This is the last inequality I'll post here.
I guess you've got enough inequalities now ... :D
So enjoy this one ! It's far from easy I must admit ... :wink:

Let a, b, c be positive reals. Prove that

a/:sqrt:(a + 8bc) + b/:sqrt:(b + 8ca) + c/:sqrt:(c + 8ab) :ge: 1.

:mrgreen::^3:

This is difficult. Here are three possible approaches (HINT) :

(1) Determine a real number k such that for a, b, c > 0 :

a/:sqrt:(a + 8bc) :ge: (a^k)/(a^k + b^k + c^k).

Tricky !
I wouldn't have thought of this. This is the official solution ...

(2) Use CAUCHY ! (Be inventive, it's not obvious how to apply it.)

(3) Use Jensen for the convex function f(x) = 1/:sqrt:x.

Happy solving :)

Well, I for one have seen so many solns I'm not even sure I CAN come up with sth original. The idea of having every term n the left >=a^k/(a^k+b^k+c^k) (or b^k/(blah) or c^k/(blah) ie really cool. I tried applying it myself for some ineqs and it kind of worked, but I don't think you can apply it that often. Anyway, it's really nice and elegatn. In this case k is, I think, 4/3, but I'm not sure. It can't be actually fund like when you solve an eqn, you have to do a bit of guessing..

OK I'll post the solutions myself.

[1st solution] Proving that
a/:sqrt:(a + 8bc) :ge: a^(4/3)/(a^(4/3) + b^(4/3) + c^(4/3)
is quite straightforward.
(The proof is straigthforward, the idea isn't straightforward at all :mrgreen:)
Now the result follows from adding the three similar inequalities.

[2nd solution] Cauchy gives :

(1) LHS x (:Sigma:a:sqrt:(a + 8bc)) :ge: (a + b + c)

(2) :Sigma:a:sqrt:(a + 8bc) = :Sigma::sqrt:a:sqrt:(a + 8abc)
:le::sqrt:((a + b + c)(a + b + c + 24abc)
:le::sqrt:((a + b + c)(a + b + c)) = (a + b + c)
where a + b + c + 24abc :le: (a + b + c) follows from AM - GM.

Now combining (1) and (2) gives the result.

[3rd solution] WLOG assume a + b + c = 1.
Then Jensen applied to the convex function 1/:sqrt:x gives :
:Sigma:af(a + 8bc) :ge: f(:Sigma:a(a + 8bc)) = f(a + b + c + 24abc)
:ge: f((a + b + c)) = f(1) = 1.
since a + b + c + 24abc :le: (a + b + c) and f is decreasing.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac{\pi ^2}{4}} (2\sin \sqrt{x}\plus{}\sqrt{x}\cos \sqrt{x})\ dx$.

Let $ u \equal{} \sqrt{x}$. Then we have:
$ \int_0^{\frac {\pi ^2}{4}} (2\sin \sqrt {x} \plus{} \sqrt {x}\cos \sqrt {x})\ dx$
$ \equal{} 2\int_0^{\frac {\pi}{2}} (2u\sin u \plus{} u^{2}\cos u)\ du$
$ \equal{} 2\int_0^{\frac {\pi}{2}} (u^{2}\sin u)'\ du$
$ \equal{} \frac {\pi ^2}{2}$.

That's right! The problem is very easy for you. :lol:


Consider 7 distinct  positive integers not exceeding 1706. Prove that there are three of them say a,b,c, such that a < b+c < 4a.

Suppose there exist positive and pairwise distinct integers $x_1, x_2, \ldots, x_7 \le 1706$ not fulfilling the condition postulated by the problem. Wlog, you may suppose $1 \le x_1 < x_2 < \ldots < x_n \le 1706$. Then necessarily $4x_i \le x_{i+1} + x_1$, i.e. $x_{i+1} \ge 4x_i - 1$, for any $i = 2, 3, \ldots, 6$. This means $x_3 \ge 8 \cdot 2 - 1 = 7$; $x_4 \ge 4 \cdot 7 - 1 = 27$; $\ldots$;  $x_7 \ge 4 \cdot 427 - 1 = 1707$, absurd! Now conclude...


original image from http://www.dlsonline.net
hat image from http://www.neowin.net

Something must seriously wrong with my eyes , I see only one Image.

Please reply only to threads that have been active recently. You may start your topic if the subject of your discussion is not on one of those threads.


Zdravo svima....
Postavljam  ovaj topic, jer mislim da je predizvik za sve koji vole matematiku.
Najcesci dokaz o ovu teoremu je onaj poznati dokaz koristeci nize .
Pa zeljela bi da zajedno dokazemo Teorema Picarda (za postojanje i jedinstvenost resenije diferencijalnih jednacine), us pomoc Teoremi Banach o fiksnoj tacki.
Mozda je ovo neko vec dokazao???
Molim vas pisite....
Pozdrav....

Nadam se da je [url=http://www.mathlinks.ro/viewtopic.php?t=258706]odgovor profesora Merryfielda[/url] bio zadovoljavajući. Ako ne, pogledaj još i ovo

http://www.its.caltech.edu/~awrichar/HW/ACM100_Picard_ExistenceUniqueness.pdf

a možda se još neko pojavi sa svojim dokazom  :)

Dobrodošla na Mathlinks!


Can anyone point me to a good link explaining what is the infimum of a set or can anyone explain it to me ?

Thank you!  :)

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

mathworld is usually a good place to look for these things..

It's the greatest lower bound.

For example, $0$ is the infimum of the set $1,\frac{1}{2},\frac {1}{3},\frac {1}{4}$, etc.  This can be seen because if we pick any $x>0$, by the Archimedian property, there exists some $t$ such that $\frac {1}{t}<x$.  Since $x>0$ is not a possibility and $x=0$ works, we conclude that $0$ is the inf of the set.


Prove that:
$ (\frac{\pm\sqrt{\sqrt[3]{4}\plus{}3\sqrt{\frac{3}{\sqrt[3]{4}}}}\minus{}\sqrt{\sqrt[3]{4}\minus{}\sqrt{\frac{3}{\sqrt[3]{4}}}}}{2})^3\plus{}1\equal{}\pm\sqrt{\frac{3\sqrt{3}}{2}\plus{}1}$
Good luck!.  :lol:

[quote="Allnames"]Prove that:
$ (\frac {\pm\sqrt {\sqrt [3]{4} \plus{} 3\sqrt {\frac {3}{\sqrt [3]{4}}}} \minus{} \sqrt {\sqrt [3]{4} \minus{} \sqrt {\frac {3}{\sqrt [3]{4}}}}}{2})^3 \plus{} 1 \equal{} \pm\sqrt {\frac {3\sqrt {3}}{2} \plus{} 1}$
Good luck!.  :lol:[/quote]

In LHS, multiply upper and lower part of the fraction by $ \sqrt[6]2$. When entering inside the square roots, this becomes $ \sqrt[3]2$ and what we get is :

$ (\frac {\pm\sqrt {2 \plus{} 3\sqrt 3} \minus{} \sqrt {2 \minus{} \sqrt 3}}{2\sqrt[6]2})^3 \plus{} 1 \equal{} \pm\sqrt {\frac {3\sqrt {3}}{2} \plus{} 1}$ $ \iff$ $ (\pm\sqrt {2 \plus{} 3\sqrt 3} \minus{} \sqrt {2 \minus{} \sqrt 3})^3 \plus{} 8\sqrt 2 \equal{} \pm 8\sqrt {2\plus{}3\sqrt 3}$

Using the fact that this equality must be true for the two values of $ \pm$,  it would be enough to verify :

$ (i)$ : $ (\sqrt{2\plus{}3\sqrt 3})^3\plus{}3\sqrt{2\plus{}3\sqrt 3}(2\minus{}\sqrt 3)\equal{}8\sqrt{2\plus{}3\sqrt 3}$ $ \iff$ $ (2\plus{}3\sqrt 3)\plus{}3(2\minus{}\sqrt 3)\equal{}8$, which is obviously true

$ (ii)$ : $ \minus{}3(2\plus{}3\sqrt 3)\sqrt{2\minus{}\sqrt 3}\minus{}(\sqrt{2\minus{}\sqrt 3})^3\plus{}8\sqrt 2\equal{}0$ $ \iff$ $ \sqrt{2\minus{}\sqrt 3}(1\plus{}\sqrt 3)\equal{}\sqrt 2$

Squaring both sides (since positive), $ (ii)$ $ \iff$ $ (2\minus{}\sqrt3)(4\plus{}2\sqrt 3)\equal{}2$, which is obviously true.

Q.E.D.

But ..... what could be the interest of such a problem ?

[quote="pco"][quote="Allnames"]Prove that:
$ (\frac {\pm\sqrt {\sqrt [3]{4} \plus{} 3\sqrt {\frac {3}{\sqrt [3]{4}}}} \minus{} \sqrt {\sqrt [3]{4} \minus{} \sqrt {\frac {3}{\sqrt [3]{4}}}}}{2})^3 \plus{} 1 \equal{} \pm\sqrt {\frac {3\sqrt {3}}{2} \plus{} 1}$
Good luck!.  :lol:[/quote]

In LHS, multiply upper and lower part of the fraction by $ \sqrt [6]2$. When entering inside the square roots, this becomes $ \sqrt [3]2$ and what we get is :

$ (\frac {\pm\sqrt {2 \plus{} 3\sqrt 3} \minus{} \sqrt {2 \minus{} \sqrt 3}}{2\sqrt [6]2})^3 \plus{} 1 \equal{} \pm\sqrt {\frac {3\sqrt {3}}{2} \plus{} 1}$ $ \iff$ $ (\pm\sqrt {2 \plus{} 3\sqrt 3} \minus{} \sqrt {2 \minus{} \sqrt 3})^3 \plus{} 8\sqrt 2 \equal{} \pm 8\sqrt {2 \plus{} 3\sqrt 3}$

Using the fact that this equality must be true for the two values of $ \pm$,  it would be enough to verify :

$ (i)$ : $ (\sqrt {2 \plus{} 3\sqrt 3})^3 \plus{} 3\sqrt {2 \plus{} 3\sqrt 3}(2 \minus{} \sqrt 3) \equal{} 8\sqrt {2 \plus{} 3\sqrt 3}$ $ \iff$ $ (2 \plus{} 3\sqrt 3) \plus{} 3(2 \minus{} \sqrt 3) \equal{} 8$, which is obviously true

$ (ii)$ : $ \minus{} 3(2 \plus{} 3\sqrt 3)\sqrt {2 \minus{} \sqrt 3} \minus{} (\sqrt {2 \minus{} \sqrt 3})^3 \plus{} 8\sqrt 2 \equal{} 0$ $ \iff$ $ \sqrt {2 \minus{} \sqrt 3}(1 \plus{} \sqrt 3) \equal{} \sqrt 2$

Squaring both sides (since positive), $ (ii)$ $ \iff$ $ (2 \minus{} \sqrt3)(4 \plus{} 2\sqrt 3) \equal{} 2$, which is obviously true.

Q.E.D.

But ..... what could be the interest of such a problem ?[/quote]
Thank for your fast reply. As I wrote. It is time for relax. This problem is posted for funny purpose (but not spam  :lol: ). Did you surprise on it ?. And I think that is interested point!.  :)
I am waiting for another proof, my friends!

If you are interested in stange equalities, a classical way to build some is Cardan's solution for $ 3^{rd}$ degree equations :

Let $ x^3 \plus{} 3bx \minus{} 8a^3 \minus{} 6ab \equal{} 0$; with $ b\geq 0$.

This equation has a unique real root $ 2a$ and Cardan's formulas gives an unique real solution for this equation $ \sqrt [3]{4a^3 \plus{} 3ab \plus{} \sqrt {a^2(4a^2 \plus{} 3b)^2 \plus{} b^3}}$ +$ \sqrt [3]{4a^3 \plus{} 3ab \minus{} \sqrt {a^2(4a^2 \plus{} 3b)^2 \plus{} b^3}}$

So the identity : $ 2a \equal{} \sqrt [3]{4a^3 \plus{} 3ab \plus{} \sqrt {a^2(4a^2 \plus{} 3b)^2 \plus{} b^3}}$ +$ \sqrt [3]{4a^3 \plus{} 3ab \minus{} \sqrt {a^2(4a^2 \plus{} 3b)^2 \plus{} b^3}}$ $ \forall a,\forall b\geq 0$

With $ a \equal{} b \equal{} 1$, for example, you get $ 2 \equal{} \sqrt [3]{7 \plus{} 5\sqrt {2}}$ +$ \sqrt [3]{7 \minus{} 5\sqrt {2}}$

With $ a \equal{} 1$ and $ b \equal{} 2$, you get $ 2 \equal{} \sqrt [3]{10 \plus{} 6\sqrt {3}}$ +$ \sqrt [3]{10 \minus{} 6\sqrt {3}}$ which may also be written $ \sqrt [3]4 \equal{} \sqrt [3]{5 \plus{} 3\sqrt {3}}$ +$ \sqrt [3]{5 \minus{} 3\sqrt {3}}$

[quote="Allnames"]I am waiting for another proof, my friends![/quote]

If you want another solution, here is one :

It's easy to check that $ (x\minus{}a)^3\minus{}8x\plus{}8a(3\minus{}a^2)\equal{}(x^2\plus{}3a^2\minus{}8)(x\minus{}3a)$ $ \forall x,a$

And so $ (\pm\sqrt{8\minus{}3a^2}\minus{}a)^3\plus{}8a(3\minus{}a^2)\equal{}\pm 8\sqrt{8\minus{}3a^2}$ $ \forall a$ such that $ 3a^2\leq 8$

Setting $ a\equal{}\sqrt{2\minus{}\sqrt 3}$ is this identity gives the result $ (\pm\sqrt {2 \plus{} 3\sqrt 3} \minus{} \sqrt {2 \minus{} \sqrt 3})^3 \plus{} 8\sqrt 2 \equal{} \pm 8\sqrt {2 \plus{} 3\sqrt 3}$ which is equivalent to the required result (see my previous post).

