What is depicted in this diagram? It looks definitely modular...

A sprial of consecutive numbers from $ \minus{}9$ to $ 11$, in each "ray" the numbers increased by $ 6$ outward.  Unless there is something more?

It is from a cover of a Springer-Verlag algebra book. I am interested to know what the black line which overlaps with the spiral is..

The [color=violet]magenta[/color] line is the set of integers $ \mathbb{Z}$ and the black line highlights a cycle of 6 numbers, that is $ \mathbb{Z}_6$.

[quote="Los"]why is the black line sometimes in the magenta line but sometimes going outside of it?[/quote]If you follow around the black line, you get 0 1 2 3 4 5 0 1 2 3 4 5 0 ..... it repeats the cycle 0 1 2 3 4 5, like mod 6 or $ \mathbb{Z}_6\equal{}\{0,1,2,3,4,5\}$ <-----(this is a set). Same thing, different notations. Not only that, each of the numbers lined up they belong to the same [i][[equivalence class]][/i] because [[congruence]] is an [i][[equivalence relation]][/i].

[quote="Los"]why is the black line sometimes in the magenta line but sometimes going outside of it?[/quote]

Actually I think the explanation for that is very simple. The black line is probably a circle, while the radius of the spiral is increasing as it moves outwards, so that the black line will have to end up going through the magenta line (in essence, the region bounded by the line has no special meaning, just serves as a number line). The markers on the black line correspond to $ \mod{6}$.

The numbers are supposed to continue below -9 and above 11, but there is an end to the negative numbers. The loops of the spiral are evenly spaced at 0.15 times (by my measurement) the diameter of the circle (3/10 of the radius). If we move in from 0, we can only go 10/3 of a turn to -20. Not all integers then.

Also, if we take the radius of the circle as 1, the position of number t can be written as a function of t.
In polar coordinates, its distance to the center of the circle is 1+0.05t. The angle is (pi/3)t.
We could give the position as (x,y) too, involving cos((pi/3)t) and sin((pi/3)t).
Does the same picture appear in the chapter on trigonometric functions?

Maybe all they meant is that math is beautiful, or that math and art are related.


Why do people always say Chemistry is easier than Physics?

I find it the exact opposite.  Physics seems to be more intuitive in some ways.  Is there a certain point when it all connects?  Chemistry, especially some of the equilibrium stuff, is a little strange.

I think Physics is conceptually easier. Until you get into Modern Physics, which is often related to chemistry, too. The thing is, at least with the AP classes I've taken, the Chemistry questions rely a lot less on problem-solving skills than simple application skills. You see the problem, find the formula, plug it in. Physics requires more thinking in my opinion, and that makes it more complicated to a lot of people.

Yea, to me it's mostly that Physics requires actually working a little bit to get towards the answer; most of chemistry is either you know it or you don't...but I only view it that way cause the only questions I consistently miss on any chemistry test are the lab/qualitative ones.

I think chemistry is easier.  Perhaps the reason that's what so many think is that physics requires far more math, another subject plenty of students dislike.  Trig is used only in physics (at least at the high school level) and the most mathematically "difficult" thing i encountered in ap chem were logs (whooaaa...).  I think classes like calculus based physics in high school are just more intimidating than even ap chem.

[quote]Is there a certain point when it all connects?[/quote]

Yes. You're not going to find this in an AP course, but in later courses you will.


Let $ a,b,c,d$ be positive real numbers. Prove that:

$ \frac{d\plus{}a^2}{d\plus{}bc}\plus{}\frac{d\plus{}b^2}{d\plus{}ca}\plus{}\frac{d\plus{}c^2}{d\plus{}ab} \ge 3.$

$ \frac{d}{d\plus{}bc} \plus{} \frac{d}{d\plus{}ca}\plus{}\frac{d}{d\plus{}ab} \ge \frac{9d}{3d\plus{}ab\plus{}bc\plus{}ca}$
$ \frac{a^2}{d\plus{}bc}\plus{} \frac{b^2}{d\plus{}ca} \plus{} \frac{c^2}{d\plus{}ab} \ge \frac{(a\plus{}b\plus{}c)^2}{3d\plus{}ab\plus{}bc\plus{}ca}$
Adding the two we obtain $ LHS \ge \frac{9d\plus{}(a\plus{}b\plus{}c)^2}{3d\plus{}ab\plus{}bc\plus{}ca} \ge 3$
equality when $ a\equal{}b\equal{}c$.

It would not be a surprise since the following inequality is true for all a,b,c and $ d\ge 0$

$ (a^2\plus{}d)(b^2\plus{}d)(c^2\plus{}d) \ge (ab\plus{}d)(bc\plus{}d)(ca\plus{}d)$  :wink:

[quote="great math"]It would not be a surprise since the following inequality is true for all a,b,c and $ d\ge 0$

$ (a^2 \plus{} d)(b^2 \plus{} d)(c^2 \plus{} d) \ge (ab \plus{} d)(bc \plus{} d)(ca \plus{} d)$  :wink:[/quote]

Proof:
By Cauchy-Schwarz inequality we have
$ (a^2\plus{}d)(b^2\plus{}d)\ge (ab\plus{}d)^2$
$ (b^2\plus{}d)(c^2\plus{}d)\ge (bc\plus{}d)^2$
$ (a^2\plus{}d)(c^2\plus{}d)\ge (ac\plus{}d)^2$

Multiply those inequalities will prove the problem
 :D


You have 101 coins. 50 are X and the rest are Y. X coin weights 1g more than Y coin. You have a machine that can calculate the difference of the weight of two sides in gram. If I choose one coin from this 101 coin, how can you know whether it's X or Y? and what's the minimun times of using the machine?

[hide]Without loss of generality suppose that the X-coin weights 1g and the Y-coin weights 0g
So the total weight of the 101 coins is 50g.

Suppose we take a Y-coin (The total weight of the rest coins is still 50g).
We divide the rest in two groups of 50 coins and we put them on the machine.

Let $x$ be the number of X-coins on the left pan. Then we have $50-x$ X-coins on the right pan.
The difference is $|50-2x|$, which is an even number.

Now suppose that we take an X-coin (The total weight of the rest coins is  now  49g).
We divide the rest in two groups of 50 coins and we put them on the machine.

Let $x$ be the number of X-coins on the left pan. Then we have $49-x$ coins on the right pan.
The difference is $|49-2x|$, which is an odd number.[/hide]


Choose three points $ A_1;B_1;C_1$ on three sides of a triangle $ ABC$($ A_1 \in BC;B_1 \in CA;C_1 \in AB$) such that $ AA_1;BB_1;CC_1$ meet at a common point.Again,choose three points $ A_2;B_2;C_2$ on the sides of triangle $ A_1B_1C_1$$ (A_2 \in B_1C_1;B_2 \in C_1A_1;C_2 \in A_1B_1)$
Prove that $ AA_2;BB_2;CC_2$ meet at a common point if and only if so do $ A_1A_2;B_1B_2;C_1C_2$

Please delete my post and the hint of ¬[ƒ(Gabriel)³²¹º]¼
here is a problem in MYM this month,and the contest is running
Warn negative!,please


Hola! Posteo estos problemas para ver si alguien me puede ayudar o dar alguna pista sobre cómo solucionarlos  :(  Gracias.

1. ¿Existe algún $ a\in\Re$ para que el polinomio $ P(X)\equal{}X^5 \minus{}((10*a^2)/3)*X^3 \plus{} 5*a^4*X\minus{}8*a^3$ tenga un cero de multiplicidad exactamente tres?

2. Probar que el polinomio 1 + X + ... + $ X^(p\minus{}1)$ es irreducible sobre Q (los racionales) para cualquier primo p

3.Sean m , n $ \in N$ y consideremos el polinomio $ P(x) \equal{} 1 \plus{} (n\plus{}1)*X \plus{} 2(m \plus{} 1)*X^2 \plus{} n*X^3 \plus{} x^4 \in Z[X]$. Probar que p(x) es irreducible sobre Q (los racionales).

4. Desmostrar que existen infinitos enteros no representables como suma de tres cuadrados.

5. Demostrar que si (n-1)! $ \equiv$ 0 mod n entonces n es primo.

6. Demostrar que si p es primo ( p distinto de 3) entonces $ (p\minus{}2)*2^(p\minus{}2) \plus{} 1$ no es primo

7. Probar que $ (n^5)/5 \plus{} (n^3)/3 \plus{} (7*n)/15$ es un entero para todo n.

8. Sea (a, b, c) una terna pitagórica. Demostrar :
a. Al menos uno de los valores a, b, c es múltiplo de 3.
b. abc es múltiplo de 4
c. al menos uno del os valores es múltiplo de 5

9. Demostrar que si  mcd (a, n) = 1 y mcd (b, n) = 1, la ecuación ax + by = c tiene exactamente n soluciones en Z/n (el grupo de enteros Z módulo n)

10. Demostrar que $ (n^5 \minus{}1)*n*(n^5 \plus{} 1)$ es divisible por 22.

[quote="jaden"]
4. Desmostrar que existen infinitos enteros no representables como suma de tres cuadrados.

7. Probar que $ (n^5)/5 \plus{} (n^3)/3 \plus{} (7*n)/15$ es un entero para todo n.

10. Demostrar que $ (n^5 \minus{} 1)*n*(n^5 \plus{} 1)$ es divisible por 22.[/quote]

4. Notar que todos los numeros de la forma $ 8m\plus{}7$ no son representables como suma de tres cuadrados ya que los residuos cuadraticos modulo $ 8$ son $ 0,1,4$

7. $ (n^5)/5 \plus{} (n^3)/3 \plus{} (7*n)/15\equal{}\frac{3n^5\plus{}5n^3\plus{}7n}{15}$ entonces habria que probar que $ 3|3n^5\plus{}5n^3\plus{}7n$ y $ 5|3n^5\plus{}5n^3\plus{}7n$
$ 3| 5n^3\plus{}7n$ o $ 3| 2n^3\plus{}n$ o $ 3| (2n^2\plus{}1)n$ lo que es verdadero para todo $ n$
$ 5| 3n^5\plus{}7n$ o $ 5|n(3n^4\plus{}2)$ lo que es verdadero para todo $ n$

10. $ 22|(n^5 \minus{} 1)*n*(n^5 \plus{} 1)$ entonces habria que probar que $ 2|(n^5 \minus{} 1)*n*(n^5 \plus{} 1)$ y $ 11|(n^5 \minus{} 1)*n*(n^5 \plus{} 1)$
$ 2|(n^5 \minus{} 1)*n*(n^5 \plus{} 1)$ lo que es verdadero ya que $ n$ y $ n^5\plus{}1$ tienen distinta paridad
$ 11|(n^5 \minus{} 1)*n*(n^5 \plus{} 1)$ notar que los residuos modulo $ 11$ de $ n^5$ son $ \minus{}1,0,1$

:)  Muchas gracias, creo que poco a poco voy entendiendo la dinámica de estos problemas.

Hola [b]jaden[/b], quería ver si te ayudaba con algunos de los ejercicios... :)

[quote="jaden"]
5. Demostrar que si (n-1)! $ \equiv$ 0 mod n entonces n es primo.
[/quote]

Esto no es correcto (por ejemplo $ n \equal{} 6$); de hecho, lo correcto es decir "si $ (n \minus{} 1)! \equiv \minus{} 1 \pmod{n}$, entonces $ n$ es primo". Esto es equivalente a probar que si $ n$ es compuesto, entonces $ n$ no divide a $ (n \minus{} 1)! \plus{} 1$, lo cual es intuitivo ya que si $ n$ es compuesto, todos sus factores primos estarán en la descomposición de $ (n \minus{} 1)!$, así que $ n$ divide a $ (n \minus{} 1)!$, y por tanto no puede dividir a $ (n \minus{} 1)! \plus{} 1$.

[quote="jaden"]
8. Sea $ (a,\, b,\, c)$ una terna pitagórica. Demostrar que:
a. Al menos uno de los valores $ a$, $ b$, $ c$ es múltiplo de 3.
b. $ abc$ es múltiplo de 4.
c. al menos uno del los valores es múltiplo de 5.
[/quote]

La ecuación pitagórica es $ a^2 \plus{} b^2 \equal{} c^2$, y lo que hacemos es ver los posibles de los cuadrados perfectos módulo 3, 4 ó 5. De hecho, $ a^2 \equiv 0\ \mathrm{\acute{o}}\ 1 \pmod{3}$, $ a^2 \equiv 0\ \mathrm{\acute{o}}\ 1 \pmod{4}$ y $ a^2 \equiv 0,\ 1,\ \mathrm{\acute{o}}\ \minus{} 1 \pmod{5}$, y de aquí es fácil probar cada propiedad verificando todos los casos posibles.

[quote="jaden"]
10. Demostrar que $ n(n^5 \plus{} 1)(n^5 \minus{} 1)$ es divisible por 22.[/quote]

Hay que probar que $ n(n^5 \plus{} 1)(n^5 \minus{} 1)$ es divisible por 2 y por 11. Es claro que el producto es par dado que hay siempre un par entre los números $ n$, $ n^5 \plus{} 1$ y $ n^5 \minus{} 1$. Y como $ (n^5 \plus{} 1)(n^5 \minus{} 1) \equal{} n^{10} \minus{} 1$, la divisibilidad por 11 se sigue del teorema de Fermat $ n^{10} \equiv 1 \pmod{11}$.


linear algebra
matrix
limit
linear algebra unsolved

let A matrix in M(n,R)

prove the equivalence

1) rg(A)=rg(A^2)

2) x(A+xI_n)^{-1} has a limit when x tend to zero

it seems very easy with Jordan form  :maybe:

It is very easy using Jordan form. However, don't forget that you are not allowed to use it, so the problem of finding an elementary solution should be put. I have one, but it is long and very boring. If I don't see another, nicer thing, I will try to write my own.

