geometry
3D geometry
combinatorics unsolved
combinatorics

I haven´t been able to prove this problem  :P. Can someone prove it  :) ?.

On the vertices of a cube the numbers from 1 to 8 are placed without repetition. Prove that it doesn't matter how you arrange the numbers there always exist two different edges whose vertices sum up a multiple of 3.

Assume the contary..

Take a vertice $A$ which labeled a no. $\equiv 1 mod 3$. The vertice is connected to three other vertices $B,C,D$ which each can't be labeled with no. $\equiv 2 mod 3$, there're $3$ no. in this form. So the remaining no. to be placed in the $3$ vertices are $2$ no. $\equiv 0 mod 3$, and $2$ no. $\equiv 1 mod 3$. Consider these cases of what no. will be placed in the $3$ vertices:

i) $1$ no. $\equiv 1mod 3$ (say $B$ and $C$) and $2$ no. $\equiv 0 mod 3, (D)$
The same thing happened with $D$ ,there're $2$ edges remaining emanating from $D$, with only $1$ no. $\equiv 1 mod 3$ can be placed (because $BC, BD$ aren't edges).

ii) $2$ no. $\equiv 1 mod 3$ (say $E$ and $F$) and $1$ no. $\equiv 0 mod 3, (G)$
$E$ and $F$,each must be connected to the remaining $1$ no. $\equiv 0 mod 3$ and $2$ no. $\equiv 1 mod 3$ which is impossible.

iii) $3$ no. $\equiv 1 mod 3$
Then, these three vertices must be conected to the remaining $2$ no. $\equiv 0 mod 3$ which is impossible, too...

I write number as no.
N I hope I'm right 'cause I ain't that good in combin.   :P

A bit shorter solution:

First note that 2 vertices are $0\mod 3$, 3 vertices are $1\mod 3$, and 3 vertices are $2\mod 3$.  Consider a vertex $v$ which is $1\mod 3$ and assume $v$ is not adjacent to a $2\mod 3$ (or we're done).  Since $v$ has 3 neighboors, one of the neighboors must be $1\mod 3$ as well, say $u$.  Together $v$ and $u$ are adjacent to all the vertices except 2, hence one of them must be adjactent to a $2\mod 3$ and we're done.


Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
I'm familiar with affirmative equation symbols like,
$ { = , > , \geq,...,\mbox{etc.}}$
Additionally, I'm familiar with negation equation symbols like,
$ {\neq, \ngtr, \ngeq,...,\mbox{etc.}}$

However, what is the negation symbol for the set notation $ \in$ (in)[b]?[/b] I would imagine it's the same except with a diagonal line through it (ie. not in), but I couldn't find that symbol in LaTeX.

[b]2. Relevant equations.[/b]
Knowledge of $ {\text{\LaTeX}}$ formatting.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
Yea, I looked through my $ {\text{\LaTeX}}$ symbol list and couldn't find it. :maybe:
[/hide]
Thanks,

-APSStudent

As you'd expect, the notation is "not in", which is $ \notin$.  The syntax is quite simple:
[code]\notin[/code]

Hey,
[quote="JRav"]
As you'd expect, the notation is "not in", which is $ \notin$. The syntax is quite simple:
[code]
\notin
[/code]
[/quote]
[quote="jmerry"]
I've always used \not\in: $ \not\in$. It's actually slightly different, which I didn't realize until now.
[/quote]
Thanks for the information. What are the differences (besides the slight difference in appearance) between using,
$ {\notin},{\,}{\,}{\mbox{ \minus{} notin}}$
versus
$ {\not\in},{\,}{\,}{\mbox{ \minus{} not \minus{} in}}$
Where, "-" is in place of "" since I couldn't figure out how to add the text directly as "\notin"/"\not\in".

More generally what are the differences between using single negation equations symbols like,
$ {\neq{\,}\ngtr{\,}\ngeq,...,\mbox{etc.}}$
versus using negation symbols with the "\not" code like,
$ {\not \equal{} {\,}\not>{\,}\not\geq,...,\mbox{etc.}}$

Thanks,

-APSStudent

It seems like the \not command can give you extended capabilities.  For example, there is no
[code]\notsubset[/code]
But if I type
[code]\not\subset[/code]
I get $ \not\subset$.

Hey,
[quote="jmerry"]
It's pretty clear that that's what \not does: it overlays a slash on the symbol that follows. $ \not\alpha$, for a nonsensical example. The spacing doesn't work well for large symbols.

If a specific negated symbol already exists, you should probably use it.
[/quote]
Thanks for the clearing that up.

Thanks,

-APSStudent


Could you help me please on solving that problem:

Charles deposited 50€ in his bank. After five years he could take 62€ out of his account. What was the average interest of the account during these five years when each annual accumulated interest was compounded with the balance of the account?

Should I use compund or simple interest here? I'm not sure. :?:

[quote="valeriummaximum"]the answer is 11.35% per annum...[/quote]

How did you solve that? I guess you didn't use the compund interest.

It seems like it would be more than 62 euro in five years if the interest would be so high - 11,35%.

Sorry a calculation mistake..

this is very simple sum...

just use the formula
Let amount be A
Principal be P
and rate be r

$ {A}{ \equal{} }{P}{(1 \plus{} \frac{r}{100})^5}$......

r comes out to be 4.396%per annum

[quote="valeriummaximum"]Sorry a calculation mistake..

this is very simple sum...

just use the formula
Let amount be A
Principal be P
and rate be r

$ {A}{ \equal{} }{P}{(1 \plus{} \frac {r}{100})^5}$......

r comes out to be 4.396%per annum[/quote]

Can I ask you where you got this formula from? Just can't figure out what it means... It doesn't look like formula for compound interest...

see this link....

[url]http://en.wikipedia.org/wiki/Compound_interest[/url]


superior algebra
superior algebra unsolved

Prove that for any natural numbers $ a,b$  $ Z_a X Z_b \sim Z_{(a,b)} X Z_{[a,b]}$

I think you mean $ \mathbb{Z}/a \times \mathbb{Z}/b \cong \mathbb{Z}/gcd(a,b) \times \mathbb{Z}/lcm(a,b)$. this is trivial if $ a,b$ are powers of the same prime number. deduce the general case with the chinese remainder theorem.