Using other values for $ a$ would lead to other "identities" ...

Thank you; dear pco. I got that equality when I solved the equation $ \sqrt[3]{x\minus{}1}\plus{}\sqrt[3]{x\plus{}1}\equal{}\sqrt[3]{2}x$.
I thought that everyone would work hard to get that equation above. But what you did makes my thinking be silly.
Thank you!


[quote="bunhiacovski"][quote="tranquoc"]let a,b,c>0 such that a+b+c=1  prove that
$ \sum\sqrt {a \minus{} b \plus{} c^3} \leq 1$[/quote]

$ \sum\sqrt {a \minus{} b \plus{} c^3}\le \sqrt {2(a \minus{} b \plus{} b \minus{} c \plus{} c \minus{} a) \plus{} (\sum a\sqrt {a})^2} \equal{} \sqrt {\sum (a\sqrt {a})^2}$

we need prove that

$ \sum (a\sqrt {a})\le 1$

we have $ \sum (a\sqrt {a})\le \sqrt {(\sum a^2)(\sum a)} \equal{} \sqrt {\sum a^2}\le \sqrt {\sum a} \equal{} 1$

(because $ a\le 1,b\le 1,c\le 1$)

done~![/quote]
When does equal occur ?

[quote="Marius Mainea"][quote="tranquoc"]let a,b,c>0 such that a+b+c=1  prove that
$ \sum\sqrt {a \minus{} b \plus{} c^3} \leq 1$[/quote]

  Try    $ a\rightarrow0, c\rightarrow0, b\rightarrow1$[/quote]
Any condition,tran quoc,
This ineq is quite nice and i thank if we have a suitable condition,this ineq will be true (i cant find it now,sorry)


geometry
3D geometry
number theory proposed
number theory

Show that 1997^17 cannot be a sum of consecuative cubes.I think it's easy!!

[quote="a_vakilian"]I think it's easy!![/quote]
It seems that it is not so esay because no one solved it,but I know it is realy simple :?

Let me tell you a common fact among mathlinks to you...problem remaining unsolved is always because of two reasons:
-extremely extraordinary hard
-too easy

You problem definitely belongs to the latter type.

Anyway, the essential idea for the solution is just the formula $1^3+2^3+...n^3=(\frac{n(n+1)}{2})^2$.


AMC
USA(J)MO
USAMO
number theory proposed
number theory

Find the lesat positive integer $n(n>1)$ such that $\frac{1^{2}+2^{2}+...+n^{2}}{n}$ is a perfect square number.

I don't understand.Can you post full solution?

its a USAMO problem and i think the answer is $n=13$.  :)

ADD: oh Rust used the theory of pell equations. if you're not familiar, you can get it by trial. :wink:


Hello,

Let $ a \geq b \geq c>0$
Prove that : $ a^2 \times \frac{b}{c}\plus{}b^2 \times \frac{c}{a}\plus{}c^2 \times \frac{a}{b} \geq a^2\plus{}b^2\plus{}c^2$

mhdi

Hello,

I think there is a mistake here :[quote]$ (b \minus{} c)(\frac {a^2}{c} \minus{} \frac {b^2}{a})$[/quote]
If i expand i'll find this term : $ \frac {b^3}{a}$ which doesn't exist!

Mhdi

You'll find a term $ \minus{} \frac {b^3}{a}$ and term $ \frac {b^3}{a}$ ;)

$ \Leftrightarrow$ $ a^3b^2\plus{}b^3c^2\plus{}c^3a^2\geq a^3bc\plus{}ab^3c\plus{}abc^3\geq 0$
      $ \Leftrightarrow$
      $ a^3b(b\minus{}c)\plus{}b^2c^2(b\minus{}c)\plus{}z^3(a^2\plus{}b^2\minus{}2ab)\minus{}abc(b^2\minus{}c^2)\geq0$
      $ \Leftrightarrow$
      $ (b\minus{}c)(a\minus{}c)(a^2b\plus{}bc(b\minus{}c))\plus{}a^3(b\minus{}c)^2\geq 0$


find all function $ f: \mathbb{R}\to \mathbb{R^\plus{}}$ such that:
$ \sum \sqrt{f(x)^2\plus{}f(x)f(y)\plus{}f(y)^2}\sqrt{f(y)^2\plus{}f(y)f(z)\plus{}f(z)^2}\equal{}(\sum f(x))^2$
for all $ x,y,z \in \mathbb{R^\plus{}}$.(Symbol $ \mathbb{R^\plus{}}$ denotes the set of all positive real number.)

I guess it's just an inequality.
    L.H.S.≥R.H.S. and equality holds only when f(x) is a constant.


geometry
advanced fields
advanced fields unsolved

\documentclass{article}
\usepackage{latexsym}

\begin{document}

Let $\omega$ be a symplectic differentiable 2-form which is globally
defined on ${\rm R}^{4}$. Prove that there exist four 1-forms $\theta_{1},\;\theta_{2},\;\theta_{3},\;\theta_{4}$ defined on ${\rm R}^{4}$ so that the
following equality holds (globally) on ${\rm R}^{4}$: $\omega =\theta_{1}\wedge \theta_{2}+\theta_{3}\wedge \theta_{4}$.

If its true, where could I find the proof? If not, is there a
counterexample?

Thanks

\end{document}

If $\omega$ had constant coefficients, you would find $\theta_{i}$ using linear algebra. (Namely, the block-diagonal form of a skew-symmetric matrix.) Apply the same process to $\omega(x)$ pointwise; the resulting 1-forms $\theta_{i}(x)$ will be differentiable by virtue of nondegeneracy of $\omega$.


calculus
integration
function
geometry
inequalities
calculus computations

Let $ f(x)$ be a continuous function such that $ f(x) \equal{} \frac {f(x \plus{} y) \plus{} f(x \minus{} y)}{2}$ for all $ x,\ y\in\mathbb{R}$.

Prove that $ \int_a^b f(x)\ dx \equal{} \frac {b \minus{} a}{2}\{f(a) \plus{} f(b)\}\ (a < b)$.

It looks like the equation defines a straight line, the mid-value being the mid-point of the line segment joining $ f(x\minus{}y)$ to $ f(x\plus{}y)$ and so the integral is the area of a trapezium.

Yeah its Jensens functional equation with solution f(x)=ax+b

[quote="kunny"]Could you show your work in detail, guys? :)[/quote]

Hello kunny! Here is a simple way :

Let $ P(x,y)$ be the assertion $ f(x)\equal{}\frac{f(x\plus{}y)\plus{}f(x\minus{}y)}2$
Let $ g(x)\equal{}f(x)\minus{}f(0)$

Obviously, $ g(x)$ is too a solution and so matches $ P(x,y)$ and, since $ g(0)\equal{}0$ :

$ P(0,x)$ $ \implies$ $ g(\minus{}x)\equal{}\minus{}g(x)$
$ P(x,x)$ $ \implies$ $ g(2x)\equal{}2g(x)$

$ P(x,x\plus{}2y)$ $ \implies$ $ g(x)\equal{}\frac{g(2x\plus{}2y)\plus{}g(\minus{}2y)}2$ $ \equal{}\frac{2g(x\plus{}y)\minus{}2g(y)}2$ and so $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$

This is a Cauchy equation whose only continuous solutions are $ g(x)\equal{}ax$, which indeed are solutions of the original equation.

Hence the general solution $ f(x)\equal{}ax\plus{}b$

Hence the result.


Hanna runs faster than Jenny but slower than Lim. They started at
the same time from the same place and ran around a circular path. After a while they stopped simultaneously at the same place from which they started. It turned out that Lim overtook Jenny ten times during this run. How many times did it occur that one of the girls overtook another? Assume that each of the girls ran at a constant speed. (The final stop together is not counted as an overtake.)

okay, if lim over took jenny ten times then he over took hanna 5 times hanna overtook jenny 5 times too. jenny never over took any one because she's the slowest so m guess is that all together they over took 20 times????? :?:
is that right???? :thumbup:
or not???? I'm not sure because i was assuming 5 :blush:
it might have some thing to do with the circumference of the circle.and thats how u find the real answer, but i hope my answer is correct.

has nothing to do with dircumference or even the track shape

as long as it is  track

so lim overtook Jenny 10 times which means lim did 11 more laps than Jenny *think about it yourself, it is commonsense after you think about it*

hanna is somewhere between them, say she did x laps more than Jenny and 11-x laps less than lim
she overlapped jenny x-1 times and was lapped by lim 10-x times for a total of 9 times
so 10+9=19
im not sure about the number theaory but the point is as long as she is faster than Jenny and slower than lim (and ends on the finish line), it doesnt metter how much faster or slower.

Let's say that Jenny ran one lap. That means that Lim ran 12 laps. Let's say that Hanna ran 5 laps. Lim passed Jenny 10 times. Lim passed Hanna 6 times. Hanna passed Jenny 3 times. Therefore, a girl overtook another 19 times in all.


function
inequalities proposed
inequalities

[quote="kunny"]For $ n\geq 3$, Let $ x_k\ (k \equal{} 1,2,\ \cdots n)$ be real numbers such that $ \sum_{k \equal{} 1}^n x_k \equal{} 0,\ \sum_{k \equal{} 1}^n x_k^2 \equal{} 1$.
Find the maximum value of $ \sum_{k \equal{} 1}^n x_k^3$.[/quote]By Lagrange multipliers, :blush:  it is easy to find $ max \equal{} \frac {n \minus{} 2}{\sqrt {n(n \minus{} 1)}}$,

which is attained for $ x_1 \equal{} ... \equal{} x_{n \minus{} 1} \equal{} \minus{} \frac {1}{\sqrt {n(n \minus{} 1)}}$ and $ x_n \equal{} \sqrt {\frac {n \minus{} 1}{n}}$  :lol:

I wonder if there is a better method :maybe:

We can prove this problem using the following extension of EV-Theorem to real numbers.

[b]EV-Theorem Extension.[/b]
[i]Let $ p$ be an even positive integer, and let  $ f(u)$ be a differentiable function on [b]R[/b] such that $ g(x)\equal{}f'(x^\frac{1}{p\minus{}1})$ is strictly convex on [b]R[/b].
Let $ a$ and $ b$ be real numbers such that $ |a|<b$, and let $ x_1,x_2,...,x_n$ ($ n\ge 3$) be real numbers such that $ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}na$ and $ x_1^p\plus{}x_2^p\plus{}...\plus{}x_n^p\equal{}nb^p$.
Then, the sum $ f(x_1)\plus{}f(x_2)\plus{}...\plus{}f(x_n)$ is maximal for $ x_1\equal{}x_2\equal{}...\equal{}x_{n\minus{}1}\le x_n$, and is minimal for $ x_1\le x_2\equal{}x_3\equal{}...\equal{}x_n$.[/i]

[quote="Vasc"]We can prove this problem using the following extension of EV-Theorem to real numbers.

[b]EV-Theorem Extension.[/b]
[i]Let $ p$ be an even positive integer, and let  $ f(u)$ be a differentiable function on [b]R[/b] such that $ g(x) \equal{} f'(x^\frac {1}{p \minus{} 1})$ is strictly convex on [b]R[/b].
Let $ a$ and $ b$ be real numbers such that $ |a| < b$, and let $ x_1,x_2,...,x_n$ ($ n\ge 3$) be real numbers such that $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} na$ and $ x_1^p \plus{} x_2^p \plus{} ... \plus{} x_n^p \equal{} nb^p$.
Then, the sum $ f(x_1) \plus{} f(x_2) \plus{} ... \plus{} f(x_n)$ is maximal for $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{n \minus{} 1}\le x_n$, and is minimal for $ x_1\le x_2 \equal{} x_3 \equal{} ... \equal{} x_n$.[/i][/quote]
Hey VasC,can you share with us some document about your theorem -EV,and its proof,I saw your had used it many time ago without any reason,so I really want to know your theorem  ,please help me
Thank you very much :)

wait! I want you to solve the problem in a basic way, because the problem is to solve by high school students at the age of 16~17.

What do you think how difficult the problem is?

[quote="Allnames"] Hey VasC,can you share with us some document about your theorem -EV,and its proof,I saw your had used it many time ago without any reason,so I really want to know your theorem  ,please help me
Thank you very much :)[/quote]
See
http://www.emis.de/journals/JIPAM/article828.html?sid=828

Can anyone solve the problem in basic way without using EV theorem and Lagrange multipliers?


linear algebra
matrix
search
abstract algebra
algorithm

I know this isn't a right thing to do ;), but can please someone give me some hint as to how to crack this problem? :?

My current solution involves this:
1)Find Minimum distance from each square to another on the board for both, the knights and the king
2)Find the Minimum cost of rendezvous at all the points
3)For each square find the minimum cost using the info above.

Actually it's perfectly ok to ask.  I ask the coaches for help sometimes too.

But I can't help you with your problem.  Sorry  :(   I'm only on 3.2, and probably won't get any farther for a while.

[quote="h_s_potter2002"]Actually it's perfectly ok to ask.  I ask the coaches for help sometimes too.[/quote]

     Aah, so do they reply back? Anyways, I am a non-US student.

