Eστω $ p$ πρωτος και $ n,a,b$ θετικοι ακεραιοι
Εφαπτεται στον $ x'x$  η παραβολη $ p^nx\equal{}\frac{a}{2b}x^2\plus{}\frac{a\plus{}1}{2(b\plus{}1)}$;

[quote="xp"]Eστω $ p$ πρωτος και $ n,a,b$ θετικοι ακεραιοι
Εφαπτεται στον $ x'x$  η παραβολη $ p^nx \equal{} \frac {a}{2b}x^2 \plus{} \frac {a \plus{} 1}{2(b \plus{} 1)}$;[/quote]

To $ y$ είναι στη θέση του $ x$ της LHS? :huh:

Cheerio,

Durandal 1707

Eχεις δικιο στις 2 το βραδυ αλλα βλεπουν τα ματια και αλλα επεξεργαζεται ο νους....
Η παραβολη ειναι $ y\equal{}\frac{a}{2b}x^2\minus{}p^nx\plus{}\frac{a\plus{}1}{2(b\plus{}1)}$

Καλησπερα σας.
Η απαντηση ειναι οχι.
Η αποδειξη που βρηκα ειναι η εξης.
λοιπον για να εφαπτεται πρεπει η διακρινουσα να ειναι ιση με το 0.
δηλαδη η εξισωση
$ a(a \plus{} 1) \equal{} p^{2n}b(b \plus{} 1)$(1)
να εχει λυση
Η εξισωση μας γραφεται ισοδυναμα
(- Η σκεψη μου ειναι να σχηματισω τετραγωνα ωστε να παραγοντοποιηθει η εξισωση)
$ p^{2n} \minus{} 1 \equal{} (p^n(2b \plus{} 1) \plus{} (2a \plus{} 1))(p^n(2b \plus{} 1) \minus{} (2a \plus{} 1))$
αρα
$ p^{2n} \minus{} 1 \geq p^n(2b \plus{} 1) \plus{} (2a \plus{} 1) > 2a \plus{} 1 \leftrightarrow p^{2n} > 2(a \plus{} 1)$(2)
Oμως, η (1) λογω του οτι $ gcd(a,a \plus{} 1) \equal{} 1$ μας δινει
ειτε
$ p^{2n} | a \plus{} 1$
ειτε
$ p^{2n}|a$
και οι δυο σχεσεις ειναι αδυνατες λογω της (2).
QED  :)


Is it true that if $ f_n\rightarrow f$ in $ L^1(K)$ where K is a compact set (for example in $ \mathbb{R}$ for $ K \equal{} [0,1]$)
then $ f_n\rightarrow f$ almost everywhere ? Ok its not, it was too late :)

No. Standard counterexample: enumerate the double sequence $ f_{i j}\equal{}\chi_{[(i\minus{}1)2^{\minus{}j},i2^{\minus{}j}]}$, where $ j\ge 1$, $ i\equal{}1,\dots,2^j$.


this was question number 2 on a 1995 high school math league test.

What is the integer n for which 5^n+5^n+5^n+5^n+5^n=5^25

if you want harder problems, just ask. or if you want easier problems, ask also. make sure to put it in spoiler font also.

this one is open to everybody i guess, since its a high school mathleague. but give the little uns a chance.

Wait, is this for people who were in 8th grade, or people who are going into 8th grade? Well, technically, I'm not a 9th grader yet...so without further adue...

[hide]

5^n+5^n+5^n+5^n+5^n = 5* 5^n = 5^ n+1 = 5^25

Thus n= 24

[/hide]

i dont really care. its for the purpose of learning. i think ill boost the difficulty up next problem, unless a pm tells me otherwise. good job though

[hide="clik dis cause its da answer"]5^n+5^n+5^n+5^n+5^n = 5* 5^n = 5^ n+1 = 5^25
so n=24
[/hide]

:mad: Barnacle, please stop bringing back topics from 2003 and 2004!  :roll:

[quote="Syntax Error"]

What is the integer n for which 5^n+5^n+5^n+5^n+5^n=5^25[/quote]

[hide="um...i guess"] well u have 5^n power 5 time so you van right the equaton out like this
5(5^n)=5^25
but you added another power to 5 so...
5^n+1=5^25
the bases are the same so u can now ignore them.
n+1=25
-1      -1
----------
n=24

24 YAY!!!!!! :cool: i rock[/hide]


Another variable is not set in the portal page too.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

If $ e : R \to k$ is a homomorphism of $ k \minus{}$algebras and $ \varphi : R \to k[x_1,\dots,x_n]$ is an injective homomorphism of $ k$-algebras, does there exist a homomorphism $ \psi : k[x_1,\dots,x_n] \to k$ such that $ e \equal{} \psi \varphi$.? (Here $ k$ is an algebraically closed field).

No: Take $ \varphi: R\equal{}k[X]\to k[X,Y]/(XY\minus{}1)$ with $ \varphi(X)\equal{}X$ and $ e(X)\equal{}0$.

In abstract terms this means in particular that not every dominant morphism of varieties is surjective.

But then $ \varphi$ isn't a map to $ k[x_1,x_2]$, but a quotient of it instead. I suspect that there's something special about affine $ n$-space in particular.

Although this breaks it, instead: let $ k[r,s] \to k[x,y]$ by $ r \mapsto x$ and $ s \mapsto xy$. Then $ e(r) \equal{}0, e(s) \equal{} 1$ does not extend.

However, I suspect that if $ k[x_1,\dots,x_n]$ can be obtained as an integral extension of $ R$, then the claim is true.

Proof: the map extends iff the extension of $ \ker e$ into $ k[x_1,\dots,x_n]$ is not the unit ideal, using Hilbert's Nullstellensatz. By the going-up theorem, since $ \ker e$ is prime, there's a prime ideal in $ k[x_1,\dots,x_n]$ lying above $ \ker e$. Therefore the extension is not the unit ideal.

So, a nice generalization:

If $ R \subset S$ is an integral extension of finitely generated $ k$-algebras, and $ S$ is a domain, then every homomorphism $ R \to k$ extends to a homomorphism $ S \to k$.

Note that the assumptions on your generalization can be slightly more general: If $ A\subset B$ is an integral ring extension, then every homomorphism $ A\to k$ where $ k$ is an algebraically closed field extends to $ B$.


function
integration
calculus
derivative
calculus computations

Give an  example of a Riemann integrable function $ f$ on $ J$ such that the function $ F$, defined for $ x\in J$ by

$ F(x) = \int_{a}^{x} f$,

does not have a derivative at some points of $ J$. An example of an integrable function $ f$ such that $ F$ is not continuous on $ J$?

My question is would something like a step function ($ f(x) =[x]$) work? or even some variation of a Dirchlet function for both examples asked?

Also, one curiosity about step functions such as $ f(x) = [x]$. I 'm pretty sure $ f$ is Riemann integrable, for example on some interval from [0,2] let' say, but is it a derivative of any function, since right now, none come to mind. why is it or is it not a derivative of any function?

thanks!

Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=173651]your own thread[/url] from a month ago.

The step function idea is fine.

Let's do it this way: let $ f(x)\equal{}\text{sgn}(x)\equal{}\begin{cases}1&x>0\\0&x\equal{}0\\\minus{}1&x<0\end{cases}.$

Then $ F(x)\equal{}\int_0^xf(t)\,dt\equal{}|x|.$ You can show quite directly from the definition of the derivative that $ F$ is not differentiable at $ 0.$

However, you're not going to find any examples in which $ F$ is itself discontinuous. That simply can't happen, and the proof is quite simple.

Any function which is Riemann integrable on an interval must be bounded on that interval; else we can find a sequence of Riemann sums that tends to $ \infty.$ So we have that there exists some $ M$ such that $ |f(x)|\le M$ on $ J.$

WLOG, let $ y>x.$ Then

$ |F(y)\minus{}F(x)|\equal{}\left|\int_x^yf(t)\,dt\right|\le\int_x^y|f(t)|\,dt\le M(y\minus{}x).$

Hence, $ F$ is not merely continuous, but satisfies something stronger: a Lipschitz condition.


There are given three points P,Q,R.Build the quadrilateral ABCD for wich those three points are the midlles of the sides AB,BC and CD,and AB=BC=CD

[hide="Solution"]Because $ AB \equal{} BC$ and $ P,Q$ are midpoints of $ AB,BC$ then $ BP \equal{} BQ$, it mean $ B$ must be on the perpendicular bisector $ a$ of $ PQ$.
Similarly $ C$ must be on the perpendicular bisector $ b$ of $ QR$.
We must to contruct $ B\in a, C\in b$ so that $ Q$ is midpoint of $ BC$: Delate $ V^{ \minus{} 1}_Q: a\rightarrow a',B\rightarrow C$ then $ C\in a'$, it mean $ C \equal{} a'\cap b$!
When we have $ C$, it is very easy to finish your problem![/hide]


induction
number theory unsolved
number theory

I agree, it's nice!
(but this is not a good title, see http://www.mathlinks.ro/Forum/viewtopic.php?t=135915  :wink: )

Let's write $a_{i}= 2005 b_{i}+r$.
From $a_{1}> a_{2}> \ldots > a_{n}> 0$ we get $b_{1}> b_{2}> \ldots > b_{n}\ge 0$ and so, by induction, $b_{i}\ge n-i$.

From (c) we derive: $x_{1}(2005b_{1}+r)+\ldots+x_{n}(2005b_{n}+r) = 2005$ and, dividing by 2005, $x_{1}b_{1}+\ldots+x_{n}b_{n}+r'(x_{1}+x_{2}+\ldots+x_{n}) = 1$ where $0 \le r' < 1$ is $\frac r{2005}$.
Now we subtract (b) from this and get:
$x_{1}(b_{1}-(n-1))+\ldots+x_{n-1}(b_{n-1}-1)+x_{n}b_{n}=-r'(x_{1}+\ldots+x_{n})$. But $-r'(x_{1}+\ldots+x_{n}) \le 0$.

From (a) it is easy to understand that every linear combination of $(x_{i})$ with non negative coefficients is non negative. But $x_{1}(b_{1}-(n-1))+\ldots+x_{n-1}(b_{n-1}-1)+x_{n}b_{n}$ [b]is[/b] a linear combination of non negative coefficients, and it is $\le 0$.
From this we conclude that $b_{1}-(n-1) = \ldots = b_{n-2}-2 = b_{n-1}-1 = b_{n}= 0$, then $b_{i}= n-i$ and $a_{i}= 2005(n-i)+r$.   :)


algebra
polynomial
inequalities
inequalities proposed

If $ a,b,c$ are nonnegative real numbers, then

$ \sum a(a\minus{}b)(a\minus{}c)(a\minus{}2b)(a\minus{}2c) \ge 0$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a(a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c) \ge 0$.[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$ Hence, $ t\geq1$ and
$ \sum_{cyc} a(a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c) \ge 0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5 \minus{} 3a^4b \minus{} 3a^4c \plus{} 2a^3b^2 \plus{} 2a^3c^2 \plus{} 9a^3bc \minus{} 8a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \plus{} (13u^2 \minus{} 16v^2)w^3\geq0\Leftrightarrow$
$ \Leftrightarrow3u(3u^2 \minus{} 4v^2)^2 \plus{} (13u^2 \minus{} 16v^2)w^3\geq0.$
Hence, it remains to prove our inequality for $ 1\leq t\leq\frac {16}{13}.$
But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Hence, we need to prove only that $ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \plus{} (13u^2 \minus{} 16v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\geq0.$
But
$ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \plus{} (13u^2 \minus{} 16v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow u^5 \minus{} u^3v^2\geq2(16v^2 \minus{} 13u^2)\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow u^3\geq2(16v^2 \minus{} 13u^2)\sqrt {u^2 \minus{} v^2}\Leftrightarrow$
$ \Leftrightarrow t^3\geq4(16 \minus{} 13t)^2(t \minus{} 1)\Leftrightarrow(15t \minus{} 16)^2(3t \minus{} 4)\leq0,$
 which is true for $ 1\leq t\leq\frac {16}{13}.$ :)
An equality occurs when $ b \equal{} c$ and $ t \equal{} 1$ or $ t \equal{} \frac {16}{15}$ id est,
when $ a \equal{} b \equal{} c$ or when $ a \equal{} 2b \equal{} 2c$ and for cyclic permutations of the last.

:)  Prove for degree 7:

For real $ k,\ell<2$ such that $ (2 \minus{} k)(2 \minus{} \ell)\ge1$, we have
\[ \sum a(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) (a \minus{} \ell b)(a \minus{} \ell c) \ge 0
\]
for all $ a,b,c\ge0$.

[quote="spanferkel"]:)  Prove for degree 7:

For real $ k,\ell < 2$ such that $ (2 \minus{} k)(2 \minus{} \ell)\ge1$, we have
\[ \sum a(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) (a \minus{} \ell b)(a \minus{} \ell c) \ge 0
\]
for all $ a,b,c\ge0$.[/quote]
My computer shows that it is not true.
For instant, the inequality below ($ k\equal{}0$ and $ l\equal{}3/2$) is false:
$ \sum a^3(a\minus{}b)(a\minus{}c)(2a\minus{}3b)(2a\minus{}3c) \ge 0$.