[quote="Dr Sonnhard Graubner"]hello, this equation has only two real solutions.
Sonnhard.[/quote]

can you explain why or how?

Hello !!

let : $ f(x)\equal{}x^6\minus{}2x^5\plus{}0.98$ for $ x \in  \mathbb{R}$

$ f'(x)\equal{}x^4(6x\minus{}10)$

by using  TVI (Théorème des valeurs intermédiaires) , and evaluating the variation of f

we can prove that the équation $ f(x)\equal{}0$ has only two real solutions.

[hide="Descartes Rule of Signs"]$ f(x) \equal{} x^{6} \minus{} 2x^{5} \plus{} 0.98$

- By inspection, there are no negative roots.
- By Descartes' Rule of Signs, this equation may have either 2 positive roots or 0 positive roots. We need to figure out which of these is true.

$ f(0) \equal{} 0.98 > 0$, $ f(1) \equal{} \minus{} 0.02 < 0$, and $ f(2) \equal{} 0.98 > 0$
So there must be a root at $ x \in (0,1)$ and another in $ (1,2)$.

Since there were at most 2 positive roots, we know there are no other roots apart from these 2.[/hide]


1.the sum and product of three numbers =585.find the numbers.

2.calculate 7^99 without using calaulator.
3.how many triangles are there in a grid of triangle 15units on each side.

1.p and q are the numbers p+q =585 pq=585
$ x^2 \minus{} 585x \plus{} 585 \equal{} 0$
so solve it by the general solution $ \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$
$ ax^2 \plus{} bx \plus{} c$ ,$ a\neq0$

@asintota:  The question mentioned the sum and product of [b]three[/b] numbers, not two.

Stop asking us to "calculate 7^99 without using calaulator.".

You already posted that in the other thread.  Also, you did not ask us what you wanted.  If you want all the digits of 7^99, go buy mathematica, install it, and plug it in there.

If I understood correctly.

The grid has then $ \frac{15*16}{2}\equal{}15*8\equal{}120$ points.

$ \binom{120}{3}$ are the number of ways in wich we can choose three points.

So the number of disctint triangles is $ \frac{1}{3}\cdot\binom{120}{3}$

 :)

[quote="math_explorer"]I don't quite understand how $ x^2 \minus{} 585x \plus{} 585 \equal{} 0$ follows from pq = p+q = 585, but perhaps that method would work if we set one number = 1 and let pq = 585 while p+q = 584?
Well anyways, 585 = 3^2 * 5 * 13.

Oh, and are the triangles nondegenerate?[/quote]No, you can't just set a number like 1 randomly without any reason for it. If you set $ pq\equal{}585$ and $ p\plus{}q\equal{}585$ and solve for one of the variables, you'll get that quadratic. He used the so-called Vieta's (or Viete's) formulas. Whatever the case, that's not the answer. The question doesn't really make sense because what type of numbers are we talking about here? Positive integers? Integers? Rationals? Reals?

@11273: You have posted a total of 3 posts and all of them are basically the [u]same[/u] questions. That is SPAM, stop posting multiple posts of the same content. It's boring and not intellectually smart.

There is no need to answer #2. Computational problem has no content. Leave that for Maple or Mathematica.

For #3 to make sense, triangles are non-degenerate, obviously.

If you have three numbers $ a,b,c$ such that $ a\plus{}b\plus{}c\equal{}abc\equal{}585$, then it is impossible to determine the numbers.  You need another equation.

That's pretty much what I thought. I doubt it's possible with integers, though.


geometry
circumcircle
incenter
trigonometry
power of a point
radical axis
geometry unsolved

[i]Proof:[/i]
$ (a)$ Because $ A_2B_1A_2C_1$ is harmonic quadrilateral $ \Longrightarrow \frac {A_2B_1}{A_2C_1} \equal{} \frac {A_1B_1}{A_1C_1}$. But $ A_3A_1$ is the bisector of $ \angle B_1A_1C_1$, $ A_3\in B_1C_1\Longrightarrow \frac {A_1B_1}{A_1C_1} \equal{} \frac {A_3B_1}{A_3C_1}$, which implies that $ \frac {A_2B_1}{A_2C_1} \equal{} \frac {A_3B_1}{A_3C_1}\Longrightarrow A_2A_3$ is also the bisector of $ \angle B_1A_2C_1$.

$ (b)$ Let $ A_1A_4$ be the external bisector of $ \angle B_1A_1C_1$, $ A_4\in B_1C_1$. It is so obvious that $ (B_1C_1A_3A_4) \equal{} \minus{} 1$. But $ A_2A_3$ is the internal bisector of $ \angle C_1A_2B_1\Longrightarrow A_4A_2\bot A_2A_3$, which implies $ \angle A_4A_2A_3 \equal{} 90^{\circ}$. We also have $ \angle A_4A_1A_3 \equal{} 90^{\circ}$. Therefore $ A_4,A_2,A_3,A_1$ are concyclic. Let $ O_a$ be the circumcenter of $ (A_2A_3A_1)\Longrightarrow O_a$ is the midpoint of $ A_4A_3\Longrightarrow O_aA_3^2 \equal{} O_aA_2^2 \equal{} O_aA_1^2 \equal{} \overline {O_aC_1}\cdot \overline {O_aB_1}\Longrightarrow O_aA_2\bot IA_2$;  $ O_aA_1\bot IA_1$, which implies that $ (O_a)$ is perpendicular to $ (I)\Longrightarrow \mathcal {P}_{I/(O_a)} \equal{} r^2$, where $ r$ is the radii of $ (I)$. Similar for $ O_b$, we also have $ \mathcal {P}_{I/(O_b)} \equal{} r^2 \equal{} \mathcal {P}_{I/(O_a)}\Longrightarrow I\in PQ$- which is the radical axes wrt $ (O_a),(O_b)$.
Our proof is completed then.