Well consider this problem:
   Given a R(1 <= R <= 60) x C(1 <= C <= 26) chessboard, how can I find the fastest way to move a knight from each square to every other square?

   My current method involves saving the reachable squares in 1 move for all squares, and then running BFS for each square. The problem requires that I have a Cost Matrix Like: CostKnight[MAXRow][MAXCol][MAXRow][MAXCol]

  Those of you have working progs for this Q, can you tell me whether this method is fast enough for the given range of R & C?

Well i solved this problem quite a long time ago. the actual IOI problem did not have such a vast searching space! if i can remember that was only 8*8. But in usaco it is very big enough!

Well my memory is quite bad to remember anything. if i am not wrong then i first made boards for each knights which says the shortest distance to a position. Then run a search code which selects a meeting point and then calculate the minimum possible steps to reach the goal!

if u need more help tell me! :P

My BFS module to find shortest distances from each square to another square
takes about 3sec for the largest test-case of(40Rows * 26Col) on my 550MHz
P3 machine....

How can I speed it up? Is it necessary to find distances from each square to
each square (or is there another solution)? Or is BFS the only way to find
these distances, or is there a mathematical closed formula(or any faster
way) to calculate this?

sorry i lost my code! anyhow here is the description given in the official soln. see if it helps:

This is a modification of the shortest path algorithm. If there was no king, then the shortest path algorithm can determine the distance that each knight must travel to get to each square. Thus, the cost of gathering in a particular square is simply the sum of the distance that each knight must travel, which is fairly simple to calculate.

In order to consider the king, consider a knight which 'picks-up' the king in some square and then travels to the gathering spot. This costs some number of extra moves than just traveling to the gathering spot. In particular, the king must move to the pick-up square, and the knight must travel to this square and then to the final gathering point. Consider the number of extra moves to be the `cost' for that knight to pick-up the king. It is simple to alter the shortest path algorithm to consider picking-up the king by augmenting the state with a boolean flag stating whether the knight has the king or not.

In this case, the cost for gathering at a particular location is the sum of the distance that each knight must travel to get to that square plus the minimum cost for a knight picking up the king on the way.

Thus, for each square, we keep two numbers, the sum of the distance that all the knights that we have seen thus far would have to travel to get to this square and the minimum cost for one of those knights picking up the king on the way (note that one way to 'pick-up' the king is to have the king travel all by itself to the gathering spot). Then, when we get a new knight, we run the shortest path algorithm and add the cost of getting that knight (without picking up the king) to each square to the cost of gathering at that location. Additionally, for each square, we check if the new knight can pick-up the king in fewer moves than any previous knight, and update that value if it can.

After all the knights have been processed, we determine the minimum over all squares of the cost to get to that square plus the additional cost for a knight to pick-up the king on its way to that square.

aah, the second part(calculating the costs) seems ok.....It's the first part: finding shortest distances thats troubling me. Which is the fastest algorithm?

Maybe Floyd Warshall or Dijkstra?

Floyd Warshall is probably "touching the limits", but should still be possible, if you optimize it...

But, you don't actually need to have "all pair shortest path"... My solution loops from the first knight to the last, and for every knight calculates the "shortest path from that knight to all squares" using BFS (i.e. changing from O(n^3) to (k * n ^ 2), n is the number of squares and k is number of knights)... Should be faster if the board is not completely filled with knights...  :lol:

Wrote my solution a year ago or so... Can hardly understand it anymore. :D

Weeblie, please try your code, coz I heard that USACO decreased time limit to 1sec/test.

Well, msn, have not u solved the problem?

Yeah! But after some illegal help though. :(

Some time that is needed! bcoz some test cases are so tough that solving those are the work of a great coder :D


analytic geometry
function
number theory
relatively prime

Prove that the set of all the points with both coordinates begin rational numbers can be written as a reunion of two disjoint sets $ A$ and $ B$ such that any line that that is parallel with $ Ox$, and respectively $ Oy$ intersects $ A$, and respectively $ B$ in a finite number of points.

Define the function $ H: \mathbb{Q}\to\mathbb{N}$ as $ H\left(\frac{m}{n}\right)\equal{}|m|\plus{}|n|$ where $ m$ and $ n$ are relatively prime integers. (By convention $ H(0)\equal{}H\left(\frac{0}{1}\right)\equal{}1$). We define $ A$ and $ B$ by

$ (x,y)$ is in $ A$ if and only if $ H(x)\le H(y)$.
$ (x,y)$ is in $ B$ if and only if $ H(x)>H(y)$.

Now I will show that this satisfies the conditions.

Lemma. For any number $ a$, there are finitely many values $ b$ such that $ H(b)\le a$.

Proof. First look only at positive $ b$. If $ b\equal{}\frac{m}{n}$ then $ m\plus{}n\le a$, so $ m\le a$ and $ n\le a$. Each of $ m$ and $ n$ will be of $ 0,1,2,...,a$ so there are $ a\plus{}1$ possible values for each, so there are at most $ (a\plus{}1)^2$ possible values of positive $ b$. The same applies for negative $ b$, so there are at most $ 2(a\plus{}1)^2$, a finite amount, of values $ b$ satisfying $ H(b)\le a$. This completes the lemma.

Now, for a given value of $ y$, there are finitely many values of $ x$ such that $ H(x)\le H(y)$, so on a given horizontal line, there are finitely many rational points in $ A$. Similarily, for a given value of $ x$, there are finitely many values of $ y$ such that $ H(x)>H(y)$, so on a given vertical line, there are finitely many rational points in $ B$, so this partition satisfies the conditions. QED.


function
real analysis
real analysis unsolved

How is it that the following definitions are equivalent?:

1. [i]A function $ f: \left(X,\mathcal{M}\right)\rightarrow\left(\mathbb{R},\mathcal{B}_{\mathbb{R}}\right)$ is measurable if for each $ E\in\mathcal{B}_{\mathbb{R}}$, $ f^{\minus{}1}\left(E\right)\in\mathcal{M}$,[/i]

2. [i]A function $ f: \left(X,\mathcal{M}\right)\rightarrow\left(\mathbb{R},\mathcal{B}_{\mathbb{R}}\right)$ is measurable if the set $ \left\{x\in X\ : \ f\left(x\right)>a\right\}$ is measurable for each $ a\in\mathbb{R}$.[/i]

$ \mathcal{B}_{\mathbb{R}}\equal{}\sigma (J_{oc} )\equal{} \sigma (J_{oo}) \equal{}\sigma (J_{co}) \equal{}\sigma (J_{cc})$

$ J_{oc} \equal{} \{ (a,b]: a,b \in \mathbb{R} \}$
$ J_{oo} \equal{} \{ (a,b): a,b \in \mathbb{R} \}$
$ J_{co} \equal{} \{ [a,b): a,b \in \mathbb{R} \}$
$ J_{cc} \equal{} \{ [a,b]: a,b \in \mathbb{R} \}$
as well as the open and closed rays, which generate the same $ \sigma$-algebra


Use the following information: Q is in the interior of <ROS. S is in the interior of <QOP. P is in the interior of <SOT. m<ROT=60 degrees, m<SOT= 100 degrees, and m< ROQ = m<QOS = m<POT.

I have to find the measures (in degrees) for m<QOP, m<QOT, m<ROQ, and m<SOP. How do I do it?[/hide][/b]

this isn't very clear to me...  :(
i'm sorry if i'm being stupid, or pushy.  :oops:

The problem cannot be solved unless it specifically states where point T is...Otherwise the angles can change.


real analysis
limit
analytic geometry
geometry
3D geometry
search
function

There is the following fact, which is more general then the usual LDT:

if $ A\subset \mathbb{R}^d$ is a set of positive Lebesgue measure ($ \mu(A)>0$), then for almost all points $ x\in A$ the following holds:
\[ \lim \mu(A\cap P)/\mu(P)\equal{}1,\] where $ P$ is a coordinate parallelepiped, containing $ x$, and diameter of $ P$ tends to 0.  Here coordiante parallelepiped is the parallelepiped with edges parallel to coordinate aces (i.e. $ P\equal{}\{(t_1,t_2\dots,t_n): a_i\leq t_i\leq b_i\}$).

The difference with the usual LDT is that we take not only regular guys like cubes or balls, but parallelepipeds with arbitrary different edges, the only requirement is that edges are small.

It is easy to prove, but I need a reference. Quite surprisingly, it is not well known and usually is absent in measure theory books and lecture curses.

Thanks!

Search Google.  I did one search and came up with a lot of sites.

Thanks! Really, of course I tried with google, but without success. It suggests a book "Geometry of Sets and Measures in Euclidean Spaces. Fractals and Rectifiability" by Pertti Mattila, but it is not available neither online nor in my Institute. So I do not know, whether it contains such modification.

Would you please give me a hint, what should I search for?

You're looking for the "Strong Maximal Function" and the theorem of Jessen, Marcinkiewicz, and Zygmund.

For a locally integrable function $ f,$ the strong maximal function is this:
\[ M_S(f)(x) \equal{} \sup_{P\ni x}\frac1{|P|}\int_P|f|\]
where $ P$ ranges over the parallelepipeds that you mention.

If the dimension $ n$ equals $ 1,$ then this is the Hardy-Littlewood maximal function. The Hardy-Littlewood maximal operator maps $ L^p$ to $ L^p$ for $ p > 1$ and $ L^1$ to weak $ L^1.$ Because of the latter, we have the Lebesgue theorem on the differentiation of the integral, which works for every (locally) $ L^1$ function.

The key to the Jessen-Marcinkiewicz-Zygmund result is to somehow make the strong maximal operator appear as if you composed $ n$ sequential one-dimensional Hardy-Littlewood operators. As a result, we say that $ M_S$ maps $ L^p$ to $ L^p$ for $ p > 1.$ As a result, if $ f\in L^p,\ p > 1,$ we have
\[ \lim_{P\text{ shrinks to }x}\frac1{|P|}\int_Pf \equal{} f(x)\text{ for almost every }x.\]
Since you're applying this to the characteristic function of $ A,$ and that characteristic function is in $ L^{\infty},$ hence locally $ L^p$ for any $ p,$ that's enough for your result.

But note that the strong maximal function does not send $ L^1$ to weak $ L^1,$ and as a result, strong differentiation (that is, this kind of differentiation, through rectangles), fails for some $ f\in L^1.$ What we can say (and this is in the original J-M-Z paper) that if $ f\in L\log^{n \minus{} 1}L,$ then $ M_S(f)$ is in weak $ L^1$ and the differentiation theorem applies.

This is all assuming that the edges of your paralellepiped are parallel to the axis. If you allow arbitrary orientations, then you're walking into the "Kakeya problem" and the results are far worse.

What exactly are you looking for?  Just books that contain references to it?  Here are two:

"Measure and Integral" by Wheeden & Zygmund.  They don't explicitly name it but it looks like they discuss it (see chapter 7).

"Real Analysis" by Carothers. It mentions it by name and proves it (see chapter 20).

Dear Kent, thank you for the valuable comment. But I rather need the reference to this particular result, not the proof of it (I may prove it myself :blush: )...

JRav, alas, I could not find it neither of these two books, only one-dimensional or "uniform" versions (with balls or cubes instead parallelepipeds).

Jessen, B., Marcinkiewicz, J. and Zygmund, A., Note on the differentiability of multiple integrals, [i]Fundamenta Mathematicae[/i] [b]25[/b] (1935), 217-34.

(I have the paper itself as reprinted in the collected works of Marcinkiewicz.)

You might also see what searching for the words "strong density" gets you.


AMC
AIME
USA(J)MO
USAMO
AMC 12
function
inequalities

Post your tenative scores and discuss problems here. I think I got 8 at this point... Not sure about number 12.

1)252
2)025
3)314
4)080
5)014
6)017
8)047
12)375

8. Because I interpreted #5 weirdly just for the fun of it, off-by-one'd #6, and didn't attempt #12, #14, or #15 even though they were painfully easy.

7 because I didn't think much on 7 since I thought it was hard, and 12 because it looked beastly. Messed up number 9 because of addition error, was being dumb on number 11 :( I could've had a 9 at least...if I wasn't being dumb at all then 11. Congratulations to Albert for miserably failing.

[quote="cognos599"]1. 252
2. 025
3. 314
4. 080
5. 014
8. 047
9. 190

for me.[/quote]
Yo at least like have the courtesy of saying "walks over to CA" if you're gonna help spam up our forum. Don't you Indiana people have your own forum to spam up? I see you doing a pretty good job with that so I don't see why we need any help here.

i think i got a 6 woot woot!
1-3, 5, 12
on 4 i thought the radius of the big circle was the height, not the slant height  :wallbash:
and on 7 i got $ \frac {570}{3} \equal{} 170$... :cursing:  :cursing:
but im happy! i probably wont make USAMO but im only in 8th grade so its cool  :D

hehe i totally copied this post from my one in the other forum  :P

here, in good spirit i edited my post and i am sorry for spamming the CA forum.  :)

Wow, you two, just shut up. What does saying "walking over to CA" accomplish?
Sure, posting in a CA forum when you're in IN is a bit weird, but whatever, it's cool. It's not like it's taking up space on your computer, and it hardly qualifies as "spamming".

come on guys
itsh all cool
9 by the way
gahhh missed 14
thought that $ (9 \plus{} x)^2 \minus{} 81 \equal{} 81 \plus{} 18x$
oh dude
Andrew, I think you mean "miserably failing"
yeah quotes help
holy bananas... daermon's in 8th
I thought you were in 9th dude

Assuming the bubbling was okay,  got a 7, and I doubt my error-ridden AMC will get me to USAMO.
The first five were okay.    THen I got the right strategy for #6 but miscounted the number of entries in the vertical column.  Then I missed #7 by one, because $ S_{999}$ obviously implied that there were 999 sets.  8 was very familiar, 9 was ok as well.  I died on 10, messed up my casework for 11, and didn't really look at 12-15.

