{x}$ :

$ K(k)\equal{}\frac12 \int_0^\infty \frac{dx}{\sqrt{x}\, \sqrt{x^2\plus{}\sqrt{3}\, x\plus{}1}}$

[i]3.Substitution[/i] $ x\mapsto\frac{x}{\sqrt{3}}$ :

$ K(k)\equal{}\frac12 \int_0^\infty \frac{\frac{dx}{\sqrt{3}}}{\frac{\sqrt{x}}{\sqrt[4]{3}}\, \sqrt{\frac{x^2}{3}\plus{}x\plus{}1}}
\equal{}\frac{\sqrt[4]{3}}{2}\int_0^\infty \frac{dx}{\sqrt{x^3\plus{}3x^2\plus{}3x}}$

$ \equal{}\frac{\sqrt[4]{3}}{2}\int_0^\infty \frac{dx}{\sqrt{(x\plus{}1)^3\minus{}1}}$

[i]4.Substitution[/i] $ x\mapsto x\minus{}1$ :

$ K(k)\equal{}\frac{\sqrt[4]{3}}{2}\int_1^\infty \frac{dx}{\sqrt{x^3\minus{}1}}$

[i]5.Substitution[/i] $ x\mapsto\frac{1}{x}$ :

$ K(k)\equal{}\frac{\sqrt[4]{3}}{2} \int_0^1 \frac{dx}{x^2\, \sqrt{\frac{1}{x^3}\minus{}1}}\equal{}\frac{\sqrt[4]{3}}{2} \int_0^1 \frac{dx}{\sqrt{x}\, \sqrt{1\minus{}x^3}}$

[i]6.Substitution[/i] $ x\mapsto x^\frac13$ :

$ K(k)\equal{}\frac{\sqrt[4]{3}}{2} \int_0^1 \frac{\frac13\, x^{\minus{}\frac23}\, dx}{x^\frac16\, \sqrt{1\minus{}x}}\equal{}\frac{3^{\minus{}\frac34}}{2} \int_0^1 x^{\frac16\minus{}1}\, (1\minus{}x)^{\frac12\minus{}1}\, dx$

$ \equal{}\frac{3^{\minus{}\frac34}}{2} \cdot B\left(\frac16,\frac12\right)\equal{}\frac{3^{\minus{}\frac34}}{2} \cdot \frac{\Gamma\left(\frac16\right)\, \Gamma\left(\frac12\right)}{\Gamma\left(\frac23\right)}$

[i]After Legendre's duplication formula we can replace[/i] $ \Gamma\left(\frac16\right)$ [i]by[/i] $ 2^\frac23\, \sqrt{\pi}\,\; \frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac23\right)}$ :

$ K(k)\equal{}2^{\minus{}\frac13}\cdot 3^{\minus{}\frac34}\cdot\pi\cdot \frac{\Gamma\left(\frac13\right)}{\Gamma^2\left(\frac23\right)}$

[i]After Euler's reflection formula[/i] $ \Gamma\left(\frac13\right)\,\Gamma\left(\frac23\right)\equal{}\frac{\pi}{\sin\frac{\pi}{3}}\equal{}\frac{2\pi}{\sqrt{3}}$

[i]we can replace[/i] $ \frac{1}{\Gamma^2\left(\frac23\right)}$ [i]by[/i] $ \frac{3}{(2\pi)^2}\Gamma^2\left(\frac13\right)$ :

$ K(k)\equal{}2^{\minus{}\frac73}\cdot 3^{\frac14}\cdot \pi^{\minus{}1}\cdot \Gamma^3\left(\frac13\right)$

For $ k\equal{}\cos\frac{\pi}{12}\equal{}\frac{\sqrt{6}\plus{}\sqrt{2}}{4}$ , and hence $ k^2\equal{}\frac{2\plus{}\sqrt{3}}{4}$ ,

we get with the same substitutions

$ K(k)\equal{}\int_0^\infty \frac{dx}{\sqrt{x^4\minus{}\sqrt{3} \, x^2\plus{}1}}\equal{}\frac{\sqrt[4]{3}}{2}\int_0^\infty \frac{dx}{\sqrt{x^3\minus{}3x^2\plus{}3x}}$

$ \frac{\sqrt[4]{3}}{2}\int_0^\infty \frac{dx}{\sqrt{(x\minus{}1)^3\plus{}1}}\equal{}\frac{\sqrt[4]{3}}{2}\int_{\minus{}1}^\infty \frac{dx}{\sqrt{x^3\plus{}1}}$ ,

where $ \int_{\minus{}1}^\infty \frac{dx}{\sqrt{x^3\plus{}1}}\equal{}\int_0^1 \frac{dx}{\sqrt{1\minus{}x^3}}\plus{}\int_0^\infty \frac{dx}{\sqrt{x^3\plus{}1}}$

$ \equal{}\frac13 B\left(\frac13,\frac12\right)\plus{}\frac13 B\left(\frac13,\frac16\right)\equal{}\frac{\sqrt{3}}{3} B\left(\frac12,\frac16\right)\equal{}\sqrt{3} \int_1^\infty \frac{dx}{\sqrt{x^3\minus{}1}}$.

So $ K\left(\cos\frac{\pi}{12}\right)\equal{}\sqrt{3}\, K\left(\sin\frac{\pi}{12}\right)$ .

Can you show

$ K(3\minus{}2\sqrt{2})\equal{}\frac{\sqrt{\pi}}{8}\cdot \frac{2\plus{}\sqrt{2}}{\Gamma^2\left(\frac34\right)}$ ?

Now I know how to show it. Set $ k\equal{}(\sqrt{2}\minus{}1)^2\equal{}3\minus{}2\sqrt{2}$ .

$ 2\, K(k)\equal{}\int_0^\pi \frac{dx}{\sqrt{1\minus{}k^2 \cos^2 x}}$ is after substitution $ x\mapsto 2\arctan x$

$ \int_0^\infty \frac{\frac{2\, dx}{1\plus{}x^2}}{\sqrt{1\minus{}k^2\, \left(\frac{1\minus{}x^2}{1\plus{}x^2}\right)^2}}\equal{}2\cdot\int_0^\infty \frac{1}{\sqrt{(1\plus{}x^2)^2\minus{}k^2 (1\minus{}x^2)^2}}$ ,

where $ (1\plus{}x^2)^2\minus{}k^2 (1\minus{}x^2)^2\equal{}4\cdot \sqrt{2}\cdot k\cdot (x^2\plus{}\sqrt{2}) \left(x^2\plus{}\frac{1}{\sqrt{2}}\right)$ .

So $ K(k)\equal{}\frac{1}{2\, \sqrt[4]{2}\, (\sqrt{2}\minus{}1)}\, \int_0^\infty \frac{dx}{\sqrt{(x^2\plus{}\sqrt{2}) \left(x^2\plus{}\frac{1}{\sqrt{2}}\right)}}$

Now we substitute $ x\mapsto \frac{1}{\sqrt[4]{2}}\, \tan x$ .

$ K(k)\equal{}\frac{1}{2\, \sqrt[4]{2}\, (\sqrt{2}\minus{}1)}\, \int_0^{\frac{\pi}{2}} \frac{\frac{dx}{\sqrt[4]{2}\cos^2 x}}{\sqrt{\left(\frac{1}{\sqrt{2}}\, \tan^2 x\plus{}\sqrt{2}\right)\left(\frac{1}{\sqrt{2}}\tan^2 x\plus{}\frac{1}{\sqrt{2}}\right)}}$

$ \equal{}\frac{1}{2\sqrt{2}\, (\sqrt{2}\minus{}1)}\, \int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1\plus{}\frac12 \tan^2 x}\, \cos x}\equal{}\frac{1\plus{}\sqrt{2}}{2\sqrt{2}}\, \int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{\cos^2 x\plus{}\frac12 \sin^2 x}}$

$ \equal{}\frac{2\plus{}\sqrt{2}}{4}\, \int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1\minus{}\frac12 \sin^2 x}}$ , where $ K\left(\frac{1}{\sqrt{2}}\right)\equal{}\frac12 \cdot \frac{\pi^\frac32}{\Gamma^2\left(\frac34\right)}$ .

So $ K(k)\equal{}\frac{\pi^\frac32}{8} \, \frac{2\plus{}\sqrt{2}}{\Gamma^2\left(\frac34\right)}$ . (I can't correct factor $ \sqrt{\pi}$ to $ \pi^\frac32$ in the last post)

Define $ E(k)\equal{}\int_0^{\frac{\pi}{2}} \sqrt{1\minus{}k^2 \sin^2 x} \, dx$ .

I know how to compute $ K(k)$ and $ E(k)$ for $ k\equal{}\frac{1}{\sqrt{2}}$ .

$ K\left(\frac{1}{\sqrt{2}}\right)\equal{}\int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1\minus{}\frac12 \sin^2 x}}$

$ \Rightarrow \frac{1}{\sqrt{2}}\, K\left(\frac{1}{\sqrt{2}}\right)\equal{}\int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{2\minus{}\sin^2 x}}\equal{}\int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1\plus{}\cos^2 x}}$

This is after substitution $ x\mapsto \arccos x$

$ \int_0^1 \frac{1}{\sqrt{1\plus{}x^2}}\, \frac{dx}{\sqrt{1\minus{}x^2}}\equal{}\int_0^1 \frac{dx}{\sqrt{1\minus{}x^4}}$ .

Then substitute $ x\mapsto x^\frac14$ to get

$ \frac14 \int_0^1 \frac{x^{\frac14\minus{}1}}{\sqrt{1\minus{}x}}\, dx\equal{}\frac14 B\left(\frac14,\frac12\right)\equal{}\frac14 \, \frac{\Gamma^2\left(\frac14\right)\,\sqrt{\pi}}{\sqrt{2}\, \pi}$ .

So $ K\left(\frac{1}{\sqrt{2}}\right)\equal{}\frac14 \, \frac{\Gamma^2\left(\frac14\right)}{\sqrt{\pi}}$ .

$ E\left(\frac{1}{\sqrt{2}}\right)\equal{}\int_0^{\frac{\pi}{2}} \sqrt{1\minus{}\frac12\sin^2 x}\, dx$

$ \Rightarrow \sqrt{2}\, E\left(\frac{1}{\sqrt{2}}\right)\equal{}\int_0^{\frac{\pi}{2}} \sqrt{2\minus{}\sin^2 x}\, dx\equal{}\int_0^{\frac{\pi}{2}} \sqrt{1\plus{}\cos^2 x}\, dx$

This is after substitution $ x\mapsto \arccos x$

$ \int_0^1 \frac{\sqrt{1\plus{}x^2}}{\sqrt{1\minus{}x^2}}\, dx\equal{}\int_0^1 \frac{1\plus{}x^2}{\sqrt{1\minus{}x^4}}\, dx$ .

Then substitute $ x\mapsto x^\frac14$ to get

$ \frac14 \int_0^1 \frac{x^{\frac14\minus{}1}\plus{}x^\frac12 \cdot x^{\frac14\minus{}1}}{\sqrt{1\minus{}x}} \, dx\equal{}\frac14 B\left(\frac14,\frac12\right)\plus{}\frac14 B\left(\frac34,\frac12\right)\equal{}\frac14 \frac{\Gamma^2\left(\frac14\right)}{\sqrt{2}\, \sqrt{\pi}}\plus{}\frac{\sqrt{2}\, \pi^\frac32}{\Gamma^2 \left(\frac14\right)}$ .

So $ E\left(\frac{1}{\sqrt{2}}\right)\equal{}\frac18 \frac{\Gamma^2\left(\frac14\right)}{\sqrt{\pi}}\plus{}\frac{\pi^\frac32}{\Gamma^2 \left(\frac14\right)}$ .

But how can I show

$ E\left(\frac{\sqrt{6}\minus{}\sqrt{2}}{4}\right)\equal{}\frac{2^\frac13\, \pi^2}{3^\frac34 \, \Gamma^3\left(\frac13\right)}\plus{}\frac{(\sqrt{3}\plus{}1)\, \Gamma^3\left(\frac13\right)}{2^\frac{10}{3}\, 3^\frac14\, \pi}$

and

$ E\left(\frac{\sqrt{6}\plus{}\sqrt{2}}{4}\right)\equal{}\frac{2^\frac13\, \pi^2}{3^\frac14 \, \Gamma^3\left(\frac13\right)}\plus{}\frac{(\sqrt{3}\minus{}1)\, \Gamma^3\left(\frac13\right)}{2^\frac{10}{3}\, 3^\frac14\, \pi}$ ?

Hey Doe John,
You're posts on this subject have been interesting. I have nothing of real value to add to this discussion but I am trying to figure out if there is any underlying structure that dictates which values of these elliptic integrals will give us a closed form solution. So far you have provided solutions for $ \cos(\frac{\pi}{12})$ and $ \sin(\frac{\pi}{12})$ for the complete Elliptic integrals of the first kind. Your solutions seem to hinge on particular substitutions in the integrals working out, but is there anything deeper than that going on?

Also, are there any other non trivial examples where this integral will work out to something nice as it does in the above mentioned cases?

Cheers,

Nick


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra theorems

Let $ F$ be a field, and let $ F^*$ be the multiplicative group consisting of the nonzero elements of $ F$. Let $ G$ be a subgroup of $ F^*$ of finite order. Show that $ G$ is cyclic.

The proof of this is essentially the same as what you can find in an elementary number theory textbook for the existence of primitive roots mod $p.$

Since $G$ is finite, any element of $G$ must have finite order, and in particular, if $x\in G,$ then $o(x)|o(G).$

If $x$ has order $m,$ then $x$ is a root of the equation $x^m-1=0$. But that equation has at most $m$ roots, of which at most $m-1$ are different from $1.$

Let $k(m)$ be the number of elements of $F^*$ of order exactly $m.$ If $k(m)>0$ then there exists an element $y$ of order $m$, and $o(y^j)=m$ iff $j$ is relatively prime to $m.$ Furthermore, the powers of $y$ are all of the roots of $y^m-1=0.$ That is, either $k(m)=0$ or $k(m)=\phi(m).$

Let $o(G)=n.$ Every element of $G$ has an order that divides $n.$ For any $m|n$, let $q(m)$ be the number of elements of $G$ with order exactly $m$. Clearly, $q(m)\le k(m).$

$n=\sum_{m|n} q(m) \le \sum_{m|n} k(m) \le \sum_{m|n}\phi(m)=n$

(The last equality being a well-known theorem in number theory.)