Can somebody tell me what is the mean of "rg(A)"?

Thank you,
but how can we show (A+xI_n) is invertable for small x?

because there is only a finite number of $x$ such that $A+xI$ is not invertible !

[quote="spix"]I suppose it stands for the rank of the matrix.[/quote]rank or range, it's all the same.

Observe that $rg(A) = rg(A^{2})$ is equivalent to saying that $A$ acts as an isomorphism on $col(A)$, the column space of $A$. This is in turn equivalent to saying that $\mathbb{R}^{n}= col(A) \oplus nul(A)$, where $nul(A)$ is the null space of $A$. Then a computation shows that if $u = v+w$ with $v \in col(A), \, w \in nul(A)$, we have
\[\lim_{x \to 0}x(A+xI)^{-1}u = w \, . \]
Reversely, if this limit exists, it is in $nul(A)$, since $A \left(x(A+xI)^{-1}u\right) = x(u-x(A+xI)^{-1}u)$ which converges to 0. Then $u-w = \lim_{x \to 0}A(A+xI)^{-1}u \in col(A)$. So $col(A) \oplus nul(A) = \mathbb{R}^{n}$ (remember that their dimensions must add to n).  :lol:


[hide="Solution."]
Let the value to which the sequence converges to be $ x$.
$ x \equal{} 2 \plus{} \frac {1}{2 \plus{} \frac {1}{2 \plus{} ...}}$

$ \implies x \equal{} 2 \plus{} \frac {1}{x}$

$ \implies x \equal{} \boxed{1\plus{}\sqrt{2}}$

[/hide]

$ L\equal{}2\plus{}\frac{1}{L}$ gives $ L\to\sqrt 2\plus{}1$ :(

[quote="misan"]$ L \equal{} 2 \plus{} \frac {1}{L}$ gives $ L\to\sqrt 2 \plus{} 1$ :([/quote]

Yes, I caught my error right after you did.  :blush:

Solve $ a_{n \plus{} 1} \equal{} 2 \plus{} \frac {1}{a_n}$, let $ \alpha \equal{} 1 \minus{} \sqrt {2},\ \beta \equal{} 1 \plus{} \sqrt {2}$, we have $ \frac {a_{n \plus{} 1} \minus{} \beta}{a_{n \plus{} 1} \minus{} \alpha} \equal{} \frac {\alpha}{\beta}\cdot \frac {a_n \minus{} \beta}{a_n \minus{} \alpha}\Longleftrightarrow \frac {a_n \minus{} \beta}{a_n \minus{} \alpha} \equal{} r^n$ where $ r \equal{} 2\sqrt {2} \minus{} 3$, yielding $ a_n \equal{} \frac {\beta \minus{} \alpha r^n}{1 \minus{} r^n}\longrightarrow \beta \equal{} 1 \plus{} \sqrt {2}\ (n\to\infty)$.


integration
trigonometry
calculus
calculus computations

Without using complex method, evaluate $ \int_{0}^{2\pi}\frac {1}{a \plus{} b\cos{\theta}}d\theta$($ 0 < b < a$).

Thanks.

$ \int_{0}^{2\pi }\frac{1}{a\plus{}b\cos \theta }d\theta \equal{}\allowbreak \int_{\minus{}\pi }^{\pi }\frac{1}{a\minus{}b\cos \theta }\,du$

$ \equal{}2\int_{0}^{\pi }\frac{\sec ^{2}\frac{\theta }{2}}{(a\minus{}b)\plus{}(a\plus{}b)\tan ^{2}\frac{\theta }{2}}d\theta$

$ \equal{}\left[ \frac{4}{\sqrt{a^{2}\minus{}b^{2}}}\tan ^{\minus{}1}\left( \sqrt{\frac{a\plus{}b}{a\minus{}b}}\tan \frac{\theta }{2}\right) \right] _{0}^{\pi }\equal{}\frac{2\pi }{\sqrt{a^{2}\minus{}b^{2}}}$

At first, thanks again for a nice solution. It was easy than I thought... Here is my approach, rather complicated.

$ \int_{0}^{2\pi}\frac{1}{a+b\cos{\theta}}d\theta=\frac{1}{a}\int_{0}^{2\pi}\frac{1}{1+\varepsilon\cos\theta}d\theta(0<\varepsilon=\frac{b}{a}<1)$.

 We have $ |\varepsilon\cos\theta|<1$, so the integrand can be rewritten as $ \frac{1}{1+\varepsilon\cos\theta}=\frac{1}{1-(-\varepsilon\cos\theta)}=1-\varepsilon\cos\theta+\varepsilon^2\cos^2\theta-\varepsilon^3\cos^3\theta+\cdots$.

 $ \int_{0}^{2\pi}\cos^n\theta{d\theta}=2\int_{0}^{\pi}\cos^n\theta{d\theta}(\because \cos(\pi-\theta)=\cos(\pi+\theta))$, $ n$ is natural number.
This value is $ 0$ when $ n$ is odd, and $ 4\int_{0}^{\frac{\pi}{2}}\cos^n\theta{d\theta}$ otherwise${ (\because \cos(\frac{\pi}{2}+\theta)=-\cos(\frac{\pi}{2}-\theta}))$. Applying Tonelli's theorem and this, we get

$ \int_{0}^{2\pi}\frac{1}{a+b\cos{\theta}}d\theta=\int_{0}^{2\pi}1-\varepsilon\cos\theta+\varepsilon^2\cos^2\theta-\varepsilon^3\cos^3\theta+\cdots{d\theta}=\int_{0}^{2\pi}\sum_{n=0}^{\infty}(-\varepsilon)^n\cos^n{\theta}d\theta=\sum_{n=0}^{\infty}\int_{0}^{2\pi}(-\varepsilon)^n\cos^n{\theta}d\theta=4\sum_{m=0}^{\infty}\varepsilon^{2m}\int_{0}^{\frac{\pi}{2}}\cos^{2m}\theta{d\theta}$.

I want to evaluate this series by using Wallis's product, but I've got stuck here. Please, help me. Thanks.


Vieta
complex analysis
complex analysis unsolved

Let $ X_k = e^{i\frac {2k \pi}{n}}$

Then prove : $ \forall X \in \mathbb{C}$ $ \boxed{\prod_{k = 0}^{n - 1} (X - X_{k} ) = X^n - 1}$

[hide="Hint:"]
Using Vieta's formula it turns to prove:

$ \sum_{i < j} X_iX_j = 0$

$ \sum_{i < j < k} X_iX_jX_k = 0$
...[/hide]

Is not it obvious, since $ X_k$ are the roots of the equation $ X^n\minus{}1\equal{}0$ and moreover for $ k\equal{}0,1,\ldots,n\minus{}1$ they are all the roots?


A tower is 300 feet high. How long should a guy wire be if it is to be attached to the tower 20 feet from the top and is to make an angle of 23 degrees with the ground?


probability
number theory
least common multiple

7. Two fair dice, each with faces numbered 1 through 6, are rolled at the same time. Each die has five exposed faces, which are summed. Express as a common fraction the probability that the least common multiple of the sums of the exposed faces is a multiple of 6.

[hide="Answer"]$ \frac5{12}$[/hide]

[hide]The exposed faces of a die either equal 20, ,19, 18, 17, 16, or 15. Each of these is like saying a certain number is on the bottom, so each have a 1/6 chance of occurring. To have an LCM of 6, either (or both) must be a multiple of 6, or one is a multiple of 2 and the other of 3.
One multiple of 6:
18 with any of the other 5
any other 5 with 18
2 multiples of 6:
18 and 18
mult. of 2, then mult. of 3 (cannot use 18)
2*1 (20 or 16 and 15)
mult. of 3, then mult. of 2 (not use 18)
1*2
Totaling them up, there are 15 possibilities, divided by 36 possible outcomes, so answer is
5/12
[/hide]


Check the item that has the word you do not have the same characteristics as the other, on the feet.

a) ANT
b) SPIDER
c) BEE
d) BOOKWORM
e) BUTTERFLY

This should be in a different forum....

In any case, the answer's spider, since it's not an insect.

This seems appropriate to the "Puzzles..." section.

I believe it is choice C) because I am looking at the bottom
of the SHAPE of the letters, or "the feet."

All of the other choices have at least a letter with a straight
section meeting the bottom.  However, with "BEE," there is the
rounded portion of the first letter and then the flat portion of the
next two repeated letters.  It has no letters with free stand-alone
straight portions coming down.

To contrast,
choice a) has an "N" and a "T",
choice b) has a "P" and an "R",
choice d) has a "K", "R", and an "M",
choice e) has a "T", a "T", an "R", an "F', and a "Y".

I do not know for sure.

I don't understand this question too.
I think it's BOOKWORM because it don't have feet  :D
If anyone knows how to do this question, please explain me.  :wink:

luiseduardo,

as a suggestion for other problems of this type,

please go to the section

"Other Problem Solving Topics" and then post it under

"Puzzles & Brainteasers."


How many three digit numbers are left if we exclude all three digit numbers in which there are exactly two digits the same, but these two digits are not adjacent?

The number in the second case must be in form of $ \overline{aba}$, where $ a$ is between $ 1$ and $ 9$, and $ b \neq a$.

Two cases to consider:

[i]Case 1:  $ b \equal{} 0$[/i]

This is simply $ 9 \times 1$, or $ 9$.

[i]Case 2:  $ b \neq 0$[/i]

Well, since $ b \neq a$, we have $ 9 \times 8$, or $ 72$.

Adding, we get $ 81$.  Subtracting from the number of three digit numbers ($ 9 \times 10 \times 10$), we get $ \boxed{819}$.[/i]

Why is there only 9 options for B? Shouldn't there be 10?


simplify,

$ \cos(x)\plus{}\cos(2x)\plus{}\cos(3x)\plus{}\cos(4x)\plus{}...\plus{}\cos(nx)$

I know that there are several forms of resolution.

which is easiest?

good luck!

___________________

- kl -

hello, prove by induction that
$ {\sum_{i=1}^n\cos(ix)=-1/2\,\cos \left(  \left( n+1 \right) x \right) +1/2\,{\frac { \left(
\cos \left( x \right) +1 \right) \sin \left(  \left( n+1 \right) x
 \right) }{\sin \left( x \right) }}-1/2}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, prove by induction that
$ {\sum_{i = 1}^n\cos(ix) = - 1/2\,\cos \left( \left( n + 1 \right) x \right) + 1/2\,{\frac {\left( \cos \left( x \right) + 1 \right) \sin \left( \left( n + 1 \right) x \right) }{\sin \left( x \right) }} - 1/2}$
Sonnhard.[/quote]

as I do?

Sum of Sine Arithmetic Progression
http://www.mathlinks.ro/viewtopic.php?t=249034


find all pair of [b](x,y)[/b]  in natural number:
[b]x x+y=(x+y) y.[/b]

Solutions are the pairs $(x,y) = ((k+1)^k,k(k+1)^k)$ where $k$ is any nonegative integer.

Clearly, if $x=1$ then $y=0$, thus we assume that $x > 1.$
Let $p$ be a prime which divides $x$ and $p^a$ be the greatest power of $p$ which divides $x$. It is clear that $p$ must divides $y$. Let $p^b$ be the  greatest power of $p$ which divides $y$.
Suppose that $b<a$ : then $p^a$ is the greatest power of $p$ which divides $x+y$, so that we must have $a(x+y) = ay$ which is absurd. Thus $ a \leq b$.
Since $p$ is an arbitrary prime which divides $x$, we deduce that $x$ divides $y$. Let $y=kx$, where $k$ is a positive integer. The given equation leads immediately to $x=(k+1)^k$ so that $y=k(k+1)^k$, as claimed.
Conversely, it is easy to verify that these pairs are indeed solutions.

Pierre.


superior algebra
superior algebra unsolved

Is there an uncountable collection $C$ of sets such that :

  i) Every set in $C$ is of countable cardinal
 ii) For every pair of sets $a,b$ in $C$ we have either $a \subset b$ or $b \subset a$
iii) The union of $C$ is just countable

An example: For each real number $\alpha$, let $S_{\alpha}$ be the set of all rational numbers greater than $\alpha$.

What if we change the question a little to

iii) The union of $C$ is continuum

Oh, there is one example which immediately comes to my mind, but the union does not necessarily have continuum cardinality; it's just not countable. It's a famous example which provides a starting point for many counterexamples in topology.

Call a subset of an ordered set of the form $S_{\alpha}=\{x|x<\alpha\}$ a section (I think it's the usual terminology). It can be shown that there is a well-ordered set (although we don't need the well-ordering) which is callede "the minimal uncountable well-ordered set", that is an uncountable ordered set in which every section is countable. Let your sets be the sections of this set.

Their union is the entire set, which is uncountable, every set is, of course, countable, and they're totally ordered because any well-ordering is, by definition, a total ordering.

I'll think about your original question.

[quote="grobber"]Call a subset of an ordered set of the form $S_{\alpha}=\{x|x<\alpha\}$ a section (I think it's the usual terminology).[/quote]
Erdos calls it segments in his "Proofs from the BOOK".