Good job everyone!?
Got #1-7, 9, 12 correct
Got the correct answer for 8, bubbled it in, thought it was wrong and changed it... (yes, i'm a failure)  :(
Was just being stupid on 10, 14, 15...and gave up on 13 after looking at it once  :lol:
Hopefully I qualified for USAMO, keeping my fingers crossed  :roll:

Fine fine fine sorry. Dude David the "walks over to CA" thing was a joke...
What...archimedes you failed too? Dude this is really awesome for me, if I don't make USAMO, at least I have consolation that TWO beastly people didn't make USAMO (Brian Zhang has 126 AMC12, 7 AIME) :D But then its not as awesome as getting a good score and making USAMO so...

8 i think
maybe 9 or 10 if i have good luck
mistakes
23+24=57 on #8
misbubbled (maybe) on 12
gah were 10,11 hard?
9) 190
12) 375?
13) 012?????

ill hope for teh best

[quote="vishalarul"]As mentioned before, I got #1 wrong. :noo: But to make up for that, I got 2-10, 14. So I got a score of [b]10 (assuming no bubbling errors)[/b]. On #10, it was weird since I assumed that AB=BC=CD without even thinking about it.

This is surprising. I did better than most people here, though I really didn't expect it.

And... After the test, I got 1, 11, 13, and 15. Technically, if I sped up a bit, I could've pulled off a 14. But then again, there are many, many people who could've scored a 15 "if they just had some more time".

Did anyone think that #12 was wierd? When working on it, I got something like a constant over a linear function of $ n$, and I thought, "that can't be it. AIME never tests you on limits", but that's exactly what they did on the solution. Maybe there's a nice AM-GM way or some inequality I'm not familiar with.[/quote]
Hmmmm...didn't it say 4 feet for each 15mph OR FRACTION THEREOF? So its like, "stepped" and the last step is at the asymptote.
I could've gotten 11 but eh...I suck. 7, 12 literally took me 2 seconds each after the test...too bad it was after.
You're lucky you had calc and other bashing methods at your disposal for number 14...I attempted it but it quickly turned into stuff that was too advanced for my limited knowledge :( I really need to work on getting the simple solutions heh.

[quote="vishalarul"]As mentioned before, I got #1 wrong. :noo: But to make up for that, I got 2-10, 14. So I got a score of [b]10 (assuming no bubbling errors)[/b]. On #10, it was weird since I assumed that AB=BC=CD without even thinking about it.

This is surprising. I did better than most people here, though I really didn't expect it.

And... After the test, I got 1, 11, 13, and 15. Technically, if I sped up a bit, I could've pulled off a 14. But then again, there are many, many people who could've scored a 15 "if they just had some more time".

Did anyone think that #12 was wierd? When working on it, I got something like a constant over a linear function of $ n$, and I thought, "that can't be it. AIME never tests you on limits", but that's exactly what they did on the solution. Maybe there's a nice AM-GM way or some inequality I'm not familiar with.[/quote]

Yeah, I wasted so much time on #12. Anyways, the way you did it isn't entirely correct, although a lot of people did that - there are no such things as limits on this problem. Cars have a finite speed (the phrase "cars can go at any speed" doesn't mean they can go at infinite speed), so clearly limits have nothing to do with the problem. The correct solution is what serial pointed out, that you need to have a car start off next to the eye, and I missed that :( .

So I didn't make USAMO :( :( :( .

Also, how did you do #13?

[quote="davidyko"]Dude, Vishal, seriously, nice job.

Hmm, perhaps I'll go into more detail about my AIME experience.
So at the 2-hour mark, I'd done #1-5, 7-9. (having skipped #6 temporarily because I was being dumb) So, as my goal was 8 anyway and I had done 8 problems, I was like, "oh, I should check now."

[b]Which was a good thing, because I had somehow managed to mess up #1 and #3.[/b]

So that left about 45 minutes left. Got #6 in about another 10 minutes, and then went to the bathroom, as I was feeling quite accomplished (after all, I had only got 5 on last year's AIME).

30 minutes till the end of the test: Tried #10 without much luck because I couldn't figure out where point E was, and tried #12 half-heartedly, but made some random error and got a nonsensical answer. So, with 15 minutes left, I decided to check my answers again.

[b]Which I'm glad I did, because it turns out that I had made an off-by-one on #6.[/b]

Finally, the last five minutes were devoted to bubble-checking and guessing.  :P

So in short, I got 9.

P.S. Yeah, I was kinda pissed about other stuff when I made that other post above. I'm over it now, sorry about that.  :wink:[/quote]
Nice, I got everything I actually did seriously and I was completely sure of correct on the first try.
Vishal that is pretty beast, forgot to mention that.
Oh and regarding a timeline of events, this is about how it went:
Start: 8:14
Finished 1-6 : 8:45ish, feeling pretty good about myself, because I thought that 7 was the usual cutoff, so I'm like YESH, I have plenty of time, gonna make USAMO.
Finished 8: 8:50, feeling really great, I THINK I made USAMO already with my 7, even though a 7 in actuality is very dangerous, I go to bathroom.
In bathroom I start laughing, cuz I'm like, WAHOOO I MADE USAMO!!! ROFL THAT WAS EASY!!! and I keep that mentality for the rest of the test.
8:55ish I come back, and I'm like, okay I'm ready to solve a few more and get an awesome score!!!
[looks at 9, makes mistake, dies, but not aware]
[looks at 10, draws diagram, decides too complicated for my small brain, gives up]
[looks at 11, tries until about 10:00 with casework :O, thinks he's got the right answer, put its down, feeling really good]
[looks at 12, thinks its beastly, gives up :mad:]
[looks at 14, weeps at crap geo skill, moves on]
[looks at 15, weeps at crap geo skill, goes to 13]
[looks at 13, has no idea because of no experience on the type of problem, weeps, but still feeling good about qualifying for usamo (which was a false belief)]
9:55 Hmmm...I should check some more. And of course start making up BS for what I'm gonna put for all the ones I have no idea how to do. I decide that all my answers are correct  :roll: and move on to BSing 10, 12, 13, 14, 15. I decide that 159 looks like a cool number because it starts with a 1, and is an arithmetic progression! (the digits), put that for 15. Similar BS went to the other ones.
Heh interesting eh. I think the issue was that I got way too excited over my false USAMO qualifying idea, and so I got every single problem after that wrong.

So, never mind. Official answers are up, and I actually got 9!
1)252
2)025
3)314
4)080
5)014
6)017
8)047
10)032
12)375

I was working on 7, 14, and 15... but I ran out of time on all 3 and ended up guessing. Oh well.

Short: I got 1-7, 9, 12, off-by-1 on 10, guessed on 8, 11, 13, and 14, and thought 15 was more simple than it was.

Long: So first I did 1-3, skipped 4 because I thought it looked hard, did 5-7. Then I skipped 8, took far too long on 9, misread 10 (I thought 'diagonals' was 'legs' for some reason) so I didn't get anywhere.

Checked my answers a buncha times, which took about 40 minutes.

Then for the next hour or so I was just working randomly through the rest of the test. I realized 4 wasn't that hard, and re-read 10, and solved it... except I thought 25+7=31 :dry:
I was thinking wrong on 12 and thought they had to go at an infinite speed, but luckily I got 375 anyways and left it even though I thought it was strange. On 15 I thought there was a tetrahedron so I got 2rt17 (019) as my answer, except there is no tetrahedron so....

Checked my answers again, 30 minutes left. Somehow I didn't catch the 25+7.

I worked randomly on 11 and 13 but didn't really get anywhere. I thought about 14 for a long time and figured out the answer had to be somewhere in the 400's, so I just guessed 450. And then with no more time left I just filled 8, 11, and 13 with 042 :P

Let's just say I'm lucky all the questions are weighted equally

also for 15 i got the answer, but i couldn't put it into the form they wanted.

I actually did decently on this AIME. It was good. :)

Meh. And then I tried the problems I didn't get after the test and I got 7, 9, 11,14,15...all the ones I missed on the actual thing. I just ran out of time :P Sheesh, I spend like an hour on 1,3, and 4... The only one I couldn't get in a reasonable amount of time was 13... :(

I did 1-5, 12, and 15 for a total of 7.

I couldn't get 10 for quite a while, so I eventually just put down a random guess of 032. Right before the end of the 3 hours, I changed it to 023 for absolutely no reason. :|

If I don't make USAMO, I blame it on not trusting my instincts, even though that's quite stupid, considering that instincts won't tell you the correct answer out of 1000 possible choices.

[quote="archimedes1"]If the floor happens to be 7, then there'll definitely be a cutoff aplied to under-10.  So paired with a rock-bottom official 120 for AMC 10, (five mistakes?? 5!?!), I have no USAMO chance.
[quote="pythag011"]Got a 9. Did 1-9, 11-12. My #5 was wrong because I added 1 instead of subtraacting. This year's #15 was actually pretty easy.[/quote]
All you need to do is improve by 2 points a year and you'll have a freshman 15.
hahahahaha ..ouch[/quote]

Dude TWO OFF BY ONE ERRORS ON THE AIME.

Holy crap archimedes, you managed to fail worse than me at AMC10? I had 4 mistakes, but I decided to leave one blank because I caught it but had no idea how to get the right answer after.
Albert you shouldn't feel that bad. The Zuton Force made 8 computation errors but could've had perfect...now his USAMO chances are in jeopardy (and he's crazy beast)

[quote="archimedes1"][quote="serialk11r"]Holy crap archimedes, you managed to fail worse than me at AMC10? [/quote]
Yes.  I'm honored  :P

Wait, what happened to ubemaya?[/quote]
EDIT: By request of Ubeymaya :)

My birthday (24th, yesterday) was the Math Jam where I got to see how close I was to solving #15 (drew the correct lines, got the lengths, but forgot about one!!! AHH) & 14 :( that was pretty...pleasant.

[quote="Ubemaya"][quote="archimedes1"][quote="serialk11r"]Holy crap archimedes, you managed to fail worse than me at AMC10? [/quote]
Yes.  I'm honored  :P

Wait, what happened to ubemaya?[/quote]

Failed, two off by one errors :( .

Wait, who got the 12 in the poll?[/quote]
For all who don't know, that would be Albert...hence the smiley in my previous edited post.

I got an 8, but it's all good  :) , since I'm only a 9th grader, and I got 135 on the 10B.

I got #1-5 (which were way to easy), 7, 8, and 10. I skipped some easy problems, so that sucks. I also miscounted on #6. Oh well, I doubled my score from last year.

Well, I don't really expect to do very well on the USAMO, since I'm really bad at proof problems. I would consider myself lucky if I could even solve one of those problems. Besides, I'm very used to those types of competitions. The only sort of olympiad-style contest I've done was the BAMO last year, and I think I got like 7 or 8.

Well, my original goal on USAMO was "USAMO:0"

Meh. You never know. I might do so bad they might give me a negative score :P

My 12A score is 114 (fail) for a USAMO index of 204. Yay, its actually over 200! :)

[quote="serialk11r"]Hmmm I find it very wierd how tons of people have high AIME goals, and then they're like "USAMO: 1". Come on I didn't even qualify for AIME last year (111 phail :() and if I make USAMO my goal is like 12ish...at least aim higher even if you can't score that high.[/quote]

Who cares? my aime goal was low because I didn't expect much. My usamo score is around 7 because I don't expect I'd do well either. Maybe they'd jack up usamo like they did on the aime.

[quote="moogra"][quote="serialk11r"]Maybe they'd jack up usamo like they did on the aime.[/quote][/quote]

I hope so.

number 15 on the AIME was f------ easy. Unfortunately, I wasted a lot of time by attempting problems that I got wrong, including number 11, where I was off by two. Unfortunately, for most of the problems, I was swamped by calculations (which means I got it wrong).  I think I got a 4 or a 5, and I'm in 7th grade. Besides, I'm the only kid that took any AMC competitions.

agh... pythag is too pro... must beat him next year... lol

anyways i got a 9 in 7th grade! (darnit pythag i can see you laughing) :mad:

I'm getting a 0 cause I'll be in Yosemite while I was SUPPOSED to be taking AIME.

Epic phail.


Prove that there exists one and only one function $ f: \mathbb{Q}^ \plus{} \to \mathbb{Q}^ \plus{}$ which satisfies the following conditions:
i. For $ 0 < q < \frac {1}{2}$, \[ f(q) \equal{} 1 \plus{} f(\frac {q}{1 \minus{} 2q}).\]
ii. For $ 1 < q < 2$, \[ f(q) \equal{} 1 \plus{} f(q \minus{} 1).\]
iii. For every $ q\in\mathbb{Q}^ \plus{}$,
\[ f(q)f\left(\frac {1}{q}\right) \equal{} 1\].

[quote="Mashimaru"]Let $ \mathbb{Q}^ \plus{}$ denoted the set of positive rational numbers. Prove that there exists one and only one function $ f: \mathbb{Q}^ \plus{} \to \mathbb{Q}^ \plus{}$ satisfies the following condition:
i. For $ 0 < q < \frac {1}{2}$ then $ f(q) \equal{} 1 \plus{} f(\frac {q}{1 \minus{} 2q})$.
ii. For $ 1 < q < 2$ then $ f(q) \equal{} 1 \plus{} f(q \minus{} 1)$.
iii For every $ q\in\mathbb{Q}^ \plus{}$, $ f(q)f(\frac {1}{q}) \equal{} 1$.[/quote]

I think the result is wrong.