The only way this can be true is for $q(m)=\phi(m)\,\forall\,m|n$ and in particular, $q(n)=\phi(n)>0.$ There is an element of $G$ of order $n$ and thus  $G$ is cyclic.

Among the consequences of this: if we're looking for finite multiplicative subgroups of $\mathbb{C}$, then all we ever get is contained in the roots of unity.

It follows rather easily from the fact that for any $d|n$, where $n$ is the order of the group, there are at most $d$ solutions to $x^d=1$ in our group.

Edit: I hadn't seen Kent's post.

Thanks,Kent,grobber. I think that grobber meant the following solution:
It is known that the group is cyclic iff it's exponent $m$ is equal to $|F^*|$. Let's assume $m<|F^*|$. Thus for every $x\in{F^*}$ holds $x^m=1$ but such an equations can't have more than $m$ solutions in field $F$. So our supposion with $m<|F^*|$ is false.

Another proof is possible using a form of the classification of finite abelain groups, that every abelian
group G is
$C_{i1} * ...*C_{ir}$ where i1 divides i2,......

if G is a finite subgroup of the multiplicative group, it is finite and abelian and thus of above form
now consider this equation $x^{ir}=1$, it has at most ir solutions in G, as G is part of a field, but it has the entire group G as solution, thus the order of G is ir, thus G is isomorphic with $C_{ir}$

Here is a proof.

We can factorize $F^{*}$ as:
$F^{*}\simeq \mathbb{Z}_{p_{1}^{a_{1}}}\cdot ... \cdot \mathbb{Z}_{p_{n}^{a_{n}}}$

Let $m=\mbox{lcd}(p_{1}^{a_{1}},...,p_{n}^{a_{n}})$. Note $m\leq |F^{*}|$

If $b_{i}\in \mathbb{Z}_{p_{i}^{a_{i}}}$ then $b_{i}^{p_{i}}= 1$ since $p_{i}| m$ we have $b_{i}^{m}=1$

Thus, $x^{m}=1 \to x^{m}-1=0$.
Hence, all elements in $F^{*}$ are zeros of $x^{m}-1 \in F[x]$
But it can have [b]at most[/b] $m$ elements.
Thus, $m\geq |F^{*}|$.
Together we have $m = |F^{*}|$
Which tells us the primes in the factorization are all distinct.
Hence $F^{*}\simeq \mathbb{Z}_{m}$
And the proof is complete.

[quote="sylow_theory"]We can factorize $F^{*}$ as:
$F^{*}\simeq \mathbb{Z}_{p_{1}^{a_{1}}}\cdot ... \cdot \mathbb{Z}_{p_{n}^{a_{n}}}$.[/quote]
I really don't think we can.


Find all a;b are positive integer such that: $ ab^2\plus{}9|ba^2\plus{}b$
I need find nice solution  :)


geometry
incenter
circumcircle
parallelogram
geometry proposed

[quote="babylon"]Triangle has incentre $ I$ and $ M,P$ are the tangency points of the excircles with $ AC,AB$ , respectively.Let $ AI$ meet the circumcircle of $ \triangle ABC$ in $ L$.A penpedicular line of $ AL$ is erected through $ L$ and intersects $ BC$ in $ T$.Prove $ MP\parallel{}IT$.[/quote]
The problem seem no right at all  :maybe:

[quote="babylon"]Yep , it should't be true for isosceles triangle. See the sketch below :[/quote]
I like this problem.
Here my solution:
$ ID\bigcap KL$ at $ H$.Construct parallelogram $ MNQC$.
$ \Rightarrow BEFQ$ is parallelogram.
We have $ \frac {ID}{IL}\equal{}\frac {IT}{IH}\equal{}\frac {EF}{BC}\equal{}\frac {BQ}{BC}$.
$ \Rightarrow \triangle ITH\sim \triangle BQC$.
$ \Rightarrow HT\perp QC$.
From $ HT\perp KI$ we have $ KI\parallel QC\parallel MN$.$ \oplus$.

[b]An equivalent problem:[/b]
Given triangle $ ABC$ with 3 altitudes $ AA_1, BB_1, CC_1$, orthocenter $ H$, circumcenter $ O$. Let $ J$ be the midpoint of $ AH$. A perpendicular line to $ AH$ through $ J$ intersects $ B_1C_1$ at $ K. OB\cap A_1C_1\equal{}\{B_2\}, OC\cap A_1B_1\equal{}\{C_2\}$. Prove that $ B_2C_2//HK$
[b]Solution:[/b]
Without lost of general we can assume that $ \angle B>\angle C$
Denote $ N\equal{}JK\cap (A_1B_1C_1)$ the Nine-point circle of triangle $ ABC, P$ the midpoint of $ BC$.
$ \angle JKC_1\equal{}\angle (B_1C_1,BC)\equal{}180^o\minus{}2\angle C\minus{}\angle A\equal{}\angle B\minus{}\angle C$
$ \angle JC_1N\equal{}\angle JA_1N\equal{}\angle A_1JP\equal{}90^o\minus{}\angle A_1PJ\equal{}90^o\minus{}\angle JB_1A_1\equal{}90^o\minus{}(180^o\minus{}90^o\plus{}\angle C\minus{}\angle B)\equal{}\angle B\minus{}\angle C$
So $ \angle JC_1N\equal{}\angle JKC_1 \Rightarrow JN.JK\equal{}JC_1^2\equal{}JA^2 \Rightarrow \angle JAN\equal{}\angle AKJ\equal{}\angle JKH$
Therefore $ AN\perp HK (1)$
On the other side, $ B_2,C_2,P\in (OA_1)$ then triangle $ B_2C_2P$ is isosceles. Denote $ M$ the midpoint of $ OA_1$ hence $ PM\perp B_2C_2$
But $ \Delta A_1PO\equal{}\Delta NJA$ we get $ \angle MPO\equal{}\angle A_1OP\equal{}\angle NAJ\equal{}\angle ONQ$
Thus $ MP//AN$ or $ B_2C_2\perp AN (2)$
From $ (1)$ and $ (2)$ we are done.

Denote  R  is the intersection of   IaIc
Because AB=BC we have AC//IaIc
It is easy to see that ∠IRL=∠IaRL=∠IbIcIa=90-1/2∠C   =>  ∠IRIa=180-∠C=∠ABIa
=> IR//AB
Notice FM//TL ∠PFM=∠IRT=90+1/2∠A=90+1/2∠C  =>  ⊿FMP∽⊿RTI  =>  MP//IT

[quote="plane geometry"]
Because AB=BC we have AC//IaIc
[/quote]

Dear [b]plane geometry[/b],
This problem is still right if $ ABC$ is a scalene triangle.


abstract algebra
group theory
number theory
least common multiple
superior algebra
superior algebra solved

If $G$ is a finite abelian group, $G$ contains an element $g$ whose order is divisible by the othe order of every element of $G$.

Does someone have an elegant proof for this :) It's intrigueing

From the structure theorem for finitely generated abelian groups, $G$ is isomorphic to $(\mathbb Z/n_{1}\mathbb Z)\times\ldots(\mathbb Z/n_{k}\mathbb Z)$ for some $k\in\mathbb N$ and $n_{k}|n_{k-1}|\ldots|n_{1}$.
The conclusion easily follows

you don't need to invoke classification theorem for finite abelian groups: just take elements $x_{1}, \dots, x_{k}$ of pairwise coprime orders, such that the least common multiple of their orders is the exponent of $G$ (of course, you can do it). then $x_{1}x_{2}\dots x_{k}$ has the desired property.

How about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27505]this[/url]? The argument works in essentially the same way if we replace orders of subgroups with orders of cyclic subgroups.

[quote="amfulger"][quote="ma_go"](of course, you can do it)[/quote]
Really!?[/quote]
yes: take $k=|G|$, and $\{a_{1}, \dots, a_{k}\}=G$.
now, define $x_{1}=a_{1}$, $x_{n+1}=a_{n+1}^{m_{n}}$, where $m_{n}= \text{lcm}(o(x_{1}),\dots,o(x_{n}))$.
now the sequence $x_{1}, \dots, x_{k}$ has the desired property.
of course, you can refine it by removing all identities, but that doesn't actually matter.. (i didn't ask for such a minimal sequence).

anyway, another way to see it: take elements such that the lcm $N$ of their orders is the biggest possible. now for every maximal prime power $p^{m}$ dividing $N$, consider an element $y_{p}$ which order is divisible by $p^{m}$ (and you don't need this elements to be distinct).
now consider the sequence $x_{1}, \dots, x_{k}$, where $p_{1}, \dots, p_{k}$ are prime divisors of $N$, and $x_{i}$ is a power of $y_{p_{i}}$ of order $p_{i}^{m}$.

[size=75][color=red]EDIT:[/color] typos fixed.[/size]

Sorry ma-go! For a moment there I thought you were also using some strong results, such as Sylow's Theorems. I am sorry. I was wrong.


Let positive integer m(m>=2) .
Positive integer X satisfy:
            [tex]      1+4^m+4^X =a^2, \ a\in \mathbb{N}[/tex]
Find max(X).


vector
linear algebra
matrix
abstract algebra
linear algebra unsolved

Is the set of rational points on the unit circle an infinite dimensional vector space over $ \mathbb{Z}$?
What is its span?

That's not what "dimension" means for vector spaces.

First of all, vector spaces are not defined over non-fields.  Second of all, the rational points on the unit circle are not closed under addition.  So I don't understand the intent of this question; could you phrase it more clearly?

It could be the question is to characterize the $ \mathbb{Z}$-module generated by $ S^1 \cap \mathbb{Q}^2$.


inequalities
function
inequalities unsolved

$ a,b,c\in R\plus{}$
$ a\plus{}b\plus{}c\equal{}3$ show that
$ \sum_{cyc} \frac{1}{\sqrt{a^2\minus{}3a\plus{}3}}\leq 3$

Thanks in advance for your nice solution   :)

I'm not sure

$ f(a,b,c) \equal{} \sum_{cyc}\frac {1}{\sqrt {a^{2} \minus{} 3a \plus{} 3}}$

$ \implies$  $ f(a,b,c)\leq f(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c)$

[quote="MathUniverse"]$ a \plus{} b \plus{} c \equal{} 3$
$ \frac {1}{3} \sum_{cyc}\frac {1}{\sqrt {a^{2} \minus{} 3a \plus{} 3}}\leq 1$ .
Function $ f(x) \equal{} \frac {1}{\sqrt {x^{2} \minus{} 3x \plus{} 3}}$ is concave  so by applying Jensen's inequality we get $ \frac {f(a) \plus{} f(b) \plus{} f(c)}{3} \leq f(\frac {a \plus{} b \plus{} c}{3})$.
$ f(\frac {a \plus{} b \plus{} c}{3}) \equal{} f(1) \equal{} 1$. So we get the inequality we wanted to prove.[/quote]

I think f(x) is not concave in [0,3]

$ f''(x)\equal{}\frac{8 x^2\minus{}24 x\plus{}15}{4 \left(x^2\minus{}3 x\plus{}3\right)^{5/2}}$

@cnyd
I don't think that SMV can solve this problem easily...It will be an extremely tough work for this one
@kuing
Thanks for your $ f''(x)$ so that people stop using jensen incorrectly  :) I'm waiting for your nice solution  :lol:

 :(  I have worked for this problem for a day and I couldn't solve it please help...I'm really want to see how it can be done?
Thanks in advance

$ f(x) \equal{} \frac {1}{\sqrt {x^{2} \minus{} 3x \plus{} 3}}, x\in [0,3]$

$ f'(x) \equal{} \frac {3 \minus{} 2x}{2\left(x^{2} \minus{} 3 x \plus{} 3\right)^{3/2}}$

$ f''(x) \equal{} \frac {8 x^{2} \minus{} 24 x \plus{} 15}{4\left(x^{2} \minus{} 3 x \plus{} 3\right)^{5/2}}$

when $ x\in[0,\frac {3}{2}]$, $ f'(x)\ge0$, when $ x\in[\frac {3}{2},3]$, $ f'(x)\le0$

when $ x\in[0,\frac {1}{4}(6 \minus{} \sqrt {6})]$ or $ x\in[\frac {1}{4}(6 \plus{} \sqrt {6}),3]$, $ f''(x)\ge0$, when $ x\in[\frac {1}{4}(6 \minus{} \sqrt {6}),\frac {1}{4}(6 \plus{} \sqrt {6})]$, $ f''(x)\le0$
($ 0 < \frac {1}{4}(6 \minus{} \sqrt {6}) < 1 < 2 < \frac {1}{4}(6 \plus{} \sqrt {6}) < 3$)

WLOG, assume $ a\le b \le c$, that we have $ c\ge1$.

if $ c\ge2$, that $ a\le1$ and $ b\le1$, we have $ f(a) \plus{} f(b) \plus{} f(c)\le f(1) \plus{} f(1) \plus{} f(2) \equal{} 3$;

if $ 1\le c\le 2$, that $ a,b,c\in [0,2]$ and $ [0,\frac {1}{4}(6 \minus{} \sqrt {6})]$ is convex, $ [\frac {1}{4}(6 \minus{} \sqrt {6}),2]$ is concave,
so, use the "[url=http://www.mathlinks.ro/viewtopic.php?t=64933]Half concave and half convex theorem[/url]", we just need to prove that
$ f(0) \plus{} f(3 \minus{} c) \plus{} f(c)\le3 \qquad (1)$
and
$ f(3 \minus{} 2c) \plus{} 2f(c)\le3,\qquad (2)$

$ (1) \Leftrightarrow \frac {1}{\sqrt {3}} \plus{} \frac {2}{\sqrt {\frac {3}{4} \plus{} (c \minus{} \frac {3}{2})^2}}\le3$
obvious true;

$ (2) \Leftrightarrow$

...to be continue......