[quote="arqady"] An equality occurs when $ b \equal{} c$ and $ t \equal{} 1$ or $ t \equal{} \frac {16}{15}$ id est,
when $ a \equal{} b \equal{} c$ or when $ a \equal{} 2b \equal{} 2c$ and for cyclic permutations of the last.[/quote]
In addition, equality occurs for $ a\equal{}0$ and $ b\equal{}c$, or any cyclic permutation. :wink:

[quote="Vasc"][quote="arqady"] An equality occurs when $ b \equal{} c$ and $ t \equal{} 1$ or $ t \equal{} \frac {16}{15}$ id est,
when $ a \equal{} b \equal{} c$ or when $ a \equal{} 2b \equal{} 2c$ and for cyclic permutations of the last.[/quote]
In addition, equality occurs for $ a \equal{} 0$ and $ b \equal{} c$, or any cyclic permutation. :wink:[/quote]
Yes! We obtain this case from $ w^3 \equal{} 3u^2 \minus{} 4v^2 \equal{} 0.$

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a(a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c) \ge 0$.[/quote]
This is my proof: :)
Because $ (a\minus{}2b)(a\minus{}2c)\equal{}(a\minus{}b\minus{}c)^2\minus{}(b\minus{}c)^2,$ and $ \sum a(b\minus{}c)^2(a\minus{}b)(a\minus{}c)\equal{}0,$ we may rewrite the inequality as
$ \sum a(b\plus{}c\minus{}a)^2(a\minus{}b)(a\minus{}c) \ge 0.$
Without loss of generality, we may assume that $ a \ge b \ge c \ge 0.$ Then, we have
$ a(b\plus{}c\minus{}a)^2(a\minus{}b)(a\minus{}c) \ge b(b\plus{}c\minus{}a)^2(a\minus{}b)(b\minus{}c),$
$ c(a\plus{}b\minus{}c)^2(c\minus{}a)(c\minus{}b) \ge c(a\plus{}b\minus{}c)^2(a\minus{}b)(b\minus{}c).$
So, the inequality can be reduced to
$ b(b\plus{}c\minus{}a)^2\plus{}c(a\plus{}b\minus{}c)^2\minus{}b(a\plus{}c\minus{}b)^2 \ge 0,$
which can be simplified to
$ c(b\plus{}c\minus{}a)^2\plus{}4bc(b\minus{}c) \ge 0,$
which is obviously true.
This completes our proof. Equality holds for $ (a,b,c)\sim (1,1,1)$ or $ (a,b,c)\sim (1,1,0)$ or $ (a,b,c)\sim (2,1,1)$. :)

Very nice solution, Can_hang. Congratulations! :lol:
An other interesting inequality of degree 5 is the following.

Find the maximum value of $ k$ such that

$ \sum a^5\plus{}(2k\minus{}1)abc\sum ab \ge k\sum ab(a^3\plus{}b^3)$

for all nonnegative real numbers $ a,b,c$.

[quote="Vasc"]Very nice solution, Can_hang. Congratulations! :lol:
An other interesting inequality of degree 5 is the following.

Find the maximum value of $ k$ such that

$ \sum a^5 \plus{} (2k \minus{} 1)abc\sum ab \ge k\sum ab(a^3 \plus{} b^3)$

for all nonnegative real numbers $ a,b,c$.[/quote]
Here is my solution:
We will prove that the inequality is valid for $ k \equal{}\sqrt{2} \minus{}\frac{1}{2}.$ Indeed, rewrite our inequality as
$ 2\sum a^5 \minus{}\sum ab(a^3\plus{}b^3) \ge (2k\minus{}1)\left[ \sum c(a^4\plus{}b^4)\minus{}2\sum a^2b^2\right],$
which is equivalent to
$ \sum (a^4\minus{}b^4)(a\minus{}b) \ge (2k\minus{}1) \sum c(a^2\minus{}b^2)^2.$
This inequality might write in the form $ x(b\minus{}c)^2(b\plus{}c)\plus{}y(a\minus{}c)^2(a\plus{}c)\plus{}z(a\minus{}b)^2(a\plus{}b) \ge 0,$ in which
$ x\equal{} b^2\plus{}c^2 \minus{}(2k\minus{}1)a(b\plus{}c),$ and $ y,z$ are similar.
Without loss of generality, we may assume that $ a \ge b \ge c,$ then with note at $ 2k\minus{}1>0,$ we can easily check that $ x \le y \le z.$ On the other hand, we have
$ x\plus{}y \equal{}a^2\plus{}b^2\plus{}2c^2 \minus{}2(2k\minus{}1)ab \minus{}(2k\minus{}1)c(a\plus{}b) \ge 2t^2\plus{}2c^2 \minus{}2(2k\minus{}1)t^2 \minus{}2(2k\minus{}1)tc \quad \left( t\equal{} \frac{a\plus{}b}{2}\right)$
$ \equal{}2[2(1\minus{}k)t^2 \minus{}(2k\minus{}1)tc\plus{} c^2] \ge 4\left[ 2\sqrt{2(1\minus{}k)} \minus{}2k\plus{}1\right] tc \equal{}0.$
So, $ z \ge y \ge 0.$ Now, with note that $ (a\minus{}c)^2(a\plus{}c) \ge (b\minus{}c)^2(b\plus{}c),$ we obtain
$ x(b\minus{}c)^2(b\plus{}c)\plus{}y(a\minus{}c)^2(a\plus{}c)\plus{}z(a\minus{}b)^2(a\plus{}b) \ge x(b\minus{}c)^2(b\plus{}c)\plus{}y(a\minus{}c)^2(a\plus{}c) \ge (x\plus{}y)(b\minus{}c)^2(b\plus{}c) \ge 0.$
Our proof is completed. :)

Nice your solution, Can_hang. :lol:
The following inequality of degree 5 is also interesting in my opinion.

If $ a,b,c$ are nonnegative real numbers, then

$ \sum a(a\minus{}2b)(a\minus{}2c)(a\minus{}5b)(a\minus{}5c) \ge 0$.

[quote="Vasc"]

If $ a,b,c$ are nonnegative real numbers, then

$ \sum a(a \minus{} 2b)(a \minus{} 2c)(a \minus{} 5b)(a \minus{} 5c) \ge 0$.[/quote]

Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$ Hence, $ t\geq1$ and

$ \sum a(a \minus{} 2b)(a \minus{} 2c)(a \minus{} 5b)(a \minus{} 5c) \ge 0\Leftrightarrow$

$ \Leftrightarrow(a^3\plus{}b^3\plus{}c^3)(a^2\plus{}b^2\plus{}c^2)\plus{}9\sum a^3(b^2\plus{}c^2)$

$ \minus{}7\sum a^4(b\plus{}c)\plus{}49abc(a^2\plus{}b^2\plus{}c^2)\minus{}40abc(ab\plus{}bc\plus{}ca)\geq0\Leftrightarrow$

$ \Leftrightarrow81u(u^2 \minus{} 2v^2)^2 \plus{} (123u^2 \minus{} 188v^2)w^3\geq0.$

Hence, it remains to prove our inequality for $ 1\leq t\leq\frac {188}{123}.$

But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\geq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$

Hence, we need to prove only that

$ 81u(u^2 \minus{} 2v^2)^2 \plus{}(123u^2\minus{}188v^2)(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}) \geq 0$

$ \Leftrightarrow t(165t^2\minus{}421t\plus{}240)^2 \geq (246t\minus{}376)^2(t\minus{}1)^3$

Which is true for $ 1\leq t\leq\frac {188}{123}.$

We have done :)

It's also my proof! :lol:
The equality holds when $ t\equal{}2$ and $ w^3\equal{}0.$
Id est, for $ a\equal{}(2\plus{}\sqrt3)b$ and $ c\equal{}0$ and for all permutations of this.


Hi everyone. I'm in Secondary 1 this year (that equals Grade 7 in US schools) but apparently my school is NUS [b]High School[/b], which is weird. So I have placed this problem under the Middle School section. (I'm new to this forum.)

Apparently, this question is under the topic of [b]Inequalities[/b] but I don't see why it's even related to inequalities.

[hide="Question"]
One of the integers among $ 1,2,3,...,n$ is deleted. The average of the remaining $ n\minus{}1$ numbers is $ \frac{602}{17}$. Which number was deleted? (SMO 2000)

Edit: SMO stands for Singapore Mathematical Olympiad, a math contest for secondary school students.
[/hide]

BTW, how is it possible to write LaTeX in the forums? [Sorry for not doing so. Anyway, the numbers are simple.]

\frac{602}{17}

I have learnt how to write in the LaTex today. How do I input it here?

Once you know the syntax, you just put it between dollar-signs.  As for your problem,
$ (n\minus{}1)\frac{602}{17}\plus{}a\equal{}\frac{n(n\plus{}1)}{2}$ is the sum of the numbers where $ a$ is the number deleted.  Then $ \frac{n(n\plus{}1)}{2}\minus{}a\equal{}\frac{602}{17}$.  Now just try solving for $ n$ and $ a$.

I'd have to disagree.

The sum of numbers from $ 1$ to $ n$ is $ \frac{n(n\plus{}1)}{2}$.
Therefore the sum of the numbers from $ 1$ to $ n$ without $ a$ is $ \frac{n(n\plus{}1)}{2} \minus{} a$.

And so the average of the numbers is $ \frac{\frac{n(n\plus{}1)}{2} \minus{} a}{n\minus{}1} \equal{} \frac{602}{17}$.

$ n\minus{}1$ must be a multiple of $ 17$. Testing various integers which are $ 1 \mod17$, we find that the least (only? I think) value of $ n$ to work is $ 69$. Plugging it in, we have
$ \frac{2415 \minus{} a}{68} \equal{} \frac{602}{17}$.
$ 17(2415\minus{}a) \equal{} 68 \cdot 602$
$ 2415\minus{}a \equal{} 4 \cdot 602 \equal{} 2408$
$ a \equal{} \boxed{7}$

Yes, I seem to have forgotten that step. :roll:


trigonometry
geometry
geometric transformation
reflection
geometry proposed

Let a circle $ (O)$ with diameter $ AB$. A point $ M$ move inside $ (O)$. Internal bisector of $ \widehat{AMB}$ cut $ (O)$ at $ N$, external bisector of $ \widehat{AMB}$ cut $ NA,NB$ at $ P,Q$. $ AM,BM$ cut circle with diameter $ NQ,NP$ at $ R,S$.
Prove that: median from $ N$ of triangle $ NRS$ pass over a fix point.

This problem is very nice and interesting, some Vietnamese students has killed it by trigonometry, which is the best they can do when they were in the classroom, but I'd refer a pured geometry solution, I will be really greatful if some one solved this in the most beautiful way.
Thank you very much :blush:

@mathvnpro:Please don't spam too much like that,it will make MLkers think badly about Vietnamese.
My proof for this problem:
Denote the midpoint of RS be G.
Let J be the reflection point of M through NQ,K be the reflection point of M through NP.Then N is the midpoint of KJ
By angle-chasing we have K,A,S collinear,J,B,R collinear.
First we will prove that:$ \frac {\overline{BR}}{\overline{BJ}} \equal{} \frac {\overline{AS}}{\overline{AK}} \equal{} k$(1)
We have:$ \frac {\overline{BR}}{\overline{BJ}} \equal{} \minus{} \frac {S_{NRQ}}{S_{NJQ}}$
$ \frac {\overline{AS}}{\overline{AK}} \equal{} \minus{} \frac {S_{NSP}}{S_{NKP}}$
Triangle NRQ and triangle NSP are similar,so $ \frac {S_{NRQ}}{S_{NSP}} \equal{} \frac {NQ^2}{NP^2}$
$ \frac {S_{NJQ}}{S_{NKP}} \equal{} \frac {MQ}{MP}$
Because $ \frac {MQ}{NQ}.\frac {NP}{MP} \equal{} \frac {NQ}{PQ}.\frac {PQ}{NP} \equal{} \frac {NQ}{NP}$ so we have proved (1).
All we need to do now is prove that:midpoint of KJ,midpoint of RS,midpoint of AB are collinear.
We have:$ 2\vec{ON} \equal{} \vec{AK} \plus{} \vec{BJ}$
              $ 2\vec{OG} \equal{} \vec{AS} \plus{} \vec{BR} \equal{} k(\vec{AK} \plus{} \vec{BJ})$
Then N,O,G are collinear.
Q.E.D
Figure:[url]http://img511.imageshack.us/img511/1876/bai5tst.jpg[/url]
PS:May be there are some typos because this net cafe blocked Imageshack so I can't see my figure now.I will edit when I go home.