Let $ q\equal{}\frac ab$ with $ \gcd(a,b)\equal{}1$

If $ 0<q<\frac 12$, we get $ 2a<b$ and  $ f(\frac ab)\equal{}1\plus{}f(\frac a{b\minus{}2a})$
If $ \frac 12<q<1$, we get $ 1<\frac 1q<2$ and so $ f(\frac ab)\equal{}\frac 1{1\plus{}f(\frac {b\minus{}a}a)}$
If $ 1<q<2$, we get $ f(\frac ab)\equal{}1\plus{}f(\frac{a\minus{}b}b)$
If $ 2<q$, we get $ 0<\frac 1q<\frac 12$ and so $ f(\frac ab)\equal{}\frac 1{1\plus{}f(\frac b{a\minus{}2b})}$

So, for any $ q\equal{}\frac ab>0\notin\{\frac 12,1,2\}$, we can calculate in a unique manner from another rational $ \frac cd$ such that $ c\plus{}d<a\plus{}b$

And so, any $ f(\frac ab)$ is uniquely determined by the knowledge of $ f(1),f(\frac 12)$ and $ f(2)$

We obviously have $ f(1)\equal{}1$ but we can freely choose $ f(2)\equal{}u$ and $ f(\frac 12)\equal{}\frac 1u$ so that all rules are always fullfilled.

And, in my opinion, there are as many solutions as we have different values $ u\equal{}f(2)$


For math-liking person, you would want to consider

Phillips Exeter Academy (Exeter, New Hampshire)

http://www.exeter.edu

(there are some AoPSers there)

Phillips Academy (Andover, Massachusetts)

http://www.andover.edu

and probably

Choate Rosemary Hall

http://www.choate.edu

There are some other reasonable choices, but those are all schools young people interested in math apply to in large numbers. I know one young man here in my state who is applying to the first two schools I mentioned this year; he should know about the admission decisions next week.

I see a Chinese flag in your personal profile; are you a citizen of China or a citizen of the United States? Where will you be applying from?

I hope some of the current students of the schools you are interested in describe their experiences at boarding school. Maybe today some of them are busy taking the AIME.

St. Paul's School looks pretty nice

But i have to say Andover and Exeter are the best, no question or arguement about that

If you are really dedicated to math, go to Exeter. Otherwise ... it's all a wash. Well, not exactly, but the differences aren't as striking beyond high level Olympiad math.

I do have learnt some about the schools you mentioned.  I'm a citizen of China. So,  I could only apply for the boarding schools. There are two students in a near-by school went to Andover two and three years ago. Can you list some of the best boarding schools which receive foreign students?  With friendly atmosphere, great sports, clubs and courses?
Plus, I'm taking the AIME in China on March 22th.
I must say that I like math a lot, but I see liberal education just as much important. I enjoy writing and studying History. In this case, the schools you recommend has no need to be the top one in science and math education (good school are always good in both liberal and schience, aren't they?). Being one of the several bests is good enough.

Can you give me some more ideas? :roll:

Phillips Exeter, or Thomas Jefferson is pretty good too.  I got a scholarship offer from Exeter, but I turned it down.  I want to stay in my state and have a good fun 4 yrs of high school.  I don't know if that was the right decision now, however.

[quote="juicybooty911"]I got my letters today from Andover and Exeter, reject by both :([/quote]
Is it that hard to get into Andover and Exeter?

tokenadult, did A+ get accepted?

Yes, those are the best out of the best.

I applied for 11th grade, which is the most competitivie, there was only about 20 spots open, the 9th grade have several hundred. ;)

[quote="qvc122"]tokenadult, did A+ get accepted?[/quote]

A+MATH decided not to finish submitting his application. We had some required forms and recommendation letters in before we decided not to pursue boarding school for next year. We both think that Exeter and Andover have a lot to offer, but we have local resources here that are rare, and on balance he would rather stay here next year.

I have heard that one of my son's friends in this state (also an AoPSer) has been admitted to both schools. I'm not sure how the financial aid offers will work for his family--I'm not up to date on that news. It does look like Exeter will have an awesome math team, as usual.

Who else is just getting boarding school admission news? How many AoPSers will be at those schools next year?

[quote="tokenadult"][quote="qvc122"]tokenadult, did A+ get accepted?[/quote]

A+MATH decided not to finish submitting his application. We had some required forms and recommendation letters in before we decided not to pursue boarding school for next year. We both think that Exeter and Andover have a lot to offer, but we have local resources here that are rare, and on balance he would rather stay here next year.
[/quote]

Exactly what I was trying to describe above.

[quote="tokenadult"]I have heard that one of my son's friends in this state (also an AoPSer) has been admitted to both schools. I'm not sure how the financial aid offers will work for his family--I'm not up to date on that news. It does look like Exeter will have an awesome math team, as usual.

Who else is just getting boarding school admission news? How many AoPSers will be at those schools next year?[/quote]

Exeter just announced a that its financial aid initiative reached completion or something, so they're going to cover all financial aid I believe.  But I don't know the exact details...no doubt the Exeter website has all sorts of fanfare.

Yes.  Exeter has completed that stage.  Check their website [url=http://www.exeter.edu/news_events_859.aspx]here[/url].  Basically, about 40% of people received outright grants.  Check it out!

[quote="liumnina"]I'm gonna apply for a boarding school next year, in 9th grade. Do any of you guys have advice about which school to go to? I'd like to apply for four schools, enough then?
Can you help to list the top 5 boarding school in your mind?  kind of friendly, having lots of activities, good courses and so on..... I like Phillips Academy a lot.
Thank you beforehand![/quote]

i applied to Exeter/Andover/Choate for next year (11th grade) and got accepted by Ex/Ch (waiting list for Andover). Maybe i'll meet you (if i go)! lol i live in china right now -- us citizen though.


Let $a,b,c$ be positive real numbers with sum $3$. Prove that
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{3}{2a^{2}+bc}+\frac{3}{2b^{2}+ac}+\frac{3}{2c^{2}+ab}.\]

This inequality is weak and I do not like this form.

It is weak, I am agree, but I think it's nice.

To Thuan. If you don't like it, that's your own work. So it's not very necessary to say that.


The Gene for Blood Type A is dominant over the one for Type O blood. To have type O blood, a child must inherit type O genes from both parents. If the parents of a child both have one blood type A gene and one blood type O gene, what are the odds in favour of the child having type O blood?

Can anyone help

Since the parents are both AO, draw a 2x2 table to see all the possibilities.

They are
AA
AO
OO

There are two cases of getting AO, and the total number of cases is 4.

Thus the probability to get OO (one possibility out of four), is $ \frac{1}{4}$


function
vector
real analysis
real analysis solved

Let f be a function so that there exists f' and f'' on [a,b].if f+f''=0 for any x in [a,b] find f;
Obs:it is obviousely that f=a*sinx+b*cosx is a solution;
are there any more?

The set of solutions of the diffrential equation is a vector space with dimension 2.
The characteristic equation is x 2  + 1= 0, thus its roots are i and -i.
It follows that the set of solutions is the set of functions of the form f(x) = a*cos(x) + b*sin(x), where a,b are constants.

Pierre.

f+f''=0 is a standard ODE. we can either use the fact that x^2+1=0 has i and -i as roots, and therefore the general solution is that you mentioned.

Ordinal Differential Equation. Belive me. You don't want to know more. :D

[quote="Valentin Vornicu"]he has too, now they can be subjects at the NMO in 12th grade :D:D[/quote]

very incredible...
and what's the diff. eqn. you are learning at university now? ODE or PDE?


Can newton sums be generalized to calculate the sum of the powers of the roots higher than the degree of the polynomial?

Yes, I'm sure we can do that. So, if we have $f(x) = x^{2}-4x+3,$

$s_{1}-4 = 0 \Leftrightarrow s_{1}= 4$
$s_{2}-4s_{1}+3(2) = 0 \Leftrightarrow s_{2}= 10$
$s_{3}-4s_{2}+3s_{1}= 0 \Leftrightarrow s_{3}= 28 = 3^{3}+1^{3}$

That's not nearly a proof, but it's an example. I think other examples are shown in AoPS II.

When the powers of the roots are higher (or just as high) as the degree of the polynomial, you can reduce it to powers that are less than the degree.

Example: $a,b,c$ are roots of $t^{3}-t^{2}-2007$, find $a^{4}+b^{4}+c^{4}$.

We have $a^{4}=a^{3}+2007a$ and $a^{3}=a^{2}+2007$, so the sum in question is actually
\[(a^{2}+2007a+2007)+(b^{2}+2007b+2007)+(c^{2}+2007c+2007)\]
\[=(a^{2}+b^{2}+c^{2})+2007(a+b+c)+6021\]
Now it just reduces to computing $a+b+c$ and $a^{2}+b^{2}+c^{2}$. From $a+b+c=1$, $ab+bc+ca=0$, we get $a^{2}+b^{2}+c^{2}=1$, therefore $a^{4}+b^{4}+c^{4}=\boxed{8029}$.


Let $ a,b,c$ positive angles with $ a\plus{}b\plus{}c\equal{}90$   prove that

$ \frac{cosa^{2}cosb^{2}cosc^{2}}{sinasinbsinc} \geq \frac{27}{8}$


Let $ ABC$ be an equilateral triangle and let $ D$ be a point on the minor arc $ \widehat{BC}$ of the circumcircle. Lines $ BD$ and $ AC$ meet at point $ M$, and lines $ CD$ and $ AB$ meet at point $ N$. Show that $ (AD$ is a simedian in the triangle $ AMN$.

Let $ H,K$ be the projection of $ D$ onto $ AB, AC.$
$ AD$ is the symmedian of triangle $ AMN$ iff $ \frac{AN}{AM}\equal{}\frac{DH}{DK}$
So we will prove that $ \Delta ADN\sim \Delta MDA$
$ \angle CDM\equal{}60^o\equal{}\angle ADB, \angle DCM\equal{}\angle ABD$ thus $ \angle DMC\equal{}\angle BAD$
Similarly, $ \angle BND\equal{}\angle DAC$
Therefore $ (*)$ is true.
We end the proof.


Let $f(1)=1$ and for all natural numbers n, $f(1)+f(2)+...+f(n)=n^2f(n)$. What is $f(2006)$?

[hide="Solution"]All I can think of is induction.

The equation is equivalent to

$f(1)+\dots+f(n-1)=(n^2-1)f(n)\iff f(n)=\frac{1}{n^2-1}(f(1)+\dots+f(n-1))$

Calculating the first few terms, we get $f(2)={1\over 3}, f(3)={1\over 6}, f(4)={1\over 10}, f(5)={1\over 15}$. We note that the denominators are the triangular numbers, hence we assume $f(n)=\frac{2}{n(n+1)}$

For $n=1$ we have $f(1)=\frac{2}{1\cdot 2}=1$

Inductive step:

\begin{eqnarray*}f(n+1) &=& \frac{1}{(n+1)^2-1}\left(\frac{2}{1\cdot 2}+\frac{2}{2\cdot 3}+\dots+\frac{2}{n(n+1)}\right)\\ &=& \frac{2}{(n+1)^2-1}\left(1-{1\over 2}+{1\over 2}-{1\over 3}+\dots+\frac{1}{n}-\frac{1}{n+1}\right)\\ &=& \frac{2}{(n+1)^2-1}\left(1-\frac{1}{n+1}\right)\\ &=& \frac{2}{n(n+2)}\cdot\frac{n}{n+1}\\ &=& \frac{2}{(n+1)(n+2)}\end{eqnarray*}

Hence $f(2006)=\frac{1}{1003\cdot 2007}$[/hide]


Hello,

This might've been asked before, I don't know though.  For the USAMTS entry forms and permission forms, I require a parent or guardian to sign for me.  However, I can't get my parents to sign as they're a country away, and I don't really have a guardian here.  What should I do?

Thanks.

Send a blank one now with your stuff filled out but without your parent's signature, and send one with your parent's signature with your Round 2 solutions.


AMC
USA(J)MO
USAMO
AIME
inequalities
AIME I
\/closed

[quote="bubala"]If you expect to qualify for the USAMO this year, but don't expect to do well (you got a 136 on the AMC 12 and a 10 on the AIME I), are you ready for intermediate courses or olympiad courses?  I find from the Do You Know It? and Are You Ready? that I'm somewhat in the middle.[/quote]

Probably either.

Really it depends mostly on your weak and strong subjects.  Any subject in which you feel weak, the Intermediate subject courses would be better for you.  Any subject in which you feel strong (like you could answer easier Olympiad problems -- and maybe even a few harder ones), then the Olympiad class in that subject would benefit you more.

There are a couple of exceptions that I would currently mention.

The Olympiad Inequalities class is really 35% Intermedatie and 65% Olympiad.  We'll eventually have 2 levels (one day...), but if you're very strong in Algebra, but don't have experience with Olympiad level Inequalities, that's still a good class.

The Intermediate Complex Numbers/Trig class is harder than our other Intermediate classes and gets into some Olympiad level material.  That's another subject that might one day become two levels of classes.


calculus
derivative
Vieta
algebra
polynomial
function
analytic geometry

If a and b are real, prove that $ x^4\plus{}ax\plus{}b \equal{}0$ cannot have only real roots.