This example also came to my mind.

I guess answer may depend on continuum hypotesis.

Well, if we assume the Continuum Hypothesis, the answer we talked about works.

All the guessing are correct. All we need now is a proof!! :)


While on your galacti holiday, you decide to stop at planet X, a planet known for its many mathematical milestones.

While there, you decide to pick up on some of the X-an mathematical quirks. The X-ans won't tell you exactly how itheir system works, only that it has some relationship to the standard Earthly math. They also give you six, true, X-an math problems. They are:

2+2=11
6-4=28
4*5=100
10/5=10
3-4=13
4+3*5-6=9497

So, knowing that those are all true, what's 3*6?

is it something about maths or is there a word game or kinda trick?

If 2 + 2 = 11

Does that mean 2 = 11 - 2 ?

Are the transitive, associative, communitive, etc. properties still properties?

not necessarily.

On second thought, I think you should define exactly what kind of a number system this is. Are we working with a field? A ring? I am not sure what I am allowed to do.

Sorry, I do not have an answer

For 2+2 = 11

this might give it away, but if it's hard to understand without, then.... basically, there is an algorithm you apply to 2 and 2 to yield 11 using operations from the English math system.  the transitive, distributive, etc dont necessarily apply

[quote="EFuzzy"][quote]They also give you[b] four[/b], true, X-an math problems. They are:

2+2=11
6-4=28
4*5=100
10/5=10
3-4=13
4+3*5-6=9497

So, knowing that those are all true, what's 3*6?[/quote]

So wait, 2 of those math problems are wrong?? Or does 4=6...[/quote]

Wow, I feel so embarrassed. :blush:  I didnt catch that. Confused :wacko:

OOPS. just kidding  :D , i meant six... i've become too accustomed to non-earthly math  :P sorry for the confusion!!

[quote="pkothari13"]2+2=11
6-4=28
4*5=100
10/5=10
3-4=13
4+3*5-6=9497

So, knowing that those are all true, what's 3*6?[/quote]

From
6 - 4 = 28  and  3 - 4 = 13
we can suppose that $a - b = 5a - \frac{b}{2}$.  Applying this to  4+3*5-6=9497, we have $5(4 + 3*5) - 3 = 9497$ and so $4 + 3*5 = 1900$.  Now, this is very large.  In addition, 4*5 takes two smallish numbers and spits out a largish numbers, so I will guess that $a*b = a\cdot b^2$, in which case we need to find an operation + such that $2 + 2 = 11$ and $4 + 75 = 1900$ are both valid.  We could find such an operation, for example, by solving the system $2x + 2y = 11$ and $4x + 75y = 1900$ and then defining $a + b = ax + by$, but I won't even bother as I've already decided that $a*b = a\cdot b^2$ and thus $3*6 = 108$.

JBL, you were pretty much on the right track with the $5a-\displaystyle\frac{b}{2}$.

The true answer is you view the operation as each number being a place value in a base 10 number, carry out the given operations, and then divide by two, rounding down when necessary.  In other words, for 6-4 = 28, you view 6 as the 10's place and 4 as the 1's place.  Therefore, it will be 60-4 = 56, and then you divide by 2, or 28.

2+2=11; 20+2 =22/2 = 11.
4*5 = 100; 40*5 = 200/2 = 100.
10/5 = 10; 100/5 = 20/2 = 10.
3-4 = 13; 30-4 = 26/2 = 13.
4+3*5-5=9497 = 4000+300*50-5 = 18995/2 = 9497.5 = 9497.

Did that make any sense?

From this, you will see that 3*6 will be 30*6/2, or just 90.

hahaha, i would have been thoroughly amazed if anyone solved it O_O


\[ \begin{array}{l}
 \Delta ABC\_is\_sc{\bf{a}}lene., < A\_is\_obtuse.Deter\min e\_the\_set\_of\_po{\mathop{\rm int}} s\_which\_lie\_on\_the\_extended\_line\_BC \\
 such\_th{\bf{a}}t\_\left| {AD} \right| \equal{} \sqrt {\left| {BD} \right|.\left| {CD} \right|}  \\
 \end{array}
\]

[quote="abacadaea"]wait, ABC is scalene, but A is obtuse?[/quote]

3 unequal sides?

WOW. i can be an idiot sometimes.

[hide]If they are on line BC, there are only 2 points. Either $ \angle DAB\equal{}90$ or $ \angle CAB\equal{}90$/hide]


[quote="silenthero_k09 tặng Huyền "]
[b]Problem:(Phan Hồng Sơn-11T-THPTNK QB)[/b]
[b][i]Let[/i][/b] $\large a,b,c>0$[b][i].Prove that[/i][/b]
$\large(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a})^{2}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{21}{4}$
...............

 [/quote]
[quote="Dành tặng em yêu "]
[b]My Solution [/b] :D
[b]Vì[/b] $\large a,b,c>0$[b] nên chúng ta có quyền đặt[/b]./ $\large\frac{b}{a}=x,\frac{c}{b}=y, \frac{a}{c}=z$.[b]Trong đó [/b]$\large x,y,z>0$ [b]và[/b] $\large xyz=1$.[b]Vậy bài toán chuyển thành chứng minh bất đẳng thức sau:/[/b]
[b]cho[/b] $\large x,y,z>0$ [b]và [/b]$\large xyz=1$
[b]C/m[/b] $\large(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+y})^{2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{21}{4}$
[b]Vì vai trò của[/b] $x,y,z$ [b]là bình đẳng nên không mất tính tổng quát ta có thể giả sử[/b] $\large x \leq y \leq z$ [b]suy ra[/b] $\large x \leq 1, yz\ge 1$
[b]Ta sẽ chứng minh bất đẳng thức này bằng phương pháp dồn biến [/b]
[b]Đặt[/b] $\large f(x,y,z)$[b] là biểu thức cần chứng minh .Khi đó ta chứng minh lần lượt các bất đẳng thức sau[/b]:
$\large f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz})$ [b](1)[/b]
$\large f(x,\sqrt{yz},\sqrt{yz})\ge\frac{21}{4}$ [b](2)[/b]
[b]C/m [/b][b](1)[/b] $\large f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz})$
[b]Ta có[/b] $\large f(x,y,z)-f(x,\sqrt{yz},\sqrt{yz})$
$\large =(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+y})^{2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-(\frac{1}{1+x}+\frac{2}{1+\sqrt{yz}})^{2}-\frac{1}{x}-\frac{2}{\sqrt{yz}}$
[b]Ta sẽ chứng minh [/b]
[b](+)[/b] $\large \frac{1}{1+y}+\frac{1}{1+z}\ge\frac{2}{1+\sqrt{yz}}$
[b](++)[/b] $\large \frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}$
[b]Chứng minh[/b][b] (++)[/b][b] khá đơn giản.Theo [/b][b]CauChy[/b] :lol:
[b]Bất đẳng thức [/b][b](+)[/b]
\[\Leftrightarrow\frac{1}{1+y}+\frac{1}{1+z}-\frac{2}{1+\sqrt{yz}}\ge 0 \]

\[\Leftrightarrow \frac{(\sqrt{yz}-1)(\sqrt{y}-\sqrt{z})^{2}}{(1+\sqrt{yz})(1+y)(1+z)}\ge 0 \]
[b]Nhưng bất đẳng thức này hiển nhiên đúng vì [/b]$\large yz\ge1$
[b]Từ đây ta suy ra được[/b] $\large f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz})$
 [b]Do đó theo định lý " Dồn biến không xác định[/b]-[b]UMV[/b]"( [b]Undefined Mixing Variables[/b]) [b]ta chỉ cần chứng minh trong trường hợp[/b] $\large y=z$.
[b]Với chú ý rằng[/b] $\large\sqrt{\frac{1}{x}}=\sqrt{yz}$([b]do[/b] $xyz=1$)
[b]cuối cùng chỉ cần chứng minh[/b]

 $f(x,\sqrt{\frac{1}{x}},\sqrt{\frac{1}{x}})\ge\frac{21}{4}$([b]với[/b] $x \leq1$ )
[b]Chúng ta có [/b] $\large f(x,\sqrt{\frac{1}{x}},\sqrt{\frac{1}{x}})-\frac{21}{4}$
$\large=(\frac{1}{1+x}+\frac{2\sqrt{x}}{\sqrt{x}+1})^{2}+\frac{1}{x}+2\sqrt{x}-\frac{21}{4}$
$\large=((\frac{1}{1+x}+\frac{2\sqrt{x}}{\sqrt{x}+1})^{2}-\frac{9}{4})+(\frac{1}{x}+2\sqrt{x}-3)$
$\large=\frac{(\sqrt{x}-1)(\sqrt{x}-1)^{2}(7x\sqrt{x}+3x+9\sqrt{x}+5)}{4(x+1)^{2}(\sqrt{x}+1)^{2}}+\frac{(1-\sqrt{x})^{2}(2\sqrt{x}+1)}{x}$
$\large=(\sqrt{x}-1)^{2}(\frac{(\sqrt{x}-1)(7x\sqrt{x}+3x+9\sqrt{x}+5)}{4(x+1)^{2}(\sqrt{x}+1)^{2}}+\frac{2\sqrt{x}+1}{x})$
[b]chú ý rằng[/b] $\large(\frac{(\sqrt{x}-1)(7x\sqrt{x}+3x+9\sqrt{x}+5)}{4(x+1)^{2}(\sqrt{x}+1)^{2}}+\frac{2\sqrt{x}+1}{x})\ge 0$ ([b]do[/b] $x\leq1$)
[b]Từ đây suy ra được điều phải chứng minh.Dấu đẳng thức xảy ra khi và chỉ khi do[/b] $x=y=z=1$ [b]hay là[/b] $a=b=c$
[b]We are done[/b]  :P
[/quote]
[quote="Lời tâm tình"]
[b]Đây là bài toán anh sáng tác ra tặng em.Người mà anh đả thích ngay lần nói chuyện đầu tiên(có thể em không tin vào điều này,nhưng thực sự là như vậy).Từ lần nói chuyện đấu tiên với em,trong tim anh đả có hình bóng em.Người yeu ơi có biêt anh nhớ em nhiều lắm không,anh luôn thấy em trong những giấc mơ,anh nhớ mái tóc óng ánh của em,nhớ những lúc em cười mà làm anh thấy thích em hơn.Huyền ơi? hãy đón nhận tình cảm của anh nhé.Chỉ một lần thôi cũng được.
Tặng em bài toán này với mục đích chúc em học giỏi hơn môn toán để tham gia kỳ thi bất đẳng thức của tỉnh ta em nhé.  [/b]
[/quote]

[color=purple]Please don't post in ML by Vietnamese !
            You must click here  http://www.mathlinks.ro/Forum/viewtopic.php?t=138774
 [/color]

Let he say his desired    :roll:  he a young boy is alway like that  :lol:    I hope his lover can read this message  :lol:  If it is possible, I will translate these words  :P

Ok. Here is my solution to this problem. I think it is better.

Notice that
\[(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z})^{2}\ge \frac{3}{(1+x)(1+y)}+\frac{(1+y)(1+z)}+\frac{3}{(1+z)(1+x)}. \]
By AM-Gm, we deduce
\[\frac{3}{(1+x)(1+y)}+\frac{3(1+x)}{8}+\frac{3(1+y)}{8}\ge \frac{9}{4}\]
therefore
\[3\sum \frac{1}{(1+x)(1+y)}+\frac{3}{4}\sum x \ge \frac{9}{2}. \]
So we have done because $\sum x\ge 3$.

Nice for your lover but easy  :D


Hello!

I have the following problem, when writting in subscript of \lim, I can't write anything in more then one row, for expample;

$ \lim_{n\rightarrow \infty, n\geq n_0}$

I would like to put $ n\geq n_0$ in the row bellow $ n\rightarrow \infty$, but that it's all still in subscript of \lim! I tried the following;

\lim_{n\rightarrow \infty, \newline n\geq n_0} but it still turns up the same; $ \lim_{n\rightarrow \infty, \newline n\geq n_0}$

I also tried;

\lim_{n\rightarrow \infty, \\ n \geq n_0}, but with no change; $ \lim_{n\rightarrow \infty, \\ n \geq n_0}$   .

Thats about all I tried, cause I don't know \Latex that well.

Use \substack (making sure you have the amsmath package loaded) $ \lim_{\substack{n\rightarrow \infty\\ n \geq n_0}}$


Its started! You get 48 hours to send me your answers.

Wait, even though I'm nominated, can I still participate?

the existence of this topic is totally unnecessary. we are all able to check the thread by ourselves, thank you very much.

Alright the tiebreaker has started check the thread to see if you made it


[quote="t0rajir0u"]You mean find the sum

$5+55+555+...+\underbrace{...555...}_{n \text{ times}}$

?[/quote]
 5+55+555+5555+55555+555555,......... i mean write equation of sum(when n=1 sum will be 5 n=2 sum will be 60 ) :)


Hello

Can anyone help with this exercice ?

$f$ is an application :

$f: \mathbb{R^{*}}\to  \mathbb{R}-\{2\}$
$x \to \frac{2x-1}{x}$

1. Find $f([-1;2]-\{0\})$ and $f^{-}([1;2[)$
2. Show that $f$ is a bijective application and find its inverse application.

Thanks.

This might be a little too hard for high school basics.