This is my solution:
Let I be the midpoint of $ SR, F \equal{} MA\cap (PMN), E \equal{} MB\cap (QMN)$
It's easy to see F is the reflection point of S through PN, E is the reflection point of R through NQ
We will prove $ AB//EF$
$ \Leftrightarrow \frac {MA}{AF} \equal{} \frac {MB}{BE}$
$ \Leftrightarrow \frac {S_{PMN}}{S_{PFN}} \equal{} \frac {S_{PMN}}{S_{PSN}} \equal{} \frac {S_{NMQ}}{S_{NEQ}} \equal{} \frac {S_{NMQ}}{S_{NRQ}}$
$ \Leftrightarrow \frac {S_{PMN}}{S_{NMQ}} \equal{} \frac {S_{PSN}}{S_{NRQ}}$
$ \Leftrightarrow \frac {PM}{QM} \equal{} \frac {PN^2}{NQ^2}$ (because $ \Delta PSN$~$ \Delta NRQ)$
$ \Leftrightarrow \frac {PM.PQ}{QM.PQ} \equal{} \frac {PN^2}{NQ^2}$ (It's true)
So $ AB//EF$
Let $ J \equal{} EF\cap AN$ then $ \angle JFN \equal{} \angle JSN$
We will prove $ JS//ON$:
We have $ \angle AJF \equal{} \angle NJE \equal{} \angle NAO \equal{} \angle ANO$, moreover $ \angle AJF \equal{} \angle AJS$ hence $ \angle AJS \equal{} \angle ANO$
Thus $ JS//ON \Rightarrow \angle JSN \equal{} \angle SNO$
Then $ \angle JFN \equal{} \angle SNO$, similarly $ \angle RNO \equal{} \angle NEF$
We have $ \frac {SN}{RN} \equal{} \frac {NF}{NE} \equal{} \frac {sin \angle FEN}{sin \angle EFN} \equal{} \frac {sin \angle ONR}{sin \angle ONS} \equal{} \frac {sin \angle INR}{sin \angle INS}$
Therefore $ O,N,I$ are collinear

Here is my solution for this problem
[b]Solution[/b]
Let $I$ $\equiv$ $NS$ $\cap$ $PQ$; $J$ $\equiv$ $NR$ $\cap$ $PQ$; $E$ $\in$ $NR$, $F$ $\in$ $NS$ which satisfy $AE$ $\perp$ $MN$, $AF$ $\perp$ $MN$
We have: $(MA; MI) \equiv (MA; MB) + (MB; MI) \equiv (MA; MB) + (MP; MA) \equiv (MP; MB) \equiv (NA; NI)$ (mod $\pi$)
Then: $M$, $I$, $N$, $A$ lie on a circle
Similarly: $J$, $M$, $N$, $B$ lie on a circle
So: $(NI; NA) \equiv (MB; MQ) \equiv (NQ; NJ) \equiv - (NJ; NB)$ (mod $\pi$) or $\triangle IAN$ $\stackrel{-}{\sim}$ $\triangle IBN$
Hence: $\dfrac{AE}{BF} = \dfrac{NA . PJ . NQ}{NP . QI . NB} = \dfrac{NI . PJ . NQ}{NP . QI . NJ} = \dfrac{2 S_{NQI} . \sin \angle{QNI} . S_{NPJ}}{2 S_{NPJ} . \sin \angle{PNJ} . S_{NQI}} = 1$ (With $S_{XYZ}$ is area of $\triangle XYZ$)
Combine with: $AE$ $\parallel$ $BF$, we have: $AEBF$ is parallelogram
Then: $O$ is midpoint of $EF$
We have: $(BF; BN) \equiv (QI; QN) \equiv (RA; RN)$ (mod $\pi$)
So: $\triangle NFB$ $\stackrel{+}{\sim}$ $\triangle NAR$ or $\dfrac{NF}{NB} = \dfrac{NA}{NR}$
Hence: $NF . NR = NA . NB$
Similarly: $NE . NS = NA . NB$
Then: $NF . NR = NE . NS = NA . NB$ or $\dfrac{NF}{NS} = \dfrac{NS}{NR}$
So: $EF$ $\parallel$ $SR$ or $N$ - median of $\triangle NSR$ passes through fixed point $O$


calculus
integration
calculus computations

For nonzero real numbers $ r,\ l$ and the positive constant number $ c$, consider the curve on the $ xy$ plane : $ y \equal{} \left\{ \begin{array}{ll} x^2 & (0\leq x\leq r)\quad \\
r^2 & (r\leq x\leq l \plus{} r)\quad \\
(x \minus{} l \minus{} 2r)^2 & (l \plus{} r\leq x\leq l \plus{} 2r)\quad \end{array} \right.$

Denote $ V$ the volume of the solid by revolvering the curve about the $ x$ axis. Let $ r,\ l$ vary in such a way that $ r^2 \plus{} l \equal{} c$. Find the values of $ r,\ l$ which gives the maxmimum volume.

The volume of the solid in the $ xyz$-plane can be divided with the planes $ x \equal{} r$, $ x \equal{} r \plus{} l$ into three subsolids with the volumes $ V_1$ (most left), $ V_2$ and $ V_3$(most right) respectively. There is $ V \equal{} V_1 \plus{} V_2 \plus{} V_3$ with $ V_1 \equal{} V_3$. Hence, $ V \equal{} 2V_1 \plus{} V_3$. On the other hand, we get
\[ V \equal{} 2\pi\int_{0}^{r}x^4\,\mathrm{d}x \plus{} \pi\int_{r}^{r \plus{} l}r^4\,\mathrm{d}x \equal{} \frac {\pi r^4}{5}(2r \plus{} 5l).\]
Using the constraint $ r^2 \plus{} l \equal{} c$, where $ c$ is given, we obtain
\[ V \equal{} \frac {\pi r^4}{5}( \minus{} 5r^2 \plus{} 2r \plus{} 5c) \equal{} : V(r,c).\]
This gives
\[ \frac {\partial}{\partial r}V(r,c) \equal{} 2\pi r^3( \minus{} 3r^2 \plus{} r \plus{} 2c).\]
Now, we can deduce that the only maximum occurs when $ r \equal{} \frac {1}{6}(1 \plus{} \sqrt {1 \plus{} 24c})$. The corresponding value of $ l$ can be computed using the constraint. One can obtain the fact that $ l \equal{} \frac {c}{3} \minus{} \frac {1}{18}(1 \plus{} \sqrt {1 \plus{} 24c})$. Assuming that $ l>0$ we deduce that $ c> 1$. For $ c\in (0,1]$ we take $ l \equal{} 0$.

[quote="kunny"]Perfect! :)[/quote]
Thanks, may be perfect, but you suppose in your problem that $ r$ and $ l$ are non-zero. However, my solution admit also the value $ l\equal{}0$ which contradicts the assumptions in your problem.


geometry
geometric transformation
circumcircle
symmetry
homothety
angle bisector

Incircle $ (I,r)$ of $ \triangle ABC$ is tangent to $BC,CA,AB$ at $X_a,X_b,X_c.$ Perpendiculars from $ I$ to $BC,CA,AB$ meet external bisectors of angles $ \angle A,$ $ \angle B,$ $ \angle C$ at $ A',B',C',$ respectively. Circles $ \odot(IBB'),\odot(ICC')$ meet at $I,Y_a$ and define cyclically $Y_b,Y_c.$ Let $ M,N,L$ be the midpoints of $ \overline{X_bX_c},$ $\overline{X_cX_a},$ $\overline{X_aX_b}.$

1) Show that $\odot (AIX_a), \odot (BIX_b), \odot(CIX_c), \odot(MIY_a),\odot (NIY_b), \odot(LIY_c)$ meet at the Schroder point $X_{1155}$ of $ \triangle ABC,$ i.e. the inverse of the insimilicenter of $(I) \sim (O)$ in the circumcircle.

2) Let $ R_a,R_b,R_c$ be the radii of  $\odot (IAA'), \odot(IBB'), \odot(ICC')$ and $ \Psi$ denotes the sum of the distances from the orthocenter of $ \triangle X_aX_bX_c$ to the sides of $ \triangle ABC.$ Show that

\[ \frac {1}{R_a} \plus{} \frac {1}{R_b} \plus{} \frac {1}{R_c} \equal{} \frac {\Psi}{r^2}\]

Dear Luis and Mathlinkers,
nice result.
I think that you know the great article written by Dary Grinberg
http://de.geocities.com/darij_grinberg/Schroeder/Schroeder.html
Sincerely
Jean-Louis

Dear Jean Louis, Darij's article is indeed a great material. The concurrency of the circles $ \odot(AIX_a),$ $ \odot(BIX_b)$ and $ \odot(CIX_c)$ at the Schröder point $S$ is well-known. Basically, I'm pointing out  the concurrency of the circles $ \odot(MIY_a),\odot(NIY_b)$ and $ \odot(LIY_c)$ at the inverse  of the insimilicenter of $(I) \sim (O)$ in the circumcircle.

Dear Luis,
sorry, I have read to quickly your message.
Your result is very interesting.
How do you prove it?
Sincerely
Jean-Louis

Sorry for my late reply dear Jean-Louis, I stopped thinking about Schröder point for a while.

A-mixtilinear incircle $\omega_a$ touches $AB,AC$ and the circumcirle $(O)$ at $P,Q,U.$ It is well-known that the A-excenter $I_a$ is the midpoint of $PQ.$ Inversion through pole $A$ with power ${AI_a}^2$ swaps $(I_a)$ and $\omega_a$ and takes $(O)$ into a straight line $\tau$ tangent to $(I_a)$ through the inverse $U'$ of $U.$ $\tau$ cuts $AB,AC$ at the inverses $B',C'$ of $B,C.$ Since $B'C'$ is antiparallel to $BC,$ then there exists the composition of the axial symmetry accross the angle bisector of $\angle BAC$ and a positive homothety with center $A$ taking $\triangle AC'B'$ with incircle $\omega_a$ into $\triangle ABC$ with incircle $(I)$ $\Longrightarrow$ $AX_a$ and $AUU'$ are isogonals WRT $\angle BAC.$ Since $U$ is the insimilicenter of $(O) \sim \omega_a$ and $A$ is the exsimilicenter of $(I) \sim \omega_a,$ it follows that $AU \cap IO$ is the insimilicenter $X_{55}$ of $(I)$ and $(O).$ By similar reasoning, we conclude that $X_{55}$ is the isogonal conjugate of the Gergonne point $G_e$ of $\triangle ABC \ (\star).$

Inversion WRT $(I)$ takes $X_a,X_b,X_c$ into themselves and $M,N,L$ into $A,B,C,$ respectively. Thus, $\odot(AIX_a),$ $\odot(BIX_b),$ $\odot(CIX_c)$ are taken into median lines $X_aM,$ $X_bN,$ $X_cN$  of the intouch triangle $\triangle X_aX_bX_c$ $\Longrightarrow$ $\odot(AIX_a),$ $\odot(BIX_b),$ $\odot(CIX_c)$ meet at $I$ and the inverse of the centroid $V$ of $\triangle X_aX_bX_c.$ Since $IA'$ is a diameter of $\odot(IAA'),$ it follows that the inverse line $\ell_a$ of $\odot(IAA')$ is the parallel to $BC$ passing through $M,$ i.e the tangent at $M$ of the 9-point cirle $\odot(MNL)$ of $\triangle X_aX_bX_c.$ Likewise, inverse lines $\ell_b,\ell_c$ of $\odot(IBB'),$ $\odot(ICC')$ are tangent to $\odot(MNL)$ through $N,L.$ Therefore, pairwise lines $\ell_a,\ell_b,\ell_c$ meet at the inverses $Z_a,Z_b,Z_c$ of $Y_a,Y_b,Y_c$ $\Longrightarrow$ $AZ_a,$ $BZ_b,$ $CZ_c$ are the inverses of $\odot(MIY_a),$ $\odot(NIY_b),$ $\odot(LIY_c).$ But since $\triangle ABC \sim \triangle Z_aZ_bZ_c$ are homothetic, it follows that $AZ_a,$ $BZ_b,$ $CZ_c$ concur at the insimilicenter of their incircles $(I) \sim \odot(MNL),$ i.e. the centroid $V$ of $\triangle X_aX_bX_c.$ Consequently, $\odot(MIY_a),$ $\odot(NIY_b),$ $\odot(LIY_c)$ also pass through the inverse of $V$ in the incircle $(I).$ Now, according to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=315750]Concurrent 2 (PC point again)[/url] (see the remark between bold lines) the circles $\odot(AIX_a),$ $\odot(BIX_b),$ $\odot(CIX_c)$ meet at $I$ and the inverse of the isogonal conjugate of  the Gergonne point $G_e$ in the circumcircle $(O).$ Together with $(\star),$ we conclude that $\odot(AIX_a),$ $\odot(BIX_b),$ $\odot(CIX_c),$ $\odot(MIY_a),$ $\odot(NIY_b),$ $\odot(LIY_c)$ concur at the Schröder point of $\triangle ABC.$


This a question from an Olympiad that I can't solve and it's annoying. Sorry, I can't remember the year.
'A, B, C are positive reals with product 1. Prove that (A - 1 + 1/B)(B - 1 + 1/C)(C - 1 + 1/A) ≤ 1.'
Any help greatly appreciated.

Let $ a \equal{} \frac {x}{y},b \equal{} \frac {y}{z},c \equal{} \frac {z}{x},x,y,z > 0$

The inequality is equivalent to :

$ (x \plus{} y \minus{} z)(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)\le xyz\\\Leftrightarrow x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz\ge x^2y \plus{} x^2z \plus{} y^2z \plus{} y^2x \plus{} z^2x \plus{} z^2y$

Which is Schur - inequality  :wink:

It is an IMO 2000.You should post it in Inequality section.

quangpbc

Thanks for the repsonse.
I had no idea there were all these inequalities out there like Schur's. Thanks for opening my eyes.

jimi


inequalities
trigonometry
algebra unsolved
algebra

Prove that this equation does not have any solution for real x.
cos(cos(cos(cos(x)))=sin(sin(sin(sin(x)))
I proved that for x less than pi/4, but I think the problem is difficult if x is greater than pi/4. :mad:
Thank you

It posted before. But I don't know solved or not.
If $-1\le sinx \le 0$, then $f(x)=cos(cos(cos(cos x)))>0\ge sin(sin(sin(sin x)))=g(x)$.
If $0<sin x\le 1$, then $g(x)\le sin(sin(sin 1))<f(x)$ if $\pi/4+2k\pi \le x\le 3\pi/4+2k\pi$, else $g(x)<sin(sin(sin\frac{\sqrt 2}{2}))\le g(x).$

[quote="t0rajir0u"]Try to prove an inequality of the form

$\cos \cos \cos \cos x > k > \sin \sin \sin \sin x$[/quote]
Are you sure? According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cos&t=91594]Darij's post[/url], such a constant doesn't exist.

By the way, in that topic Peter Scholze solved the problem.