[b]Clarification of question:[/b]
- $ a$ and $ b$ are not both $ 0$
- Solution should not use differentiation

[b]An approach:[/b]
[hide]Equivalently, $ x^4 \equal{} Ax \plus{} B \; :  \; A,B \in \mathbb{R}$.

It suffices to show that no line in the Cartesian plane can intersect the curve $ y \equal{} x^4$ more than twice.
[/hide]
[b]Remark:[/b] A less obvious way to write the problem would have been to say, prove that there does not exist an ordered quadruple of non-zero real numbers $ (p,q,r,s)$ such that $ \sum_{sym} p \equal{} \sum_{sym} pq \equal{} 0$, or something along those lines.

Letting the roots be $ r_1,r_2,r_3,r_4$, we have using Vieta's,

$ r_1^2\plus{}r_2^2\plus{}r_3^2\plus{}r_4^2\equal{}(r_1\plus{}r_2\plus{}r_3\plus{}r_4)^2\minus{}2(r_1r_2\plus{}r_1r_3\plus{}r_1r_4\plus{}r_2r_3\plus{}r_2r_4\plus{}r_3r_4)\equal{}0\Rightarrow r_1\equal{}r_2\equal{}r_3\equal{}r_4\equal{}0$ or one of $ r_i\not\in\mathbb{R}$.

Obviously, we can't have $ r_1\equal{}r_2\equal{}r_3\equal{}r_4\equal{}0$ since then the polynomial becomes $ x^4\equal{}0$.

how do you show that no line in the Cartesian plane can intersect the curve  more than twice?

Yeah! because i was thinking if you have a function y=ax, a being any positive number - it should intersect f(x)=x^4 at 0 and at a^(1/3) - and the addition of any constant to y show give you 2 non-zero intersections

The curve $ y \equal{} x^4$ is [url=http://en.wikipedia.org/wiki/Convex_function]strictly convex[/url], so the portion of a curve over some x interval is below the secant line that spans that interval. So if your interval is the leftmost intersection to the rightmost intersection, we can't have any more intersections in between.


The walls of standard houses are constructed with two-by-four
studs placed 16 inches apart, center to center. How many studs are needed for a wall that is 48 feet wide?

$ 16$ inches is $ 1 \frac{1}{3}$ feet. $ 48$ feet divided by $ 1 \frac{1}{3}$ feet is 36. However, since the

studs are from center to center, only one of the ends should be calculated within 48. Thus, the answer is $ 36\minus{}1\equal{}\fbox{35}$, unless I misinterpreted this hard-to-understand problem.

The answer says $ \boxed{37}$ though..... I think that you divided 48 by $ 1\frac{1}{3}$, and then add on one for on of the end posts.

Think about it like this, 16 inches is 1 1/3 feet. the width of the wall is 48 feet wide. 48 ft divided by 1 1/3 feet is 36 ft. but this is just the spaces between the studs (for example in a line of studs, if there are 4 spaces between the studs there are 5 studs). So the answer is 37 studs.


It really doesn't matter, as long as you're doing problems. If you find you have weaknesses, work on those. Do some practice tests to get comfortable with the format.

[quote="batteredbutnotdefeated"]AMC 12 is beyond AoPS volume 1, but you should still read through it and master the basics. AoPS Volume 2 will help a lot, but you should also do lots of practice AMC12's and AIME's.[/quote]

Do a lot of problems. Most of the AMC 12 is covered by Intro Series, except some algebra/trig

[quote="geomeister33"]This year will be my first year taking the AMC 12. I own both the AOPS books but am not sure as to how I should spend my time studying throughout the year. At most, each day I will be able to spend 1.5 hours a day preparing/practicing for these competitions. I am a rising junior and would like advice as to what would probably be most beneficial for me to improve my math ability. My highest AMC 10 score was a 118.5 and my highest (practice) AIME score was a 5.[/quote]

I think you should study more of the Introduction series and Volume 1. Also do a lot of problems. When I finished the Introduction series, my AMC10 score raised from 115.5 to a 144!!

For AMC 12, most is covered by the introduction series, but the trig and other topics you can get from some of the Intermediate books and Precalculus possibly. The Intermediate series is how I prepare for AIME.:)

Are the intermediate courses better or the books in your opinion? If you've taken the courses that is.


geometry
incenter
trigonometry
geometry solved

The incircle of triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $P$ be any point inside the incircle of triangle $ABC$, and let $X$, $Y$, $Z$ be the points where the segments $PA$, $PB$, $PC$, respectively, meet the incircle. Prove that the lines $DX$, $EY$, $FZ$ are concurrent.

I think I managed to prove the converse, that is: if DX and the others are concurrent, then AX and the others are cincurrent, but I think the converse is enough. I'll post more tonight.

hey
i have only managed to prove it for point P being the incenter itself using pure trig.but it's getting difficult for the general case of P being any point inside the incircle.will post more later(if i make sum progress).ne1 else who has solved this one??
cheers

Here's how I think the converse of the problem can be proved (that is: if X, Y, Z are points on the incircle such that DX, EY, FZ are concurrent, then AX, BY, CZ are also concurrent).

First of all we use Ceva's theorem (trigonometric form) with sines on triangle DEF, and we get: (sinFDX/sinXDE)*(sinEFZ/sinZFD)*(sinDEY/sinYEF)=1 (*). But by applying sine law we get FX=2r*sinFDX, XE=2r*sinXDE etc., where r is the inradius. By replacing these in (*) we get FX/XE*EZ/ZD*DY/YF =1. So product of (AX/FX)/(AX/XE) and the other 2 analogous expressions is 1 (**). But AX/XF=sinAFX/sinFAX and because AF is tangent to the incircle AFX=FDX. Just use all these things in (**) and we get sinFAX/sinXAE*sinECZ/sinZCD*sinDBY/sinYBF=1, and fromthis we derive by the reciprocal of Ceva's thm with sines that AX, BY, CZ are concurrent at a point P. This, of course, is the reciprocal of the problem Alekk posted, but I think it can be used to prove the actual problem, or we could use a similar method.

hey, nice proof grobber.

I think we have to use the fact that (AD),(CF),(BE) are
concurrent to finish the problem. I saw a similar problem (but much
more easy) wich is:
**
let ABC be a triangle and D,E,F points on sides BC,CA,Ab such that
the cevians AD,BE,CF are concurrent . Show that if M,N,P are points
on sides EF,FD,DE then the lines Am,BN,CP concur iff the lines DM,EN,FP
concur.
**
bye.

hey, i used grobber method and i found a "solution"(i'm not sure):

LEt G=sin(XAE)/sin(FAX)*sin(DCZ)/sin(ECZ)*sin(YBF)/sin(DBY)
we want to prove G=1
by grobber method we have:
sin(FDX)/sin(XDG)*sin(FZ)/sin(ZFD)*sin(DEY)/sin(YEX)
  =FX/XE*EZ/ZD*DY/YF=...
  =[sin(AFX)/sin(AEX)*sin(CEZ)/sin(CDZ)*sin(BDY)/sin(BFY)]*G
    or AFX=FDX etc..
  =[sin(FDX)/sin(XDG)*sin(FZ)/sin(ZFD)*sin(DEY)/sin(YEX)]*G

then
  G=sin(XAE)/sin(FAX)*sin(DCZ)/sin(ECZ)*sin(YBF)/sin(DBY)=1
and we have what we want.  (is it right? I'm not sure)

regards.

sorry it was false.
here what i found (it's ugly then i'm not sure)

LEt G=sin(XAE)/sin(FAX)*sin(DCZ)/sin(ECZ)*sin(YBF)/sin(DBY)
    H=sin(FDX)/sin(XDG)*sin(FZ)/sin(ZFD)*sin(DEY)/sin(YEX)
we want to prove H=1
by grobber method we have:
H = sin(FDX)/sin(XDE)*sin(EFZ)/sin(ZFD)*sin(DEY)/sin(YEX)
  =FX/XE*EZ/ZD*DY/YF
  =(FX/AX)/(XE/AX)*(EZ/CZ)/(ZD/CZ)*(DY/BY)/(YF/BY)
  =..
  =[sin(FAX)/sin(XAE)*sin(ECZ)/sin(DCZ)*sin(DBY)/sin(BFY)]/H

1/G=H^2  or G=1

then
  H=1
and we have what we want.(is it still false?)

[quote="amfulger"]I think the problem can be extended from the triangle ABC to a hexagon AECDBF circumscribed to circle W. A point P inside W. PA,PB,PC intersect W at X,Y,Z. Prove DX,... are concurent.
I didn't prove this problem nor the first but I have no reason to belive otherwise.[/quote]

Unfortunately, my dynamic geometry sketch proves this theorem wrong. However, for the original (triangle) theorem, see also Theorem 2 of my note "Variations of the Steinbart Theorem" at http://www.cip.ifi.lmu.de/~grinberg/ .

  Darij Grinberg

Dear Mathlinkers,
for an overlook of this situation, you can see

http://perso.orange.fr/jl.ayme   vol. 3  Les points de Steinbart et de Rabinowitz

Sincerely
Jean-Louis


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

Let ABCD be a cyclic quadrilateral with AB=BC. If P is a point on CD such that AP=PB=BA, prove that AD=R, the circumradius of $ \triangle ABC$.


limit
real analysis
real analysis unsolved

Let $f,g: \Re\to\Re$ such that $g$ is strictly monotonic. If $\lim_{t\to x} f(t)=g(f(x))$ then $f$ is continous.

The main problem was:
   [quote="gm"]Let $f: \Re\to\Re$ a non-constant function such that there exists $g$ continous and strictly monotonic such that $\lim_{t\to x} f(t)=g(f(x)),\forall x \in \Re$. Prove that there exists a non-degenerated interval formed by fix points of $g$.[/quote]

 Altough it may seem easier this is the original text the first post being given as a variation :roll:

I don't think it's easy...been working on it for a couple of days and NOTHING!!! there must be something i'm missing! :wallbash:

Here's some ideeas I have reached to. Maybe they will help you.

1)I proved that $\lim_{t\to x}f(t)$ is continuous. Here http://www.mathlinks.ro/Forum/viewtopic.php?p=405226#p405226 i made a proof for this fact. Hence $gof$ is continuous.

2) Assume that $g(a)> a+\epsilon$ and prove that there exist $b>a$ such that $g(x) \ge x+\epsilon \forall x\in [a,b]$. Indeed, $\forall x\in [a,g(a)-\epsilon]$ we have that $x>a \Rightarrow g(x) > g(a) \ge x-\epsilon$. Similarly, If we have $g(a)<a-\epsilon$, there will exist an interval $[b,a]$ on which $g(x)<x-\epsilon$.

3) If there exist $a\in int(Imf)$ (i.e. an interior point of the image of $f$) and $g(a)\neq a$, from 2) we can find an interval $[a,b]\subset Imf$ on which $f(x) >\lim_{t \to x}f(t)+\epsilon$(I took the ">" sign without loss of generality). Hence, on this interval, for every $x$ and $\delta$ there exists an $y$ such that $|x-y|\le \delta$ and $f(x)\ge f(y)+\epsilon$.

4) For any $n$ we can define $x_1,x_2,..x_n$ by the method described in 3) such that $x_{i+1}-x_i \le \frac{max (|x_1-a|,|b-x_1|)}{n^2}$. This sequence is contained in interval $[a,b]$ hence it must have a limit point, name it $d$, and $f(x_k)$ is unbounded. Hence $g(f(d))=\lim_{x\to d}f(x)=\lim_{n\to infty} {x_{k_n}}=\infty$. Contradiction.
Hence for any interior point of $Imf$ we have that $g(x)=x$.

5) If $f(x_0)=a_0$ is boundary point of $Imf$ then there exist a sequence $x_k$ such that $f(x_k)\to a_0$ and $f(x_k)> (<)a_0$ (the same sign holds for all $x_k$. I'll take it ">" WLOG), or $a_0$ is a isolated point.

Maybe the proof can be made even for these points (probably not so hard), but I'll give it a try another day.

I think it is something similar to 3). Suppose there is some $x \in {\mathbb R}$ such that $f(x) \neq g(f(x))$. WLOG $f(x)>g(f(x))$ and $g$ is strictly increasing. Then there is $\epsilon>0$ and $a>0$ such that $f(x)>f(t)+a$ for all $t \in (x-\epsilon,x+\epsilon)$, $t \neq x$. We have \[ f(x)>f(t)+a \Rightarrow f(x)-a>f(t) \Rightarrow g(f(t))<g(f(x)-a) \Rightarrow g(f(x)) > g(f(t))+\left(g(f(x))-g(f(x)-a)\right), \forall t \in (x-\epsilon,x+\epsilon),t \neq x,  \] with $g(f(x))-g(f(x)-a)>0$, which contradicts the fact that $\displaystyle g \circ f = \lim_{t \rightarrow x} {f(t)}$ is continuous in $x$, so $\displaystyle f(x) = g(f(x)) = \lim_{t \rightarrow x} {f(t)}$ for all $x \in {\mathbb R}$, hence $f$ is continuous.

I see that the same method works even for boundary points which are not isolate. (because the interval [a,b] mntioned in 3),4) lies inside $Imf$).

Assume that $Imf$ have an isolated point $a$. Hence, $f$ has a jump discontinuity in $f^{-1}(a)$ which implies that $gof$ has a jump discontinuity in $f^{-1}(a)$, Contradiction with the continuity of $gof$. So, the solution is finished. Pftiu!!