if $ x \ge \frac{\sqrt{5}\minus{}1}{2}$ then we prove that:
$ \frac{1}{\sqrt{x^{2}\minus{}3x\plus{}3}}  \le \frac{x\plus{}1}{2}$
since both parts of the inequality is positive we can say

$ \frac{1}{\sqrt{x^{2}\minus{}3x\plus{}3}}  \le \frac{x\plus{}1}{2}$

$ \leftrightarrow (\frac{1}{\sqrt{x^{2}\minus{}3x\plus{}3}})^2  \le (\frac{x\plus{}1}{2})^2$

$ \leftrightarrow 4 \le x^4\minus{}x^3\minus{}2x^2\plus{}3x\plus{}3 \leftrightarrow 0 \le (x\minus{}1)^2(x^2\plus{}x\minus{}1)$

but we asuumed $ x \ge \frac{\sqrt{5}\minus{}1}{2}$ which implies $ x^2\plus{}x\minus{}1 \ge 0$ and since $ (x\minus{}1)^2 \ge 0$ so the inequality is true.
suppose that $ a,b,c$ are $ \ge  \frac{\sqrt{5}\minus{}1}{2}$ then by using the above inequality for each of the fractions we are done,
since all of the variables can't be lesser than $ \frac{\sqrt{5}\minus{}1}{2}$ so we have to check two cases ,
1-suppose that the two variable lesser than $ \frac{\sqrt{5}\minus{}1}{2}$ are $ a,b$.
then

$ 3\minus{}3a,3\minus{}3b > 3\minus{}3\frac{\sqrt{5}\minus{}1}{2} > 1,1457$

which implies $ a^2\plus{}3\minus{}3a,b^2\plus{}3\minus{}3b > 0\plus{}1,1457$

or in other words

$ \frac{1}{\sqrt{a^2\plus{}3\minus{}3a}}\plus{}\frac{1}{\sqrt{b^2\plus{}3\minus{}3b}} < 2 \times \frac{1}{\sqrt{1,1457}} < 1,86851$

so it is sufficient to prove that :

$ \frac{1}{\sqrt{c^2\plus{}3\minus{}3c}} \le 3\minus{}1,86851\equal{}1,13149$

$ \leftrightarrow (\frac{1}{1,13149})^2 \le c^2\plus{}3\minus{}3c$

$ \leftrightarrow 0,78108... \le c^2\plus{}3\minus{}3c$

note that $ a,b < \frac{\sqrt{5}\minus{}1}{2}$ which implies $ c > 3\minus{}(\sqrt{5}\minus{}1) > 1,763931$

but since $ 1,763931 > \frac{3}{2}$ and the function $ x^2\minus{}3x\plus{}3$ is strictly increasing in $ [\frac{3}{2},\plus{} \infty)$

so we have $ c^2\minus{}3c\plus{}3 > (1,763931)^2\minus{}3(1,763931)\plus{}3\equal{}0,8196... > 0,78108...$

so the first case is proven,

2-now suppose that the only variable lesser than $ \frac{\sqrt{5}\minus{}1}{2}$ is $ a$ and both $ b$ and $ c$ are at least $ \frac{\sqrt{5}\minus{}1}{2}$
('didn't succeed for solving this case but i think it is easy...)

continue the 6#:

$ (2)\Leftrightarrow \frac{2}{{\sqrt {c^2  \minus{} 3c \plus{} 3} }} \plus{} \frac{1}{{\sqrt {4c^2  \minus{} 6c \plus{} 3} }} \le 3,$

see image:

[img]http://bbs.pep.com.cn/attachments/month_0911/20091119_e8df97eb087e6076abc8Xnz5Oqo7Hv2O.jpg[/img]


Find the area of a triangle whose medians have measures $39$, $42$, and $45$.

[hide]

i used stewarts theorem three times, got three equations, and solved for the sides...and if i didn't make a mistake, the sides are  2 :sqrt: 673, 2 :sqrt: 505, and 2 :sqrt: 592.

so the area is....
1008
[/hide]

[quote="isaacchao"][hide]
s = (m1 + m2 + m3) / 2   where m1,m2,m3 are the 3 mediums
A = 4/3sqrt(s*(s-m1)(s-m2)(s-m3))
[/hide][/quote]

Looks like you just copied from [url=http://mathworld.wolfram.com/TriangleMedian.html]here[/url].   You could have at least put the formula in $\text{\TeX}$  :huh: .

There is an easier way than the one that d343seven used.


[hide][i]in[/i] the 5th bounce, the ball will have travelled 7.776m, one way.  i am not sure if i interpreted the question properly[/hide]

[quote="pascal_1623"][quote="math92"]If a ball is dropped from a height of 100m, each time it hits the ground it bounces 3/5 of the height it fell. How far will the ball have travelled in the 5th bounce?[/quote]

[hide=answer]
100+60+36+21.6+12.96+7.776=238.336
But since the ball also has to rise up, you subtract 100 from it, because it only fell 100m, it did'nt also rise up 100m.
238.336+138.336= $\boxed{376.672}$[/quote]

Don't forget to end your hide tag :D


Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others

Why that¿ (the desired number and another number could be divisible by $7$)

Ops. I did not finish my proof.
We need to show that there is a number not divisible by 2, 3, 5, or 7.

First we show there there are at least 2 numbers not divisible by 2, 3, or 5.
Among 10 consecutive numbers there are
5 divisible by 2
3 or 4 divisible by 3
2 divisible by 5
1 or 2 divisible by 6
1 divisible by 10
Moreover, if there are 4 divisible by 3, then there are 2 divisible by 6.
Therefore, they form a set of size at most 5+3+2-1-1 or 5+4+2-2-1 which is 8.

Now consider two numbers that are not divisible by 2,3,5. If they both are divisible by 7, then one of them is even that is impossible.
So there is a number that is not divisible by 2,3,5,7. It is co-prime to the product of the others.

But I think Zetax' one can't hold for the generalization one:Replace 10 by one arbitrary number formed 4k+2 for all k


AMC
USA(J)MO
USAMO
AIME
geometry
AMC 10
AMC 12 B

Did you hear that from Zuming, or did you figure that out from scores? And are you a winner?

-Adam Hesterberg

Shoot ... that's 5 points higher than last year, putting the MOP cutoff at like 19 or 20.  Please please please extend Red MOP to sophomores like the rumors said ...

JB

What about AIME, can someone like me with a 13 on AIME, 4 points above the minimum cutoff for fish and sophomores make MOP even my USAMO score is like 3/4 points below cutoff?

[quote="darktreb"][quote]heard it from zuming[/quote]

Yo Zach, does that mean you made top 12?[/quote]

honorable mention with a 28 (he didn't tell me how many 28s there were...)

Either way, it'll be good to see you at MOP man.

[quote="darktreb"]I heard it from my parents, who called me at school (I live at a boarding school). Interestingly enough, they did not bother to ask for my actual score, so all I know is that I made top 12 (though I had originally predicted a 29 for myself so I'm willing to bet that's my score).[/quote]Congrats!

AFAIK only the winners can participate in the TST, is that correct?

[quote="Valentin Vornicu"]
AFAIK only the winners can participate in the TST, is that correct?[/quote]

If some of the winners are not able to participate, then they extend the TST invitation to some of the honorable mentions.

(btw, what does AFAIK mean?)

As far as I know, AFAIK = as far as I know :D.  Do you think the MOP cutoff will be 17 or lower?  I hope so.  :?

haha, I HOPE SO TOO!!! and that they extend Red Group to sophomores.... :(

DPopov --

What score do you think you got?

JB

BTW I'm completely with you on this...please please please extend the red group

[quote="beta"][quote="Magnara"]It would almost certainly be in their interest to make Red Group Freshmen and Sophomores and raise the bar a little, maybe to a 10 or so.  Or 7...that'd make a little more sense.  Maybe 7 for freshmen, 10 for sophomores.[/quote]

Why?[/quote]

Well, the idea is that anyone who made USAMO as a freshman has great potential to do IMO in a few years.  I would argue that any sophomore who gets a 10 has just as much potential, if not more.  Especially since some people are just late bloomers with this sort of thing.

What will they do with the gigantic tie at 28 then?  Will they administer more TSTs?

[quote="paladin8"]Yeah, like, umm... perhaps, me. :lol: I like your thinking, Magnara, and I hope the AMC thinks the same way.[/quote]

yea... i think that tooo.... but just for thisyear ;)  because i am a soph. But the proposed cutoff by Magnara seems too small... i mean i suck at proof writing, but i know i got at least 10.

Well, I'm of course biased too.  A friend of mine (jackb#)  got a 14ish, so I'm rooting for him.  Speaking of which, John, if you do get a call, then you have to call me immediately and tell me, or pay the consequences.

well i know a lot of sophs who did well...(double digits) so i hope it works out okay. maybe in the midst of all these rumors, i will get an invite as well :D

What's sad is how he could very well be correct.

[quote="Valentin Vornicu"][quote="billzhao"][quote="darktreb"] However, if Canadians get spots in the top 12, they will extend it so 12 Americans compete for the 6 spots. (at least I think they did last year)[/quote]

Rest assured, at least 2 Canadians got into top 12.[/quote]You are so modest billzhao :D :D :D[/quote]

Even more modest when he counts himself as two spots in the top 12... j/k :D

Wait ... are the results available online?  :?  :?

When do they get posted?  Is it next friday?  I seem to remember reading that somewhere...

[quote="MithsApprentice"]Friday, Valentin. Sheesh, you're more anxious than we are... ;)[/quote]Well as we have just released the Romanian Team I was anxious to see who's going to compete against them :D :D

lol

Valentin--that picture of you makes you seem all the more nervous :D

Does no one know what the HM cutoff is? Come on guys...

[quote="DPopov"]I predict maybe around a 14 +/- 2 points for my score JB and they do need to invite sophomores to Red Group because like Magnara said, we have potential to do well in future years just as the freshmen do.[/quote]

DPopov, I think I'm right in your area on the score: 6 or 7 on one, 6 or 7 on 4, and probably 1 or 2 on 5.

Obviously I have to agree with you, but I of course am biased as well.

I completely agree with Magnara about the possibility of "late bloomers."  Last year I scored a 114 on the AMC 10 and a 3 on the AIME, this year I got 132.5 on AMC 12, 10 on AIME, and 13-16 on USAMO.  Not that the AIME wasn't easier this year, but it goes to show that it *clicks* later for some people.

I still don't know where this rumor originated or how valid this possibility is.  Can anyone who knows the full story explain this to me?

JB

And by the way, still no one from the red group has gotten a call???

That is almost the same story with me, 103.5 on the AMC 12 B last year and a 1 on the AIME. This year I went to a 126.5 on the AMC 12 A and a 13 on the AIME I. So I agree with the "late bloomer" statement. As for USAMO I should have around 14 points (I hope).

I'm dying to know what HM cutoff is as well.

[quote="Mildorf"][quote="darktreb"]AFAIK, AFAIK means as far as I know.  ;)

The only people who can take the TST and have the score count are the top 12. I'm not sure if they always invite more people if somebody in the top 12 is unable to take it. I remember 2003 when Tiankai chose to do RSI over IMO, it was 11 people competing for 6 spots. However, if Canadians get spots in the top 12, they will extend it so 12 Americans compete for the 6 spots. (at least I think they did last year)[/quote]

Actually, I know for a fact that they will officially extend the TST to some people outside of top 12.  I happen to be one of those people.[/quote]
Congrats, Mildorf!


Here are the problems:

1: [url]http://www.mathlinks.ro/viewtopic.php?p=874747#874747[/url]

2: [url]http://www.mathlinks.ro/viewtopic.php?p=874749#874749[/url]

3: [url]http://www.mathlinks.ro/viewtopic.php?p=874750#874750[/url]

4: [url]http://www.mathlinks.ro/viewtopic.php?p=874754#874754[/url]

How can we learn the proposer countries?

The problems 1,2 was proposed by Greece . A very nice geometry  :D

Does anybody know where can we find the results of JBMO 2007?

Does anybody knows where JBMO 2008 will take place?

Thanks for link But still I didn't get information that I need ( where JBMO 2008 takes place)

Salem


Let $ S_n\equal{}\{1,2,...,n\}$.  $ A$ is subsets of $ S_n$ which contains $ k$ elements. Denote $ d(n,k)$ is the difference of number of sets in $ A$ which have sum of all elements is even and number of sets in $ A$ which have sum of all elements is odd.
Evaluate :
$ i) d(2n,k)$
$ ii) d(n,k)$

this may give you some ideas: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321131


geometry
trigonometry
cyclic quadrilateral
trig identities
AMC

A circle has center on the side AB of the cyclic quadrilateral ABCD. The other three sides are tangent to the circle. Prove that AD+BC=AB. [hide]IMO 1985[/hide]

<spoiler>Let O be the center of the circle. Let the perpendicular from O to BC be E, from O to AD be F, and from O to CD be G. Since triangles OCG and OCE are congruent, <OCG and <OCE are equal (call them each x), and similarly, <ODF and <ODG are equal (call each y).

Then since ABCD is cyclic,

BC = r ( cot (180-2y) + cot x)
AD = r ( cot (180-2x) + cot y)
AB = r ( csc (180-2y) + csc (180-2x))

BC + AD = r ( cot x + cot y - cot 2x - cot 2y)
AB = csc 2x + csc 2y

Using simple trig identities shows that AB = BC + AD (in other words, I worked this out on my paper but am far too lazy to write it out here, and I'm sure all of you are smart enough to figure out that they're equal cuz I assure you it's nothing more than simple double angle rules) </spoiler>

Why is ABCD cyclic? For the spoiler, use these: [ and ], not < and >. Thus, you would write spoiler as [ spoiler ] except without the spaces.

haha let's see if i can do spoiler right this time (btw, when i do italics, the < work just fine...it's odd that they don't with spoiler):

[hide] you said in your original statement of the problem that ABCD is cyclic: "A circle has center on the side AB of the cyclic quadrilateral ABCD. The other three sides are tangent to the circle. Prove that AD+BC=AB."

=)

[/hide]

cyclic means there is a circle in which ABCD is inscribed. unfortunately, that's not exactly what the problem statement said.