How many times does the digit 1 occur in the result of
$ 1+11+111+1111+...+111...111$?  (last number has 2002 digits, i.e 2002 1's)

[quote="red_dog"][hide="solution"]
$ 1+2+3+4+\ldots+2002=\frac{2002\cdot 2003}{2}=1001\cdot 2003$
[/hide][/quote]

I dont get why you did thats...its the sum.

I was just reading a book and it had a similar problem to this,,,

[hide="solution"]
Let $ t_{n}=111...111$ (n 1's)
Let $ S_{n}=1+11+111+...+t_{n}$
          $ =1+(10+1)+(110+1)+(1110+1)+...$
          $ =10S_{n-1}+n$
$ \Rightarrow S_{n}=10(S_{n}-t_{n-1})+n$
$ \Rightarrow 9S_{n}=10t_{n-1}-n$
$ \Rightarrow S_{n}=\frac{10t_{n-1}-n}{9}$
Now to solve the problem, let $ n=2002$
$ \Rightarrow S_{2002}=\frac{10t_{2001}-2002}{9}$
$ \Rightarrow S_{2002}=\frac{100000t_{1998}+9108}{9}$
After some fiddling with the numbers, $ \frac{100000t_{1998}+9108}{9}=\frac{111...11100000+9108}{9}$
You get the answer,,, i got not enought time to post up the solution... sorry
My answer is $ 224$
[/hide]


algebra
polynomial
vector
algebra proposed

Let $ p(x)\equal{}a_0 \plus{}a_1 x\plus{}...\plus{}a_n x^n$ be a polynomial with nonnegative real coefficients. Suppose that $ p(4)\equal{}2$ and $ p(16)\equal{}8$. Prove that $ p(8) \le 4$ and find all such $ p$ with $ p(8)\equal{}4$.

$ p\left(\sqrt{ab}\right) \leq \sqrt{p\left(a\right)}\sqrt{p\left(b\right)}$

By Cauchy Schwarz, \[ \sum 4^na_n \sum 16^na_n \geq \left(\sum 8^na_n\right)^2\]
Hence the result, with equality when the two vectors are parallel.


quadratics ratio search MATHCOUNTS algorithm LaTeX algebra

A kid walks up to you envious of your math skills.,
He decides to challenge you to an "impossible" math problem.

You stare at the problem:

Solve for x if:

$x = 1 + \frac{1}
{{1 + \frac{1}
{{1 + \frac{1}
{{1 + \frac{1}
{{1 + ...}}}}}}}}
$
Some guy runs by and gives you
[hide="a hint"]Vegiterian hot dogs are made with meat [b]substitute![/b][/hide]

Try, if you dare  :D  :lol:  ;)  :P

[hide]
Ooh! I've heard this one before, so sorry to spoil your fun.

IIRC, $x=\phi$[/hide]

Lol, metsfan001 is obsessed with phi...i believe that equation leads to a quadratic which leads to phi.

I am more obsessed with it than he is...lol... I gave the problem after all and have you read my signature?

Anyway I am just posting this as a challenge.  If you know the answer, write up a proof for it.

But I will tell you that you are both wrong.   Now here is YOUR challenge.  Tell me why...
 :D

[quote="GoBraves"][i][b][hide]
 Since it keeps repeating, we can replace the whole bottom part with x.
So
     x=1+1/x
    After we simplify we get x^2-x-1=0

Solve using the quadratic equation and we get

x= (1+(sqrt 5))/2        OR          x= (1-(sqrt 5))/2

Sorry i know this looks confusing, but i dont know how to do all the html and stuff.
[/hide][/b][/i][/quote]

=phi.

[quote="GoBraves"][i][b][hide]
 Since it keeps repeating, we can replace the whole bottom part with x.
So
     \[x=1+1/x\]
    After we simplify we get \[x^2-x-1=0\]

Solve using the quadratic equation and we get

\[x= {1+\sqrt 5}{2}\]        OR          \[x= \frac{1-\sqrt{5}}{2}\]

Sorry i know this looks confusing, but i dont know how to do all the html and stuff.
[/hide][/b][/i][/quote]

Most of us don't use html to do math expressions, but $\text{\TeX}$. Click on the link on the side to read about it.

Notice that because both of the answers work, Farazy is right.

Come on, you all are EXTREMELY close.  Try and find out what is wrong.

sorry about posting on such an old topic but, in response to Farazy:

the answer has to be positive...
(1+ :sqrt: 5)/2 is the only answer

our class has done stuff like this but not to infinite

Its good to see how these work.  This uses the basic principles of manipulations to create quadratics.  In my opinion, it would make a really nice addition to the Section 6.4.2 in the first art of problem solving (that section deals with substitutions with quadratics).

Using that principle, you can solve problems like that one and others:

$x=\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6...}}}}}}}}}}$
$x=?$

[hide="solution"]
x= :sqrt: 6x
x 2 =6x
x=6
Notice x cannot be negative, because you would have a nonreal as a result of the root.[/hide]

[hide]
x=6 :sqrt: x
x 2=36x
x 2 -36x=0
x(x-36)=0
x cannot be 0, so
[b]x=36[/b]
[/hide]

fairly easy problem, if you know how to do it

sry

i read the problem wrong

[hide]
x= :sqrt: 6x
x 2 =6x
x 2 -6x=0
x(x-6)=0
x cannot be 0, so
[b]x=6[/b] answer[/hide]

If its less than 1, then it soon tends and becomes 1.  If its more than one it, it just keeps increasing so its impossible.

The solution is $x=\sqrt2$, and actually it works, you can try a few thosands of terms in basically any programmable calculator, and you will notice that if you define $a_0=\sqrt2$ and ${a_{n+1}=\sqrt2^{a_n}}$, then $a_n$ converges to 2. It works also if you do $x^{x^{x^{x^{\dots}}}}=3$, and you get $x=\sqrt[3]{3}$, and it also works. Unfortunatelly, if you do it with $x^{x^{x^{x^{\dots}}}}=4$ it doesn't, because if you try with the same algorithm you'll get $x=\sqrt[4]{4}=\sqrt2$, but we already said that that converges to $2$, therefore it cannot converge to $4$.

I really don't have an answer why it is that way, but it is interesting anyway! :D

Best,

In 57 terms, it is within 9 points precision. on my calculator

[quote="GoBraves"][quote="djimenez"]The solution is $x=\sqrt2$, and actually it works, you can try a few thosands of terms in basically any programmable calculator, and you will notice that if you define $a_0=\sqrt2$ and $a_{n+1=\sqrt2^{a_n}}$, then $a_n$ converges to 2. It works also if you do $x^{x^{x^{x^{\dots}}}}=3$, and you get $x=\sqrt[3]{3}$, and it also works. Unfortunatelly, if you do it with $x^{x^{x^{x^{\dots}}}}=4$ it doesn't, because if you try with the same algorithm you'll get $x=\sqrt[4]{4}=\sqrt2$, but we already said that that converges to $2$, therefore it cannot converge to $4$.

I really don't have an answer why it is that way, but it is interesting anyway! :D

Best,[/quote]

I dont think thats right, or atleast my calculator dosnt think its right....[/quote]

Actually, it is...

Try programming this on a TI-83+, or another programmable calculator.

:sqrt: 2 [store] A
For(X,1,100)
 :sqrt: 2^A [store] A
Disp A
End

You'll notice that the calculator displays a value close to 2...eventually, it becomes 2.

I did that too, except I did

While A  :!=: 2

 instead of

For(A,1,100)

and I put a iteration-counter in it too.

Sorry to bring back an old topic, but nobody ever said [b]how[/b] to solve it, did they?

1st Problem:

$\frac{1+\sqrt{5}}{2}$

Wow! my $\LaTeX$ worked!

it is possible.  If you have x to the power of x infinity times with no parentheses, you basically have the inverse of x^(1/x) so it will never be larger than e or about 2.71.
1.4446*... to the power of itself an infinite amount of times will result in a finite number all higher will be infinite.

*e^(1/e)

What about the point that djimenez made?  You say it works for 2.  But it can't work for all numbers, example 4 since it already is 2

[quote="Aryth"]The answer to the problem is:
[hide](One plus/minus SQRT-3)/2, the equation can turn into X=1+1/X which is then X-1 = 1/X or X^2-X-1 = 0. Using the quadratic formula, you get the answer.[/hide][/quote]
$\frac{1\pm\sqrt3}{2}$
$x=1+\frac1x$
$x-1=\frac1x$
$x^2-x-1=0$


geometry
combinatorics unsolved
combinatorics

A lattice point $X$ in the plane is said to be [i]visible[/i] from the origin $O$ if the line segment $OX$ does not contain any other lattice points. Show that for any positive integer $n$, there is square $ABCD$ of area $n^{2}$ such that none of the lattice points inside the square is visible from the origin.

The point (a,b) is not visible if and only if a and b are not coprime.
The square with vertices (a,b) ,(a+n,b),(a+n,b+n),(a,b+n) satisfies the condition of the problem if for each i and j  a+i and b+j are not coprime but by chinese remainder theorem if we pick $p_{ij}$ to be different primes we can set $p_{ij}$ to divide a+i and b+j for all i,j=0,1,2...n.

So no points in this square is visible from the origin

How does the above construction ensure that the visible points when scaled down will also be contained inside a square with area $n^2$?

The key is to see that if $\gcd(x_i,y_i)\ne 1$ then a point is invisible. Then, we know by CRT there exists a solution with primes $p_l\forall 1\le l\le n^2$ s.t. if $x_1,y_1$ is the coordinate of the bottom left corner, we have $$x_1+i\equiv0\pmod{\prod_{k=1}^{n}p_{k+in}},y_1+i\equiv0\pmod{\prod_{j=0}^{n-1}p_{i+1+jn}}\forall 0\le i\le n-1.$$


abstract algebra
vector
superior algebra
superior algebra unsolved

Let G be a finite group such that Aug(G) acts transitively on the set G\{e}. Show that G is a p-group for some prime p and G is abelian.

$ G$ is even an elementary abelian $ p$-group, i.e. a vector space over $ \mathbb{F}_{p}$.

we may assume that $ G$ is nontrivial. then there is a element of prime order $ p$. since $ Aut(G)$ is transitive on $ G\minus{}\{e\}$, the order of elements $ \neq e$ is constant, thus $ p$. since $ G$ is a nontrivial $ p$-group, there is a nontrivial element in the center of $ G$. by transitivity, this already implies that every nontrivial element lies in the center, because this is a characteristic subgroup. thus, $ G$ is abelian.


Find all positive integer solutions to
a^2 + b^2 + c^2 = 2d^2
where d < 50

Hmm.. from does it come? ;)
Since there are only finite cases you just have to use a computer....

Pierre.

Are there any easy ways not using a computer?


Let $ S$ be the set of all triangles $ ABC$ which have property: $ \tan A,\tan B,\tan C$ are positive integers. Prove that all triangles in $ S$ are similar.

$ \arctan 1 \equal{} 45^\circ$, $ \arctan 2 \approx 63^\circ$, $ \arctan 3 \approx 72^\circ$, $ \arctan 4 \approx 76^\circ$. We have $ \tan A \tan B \tan C \equal{} \tan A \plus{} \tan B \plus{} \tan C$
then we can't have 1,1,x because $ x < 1\plus{}1\plus{}x$ and we can have $ 1,2,x$ with $ x<4$ or 2,2,2 it's impossible and also for the other that are bigger.
It's also simple to prove that 1,2,3 is possible and 1,2,2 not.
So all triangles have $ \tan A \equal{} 1$, $ \tan B \equal{}2$, $ \tan C \equal{}3$ or permutation then thay are similar.

[quote="crazyfehmy"]This problem is 2004 Turkey 1. Round question . Isn't it too easy for 3. question :maybe:[/quote]Indonesia Olympiad problems are indeed very easy. :P


[quote="tranquoc"]Prove If a,b,c >0 such that  $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$ then
$ a^2 \plus{} b^2 \plus{} c^2 \geq a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$

[color=red]Edited by N.T.TUAN[/color][/quote]
This is Solution
[img]http://k2pi.net/Qupload/uploads/k2pi.net---59575683898434260156.jpg[/img]

The problem was solved in this forum; see
 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=511272&p=2875031#p2875031
__________________________________________________
Sandu Marin, Bucuresti, Romania


group theory
abstract algebra
superior algebra
superior algebra solved

let a,b two positive integers greater than 1. Prove that the group Sa+b of permutations of the set {1,2,...,a+b} contains a commutative subgroup of order ab.

source: [i]Mircea Becheanu[/i]

I'm not sure about this because it's 6 AM and I'm noit what you would call awake right now :D, but doesn't <(1, 2, .., a), (a+1, a+2, .., a+b)> work?

I'm referring to the group generated by the cycle of the first a elements and the last b elements. Let the first cycle be u and the second one be v.

The 2 cycles commute because they're disjoint, so the group is commutative. Further more, it has ab elements because ua = vb =1, so every element can be written as um*vn, with m from 0 to a-1 and n from 0 to b-1.

obs: this problem was given this year at the "gheorghe vranceanu'' contest in bacau, romania.


Study the convergence a. s. of $ Y_n\equal{}\frac{\sum_1^n(X_r\plus{}X_{r\plus{}1})^2}{n}$,
where $ (X_n)$ are i.i.d. uniform on $ (0,1)$ r. v. .

If you split the sum in two, with even $ r$ and with odd $ r$, then you again have i.i.d. variables, albeit not uniformly distributed.


function
integration
real analysis
real analysis unsolved

suppose that $m$ is the measure that is uniformly distributed on the usual Cantor set (its characteristic function is the Devil's staircase function). Compute
$F(x) = \int_{[0;1]}e^{itx}dm(t)$

I think the answer is $ \prod_{k=1}^{ \infty}\frac{1}{2}[1+\exp(\frac{-2xi }{3^{k}})]$.