Anyway, thank to dzeta for the beautifull solution.

rats...it was rather accessible .... anyway thanx dzeta and xirti :)


geometry
parallelogram
symmetry
geometry proposed

Dear All My Friends,
Cutting one convex quadrilateral by two segments each connected midpoints of quadrilateral opposite sides we can divide the quadrilateral into four small convex quadrilaterals $ Q_1$, $ Q_2$, $ Q_3$, $ Q_4$. Suppose $ I_1$, $ I_2$, $ I_3$, $ I_4$ are intersections of diagonals of $ Q_1$, $ Q_2$, $ Q_3$, $ Q_4$ respectively and $ S_{1}$ is area of quadrilateral $ I_1I_2I_3I_4$

By rearranging four small convex quadrilaterals $ Q_1, Q_2, Q_3, Q_4$ (only moving, not upturn) we can bound one parallelogram (without gap, not overlapping). In this case their intersections of diagonals $ I_1$, $ I_2$, $ I_3$, $ I_4$ bound another convex quadrilateral with area $ S_{2}$

Please prove that $ S_{1}\equal{}S_{2}$
Thank you and best regards,
Bui Quang Tuan

[quote="Quang Tuan Bui"]
By rearranging four small convex quadrilaterals $ Q_1, Q_2, Q_3, Q_4$ (only moving, not upturn) we can bound one parallelogram (without gap, not overlapping). In this case their intersections of diagonals $ I_1$, $ I_2$, $ I_3$, $ I_4$ bound another convex quadrilateral with area $ S_{2}$
[/quote]

I don't understand this text but if I understand truely (as my figure) then two quarilateral are equal, here we use point symmetry only.

[quote="encyclopedia"]
I don't understand this text but if I understand truely (as my figure) then two quarilateral are equal, here we use point symmetry only.[/quote]
Dear encyclopedia,
As your figure you understand me truely! In any case two quarilaterals are not equal: they have equal area only!
Best regards,
Bui Quang Tuan

oh you are right, thank you. I didn't see teh figure carefully, nice problem, I will try  :) .


analytic geometry
geometry unsolved
geometry

[quote="Lovasz"]Let $ABCD$ be a tetrahedron. Find the point $M$ in space satisfying the expession $P=MA^2+MB^2+MC^2+MD^2$ gets the least value.[/quote]

This point is the centre of gravity of the tetrahedron.

Let's concider the coordinates:
$P=\sum (x_{A}-x_{M})^2 +\sum (y_{A}-y_{M})^2 +\sum (z_{A}-z_{M})^2 =$
$4*( (x_{M}-\frac{x_{A}+x_{B}+x_{C}+x_{D}}{4})^2 + (y_{M}-\frac{y_{A}+y_{B}+y_{C}+y_{D}}{4})^2+$
$(z_{M}-\frac{z_{A}+z_{B}+z_{C}+z_{D}}{4})^2 + t)$
where $t$ is some constant.
So P is the least when equals to $4t \Rightarrow$
$x_{M}=\frac{x_{A}+x_{B}+x_{C}+x_{D}}{4}$
$y_{M}=\frac{y_{A}+y_{B}+y_{C}+y_{D}}{4}$
$z_{M}=\frac{z_{A}+z_{B}+z_{C}+z_{D}}{4}$

So we can see that this point is the centre of gravity of ABCD.
V:)

Yes :)
But there's not any proof without using coordinate?!?!?!?


find all functions wich satisfy $|f(x+y)-f(x)-f(y)| \leq 1$.
for all reel nubers $x$ and $y$.

If $g: \mathbb{R}\to\mathbb{R}$ is arbitrary solution of Cauchy equation $g(x+y)=g(x)+g(y)$ and $h: \mathbb{R}\to [-{1\over 3},{1\over 3}]$ arbitrary,
then $f(x)=g(x)+h(x)$ for $x\in\mathbb{R}$ defines a solution of above problem.


trigonometry function algebra domain Gauss integration search

why can't a judge be disturbed at dinner?

A mathematician, an engineer, and a chemist were walking down the road when they saw a pile of cans of beer. Unfortunately, they were the old-fashioned cans that do not have the tab at the top. One of them proposed that they split up and find can openers. The chemist went to his lab and concocted a magical chemical that dissolves the can top in an instant and evaporates the next instant so that the beer inside is not affected. The engineer went to his workshop and created a new HyperOpener that can open 25 cans per second.

They went back to the pile with their inventions and found the mathematician finishing the last can of beer. "How did you manage that?" they asked in astonishment. The mathematician answered, "Oh, well, I assumed they were open and went from there."

Whenever I have a trig test, I always look for a $ sin$ from above.

[quote="7h3.D3m0n.117"]This might be better off on the Puzzles and Brainteasers  :wink:. Moved.[/quote]

Actually, shouldn't this go in the G&FF?
They should just make you a moderator so that you don't have to switch accounts all the time.

Well I assumed that we'd be swapping answers and jokes/brainteasers so I thought this might be better suited.

[quote="Temperal"]They should just make you a moderator so that you don't have to switch accounts all the time.[/quote]

 :?:

What is the volume of a pizza with z radius and a height?

Here is a similar one:

Two functions are walking in a domain, suddenly another robber function comes and says:
-I am the fourier transformer, so throw me your ranges or I will fourier transform you two.
One of the functions starts laughing and says:
-I am the Gauss Curve hahaha.

 :rotfl:  :rotfl:

[quote="Thaakisfox"]I think this is very well known, but still I will write here since no one has yet :D  :D

Two functions are walking in a domain, suddenly a robber function comes, and says:
-I am the differentiator. you two there, throw me your ranges or else I will differentiate both of you.
One of the functions starts laughing really badly, so the robber function asks:
-Hey why are you laughing, arent you scared?
So the function says:
- hel no I aint scared, I am $ e^x$.

 :rotfl:  :rotfl:[/quote]

What's wrong with being reduced to a constant?

The differentiator says, "I'm $ \frac{d}{dy}$ !"  :rotfl:

What is
$ \displaystyle\int\frac {1}{cabin}\,d(cabin)$?

[hide="click here after you found the answer"]
If you forgot the $ \plus{} C$, shame on you![/hide]

[quote="Thaakisfox"]I think this is very well known, but still I will write here since no one has yet :D  :D

Two functions are walking in a domain, suddenly a robber function comes, and says:
-I am the differentiator. you two there, throw me your ranges or else I will differentiate both of you.
One of the functions starts laughing really badly, so the robber function asks:
-Hey why are you laughing, arent you scared?
So the function says:
- hel no I aint scared, I am $ e^x$.

 :rotfl:  :rotfl:[/quote]

Slight variation
Robber: Who says I differentiate on X

Well, we wouldn't have a problem if the function was

\[ e^{(\prod\limits_{cyc}a_i})\]

which covers pretty much every variable possibly dnoted by one letter and a subscript.

[quote="Temperal"]Well, we wouldn't have a problem if the function was
\[ e^{(\prod\limits_{cyc}a_i})
\]
which covers pretty much every variable possibly dnoted by one letter and a subscript.[/quote]
Well that doesn't differentiate to zero, no... but it doesn't differentiate to itself either, which is still a problem.

(unless that product symbol was supposed to be summation ;))

---

Robber: I'll differentiate you until you're zero!

Function: Too bad, I'm f(x)=1 if x is rational, 0 otherwise

[quote="Hamster1800"][quote="lingomaniac88"]If and only if $ |\text{for Dummies}| < 1$.  Otherwise it doesn't converge.[/quote]

What?

$ 1 \plus{} 2 \plus{} 4 \plus{} 8 \plus{} 16 \plus{} \cdots \equal{} \minus{} 1$. How can you say that doesn't converge?[/quote]

 :rotfl:
but can be demostrated:

let C=1+2+4+...

C=1+2+4+8+...=1+(2+4+8+...)=1+2(1+2+4+...)=1+2C
C=1+2C
0=1+C
-1=C
QED :D

Why don't we transform it in the "rightful demostration of wrong things"?

Why do you say that $ 1\plus{}2\plus{}4\plus{}8\plus{}\cdots\equal{}\minus{}1$ is a "wrong thing"? It is very much a right thing.

Now let's compute $ \sum_{n\equal{}0}^{\infty} n2^n$. Well, we simply divide by $ 2$ and then integrate with respect to $ 2$ to get the following equalities:

$ \sum_{n\equal{}0}^{\infty} n2^n \equal{} 2*\frac{d}{d2}\left(\sum_{n\equal{}0}^{\infty} 2^n\right) \equal{} 2*\frac{d}{d2}\frac{1}{1\minus{}2} \equal{} \frac{2}{(1\minus{}2)^2} \equal{} 2$

[quote="Hamster1800"]$ 1 \plus{} 2 \plus{} 4 \plus{} 8 \plus{} 16 \plus{} \cdots \equal{} \minus{} 1$. How can you say that doesn't converge?[/quote]
Please tell me how a bunch of positive numbers, when added together, can result in a negative number without bending the rules of mathematics for the purpose of humor.

Then again, it is well-known that $ 2\equal{}1$.

It's the same way a bunch of rational numbers can be added together to get an irrational number.

[quote="Lawrence Wu"]Compute: $ \dfrac{1}{0}\times{0}$.
Any number multiplied by 0 is 0. Therefore it is 0.
The 0 cancels the denominator of 1/0. Therefore it is also 1.
0=1[/quote]

also, you can never divide by 0. Thus, it is also undefined

0=1=undefined

x^2=x+x+x+x..... (x times). Taking the derivative of both sides gives 2x=1+1+1...(x times)=x, so 1=2.

easier

let A=B=1

$ A\equal{}B$
$ A^2\equal{}AB$
$ A^2\minus{}B^2\equal{}AB\minus{}B^2$
$ (A\minus{}B)(A\plus{}B)\equal{}B(A\minus{}B)$
$ A\plus{}B\equal{}B$

so replacing A=B=1,

2=1

wowwww... (a-b)=(1-1)=0...

no dividing by zero. i've seen that bogus proof like fifty times.
so has everyone else.. xD

$ \sin{\pi}\equal{}\sin{2\pi}$
lol, yeah that doesn't work but thats as close as I can get.

[quote="Temperal"]here's one of the cooler ones:

consider a wheel with circumference k, and a wheel with circumference k/2 that's fixed to the first wheel's center so that it turns with it. when the first wheel makes a complete revolution, the second wheel does the same, so the circumferences are equal.
k=2k, so 2=1

there's another one involving trig and integrals, but i forget it.[/quote]

Here are the calculus ones: http://en.wikipedia.org/wiki/Invalid_proof#Proof_that_2_.3D_1_2

[hide="All numbers are equal"]To prove any two real (or complex, I haven't tried) numbers equal...
Let the two numbers be $ a$ and $ b$.  We have...
\begin{align*}a^2-ab+b^2&=a^2-ab+b^2\\
a^2-a^2-ab&=b^2-b^2-ab\\
a^2-2a\left(\frac{a+b}{2}\right)&=b^2-2b\left(\frac{a+b}{2}\right)\\
a^2-2a\left(\frac{a+b}{2}\right)+\left(\frac{a+b}{2}\right)^2&=b^2-2b\left(\frac{a+b}{2}\right)+\left(\frac{a+b}{2}\right)^2\\
\left(a-\frac{a+b}{2}\right)^2&=\left(b-\frac{a+b}{2}\right)^2\\
a-\frac{a+b}{2}&=b-\frac{a+b}{2}\\
a&=b
\end{align*}[/hide]

the trouble with the proof is  that in  the third to last line the term s inside the square are additive inverses to each other: when  square rooting there has to be a plus-minus sign(in this case minus sign)

http://www.kent.k12.wa.us/staff/DavidWright/calculus/book/55/jokes.html

http://www.workjoke.com/projoke22.htm

Some of these are good.


Find the value of $ c$ such that $ \sum_{n \equal{} 1}^{11} (n \minus{} c)(2n \minus{} c)$ is minimal and find the minimum value $ m$.

note that $ \sum_{n\equal{}1}^x n\equal{}x(x\plus{}1)/2$ and $ \sum_{n\equal{}1}^x n^2 \equal{} x(x\plus{}1)(2x\plus{}1)/6$ (both well known facts which are easily proven by induction)

$ \sum_{n\equal{}1}^{11} (n\minus{}c)(2n\minus{}c) \equal{} \sum_{n\equal{}1}^{11} 2n^2\minus{}3nc\plus{}c^2 \equal{} 2*11*12*23/6 \minus{} 3*11*12/2*c \plus{} 11c^2$
$ \equal{} 1012 \minus{}198c\plus{}11c^2$.
$ \equal{} 11(92\minus{}18c\plus{}c^2)$
$ \equal{} 11((c\minus{}9)^2 \plus{}11)$

Thus the mininum occurs at $ c\equal{}9$ and is $ m\equal{}121$.

That's correct answer.
Let me say one thing, $ \sum_{n \equal{} 1}^{11} 2n^2 \minus{} 3nc \plus{} c^2$ should be written as $ \sum_{n \equal{} 1}^{11} (2n^2 \minus{} 3nc \plus{} c^2)$.


[quote="nameisong"]hmmmm, I got marked down for not showing the diagram in the first round of the USAMTS. could someone please show me how to position latex diagrams properly. so the text fits nicely around them? thanks.  :D basically, all i've done is copy and paste from the original latex file of the problem set, obviously, the diagram doesn't go where i want it to.[/quote]

None of the Round 1 solutions required a diagram. For example, in Problem 1/1/16, a solution could have said that the numbers 1, 10, 3, 6, 5, 7, 2, 8, 4, 9 wrap around the pentagon in that order with the 1, 3, 5, 2, and 4 at the vertices, without drawing the pentagon. However, many confusing descriptions can be made clearer by adding a diagram. Perhaps that is what your USAMTS grader meant. An alternative to mastering diagrams would be to instead write a clear, well-organized solution.