[b]REMARK-[/b] As the matter of fact, $ (A_1A_2A_3),(B_1B_2B_3),(C_1C_2C_3)$ are a pencil of circles, with $ PQ$ is the radical axes wrt them. Indeed, let $ G_e$ be the $ Gergone$ point wrt $ \triangle ABC$- which is also the $ Lemoine$ wrt $ \triangle A_1B_1C_1$, then $ \overline {G_eA_1}\cdot \overline {G_eA_2} \equal{} \overline {G_eB_1}\cdot \overline {G_eB_2} \equal{} \overline {G_eC_1}\cdot \overline {G_eC_2} \equal{} \mathcal {P}_{G_e/(I)}$, which implies that $ \mathcal {P}_{G_e/(O_a)} \equal{} \mathcal {P}_{G_e/(O_b)} \equal{} \mathcal {P}_{G_e/(O_c)}$, where $ (O_a)$ is the circumcircle of $ \triangle A_1A_2A_3$, $ (O_b),(O_c)$ are defined similarly. But in the problem, we have already known that $ \mathcal {P}_{I/(O_a)} \equal{} \mathcal {P}_{I/(O_b)} \equal{} \mathcal {P}_{I/(O_c)}$. Hence $ IG_e$ is respetively perpendicular to $ O_aO_b,O_bO_c\Longrightarrow O_a,O_b,O_c$ are collinear, which implies to the result at the biginning. Further, $ IG_e\equiv PQ$, which means that $ IG_e$ is also the radical axes of pencil of circles $ (O_a),(O_b),(O_c)$.

[quote="babylon"]The incircle $ k$ of a non-isosceles triangle $ ABC$ touches the sides $ AB,BC,CA$
at $ C_{1},A_{1},B_{1}$ , respectively. Lines $ AA_{1},BB_{1},CC_{1}$ intersect $ k$ again at
$ A_{2},B_{2},C_{2}$, respectively. Let $ A_{1}A_{3}$ and $ B_{1},B_{3}$ be the bisectors of the
angles in △$ A_{1}B_{1}C_{1}$ ($ A_{3}$ ∈ $ B_{1}C_{1}, B_{3}$ ∈ $ A_{1}C_{1}$). Prove that:

$ (a)$  $ A_{2}A_{3}$ is a bisector of $ \angle B_{1}A_{2}C_{1}$;

$ (b)$  If the circumcircles of △$ A_{1}A_{2}A_{3}$ and △$ B_{1}B_{2}B_{3}$ intersect at $ P$ and
$ Q$, then the incenter $ I$ of △$ ABC$ lies on the line $ PQ$.[/quote]
Nice problem :) . Here is my solution.
[b]Part (a)[/b]

We have $ \frac{C_1A_3}{A_3B_1}\equal{}\frac{C_1A_1}{A_1B_1}\equal{}\frac{\cos \frac B2}{\cos \frac C2}$. So, it is enough to prove that, $ \frac{\cos \frac B2}{\cos \frac C2}\equal{}\frac{C_1A_2}{A_2B_1}$
Using the fact $ AC_1\equal{}AB_1$ and then by sine law we get,
\[ \frac{C_1A_2}{A_2B_1}\equal{}\frac{\sin C_1A_1A}{\sin B_1A_1A}
\equal{}\frac{\frac{\sin C_1A_1A}{AC_1}}{\frac{\sin B_1A_1A}{AB_1}}
\equal{}\frac{\frac{\cos \frac B2}{AA_1}}{\frac{\cos \frac C2}{AA_1}}
\equal{}\frac {\cos \frac B2}{\cos \frac C2}
\]So, we are done.

[b]Part (b)[/b]
Just simple angle chasing; we shall show that $ IA_1$ and $ IB_1$ are the tangents to the respective circumcircles. We have $ \angle A_1A_2B_1\equal{}90^{\circ}\minus{}\frac C2$ and using part (a) we have $ \angle A_3A_2B_1\equal{}\frac{90^{\circ}\plus{}\frac A2}{2}$. After some calculation we get that, $ \angle A_3A_2A_1\equal{}\left |\frac B4 \minus{}\frac C4 \right |$
Also, $ \angle IA_1A_2\equal{}\frac B2$ and $ \angle A_3A_1C_1\equal{}\frac{90^{\circ}\minus{}\frac A2}{2}$. Analogous calculation gives, $ \angle IA_1A_3\equal{}\left |\frac B4 \minus{}\frac C4 \right |$
So, $ \mathcal{P} (I)\equal{}IA_1^2\equal{}r^2\equal{}IB_1^2$. Which implies that, $ I$ is on the radical axis $ PQ$ of the circumcircles.

And we are done again.  :wink:


Let $ a,b,c\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ (2\minus{}ab)(2\minus{}bc)(2\minus{}ca)\geq1$

By the way,this is another inequality from Vasc

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ (2 \minus{} ab)(2 \minus{} bc)(2 \minus{} ca)\geq1$[/quote]
After expanding use $ a^2b^2c^2\leq\frac{1}{3}\sum_{cyc}a^2bc$ and use SOS.

[quote="Inequalities Master"]Sorry,but how to apply SOS?  :?:[/quote]
OK.
$ (2 \minus{} ab)(2 \minus{} bc)(2 \minus{} ca)\geq1\Leftrightarrow7 \minus{} 4(ab \plus{} ac \plus{} bc) \plus{} 2abc(a \plus{} b \plus{} c) \minus{} a^2b^2c^2\geq0.$
Since, $ a^2b^2c^2\leq\frac {1}{3}\sum_{cyc}a^2bc$ is true then it remains to prove that
$ 21 \minus{} 12(ab \plus{} ac \plus{} bc) \plus{} 5abc(a \plus{} b \plus{} c)\geq0.$
But $ 21 \minus{} 12(ab \plus{} ac \plus{} bc) \plus{} 5abc(a \plus{} b \plus{} c)\geq0\Leftrightarrow$
$ \Leftrightarrow7(a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 12(ab \plus{} ac \plus{} bc)(a^2 \plus{} b^2 \plus{} c^2) \plus{} 15abc(a \plus{} b \plus{} c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(7a^4 \minus{} 12a^3b \minus{} 12a^3c \plus{} 14a^2b^2 \plus{} 3a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(14a^4 \minus{} 24a^3b \minus{} 24a^3c \plus{} 28a^2b^2 \plus{} 6a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow7\sum_{cyc}(a^4 \minus{} a^3b \minus{} ab^3 \plus{} b^4) \minus{} 17\sum_{cyc}(a^3b \minus{} 2a^2b^2 \plus{} ab^3) \minus{}$
$ \minus{} 3\sum_{cyc}(c^2a^2 \minus{} 2c^2ab \plus{} c^2b^2)\geq0\Leftrightarrow\sum_{cyc}(a \minus{} b)^2(7a^2 \minus{} 10ab \plus{} 7b^2 \minus{} 3c^2)\geq0.$
Let $ a\geq b\geq c.$ Then $ S_c\equal{}7a^2 \minus{} 10ab \plus{} 7b^2 \minus{} 3c^2\equal{}7a^2\minus{}10ab\plus{}4b^2\plus{}3(b^2\minus{}c^2)\geq0,$ $ S_b \equal{} 7a^2 \minus{} 10ac \plus{} 7c^2 \minus{} 3b^2 \equal{} 4a^2 \minus{} 10ac \plus{} 7c^2 \plus{} 3(a^2 \minus{} b^2)\geq0$
and $ a \minus{} c\geq b \minus{} c.$
Thus,$ \sum_{cyc}(a \minus{} b)^2(7a^2 \minus{} 10ab \plus{} 7b^2 \minus{} 3c^2)\geq(a \minus{} c)^2S_b \plus{} (b \minus{} c)^2S_a\geq(b \minus{} c)^2(S_b \plus{} S_a) \equal{}$
$ \equal{} (b \minus{} c)^2(4a^2 \plus{} 4b^2 \plus{} 14c^2 \minus{} 10bc \minus{} 10ac) \equal{} 2(b \minus{} c)^2(7c^2 \minus{} 5(a \plus{} b)c \plus{} 2a^2 \plus{} 2b^2)\geq0.$
The last inequality is true because $ \Delta \equal{} 25(a \plus{} b)^2 \minus{} 56(a^2 \plus{} b^2)\leq0.$