If we represent the Lebesgue Stelties measure, defined by the Cantor function $ \mu$ as a convolution product $ * \prod \frac{1}{2}(\delta_{0}+\delta_{\frac{2}{3^{k}}})$, then we obtain the result evaluating
$ F[ \frac{1}{2}(\delta_{0}+\delta_{ \frac{2}{3^{k}}}) ] = \frac{1}{2}[1+\exp ( \frac{2xi }{3^{k}}) ]$.

Here $ F$ stands for the Fourier transform operator.


Show that if $ \alpha$ is a root of the equation $ 5 \cos x \plus{} 12 \sin x \equal{} 7$, then either \[ \cos \alpha \equal{} \frac{35 \minus{} 12\sqrt{120}}{169}\] or $ \cos \alpha$ has one other value you should find.

Prove carefully that if $ \frac {\pi}{2} < \alpha < \pi$, then $ \alpha < \frac {3\pi}{4}$.

[hide="The other value ..."]
The other value of $ cos \alpha$ will be  $ \frac{35 \plus{} 12\sqrt{120}}{169}$

Form the quadratic equation in $ cos \alpha$ ,

you get

$ 169 cos^2\alpha \minus{} 70 cos\alpha \minus{} 95 \equal{} 0$

its easy to find the roots now .... one you have obtained and the other will be its conjugate

$ \frac{35 \plus{} 12\sqrt{120}}{169}$

[/hide]


how can u find the number of triangles determined by n points?
(and i dont think its the number of points divided by 3)

$ \frac{n(n\minus{}1)(n\minus{}2)}{6}$ unless there are like 3 collinear points.

what if there are 3 collinear points? or will that never be true?

IF there are only 3 collinear points, then there is 1 triangle that we have ounted that cant be made (the one with those 3 points)

what if there are like 30 points and the question says 6 are collinear?

then you subtract the "triangles" made from the 6 collinear points:
$ \frac{(30)(29)(28)}{6}\minus{}\frac{(6)(5)(4)}{6}$


Sa se arate ca intr-un tetraedru echifacial, centrul sferei inscrise in tetraedru coincide cu centrul de greutate al tetraedrului.

Problema M9 din Kvant:
Următoarele condiţii sunt echivalente:
1. Toate feţele au aceeaşi arie
2. Muchiile opuse sunt congruente
3. Feţele sunt congruente
4. Centrul sferei înscrise coincide cu centrul sferei circumscrise
5. Sumele unghiurilor plane de la fiecare vârf sunt egale
 :)

Haha, ok imi retrag titlul de proprietate asupra problemei :D
Dar de descoperit am facut-o singur la lumina monitorului! :)


It's almost readable in English as-is -- "montrer" means "show" or "prove."  Perhaps a moderator could move this to the French forum?

[quote="JBL"]It's almost readable in English as-is...[/quote]
I'll give it a shot, even though I don't actually read French:

1) Show that the equation $ \sum_{k\equal{}1}^nx^k\equal{}1$ admits a unique solution $ u_n\in[0.1].$

2) a) Show that for every $ n\in\mathbb{N},$ there exists $ k\in\{1,2,\dots,n\}$ such that $ u_{n\plus{}1}^k\minus{}u_n^k\ge0$

b) Show that $ (u_n)$ converges to a limit $ h.$

c) Show that there exists $ N$ such that for all $ n\ge N,\ u_n\le\frac34.$

3) Show that $ 2u_n\minus{}1\equal{}u_n^{n\plus{}1}.$

4) Find the limit $ h.$

5) Setting $ \nu_n\equal{}u_n\minus{}\frac12,$ and noting that $ 2n\nu_n\equal{}\left(\nu_n\plus{}\frac12\right)^{n\plus{}1},$ show that $ \lim_{n\to\infty}n\nu_n\equal{}0.$

6) (Sorry - someone's going to have to help me with "Déduire.")

6. Deduce that $ u_n \approx (1/2)^{n\plus{}1}$


Suppose $ U$ is the subspace of $ P(F)$ consisting of all polynomials $ p$ of the form

$ p(z) \equal{} az^2 \plus{} bz^5$

where $ a,b \in F$. Find a subspace $ W$ of $ P(F)$ such that $ P(F) \equal{} U \oplus W$.

All polynomials with zero coefficients of $ z^2$ and $ z^5$?

yea a lot of these problems in my textbook seem trivial... which makes me doubt myself...


Let {b[size=67]n[/size]} be a bounded sequence and suppose that a[size=67]n[/size] -> 0 . Prove that a[size=67]n[/size]b[size=67]n[/size] -> 0

thanks in advance

What does your problem really say? What you posted is so obviously false that it cannot be the statement you meant, but I'm not willing to guess what it should be.

that is what it says
it's from the textbook: Fundamental Ideas of Analysis by Michael Reed
page 39 problem # 4.

Oh, sorry. I misunderstood your notation, as I was reading the symbol you meant to be $\to$ as $\ge$, which changes the meaning.

One thing you'll find around here: knowing a little $\LaTeX$ greatly improves your ability to communicate. Here's your problem. Just wave your mouse over it or click on the coded sections to see what the code looks like.

Let $\{b_n\}$ be a bounded sequence and suppose that $a_n \to 0$. Prove that $a_n b_n \to 0.$

Let's try walking through this. What does "$a_n\to 0$ mean? It means this:

$\forall \epsilon>0\,\exists N$ such that if $n>N,\,|a_n|<\epsilon.$

What does "$\{b_n\}$ is bounded" mean? It means this:

$\exists\,M$ such that $|b_n|\le M$ for all $n.$

Now: you want to show that $|a_nb_n-0|=|a_nb_n|<\epsilon$ for some $\epsilon.$ What are the resources you can bring to bear? What facts do you know about $a_n$ and $b_n$ that you can use?


The king receives $ n$ couples around his round table having $ 2n\plus{}1$ seats. For $ i\equal{}1,\dots,n$, the members of the $ i\minus{}$th couple want to be seated at distance $ d_i\in\{1,\dots,n\}$ from each other (ie. separated by exactly $ d_i\minus{}1$ chairs) during dinner.

If $ 2n\plus{}1$ is not prime, this is not always possible. (A counterexample is given by a set of couples
wanting all to be seated at distance $ k$ for $ k$ a non-trivial divisor of $ 2n\plus{}1$).

A computer-program checked that there are no restrictions on the distances for $ 2n\plus{}1$ a prime number $ \leq 23$.  This leads to the conjecture that it is always possible to have a seating arrangement if $ p\equal{}2n\plus{}1$ is a prime number.

A stronger conjecture is the following: It seems always possible to have a seating arrangement if all $ n$ distances $ d_i$ are invertible in the ring $ Z/(2n\plus{}1)Z$.

Can somebody give a proof of the first, or even better, of the second conjecture?

Remarks: The invertible elements modulo $ 2n\plus{}1$ act on the set of distances. This can be used
to reduce the number of cases.

In particular, it is easy to proof the conjecture if there are at least $ n\minus{}1$ different values for the distances or at most $ 2$ different values for the distances.

Emmanuel Preissmann found a beautiful proof that there are no obstruction for
placing $ n$ couples at arbitrarily prescribed  distances $ d_1,\dots,d_n\in\{1,\dots,n\}$,
provided that $ 2n\plus{}1$ is a prime.


function
logarithms
calculus
calculus computations

I was trying to work out $\sinh^{-1}(x)$, so I write down
$\frac{e^{y}-e^{-y}}{2}= x$
Solving for y, this gives two roots
$y = \ln{x \pm \sqrt{x+1}}$
Which is $\sinh^{-1}(x)$, then?

$y = \ln(x+\sqrt{x^{2}+1})$, since $-$ would give a negative number under the log.

What if we don't limit outselves with real numbers and expand the region to complex numbers?

Then the inverse is multi-valued -- there are infinitely many complex numbers $y$ such that $\sinh y=x$.


induction
number theory proposed
number theory

From $f(f(n))=n$ we easily deduce that $f$ is bijective. It follows that, using $f(0)=1$ we have $f(n+2)+2 = f(n)$ for all $n$.

Moreover $f(1)=f(f(0))=0$ and $f(-1) = f(-1+2)+2 = f(1)+2 = 2$.
Now we prove that $f(n) = 1-n$ by induction on $|n|$.
Assume that it holds for all $k \in \{-n, \cdots, n \}$.
Then :
$f(n+1) = f(n-1) - 2 = 1-(n+1)$
and
$f(-n-1) = f(-n+1)+2 = 1-(-n-1)$.
This ends the induction.

Conversely, it is easy to verify that $f(n)=1-n$ is indeed a solution.

Pierre.


function
real analysis
limit
real analysis unsolved

if $ f$ is an arbitrary measurable function with period $ 1$ then prove that one has almost every where the relations:$ \overline{lim}_{n\rightarrow \infty}f(nx)={\text{essential upper bound}\ }\ f(x)\ in\ [0,1]$
$ \underline{lim}_{n\rightarrow \infty}f(nx)={\text{essential lower bound}\ }\ f(x)\ in\ [0,1]$

I think that one can even prove that it works for $ \limsup f(2^{n}x)$

Yes, the stronger version holds. Another way to state the problem:

Let $ I$ be the unit interval with the usual Lebesgue measure, and let $ T: I\to I$ be the "doubling modulo $ 1$" transformation, i.e. $ T(x)=2x$ if $ 0\le x\le\frac{1}{2}$, and $ T(x)=2x-1$ otherwise. Also, let $ f: I\to\mathbb R$ be a measurable function. Then, for almost all $ x\in I$, the upper limit $ \overline{\lim}f\left(T^{n}x\right)$ equals the essential supremum of $ f$ on $ I$.

This is almost obvious once you notice that $ T$ is measure-preserving and ergodic on $ I$.


ABCD is a quadrilateral such that the angles

ABD, DBC, and ADC are all 45 degrees.

Prove angle ACB is twice angle ADB

I think i got a proof, will post it later :D .

Consider the point $O$ on $BD$ such that $ABCO$ is cyclic. Then as $\angle ABO = \angle CBO$, $AO = CO$ and $\angle AOC = 90^{\circ} = 2 \angle ADC$. Therefore $O$ is the centre of the circumcircle of $\triangle ACD$.

Thus $\angle ACB = \angle AOB = 2 \angle ADB$.

Wow, really nice¡¡  :P   My solution is longer, because i didn't notice $O$ was the circuncentre :blush:


geometry
perimeter
Putnam
geometric transformation
reflection
inequalities
triangle inequality

Given a point $ (a,b)$ with $ 0<b<a$, determine the minimum perimeter of a triangle with one vertex at $ (a,b)$, one vertex on the x-axis, and one vertex on the line $ y\equal{}x$.  Assume a minimum perimeter exists, and find it in terms of $ a$ and $ b$.

Actually, there's a pretty elegant solution using simple reflections.

[quote="ZzZzZzZzZzZz"]Given a point $ (a,b)$ with $ 0 < b < a$, determine the minimum perimeter of a triangle with one vertex at $ (a,b)$, one vertex on the x-axis, and one vertex on the line $ y \equal{} x$.  Assume a minimum perimeter exists, and find it in terms of $ a$ and $ b$.[/quote]
What this was a Putnam problem in Zeitz's book?
[hide="Solution"]
Allow us to label $ (a,b)$ as $ A$.  Now, we reflect $ (a,b)$ over the X-Axis and over $ x \equal{} y$ to get $ A'$ and $ A''$ respectively.  We can find that $ A'' \equal{} (b,a)$ and $ A' \equal{} (a, \minus{} b)$.  Let $ B$ be any other point on the $ X$ axis and $ C$ be any other point on $ x \equal{} y$.  Hence, we notice that $ BA \equal{} BA'$ and $ AC \equal{} A''C$.  Thus, $ AB \plus{} BC \plus{} AC \equal{} A''C \plus{} BC \plus{} A'B\ge A''A'$.  Using the distance formula, we find that $ A'A'' \equal{} \boxed{\sqrt {2a^2 \plus{} 2b^2}}$.  [/hide]

EDIT: Thanks, [b]johan.gunardi[/b]!  I edited my solution

[quote="The QuattoMaster 6000"][quote="ZzZzZzZzZzZz"]Thus, $ AB \plus{} BC \plus{} AC \equal{} A''C \plus{} BC \plus{} A'B\le A''A'$[/quote][/quote]

I think it should be $ AB \plus{} BC \plus{} AC \equal{} A''C \plus{} BC \plus{} A'B\ge A''A'$

Algebraically, the problem comes down to a clever use of [url=http://mathworld.wolfram.com/MinkowskisInequalities.html]Minowski's inequality[/url] (when $ p \equal{} 2$), which is really the triangle inequality in disguise.

That's pretty much what I did, with the reflections and all.  And yes, it was a Putnam problem in Zeitz's book.  Amazingly simple for Putnam, I thought, but it was in a chapter where reflections were discussed, which simplified it a little.


Given that $\log_{4n}40\sqrt{3}= \log_{3n}45$, find $n^{3}$.

Plz, give a hint.

You can use $\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$.

I did
\[\frac{\log 40\sqrt{3}}{\log 4n}= \frac{\log 45}{\log 3n}\\ \\ \leftrightarrow \frac{\log 40\sqrt{3}}{\log 45}=\frac{\log 4n}{\log 3n}= \frac{\log 4+\log n}{\log 3+\log n}\]

and then solved for log n; but the expression I get for n is 'messy'. I think there must be a more compact answer. Another hint? Thanks.