When I typeset last year's USAMTS (Year 15) in LaTeX, I put the diagrams to the right of the paragraphs mainly because I could not afford to waste room on the problem sets. A problem set is easier to look over when it fits onto a single page. Richard Rusczyk (or someone else at AoPS) did likewise for this year's USAMTS. However, when you submit a solution, you may take several pages. So feel free to put your diagrams between paragraphs and not worry about making the text flow around a diagram.

Let's get back to your LaTeX question. The LaTeX code for the pentagon in Problem 1/1/16 is:
[quote]\USprob{1/1/16.}{
\rightskip 2in
The numbers 1 through 10 can be arranged along the vertices and sides of a pentagon so that the sum of the three numbers along each side is the same. The diagram on the right shows an arrangement with sum~16. Find, with proof, the smallest possible value for a sum and give an example of an arrangement with that sum.
\vskip-1.2in\hskip4.5in\raisebox{-70pt}[0pt]{\setlength{\unitlength}{1pt}
\begin{picture}(90,80)
\put(15,46){\line(1,-3){12}}
\put(27,10){\line(1,0){38}}
\put(65,10){\line(1,3){12}}
\put(15,46){\line(4,3){31}}
\put(77,46){\line(-4,3){31}}
\put(20,0){4}
\put(43,0){9}
\put(65,0){3}
\put(73,22){6}
\put(80,44){7}
\put(63,59){8}
\put(43,71){1}
\put(22,59){5}
\put(1,44){10}
\put(12,22){2}
\end{picture}}}[/quote]

The portion that begins with \begin{picture} and ends with \end{picture} is the diagram itself. LaTeX treats it as a solid rectangle (a "box" in TeX nomenclature) measuring 90 units wide and 80 units high, due to the (90,80) after the \begin{picture}. The \setlength command before the diagram specifies that 1 unit equals 1 point (0.013837 inch), so the box is 90 points wide and 80 points high. That is all you need to put a copy of the diagram into your LaTeX solution:
[quote]\setlength{\unitlength}{1pt}
\begin{picture}(90,80)
\put(15,46){\line(1,-3){12}}
\put(27,10){\line(1,0){38}}
\put(65,10){\line(1,3){12}}
\put(15,46){\line(4,3){31}}
\put(77,46){\line(-4,3){31}}
\put(20,0){4}
\put(43,0){9}
\put(65,0){3}
\put(73,22){6}
\put(80,44){7}
\put(63,59){8}
\put(43,71){1}
\put(22,59){5}
\put(1,44){10}
\put(12,22){2}
\end{picture}[/quote]

All the other LaTeX code is to make the text and diagram fit together. The \rightskip 2in at the beginning of the paragraph tells LaTeX to leave 2 inches of white space to the right of the paragraph before the right margin. The left bracket "{" before the \rightskip command matches to the last right bracket "}" after the \end{picture} command. The purpose of that pair of brackets it to make sure the \rightskip 2in command does not apply to any other paragraph. The \vskip-1.2in\hskip4.5in\raisebox{-70pt}[0pt]{...} serves to move the diagram into that white space to the right of the paragraph. The exact size, 1.2 inches, of the \vskip depends on the length of the paragraph and the spacing between the lines of text, but it is negative because we are moving the diagram upward. The \raisebox command does the same thing as the \vskip command, but without leaving horizontal mode (LaTeX calls this LR mode). It is sloppy to use both: that problem had been edited a few times and accumulated redundant commands. The \hskip command moves the diagram rightward 4.5 inches because the paragraph is 4.5 inches wide.

Do you have any other questions? The picture environment is complicated. An alternative is to create a diagram with other software and save it as a JPEG or PDF image file. The \includegraphics command can insert the image into the PDF file created from the LaTeX, but you still have to mess with \vskip and \hskip to postion the image. Furthermore, when you e-mail your LaTeX solution to the USAMTS, you have to e-mail the image file too.

Erin Schram
Former typesetter of the USAMTS

Thanks a lot for this clear, informative explanation! props   :P


Point P lies on minor arc AB of the circumcircle of square ABCD. AB=5 and AP=4.  Find PD.

the picture is easy to draw.

[hide="hint"]
Law of Cosines on APB to find PB, then use the fact that DPB is a right triangle
[/hide]

i would hate to use trig in a geometry problem  :P

[quote="timwu"]Non-trig approach
[hide]
Connect $ AC, PC$. Find $ PC$ in the right triangle $ APC$, then use Ptolemy's Theorem to find $ DP$ in the cyclid quad $ ACDP$.[/hide][/quote]

wow so easy... :)  :o


Any one able to help me with this.
Use newtons method to find as many rootd as you can of the nonlinear system:
x+y+z=6
x^2 + y^2 + Z^2=14
x^3 + y^3 + z^3=37
produce 3m-fikes
@ each inteation step solve the equation
Jd=-f
where d=x^(k+1)-x^k in matlab by:
d=-J/f

This is an interesting question, but this probably isn't the best forum to post it on. Maybe I should move this thread to another forum: which forum would fit better?

hey for matlab questions go to http://www.mathworks.com and go to the user community.  they are really good at answering matlab questions as they are professional programmers :)

Other problem solving topics or the calculus computations and tutorials, I say.


geometry romania incenter trigonometry conics hyperbola

This one is easy.

Move $M$ on $AI$. Then map $BM \mapsto BN$ is projective from pencil of lines through $B$ onto it self, since it is rotation for angle ${1\over 2}\angle ABC$ around $B$. There for map $M \mapsto N$ is projective form line $AI$ to line $CI$. Now this map induces projective map $\rho: DM \mapsto DN$ from pencil of lines through $D$ onto itself.

Now, when $M= A$ then $N= I$, so $\angle MDN = {1\over 2} \angle ADC$
and when $M= I$ then $N= C$ so again $\angle MDN = {1\over 2} \angle ADC$,
and when $M$ is incenter of triangle $ABD$, $N$ is incenter of triangle $BCD$, from where we get again $\angle MDN = {1\over 2} \angle ADC$.

This implys that map $\rho$ is actualy rotation around $D$ for fixed angle ${1\over 2}\angle ADC$ since every projective map is exactly determined with 3 points.

my solution :

let $\alpha$ be $\angle ABM$ and let $\beta$ be angle $CBN$. Then $\angle IBM=\beta$ and $\angle IBN=\alpha$. Then let $p,q,r,s$ be the angles in $D$. By the sine theorem we will get :
$\displaystyle\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin BIA\sin BIC}{\sin A/2\sin C/2}$.
And :

$\displaystyle\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin p\sin s}{\sin r\sin q}\cdot\frac{\sin DIA\sin DIC}{\sin A/2\sin C/2}$.

But $\sin BIA\sin BIC=\sin DIA\sin DIC$, because it will be the same with $\cos\frac{A-C}{2} +\cos(D+\frac{A+C}{2})=\cos\frac{C-A}{2} +\cos(B+\frac{A+C}{2})$, but this is true because $\cos x=\cos (-x)=\cos (2\pi-x)$.

we will obtain that $\sin p\sin s=\sin r\sin q$ and because $p+q=r+s=D/2$, we transform into sum of cosines and we will obtain easy that $p=r$ and this solves the problem.

[quote="maky"]my solution :

let $\alpha$ be $\angle ABM$ and let $\beta$ be angle $CBN$. Then $\angle IBM=\beta$ and $\angle IBN=\alpha$. Then let $p,q,r,s$ be the angles in $D$. By the sine theorem we will get :
$\displaystyle\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin BIA\sin BIC}{\sin A/2\sin C/2}$. [/quote]

Can you show this step by step how to get
Abdurashid

well, first :

$\alpha=\angle ABM$ and $\beta=\angle CBN$ (notations).
then, because $\angle BMN$ is half of $\angle B$, we get that $\angle MBI=\beta$ and $\angle NBI=\alpha$.

also,

$p=\angle CDN$, $q=\angle NDI$, $r=\angle IDM$ and $s=\angle MDA$.

Then, sine theorem in $AMB$ and we get :

$\frac{AM}{MB}=\frac{\sin\alpha}{\sin A/2}$.

Sine theorem in $BMI$ :

$\frac{MB}{MI}=\frac{\sin BIA}{\sin\beta}$.

Sine theorem in $BNI$ :

$\frac{BN}{IC}=\frac{\sin BIC}{\sin\alpha}$.

Sine theorem in $BNC$ :

$\frac{CN}{BN}=\frac{\sin\beta}{\sin C/2}$.

We multiply this four relations and there goes the desired result :

$\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin BIA\sin BIC}{\sin A/2\sin C/2}$

In a similar way, we get the other part :

$\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin p\sin s}{\sin r\sin q}\cdot\frac{\sin DIA\sin DIC}{\sin A/2\sin C/2}$.

Now we show that $\sin BIA\sin BIC=\sin DIA\sin DIC$. (*)
Let $A,B,C,D$ be the angles of $ABCD$.

We get $\sin BIA=\sin\frac{A+B}{2}$, and similar relations.

Then (*) becomes equivalent with :

$\cos(\frac{A+B}{2}-\frac{C+B}{2})-\cos(\frac{B+A}{2}+\frac{B+C}{2})$ $=$ $\cos(\frac{A+D}{2}-\frac{C+D}{2})-\cos(\frac{D+A}{2}+\frac{D+C}{2})$.

But we have $\cos\frac{A-C}{2}=\cos\frac{C-A}{2}$ and $\cos(B+\frac{A+C}{2})$ $=$ $\cos(2\pi-B-\frac{A+C}{2})=\cos(A+B+C+D-B-\frac{A+C}{2})=\cos(D+\frac{A+C}{2})$.

But using the above facts, this becomes $\sin p\sin s=\sin q\sin r$. (**)
Now let`s look at the angles $p,q,r,s$. We have $p+q=r+s=D/2$, and because $ABCD$ is convex, $D/2<\pi/2$, so $p+q<\pi/2$ and $r+s<\pi/2$. Also, $p,q,r,s$ are all smaller than $D/2$.
Let $t=D/2$.
Then $q=t-p$ and $s=t-r$.
(**) means $\sin p\sin(t-r)=\sin r\sin(t-p)$, equivalent with $\cos(p-t+r) - \cos(p+t-r)=\cos(r-t+p)-\cos(r+t-p)$, which means $\cos(t+r-p)=\cos(t+p-r)$. We also have $0<t+r-p<2t<\pi$, and $0<t+p-r<\pi$.
Then we must have $t+r-p=t+p-r$, which gives us $p=r$ and from here we get that $\MDC=q+r=q+p=D/2$.

[quote="Valentin Vornicu"]The circle of center $I$ is inscribed in the convex quadrilateral $ABCD$. Let $M$ and $N$ be points on the segments $AI$ and $CI$, respectively, such that $\angle MBN = \frac 12 \angle ABC$. Prove that $\angle MDN = \frac 12 \angle ADC$.[/quote]

Nice solution by inversion on the way

Let $E$, $F$, $G$ and $H$ be the respective points of tangency of the circle, with sides $AB$, $BC$, $CD$, $DA$.  Inversion, center $I$, radius $IE$. The images of $A$, $B$, $C$ and $D$ are $A'$, $B'$, $C'$ and $D'$, respective midpoints of segments $HE$, $EF$, $FG$, $GH$. Let $M'$ and $N'$ be the images of points $M$ and $N$. The inverted problem is formulated as follows:
If $\angle B'M'I + \angle B'N'I = \angle B'A'I$ then $\angle D'M'I + \angle D'N'I = \angle D'A'I$.
Now
1) $\angle B'A'I = \angle B'C'I$ properties of inversion, angle chasing or whatever
2) Hence $\angle M'B'A' = \angle B'N'I$, $\angle N'B'C' = \angle B'M'I$, hence triangles $M'B'A'$ and $N'B'C'$ are similar
3) Therefore $M'A' \cdot N'C' = A'B' \cdot B'C'$
4) But $A'B' = C'D'$ and $B'C' = A'D'$ this is because all those points are midpoints of sides of a quadrilateral
5) Hence $M'A' \cdot N'C' = A'D' \cdot D'C'$
6) Not hard to prove that $\angle M'A'D' = \angle N'C'D'$ hence triangles $M'A'D'$ and $N'C'D'$ are similar and hence result

Daniel

I have two solution this beautiful problem:
Proof 1: After some sine law , this problem is equivalent to $  \frac{AB \cdot BC}{AD \cdot DC}=\frac{BI^2}{DI^2} $ ,but it's well-known identity.
Proof 2: Let the circles $ w_m=(M,r_n) $ and $ w_n=(N,r_n) $  such that $ w_m $ , is tangent to $ AB,AD $ and $ w_n $ tangent to $ BC,DC $ ,respectively.It's easy to see that $ 2 \angle MBN=ABC $ if and only if $ 2 \angle MDN=ADC $ if and only if $ B $ lies on one of two internal common tangent of $ w_m $ and $ w_n $.

My solution:

Since $ AB-AD=CB-CD $ ,
so $ A, C $ lie on a hyperbola $ \Omega $ with focus $ B $ and $ D $ .
Since $ \angle BAI=\angle IAD, \angle DCI=\angle ICB $ ,
so $ IA, IC $ is tangent to $ \Omega $ at $ A, C $, respectively .
From $ \angle MBN=\frac{1}{2} \angle ABC $ we get $ MN $ is tangent to $ \Omega $ ,
so we get $ \angle MDN=\frac{1}{2} \angle ADC $ .

Q.E.D