Thanks a lot, arqady!!!I want to be good at mathematics like you!

[quote="tardi"][quote="arqady"]

Since, $ a^2b^2c^2\leq\frac {1}{3}\sum_{cyc}a^2bc$ is true
[/quote]

Dear arqady, can you post your solution[/quote]
It's true by AM-GM:
$ a^2b^2c^2\leq\frac {1}{3}\sum_{cyc}a^2bc\Leftrightarrow(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)abc\geq9a^2b^2c^2.$


If the equation $ ax^2\minus{}by^2\equal{}1$ has a solution in $ \mathbb {N}^*$ then it has an infinity of solutions in $ \mathbb{N}^*$.

[hide="Solution"]
We have that $ (x\sqrt {a} \minus{} y\sqrt {b})(x\sqrt {a} \plus{} y\sqrt {b}) \equal{} 1$.  Cubing gives us that\[ [(x^3a\sqrt {a} \minus{} 3x^2ya\sqrt {b} \plus{} 3xy^2b\sqrt {a} \minus{} y^3b\sqrt {b}][x^3a\sqrt {a} \plus{} 3x^2y\sqrt {b} \plus{} 3xy^2b\sqrt {a} \plus{} y^3b\sqrt {b}] \equal{} 1\]and hence\[ [(x^3a \plus{} 3xy^2b)\sqrt {a} \minus{} (3x^2ya \plus{} y^3b)\sqrt {b}][(x^3a \plus{} 3xy^2b)\sqrt {a} \plus{} (3x^2y \plus{} y^3b)\sqrt {b}] \equal{} 1\]which is in the desired form, so if $ (x, y)$ is a solution, then $ (x^3a \plus{} 3xy^2b, 3x^2y \plus{} y^3b)$ is also a solution.  Clearly, if $ x$, $ y$, $ a$, and $ b$ are natural, then both terms are greater than $ x$ and $ y$.  Thus, we have found another pair of solutions and can keep on repeating this procedure to get infinitely many solutions.  [/hide]


geometry
3D geometry
octahedron
probability
function

Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Stimultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?

seems to be a tough problem, because no one has answered it! :P

Because each ant can move from its vertex to any of four adjacent vertices, there are 4^6 possible combinations of moves. In the following, consider only those combinations in which no two ants arrive at the same vertex. Label the vertices as A, B, C, A', B' and C', where A', B' and C' are opposite A, B and C, respectively. Let f be the function that maps each ant's starting vertex onto its final vertex. Then neither of f(A) nor f(A') can be either A or A', and similar statements hold for the other pairs of opposite vertices. Thus there are 4\cdot 3 = 12 ordered pairs of values for f(A) and f(A'). The vertices f(A) and f(A') are opposite each other in four cases and adjacent to each other in eight.
Suppose that f(A) and f(A') are opposite vertices, and, without loss of generality, that f(A) = B and f(A') = B'. Then f(C) must be either A or A' and f(C') must be the other. Similarly, f(B) must be either C or C' and f(B') must be the other. Therefore there are 4\cdot 2\cdot 2 = 16 combinations of moves in which f(A) and f(A') are opposite each other.
Suppose now that f(A) and f(A') are adjacent vertices, and, without loss of generality, that f(A) = B and f(A') = C. Then one of f(B) and f(B') must be C' and the other cannot be B'. So there are four possible ordered pairs of values for f(B) and f(B'). For each of those there are two possible ordered pairs of values for f(C) and f(C'). Therefore there are 8\cdot 4\cdot 2 = 64 combinations of moves in which f(A) and f(A') are adjacent to each other.
Hence the probability that no two ants arrive at the same vertex is $ \frac{16 \plus{} 64}{4^6} \equal{} \frac{5\cdot 2^4}{2^{12}} \equal{} \frac{5}{256}$.


Let $ a,b,c\geq0$.Prove that:
$ \dfrac{a^2}{b\plus{}c}\plus{}\dfrac{b^2}{a\plus{}c}\plus{}\dfrac{c^2}{a\plus{}b}\geq\dfrac{3(a^3\plus{}b^3\plus{}c^3)}{2(a^2\plus{}b^2\plus{}c^2)}$