I'm confused....so if the problem didn't say that ABCD is cyclic, then why did it say "cyclic quadrilateral ABCD"?

lol so is ABCD cyclic or not...and then what does the problem say and what can I assume?

[color=cyan]froufrou, you appear to be right.  Mystic just didn't look at that part of the problem -- he thought youwere using the circle O to prove ABCD was cyclic, not the fact that it was given as cyclic.  I did the same thing too, at first.  The word "cyclic" is sort of snuck in there, I guess.

There must be a way without trig, though . . . synthetic methods are nice.[/color]

sigh. ok, i'm stupid. sorry about that. yes, you are right, it is cyclic. i will now bonk myself on the head several times. hmmm...i suppose i should probably work through the problems before i post. but...hmm....i can't do the problems. erg.

A non-geometric solution can be found at

http://www.kalva.demon.co.uk/imo/isoln/isoln851.html

It's also a good solution. Trig was just the first thing that occured to me =).


find an $ n_{th}$ formula for this sequence $ 1,2,5,14,41,122,\cdots$

[hide]To get the next term, add the next power of 3.

$ a_n\equal{}1\plus{}\sum_{k\equal{}0}^{n\minus{}1}3^{k}$[/hide]

[quote="Temperal"]
If you let $ a_1 \equal{} 1$, $ a_n \equal{} \sum_{k \equal{} 0}^{n \minus{} 1}3^{k}$ is the formula.[/quote]

I'm pretty sure I_like_pie is correct.  According to your sequence, we get:
1, 4, 13, ...

You'll notice that if we add 1 to your sequence, we get the correct sequence starting at the second term.

Fixed. I thought something fishy was up....

@i_like_pie how did you get that ? can you explain the steps please

i turned your formula into this form , is this correct $ a_n\equal{}1\plus{}\frac{(3^{n\minus{}1}\minus{}1)}{2}\equal{}\frac{(1\plus{}3^{n\minus{}1})}{2}$

I_like_pie's formula is slightly off. See my post for the correct one.

$ a_{n\plus{}1}\minus{}a_n\equal{}3^{n\minus{}1}$ - you can get this by testing it out and seeing. Now can you figure out how to get it?

We also have $ a_1\equal{}1, a_{n\plus{}1}\equal{}3a_n\minus{}1, n \in \mathbb{N}$.

[quote="Temperal"]I_like_pie's formula is slightly off. See my post for the correct one.

$ a_{n \plus{} 1} \minus{} a_n \equal{} 3^{n \minus{} 1}$ - you can get this by testing it out and seeing. Now can you figure out how to get it?[/quote]

no i couldnt figure out, and how did you get this $ a_{n \plus{} 1} \minus{} a_n \equal{} 3^{n \minus{} 1}$
it might be very easy for you lot, but i am completly lost, so if you explain how you got that formula step by step ill be grateful.

So you have the numbers 1, 2, 5, 14, 41, 122

2-1 = 1 = 3^0
5-2 = 3 = 3^1
14-5 = 9 = 3^2
41-14 = 27 = 3^3
122-41 = 81 = 3^4

Generally, it appears that if we subtract the nth term from the (n+1)th term, we get 3^(n-1)

[quote="t0rajir0u"][b]Follow-up:[/b]  Find a recursive formula without using $ 3^n$.[/quote]
[hide]
The only thing I can think of is $ a_n \equal{} a_{n \minus{} 1} \plus{} 3^{n \minus{} 2}$ but that does use $ 3^n$.[/hide]

[quote="Temperal"][hide="Not quite"]
If you let $ a_1 \equal{} 1$, $ a_n \equal{} 1 \plus{} \sum_{k \equal{} 0}^{n \minus{} 1}3^{k \minus{} 1}$ is the formula.[/hide][/quote]Not quite correct either after 2 edits, your initial index needs to start with $ k\equal{}1$.

well the recursion formula would be a_{n+1}= a_n+3^(n-1)
sorry don't know how to write subscripts so that's how he got a_{n+1} - a_n= 3^(n-1)
edit: it should be 3^(n-2) not n-1, sorry :oops:

[quote="Brut3Forc3"]We also have $ a_1 \equal{} 1, a_{n \plus{} 1} \equal{} 3a_n \minus{} 1, n \in \mathbb{N}$.[/quote]

This is a good answer to my question.  The other answer I was thinking of is $ a_{n\plus{}1} \equal{} 4a_n \minus{} 3a_{n\minus{}1}$.  :)

[hide="Formula"]If $ a_0$ is the first term, then $ a_n\equal{}1\plus{}\sum_{k\equal{}1}^{n}3^{k\minus{}1}$.[/hide]

[hide="Recursive Formula"]If $ a_0$ is the first term, then $ a_n\equal{}3a_{n\minus{}1}\minus{}1$[/hide]

[quote="t0rajir0u"][quote="Brut3Forc3"]We also have $ a_1 \equal{} 1, a_{n \plus{} 1} \equal{} 3a_n \minus{} 1, n \in \mathbb{N}$.[/quote]

This is a good answer to my question.  The other answer I was thinking of is $ a_{n \plus{} 1} \equal{} 4a_n \minus{} 3a_{n \minus{} 1}$.  :)[/quote]

Wait - I had that, but somehow when I tested it it didn't seem to work (I probably multiplied wrong), so I didn't post it....  :( Since it's a linear recurrence, doesn't that mean there's a closed form exponential?

There is, and it's been given already.  :)  Read the thread more carefully.

I may just be being thick, but I can't find it. I mean a formula like Binet's formula, with no summations nor previous terms.

[quote="binomial_4eva"]i turned your formula into this form , is this correct $ a_n \equal{} 1 \plus{} \frac {(3^{n \minus{} 1} \minus{} 1)}{2} \equal{} \frac {(1 \plus{} 3^{n \minus{} 1})}{2}$[/quote]

It's right here, guys.  :)

Aha, thanks. Sorry for not seeing it on my own.  :oops:


Prove or disprove the following inequality:
$ \begin{gathered}
  \left( {\frac{{a \plus{} b}}
{{b \plus{} c}}} \right) \plus{} \left( {\frac{{b \plus{} c}}
{{c \plus{} a}}} \right) \plus{} \left( {\frac{{c \plus{} a}}
{{a \plus{} b}}} \right) \geqslant a \plus{} b \plus{} c,\forall a,b,c{\text{ }} \hfill \\
  {\text{such that }}0 < abc \leqslant 1 \hfill \\
\end{gathered}$
(Inspired by Potla)

by am gm rhs>=3.so $ a\plus{}b\plus{}c>\equal{}3$ means that $ abc\equal{}1$

@aadil: Your intentions are obscure.Would you mind corroborating your claims? :wink:

Hello,

I belive it can be done this way.

Please tell me if I've made a mistake.

WLOG we can assume that $ a >\equal{} b >\equal{} c$

Then we get : $ \displaystyle\sum_{cyc} (a \plus{} b)/(b \plus{} c)$ $ >\equal{}$ $ \displaystyle\sum_{cyc} (a \plus{} b)/2a$

From $ \displaystyle\sum_{cyc} (a \plus{} b)/2a$ = $ (a \plus{} b \plus{} c)$ we are done.

P.S. Sorry for ugly Latex using  :) .

[quote="Tomekk"]Hello,

I belive it can be done this way.

Please tell me if I've made a mistake.

WLOG we can assume that $ a > \equal{} b > \equal{} c$

Then we get : $ \displaystyle\sum_{cyc} (a \plus{} b)/(b \plus{} c)$ $ > \equal{}$ $ \displaystyle\sum_{cyc} (a \plus{} b)/2a$

From $ \displaystyle\sum_{cyc} (a \plus{} b)/2a$ = $ (a \plus{} b \plus{} c)$ we are done.

P.S. Sorry for ugly Latex using  :) .[/quote]
sry tommek but this method fails as we cant get all terms, i.e we cant get c from any of -$ a\plus{}b,b\plus{}c,c\plus{}a$ besides maybe u meant something else but how is -$ \sum_{cyc} (a \plus{} b)/2a \equal{} a\plus{}b\plus{}c$?(am i jus too sleepy now? or u r wrong?)

Yes sorry,

I really dunno what I was thinking by that, like I've been drugged lol. It's a mistake :).


geometry
Euler
ratio
projective geometry

Para quien esta leyendo este post: La discusion es acerca de temarios en el area indicada por el titulo. Se pretende que si usted pone algo aqui, que sea referente al temario, discutiendolo (criticas completamente aceptadas), haciendo preguntas sobre el o proponiendo otros posibles resultados dentro del tema. Por favor, solo por mantener el orden, si su post no cae en esta categoria, pues inicie un nuevo topic para ello. Esta peticion es unicamente para conservar el orden para quienes estamos interesados en seguirle la pista al temario!!

Solo cuestion de notacion, las secciones, subsecciones y subsubsecciones

estan indicadas asi:
- Seccion
* Subseccion
** Subsubseccion

[b]Geometria[/b]
- Construcciones Basicas (Bisectriz, mediatriz, y un monton mas)
- Formulas basicas (Areas, etc)
- Definiciones Basicas (Punto, Recta, Segmento)
- Distintas formulas para el area
  * Heron
- Triangulos
  * Desigualdad Triangular
  * Semejanza y Congruencia
  * Teorema de Thales
  * Teorema de Pitagoras
  * Teorema de la Altura
  * Triangulos especiales
  * Colinearidad y Congruencia
  * Puntos especiales y rectas especiales
    ** Bisectrices e Incentro
    ** Mediatrices y Circumcentro
    ** Alturas y ortocentro
    ** Medianas y Baricentro (o centroide)
    ** Excentros
  * Linea de Euler
  * Circulo de los nueve puntos
  * Teorema de Ceva
  * Teorema de Menelao
  * Teorema de Stewart o Apolonio
  * Circulo de Apolonio
  * Punto de Apolonio
  * Teorema de Viviani
  * Teorema de Ptolomeo
- Circulos
  * Identidades acerca de angulos y semiangulos
  * Tangentes
  * Potencia Punto
  * Recta Radical
  * Recta de Simpsom
  * Teorema de Pitot (Un cuadrilatero convexo $ABCD$ tiene un circulo inscrito si y solo si $AB + CD = BC + DA$)
- Transformaciones
  * Rotaciones
  * Translaciones
  * Homotecias
  * Inversion
- Lugar Geometrico
  * Circulo de Apolonio
  * Circulos Secantes
  * Circulos Tangentes
- Geometria Analitica
- Estereometria.

Quisiera agregar lo siguiente

-Geometria Proyectiva
*Conjugados Harmónicos
*Cross Ratio (no se la traduccion en español)
*Polos y Rectas Polares
**Transformacion Polar - Dualidad
*Teorema de Pascal
*Teorema de Brianchon

[quote="Leonardo"]*Teorema de Pascal
*Teorema de Brianchon[/quote]
Tenes algun ejercicio en el cual se utilicen estos teoremas? Digo, como para tener una idea de ejemplos.

AAH, y me falto mencionar un ultimo teorema importante

Teorema de Desargues

Hmm, respecto a lo que dices David, hay muchos problemas de concurrencias y colinealidades que salen con esos teoeremas. Por ejemplo, el problema 2 de la olimpiada China de este anho salia con el Teorema de Desargues.Tambien hay muchos otros problemas clasicos que salen con eso, en este momento no tengo mis libros conmigo, pero te digo que suele ser bastante util :D

[quote="Leonardo"]*Cross Ratio (no se la traduccion en español)
[/quote]

Se dice razón cruzada

Un ejemplo bastante simple sobre el uso del teorema de Pascal es el siguiente:

(IMO Shorlist 1991\1) Sean un triángulo ABC y P un punto en su interior. Sean E y F los pies de las perpendiculares a AC y BC desde P, respectivamente, y sean G y H los pies de las perpendiculares desde C a AP y BP, respectivamente. Mostrar que BC, FG y EH son concurrentes.

Dado que <CEP=<CHP=<CFP=<CGP=90°, resulta que E, F, G y H están sobre la circunferencia de diámetro PC. Ahora bien, por el teorema de Pascal aplicado al cuadrilátero cíclico CEHPGF, podemos concluir que EC :cap: PG=A,
PH :cap: FC=B y FG :cap: EH están alineados, como queríamos.

cross ratio es razon cruzada y quisiera que me indiquen donde puedo conseguir amplio naterial de dualidad y donde puedo encontrar "la regla geometrica de la falsa posicion"

[b]0. Introduccion: [/b] 0.1 Conjuntos 0.2 Conceptos primitivos: pounto, recta y plano 0.3 Numeros Reales 0.4 Desigualdades 0.5 La recta Real

[b]1. Rectas, planos, separacion:[/b] 1.1 Definiciones 1.2 Conjuntos Convexos

[b]2. Angulos y triangulos:[/b] 2.1 Definiciones Fundamentales. 2.2 Medida Angular. 2.3 Ángulos Rectos, Perpendicularidad. 2.4 Isometrias y Congruencias. 2.5 Clasificación de Triángulos. 2.6 Líneas Notables de un Triangulo. 2.7 Perpendiculares. 2.8 Desigualdades Geométricas. 2.9 La Distancia Entre Una Recta y Un Punto. La Desigualdad Del Triangulo. 2.10 Alturas de Triángulos.

[b]3. Paralelas:[/b] 3.1 Traslaciones. 3.2 Ángulos Correspondientes, suma de angulos de un triangulo. 3.3 Punto de Fermat. 3.4 Polígonos. 3.5 Cuadriláteros. 3.4 Secantes a Varias Rectas Paralelas

[b]4. Circunferencia:[/b] 4.1 Definiciones. 4.2 Arcos de Circunferencia. 4.3 Posiciones relativas de dos circunferencias. 4.4 Posiciones Relativas de un Triangulo y una Circunferencia. 4.5 Cuadriláteros circunscriptibles e inscriptibles. 4.6 Triángulos de Napoleón y Teorema de Miquel.

[b]5. Geometria del triangulo I:[/b] 5.1 Introducción. 5.2 Medianas, Baricentro 5.3 Alturas, ortocentro. 5.4 Mediatrices y Circuncentro 5.5 Bisectrices y centros tritangentes. 5.6 Recta de Euler. 5.7 Circunferencia de Euler. 5.8 La recta de Simson. 5.9 Geometria del Cuadrilatero.