[quote="pedrito"]the expression I get for n is 'messy'. I think there must be a more compact answer.[/quote]

First of all, no there mustn't.  However, in this case, there actually is.  What you should do is exactly what Kunny is showing you: after solving for $\log n$, break each of your logarithms down into its prime components.  There's really nothing more to say than that.

[quote="kunny"]$\log 40\sqrt{3}=\log 2^{3}\cdot 3^{\frac{1}{2}}\cdot 5$[/quote]

which equals

$\log 4\sqrt{3}$

1. No, it doesn't, not in any base.
2. There's no particular reason to use base 10 here (although it will work, just as any other base would).


please write correctly!
i think the answer of b is no!

[quote="tkhtn"](a)Prove that:${n\sqrt{3}}>\frac{1}{n\sqrt{3}}$ for every positive integer n. where${x}$ denotes the fractional part of x.
(b)Does there exits a constant c>1 such that ${n\sqrt{3}}>\frac{c}{n\sqrt{3}}$ for every positive integer n.[/quote]
"](a)Prove that:$\{n\sqrt{3}\}>\frac{1}{n\sqrt{3}}$ for every positive integer n. where$\{x\}$ denotes the fractional part of x.
(b)Does there exits a constant c>1 such that $\{n\sqrt{3}\}>\frac{c}{n\sqrt{3}}$ for every positive integer n.


calculus
integration
limit
logarithms
calculus computations

At first sight,it doesn't look like an integral,but actually it is.
Right?

That's right, but Few Japanese high school student solve like this.

They don't know of Stirling formula?
Well,neither do I  :|
With Integral the solution is shorter kunny?
Nice solution Solar Plexus. :coolspeak:

Stirling's formula usually appears in university lesson.

Another approach, you are to use Riemman's sum. :D


calculus
integration
trigonometry
logarithms
calculus computations

Can you prove:
$ \int_0^{ \plus{} \infty} \frac {\sin x}{x^{1 \minus{} p}} \, \textbf{d}x \equal{} \boxed{\sin \left( \frac {p \pi}{2} \right) \Gamma \left(p \right)}, \qquad p \in (0,1)$

 :?:  :)

$ \textbf{I} = \int_0^{ + \infty} \frac {\sin x}{x^{1 - p}} \, \textbf{d}x = \int_0^{ + \infty} x^{p - 1}{\sin x} \textbf{d}x = \textbf{Im}\int_0^{ + \infty} x^{p - 1}e^{ix} \textbf{d}x$
Now put here $ x = it$.
${ \textbf{I} = \textbf{Im}\int_0^{ + \infty} i^{p}\cdot t^{p - 1}e^{ - t} \textbf{d}t = \textbf{Im}\int_0^{ + \infty} \sin\left(\frac {p\pi}{2}\right)\cdot t^{p - 1}e^{ - t} \textbf{d}t = \sin \left( \frac {p \pi}{2} \right) \Gamma \left(p \right)}$
Because $ i^p = e^{p\ln i} = e^{p(\ln|i| + \frac {\pi i}{2})} = \cos \left(\frac {p\pi}{2}\right) + i\cdot\sin\left(\frac {p\pi}{2}\right)$

Nice solution.
I used $ \frac {1}{x^s} \equal{} \frac {1}{\Gamma (s)} \int_0^{ \plus{} \infty} z^{s \minus{} 1} e^{ \minus{} zx} \, \textbf{d}z$  in mine  :rotfl:
definitely too much work with that  :blush:

also here :
http://www.mathlinks.ro/viewtopic.php?t=197640&start=20

[quote="Yuriy Solovyov"]
Now put here $ x \equal{} it$.
$ \textbf{Im}\int_0^{ \plus{} \infty} i^{p}\cdot t^{p \minus{} 1}e^{ \minus{} t} \textbf{d}t \equal{} \textbf{Im}\int_0^{ \plus{} \infty} \sin\left(\frac {p\pi}{2}\right)\cdot t^{p \minus{} 1}e^{ \minus{} t} \textbf{d}t$[/quote]
I don't think, that it's correct.
when you put $ x: \equal{}it$
then interval of integration will be $ \int_{0}^{\minus{}i\infty}$
I think here  is some explanation because we take $ Im$ hence it's not important, but why? it's not so obviusly

In a similar way we can also find $ I \equal{} \int_0^{\plus{} \infty}\frac{\cos x}{x^{1\minus{}p}}\,dx$


a bag contains three red, five green, and two blue marbles. when two marbles are chosen, what is the probability that they are of different colors?

How would you do this problem?

31/45 because you find total probabilities of green and a different one, blue and a different one, red and a different one and add up

= red , then green or blue is 3/10 x 7/9 = 21/90

=blue, then red or green 2/10 x 8/9 = 16/90

= green, then red or blue 5/10 x 5/9= 25/90

add together and get 62/90 then simplify to get 31/45 which is the answer

I hope that helps if it is right.


geometry
incenter
circumcircle
rectangle
geometric transformation
homothety
ratio

Let $I$ be the incenter of a triangle $\triangle ABC$, let $(P)$ be a circle passing through the vertices $B, C$ and $(Q)$ a circle tangent to the circle $(P)$ at a point $T$ and to the lines $AB, AC$ at points $U, V$, respectively. Prove that the points $B, T, I, U$ are concyclic and the points $C, T, I, V$ are also concyclic.

By performing first an inversion of pole $I$, and then an inversion of pole $B'$, the image of $B$ through the previous inversion, I reached a configuration which looked very familiar:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=554[/url]

In fact, I got that proving that $B,T,I,U$ are concyclic is the same as proving the quoted problem in an extended version, namely showing that the fixed point it talks about is an excenter of the triangle. I'm sure there are shorter proofs though :).

I believe my proof is more or less similar to Grobber's one, anywway, here it is.
We start with a simple lemma:
Let X and Y be the points on sides AB and AC s.t. AX=AY, T the second intersection points of the circumcircles of triangles IXB and IYC. Then the circumcircle of $\triangle PXY$ is tangent to AB and AC.

After that, inverse the points wrt incircle, we see that the circumcircles of triangles BCT and PXY are tangent.

This is a lemma asked for in the threads [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127511]India 01[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28200]Rectangle with incircles[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6490]Not Easy[/url]. The replies here did not make me much wiser, but now I have a solution. It is sufficient to show that the points B, T, I, U are concyclic.

AC cuts (P) again at D. Let UV and the external bisector of the $\angle DBC$ meet at K and let UT, VT meet the circle (P) again at L, M, respectively. As the tangency point T is the internal homothety center of the circles (P), (Q), $LM \parallel UV.$ Moreover, tangents of (P) at L, M are parallel to tangents AB, AC of (Q) at U, V, respectively. Thus M is the midpoint of the arc CD of (P) containing B and $M \in BK,$ the external bisector of the $\angle DBC.$ fURTHERMORE, $\angle LTM = \angle UTV = \angle UVA.$ Let N be the midpoint of the arc CD opposite to B, BN bisecting the $\angle DBC$ internally. Then $\angle MTB = \angle MNB$ and

$\angle LTB = \angle LTM+\angle MTB = \angle UVA+\angle MNB$

Let X be intersection of the external bisector AK of the $\angle DBC$ with the tangent $t_{N}\parallel t_{M}\parallel AC$ of (P) at N. From the right triangle $\triangle MNX$ with the N-altitude NB, $\angle BXN = \angle MNB.$ If we draw a parallel to $t_{N}\parallel AC$ through K, it is clear that

$\angle UKB = \angle UVA+\angle BXN = \angle UVA+\angle MNB = \angle LTB.$

It follows that $\angle UTB = 180^\circ-\angle LTB = 180^\circ-\angle UKB$ and the quadrilateral BTUK is cyclic. Let UV meet BC at W and let the external bisector BK of the $\angle DBC$ meet CD at E. By Menelaus theorem for the triangle $\triangle ABC$ cut by the transversal UVW,

$\frac{WB}{WC}= \frac{UB}{UA}\cdot \frac{VA}{VC}= \frac{UB}{VC}$

because UA = VA. By Menelaus theorem for the triangle $\triangle EBC$ cut by the same transversal $KVW \equiv UVW,$

$\frac{KB}{KE}= \frac{VC}{VE}\cdot \frac{WB}{WC}= \frac{VC}{VE}\cdot \frac{UB}{VC}= \frac{UB}{VE}= \frac{UB}{CE-CV}$

Denote b = CD, c = DB, d = BC sides of the $\triangle BCD,$ then $CE = \frac{db}{c-d}.$ By [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67398]Purser theorem[/url] (generalized Ptolemy) for the $\triangle BCD$ with the circumcircle (P) and with the circle (Q) tangent to (P),

$d \cdot (b+CV) = d \cdot DV = c \cdot CV+b \cdot BU$

$db-(c-d) \cdot CV = b \cdot UB$

$\frac{KB}{KE}= \frac{(c-d) \cdot UB}{db-(c-d) \cdot CV}= \frac{c-d}{b}.$

Thus UV cuts BE in precisely the same ratio as the external bisector of the angle $\angle BCD,$ which means that $K \equiv UV \cap BE$ is the D-excenter of the $\triangle BCD.$ Once this is known, we can finish the angle chase. First, $\angle BUK = \angle AUV = \angle UVA.$ Since $CK \equiv CI$ is  external bisector of the $\angle BCD$ and internal bisector of the $\angle BCA,$

$\angle BIK = 180^\circ-\angle BIC = 90^\circ-\frac{\angle CAB}{2}\equiv 90^\circ-\frac{\angle VAU}{2}= \angle UVA,$

so that the quadrilateral BIUK is also cyclic and consequently, all 5 points B, T, I, U, K are concyclic.

sorry I changed the notaions.
[b]Problem[/b].a circle is tangent to an arbitrary circle which passes through points B and C of triangle ABC at T,and is tangent to AB and AC at points V and U. let $ I$ be incenter of $ \triangle ABC$ .prove that $ UTIC$ is cyclic.

[i]Proof.[/i]let's invert everything around $ B$ . after inversion point $ I$ will be B-excenter of $ \triangle A'BC'$.now if we prove that $ T'U'C'I'$ is cyclic the problem solved.let D' be midpoint of arc $ A'B$ which contains $ C'$.let $ V'T'\cap C'D'\equiv K$ .cause U' is homotecy center of circumcircles of triangles $ U'V'T'$ and $ A'BC'$ hence $ \angle V'T'U' \equal{} \angle D'C'U'$ hence $ T'U'C'K$ is cyclic therefore $ \angle U'KD' \equal{} \angle U'T'C' \equal{} \angle T'V'D'$ hence $ D'K^2 \equal{} D'U'.D'V'$ also we have $ D'A'^2 \equal{} D'U'.D'V'$(cause $ \angle U'A'D' \equal{} \angle A'V'D'$) hence $ D'A' \equal{} D'K$ therefore $ K$ is B-excenter of triangle $ A'BC'$ hence $ K\equiv I'$

Dear Valentin, Orlando, Ben_Grosso and all,
revisiting some problems, I come on the TST Roumanian 2002 problem 2 which is proposed on Mathlinks.
I propose you a new synthetic proof that you will find on my site
http://perso.orange.fr/jl.ayme/
see : volume 2 (2008) Un remarquable résultat de Vladimir Protassov
Sincerely
Jean-Louis

Dear Mathlinkers,

who is the author of this nice problem?
Any precise reference are welcome...

I saw : India MO 2001?

Any help welcome...

Sincerely
Jean-Louis

[quote=jayme]Any ideas?
I think that the author must be an indian geometer...

Sincerely
Jean-Louis[/quote]

It might even be an indian undergrad student but must be Indian.


AMC USA(J)MO USAMO AIME ARML geometry AwesomeMath

I was wondering if anyone could help me out with the following: Last year I recall reading something about a USAMO index for something with a name like the Girls Chinese Math Olympiad.  I can't remember exactly what it was, though, or how it worked.  Were girls with USAMO indexes above a certain score invited to compete?

I have a seventh grade student who has a good chance of making the USAMO if the AIME cutoff is 7.  From what I remember of the listing for the Girls Math Olympiad last year, she would have a chance of qualifying for that one, too.  I'd like to let her know about it, but I don't remember enough about the post to find it here.  If anyone can point me in that direction, I'd much appreciate it.

Thank you for any help you can give.

Catherine Asaro, Coach
Chesapeake ARML
Howard Area Homeschoolers

[quote="bookaholic"]There were eight team members and four alternates who were invited to train at MOP, and the team members competed in China in August. Our comments and results along with the 2008 CGMO problems are at http://www.msri.org/specials/gmo/2008. The cutoff for the team was 11 on the USAMO and the cutoff for the alternates was 7. I can be much more long-winded about the CGMO if you so desire :)[/quote]

Yes, please do be long-winded!  I'd be interested in anything you can tell me about it.  Thanks.

Best -- Catherine

Basically if your 7th grader can get about 10 on the USAMO, she will be able to go most likely. USAMO index doesn't count for anything though :(

Thank you!  That's great.  It gives me a much better idea.

I won't have any more questions unless she makes the USAMO, and then we'll have to see how she does on that.  But given her young age, I wanted to give her a heads up on all of this, just in case she did make any of these things, either this year or in her remaining future five years.

Thanks much for your input.

Best,
Catherine

[quote="bookaholic"]Okay, so first of all there's the training at MOP. I can't be sure that the CGMO girls will actually go to MOP again this year--in 2007 they trained at AwesomeMath--but from what people were saying it's the most likely course of action. Team members are placed into whatever student group suits them best. Last year we were all in the red group and we were treated basically like the typical red MOPper except that we had separate study sessions to focus on CGMO problems in the evenings. As with normal MOPpers, camp was free.