[hide="ugly solution"]
our inequality is equivalent to
$ LHS \minus{} \frac{a\plus{}b\plus{}c}{2} \geq RHS \minus{} \frac{a\plus{}b\plus{}c}{2}$
$ \frac{1}{2}\frac{(a\plus{}b\plus{}c)(2 \sum a^3 \minus{} \sum ab(a\plus{}b)}{(a\plus{}b)(b\plus{}c)(a\plus{}c)} \geq \frac{1}{2}(\frac{(2\sum a^3\minus{}\sum ab(a\plus{}b)}{a^2\plus{}b^2\plus{}c^2}$
equivalent to $ \frac{1}{2}(2\sum a^3\minus{}\sum ab(a\plus{}b))(\frac{a\plus{}b\plus{}c}{(a\plus{}b)(b\plus{}c)(a\plus{}c)}\minus{}\frac{1}{a^2\plus{}b^2\plus{}c^2}) \geq 0$
which means $ \frac{1}{2}(2\sum a^3\minus{}\sum ab(a\plus{}b))(\frac{a^3\plus{}b^3\plus{}c^3\minus{}2abc}{(a\plus{}b)(b\plus{}c)(a\plus{}c)(a^2\plus{}b^2\plus{}c^2)} \geq 0$obviously true[/hide]

[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ \dfrac{a^2}{b \plus{} c} \plus{} \dfrac{b^2}{a \plus{} c} \plus{} \dfrac{c^2}{a \plus{} b}\geq\dfrac{3(a^3 \plus{} b^3 \plus{} c^3)}{2(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]
By The Cauchy Schwarz Inequality, we get
\[ \sum \frac{a^2}{b\plus{}c} \ge \frac{ (\sum a^3)^2}{\sum a^4(b\plus{}c)}\]
It suffices to prove that
\[ 2(\sum a^3)(\sum a^2) \ge 3\sum a^4(b\plus{}c)\]
\[ \Leftrightarrow \sum (a^5\plus{}b^5\plus{}2a^3b^2\plus{}2a^2b^3 \minus{}3a^4b\minus{}3ab^4) \ge 0\]
It suffices to prove that
\[ a^5\plus{}b^5\plus{}2a^3b^2\plus{}2a^2b^3 \minus{}3a^4b\minus{}3ab^4 \ge 0\]
\[ \Leftrightarrow (a\minus{}b)(a^4\minus{}b^4) \minus{}2ab(a\minus{}b)^2(a\plus{}b) \ge 0\]
\[ \Leftrightarrow (a\minus{}b)^4(a\plus{}b) \ge 0\]
which is trivial. :)


uh how do you define "heads" or "tails" on a dice?

assuming you mean coins...

There are two possible outcomes on each flip, and are independent of each other, so there are
\[ 2^{10} \equal{} 1024
\]
outcomes.

There is only one way to have all heads, and only one way to have all tails, so the probability is $ \frac {2}{1024} \equal{} \boxed{\frac {1}{512}}$

omg sorry on a coin

and are you sure you are correct

That is the only answer because there are a total of $ 2^{10}$ possible outcomes and only $ 2$ ways to get $ 10$ of either tails or heads making it as RoFlLoLcOpT said $ \boxed{\dfrac{1}{512}}$..

@RoFlLoLcOpT: Btw it's just a typo but you might want to change that $ 256$ for the $ 512$

@Yankzgodzilla55: Why did you put 9 heads and 1 tail, 8 heads and 2 tails... and so on as answers? Those do not have neither 10 heads in a row nor 10 tails in a row.

[quote="Yankzgodzilla55"]Hello!

I was wondering if anyone could contradict my solution to this problem.

Probability:

Problem:

I roll a dice 10 times. What is the probability that you will get 10 tails in a row or 10 heads?

Solution:

10 h 0 t
9 h 1 t
8h 2 t
7h 3t

and so on makes 11 options

so the answer is 2/11

:lol:

Thanks![/quote]

What you did is almost right but you forgot that you can rearrange things.
So the probability of getting 10 heads is 1/2 ^10.
The probability of getting tails is 1/2 ^ 10.
So the total probability is 2 x 1/2^10 = 1/ 2^9=  1/512

The reason why fault can be found with your solution is the fact that not each combination of heads or tails is as probable to receive.  The outcomes of "all heads" and "all tails" are the two most unlikely combinations when flipping coins, simply because there is only one way for each to happen.
However, if we consider flipping 1 tails and 9 heads, this is much more likely because there are 10 ways to choose the tails coin.

You should refer to the 10th row on pascal's triangle. It reads:
$ 1$, $ 10$, $ 45$, $ 120$, $ 210$, $ 252$, $ 210$, $ 120$, $ 45$, $ 10$, $ 1$

From this, you can see the distribution of the likelihoods of each number of heads or tails. All heads or tails are each only $ \frac {1}{512}$ likelihood, while an equal number of each has $ \frac {63}{256}$ likelihood.

EDIT: 1000th post!!!

Just a style note: The best way to refute a given argument is to address, as directly as possible, where the fallacy is in the original argument; not to present a separate (correct) argument.

That being said, the problem with Yankzgodzilla55's original argument is that the formula he is basing it on, $ P(good \, outcome) \equal{} \frac{number \, of \, good \, outcomes}{total \, number \, of \, possible \, outcomes}$ is only valid if the probabilities of each of the individual outcomes that are being counted are equal. So when you count outcomes you have to be sure that they are grouped in a way so that each of them is equally likely.

As an extreme example of this error, suppose someone gave you the following argument:
"There is a pot of gold behind one of seven doors. When I pick any door there are two possibilities - either there is gold behind it or there is not. So the probability that there is gold behind any given door is one (possible outcome) out of two (possible outcomes), or $ \frac{1}{2}$"

Okay so what you're saying is it's always 1/ total [b]different[/b] outcomes. That's like saying we have an unfair coin. 100% of the time it's flipped it displays heads.

Possibilities:
10h 0 tails
9 h 1 tails
etc.

It's still 100% all heads.

The point is not all outcomes have the same weight.

I wouldn't even say "different" outcomes, because that word could be confusing in some situations. But your idea of thinking of probabilities in terms of "weighted outcomes" is correct.

[quote="gauss202"]That being said, the problem with Yankzgodzilla55's original argument is that the formula he is basing it on, $ P(good \, outcome) = \frac {number \, of \, good \, outcomes}{total \, number \, of \, possible \, outcomes}$ is only valid if the probabilities of each of the individual outcomes that are being counted are equal. [/quote]
Eh I just feel like pointing this out.
[quote="Valentin Vornicu"]Please do not use plain TEXT in LaTeX tags! Use the $ \text{\LaTeX}$ tags only for formulae, variables, symbols. Use normal writing and BBCode for normal text.[/quote]

To type text in a $ \text{\LaTeX}$ tag just input [code]\text{whatever you want to say}[/code]


function
real analysis
real analysis unsolved

suppose that [b]fn[/b] : [a,b] -> R, and [b]fn[/b] converges to [b]f[/b] uniformly as n tends to infinity. If for each n in N, [b]fn[/b] has at most ten discontinuities. Prove that [b]f[/b] has at most ten discontinuities also.