So, what should I do ?
Shall I send a private message to a moderator to move this Topic ?

You could do that, though usually one of us will notice it and move it sooner or later  :roll:

Suppose a,b are from R*, a>b>0. This implies 1/a<1/b and
2-1/a>2-1/b. Accordingly to this a>b implies f(a)>f(b), so f is monotonous for a>0.
a<b<0 implies f(a)<f(b), so f is monotonous for a<0.

Let's proove that f is surjective. For every b in R\{2}, there exists a in R* which satisfy condition b=f(a).
If b=(2a-1)/a then a=1/(2-b) and F is surjective.

So f is bijective application and has inverse application f_. (Sorry for this ugly notation.)
f(a)=(2a-1)/a=2-1/a=b implies a=1/(2-b)=f_(b) wanted inverse application.

f(-1)=3,f(2)=3/2, left limes in a=0 is +∞  and right limes is -∞, so
f([-1,2])=R\(3/2;3).
f_(1)=1,f_(2)=+∞ , so f_([1,2])=[1,+∞ ). :wink:


calculus
integration
function
logarithms
real analysis
real analysis unsolved

$ \
\begin{array}{l}
 Let\,m \in R\,and\,function\,f: [0,2] \to R,\,f(x) \equal{} \{ \frac{{x \minus{} m,\,x \in [0,1]}}{{x\ln x,x \in (1,2]}} \\
 c)\,For\,m \equal{} 1,\,to\,be\,prooved\,that,\,for\,every\,t \in (0,2)\,exists\,a,b\, \in [0,2],\,a \ne b,\,\,so\,that \\
 \int\limits_a^b {f(x)dx \equal{} (b \minus{} a)f(t)}  \\
 \end{array}
$

this is just the statement that the function takes on its average value.  For m=1 the function is continuous.  Note that the function is also strictly increasing.

I am not sure how to prove this rigorously, but the idea is going to be that $ a<t$ and $ t<b$ (always possible, since $ t$ is in an open interval) and thus $ f(a) < f(t) < f(b)$, so if we choose $ a$ close to $ t$ we can move $ b$ as close or as far as necessary for $ f(t)$ to be the average of the function on $ [a,b]$ (making b greater increases the average since f is strictly increasing; clearly $ 1/(t-a)\int_a^t f(x)\,dx$ < f(t)$ and$1/(b-t)\int_t^b f(x)\dx$ > f(t)$ so some choice of $ a$ and $ b$ must work).


calculus
trigonometry
algebra
polynomial
function
integration
complex numbers

Find the general solution of this second order differential equation:

$\frac{d^{2}y}{dx^{2}}+25y = 2e^{(1+i)x}$

[hide="never taken any differential calculus course but I will try to mess around with this:"]
The solution is: $y=\frac{e}{12}\cos(x)+\frac{e*i}{12}\sin(x)$
Which is easy to easy when you note the following:
$y''+(25)y=2e^{(1+i)x}=2e*e^{ix}=2e*\cis(x)=2e\cos(x)+2ei\sin(x)$
And now it becomes clear that $y=w\cos(x)+z\sin(x)$ because it is still nice when you take the second derivative. So now you just plug in $y$ and solve for $w,z$. And I think constant term ($ax+b$) must be 0.[/hide]

[hide="Look here, me@home"]$e^{(1+i)x}=e^{x}\cdot e^{ix}=e^{x}\cos x+ie^{x}\sin x$

You've made some algebra errors, and in general your approach does not appear to be one that would lead to the solution.[/hide]

This is a linear constant coefficient ODE, nonhomogeneous but the non-homogeneity belongs to the "exponential-polynomial class" of functions that could be roots of some constant coefficient linear differential operator. For such equations, one standard technique is [b]undetermined coefficents.[/b]

The characteristic equation is $r^{2}+25=0,$ which has roots $\pm5i.$ This tells you what the general solution of the corresponding homogeneous equation is. Since $1+i\ne\pm5i,$ no coincidence of exponents has occurred and we may proceed simply. We are looking for a particular solution. Assume that the particular solution has the form

$y=Ae^{(1+i)x}$

Then $y''=2iAe^{(1+i)x}.$ Put this in the differential equation:

$2iAe^{(1+i)x}+25Ae^{(1+i)x}=2e^{(1+i)x}$

Divide by the exponential, which is never zero:

$2iA+25A=2$

$A=\frac{2}{2i+25}=\frac{50-4i}{629}$

We have our particular solution; for the general solution we must add the solutions to the corresponding homogeneous equation. The general solution is:

$y=\frac{50-4i}{629}\cdot e^{(1+i)x}+c_{1}e^{5ix}+c_{2}e^{-5ix}.$

This can also be expressed using sines and cosines; it cannot, however, be expressed entirely in real form because the "input" function (the nonhomogeneity) was not real.

If in fact, we really wanted to solve

$y''+25y=2e^{x}\cos x,$

then take the real part of my answer. (This will take some work.)

http://www.mathlinks.ro/Forum/viewtopic.php?t=98305
Here I posted the general way to find particular solutions. I think mostly it's easier to use this than guessing.

Kalle, what you posted is essentially the method known as "variation of parameters." When I teach an ODE course, I do teach that method as well, and it does apply to a somewhat broader range of cases than undetermined coefficents. Nonetheless, I find that undetermined coefficients, when it works, is usually quicker (assuming pencil-and-paper solutions) and less prone to computational error.

You can call undetermined coefficients a "guess," but it's a very structured and predictable guess, if that. You don't have to invent a new idea each time.

I used the method of finding the Particular integral and complementary fundtion separately. I got:

$y = A\cos{(5x+\epsilon)}+\frac{50-4i}{629}e^{(1+i)x}$

Is it the same?  :)

Yes, that's the same. You could also have written the complementary solution as $A\cos(5x)+B\sin(5x).$

One word of warning: we can't count on $A$ and $B$ necessarily being real.

[quote="Kent Merryfield"]
You can call undetermined coefficients a "guess," but it's a very structured and predictable guess, if that. You don't have to invent a new idea each time.[/quote]
It depends on how long it was since the last time you had a course where linear ODEs were important :p

I'm not really one to say anything on the subject; I barely know anything about ODEs. I've never taken a course treating DEs in a systematic way. However, in my opinion just stating the integral without evaluation is enough  to give a particular solution. I realize this wouldn't net as many points on an exam, but I personally think solutions to linear equations are pretty uninteresting.

To go off on a tangent, is there some similar 'expression' that will yield solutions to linear difference equations?


The template is not working for me anymore!  :(

I've put everything in the right folder, but when I run it, it says that it doesn't find any \citation commands, \bibdata commands, or \bibstyle commands... Whats going on?

Hurm - is anyone else seeing this?  We don't use any of that for the template.  Try putting everything you need in a fresh folder.


modular arithmetic
number theory
relatively prime

hi there!
pls help me with this two basic problems.

1.) What time is 3hrs and 17mins before 2:07am? my answer is 10:50pm, but is there any technique in solving this? i solved it by counting from 2:07am and so on...

2.) The telephone and the doorbell ring together at 9:15am. The telephone rings every 12mins and the doorbell rings every 17miins. what time will they both ring together?
  i really dont have any idea of solving this. pls help\
thx in advance!  :P

These can be solved using the ideas of [url=http://en.wikipedia.org/wiki/Modular_arithmatic]modular arithmetic[/url].

[hide="Number 1"]I don't know what you mean by "is there any technique in solving this" – there's nothing to "solve" in this problem, you just go back 17 minutes and then 3 hours.[/hide]
[hide="Number 2"]Say the time they ring together is time $ 0$.  The telephone will ring at times $ 0, 12, 24, 36, \ldots$ and the doorbell will ring at times $ 0, 17, 34, 51, 68, \ldots$.  In other words, the telephone rings on multiples of $ 12$ and the doorbell rings on multiples of $ 17$.  Because $ 12$ and $ 17$ are relatively prime, the smallest positive integer that is a multiple of both is $ 12 \times 17 \equal{} \boxed{204}$.[/hide]

For (2), the question was actually what time, so you would need to add $ 204$ minutes to $ 9: 15$.  $ 204$ minutes is $ 3$ hours and $ 24$ minutes, so your time would be $ \boxed{12: 39}$.

[hide="Number 1"]
You can use some ideas of modular arithmetic.
$ 2\cdot 60\plus{}7\minus{}3\cdot 60\minus{}17\equal{}\minus{}70\equiv 50\pmod{60}$.
$ 2\minus{}(3\minus{}x)\equiv 11\minus{}x\pmod{12}$, where $ x$ is some strange quantity representing minutes. This means we're at an hour less than 11, so 10.

Putting it together, 10:50.
[/hide]


group theory
superior algebra
superior algebra unsolved

In alternating group ..how they are restricting their elements from symmetric group...
 that is S4 has 24 elements..but A4 has 12 elements ...out of 24 elements how pick 12 elements...under what condition we are restricting the elements of symmetric to alternating..


f v=18i + 450j and w=3i + 2j, then what is v*w

I dont even have a clue on this one, can someone please show me the correct answer and how to get it?  Thank you very much I appreciate it.. :)

If the asterisk between $ v$ and $ w$ stands for the [url=http://en.wikipedia.org/wiki/Dot_product]dot product[/url], then wikipedia has all you need.

im pretty sure that it stands for muliplication....which i don't know how to do.....or dot product stuff for that matter....sad i know, but its true, i suck at math...all math :ninja:

There is no such operation as "multiplication" of vectors. There are many operations- the dot product, the pointwise product, the complex product. mlok has merely suggested the most likely alternative.
Go read the relevant section of your textbook, where this problem came from. There should be a definition there.

you both are right, that just goes to show what i really know, which is nothing.........look what I did, I worked it out on my own, can someone tell me if its right and if not what I did wrong......I appreciate your help so much   :)

180(3)+450(2)=540+900=1440

am i right, am i right?????

(1) This is the dot product.

(2) You have an $ 18$ in your first post and a $ 180$ in your last post. No more than one of those can be correct - I assume one of them is correct and one is a typo, but I'm not going to guess which.

If you do mean the dot product, and if you clean up the typo, then it would be correct.

thank you very much Kent Merryfield.....its 180  :D


Let $a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e+f=6$. Prove that:
\[A\ge B\]
In here:
\[A=(a^{2}+1)(b^{2}+1)(c^{2}+1)(d^{2}+1)(e^{2}+1)(f^{2}+1)\]
\[B=(a+1)(b+1)(c+1)(d+1)(e+1)(f+1)\]
If you want see more, you should conect to [url]http://bdt.ch06k.com/[/url]  :P

hix I posted it for 2 week, but I don't receive any solution  :roll: How to solve it? Well-come everybody


search
combinatorics proposed
combinatorics

We have a n-gon such that no three diagonals intersect at the same point. Find the number of the points of intersections of the diagonals.

Sounds like a staple but it's a good introductory problem.

Do you mean a convex $ n$-gon?
Did I miss something or it is $ \binom{n}{4}$ since any intersection point is defined by the intersection of two diagonals which are determined by $ 4$ vertices. And, conversely, any choice of $ 4$ vertices gives exactly one intersection point.

Pierre.

I'm afraid it's wrong. It is a convex polygon with n vertices and no three diagonals intersect in a single point. Do you want me to post the answer (not the solution) ?

It surely is not wrong.  It has also been posted recently: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1582761869&t=212526

I have another solution. Maybe we talk about the same thing.
This is the solution I'm talking about:
[hide]
We number the vertices $ 0$ to $ N\minus{}1$. Let's take a diagonal from vertex $ 0$ to vertex $ i$. Vertices $ 1$ through $ i\minus{}1$ are on the one side of the diagonal and vertices $ i\plus{}1$ through $ N\minus{}1$ on the other. The diagonal intersects only those diagonals which have one vertex on one side and the other on the other side. So, there are $ (i\minus{}1)(N\minus{}1\minus{}i)$ diagonals intersecting it.
We count all diagonals with one vertex being 0 and multiply the number of diagonals by $ N$. We counted each intersection on one diagonal twice and each intersection four times. So we divide the result by 4.
[/hide]

Is it wrong or are we talking about the same thing?

$ \sum_{i \equal{} 1}^{n \minus{} 1}(i \minus{} 1)(n \minus{} 1 \minus{}i) \equal{} \frac{(n \minus{} 1)(n \minus{} 2)(n \minus{} 3)}{6}$, so when you multiply by $ n$ and divide by 4 you get exactly $ \binom{n}{4}$.

Oh, I should have made more calculations  :blush: Your solution is much better and much easier than mine  :blush:  :blush:


how did you get so good at math?

are those art of problem solving books really good? if I read one and understood all the concepts, would I score perfect on the AMC 10 or AMC12?

I'm not good lol.  :maybe:

Basically, it is practice combined with intuitive thought and motivation.

99% perspiration, 1% inspiration

I didn't read any books...it's largely just interest.  I got much better just doing problems here and learning from that...I suggest you start there, because books are never a fun way to learn.  Even when I try to learn from a book, I end up just perusing the book for problems to do...


That is the title of the chapter. And I need help FAST, because Mom is riding my heels, and I don't have much time.

 I will post five questions here. I will need help with them all, and PLEASE do not just solve them. I am a student, and I need to learn how to do this crap we call numbers. If you would please try to explain to me /how/ I solve the problem, it would be much appreciated.