After MOP ended at the start of July, we went home and did normal summer things until August, then we went to China for two weeks. Again, we didn't have to pay for the trip. The test itself lasted two days--four problems in four hours each day--and you can see the test we took last year at the above link as well as the 2007 test at the 2007 team page, http://www.msri.org/specials/gmo. I'd estimate that the problems range from slightly below USAMO level to easy/mid-USAMO level, and I think they tend to be slightly shorter (i.e. they require fewer major insights per problem).

Aside from the test, we spent the rest of the trip touring. We went to Hong Kong, Shanghai, and Zhongshan (where the test was hosted). Where this year's team goes will presumably depend on where the host is this year, since nobody wants to spend the whole trip on an airplane. It was very, very, very fun, but your student should be prepared for somewhat sparser accomodations than is the norm in the U.S. (Which is to say, if she can't stand the idea of sleeping on wood, she had better bring a mattress [sleeping on wood isn't THAT uncomfortable though, really.])

If you have any more specific questions I'd be happy to answer them as well :)[/quote]


geometry
inradius
circumcircle
inequalities
inequalities proposed

Prove that for any triangle of sides $a,b,c$ ,inradius $r$ and circumradius  $R$ we have the ineq

$8R^2+4r^2 \geq a^2+b^2+c^2$

This is so damn nice! And do you know why? Because it is equivalent to Poland inequality that we have discussed today. Indeed,  using well known trigonometric formulas, this reduces to $ (1-cosA)(1-cosB)(1-cosC)\geq cos A cos B cosC$ which is nothing else than the Polish inequality. Much said Polish inequality, since it seems that Laurentiu Panaitopol has found it earlier.

[quote="harazi"]This is so - nice! And do you know why? Because it is equivalent to Poland inequality that we have discussed today. Indeed,  using well known trigonometric formulas, this reduces to $ (1-cosA)(1-cosB)(1-cosC)\geq cos A cos B cosC$ which is nothing else than the Polish inequality. Much said Polish inequality, since it seems that Laurentiu Panaitopol has found it earlier.[/quote]

WOW... I must have been blind not to realize this, since that "Polish" inequality (which was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=4972 and which I have just found in a paper by Sefket Arslanagic) is one of my favorite triangle inequalities.

Anyway, I have a really idiot proof using Muirhead, I am ashamed of posting it but I would be more ashamed to delete a proof already written down, so here it goes...

If $s=\frac{a+b+c}2$ is the semiperimeter of triangle ABC, then a = y + z, b = z + x and c = x + y, where x = s - a, y = s - b and z = s - c. Note that the numbers x, y, z are all positive. Also, x + y + z = (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s. Hence, the Heron formula yields $S^2=s\left(s-a\right)\left(s-b\right)\left(s-c\right)=\left(x+y+z\right)xyz$, where S is the area of triangle ABC, and since we have $R = \frac{abc}{4S}$, short algebraic calculations show that $R^2 = \frac{\left(\left(y+z\right)\left(z+x\right)\left(x+y\right)\right)^2}{16\left(x+y+z\right)xyz}$. Also, we have $r^2 = \frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s} = \frac{xyz}{x+y+z}$. Using these formulas, we can rewrite the inequality in question as

$\frac{\left( \left( y+z\right) \left( z+x\right) \left( x+y\right) \right)^2}{2\left( x+y+z\right) xyz}+\frac{4xyz}{x+y+z}\geq \left( y+z\right)^2+\left( z+x\right) ^2+\left( x+y\right) ^2$.

For the proof, remark that

$\frac{\left( \left( y+z\right) \left( z+x\right) \left( x+y\right) \right)^2}{2\left( x+y+z\right) xyz}+\frac{4xyz}{x+y+z}-\left( \left( y+z\right)^2+\left( z+x\right) ^2+\left( x+y\right) ^2\right) =$
$\frac{1}{2\left( x+y+z\right) xyz}\left( \left(x^2y^4+x^2z^4+y^2z^4+y^2x^4+z^2x^4+z^2y^4-2x^4yz-2y^4zx-2z^4xy\right) +\right.$
$\left. 2\cdot \left( \left( yz\right) ^3+\left( zx\right) ^3+\left(xy\right) ^3+3\cdot yz\cdot zx\cdot xy-\left( \left( yz\right)^2zx+\left( yz\right) ^2xy+\left( zx\right) ^2xy+\left( zx\right)^2yz+\left( xy\right) ^2yz+\left( xy\right) ^2zx\right) \right)\right)$,

what is $\geq 0$ because of the two inequalities

$x^2y^4+x^2z^4+y^2z^4+y^2x^4+z^2x^4+z^2y^4-2x^4yz-2y^4zx-2z^4xy \geq 0$

and

$\left( yz\right) ^3+\left( zx\right) ^3+\left(xy\right) ^3+3\cdot yz\cdot zx\cdot xy-\left( \left( yz\right)^2zx+\left( yz\right) ^2xy+\left( zx\right) ^2xy+\left( zx\right)^2yz+\left( xy\right) ^2yz+\left( xy\right) ^2zx\right) \geq 0$;

the first of these inequalities follows from Muirhead with $\left(4;\;2;\;0\right) \succ \left(4;\;1;\;1\right)$, and the second one follows from Schur, applied to the numbers yz, zx, xy.

Proof complete.

  Darij

Muirhead inequality is for me one of the worst things that ever happened since the invention of the sign $\geq$. ;)  People cannot create hard and nice inequalities anymore, since they are immediately killed with this force machine! I hate Muirhead!  :D

Using  the formula  $ a^2+b^2+c^2=2(s^2-r^2-4Rr)  $  , Manlio's  problem  is  equivalent  with  $ s^2\leq4R^2+4Rr+3r^2 \Leftrightarrow IH^2\geq0  $
As  far  i know , this  problem  was  pirst  published  in  1953 (look  O.Bottema  et al. Geometric inequalities , Groningen 1969 , page 53)

The first ineq of Darij follows more directly from

$x^4(y-z)^2 + y^4(z-x)^2 + z^4(x-y)^2 \geq 0$

so you can avoid Muirhead.

However, I am agree with Harazi about Muihead. This theorem kill many inequalities in a mechanical way and so kill also the fantasy and genius of the human brain, characteristics of the thought we can use but not to understand entirely.

As Godel said : "The human brain will never be able to understand itself or , better, the mathematical formalism will never be able to capture all truths of the Universe"

Sorry, for this digression about Kurt Godel, but I consider him one of the best  brains in the history of human thought and I love to remember him.

[quote="manlio"]The first ineq of Darij follows more directly from

$x^4(y-z)^2 + y^4(z-x)^2 + z^4(x-y)^2 \geq 0$

so you can avoid Muirhead.[/quote]

You're right, Manlio, about that. I was too lazy and very much in hurry, so I just used Muirhead. But even without Muirhead, it's still a rape rather than a proof.

And you're right, Harazi, about Muirhead. There is (I hope, however) one way to avoid the possibility to use Muirhead: generic inequalities for n variables.

  Darij

Long live inequalities in n variables! Yet, they are so hard to create... :(


AMC USA(J)MO USAMO calculus inequalities induction geometry

I heard I don't have to study calculus..and have to study some number theory..anything else?

I read a syllabus of the IMO before and from what I understand is that the mathematics level for IMO is "elementary". It sounds like you don't need calculus or anything beyond that but I think you have to know the [b]key[/b] topics and use sophisticated methods to attack a problem. To answer your question, strictly speaking no, but a 'yes' would be at your advantage doesn't it? ;)  Instead of learning the 'minimum' knowledge to make it to USAMO/IMO, it is better for you to learn as much as you can, learning [u]everything[/u].

You do need to know some number theory that is not covered in standard high school curriculum.

There is no calculus on high school olympiads (except in Romania). That isn't really the same thing as being elementary; the number theory and combinatorics on olympiads could easily be college material.
The problems are all at least understandable with elementary methods and high-school algebra/geometry. You may need a lot of different tricks to solve them, and those tricks aren't always elementary.

Convexity and concavity are pretty much as college-level as you would need to get for MO problems, proving inequalities or finding maxima/minima and so forth.  It's not required, in theory, but a lot of otherwise difficult problems become nearly trivial this way.

The USAMO is definitely NOT based on the high school curriculum, and many of the concepts, ideas, theorems, and methods that you see there will be things that are not covered at all in high school.  Some of the subjects do not even appear in other math contests, simply because some subjects only work with proof problems.  Consider functional equations and inequalities, for example.  Those scared the daylights out of me when I first started preparing for USAMO, especially functional equations.  They require a completely new set of techniques.  While these techniques do not require knowledge that you normally wouldn't learn until college, they require knowledge and experience well beyond what the typical high school student has.  Picking up this knowledge and experience is what preparing is for.

You should also have a working knowledge of number theory.  A lot of the things you may need to know in NT for the USAMO are in fact covered in a typical college course.  You should know everything about modular arithmetic like the back of your hand (especially how to deal with powers), and some standard theorems (Fermat's Little for sure!).  Beyond that, what you need most is creativity.

As a side note, functional equations haven't shown up on the USAMO since 2002.  I don't know whether makes it more or less likely that they will show up this year, though.

Yep, lots of stuff arent taught in hs.  u "need" to know bunch of elementary formulas that most high schools don't teach (flt, cauchy, amgm, power of point, herons, pythagorean:)......etc.)

i believe i'm right in assuming then that "Inequalities" by Hardy, Littlewood, and Polya is IMO material... :?

[quote="ThAzN1"]Hmm... but most of that book goes WAY beyond the Olympiad syllabus.  You really don't need as much of a background as you think to be decent at solving Olympiad problems.  Personally, I just stick to doing problems / reading solutions and looking up / reading about whatever you don't know.  A sufficient background isn't too hard to develop (but takes some time, of course) -- the hard part is learning when to use what, and that's what you learn only by doing problems.  Pick up the 'tools' you need as you learn to use them![/quote]

I agree. A general knowledge in doing proofs will usually trump knowing a bunch of advanced stuff. Why? Because [i]every[/i] question involves writing a proof. But advanced knowledge only covers [i]one type[/i] of question. And there's always questions that are completely "ad hoc", like the game problems.

Also a lot of times, if you have a lot of experience proving stuff you can oftentimes break down a problem that you have no idea how to do into problems that you do know how to do (either via case analysis or just proving something in a round-about way that gets past the major difficulties of the problem).

So some of the most important things you can learn and make sure you are completely comfortable with are induction, proof by contradiction, case analysis, and probably a lot of other methods that I'm not sure have names to my knowledge (like proving that it is impossible to get from state A to state B, and to get to state C you must go through state B, so it is impossible to get from state A to state C). A lot of it, too, is logic and the willingness to be inventive.

Of course, as other people have said, it doesn't hurt to have a lot of background knowledge either.

One thing I've noticed that not a lot of people have mentioned here is that Olympiad problems use a whole ton more geometry than the standard HS curriculum. Basically, you can expect anything from Power of a Point to inversion to be on them.

In summary, I think the best way to describe what's on USAMO/IMO is just what's in AoPS!


abstract algebra
topology
group theory
advanced fields
advanced fields unsolved

Notation: Given a locally compact abelian group $ G$, $ \widehat{G}$ is defined as the set of all continuous homomorphisms $ \chi: G\rightarrow\mathbb{S}^1$.

I have the following problem:

Let $ G$ be a locally compact abelian group and $ H\leq G$ closed.  Given that $ H^*: \equal{} \{\chi\in\widehat{G}: \chi|_{H}\equiv 1\}$, prove that $ \widehat{G\slash H}\cong H^*$.

Now, my question is why does $ H$ need to be closed?

I'm not very familiar with the theory, but offhand it looks like it's not meaningful to define a topology on $ G/H$ unless $ H$ is closed.  For example, take $ G \equal{} \mathbb{R}, H \equal{} \mathbb{Q}$.  I believe the quotient topology on $ \mathbb{R}/\mathbb{Q}$ is trivial.  In any case, it certainly doesn't look locally compact.

General rule: toplogical subgroups need to be closed practically everytime; otherwise there happen very ugly things.

And yes, as already said, the quotient topology is rather useless, causing no nontrivial continuous homomorphism to exist.


[quote="pvthuan"]It interesting to note that equality holds at $(0,0,0)$, $(\tfrac13,\tfrac13,\tfrac13)$, and $(0,1,2)$.[/quote]
There is not equality at $(0,1,2)$. :P

Sorry for my computation. The following should be true. \[x(x+y)^{5}+y(y+z)^{5}+z(z+x)^{5}\geq\frac18(x+y+z)^{6}.\]

do you mean that the first inequality is not true for all reals?
[for positive ones I have a very simple solution]

No, the first inequality is true and solved. (true for all real numbers)

[quote="Rzeszut"][quote="VANCHANH123"][quote="pvthuan"]Let $x,y,z$ be real numbers, prove that \[x(x+y)^{3}+y(y+z)^{3}+z(z+x)^{3}\geq\frac8{27}(x+y+z)^{4}.\] This is just a special case of many results of inequalities in my book.[/quote]
*If xyz=0....
*xyz<>0 then: ... x+y+z=1
then $\sum x(1-x)^{3}\geq \frac{8}{27}$[/quote]
I thnk you misread $\sum x(y+z)^{3}$ instead of $\sum x(x+y)^{3}$.[/quote]
 I AM SORRY

[quote="pvthuan"] inspired from a problem of Janous[/quote]
Where is published  the "problem of Janous"?
By the way, your first refined inequality is very nice! :D

Dear Vasc:

It is Problem 1831 in CRUX MATHEMATICORUM published by Canadian Mathematical Society.