Define $\text{osc}(f)(x)=\limsup_{y\to x}f(y)-\liminf_{y\to x}f(y).$

Then $f$ is continuous at $x$ iff $\text{osc}(f)(x)=0.$

Suppose $f$ is discontinuous at $x$ - that is, $\text{osc}(f)(x)=a>0.$

For all $\delta>0$ we can choose $y,z$ such that $|y-x|<\delta,\,|z-x|<\delta,$ and $|f(y)-f(z)|>\frac a2.$

Now choose $N$ such that for all $n>N,\,|f_n(t)-f(t)|<\frac a6$ for all $t.$

Then for $n>N,$

$|f_n(y)-f_n(z)|>|f(y)-f(z)|-|f_n(y)-f(y)|-|f_n(z)-f(z)|>\frac a2-\frac a6-\frac a6=\frac a6.$

Since we can do this for any $\delta>0,$ we conclude that for every $n>N,\,\text{osc}(f_n)(x) >\frac a6$

and thus for each $n>N,\,f_n$ is discontinuous at $x.$

Rephrasing that: eventually (in $n$), $f_n$ must be discontinuous at every point at which $f$  is discontinuous. Since $f_n$ can be discontinuous no more than $10$ times, we conclude that $f$ can be discontinuous no more than $10$ times.

Or: if each $f_n$ had no more than $m$ discontinuities (for a finite $m$), then $f$ has no more than $m$ discontinuities. This applies to the case $m=0,$ so this proof contains as a special case the classical theorem that the uniform limit of continuous functions is continuous.


inequalities
combinatorics solved
combinatorics

Let n be a positive integer and {A, B, C} be a partition of {1, 2, 3, ..., 3n} such that |A| = |B| = |C| = n. Prove that there exist x from A, y from B, z from C such that one of x, y, z is the sum of the other two.

Namdung

There is a more general result, which I think has been given in the 1987 Schweizer contest :

Let {1,2,...,n} be 3-colored, such that each color is used at least n/4 times. Then, the equation x+y = z has a tricolor solution.

If you are interested, I may give a (long) solution.

Pierre.

Ofcourse I'm. Please give the proof.

Namdung

Let's go for a long solution....

Suppose for a contradiction that it is possible to color each of the integers 1,2,...,n by one of the three colors, say red, blue, green, such that there is no tricolor solution of the equation x+y = z, and such that each of the color is used at least n/4 times.

Wlog, we may assume that 1 is red. Thus, there do not exist two consecutive integers which are blue and green.
In the following, the interval [a,b] denotes {a,a+1,...,b}. Note that |[a,b]| = b-a+1.
From above the blue and green integers form some disjoint monochromatic intervals, with at least a red integer as a separator.

- First, we prove that if there is a green (resp. blue) interval A such that |A| > 1 then there is no blue (resp. green) interval B such that |B| > 1 :
Since there is a finite number of interval, we may consider the green interval A = [a_1, a_2] with maximal length among the green intervals, and a blue interval B = [b_1, b_2] with maximal length among the blue intervals.
Then, for a in A and b in B, |a-b| is not red, since we have no tricolor solution to x+y = z.
Let C = {|a-b| / a in A and b in B}. Then, C is not empty and contained in {1,2,...,n}. Moreover, it contains no red element thus, from above, it is a monochromatic interval. More precisely :
C = [b_1 - a_2, b_2 - a_1] if a_2 < b_1
C = [a_1 - b_2, a_2 - b_1] if b_2 < a_1.
Then, in both cases |C| = |A| + |B| - 1. Then, if |A|, |B|  \geq 2, then |C| > |A| and |C| > |B|, which contradicts the maximality of A and B.

Wlog, we then assume that there is no green interval A such that |A| > 1, that is there do not exist two consecutive green integers.

- Then, we prove that there exist two consecutive blue integers :
If not, since blue and green integers are never consecutive ones, and since 1 is red, it would follow that for each blue integer b, the number b-1 is red. Similarly, for each green integer g, we would have g-1 red. Thus, there would be at least n/2 red integers, which contradicts that the are more than n/2 non-red integers.

- Now, we prove that there exist g and g+2 which both are green :
First, note that, from above, if g  \leq n-1 is green, then g  \geq 2 and g-1 and g+1 are red.
Suppose, for a contradiction, that there is no green integer g such that g+2 is green too. Let N(R), N(G), N(B) be the number of red, green, blue integers respectively.
When g runs over the set of the green intgers with g \leq  n-1, the intervals [g-1, g+1] form a family of pairwise disjoints intervals, each containing 2 red integers. Thus :
- If n is not green, then N(R)  \geq  2N(G), from which we deduce that
n = N(R) + N(G) + N(B)  \geq 3N(G) + N(B) > 3n/4 + n/4 = n, a contradiction.
- If n is green, then N(R)  \geq  2N(G) - 1 and 2 is green (if not 1 has not been counted and we have the same contradiction as above). If B is the blu interval with maximal length, with the above notations, then b_2 < n. Thus b_2 + 1 is red. Moreover, since |B|  \geq  2, it follows that b_2 - 1 is blue. Thus (b_2 - 1) + 2 = b_2 + 1 is a tricolor solution of x+y = z. A contradiction again, hence result.

Now, let's suppose that g and g+2 are geen.
Then g+1 is red and, if B is the same as above, we either have b_1 > g + 2 or b_2 < g.
- If b_1 > g + 2 : Let E = {b-g / b in B} \cup  {b-g-2 / b in B}. We will prove that E is a monochromatic interval :
Let x be in E. If x = b-g, then x+g = b, where g is green, b is blue. If x = b-g-2, then x + (g+2) = b, where g+2 is green and b is blue. In both cases, we deduce that x is not red.
From B = [b_1, b_2], we deduce that E = [b_1 - g, b_2 - g]  \cup  [b_1 - g - 2, b_2 - g - 2]. And, since b_1 < b_2, we have b_2 - g - 2  \geq  b_1 - g - 1, thus E is an interval.
Then E is an interval which contaions no red integer. From above, we deduce that E is monochromatic, but not red.
Moreover |E| = b_2 - g - (b_1 - g - 2) + 1 = |B| + 2. Thus, from the maximality of B, we deduce that E is not blue. And, since |B| > 2, it follows that E is not green. A contradiction.