 And yea, forgive me if I am rude. I just broke my toe a few days back, and I am unable to ride my horse, OR do anything else productive, so I am kinda in a bad mood.  :wallbash_red:

 Sorry. Help?  :bomb:

And... I didn't know that it made me wait ten minutes... -Smashes computer-

 Sorry 'bout that.

 Solve each equation. Check your solution.

 4. $ (x\plus{}1)(x\minus{}7)\equal{}0$

 6. $ (2m\minus{}1)(m\equal{}4)\equal{}0$

 8. $ x^2\equal{}x$

 10.  The height of a triangle measures 5 centimeters more then its base. The area of the triangle is 18 square centimeters. Find the measures of the base and the height of the triangle.

 12. $ 5z(z\plus{}8)\minus{}0$

 Behold, the ebil problems. Sorry for the wait.

For number 4, I'm assuming that you meant (m - 4).

Use the following logic: if two terms multiplied together equal zero, then either one or 2 of them are equal to 0. However, if they are both 0, then that means the 2 factors are equivalent.

I don't see how this is factoring, the factoring has already been done for 4, 6, 8,  and 12.

Now all we have to do is use my first paragraph to set up equations.

4. x+1=0
$ \boxed{x=-1}$
OR
x-7=0
$ \boxed{x=7}$

6. 2m-1=0
2m=1
$ \boxed{m=\frac{1}{2}}$
OR
m-4=0
$ \boxed{m=4}$

8. $ x^2=x$
$ x^2 - x=0$
$ x(x-1)=0$
$ \boxed{x=0}$
x-1=0
${ \boxed{x=1}}$

10. Hopefully, you recall how to find the area of a triangle We can set up an equation in terms of the base.

$ \frac{b(b+5)}{2}=18$
$ b(b+5)=36$

At this point, there are 2 points to do this.

Method one:
List out the factors of 36.
Aha! 9 and 4 have a difference of 5!
Therefore,
$ \boxed{b=4\text{ and }h=9}$

Method Two:
Use the distributive property to multiply it out and then subtract 36 from both sides (we want standard form, or 0 on one side), getting
$ b^2 + 5b - 36=0$
Note that this factors to
$ (b+9)(b-4)=0$
However, you obviously can't have a negative side length, so b=4.
You know that b+5=h, so h=9.
$ \boxed{b=4\text{ and }h=9}$

12. I'm assuming you meant 5z(z+8)=0.
Use the same method for 4,6,8
Solving,
z=0,-8

and btw, if u look at your post, at the bottom there is a thing that says EDIT. If you click on that, you can edit your text without waiting 10 minutes. Just let other people know if yo;u've changed something impportant that people are discussing.  :)

[quote="Verridith"]-Prints-[/quote]

Whoa there, if this is homework, please refrain from directly copying someone else's solution (this includes printing). You can use the general methods, but it is plagiarizing someone else's work.


Which is better? DS or PSP?

i think they are both the same...

I like the gameboy series, but I don't really like the ds. Oh wow two screens... it only makes it bulkier. I like sp, because it is so [b]small[/b]. However, playing mario 64 is cool.

I have also played psp, and I have to say that it is awesome. So there. (one of my friends has both of them; I tried them out at his house).

psp x1000000000000000000000
not only can it do games, but it can do movies, music and can browse the internet. i really want one :(

For the PSP:

I'm ok with the games and music part, but who would want to watch a DVD or browse the internet on that little screen when you can do the same with other electronics with bigger screens.  Plus, you have to buy a different type of DVD because a normal DVD obviously won't fit.

For DS:

I think it's pretty cool, but I don't have one and have only played it twice.  :(

http://www.supercard.cn/eng
I think that's the website. Basically, it lets the DS to everything with media the PSP can. Of course, you'd never use it to pirate games as well as using it for legal music, ebooks, pictures, movies...so, I'd recommend the DS. Look into Phoenix Wright: Ace Attorney for a fun brain-teaser game, and Trauma Center: Under the Knife for a fun action game. :stretcher:

Something you might find interesting:

[url=http://www.gamespot.com/features/6140988/index.html]What do the critics say?[/url]

The basic gist of the article is, PSP is better if you want a sort of portable version of PS2, DS is better if you want unique games that don't exist anywhere else.

I'd rather have the PSP. I played one at my aunt's house and I played a DS at my friend's house. PSP>DS.

Oh, by the way, this was just announced:

[url=http://www.gamespot.com/news/6143136.html]A smaller version of the DS.[/url]

PSP is by far the better one. Thats not to say i will get it, but it is better. The DS is unique, yes, but that does not make it good. PSP has better graphics, and multifunctionality.

DS. MUCH better games, plus the battery life lasts for more than a few hours.

definitely psp, smaller and more stylish

Here in India, I think it will be quite some time before either of the two enter the mainstream market....Most of us here are with either the PC or the PS2 as far as games are concerned.

[quote="ToBe"]PSP has multifunctionality.[/quote]

Read my post.

PSP is too expensive, plus Nintendo games are better.

DS.


There are only three symbols in a concrete alphabet. How many 6 letter words we can make if there can't be three same letters next to each other?

[hide="someone should check"]
$ 3^6$ total, but subtract those with a group of 3 (or more) same letters. Casework size of "same letter group"
6:$ 3$
5:$ \dbinom 21 \cdot 3 \cdot 2 \equal{} 12$
4:$ \dbinom 31 \cdot 3 \cdot 2 \cdot 3 \equal{} 54$
3:$ \frac{\dbinom 41 \cdot 3 \cdot 2 \cdot 3 \cdot 3}{2} \equal{} 108$

So $ 729 \minus{} 177 \equal{} 552$ I think.[/hide]

[quote="hurdler"]3:$ \frac {\dbinom 41 \cdot 3 \cdot 2 \cdot 3 \cdot 3}{2} \equal{} 108$[/quote]  I don't understand what your arithmetic corresponds to in terms of counting (and the number you get is wrong).  In fact, your case 4 is wrong, too: when we have abbbbc, there are three choices for b and two each for a, c (so 12 total) but your solution suggests that there are 18.

Other possible approaches are with inclusion-exclusion or a recursive approach.

[hide="Casework"]
We're going to find the number of words with three letters next to each other:
6 consecutive letters: 3 (repeated letter) = [b]3[/b]
5 consecutive letters: 3 (repeated letter) * 2 (location of block) * 2 (other letter) = [b]12[/b]
4 consecutive letters: Two subcases. Case 1: AAAAXX (at end); 3 (repeated letter) * 2 (location of block) * 2 * 3 (other letters) = [b]36[/b]. Case 2: XAAAAX (in middle); 3 (repeated letter) * 2 * 2 (other letters) = [b]12[/b]
3 consecutive letters (only one block): Two subcases. Case 1: AAAXXX (at end); 3 (repeated letter) * 2 (location of block) * 2 * 3 * 3 (other letters) = [b]108[/b]. Case 2: XXAAAX (in middle); 3 (repeated letter) * 2 (location of block) * 2 * 2 *3 (other letters) = [b]72[/b]
3 consecutive letters (two blocks; AAABBB): 3 (repeated letter) * 2 (other repeated letter) = 6. Each of these has been counted twice already (3 consecutive letters, subcase 1), so this actually contributed [b]-6[/b]

Total words with three or more consecutive letters is sum of bolded numbers: 3+12+36+12+108+72-6=[b]237[/b]

Hence the number of acceptable 6 letter words is 729 - 237 = [b]492[/b]
[/hide]

Oops, thanks. What I was trying to do  was look at the group of 3,4,5 or 6 same letters (depending on which case) as a unit - lets call this group of letters U, and the number of ways to U along with the other letters (where we assume the other letters are indistinguishable for a second) would be $ \dbinom {6 \minus{} c \plus{} 1}{1}$. We multiply each case by three since that U can be 3 different letters. Then I messed up; I tried to make it so the letters adjacent to U are different from the single letter that U consists of. I think I need to do subcases if I do it this way. Then I divided by 2 in one of cases, which I now realize was another stupid mistake.

[hide="ok attempt at recursion"]
let $ a_n$ be the answer for $ n$ letters.

$ a_3 \equal{} 27 \minus{} 3$
$ a_4 \equal{} 3a_3 \minus{} 2 \cdot 3 \equal{} 66$
$ a_5 \equal{} 3a_4 \minus{} 2 \cdot 3 \cdot 3 \equal{} 180$
$ a_6 \equal{}3a_5 \minus{} (2 \cdot 3 \cdot 3 \minus{} 2) \cdot 3 \equal{} 492$[/hide]

[hide="attempt to fix my first solution"]
for case 4, it should be 18 + 18 + 12 = 48
for case 3, it should be 3x2x3x3 + 2x3x2x3 + 3x2x3x2 + 3x3x2x3 = 168
so it should be 498? hmmm I am still doing something wrong in at least one of my solutions[/hide]

The natural way to solve this problem recursively is to define more than one sequence.

[hide="Discussion"] Let the three symbols be $ A, B, C$.  Let $ x_n$ be the number of sequences ending in $ AA$ (by symmetry, also equal to the number of sequences ending in $ BB$, $ CC$).  Let $ y_n$ be the number of sequences ending in $ AB$ (by symmetry, also equal to the number of sequences ending in $ BA, AC, CA, BC, CB$).  Then

$ x_n \equal{} 2y_{n\minus{}1}$

because the only sequences ending in $ AA$ arise from a sequence ending in $ BA$ or $ CA$, and

$ y_n \equal{} x_{n\minus{}1} \plus{} 2y_{n\minus{}1}$

because a sequence ending in $ AB$ can arise from $ AA, BA$, or $ CA$.  Now pick your favorite method to solve the resulting system. [/hide]


greatest common divisor
number theory proposed
number theory

x,y,z -----naturals numbers.
h-greatest common divisor of x,y and z
To proof,that
a) hxyz
b)h(y-x)
are squares of natural numbers.

For the first look at the exponents of the primes, the second is just wrong.


Along a circle are written $ n$ numbers each being $ \plus{} 1$ or $ \minus{}1$.Our aim is to find the product of all $ n$ numbers.In one question we can find the product of $ k$ successive numbers $ a_i...a_{i \plus{} k \minus{} 1}$.Here $ a_{n \plus{} 1} \equal{} a_1$ and so on. How many questions $ q(n,k)$ are necessary to find the product?

[quote="Xaenir"]Along a circle are written $ n$ numbers each being $ \plus{} 1$ or $ \minus{} 1$.Our aim is to find the product of all $ n$ numbers.In one question we can find the product of $ k$ successive numbers $ a_i...a_{i \plus{} k \minus{} 1}$.Here $ a_{n \plus{} 1} \equal{} a_1$ and so on. How many questions $ q(n,k)$ are necessary to find the product?[/quote]

One question : $ q(1,n)$

But I surely misunderstood something  :roll:

Probably n and k are given constants. If so, this is just a generalization of http://www.mathlinks.ro/viewtopic.php?p=468325#468325

If not, pco did the problem :D

If the problem is general, for a given couple $ (n,k)$, the answer may be "impossible" in some situations : for example, for $ (2n \plus{} 1,2p)$, it is clearly impossible to know the product .


parameterization
calculus
calculus computations

Solve the following initial value problem

$ xy^2y'\equal{}y\plus{}1$ , $ y(3e^2)\equal{}2$

can this problem be solved by any elementary way?  I try variation of parameter and integrating factor but none seems to work. Thanks

[hide]
$ \frac {y^2.dy}{y\plus{}1}\equal{}\frac {dx}{x}$
$ \frac {((y\plus{}1)^2\minus{}2(y\plus{}1)\plus{}1)dy}{(y\plus{}1)}\equal{}\frac {dx}{x}$
$ ((y\plus{}1)\minus{}2\plus{}\frac {1}{y\plus{}1})dy\equal{}\frac {dx}{x}$
$ \frac {1}{2}y^2\minus{}y\plus{}ln(y\plus{}1)\equal{}ln(x)\plus{}C$
The value of $ C$ can be found by the second data in the problem.
[/hide]

OMG , i miss the easiest method ...  :blush:

Thanks


geometry trigonometry ARML probability inequalities

To be more specific, What is you worst area when it comes to math contests?  I would say that mine is Geometry.  I know that some areas are wider (in HS competition) than others but just try to be general.

I am the master of all math!  :D
I'm pretty much at the same level in all areas.

See, my problem is that when it comes to geom, I want to straight away use the trig. It used to be that I would straight away go for analytic, but I really dont like analytic, so now even when I can see straight off that analytic is a very good way to solve a problem, I still try to plow through a trig solution or attempt (not very successfully usually) to find a euclidean way.

Geometry all the way.  :-(

Edit: Though part of that may be that I just don't like geometry, so I usually get bored halfway through the question and go on to the next one.

I rock at algebra but I can't do anything else =/

Reminds me of the Texas ARML team in the early 90's.  We were given questionnaires before ARML where we rated our strengths from 0 to 4 in each area.  You always knew the rookies as the ones who put 4 for geometry and the second-years as the ones who put 0..

geometry, geometry, geometry.. seems I'm not alone.

Classical euclidean geometry.  I do alright with analytic geometry.

I can't count/do probablity, I always confuse things where order matters with things where order of items doesn't matter

Actually, my weakest area is not making careless mistakes.

Aside from careless errors, which have hurt me so many times in math, my weakest area is geometry. That seems to be the case for most people, though it's interesting for me because geometry happens to be my favorite area of math. It only goes to show that passion doesn't necessarily mean skill.