[b]6. Regiones poligonales y sus areas:[/b]  6.1 Regiones Poligonales 6.2 Áreas de Triángulos y Cuadriláteros 6.3 Relaciones en triángulos especiales.

[b]7. Semejanza y Proporcionalidad:[/b] 7.1 Conceptos y Homotecia.  7.2 Semejanza de Triángulos 7.3 Semejanza de Circunferencias, Centros de Similitud 7.4 La Circunferencia de Apolonio. 7.5 Razones de Áreas de figuras semejantes 7.6 Las Razones Trigonométricas. 7.7 Teoremas de Semejanza 7.8 Proyecciones, Proyeccion Ortogonal 7.9 Relaciones Métricas en los Triángulos. 7.10 Teoremas de áreas. 7.11 Relaciones métricas en los cuadriláteros. 7.12 Polígonos Regulares y sus Áreas. 7.13 División de un segmento en media y extrema razón. 7.14 Longitud de una circunferencia y Area del circulo

[b]8. Colinealidad y Concurrencia:[/b] 8.1 Teorema de menelao Y Ceva 8.2 Circunferencia de Spieker 8.3 Teoremas de Routh, Coordenadas Baricentricas. 8.4 Figuras en Perspectiva, Teorema de Desargues 8.5 division Armonica 8.6 Curvas ortogonales. 8.7 Cuadrangulos y Cuadrilateros Completos. 8.8 Relaciones en El cuadrilatero.

[b]9. Geometria del triangulo II:[/b] 9.1 Polos y Polares con respecto al triangulo. 9.2 Lineas isogonales y isotomicas 9.3 Las circunferencias de Lemoine 9.4 Circunferencias de Apolonio. 9.5 Puntos de Brocard, Circunferencia de Brocard, Triangulos de Brocard. 9.6 Circunferencias de Tucker: Circunferencia de Apolonio, Circunferencia Circunscrita, Circunferencia Coseno, Primera Circunferencia de Lemoine, Ccunferencia de Gallaty, Circunferencia de Kenmotu, Circunferencia de Taylor. 9.7 Circunferencia de Adams 9.8 Triangulos de Morley.

[b]10. Circulos:[/b] 10.1 Teoremas de semejanza en la circunferencia. 10.2 Relaciones en Circunferencias (Teorema de la tangente, secante, etc). 10.3 Relaciones en Cuadriláteros Inscritos (ptolomey, Furhman, Casey, etc). 10.4 Potencia, Eje radical, Circunferencias Coaxiales, Centro Radical

[b]11. Inversion:[/b] 11.1 Definiciones. 10.2 Invercion de rectas y circunferencias. 10.3 Inversion de un teorema. 10.4 Circunferencia de Asimilitud 10.5 Inversiones Especiales (Circunferencias en Circunferencias iguales, en si mismas, en dos que se intersequen en un angulo dado) 10.6 Celda de Paucellier 10.7 Porisma de Steiner

[b]12. Polos y Polares - Dualidad:[/b] 12.1 Definiciones 12.2 Relaciones Armónicas 12.3 Triángulo Polar y Autopolar 12.4 Dualidad

[b]13. Razon Cruzada:[/b] 13.1 Definiciones. 13.2 Los seis valores de la razón cruzada 13.3 Propiedades de razón cruzada de una circunferencia. 13.4 Teorema de Pascal 13.5 Teorema de Brianchon 13.6 Teorema de Pappus 13.7 Puntos Autocorrespondientes 13.8 El problema de Apolonio.

[b]14. Involucion:[/b] 14.1 Definiciones 14.2 Relaciones de razon cruzada de los seis puntos de la involución. 14.3 Haz de involución con vértice en una circunferencia. 14.4 Lineas conjugadas en ángulos rectos 14.5 Involución de puntos en una trasversal que interseca los lados de un cuadrangulo completo. 14.6 Involución de puntos en una transversal que interseca una circunferencia y los lados de un cuadrangulo inscrito. 14.7 Teoremas de Desargues.

[b]15. Geometria del triangulo III:[/b] 15.1 Teoremas Varios 15.2 Teorema de Feuerbach 15.3 Triangulos isologicos, paralelogicos, Ortologicos. 15.4 Isopolos, Ortopolo 15.5 ortocorrespondencia 15.6 Porisma de Poncelet.

[b]16. Temas Selectos: [/b] 16.1 El Arbelos 16.2 La teoria de los Poligonos Armonicos

Recien Consegui el Libro A sequel to Euclid de Casey asi que cuando lo termine de leer posiblmente modifique un poco el temario (en lo poco que ya vi, el usa el concepto de homologias y homografias a nivel elemental, me falta revisarlo), espero me menden sus sugerencias y opiniones. Cada Subdivicion tiene añlrededor de 30 teoremas

PD. Si alguien puede mandarme mas cosas acerca de Polos y Polares, Razon cruzada (que es la razon cruzada compleja??) e Involucion lo agradeceria mucho.

Hola, Hucht, donde puedo conseguir el libro de Casey.

Gracias por tu respuesta.


induction
arithmetic sequence
number theory solved
number theory

Let r(1)=1 and for for k>1 let r(k) equal
the product of the prime divisors of k.
Let the sequence   a_(n+1)=a_n+r(a_n)
with arbitraty first term.
-> Show that for any positive integer m,
the sequence a1,a2,.. contains m consecutive
terms in arithmetic progression.

Here's what I did for m=3, then I think you can easily extend it for larger m's (maybe not easily, but you can :D ):

Let's assume that for every k a_(k+1) has a prime divisor which does not divide a_k. Denote by b_k=a_k/r(a_k), so we have a_(k+1)= r(a_k)* (b_k+1). Now we assumed that r(a_(k+1))> r(a_k), and because r(a_k)/a_(k+1), we have r(a_(k+1))>= r(a_k)*p where p is a prime, so b_(k+1)<= (b_k+1)/p< b_k, so {b_k} is an infinite sequence of natural numbers which is strictly decreasing, which is impossible. We got contradiction, so there is a k such that r(a_(k+1))=r(a_k), so we got an arithmetic progression of 3 terms, a_k, a_(k+1), a_(k+2). I suppose we can do something similar to obtain a progression of 4 terms and so on (induction), but I'm not sure.


function
analytic geometry
graphing lines
slope
calculus
calculus computations

Which graph best represents the position of a particle, s(t), as a function of time, if the particle's velocity and acceleration are both positive?

The answer is B. Because the graph is concave up and the slope is positive. That means accel and velocity is positive.


A man goes to a river with two containers; a 16-pint and a 9-pint.  How can the man get 1-pint of water in the 16-pint container?  He has never learned to read, thus measure.

Masoud Zargar

Darn, I was going to post the solution, but this time t0rajir0u beat me to it  :P . Here's another problem: this time, the man has a 5-cup and an 8-cup container. How can he get exactly 9.5 cups of water? (You can put some in one container and the rest in the other container.)

[quote="mna851"]Eh...that seems fishy to me.[/quote]
Well, it's basically the same as geometric constructions: it's never perfectly accurate; you just assume that it is.

[quote="boxedexe"][quote="t0rajir0u"][hide="A little easy for Pre-Olympiad...?"] Fill the 16-pint.  Tilt it to pour out exactly half (well-known).  Fill the 9-pint.  Pour into the 16-pint.  Empty the 16-pint.  Empty 9-pint into 16-pint.

:) [/hide][/quote]
Well, the person does not have any ability to empty exactly half.  He cannot empty a certain amount out of the containers unless forced to do so.  ;)[/quote]

Well... there is another way to do it:
[hide]Fill up the 16-pint container. Pour water from it into the 9-pint container until the 9-pint container is full. There are now 7 pints in the 16-pint container. Next empty the 9-pint container. Pour the 7 pints into the 9-pint container. Then fill up the 16-pint container. Pour water from the 16-pint container into the 9-pint container until it is full. There are now 14 pints in the 16-pint container. Empty the 9-pint container. Pour water from the 16-pint container into the 9-pint conatiner until it is full. There are now 5 pints in the 16-pint container. Empty the 9-pint container. Pour the 5 pints in the 16-pint container into the 9-pint container. Fill up the 16-pint container. Pour water from the 16-pint container into the 9-pint container until it is full. There are now 12 pints in the 16-pint container. Empty the 9-pint container. Pour water from the 16-pint container into the 9-pint container until it is full. There are now 3 pints in the 16-pint container. Empty the 9-pint container. Pour the 3 pints in the 16-pint container into the 9-pint container. Fill up the 16-pint container. Pour water from the 16-pint container into the 9-pint container until it is full. There are now 10 pints in the 16-pint container. Empty the 9-pint container. Pour water from the 16-pint container into the 9-pint container until it is full. There is now 1 pint in the 16-pint container.

Whew, that was a long one  :lol: .[/hide]

[quote="mna851"]Nice...  Note also that if we had a really large bucket we would be guranteed to be able to get one pint, because there must be integers $\ x$ and $\ y$ such that $\ 16x + 9y=1$ because $\ 16$ and $\ 9$ are relatively prime.[/quote]
wow that seems interseting observation! but how do we find the solution from the equation? if we have $3x+5y=1$ what does the solution $x=2,y=-1$ mean?

You'd solve it like a regular diophantine equation.  In this case take both sides $\ \mod 9$ or $\ \mod 16$ .

[quote="nebula42"][quote="boxedexe"][quote="t0rajir0u"][hide="A little easy for Pre-Olympiad...?"] Fill the 16-pint.  Tilt it to pour out exactly half (well-known).  Fill the 9-pint.  Pour into the 16-pint.  Empty the 16-pint.  Empty 9-pint into 16-pint.

:) [/hide][/quote]
Well, the person does not have any ability to empty exactly half.  He cannot empty a certain amount out of the containers unless forced to do so.  ;)[/quote]

What you do is you hold the cup so that the top edge is horizontal with the bottom edge of the opposite side. Then half of the container water will pour out.[/quote]

What if the cup has an irregular shape?

Anyway, the main idea in this problem was to recongnize the diophantine equation $9x+16y=1$.  It is very easy to solve this equation to get $(x_0,y_0)$; just a little number theory followed by the solutions of the form $x_n=x_0+16k,\ y_n=y_0-9k\ \forall k\in\mathbb{Z}$.


after he started posting on the MO forum, buzzer11's posts began to go up at a steady rate of 3% each day. after 93 days, the posts were deleted, and he lost 10% of his posts. how many posts does buzzer have now if he started with 780 posts? (round to the nearest integer)

when you say 3%, do you mean that the same amount is added each day, or 3% of the daily post count is added? (its like compound interest vs. annual interest).

thanks.

okay, well, we have:

$ 1.03^{93}\cdot0.9\cdot780$

i dont have a calc...


trigonometry
LaTeX
integration
geometry
geometric transformation

TanB/(1+SecB)  +  (1+secB)/TanB?

Basically, adding the reciprocal.

I totally got stuck after   (2sec^2(B)+2secB)/((1+SecB)(TanB))    :?

i showed the steps for this one for ya  :D
[hide]
$\begin{eqnarray*} \frac{\tan b}{1+\sec b}+\frac{1+\sec b}{\tan b}&=&\frac{\tan^2 b+\sec^2 b+2\sec b +1}{\tan b(1+\sec b)}\\ &=&\frac{2\sec^2 b+2\sec b}{\tan b(1+\sec b)}\\ &=&\frac{2\sec b(1+\sec b)}{\tan b(1+\sec b)}\\ &=&\frac{2\sec b}{\tan b}\\ &=&2\csc b$
[/hide]

edit - oops, made a mistake on my identities  :blush:

A modification to the above solution  :)

[hide]$\begin{eqnarray*} \frac{\tan b}{1+\sec b}+\frac{1+\sec b}{\tan b} &=&\frac{\tan^2 b+\sec^2 b+2\sec b +1}{\tan b(1+\sec b)}\\ &=&\frac{2\sec^2 b+2\sec b}{\tan b(1+\sec b)}\\ &=&\frac{2\sec b(1+\sec b)}{\tan b(1+\sec b)}\\ &=&\frac{2\sec b}{\tan b}\\ &=&\frac{2}{\cos b}\cdot \frac{\cos b}{\sin b}\\ &=& 2\csc b$[/hide]

i think that the solution is the next

[hide]\begin{eqnarray*} \frac{\tan b}{1+\sec b}+\frac{1+\sec b}{\tan b}&=&\frac{\tan^2 b+\sec^2 b+2\sec b +1}{\tan b(1+\sec b)}\\ &=&\frac{2+2\sec b}{\tan b(1+\sec b)}\\ &=&\frac{2(1+\sec b)}{\tan b(1+\sec b)}\\ &=&\frac{2}{\tan b}\\ &=&2\cot b[/hide]

can somebosy help me to write my anwers like you.

You can find out how others wrote their maths expressions by clicking on the quote button for their post(s).
It will all be jumbled up and a bit difficult to make sense from, but ...

Maths eqns are enclosed in delimiters. On this forum they use the dollar-symbol, which I can't show you since the board would think I'm writing latex. So I'll use [ as a starting delimiter and ] as an ending delimiter. So you should write [ ... maths text ... ]. You could enclose all the lines of your maths expressions in [ ... ] or you could write multiple lines, each one enclosed in [ ... ].