- If b_2 < g, we use the same reasoning on E' = {g-b / b in B} \cup  {g+2-b / b in B}, to obtain the same contradiction.

And we are done....

Note that the result is optimal in the sense that for n= 4k, we may color the odd integers in red, the numbers of the form 4p+2 in blue and the numbers of the form 4p in green. Then N(R), N(B), N(G)  \geq  4 and, by parity argument, there is no tricolor solution of x+y = z.

Pierre.

[quote="jensen"]does it have a short solution?[/quote]
Yeah,of course!!! I have a shorter solution!
I'll post it when I have time.

We call $(a,b,c)$  "good" if $a\in A,b\in B,c\in C$ and have a number  equal the sum of others.We prove by contradiction way.
Stragedy:
+Assume $1 \in A$.$k \geq 1$ such that $1,2,...,k-1 \in A$and $k \notin A$.WLOG,assume $k \in B$.
+Assume there exist $A,B,C$ st there isn't "good".
We prove :If $x \in C$ then $x-1 \in A$.
...I'll come back.

this solution is well-known
jensen ask for the general one

I don't understand anhminh solution.Can someone help me?


If you dont know this... or maybe was it a 5 sec question? Order of operations, so $ 2*(2 \minus{} 4)^2$ will be $ 2* (\minus{} 2)^2$. Exponent first, $ 2*4$ and then you get $ 8$. Please note me if I am wrong, I did this in the middle of the night.

That's wrong.
$ 2*(2\minus{}4)^{2}$
Simplify inside the parentheses
$ 2*(\minus{}2)^{2}$
Exponents are before multiplication.
$ 2*4\equal{}\boxed{8}$

Actually, shouldn't it be 2(2+4)(2-4)=-24? Please correct me if i'm wrong. :)

$ (2\minus{}4)^2\equal{}(2\minus{}4)(2\minus{}4)$, not $ (2\plus{}4)(2\minus{}4)$. I'm not sure where you got the $ (2\plus{}4)$ from.

jc123 wrote:

Actually, shouldn't it be 2(2+4)(2-4)=-24? Please correct me if i'm wrong.

Only (2)^2-(4)^2 = (2+4)(2-4) not (2-4)^2 = (2+4)(2-4) because a^2-b^2 = (a+b)(a-b)

So you are wrong. The correct answer is

2*(2-4)^2 = 2*(-2)^2 = 2*4 = 8

Then also you can work out the long way which is

(2-4)(2-4)=4-16+16=4+0=4

so 2*4=8.


Four positive integers $a$,$b$,$c$, and $d$ have a product of $8!$ and satisfy

\[ab+a+b = 524\\ bc+b+c = 146, \ \text{and}\\ cd+c+d = 104.\\\]

What is $a-d$?

\[\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12\]

[hide="Hints"] Add $1$ to every equation. [hide="And then..."] Use Simon's Favorite Factoring Trick. [hide="Which is..."] The idea that $(a+1)(b+1) = ab+a+b+1$ and so forth. [/hide] [/hide] [/hide]


algebra
polynomial
algebra proposed
Kazakhstan

Let $ f(x,y,z)$ be the polynomial with integer coefficients. Suppose that for all reals $ x,y,z$ the following equation holds:
\[ f(x,y,z) \equal{} \minus{} f(x,z,y) \equal{} \minus{} f(y,x,z) \equal{} \minus{} f(z,y,x)
\]
Prove that if $ a,b,c\in\mathbb{Z}$ then $ f(a,b,c)$ takes an even value

$ f(x,y,z) =-f(x,z,y) =-f(y,x,z) =-f(z,y,x) $ ....(1)
from (1) we can see $ f(x,y,z) \equiv f(x,z,y) \equiv f(y,x,z) \equiv f(y,z,x) \equiv f(z,y,x) \equiv f(z,x,y) (mod 2) $...(2);
And from (1) x=m,y=m,z=n then $f(m,m,n)=0$
put m=n=0; then $0=f(0,0,0)\equiv f(2k,2l,2m) (mod 2)$;
put m=0,n=1; then by (2) $0=f(0,0,1)\equiv f(2k,2l,2m+1)$[and all of it's permutations]$(mod 2)$,
put m=n=1 then$0=f(1,1,1)\equiv f(2k+1,2l+1,2m+1)(mod 2)$
put m=1,n=0 then by (2) $0=f(1,1,0)\equiv f(2k+1,2l+1,2m)$[and all of it's permutations]$(mod 2)$;
so $f(a,b,c)\equiv 0(mod 2)$... :)


I needed help with this problem ASAP :)

a . b = 0 (given)
x + a(x . b) = b
find x
x, a, and b are all vectors
the "." means dot product
so obviously you can get from that x . b = b .b
but that doesnt get you anywhere cos it doesnt mean that x = b

thanks

It also has to be assumed that the vectors $\vec a, \vec b, \vec x$ are in the same plane and that $\vec a \neq 0, \vec b \neq 0$. The vectors $\vec a, \vec b$ are than the base of this plane and every vector in this plane can be expressed as their linear combination:

$\vec x = \kappa \vec a + \lambda \vec b$

Then

$\vec x \cdot \vec a = \kappa \vec a \cdot \vec a + \lambda \vec b \cdot \vec a = \kappa a^2$

$\vec x \cdot \vec b = \kappa \vec a \cdot \vec b + \lambda \vec b \cdot \vec b = \lambda b^2$

Dot-multiplying the given equation by $\vec b$

$\vec x \cdot \vec b + (\vec a \cdot \vec b)(\vec x \cdot \vec b) = \vec b \cdot \vec b$

$\vec x \cdot \vec b = b^2$

It follows that $\lambda = 1$. Dot-multiplying the given equation by $\vec a$

$\vec x \cdot \vec a + (\vec a \cdot \vec a)(\vec x \cdot \vec b) = \vec b \cdot \vec a$

$\vec x \cdot \vec a + a^2b^2 = 0$

It follows that $\kappa = -b^2$ and

$\vec x = -b^2 \vec a + \vec b$

Substituting this to the given equation as a check:

$-b^2 \vec a + \vec b + \vec a ((-b^2\vec a + \vec b) \cdot \vec b) = -b^2 \vec a + \vec b + \vec a b^2 = \vec b$

It checks.


show that if the angle bisector from A, altitude from B, and median from C have the same length, then ABC is equilateral

We have ma=lb=hc.