I wonder why it is that so many people find their geometry skills weak. Is it American education or is geometry just more difficult to get than other areas of math or what?

I think the weakness in geometry comes from lousy American education: too little geometry too late in the curriculum.

I'm bad a geometry mainly because i haven't taken it yet.  I know enough to do okay with it on the SATs or ACTs, but can barely do any of it on the AMCs.

[quote="yif man12"]Reading[/quote]

that belongs to careless mistakes

my weakest area is Geometry. No doubt if you'd know all open questions I've ever missed in our national olympiad have something to do with geometry, plus it's the area I solved fewest IMO and MathLinks problems from.

[quote="questioner"]no. thm.  is my weakest. ;)

some days ago i have started a lesson about it. i have not enough practice.[/quote]

Peter, I think thm refers to theorem, but what's no. thm?!?!

Moreover, questioner, are you one of the students attending the imo training session? "I have started a lesson about it". What lesson are you referring to?

[quote="bugzpodder"]soarer, no. thm of course is number theory, i would think[/quote]

jack, for the imo training session in hk, we had our no. theory lessons a month ago... so... i doubt he meant no. theory theorems.

Undoubtedly geometry - that's what I like least

I hate probability - well - in general applied math...  :cool:

But when I think about math contests - number theory...


Let $ n$ be positive integer.Find all natural number $ m$ such that ineq : $ \frac1{a^n}\plus{}\frac1{b^n}\geq a^m\plus{}b^m$ true all positive real number $ a,b$ such tha: $ a\plus{}b\equal{}2$

Why not answer?
We have answer $ m\equal{}0,1,..,n\plus{}1$.Easy to prove ineq true with $ m\equal{}0,1,2..,n\plus{}1$
But I can't problem that ineq not true with $ m\equal{}n\plus{}2$
can anyone help me?


Find all the real numbers $ \alpha$ satisfy the following property: for any positive integer $ n$ there exists an integer $ m$ such that $ \left |\alpha\minus{}\frac{m}{n}\right|<\frac{1}{3n}$.

If $ \alpha$ is  an  integer,  then for any positive integer $ n$ there  exists an integer $ m \equal{} \alpha n$ such that $ | \alpha \minus{} m| < \frac {1}{3n}$;

If $ \alpha$ isn't  an  integer,  there  exists  an  positive  integer  $ n$  such that $ | \alpha \minus{} m|  \geq  \frac {1}{3n}$  for any integer  $ m$  .


Let $ a,b,c$ be the side lengths of a triangle such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
\[ \frac{bc}{a^2\plus{}1}\plus{}\frac{ca}{b^2\plus{}1}\plus{}\frac{ab}{c^2\plus{}1} \ge \frac{3}{2}.\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be the side lengths of a triangle such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
\[ \frac {bc}{a^2 \plus{} 1} \plus{} \frac {ca}{b^2 \plus{} 1} \plus{} \frac {ab}{c^2 \plus{} 1} \ge \frac {3}{2}.
\]
:)[/quote]
If $ a \equal{} y \plus{} z,$ $ b \equal{} x \plus{} z$ and $ c \equal{} x \plus{} y$ then your inequality ( if I am not wrong  :D ) is equivalent to
$ \sum_{sym}(40x^5y \minus{} 25x^4y^2 \minus{} 15x^3y^3 \plus{} 25x^4yz \minus{} 18x^3y^2z \minus{} 7x^2y^2z^2)\geq0,$ which obvious.


conics
parabola
analytic geometry
graphing lines
slope

I'm sure you all know that light rays coming towards the open end of a parabola that are orthogonal to the directrix are all reflected at the focus of the parabola...

Using the fact that a parabola is defined as the locus of points equidistant from a point and a line (focus and directrix), prove this reflective property.

nobody?  :(

hmm... well I guess it requires some physics info (I got bored during physics and thought of it) so some things to use:

The angle a ray of light makes with a planar mirror as it comes into the mirror is the same as it makes after it is bounced off.

If a ray of light hits a curved mirror, all that it matters is the slope of the curve at the point it hits.

I think that should be enough.

you can do it the "brute force" way, it requires a bit of calculus.

basically, place the parabola in the xy plane, choose some lines, perferrably x=Constant or y=Constant. find out where these lines interset the parabola. find out the tangent of the parabola at that point, and then find the lines representing the reflected rays using the fact that the angles of incidence= angles of reflectsion....(hint: use some tan identity to find out the slope of the reflected rays)... then find out the intersecting of the x or y-axis of these new lines (depending on how to place the parabolas).


geometry
Princeton
college
Harvard
IMO
IMO 2006

The layout got messed up, but anyways...

so china is 37 35 31 34 35 42 with 6 gold!

Thanks for keeping us up to speed.

On that PDF what does China : 214 6 0 0 0 6 mean? (on page 11)
I'm having a hard time figuring it out exactly.

Arne does this mean no Belgian medals this year?

It seems Germany has become by far the strongest West European nation, they even beat USA?

it means their total is 214 and they got 6 gold.

Why in the Imo rezults 5.pdf you have a wrong medal written? I definetely know Ukraine have 1 gold,2 silver and 2 bronze. You can see it on the official sight of the IMO 2006.

IMO 2006 Gold Medalist: I have coloured those Mathlinker that I know. If I missed anyone, please tell me.

People's Republic of China Zhiyu Liu 7 7 7 7 7 7 42 Gold
[color=blue]Republic of Moldova Iurie Boreico 7 7 7 7 7 7 42 Gold   [/color]
Russian Federation Alexander Magazinov 7 7 7 7 7 7 42 Gold
People's Republic of China Caili Shen 7 7 7 7 7 2 37 Gold
Poland Przemysław Mazur 7 5 7 7 3 7 36 Gold
[color=blue]People's Republic of China Long Jin 7 7 7 7 7 0 35 Gold   [/color]
People's Republic of China Yu Deng 7 7 7 7 7 0 35 Gold
[color=blue]Germany Peter Scholze 7 7 0 7 7 7 35 Gold   [/color]
Republic of Korea Seok Hyeong Lee 7 7 7 7 7 0 35 Gold
Russian Federation Rostislav Devyatov 7 7 0 7 7 7 35 Gold
People's Republic of China Qingchun Ren 7 7 0 6 7 7 34 Gold
[color=blue]Romania Adrian-Ioan Zahariuc 7 5 7 7 7 1 34 Gold [/color]
People's Republic of China Wenying Gan 7 7 7 7 3 0 31 Gold
Japan Yuta Ohashi 7 7 7 7 3 0 31 Gold
[color=blue]Romania Andrei-Bogdan Ungureanu 7 7 7 7 3 0 31 Gold[/color]
France Ilia Smilga 7 7 4 5 0 7 30 Gold
[color=blue]Germany Darij Grinberg 7 1 7 7 7 1 30 Gold   [/color]
Italy Maria Colombo 7 7 0 7 7 2 30 Gold
[color=blue]Slovakia Ondrej Budáč 7 7 1 7 7 1 30 Gold[/color]
[color=blue]United States of America Arnav Tripathy 7 7 2 7 7 0 30 Gold [/color]
[color=blue]Germany Christian Sattler 7 7 1 7 7 0 29 Gold   [/color]
Japan Masaki Watanabe 7 7 0 7 7 1 29 Gold
Republic of Korea Ju Gang Nam 7 7 7 7 1 0 29 Gold
Republic of Korea In Hwan Choi 7 1 1 7 7 6 29 Gold
United States of America Zarathustra Brady 7 7 1 7 7 0 29 Gold
Viet Nam Mạnh Nguyễn Duy 7 1 7 7 7 0 29 Gold
Germany Georg Schönherr 7 7 0 7 7 0 28 Gold
[color=blue]Hong Kong Yun Pui Tsoi 7 7 1 6 7 0 28 Gold[/color]
[color=blue]Islamic Republic of Iran Nima Ahmadi Pour Anari 7 7 0 7 7 0 28 Gold[/color]
Islamic Republic of Iran Jaber Zarezadeh 7 7 0 7 7 0 28 Gold
Islamic Republic of Iran Naser Talebizadeh Sardari 7 7 0 6 7 1 28 Gold
[color=blue]Italy Simone Di Marino 7 7 0 7 7 0 28 Gold [/color]
Republic of Korea Whan Ghang 7 1 7 7 6 0 28 Gold
Mexico Pablo Soberón Bravo 7 7 0 7 7 0 28 Gold
[color=blue]Republic of Moldova Ivan Borsenco 7 6 7 7 1 0 28 Gold[/color]
Romania Eduard-Gabriel Băzăvan 7 7 0 7 7 0 28 Gold
Russian Federation Timofey Obraztsov 7 7 0 7 7 0 28 Gold
Switzerland Volodymyr Serbinenko 7 7 7 6 1 0 28 Gold
Taiwan Yi-Wei Chan 7 4 7 7 3 0 28 Gold
Thailand Phumpong Watanaprakornkul 7 7 0 7 7 0 28 Gold
Ukraine Nataliya Goncharuk 7 7 0 7 7 0 28 Gold
Viet Nam Hùng Hoàng Mạnh 7 1 7 6 7 0 28 Gold

you forgot christian sattler,adrian zahariuc,andrei ungureanu,ivan borsenco ,nima ahmadi pour etc :wink:

[quote="fredbel6"]Arne does this mean no Belgian medals this year?
[/quote]

We got one bronze medal and four honourable mentions...

[quote="Aleksandr Chuklin"]Why in the Imo rezults 5.pdf you have a wrong medal written? [/quote]

Well, there are a few (*very* few) small mistakes in that file, since it was a bit unclear whether silver would be on 19 or 20. I didn't type that file, by the way... I just put it online.

[quote="shobber"]
Italy Simone Di Marino 7 7 0 7 7 0 28 Gold
[/quote]

He's a Mathlinker too (Simo_the_wolf or something like that).

Ondrej Budac from Slovakia is a mathlinker as well (ondrob)

Excellent score for the Belgian team.
What surprises me is that Germany has three times gold plus that there is a non German west European with gold.  A Moldovan Mathlinker with a perfect paper, that is rather rare!

A few comments:

*Belgium certainly did better in the past three years.
-In 2003, we got a silver, a bronze, 3 HMs and we were 37th.
-In 2004, we got a silver, two bronzes, an HM and we were 42nd.
-In 2005, we got a silver, a bronze, 2 HMs and we were 40th.
-This year, we got a bronze, 4 HMs and we are 52nd.

...

*The German team was extremely strong this year. In fact, many people expected another perfect score from Peter Scholze, like last year, but he didn't manage to solve problem 3. Christian Sattler and Darij Grinberg were also doing their third IMO and got a gold medal, after winning silver medals in the past. This didn't really surprise me.

*Italy did a wonderful job, winning two gold medals, but that's also not that strange if you look at last years performance: they got two silvers and four bronze last year.

*I don't think anyone who knows how good Iurie Boreico is, has been surprised by the fact that he got full marks - again :P

Another question: is Yurie Boreico in university now? I saw him solve the undergraduate problems... Similar question for Darij and Scholze!

[quote="orl"][quote="darktreb"]Actually Gabriel did teach English in China during the past year from what I heard. I think Gabriel, Reid, and Brian Lawrence are all pretty fluent in Mandarin Chinese (I know Brian is for a fact as I was able to converse with him!)[/quote]

Very open-minded people. And what is Gabriel doing now?[/quote]

Honestly I have no idea. I assume math grad school (at least I would hope so!)

[quote="Lovasz"]Similar question for Darij and Scholze![/quote]

Darij is going to study at the München (LMU) University, he will start in october.

Princeton ;)

Any IMOers coming to take math classes with me next year and help lower my grades? :D

[quote="Arne"]
Indeed, that kind of things occurs quite often. Bruce Merry from South Africa, got nothing, gold, silver, silver and bronze respectively. That's strange - a possible reason is that he focused more on the IOI in his later years. Also, he became ill during IMO 2000 (his last IMO).[/quote]
I attended IMO 99 with Bruce and went through my IMO training with him and his ability definately didn't get worse since his Gold medal.  I think this kind of thing is probably more common in smaller countries.  Bruce is still the strongest IMO participant that SA has produced so when he got his Gold medal he was already stronger than anyone he was training with.  When most of your team is aiming for a bronze or a HM your training has got to have that focus.  In comparison the SA IOI team at the time had a few really top guys at the time who all probably helped each other improve.  I'm not criticising our IMO training program at all I just think its a reality of a smaller training program

I know that Iura is on MathLinks, but are any of the other two 42 boys in MathLinks??? What about the rest of the Chinese team? (I only know jin)

Maybe they only come to here as a guest,I think.

[quote="son of Skanderbeg"]Iura will be a senior in High School I think.
Maybe he will be able to reach the record of Ciprian Manulescu.

(He was admitted in Harvard when he was in his 3rd grade of high school)[/quote]

He has 1 IMO remainded?

Thanks Arne.