[ line1 \\
line2 \\
line3 \\ ]

or,

[ line1 ]
[ line2 ]
[ line3 ]

If you clicked on the quote button on my earlier post you would get,

<hide>[\begin{eqnarray*} \frac{\tan b}{1+\sec b}+\frac{1+\sec b}{\tan b} &=&\frac{\tan^2 b+\sec^2 b+2\sec b +1}{\tan b(1+\sec b)}\\ &=&\frac{2\sec^2 b+2\sec b}{\tan b(1+\sec b)}\\ &=&\frac{2\sec b(1+\sec b)}{\tan b(1+\sec b)}\\ &=&\frac{2\sec b}{\tan b}\\ &=&\frac{2}{\cos b}\cdot \frac{\cos b}{\sin b}\\ &=& 2\csc b]</hide>
Which is a bit of a mess.
You should note that the double backslash, \\, is used as a carriage return character and means the end-of-line. So, I'll reformat the above lot to make it a bit easier to read. I'll also remove the hide delimiters.
[ \begin{eqnarray*} \frac{\tan b}{1+\sec b}+\frac{1+\sec b}{\tan b} &=&\frac{\tan^2 b+\sec^2 b+2\sec b +1}{\tan b(1+\sec b)}\\
&=&\frac{2\sec^2 b+2\sec b}{\tan b(1+\sec b)}\\
&=&\frac{2\sec b(1+\sec b)}{\tan b(1+\sec b)}\\
&=&\frac{2\sec b}{\tan b}\\
&=&\frac{2}{\cos b}\cdot \frac{\cos b}{\sin b}\\
&=& 2\csc b ]
The above is made up mainly of fractions, which are done using \frac{}{}. I'll remove the contents so that it's easier to see.
[\begin{eqnarray*} \frac{}{}+\frac{}{} &=&\frac{}{}\\
&=&\frac{}{}\\
&=&\frac{}{}\\
&=&\frac{}{}\\
&=&\frac{}{}\cdot \frac{}{}\\
&=& 2\csc b]
As you can see, there is a [-sign at the beginning and end of this maths text, as required. The maths text is also an array of equations and so begins with \begin{eqnarray*}. It should also end with \end{eqnarray*}, but doesn't!! I don't know why it works like that. It may be a peculiarity of this board.
Anyway, equation arrays can (don't always need to) be enclosd in,

\begin{eqnarray*}
...
...
...
\{eqnarray*}.

The asterix, *, is there to suppress line/equation numbers for the eqns. Remove the asterix to see what happens.
The ampersand, &, is like a Tab-key, and is used to space out the equations so that the equals-signs line up underneath each other. In any eqn-array, you should always have the same number of &'s in every line.
You can get a list of maths symbols from [url=http://physicsforums.com/misc/howtolatex.pdf]here[/url], which is from another forum I go to, that you can use in your latex code.

[u]Powers and subscripts[/u]
Powers are done with ^, and subscripts with _. e.g.

[ F_i = x^3 - 2x^2] becomes $F_i = x^3 - 2x^2$
[\int \cos \theta d\theta] becomes $\int \cos \theta d\theta$
[\int^{x_1 = 3}_{x_0 = 0} x^{3.5} + \tan x dx] becomes $\int^{x_1 = 3}_{x_0 = 0} x^{3.5} + \tan x dx$

If the power, or subscipt, is more than one character, then it should be enclosed in {} to make sure evrything is affected.

There is much more to learn, but this should start you off. I'm sure others will be able to offer more comments.

[quote="joseluis"][quote="prototypex24"]You can also get a math expression input program ( I use MAthType) that has a Latex translator[/quote]

where i can find the latex traslator and thaks both of you for the help i really apreciate it[/quote]

you just needa put the " money symbol " at the beginning and end of the thing you want . Example 3^4 can be written as $3^4$

[quote="joseluis"]...where i can find the latex traslator and thaks both of you for the help i really apreciate it[/quote]
The latex translator in Latex is in the Preferences menu of MathType.

Click on Preferences > Translators and in the window that comes up select "Translation to other language"


inequalities
geometry
area of a triangle
Heron\u0027s formula
inequalities solved
Sides of a triangle

Let $A,B,C$ be three points on a circle of radius $\frac{\sqrt3}3$ and $a,b,c$ the sides of triangle $ABC$. Show that
\[(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)  \leq a^4b^4c^4\]

Homogenize this inequality by puting $a=rx, b=ry, c=rz$. You will find that
\[r=\frac1{xyz}\sqrt{\frac13\sum (2x^2y^2-x^4)}\]
Thus, we have to prove that
\[\left(2 \sum x^2y^2- \sum x^4\right)^3 \ge 27(xyz)^2\prod(x^2+y^2-z^2)\]
So, it is enough to prove that for $a,b,c$ sides of a triangle we have
\[\left(2 \sum ab- \sum a^2\right)^3 \ge 27abc(a+b-c)(b+c-a)(c+a-b)\]
Put $a=n+p, b=m+p, c=n+m$. It becomes
\[\left(\sum mn\right)^3 \ge \frac{27}8mnp(m+n)(n+p)(p+m)\]
Let $mn=u, np=v, pm=w$
Then it is equivalent to
\[8(u+v+w)^3 \ge 27(u+v)(v+w)(w+u)\]
which is just AM-GM.

Beautiful solution.

Thank you very much, Harazi.

Sorry, a clarification.

Your proof is perfect, neat and beautiful but I don't understand why you need to say:

"of course, we can suppose that the triangle is acute".

Where do you use this condition ? and why you can suppose it is acute?

Thank you very much, Harazi?

Because if the triangle has an obtuse or a right angle then the LHS is obviously negative...

Ok, if you are interested, I give the 'official' solution :
Let P denotes the LHS.
As Harazi noticed, we may assume that the triangle is acute. From Heron's formula, we deduce that :
16[ABC] 2  = a 2 (b 2 + c 2  - a 2 ) + b 2 (a 2  + c 2  - b 2 ) + c 2 (a 2  + b 2  - c 2 ).
Then from AM/GM :
16[ABC] 2   \geq 3(a 2 b 2 c 2 P) 1/3 .   (1)

Using that [ABC] = abc/(4R) and R = 1/ \sqrt 3 , we deduce that 16[ABC] 2  = 3a 2 b 2 c 2 .
Thus (1) leads to a 4 b 4 c 4  \geq P, as desired.

Note that equality occurs iff ABC is equilateral.

Pierre.

[quote="manlio"]Let A,B,C be three points of a circle of radius  \sqrt 3/3 and a,b,c the sides of triangle ABC. Show that

  (a 2 +b 2 -c 2 )(b 2 +c 2 -a 2 )(c 2 +a 2 -b 2 )  \leq a^4b^4c^4

Tutorat Animath 2001 dossier[/quote]
See here: http://can-hang2007.blogspot.com/2010/03/inequality-126-unknown-author.html


integration calculus probability algebra polynomial function search

The expression is equivalent to $ \int_0^1 x^a(1\minus{}x)^b\,dx$
$ \equal{}\frac{1}{a\plus{}1}x^{a\plus{}1}(1\minus{}x)^{b}\Big|_0^1\minus{}\int_0^1 \minus{}\frac{b}{a\plus{}1}x^{a\plus{}1}(1\minus{}x)^{b\minus{}1}\,dx$ $ \equal{}\frac{b}{a\plus{}1}\int_0^1x^{a\plus{}1}(1\minus{}x)^{b\minus{}1}\,dx$
$ \equal{}\frac{b}{(a\plus{}1)(a\plus{}2)}x^{a\plus{}2}(1\minus{}x)^{b\minus{}1}\Big|_0^1\minus{}\int_0^1 \minus{}\frac{b(b\minus{}1)}{(a\plus{}1)(a\plus{}2)}x^{a\plus{}2}(1\minus{}x)^{b\minus{}1}\,dx$ $ \equal{}\frac{b(b\minus{}1)}{(a\plus{}1)(a\plus{}2)}\int_0^1x^{a\plus{}2}(1\minus{}x)^{b\minus{}2}\,dx$
$ \equal{}\ldots\equal{}$
$ \equal{}\frac{b(b\minus{}1)\cdots (2)(1)}{(a\plus{}1)(a\plus{}2)\cdots (a\plus{}b)}\int_0^1x^{a\plus{}b}\,dx$ $ \equal{}\boxed{\frac{b!}{(a\plus{}1)(a\plus{}2)\cdots (a\plus{}b)(a\plus{}b\plus{}1)}}$

That is what I also did. However, I was wondering if there was any non-calculus way to compute the expression with the $ S$ sum interpretation above.

[hide="Combinatorially"]Pick $ t \in [0, 1]$.  What is the probability that $ a \plus{} b$ other randomly chosen points in $ [0, 1]$ have the property that $ a$ of them are to the left of $ t$ and $ b$ of them are to the right of $ t$?

For fixed $ t$, this is $ {a \plus{} b \choose a} t^a (1 \minus{} t)^b$ (the [url=http://en.wikipedia.org/wiki/Bernstein_polynomial]Bernstein polynomial[/url] $ B_{a, a\plus{}b}(x)$).  If in addition we pick $ t$ randomly, it becomes

$ {a \plus{} b \choose a} \int_{0}^{1} t^a (1 \minus{} t)^b \, dt$.

But in picking $ a \plus{} b \plus{} 1$ random points and deciding whether a distinguished point is in the $ (a \plus{} 1)^{st}$ position, we can pick the distinguished point after picking the location of the points, so this probability is $ \frac {1}{a \plus{} b \plus{} 1}$.  This gives

$ \int_{0}^{1} t^a (1 \minus{} t)^b \, dt \equal{} \frac {a! b!}{(a \plus{} b \plus{} 1)!}$. [/hide]

[quote="Tenoreoz"]Find a closed form for the expression
\[ S \equal{} \binom{b}{0}\frac {1}{a \plus{} 1} \minus{} \binom{b}{1}\frac {1}{a \plus{} 2} \plus{} \binom{b}{2}\frac {1}{a \plus{} 3} \minus{} \cdots \plus{} \binom{b}{b}\frac {1}{a \plus{} b \plus{} 1}( \minus{} 1)^{b}
\]
for $ a,b\in\mathbb{N}$.[/quote]

$ S \equal{} \sum_{k \equal{} 0}^b\ bC_k \frac {( \minus{} 1)^k}{a \plus{} k \plus{} 1}$

$ \equal{} \sum_{k \equal{} 0}^b\ bC_k\int_0^1 x^{a \plus{} k}( \minus{} 1)^k\ dx$

$ \equal{} \int_0^1 \sum_{k \equal{} 1}^b bC_kx^a( \minus{} x)^k\ dx$

$ \equal{} \int_0^1 x^a\ \left(\sum_{k \equal{} 0}^b\ bC_k( \minus{} x)^k\right)\ dx$

$ \equal{} \int_0^1 x^a(1 \minus{} x)^b\ dx$

$ \equal{} B(a,\ b) \equal{} \frac {a!b!}{(a \plus{} b \plus{} 1)!}$.

But that's the content of the problem - at least, the values of the [url=http://mathworld.wolfram.com/BetaFunction.html]Beta function[/url] aren't well-known to me.

[hide="Another solution"] Let $ G_n(x) \equal{} \sum_{k \equal{} 0}^{n} {n \choose k} t^k (1 \minus{} t)^{n \minus{} k} x^k \equal{} (xt \plus{} (1 \minus{} t))^n \equal{} ((x \minus{} 1)t \plus{} 1)^n$.  Then

$ \int_{0}^{1} G_n(x) \, dt \equal{} \frac {1}{(x \minus{} 1)(n \plus{} 1)} \left( x^n \minus{} 1 \right) \equal{} \frac {1}{n \plus{} 1} \cdot \frac {x^n \minus{} 1}{x \minus{} 1}$

which immediately gives $ \int_{0}^{1} {n \choose k} t^k (1 \minus{} t)^{n \minus{} k} \, dt \equal{} \frac {1}{n \plus{} 1}$ as before.  I suppose you could consider this the converse of differentiating under the integral sign. [/hide]

Both techniques can be applied almost without change to a certain problem from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=706115358&t=120504]Putnam 1984[/url].


Putnam
inequalities
Princeton
college
number theory
college contests

So, who all is taking the Putnam this year?  Any predictions as to the winning schools?

I will be a first timer, and I am just a freshman, so my goal is to get at least 1 and be above average ;)

[quote="dkasper"]... so my goal is to get at least 1 and be above average ;)[/quote]
Just to clarify: a score of 1 would never beat the mean, and it won't always beat the median.

The following table has some data about recent Putnam scores. For each year, I've listed:

1. The number of students who took it. Note the increase in participation over the last few years.
2. The mean score (on a 0-120 scale)
3. The median score
4. The percentage of the students taking it who scored zero
5. The minimum score required to have your name listed on a printout of the top approximately 500 students that gets mailed out (to the local exam supervisors, among others.)

[code]Year  # taking  Mean    Median   % zero  top 500
1998    2573    14.8     10      30.3%     28
1999    2900     6.3      0      60.2%     11
2000    2818     5.3      0      57.7%     11
2001    2954     8.9      1      44.9%     20
2002    3349    11.0      3      34.7%     24
2003    3615     7.1      1      27.8%     18
2004    3733     8.4      0      53.6%     22[/code]

And if you are planning to take it ...

The entry form is due in a little over a week, so if you haven't contacted the faculty member at your college to say you're planning to take it, do so today. If your college has some kind of qualifying exam, that should be happening now.

The reason there would be qualifying exam: while an unlimited number of students per college can take the Putnam, your local supervisor must list three of you as the "team" whose scores will count towards the college score. Yeah, we have to decide that now, in October. And the supervisor has no obligation to tell you whether you're one of those three or not; I think many will choose to let you guess.

I expect to have 6 participants here at CSULB. No, we're not going to be the #1 school, but we'll see how we do.

[quote="Kent Merryfield"][quote="dkasper"]... so my goal is to get at least 1 and be above average ;)[/quote]
Just to clarify: a score of 1 would never beat the mean, and it won't always beat the median.
[/quote]Kent, I think he ment to get at least 1 [i]problem[/i] :)

And it's easy enough to "get" one problem (or so you thought when you were writing it up) and wind up with 1 or 2 points for it. The Putnam graders are not noted for generosity.

[quote="dkasper"]Thanks for the clarification.  I have signed up to take the test, so I'm set.  Can hardly wait for december... okay, maybe I can, I have enough work to do before then that I need all of the time I can get.[/quote]Well you can find tons of stuff to work in the [url=http://www.mathlinks.ro/Forum/index.php?f=218]College Playground[/url], right here on AoPS - MathLinks ;) Good luck!

[quote="Valentin Vornicu"]Well you can find tons of stuff to work in the College Playground, right here on AoPS - MathLinks[/quote]
Don't limit yourself to just the College Playground - after all, some parts of it move fairly slowly. There have been plenty of Putnam problems that concern combinatorics, number theory, inequalities, and so forth - in other words, the stuff of the the Olympiad Section.