I think you and Walther Janous are independent.

Janous, are you here in this forum? Is anyone here a friend of his?

[quote="pvthuan"]It is Problem 1831 in CRUX MATHEMATICORUM published by Canadian Mathematical Society.
I think you and Walther Janous are independent.
Janous, are you here in this forum? Is anyone here a friend of his?[/quote]
Thanks. I think alike.
I published the inequality for $k=3$ in Gazeta Matematica in 1989, and Janous published a generalized version (problem 1831) in Crux Mathematicorum in 1994.
I don't know whether Janous is in this forum.

[quote="pvthuan"]Let $x,y,z$ be real numbers, prove that
\[x(x+y)^{3}+y(y+z)^{3}+z(z+x)^{3}\geq\frac8{27}(x+y+z)^{4}. \]
This is just a special case of many results of inequalities in my book.[/quote]
Anyway, we can transform this inequality into sum of square easily. :)

Can you post your solution? toanhocmuomau

[quote="pvthuan"]Let $ x,y,z$ be real numbers, prove that

$ x(x \plus{} y)^{3} \plus{} y(y \plus{} z)^{3} \plus{} z(z \plus{} x)^{3}\geq\frac8{27}(x \plus{} y \plus{} z)^{4}.$[/quote][quote="toanhocmuonmau"]Anyway, we can transform this inequality into sum of square easily. :)[/quote]$ 27\sum_{cyc}{x(x \plus{} y)^{3} } \minus{} 8\left(\sum{x}\right)^{4}$

$ \equal{} \sum_{cyc}{\left[\frac {\sqrt {29}(1708x^2 \plus{} 427y^2 \plus{} 427z^2 \minus{} 2026yz \minus{} 916zx \plus{} 380xy)}{5124\sqrt {2}} \minus{} \frac {5\sqrt {22}(427y^2 \minus{} 427z^2 \plus{} 432yz \minus{} 802zx \plus{} 370xy)}{3416}\right]^2}$

$ \geq0.$


let a and b positives numbers from $ R$  prove that:

for all n in $ N$  (a+b)^n ( a^n+b^n) =< 2^n ( a^2n + b ^2n)

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!


Stanford
college
function
geometry
combinatorics theorems
combinatorics

Dear Colleagues,

I have worked up a proof of the 4-Color Theorem, and I would very much appreciate your comments. Please have a look at the PDF at:
[url]http://wizrad.home.comcast.net/4ColorProof.pdf[/url]

Thanks, Sanford

If you really think you have a solution, put it on [url=http://arxiv.org/]arxiv[/url].

I've only partially read the manuscript...

Comments: Space is usually not a problem, at least when it comes to a mathematics paper. Thus, supplying clear, expository proofs are required, even if it is 100 pg paper. Now, from an overall view of it, read midway of the paper, but still regret not able to finish the entire paper with full absorption of the content but a mere skim-through, the theorems seem redundant and repetitive. A lot of the definitions contained are, in my opinion, not defined mathematically rigorous.

The assumptions need to be more clear. Name each one clearly, translate the intuitive notion into a mathematical object (in other words, stripping its 'intuitiveness' by generalizing).

Proofs need to be logically 'sound'. If-then-therefore is always more welcomed in the mathematical community than a prose-long explanation of common-sense.

Lastly, like JRav said, please try arXiV. I'm sure there are more authoritative mathematicians there that can lead you to the right direction.

This document is typical handwaving based on a few simple pictures -- there is extremely little content to any of the statements.  This is exactly the sort of vague reckoning that makes it easy to believe the 4CT is true, but it doesn't begin to exhibit the kind of rigor necessary to rise to the level of a proof.

The arXiv is a repository for published or potentially publishable papers.  It does no one (neither the author nor those of us who actually use the arXiv) any good to suggest that a document like the given one be placed there.

Thanks for response #2.
I'll consider posting to arxiv but want to get some feedback first.

Thanks for responses #3, #4
   However, your value judgments of lack of rigor, not rising to the “level of mathematical proof,” vagueness and admissions of not thoroughly reading the paper, bespeak of your own lack of rigor, vagueness and deficiency of proof (evidence). Your comments would be valid were you to be specific e.g. cite a definition, a logic, theorem, whilst explaining its deficiencies and the criterion. The presentation may not be in your “mathematical” vernacular, but evaluate it on its own epistemology. It’s all there: definitions, theorems, illustrations, reasoning and consistency. So please attend the substance, contend with specifics, and present your concrete reasoned criticisms.
   A non-conventional proof does not preclude its being well defined nor its “rising to the level of mathematical proof.”

OK, here's what I have after looking through it:

[quote]But this contact does not require more
than four colors because any contact country(s) is a buffer
between the non-contact country(s) and/or becomes a
component of another $ G_k$.[/quote]This is a gaping hole. When you reorganize this way, you create overlapping groups, and you never address those. Those "buffers" don't help if they're trying to connect too many pieces.

Your figure #6- well, it's not too hard to prove that five colors suffice using something like that. It doesn't work for four colors, because it branches off into so many configurations that must be checked.

Supposing we know what it means, the statement: "A common boundary between two countries
is mutually exclusive of another common boundary between another pair of countries",  is false.

A disk without some points is a """"continuouos""""area"""" and it does not have a boundary line.

¿¿¿...... "THEOREM 2 : Any set of n countries, An, composing a land mass, can be defined into the subsets: G1, G2, G3 & G4 configurations. This theorem follow from n = a(1) + b(2) + c(3) + d(4) where a, b, c, d are the number of each Gi, where i = 1,2,3,4." ....... ???'

Please give credit to all great mathematicians that have worked on this theorem (4-colors) (and in the correct definition of notions that are "intuitive").

Oh yeah. I should have mentioned the planar graph formulation, with countries as vertices. That's much easier to work with.

jmerry #6

Thank you for your attention.

Please, refer to p 5. C - 4, Theorem 2, and C - 5. Corollary 3. Each Gestalt is stated and defined as distinct from one another. Each element can be a member of only one Gestalt (k G i) of the given set k, which is indicated by the sum of the number of countries in set k , n k , equaling the total, n, of the countries of the map. The set k is then reconfigured to set j in which a country, A i , may be an element of the same or a different Gestalt ( j G m ) but not two Gestalts in the same set j (your “overlapping”). The n j = n .

Addressing the second point: a buffer country’s function is not to “connect” but to separate colors. See p. 6, D – 1. This function does not abrogate its being an element of a Gestalt. ¶ p. 7, D - 5 is meant as a summary and taken alone can be confusing. If you keep in mind that a buffer country(s) allows the two countries it intercedes (separates) to have the same color, and consider D – 2, 3, 4, hopefully you will conclude that Theorem 3 is proven.

By way of a simple example, look at Fig. 5, p. 7. Note the notation c A i (refer to B, p. 2). Here i = 1, 2, 3, 4, 5. There are five defined countries (areas) in this map, some with common boundaries, some without. A Gestalt (1 G 3 (the left subscript identifies the Gestalt)) of the three countries,  A3, A4 & A 5 , with simultaneous common boundaries can be configured (see C, p.3 & C–1, p.4). The colors 1, 2 & 3 are assigned: 3 A 3 , 1 A 4  & 2 A 5 . Another 2 G 3  can be formed, A 2 , A 3  , A 4 with simultaneous common boundaries by reconfiguring the map (set 2 ) utilizing  members of 1 G 3 of set 1: 3 A 3 & 1 A 4 with the colors C 3 & C 1 remaining.  A 3 & A 4 serve as a buffer between A 2 & A5 , & thus each can be assigned the same color, 2 A 2 ,  2 A 5 (incidentally  A5 is now a member of 3 G 1 ). To complete the coloring,  a  4 G 2  of  A 1 & A2  is formed in another reconfiguration (set 3 ) of the map with the colors assigned as before & A 1 is given C 1 : 1 A 1 , as A 2 serves as a buffer between 1 A 1 & 1A 4 .  (A1 could have been assigned C3 , as A 2 is a buffer between A 1 & A 3 also).

Fig. 6 is applicable in dealing with the potential of a specific arrangement of countries of a map requiring the use of a different color e.g. the enclosure      [MG 2 + G 1], p. 5, C-3, for other than the demand of simultaneous common boundaries (G 2, G 3, G 4).  (G 1 is not a problem as there is always at least one color available for use.) This question is considered (p. 8, E) and answered: only the enclosure [MG 2 + G 1] which requires at maximum 3 colors, Fig. 6 (there is a trivial typo in the explanation:  [MG 2 + G1]  should read [MG 2 + G 1 ] ) illustrates that two intersecting enclosures allow for reconfiguration of the countries dissipating  any “forced” arrangement. If you would attend the subscripts c A i , you would see only 3 colors are required.

Pertaining to jmerry #8.

To each his own.  My understanding is that “planar graph formulation” is successful for only a cubic planar surface. Also, retaining a map of countries is more meaningful [1 – pp 91, 183] Further, this proffered proof is inspired by Theorem 1 which is too coincidental to be merely incidental. In addition, to prove something more than one way embellishes its validity.

Thanks again and I hope my comments address your objections.

As your gestalts are a bit iffy sometimes, I would suggest starting to transcribe your paper in planar graph formulation, which is always possible for any planar graph.

It's just that you are creating all of these new theorems that are very confusing, and sometimes vague, which could be avoided by turning it into a graph, eliminating any "areas" or "touching".

Indeed, the standard graph abstractions are there for a reason. It helps other people
to state your ideas in the standard language (in this case, via graphs), so that you don't
have to establish basic results and so that others with the right background can easily
understand your writing.

To give an example, here's a simple proof of the five-color theorem:

We work by contradiction. Suppose that there is some planar graph which cannot be five-colored, and let $ G$ have a minimal number of vertices among such graphs. WLOG, we may assume that all faces of $ G$ are triangles, adding edges if necessary.
By Euler's formula, $ V\plus{}F\equal{}V\plus{}\frac23E\equal{}E\plus{}2$, so $ E\equal{}3V\minus{}6$. The average degree of vertices is less than $ 6$, so some vertex $ P$ has degree $ 5$.
Let $ G'$ be the graph $ G$ with $ P$ and its edges removed. By the minimality hypothesis, $ G'$ can be five-colored. Since $ G$ can't be, the five vertices connected directly to it must use all five colors. Let these vertices be $ V_1,V_2,V_3,V_4,V_5$ in counterclockwise order, colored $ c_1,c_2,c_3,c_4,c_5$ respectively.
Now, consider the subgraph $ G_{13}$ of all vertices in $ G'$ colored with $ c_1$ or $ c_3$, and the edges connecting them. There are two cases:
Case (a): $ V_1$ and $ V_3$ are in different connected components of $ G_{13}$. In this case, we can exchange colors $ c_1$ and $ c_3$ in the component of $ V_1$ while still leaving the coloring of $ G'$ valid. The new color of $ V_1$ is $ c_3$, and now we can color $ P$ with $ c_1$, five-coloring $ G$.
Case (b): $ V_1$ and $ V_3$ are in the same component of $ G_{13}$. In this case, we can find a path $ S$ of vertices alternating between colors $ c_1$ and $ c_3$, starting at $ V_1$ and ending at $ V_3$. Together with the edges $ PV_1$ and $ PV_3$, this divides the plane into two regions, with $ V_2$ and $ V_4$ on opposite sides. On the side containing $ V_2$, exchange colors $ c_2$ and $ c_4$; the coloring of $ G'$ is still valid, since no edges cross between the two regions. We now have $ V_2$ colored $ c_4$, and we can color $ P$ with $ c_2$, five-coloring $ G$.
We have five-colored $ G$ in both cases for a contradiction. Therefore there is no such minimal non-5-colorable graph, and all planar graphs can be five-colored.

Now, what if we try this idea with four colors? If the vertex we focus on has degree $ 4$, it works. A minimal non-4-colorable planar graph must have no vertices of degree $ \le 4$. If the vertex we focus on has degree $ 5$, there's a loophole. Suppose (WLOG) the colors are $ c_1,c_2,c_3,c_4,c_2$ for $ V_1,V_2,V_3,V_4,V_5$ in counterclockwise order. We can look for a path between $ V_1$ and $ V_3$ in colors $ c_1$ and $ c_3$, but if there is one, we can't usefully switch $ c_2$ and $ c_4$, because both colors appear on the same side adjacent to $ P$.
Trying to address this leads to hundreds or thousands of cases, and a proof too complicated for a human to verify. Recent work has greatly improved on the original proof's length, but it's still a monster.

Why is your work so easy to dismiss? You've barely even touched on the interesting cases, and what you have there (page 8) is handwaving that doesn't prove anything.

I agree with most comments, the proof is not convincing.
A related discussion started in the section:
Combinatorics > Open Questions > "Triangulated planar graph property on all but 2 vertices"
It ask for a proof or disproof of one of the two 2-lines conjectures.
If one of the two conjectures can be proved then the 4CT can be proved in a classic way, so don't expect them easy to prove or disprove.
As jmerry and also Nukalar suggest it uses the planar graph formulation.
A nice thing about the conjectures is that they are a generalisation (but only for triangulated graphs or cubic maps) of the 4CT, it means that they CANNOT be proved because the 4CT is prooved (it's indeed a monster but after all it's a PROOF).
The conjectures are based on Heawood's vertex character (for the dual of cubic graphs).
Please join the discussion there.

kweenie

PS: as this thread is't used for several months I think it's allowed to re-route it.