Performances at IMO:

IMO95:      ROM3      Ciprian Manolescu      7      7      7      7      7      7      42      Gold Medal
IMO96:      ROM4      Ciprian Manolescu      7      7      7      7      7      7      42      Gold Medal
IMO97:      ROM1      Ciprian Manolescu      7      7      7      7      7      7      42      Gold Medal

IMO03:      MDA1      Iurie Boreico      7      7      0      7      7      0      28      Silver Medal
IMO04:      MLD1      Iurie Boreico      7      4      2      7      7      7      34      Gold Medal
IMO05:      MDA1      Iurie Boreico      7      7      7      7      7      7      42      Gold Medal
IMO06:      MDA1      Iurie Boreico      7      7      7      7      7      7      42      Gold medal

[quote="José"]
IMO03:      MDA1      Iurie Boreico      7      7      0      7      7      0      28      Silver Medal
IMO04:      MLD1      Iurie Boreico      7      4      2      7      7      7      34      Gold Medal
IMO05:      MDA1      Iurie Boreico      7      7      7      7      7      7      42      Gold Medal
IMO06:      MDA1      Iurie Boreico      7      7      7      7      7      7      42      Gold medal[/quote]

 :o
 But how can he   participate one more year ?
4 times may be maximum .Do high schools in Moldova last for 4 years or not    :?:

Some people go before they even reach high school, or their high school goes through grade 13, or some such. I know one such person, Aleksandar Lishkov (Bulgaria) went 5 times. (He goes to Princeton now :D  :lol:  :D )

Oh,I didn't know that . :blush: I thought that IMO's  for high school students only.

Hey Gabriel Carroll seems to be working at the National Bureau of economic research. http://en.wikipedia.org/wiki/Gabriel_D._Carroll
Of course this might be vandalism.


Prove that $ (3\plus{}\sqrt{8})^n \plus{} (3\minus{}\sqrt{8})^n \minus{} 2$ is a perfect square.
Where $ n$ is an odd positive integer.

Let $ a \equal{} 1 \plus{} \sqrt{2}$.  Then $ a^2 \equal{} 3 \plus{} 2 \sqrt{2} \equal{} 3 \plus{} \sqrt{8}$ and $ \frac{1}{a^2} \equal{} 3 \minus{} \sqrt{8}$.  Now,

$ a^{2n} \plus{} \frac{1}{a^{2n}} \minus{} 2 \equal{} \left( a^n \minus{} \frac{1}{a^n} \right)^2$

so it remains to prove that $ a^n \minus{} \frac{1}{a^n}$ is an integer when $ n$ is odd.  But this follows straightforwardly from either binomial expansion or construction of the appropriate second-order linear recurrence.


Find the value of expression:

$ \frac{1}{\sqrt{2}\plus{}1}\plus{}\frac{1}{\sqrt{3}\plus{}\sqrt{2}}\plus{}\frac{1}{\sqrt{4}\plus{}\sqrt{3}}\plus{}...\plus{}\frac{1}{\sqrt{9}\plus{}\sqrt{8}}$

Can't really figure out how can I solve i? :huh:

Remember rationalising the denominator (think about $ a^2 - b^2 = (a + b)(a - b)$ as well)

[hide="Answer"]
${ \frac {1}{\sqrt {n} + \sqrt {n - 1}} = \frac {\sqrt {n} - \sqrt {n - 1}}{n - (n - 1)}} = \sqrt {n} - \sqrt {n - 1}$

Therefore the sum required is equivalent to

\begin{align*} \sum_{k = 2}^9 (\sqrt {n} - \sqrt {n - 1}) & = (\sqrt {9} - \sqrt {8}) + (\sqrt {8} - \sqrt {7}) + \cdots + (\sqrt {2} - 1) \\
& = \sqrt {9} + (\sqrt {8} - \sqrt {8}) + \cdots + (\sqrt{2} - \sqrt{2} ) - 1 \\
& = \sqrt {9} - 1 = 2 \end{align*}
[/hide]

Oh, yes,:idea:  I knew it's not that hard. Just couldn't figure out the method. Thanks a lot!


calculus integration trigonometry algebra polynomial logarithms function

Here's a computation of #1:

Note that $ \frac1{1 \plus{} x^2} \equal{} \int_0^{\infty}e^{ \minus{} (1 \plus{} x^2)t}\,dt \equal{} \int_0^{\infty}e^{ \minus{} t}e^{ \minus{} x^2t}\,dt$

Put that in the original integral and interchange the order of integration (justified by Tonelli - these are positive integrands) to get:

$ I \equal{} \int_0^{\infty}\frac {x^{1 \minus{} \alpha}}{1 \plus{} x^2}\,dx \equal{} \int_0^{\infty}\int_0^{\infty}x^{1 \minus{} \alpha}e^{ \minus{} t}e^{ \minus{} x^2t}\,dx\,dt.$

Make the substitution $ u \equal{} x^2t,\ x \equal{} u^{1/2}t^{ \minus{} 1/2},\ dx \equal{} \frac12u^{ \minus{} 1/2}t^{ \minus{} 1/2}\,dt$ to get

$ \int_0^{\infty}x^{1 \minus{} \alpha}e^{ \minus{} x^2t}\,dx \equal{} \frac12t^{\alpha/2 \minus{} 1}\int_0^{\infty}u^{ \minus{} \alpha/2}e^{ \minus{} u}\,du \equal{} \frac12t^{\alpha/2 \minus{} 1}\,\Gamma\left(1 \minus{} \frac {\alpha}2\right).$

Hence,

$ I \equal{} \frac12\,\Gamma\left(1 \minus{} \frac {\alpha}2\right)\int_0^{\infty}t^{\alpha/2 \minus{} 1}e^{ \minus{} t}\,dt$

$ \equal{} \quad\boxed{\frac12\,\Gamma\left(1 \minus{} \frac {\alpha}2\right)\Gamma\left(\frac {\alpha}2\right)}$

For $ \alpha \equal{} 1,$ this gives $ \frac12\,\Gamma\left(\frac12\right)^2 \equal{} \frac {\pi}2,$ which is the correct value, so that much checks. We can also see that $ 0 < \alpha < 2$ is required, else the integral diverges.

For #3, one possibility would be to expand the integrand as a geometric series.

Let $ I \equal{} \int_{0}^{\infty}\frac {x^{2}}{\left(x^{2} \plus{} a^{2}\right)^{2}\left(x^{2} \plus{} b^{2}\right)}dx$

Here i apply parts taking $ u \equal{} \frac {1}{ x^2 \plus{} b^2 }$ and $ v \equal{} \frac { \minus{} 1}{ 2(x^2 \plus{} a^2) }$

$ dv \equal{} \frac {x}{ ( x^2 \plus{} a^2 )^2 }$

$ I \equal{} uv \minus{} \int v du$

$ uv \equal{} \frac { \minus{} x}{ ( 2 ( x^2 \plus{} a^2) ) ( x^2 \plus{} b^2 ) }$  from 0 to $ \infty$ = 0

Let $ I_1 \equal{} \int v du \equal{} \int_{0}^{\infty }\frac { x^2 \plus{} b^2 \minus{} 2b^2 }{( 2 ( x^2 \plus{} a^2) ) ( x^2 \plus{} b^2 ) }$

$ I_1 \equal{} \int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 ) ( 2 ( x^2 \plus{} a^2) ) } \minus{} b^2\int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 )^{2} ( x^2 \plus{} a^2) }$

let $ I_2 \equal{} \int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 ) ( 2 ( x^2 \plus{} a^2) ) }$

This can be easily evaluated by writing the numerator as follows and spliting to two separate integrals

$ I_2 \equal{} \frac {1}{ a^2 \minus{} b^2 } \int_{0}^{\infty }\frac { (x^2 \plus{} a^2 ) \minus{} ( x^2 \plus{} b^2 )}{ (x^2 \plus{} b^2 ) ( 2 ( x^2 \plus{} a^2) ) }$

Let $ I_3 \equal{} b^2\int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 )^{2} ( x^2 \plus{} a^2) }$

Again write the numerator as follows and split the integrals .

$ I_3 \equal{} \frac { b^2}{b^2 \minus{} a^2 }\int_{0}^{\infty }\frac {(x^2 \plus{} b^2 ) \minus{} ( x^2 \plus{} a^2 )}{ (x^2 \plus{} b^2 )^{2} ( x^2 \plus{} a^2) }$

$ \equal{} \frac { b^2}{b^2 \minus{} a^2 } [ \int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 ) ( 2 ( x^2 \plus{} a^2) ) } \minus{} \int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 ) ^2 } ]$

both these integrals can be evaluated easily . the first one similar to $ I_2$ and the second one

can be evaluated by expanding the denominator to get a polynomial of 4th degree and in the

numerator make it as follows.

$ \int_{0}^{\infty }\frac {1}{ (x^2 \plus{} b^2 ) ^2 } \equal{} \int_{0}^{\infty }\frac {1}{ (x^4 \plus{} 2x^2b^2 \plus{} b^4 ) } \equal{} \frac{1}{ 2b^2 }\int_{0}^{\infty }\frac { (x^2 \plus{} b^2 ) \minus{} ( x^2 \minus{} b^2 ) )}{ (x^4 \plus{} 2x^2b^2 \plus{} b^4 ) }$

Split this into 2 integrals in the first one put $ x \minus{} b^2 / x \equal{} u$  and in the second one put $ x \plus{} b^2/x \equal{} v$  and integrate.

I know my solution is too big . so please post short solutions if any .

3)  $ \int_{0}^{2\pi}\frac {1}{a \plus{} \cos x}dx$

put $ t \equal{} \tan \frac {x}{2}$

so the integral becomes

 $ 2 \int_{0}^{\infty} \frac {dt}{ (a \minus{} 1) t^2 \plus{} (a \plus{} 1) ) }$

$ \equal{} \frac {2}{a \minus{} 1} \int_{0}^{\infty} \frac {dt}{ t^2 \plus{} \frac {(a \plus{} 1)}{a \minus{} 1} ) }$

$ \equal{} \frac {2}{\sqrt {a^2 \minus{} 1} } \arctan ( \frac {t \sqrt {a \minus{} 1} }{\sqrt {a \plus{} 1} }$ from 0 to $ \infty$

$ \equal{} \frac {\pi }{\sqrt {a^2 \minus{} 1} }$

Addendum to the computation for #1:

We can do this as a contour integral, using the residue theorem. Define $ z^{1 \minus{} \alpha}$ using a branch of the logarithm with a branch cut on the negative imaginary axis. Use a contour that has the real axis from $ \minus{} R$ to $ \minus{} \epsilon$ and $ \epsilon$ to $ R,$ plus a clockwise semicircle of radius $ \epsilon$ and a counterclockwise semicircle of radius $ R.$ For $ 0 < \alpha < 2,$ the contribution of both semicircles to the integral tends to zero as $ \epsilon\to0$ and $ R\to \infty,$ leaving us the integral over the real axis.

If we let $ I \equal{} \int_0^{\infty}\frac {x^{1 \minus{} \alpha}}{1 \plus{} x^2}\,dx,$ then the integral on $ (0,\infty)$ is $ I$ and the integral on $ ( \minus{} \infty,0)$ is $ e^{i\pi(1 \minus{} \alpha)}I.$

There's a lone simple pole at $ z \equal{} i,$ with residue $ \frac {e^{i\pi(1 \minus{} \alpha)/2}}{2i}.$ The integral is $ 2\pi i$ times that residue. Hence:

$ (1 \plus{} e^{i\pi(1 \minus{} \alpha)})I \equal{} \pi e^{i\pi(1 \minus{} \alpha)},$ or

$ I \equal{} \frac {\pi e^{i\pi(1 \minus{} \alpha)}}{1 \plus{} e^{i\pi(1 \minus{} \alpha)}} \equal{} \frac {\pi}{e^{ \minus{} i\pi(1 \minus{} \alpha)/2} \plus{} e^{i\pi(1 \minus{} \alpha)/2}} \equal{} \frac {\pi}{\cos\left(\frac {\pi(1 \minus{} \alpha)}2\right)}$

Both this and my previous computation are correct, which gives us a gamma function identity:
\[ \boxed{\Gamma\left(1 \minus{} \frac {\alpha}2\right)\Gamma\left(\frac {\alpha}2\right) \equal{} 2\pi\sec\left(\frac {\pi(1 \minus{} \alpha)}2\right)}\]

Alternately, for 2)
Write
$ \dfrac{x^2}{(x^2\plus{}a^2)^2(x^2\plus{}b^2)} \equal{} \dfrac{a^2}{(a^2\minus{}b^2)(x^2\plus{}a^2)^2}\plus{}\dfrac{b^2}{(a^2\minus{}b^2)^2(x^2\plus{}a^2)}$
                                                         $ \minus{} \dfrac{b^2}{(a^2\minus{}b^2)^2(x^2\plus{}b^2)}$
Therefore,
$ \int_0^\infty \dfrac{x^2\ \mathrm dx}{(x^2\plus{}a^2)^2(x^2\plus{}b^2)} \equal{} \dfrac{a^2}{a^2\minus{}b^2}I_1 \plus{} \dfrac{b^2}{(a^2\minus{}b^2)^2}I_2 \minus{} \dfrac{b^2}{(a^2\minus{}b^2)^2}I_3$
where
$ I_1 \equal{} \int_0^\infty \dfrac{\mathrm dx}{(x^2\plus{}a^2)^2} \equal{} \int_0^{\pi/2} \dfrac{a\sec^2\theta \ \mathrm d\theta}{(a^2(1\plus{}\tan^2\theta))^2}\equal{}\dfrac{1}{a^3}\int_0^{\pi/2} \cos^2\theta\ \mathrm d\theta\equal{}\dfrac{1}{2a^3}\cdot \dfrac{\pi}{2}$

and

$ I_2\equal{}\int_0^\infty \dfrac{\mathrm dx}{x^2\plus{}a^2} \equal{} \dfrac{1}{a}\left|\tan^{\minus{}1}\left(\dfrac{x}{a}\right)\right|_0^\infty\equal{}\dfrac{1}{a} \cdot \dfrac{\pi}{2}$

Similarly, $ I_3\equal{}\dfrac{1}{b}\cdot\dfrac{\pi}{2}$

Assimilating everything we get ultimately
$ I\equal{}\dfrac{\pi}{4a(a\plus{}b)^2}\qquad (a\neq b)$

Thank you Kent for the nice non-complex analysis solution to #1.
All of the three problems can be solved using residue theory [as in Kent's second solution to number 1].