[quote]So, who all is taking the Putnam this year? Any predictions as to the winning schools? [/quote]

I am taking it. My school (Princeton) ought to win this year, too! Last two years, if only they had chosen the right people for their team, they'd have won. But perhaps by now they've learned.

If you're taking Putnam for the first time, check to see if your school has any kind of organized practice. Princeton does, and the coach is very entertaining and a good teacher. There's a limit to how far you can go on your own (though it might be fairly large. I know I have taught myself quite a bit of math in the last year or so; hopefully it'll show, come December).

Since I like inequalities, I usually go on google and type "inequalities filetype:pdf" or anything else (but leave teh pdf there.  If you do this, you'll be able to find tons of info in pdf format.  I managed to find huge putnam excercise booklets including complete solutions.  It sometimes gives you brief information on several theorems that are very common and useful.  Enjoy!

Masoud Zargar

[quote="Valentin Vornicu"][quote="Xevarion"]I am taking it. My school (Princeton) ought to win this year, too! Last two years, if only they had chosen the right people for their team, they'd have won. But perhaps by now they've learned.[/quote]I am pretty sure that this year there are going to be at least two (if not all) Romanians on the team :D as in the past couple of years (namely Andrei Negut and Ana Caraiani). Of course if they would accept 1st year students the team would be completed by Gabriel Kreindler.[/quote]

No, they wouldn't, because they have Aaron Pixton, who is a two-time IMO gold medalist and a former Putnam fellow (in his only attempt).  It's pretty clear what their team is going to be (if you can pick between 3 in the top 15 and 2 in the top 5, the choice is clear)


Some nice counting problems I found:

1. How many distinct four letter words can be made (ordering is not important) from the word parallelogram?

2. How many distinct amounts of money can be created from the following denominations: 1 $ \$$2 bill, 3 $ \$$5 bills, 2 $ \$$10 bills, and 1 $ \$$20 bill (excluding $ \$$0)?

Enjoy!

But if they're words, doesn't order matter for #1?

We have the letters :p,a,r,l,e,o,g,m (8 leters)and with these letters we have to form four letters words.
So we have: $ 8\cdot 8\cdot 8 \cdot 8\equal{}4096$

But we can use "a" up to three times, "l" up to three times, and "r" up to two times. You are going to have to split it up into cases.  :wink:

[hide="Problem 2"]Notice that using the three $ 5$ dollar bills, two $ 10$ dollar bills, and $ 20$ dollar bill we can make all positive multiples of $ 5$ up to $ \$55$. For each of these, we can choose to add the $ 2$ dollar bill or not giving us a total of $ 11\times{2}\equal{}22$ monetary values. But we must add $ 1$ to this because we can just have the $ 2$ dollar bill so there are $ \boxed{23}$ different monetary values we can make.[/hide]


geometry trigonometry perimeter rectangle ratio inradius incenter

This thread will be designed in order to develop the problem solving skills in high school level. My first thread will begin on the most important topic of geometry: triangles. Every geometry problem can be broken down into triangle (even circle problems involve triangle often) and I feel that it is very important for beginners to cover triangle to full extent in order to have good geometry skills.

Here's how this works. It will be very similar to regular marathons except I'll post the problems now and then. No problem will "require" the use of trig which means if you covered basic geometry course, you should be able to solve these problems. Of course, some problems may involve a lot of work but basically, no trig will be involved.

You must put your solution in "hide" text. If you get the problem right first, way to go! :D  Also, once you got the correct answer, see if there is any other way to attack the problem. Geometry problem often has more than one way to solve them.

Important note: No proof problem will be in here.

So, are we ready? Let's start...

[size=142]Problem 1[/size]

Source: NC Comprehensive Exam

I'd like to introduce one of the most important theorems in triangles' history ---> Pythagorean Theorem.

Pythaogrean Theorem states that for right triangle $ABC$, we have $a^2+b^2 = c^2$. Remember this and do problem 1.

[color=blue]Consider an isosceles triangle with area 60 square units and a base of length 10. Find the length of this triangle's perimeter.[/color]

[hide]Consider an isosceles triangle with area 60 square units and a base of length 10. Find the length of this triangle's perimeter

A = (1/2)bh
60 = (1/2)(10)h
h = 12

After getting the height, draw a line that will divide the base into two forming to congruent right triangles. So the base of the triangles is 5 units each. That is 1/2 of 10

by Pythagorean theorem

c^2 = a^2 + b^2
       = 5^2 + 12^2
       = 169
   c  = sqrt(169)
       = 13

P = 10 + 13 + 13 = 36 units.[/hide]

C'mon guile, hide your answer  ;)

[hide]Since the area of the triangle is 60 units, its height must be 12 units.
It's an isosceles triangle, so the base is divided into equal segments of 5 by the height.

We see two congruent triangles from the height division, each a right triangle and each with sides 5 and 12. That means the other side (the legs) is 13.

13+13+10=36 units[/hide]

I thought only him was supposed to post problems. Since I hid my answer, I 'll post 2.  :D

[size=200]Problem 2[/size]

A rectangle has a base of 7 and a diagonal of 25. What is the area of this rectangle?

My hidden solution:  :lol:

[hide]Split the rectangle into two congruent right triangles and find the area of both triangles.

From the Pythagorean Theorem, $b^2=c^2-a^2$, so $b=\sqrt{25^2-7^2}=24.$

The area of the rectangle is thus $24 *7$ or $168$[/hide]

correct, the pythagorean theorem again  :D

[size=150]Problem# 3[/size]

Given that I is the center of $\triangle ABC$, $AB=AC=5$, and $BC=8$, find the distance $AI$.

And of course, this problem involves the Pythagorean Theorem as well.

[hide]Half the base would be 4, and the leg is 5, so the height must be 3. The center is I, at half down the height, so AI=1.5 [/hide]

[size=200]Problem 4[/size]

Without using Heron's formula or trigonometry (pythagorean theorem excepted), find the area of a triangle with sides 13, 14, and 15.

I would post the solution but I think somebody else should try.

[quote="chess64"][quote="236factorial"]Without using Heron's formula or trigonometry (pythagorean theorem excepted), find the area of a triangle with sides 13, 14, and 15.[/quote]

[hide]
You can split it into two triangles- 5,12,13 and 9, 12, 15. The base is 9+5=14, and the height is 12. $14/2 \cdot 12 = 84$.
[/hide][/quote]

interesting, continue  :D

4
[hide]
Draw the altitude to the side length 15.  The altitude divides the side into lengths of $x$ and $15-x$, and the altitude has length $h$.  So we get
$h^2 + x^2 = 13^2$
$h^2 + (15-x)^2 = 14^2$
subtracting
$30x-225 = 27$
$x = \frac{99}{15}$
Then
$h^2 = 13^2 - \left(\frac{99}{15}\right)^2$
$h = \frac{56}{5}$
area is $\frac 1 2 b h$, or $\frac 1 2 \cdot \frac{56}{5} \cdot 15 = 84$
[/hide]

edit: woah i'm late there

A little late, deej21. (Read my solution :D ) but you did get it right. :)

Problem 15 is a good problem to apply Heron's Formula.

[hide="Hint"]

Semiperimeter in this case is (5+5+x)/2 where x is the length of AC. [/hide]

I saw some posts that require the use of $ab \sin C \frac{1}{2}$ formula for area but try not to post those problems yet. I want to make this thread go over several important topics before moving into sin area formula.

[quote="chess64"][hide]From triangle inequality, we have $37-m < PR < 37+m$.
$(37+m)-(37-m)-1=2m-1$.[/hide]

[size=150]Problem 15[/size]

The area of $\triangle ABC$ is $100$ square units. If $AB=BC=5$, find $AC$.[/quote]

wasnt there a very similar problem on your elimination exam?

For #15, are you sure your problem is solvable.

I don't think it works. Here is the hidden text for why.

[hide="Why"]

We have triangle with area of 100 and side length 5 and 5. Say unknown side is $x$. Using sin area formula we get:

$5 \cdot 5 \cdot \frac{1}{2} \cdot \sin X = 100$
$\sin X = 8$

This is impossible since sin function only goes to maximum of 1. [/hide]

Edit your problem perhaps?

[hide="Solution"]

Using Heron's Formula, we have a nasty looking math equation.

$12 = \sqrt{(\frac{10+x}{2})(\frac{10+x}{2} - 5)^2 (\frac{10+x}{2} - x)}$

Expanding this will be a nasty work but here's a quick trick.

Note that the last expression equals $\frac{10-x}{2}$ which means $x < 10$. I prefer integer result so I picked $x = 8$ which gives me $\frac{10-8}{2} = 1$. Now:

$\frac{10+8}{2} = 9$
$\frac{10+8}{2} - 5 = 4$ and there are two of them so 16.

Thus, our expression now becomes:

$\sqrt{9 \cdot 16 \cdot 1}$ and this equals 12 so our answer is 8. [/hide]

Before posting a next problem, I like to introduce one of the most famous inequalities in triangle's history ---> Triangle Inequality (kind of surprising title :D )

[size=150]Triangle Inequality[/size]

This states that for $\triangle ABC$ with $AC > AB,BC$, we have $AB+BC > AC$.

The concept of the proof is pretty simple. Let's say you have segment $AC$ and point $B$ that is not on segment $AC$. The shortest way to go from $A$ to $C$ is a straight line. Thus, going $AB+BC$ should take longer than just going $AC$. In another word, $AB+BC > AC$.

Here is a theorem that will help you in next problem. Pretty easy concept but quite useful.

[size=150]Isosceles Triangle Theorem[/size]

For triangle $ABC$ where $AB = BC$, we have $\angle BAC = \angle BCA$. I won't post the proof but even it's an obvious theorem, it's one of the useful theorems you'll meet.

[size=142]Problem 16[/size]

This is a bit of angle chasing so have fun..

Let $ABCD$ be a rectangle with $BC = 2AB$, and let $BCE$ be an equilateral triangle crossing through the rectangle. If $M$ is the midpoint of $EC$, how many degrees are in angle $CMD$?

Actually, Problem 15 was 6.

[hide]
Heron's Formula is actually messier in this case.

We have $xh=24$, where $x=AC$ and $h$ is the altitude to $AC$, meeting at $D$. We know that $BD=DC=x/2$, so $(x/2)^2+h^2=25$ by the pythagorean theorem. Simplifying and substituting, we see $x^2+4h^2=100$ and $x=24/h$, so $576/h^2+4h^2=100 \Rightarrow 144 + h^4=25h^2$. Solving the quadratic, we find $h^2=16 \Rightarrow h=4 \Rightarrow x=6 \ \ \ \ \blacksquare$
[/hide]

We're "BOTH" right.

In my answer, I have triangle ABC with AC = 8 and height of 3. 8*3*0.5 = 12

In your answer, you have triangle ABC with AC = 6 and height of 4. 6*4*0.5 = 12

I should've known that Heron's Formula in this case could've had two answers but I was too lazy expanding it. Here's how there are two solutions. In chess64's work, he had this equation:

$h^4+144 = 25h^2$

Now, set $g = h^2$ then we have $g^2+144 = 25g$ and this factors into $(g-9)(g-16)$ so $g = 9,16$. Setting these these to $h^2$ gives 8 and 6 for $AC$, respectively.

Thanks for bringing this up. I should've been more careful.

[hide="Answer and Solution to 16"]75 degrees.  This is because triangle CMD is an isoceles triangle, and by finding a complementary angle of one of the angles of the equilateral triangle, you find that angle is 30 degrees.  Then you subtract that from 180 and divide by 2.[/hide]

That's the correct answer. Good job. Next problem?

I don't have any problems (Or it would take too much time that I don't have...), so whoever feels like it gets to post.

[size=150]Problem 17[/size]

Source: 1984 AHSME

This problem is a good application of Pythagorean Theorem and use of similar triangles.

A right triangle ABC with hypotenuse AB has side AC = 15. Altitude CH divides AB into segments AH and HB, with HB = 16. Find the area of the triangle ABC.

[quote="mathgeek2006"]I drew my triangle too small to begin with (I always draw diagrams).  So I then did guess and check and within 5 seconds I had it. :oops: Anyway, the answer is 150.  I might post a solution in a few minutes, but otherwise....[/quote]

Here is one "very" important tip that I learned over years.

When you come across a geometry problem, try to draw the problem accurate in large scale. It doesn't have to be great masterpiece but good enough so that you can check your answer later. I think in AIME, some people brute forced the later problems by drawing exact length given in problem and actually got them right! :lol: Of course, not "every" people and "every" problem works that way but it's a shot.

Your answer is right but please post the solution.

The two triangles (AHC and HCB) are similar because angle A + angle ACH = 90 degrees, so they are complementary.  But angle C is a right angle, so Angle ACH + angleHCB = 90 degrees making the triangles similar by the AA similarity (Pick one, postulate or theorem).  This is only a partial proof--I'm working on the rest.


so i beleive it's ok(assuming this)
let's start with organic..

1.$X_{2}$ along with some $Zn$ acts as a Protecting Group for the double bond.
halogen addition to a double bond protects it, by converting it to vicinal dihalide. Then to remove the protecting group (the two halogens) and re-obtain the double bond we can add some zinc.

EDIT;due to Carcul

was too lazy :maybe:

[b][u]Henderson-Hasselbalch Equation[/u][/b]

For a reaction: $ HA_{(aq)}+H_{2}O_{(l)}\rightleftharpoons A^{-}_{(aq)}+H_{3}O^{+}_{(aq)}$

We have: $ pH = pK_{a}+\log_{10}\frac{[A^{-}]}{[HA]}$

[b]relating $ C_{p}$ and $ C_{v}$[/b]

$ C_{p}-C_{v}=\frac{ \alpha^{2}TV}{\kappa_{T}}$

where $ \alpha$ is the linear coeffecient expansion of that substance,$ \kappa_{T}$ is the isothermal compressibilty of the substance

[b]Hoffmann's Ammonolyis reaction[/b]
on reacting $ RX$ with $ aq.KOH$ at low temperature with ammonia a mixture of quaternary ammonium salt, tertairy,secondary and primary amines are formed

[b]depression in freezing point/Elevation in boiling point[/b]

let a solution of molality $ m$ containing a non-volatile non dissasociating solute be formed then the depression in freezing point and levation of boiling point of such a solution is given by

$ \Delta T_{f}=K_{f}m$

$ \Delta T_{b}=K_{b}m$
respectively

[quote="Pardesi"]relating $ C_{p}$ and $ C_{v}$[/quote]

That relation is called the [i]Mayer Relation[/i] (or [i]Generalized Mayer Relation[/i]).