Nice solution kaymant for number 2. If $ a \equal{} b$, we can use the substitution $ x \equal{} a \tan \theta$.

vijaymenon's solution to number three is the typical solution, but Kent suggested to use power series expansion [the condition $ a > 1$ satisfies the condition for the geometric series expansion].


I just want to know if these 2 proofs are valid.

1) $ |a+b| \leq |a|+|b|$

I begin by stating that

$ 2ab \leq |2ab|$

Since $ |a| \geq 0$, $ |b| \geq 0$ and $ 2 \geq 0$ while $ a$ and $ b$ alone $ \in \mathbf{R}$. Now consider $ a^{2}= |a^{2}|$ and $ b^{2}= |b^{2}|$. This gives us:

$ a^{2}+2ab+b^{2}\leq |a^{2}|+|2ab|+|b^{2}|$

$ (a+b)^{2}\leq (|a|+|b|)^{2}$

$ |a+b| \leq \parallel a|+|b\parallel$

Since absolute value is idempotent, we have:

$ \boxed{|a+b| \leq |a|+|b| }$

2) This is the inequality I'm more worried about, the triangle inequality.

$ |a-b| \leq |a-c|+|c-b|$ proof:

Square both sides to get:

$ (a-b)^{2}\leq (a-c)^{2}+2ac+2ab+2c^{2}+2bc+(c-b)^{2}$

$ a^{2}-2ab+b^{2}\leq a^{2}-2ac+c^{2}+2c^{2}+2ac+2ab+2bc+c^{2}-2bc+b^{2}$

$ -2ab \leq 2ab+4c^{2}$

$ -4ab \leq 4c^{2}$

$ -ab \leq c^{2}$

$ -ab \leq |c^{2}|$

since $ c^{2}\geq 0$ whereas $ a$ and $ b$ $ \in \mathbf{R}$, the inequality must hold.

For the last one in particular, I'm not sure how to factor two absolute value binomials (ex: $ |x-3| \times |x+4|$) so it is likely that the proof went wrong somewhere. Can anyone tell me if/where I went wrong for both absolute value inequalities?

[quote="bibobeo"]Why don't you apply $ (1)$ to $ (2)$?  :D[/quote]

I've been trying that for hours and no luck  :(

Set $ A=a-c$ and $ B=c-b$, then $ A+B=a-b$, which fits...

You can square everything and work backwards.

$ |a\parallel b|=|ab|\geq ab$

$ 2|a\parallel b| \geq 2ab$

$ |a|^{2}+2|a\parallel b|+|b|^{2}\geq a^{2}+2ab+b^{2}$

$ (|a|+|b|)^{2}\geq (a+b)^{2}\geq 0$

$ \sqrt{|a|+|b|}\geq \sqrt{(a+b)^{2}}$

$ |a|+|b| \geq |a+b|$


earlier today when i logged on to alcumus i had a score of about 89-90 and around when i clicked report it listed a bunch of topics, but when i logged just now it said i had a score of 50.00 and only showed 4 topics and it also took away all of my badges.

the correct/incorrect/give up tallies, the karma and points earned all show accurate values.

does anyone know what has happened or how to fix it.

PS it says my score is 88/5993 below what it was earlier but much higher than it should be with a score of 50 so i think i might not be the only one having this problem

im having the same problem. right now, it says im ranked 5993/5993

before, my alcumus score was 72 ish but when i  went on alcumus today, it was 50.

have you gone on alcumus after a long time?
because i think that might have caused it to reset

i'm having the same problem too, but i'm ranked 93/5993 instead, and so is 5 of my friends.

EDIT: most of my videos are gone too! I only have like 6 videos instead of 40

i started out as ranked 95 and suddenly it said i was last.

and its really strange because it seems to remember previous incorrect and correct answers but it has forgotten which subjects ive passed, mastered and still working on.

I haven't been gone for a long time, it was fine earlier today

@all: This is probably because you have just finished the diagnostic phase, and your scores were therefore reset.

i thought the diagnostic phase was 60 questions, i had done over 400 questions when this happened

i hadnt used alcumus for about a week though.

is it possible that it "forgets" information due to lots of people using this program?

I guess this is probably a system crash. I'll ask levans to come look at what's going on here.

Hi guys,

There are some problems with Alcumus at the moment. The problems are on the server side so I have forwarded the issues on to the developers. I hope to hear back from them soon, but as of today, I haven't heard anything so far.


Consider the triangle $ ABC$ where $ B<90^{\circ}$ and $ C<90^{\circ}$. If $ E$ is the midpoint of $ (AC)$ and $ F$ is the foot from $ E$ to $ BC$, then prove that $ A\equal{}90^{\circ}$ if and only if: $ FB^2\minus{}FC^2\equal{}AB^2.$

G foot height AG.
BG=(a^2-b^2+c^2)/2a
GF=(a-BG)/2  So BF=(a+BG)/2
(a+BG)^2/4-(a-BF)^2=c^2
Substituing the BF and BG values in the last equation we find a^2=b^2+c^2

[quote="moldovan"] [color=darkred]Consider the triangle $ ABC$ where $ B < 90^{\circ}$ and $ C < 90^{\circ}$. If $ E$ is the midpoint of $ (AC)$

and $ F$ is the foot from $ E$ to $ BC$, then prove that $ A \equal{} 90^{\circ} \ \Longleftrightarrow\ FB^2 \minus{} FC^2 \equal{} AB^2$ .[/color] [/quote]
[color=darkblue][b][u]Proof[/u].[/b] Observe that $ EF\perp BC\Longleftrightarrow FB^2 \minus{} FC^2 \equal{} EB^2 \minus{} EC^2$ . Since $ 2\cdot EC \equal{} b$ and $ 4\cdot EB^2 \equal{} 2\left(a^2 \plus{} c^2\right) \minus{} b^2$ obtain :

$ FB^2 \minus{} FC^2 \equal{} AB^2\ \Longleftrightarrow$ $ \left\|\ EB^2 \minus{} EC^2 \equal{} AB^2\ \right\|\ \odot 4$  $ \Longleftrightarrow$ $ 2\left(a^2 \plus{} c^2\right) \minus{} b^2 \minus{} b^2 \equal{} 4c^2$ $ \Longleftrightarrow$ $ a^2 \equal{} b^2 \plus{} c^2$ $ \Longleftrightarrow$ $ A \equal{} 90^{\circ}$ .[/color]

Let $ AD$ be, the $ A$-altitude of $ \bigtriangleup ABC,$ with $ \angle A \equal{} 90^{o}.$

So, we have $ FB \minus{} FC \equal{} FB \minus{} FD \equal{} BD$ $ \Longrightarrow$ $ (FB \minus{} FC)\cdot (FB \plus{} FC) \equal{} (BD)\cdot (FB \plus{} FC) \equal{} (BD)\cdot (BC)$

$ \Longrightarrow$ $ (FB)^{2} \minus{} (FC)^{2} \equal{} (AB)^{2}$ and we are done.

Kostas Vittas.

[quote="Virgil Nicula"]Nice Vittasko ! And reciprocally ?[/quote]
Well, reciprocally is not so nice.  :)

We have $ (FB)^{2} \minus{} (FC)^{2} \equal{} (AB)^{2}$ $ ,(1)$ and let $ B'$ be the point on $ BC,$ such that $ \angle B'AC \equal{} 90^{o}$

So, we have also $ (FB')^{2} \minus{} (FC)^{2} \equal{} (AB')^{2}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ (AB)^{2} \minus{} (FB)^{2} \equal{} (AB')^{2} \minus{} (FB')^{2}$ $ \Longrightarrow$ $ 2(AF)\cdot (MZ) \equal{} 2(AF)\cdot (MZ')$ $ ,(3)$ where $ M$ is the midpoint of the segment $ AF$ and $ Z,\ Z'$ are the orthogonal projections of $ B,\ B'$ respectively, on $ AF.$

From $ (3)$ $ \Longrightarrow$ $ MZ' \equal{} MZ$ $ \Longrightarrow$ $ Z'\equiv Z$ $ \Longrightarrow$ $ B'\equiv B$ and we are done.

Kostas Vittas.

[quote] [color=darkred][b][u]An easy extension[/u].[/b] Consider a triangle $ ABC$ , the point  $ E\in (AC)$ for which $ \frac {EA}{EC} \equal{} m$ and $ F\in (BC)$ so

that $ EF\perp BC$ . Prove that $ FB^2 \minus{} FC^2 \equal{} AB^2\ \Longleftrightarrow\ ma^2 \equal{} b^2 \plus{} mc^2\ \Longleftrightarrow\ \cos A \equal{} \frac {(m \minus{} 1)b}{2mc}$ .[/color] [/quote][quote] [color=darkred][b][u]A nice generalization[/u].[/b] Consider a  triangle $ ABC$ , the midpoint $ E$ of $ (AC)$ , a point $ P$ for which

$ PE\perp AC$ and $ F\in BC$ so that $ PF\perp BC$ .  Prove that $ FB^2 \minus{} FC^2 \equal{} AB^2\ \Longleftrightarrow\ PA\perp AB$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ D\in AB$ for which $ PD\perp AB$ . Apply the [u][b]Carnot[/b]'s lemma[/u] : $ \left(FB^2 \minus{} FC^2\right) \plus{} \left(EC^2 \minus{} EA^2\right) \plus{} \left(DA^2 \minus{} DB^2\right) \equal{} 0$ ,

i.e. $ \left(FB^2 \minus{} FC^2\right) \plus{} \left(DA^2 \minus{} DB^2\right) \equal{} 0$ . Therefore $ FB^2 \minus{} FC^2 \equal{} AB^2$ $ \Longleftrightarrow$ $ AB^2 \plus{} \left(DA^2 \minus{} DB^2\right) \equal{} 0$ $ \Longleftrightarrow$

$ \overline{BA}^2 \plus{} \left(\overline{DA} \minus{} \overline {DB}\right)\left(\overline {DA} \plus{} \overline {DB}\right) \equal{} 0$ $ \Longleftrightarrow$ $ \overline{BA}^2 \plus{} \overline{BA}\cdot \left(\overline {DA} \plus{} \overline {DB}\right) \equal{} 0$ $ \Longleftrightarrow$ $ \overline{BA} \plus{} \overline {DA} \plus{} \overline {DB} \equal{} 0$ $ \Longleftrightarrow$

$ \left(\overline{BD} \plus{} \overline {DA}\right) \plus{} \overline {DA} \minus{} \overline {BD} \equal{} 0\ \Longleftrightarrow\ 2\cdot\overline {DA} \equal{} 0\ \Longleftrightarrow\ D\equiv A\ \Longleftrightarrow\ PA\perp AB$ .[/color]

[quote] [color=darkred][b][u]A stronger generalization[/u].[/b] Consider $ \triangle ABC$ and the points $ \left\|\begin{array}{ccc}
E\in AC & , & \overline {AE}\equal{}m\cdot\overline{AC}\\\\
P & , & PE\perp AC\\\\
F\in BC & , & PF\perp BC\end{array}\right\|$ .

Prove that $ FB^2 \minus{} FC^2 \equal{} AB^2\ \Longleftrightarrow\ AB^2\plus{}AP^2\equal{}BP^2\plus{}(2m\minus{}1)\cdot AC^2$ .[/color] [/quote]

[quote="moldovan"]Consider the triangle $ ABC$ where $ B < 90^{\circ}$ and $ C < 90^{\circ}$. If $ E$ is the midpoint of $ (AC)$ and $ F$ is the foot from $ E$ to $ BC$, then prove that $ A \equal{} 90^{\circ}$ if and only if: $ FB^2 \minus{} FC^2 \equal{} AB^2.$[/quote]
Denote $ a\equal{}BC,b\equal{}CA,c\equal{}AB$ and $ A \equal{} \angle CAB, B \equal{} \angle ABC, C \equal{} \angle BCA$.
Then in $ \triangle ABC$ we find $ \cos C \equal{} \frac{a^2\plus{}b^2\minus{}c^2}{2ab}$
In $ \triangle CFE$ since $ \angle CFE \equal{} 90 ^\circ$ we find $ CF \equal{} CE \cdot \cos C \equal{} \frac{a^2\plus{}b^2\minus{}c^2}{4a}$ since $ CE \equal{} \frac{b}{2}$.
Now $ FB^2\minus{}FC^2 \equal{} (FB\plus{}FC\minus{}2FC)(FB\plus{}FC) \equal{} a^2\minus{}2a \cdot FC$, since $ FB\plus{}FC \equal{} BC \equal{} a$.
So $ FB^2\minus{}FC^2 \equal{} AB^2\equal{}c^2 \iff a^2\minus{}2a \cdot \frac{a^2\plus{}b^2\minus{}c^2}{4a} \equal{} c^2 \iff a^2\equal{}b^2\plus{}c^2$ $ \iff A \equal{} 90^{\circ}$. QED