$ \text{N}_{2}\text{H}_{2}$ selective reduction of alkene .see [url=http://www.mathlinks.ro/viewtopic.php?p=934711#934711]here[/url] .
By-Carcul

[url=http://www.mathlinks.ro/viewtopic.php?t=172267]cis-trans naming of cycloalkanes[/url] by Carcul

General Nernst Equation:

$ E \equal{} E^{\circ} \minus{} \frac {RT}{nF} \ln Q$ where $ Q$ is the reaction quotient for the reduction reaction.

[quote="BanishedTraitor"]for the reduction reaction[/quote]

For the redox reaction.

Meaning of the symbols:

$ E^o$ is the potential of the cell at standard conditions (solutions with molarity of 1M, gases at 1 bar pressure);
$ E$ is the potential (or emf) of the cell at non-standard conditions;
$ R$ is the gas constant;
$ T$ is the absolute temperature;
$ n$ is the number of electrons transferred [i]in the balanced equation[/i];
$ F$ is Faraday's constant;
$ Q$ is the reaction quotient.

[quote="TaiPan~SP!"]I am not sure about this one but I think it works:

To determine the shape of a molecule of the form $ AX_m$

$ \text{Number of Lone Pair Electrons}$ $ \equal{} \frac {(\text{number of  valence electron}) \minus{} (\text{charge of molecule}) \minus{} m\times|\text{valency of X}|}{2}$

And in doing so we can determine the shape of $ AX_m$ as well as the hybridization of $ A$.

e.g. $ TeCl_4$

$ N \equal{} \frac {6 \minus{} 0 \minus{} 4 \times 1}{2} \equal{} 1$ and hence the shape of $ TeCl_4$ is a $ \text{See \minus{} Saw}$ and the hybridization of $ Te$ is $ dsp^{3}$.[/quote][quote="TaiPan~SP!"]I am not sure about this one but I think it works:

To determine the shape of a molecule of the form $ AX_m$

$ \text{Number of Lone Pair Electrons}$ $ \equal{} \frac {(\text{number of valence electron}) \minus{} (\text{charge of molecule}) \minus{} m\times|\text{valency of X}|}{2}$

And in doing so we can determine the shape of $ AX_m$ as well as the hybridization of $ A$.

e.g. $ TeCl_4$

$ N \equal{} \frac {6 \minus{} 0 \minus{} 4 \times 1}{2} \equal{} 1$ and hence the shape of $ TeCl_4$ is a $ \text{See \minus{} Saw}$ and the hybridization of $ Te$ is $ dsp^{3}$.[/quote]

didn't get that.
number of valence electron- of whom?
and how did you determine the shape if you know the value of N?


inequalities
function
inequalities proposed

For every $1 \leq a, b, c, d \leq 2$ prove the inequality

$\frac{4}{3} \leq \frac{a}{b+cd} + \frac{b}{c+da} + \frac{c}{d+ab} + \frac{d}{a+bc} \leq 2$  .

The inequality $\leq$ can be proved as follows:
- let $f(a)=\frac{a}{b+cd}+\frac{b}{c+ad}+\frac{c}{d+ab}+\frac{d}{a+bc}$. f(a) is a convex function (since it's a sum of a linear function, which has no influence on convexity, and three homographic functions which are convex), it achieves its maximum at $(a,b,c,d)=(x_{1},x_{2},x_{3},x_{4})$, where $x_{k}$ is either 1 or 2 (see [url=http://www.mathlinks.ro/Forum/topic-15431.html]http://www.mathlinks.ro/Forum/topic-15431.html[/url] - the third post from the top). We easily check that for (a,b,c,d)=(1,1,1,1) our function achieves 2, so it's the maximal value and thus the inequality is proved.


number theory
prime numbers
number theory unsolved

Let p,q odd prime numbers such that p=q+4x, for some integer x. Prove that
(x/p)=(x/q), where (a/b) denotes the symbol of Legendre a to b.


How many intersecting angles with measures less than $ 180^{\circ}$ are formed by the 4 intersecting lines shown?
[asy]import olympiad;
size(150);
pair A=(0,0), B = 4*dir(0), C=4*dir(40),D=4*dir(100),E=4*dir(150),F=4*dir(180),G=4*dir(220),H=4*dir(280),I=4*dir(330);
dot(A,red);
draw(A--B,EndArrow);
draw(anglemark(B,A,C,20),black);
label("$40^\circ$",anglemark(B,A,C),4*dir(20));
draw(A--C,EndArrow);
draw(A--D,EndArrow);
draw(A--E,EndArrow);
draw(anglemark(D,A,E,20),black);
label("$50^\circ$",anglemark(D,A,E),4*dir(120));
draw(A--F,EndArrow);
draw(A--G,EndArrow);
draw(A--H,EndArrow);
draw(anglemark(G,A,H,20),black);
label("$60^\circ$",anglemark(G,A,H),4*dir(250));
draw(A--I,EndArrow);[/asy]

Angles that measure less than 180 degrees are smaller than straight lines. Counting, I can find 17. Anyone have a different answer?

I got $ 8\plus{}7\plus{}6$ or $ \boxed{21}$...  :maybe:

Ernie's right.  First you count the amount of angles using one "angle", then the amount using 2, and 3.  If you used 4, the  measure would equal 180 degrees.

I got 24:

The number of ways you can choose two of the eight 28 minus the number of ways you get straight lines (the non straight lines should have one angle under 180 if they aren't 180) which is 24... are there any others that shouldn't be counted?

i asked this.

i got 24 also, but the answer is 12.

[color=#80BF40]Same I got the same answer as you! 24[/color][quote="GeorgiaTechMan"]This is a very vague question.  The answer should be 24 because it never asked how many noncongruent angles are there.[/quote]

Wow. I also got 24. Do they really expect us to count the number of distinct measures for each angle, because that doesn't seem feasible in 45 seconds (FTW Countdown round).

I got 24, I guess you are supposed to assume that two vertical angles only count as one. But as they all say, to ASSUME makes an a$$ out of u and me.


Here is one interesting problem

Find all pairs of positive integers $(a,b)$ such that number
\[\frac1a+\frac1b+\frac{a}{1+b}\] is also positive integer.

:)  :)  :)

You should only examine the case that the number $ \frac{a(b+ab)}{b^2+b+a} $ is an integer.

well is someone going to post the solution

a should be a factor of b(b+1).
If a is a factor of b,then suppose b=ka,we can have
$\frac{ka+1+ka^2}{ka(ka+1)}$
should be a fraction the denominator of which is a.
so k=1,then b=a,so a is a factor of a+1,so a=b=1.But the result isn't a integer.
If a is a factor of b+1,suppose b+1=ka,then
$\frac{1}{a}+\frac{1}{b}+\frac{1}{k}$
should be a integer.then(a,b,k)=(1,1,1)(1,2,2)(2,4,4)(2,3,6)(3,3,3)
so no (a,b) satisfy the condition.

Here's another.
\[
\frac1a + \frac1b + \frac a{1+b} = k
\]
so multiplying up
\[
kab(1+b) = b(1+b) + a(1+b) + a^2 b
\]
which means that  $b \mid a(1+b)$ and since $b$ is coprime to $1+b$, we have $b \mid a$, say $a=bc$.  Now
\[
kbc(1+b) = (1+c)(1+b) + b^2 c
\]
and since $c$ is coprime to $1+c$ we have $c \mid 1+b$; and since $1+b$ is coprime to $b$ we have $1+b \mid c$.  So $c = 1+b$.  Then
\[
kb(1+b) = 2+b + b^2 = 2 + b(1+b)
\]
so $b(1+b) \mid 2$.  This implies $b=1$, hence $c=2$ and $a=2$: which is not a solution.

That was the solution i was looking for, than you, because i solved this problem a few months ago, and i just couldn't remember that solution, i was remembering only the beginning but now i see it all, thank you


inequalities
function
inequalities solved

Given a,b,c positive such that a+b+c=1 and n natural positive prove

((1/a^n)-1)((1/b^n)-1)((1/c^n)-1) \geq (3^n-1)^3

I have a solution.

More general : Let n be a positive number .
If  {a(k)} are m positive numbers such that  a(1)+...+a(m)  = 1
then ((1/a(1)^n) - 1)*...*((1/a(m)^n)-1) >= (m^n - 1)^m
The proof is easy : ln((1/x^n) - 1) is convexe, Jensen inequality , ..etc ..

You are wrong, Diogene: ln(x<sup>-n</sup>-1) is not convex function.


induction
real analysis
real analysis unsolved

[quote="Gauß"]Let $(S_i)_{i\in \mathbb{N}}$ be a sequence of nonempty finite sets. For an arbitary $a\in S_{i+1}$ let $\phi (a)\in S_i$ ($\forall i\in \mathbb{N}$).
Proof that there exist a sequence $(a_i)_{i\in \mathbb{N}}$ with the following properties.

(1)  $a_i\in S_i$

(2) $\phi (a_{i+1})=a_i$

Does it also hold for infinity $S_i$?[/quote]

Well, if we take $S_n$ for any $n \in \mathbb{N}$, we can just pick any element $a_n \in S_n$. Then we have $a_{n-1} = \phi(a_n) \in S_{n-1}$. By induction, we can just choose the sequence by $a_{i-1} = \phi(a_i)$ because $a_i \in S_i \Rightarrow a_{i-1} = \phi(a_i) \in S_{i-1}$. Since $n$ can be arbitrarily large, this works for infinite $S_i$, I think.

[quote="Gauß"]A proof by induction is incorrect because it does only hold for arbitary large finite $n$. The Problem is, that if you had choosen a sequence ${(a_n,a_{n-1},\ldots , a_1)}$ with the property $\phi (a_{i+1})=a_i$, the element $a_n$ needn't to have a preimage.
For instance this is the case when all elements of the set $S_{i+1}$ will be mapped to one element in the set $S_i$ ($|S_i|>1$).[/quote]

Sorry, I probably didn't quite explain it clearly. The induction is down from $n$, not from $n \rightarrow n+1$. I chose $n$ first (arbitrarily) and then inductively proved that the sequence $(a_n, a_{n-1}, \ldots, a_1)$ exists. So, for example, if I choose $n = 10$, I can show that $(a_{10}, a_9, \ldots, a_1)$ exists by starting with any $a \in S_{10}$. Then if I go to $n = 11$ (or any positive integer, since it's not an induction on $n$), I choose a completely different sequence $(b_{11}, b_{10}, \ldots, b_1)$ starting with any $b \in S_{11}$.

Yes it does always work, but you have prooven that there exist such a sequence for every finite $n$.
I need a proof that there exist an infinite sequence, so that every element has a preimage, this needn't to be true.

Example:

Let $S_i=\mathbb{N}$ and $\phi(n)=n+1$ then for every $k$ there exist a finite sequence, but not an infinite sequence.

I see. I'm not sure how to extend it to the infinite case.

I have the solution.

The proof is made by contradiction.
Assume that all sequences (starting with an element in $S_1$) are finite, then the set of all sequences with property (1) and (2) are finite. I will proof this now.
Let $(a_i)$ and $(b_i)$ be two distinct sequences then there exist an $N$, so that $a_N\not = b_N$, it follows that $a_i\not = b_i \forall i\geq N$. Let the $|S_i|$ be bounded by $M$, then there are at most $M$ such sequences. Let now $(a_i)$ be the "longest" sequence, say $k$ Elements, then there exist a sequence $(\phi^i (a_{k+1}) )_{i=0}^k$ wich is longer. This contradicts the assumption that there exist a longest sequence.

The disadvance of this proof is, that the sets have to be bounded by an positive integer.


function algebra polynomial calculus integration derivative trigonometry

is it true that if $ f: \mathbb{R}\minus{}>\mathbb{R}$ is $ C^2$ than f is $ C^{\infty}$? how can I prove it or find a counterexample?

What?  Are you asking if any function $ f$ that is $ C^{2}$ is smooth?  Not true at all.  Counter-example: $ f(x)\equal{}x^{3}$.  This function is $ C^{3}$, but certainly not $ C^{\infty}$.

[quote="JRav"]What?  Are you asking if any function $ f$ that is $ C^{2}$ is smooth?  Not true at all.  Counter-example: $ f(x) \equal{} x^{3}$.  This function is $ C^{3}$, but certainly not $ C^{\infty}$.[/quote]
??? Polynomials are mega-smooth

To find a real counterexample: Take any continuous but not differentiable function. Since it is continuous it has a primitive given by an integral over $ (0, x)$. This new function also has a primitive in the same way.

Anyway, continuous differentiability is certainly not a strong enough condition to guarantee infinite differentiability on the real line. However, differentiable functions in the complex plane are infinitely differentiable. Being differentiable (in an open set) in the complex plane is a very strong property.

Maybe this is what JRav meant to say:

$ f(x)\equal{}\begin{cases}x^3,&x>0\\0,&x\le0\end{cases}.$

That function is $ C^2$ but not $ C^3.$ (It has a third derivative except at one point, and that third derivative is locally bounded.) Or you could use $ g(x)\equal{}|x|^3,$ which does the same thing.

Or you could be a little more spectacular: start with a continuous nowhere differentiable function (see Weierstrass example or related examples), then take two primitives (as Kalle suggested). That would give you a $ C^2$ function with no third derivative anywhere.

For instance, $ f(x)\equal{}\sum_{k\equal{}1}^{\infty}\frac{3^k}{4^{3k}}\cos(4^kx).$

The counterexample I always saw was $ f(x) \equal{} x^n\sin(\frac {1}{x})$ when $ x \neq 0$ and $ 0$ when $ x \equal{} 0$, for whichever $ n$ suits you.