Since hx \leq lx \leq mx ==>
1) mb \geq ma ==> b \leq a;
2) mc \geq ma ==> c \leq a.
So a = max(a,b,c) ==> ∠A \geq pi/3.

Let A1 is the midpoint of BC.

Consider A1H1 ⊥ AB, H1 \in AB. We have A1H1 = hc/2 ==> ∠A1AB = pi/6.
Consider A1H2 ⊥ AC, H2 \in AC. We have A1H2 = hb/2 \leq lb/2 = ma/2 ==> ∠A1AC \leq pi/6.

So ∠A = ∠A1AB + ∠A1AC \leq pi/3.

Finally pi/3 \leq ∠A \leq pi/3 ==> ∠A = pi/3 ==> ∠A1AB = ∠A1AC = pi/6 ==> triangle ABC is equilateral.


Hey, I did alot better than I expected on USAMO (not a 0  :lol: ). I was wondering, how do they choose who goes to MOP? Is there some sort of Index? Or is it just on the USAMO? Are there tiebreakers?

About where will that be this year? 400+?

I think that's not quite fair, because USAMO is completely different from those AIME and AMC. To score high on AIME and AMC not means to score high on USAMO. The MOP wishes to select people that may represent USA to enter IMO, and IMO is very like USAMO, so MOP needs more people who can score high on USAMO. In my opinion, just using USAMO score to judge MOP is enough.

Oh no! Mop also counts amc&aime scores?  That's very, very bad for me.  Where did it say that btw?

Well first of all, if there is a MOP index, then I'm pretty sure it's

9*AIME + 8*USAMO

based on memory (I couldn't find anything on the AMC website just now). However, most people think the MOP index doesn't actually exist. They probably just choose based purely on USAMO score, and then if there are ties, they will most likely accept all the people who tied. If that's way too many, I'm guessing they will look at AMC/AIME, although this case seems very very unlikely.

This is what the website wrote:

[quote="AMC"]
Selection of the students for the MOSP is based on the results of the United States of America Mathematical Olympiad (USAMO). The top 12 students are automatic selectees, and then the non-seniors from the next 15-16 highest scoring participants are chosen. Depending on available funding, AMC may invite additional groups of very promising students who have risen to the top on the American Mathematics Competitions to the MOSP.
[/quote]

I was just going off of what I heard once...just USAMO score would make the most sense to me though.

According to this post from last year, it seems that winners and honorable mentions are selected strictly by USAMO score. Any additional MOPpers are selected with an index.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34745

Hmm and what about freshman MOP? Everywhere I've looked says that the cutoff for that is solely based on USAMO.

Back in the days of five 5-point problems, the index was 4*USAMO+10*AIME+AHSME. I don't know what this was used for, and I think that any index that may still exist is used purely for students' interest. The AMC site (and everything else) says that MOSP selection is just by USAMO score. In fact, the site says very specifically that the USAMO index is essentially trashed after the AIME.

Hmm I thought USAMO index is only there for tiebreakers.

Hmm I thought a particularly elegant solution or a carefully stated non-trivial extension with proof gave you tiebreaker points.

Winners and Honorables were always selected solely based on USAMO score...

I think MOP index does exist because there was a person at 2003 MOP with a 14 on USAMO as well as a very high index, and I know at least one person who got a 15 (and a notably lower index) and did not get invited (both people were the same grade too)

[quote="darktreb"]I should rephrase - I think MOP index DID (past tense) exist.

I agree with most of the people that it no longer exists, especially with the 30 ninth grader system. If you're 10th grade or above, your USAMO score should be all that matters anyway...[/quote]

Why do you think it no longer exists? Clearly it did last year, for blue mops.

It didnt seem to for red mops, but by setting the score at 7 (last year), they only had about 25 kids.

I don't think it existed last year for blue MOPpers (though I didn't really talk to all of them). It seemed that every single person who was in Blue got 21 or above. I know at least one person who got a 20 on the USAMO and an index so high that I find it highly improbable every single Blue MOPper had a better index...

I could just be completely wrong. I was too busy worrying about my USAMO score (as seniors do...)

[quote="lfm"]

Why do you think it no longer exists? Clearly it did last year, for blue mops.
.[/quote]

[quote="ThAzN1"]Ironically, I think there was a person who got 20 on USAMO last year + really high index (and didn't make MOP) and a few other people who got 21 + much lower index (and did make it)

Weird... I wonder how they'll do it this year.[/quote]

Apparently not...


analytic geometry
geometry
algebra unsolved
algebra

Solve the following system in the set of the real numbers :

$ \{\begin{aligned}8y^2-4x^2+8z^2&=9a^2 \\
8x^2-4y^2+8z^2&=9b^2 \\
8x^2+8y^2-4z^2=9c^2\end{aligned}$

where $ a,b,c$ are the sidelenghts of a triangle . What represent $ x,y,z$ ?

Hello,

I belive I solved this, but I dont know how to say what x,y,z represents in english, so I'll try to expain so someone can translate the meaning only :).

In triangle with sidelenghts a,b,c, we have:

4x^2= 2b^2 + 2c^2 - a^2

4y^2= 2a^2 + 2c^2 - b^2

4z^2= 2a^2 + 2b^2 - c^2

From this three equations we easily get, for instance:

a^2=(4/9)(2y^2 + 2z^2 - x^2)                                        (1)

By doing circular switches we get equations for sidelenghts b and c too.

Now only turn  (1) into:

9a^2= 8y^2 + 8z^^2 - 4x^2 and you get your equations that you posted :).

The thing that bugs me is that I dont know how to say or pronounce what in triangle is "x", "y" or "z"....

But let's say that dot T in triangle is the dot which has coordinates (p,q), where p(t)=[p(a)+ p(b) + p(c)]/3 ; same for q.

Now make a line throught dots A and T. Distance between A-T and distance between T - "the dot where line A-T touches like B-C" are same as 2:1.

Please someone translate this :)

Edit: I hope this link can help you to understand what I meant with x,y,z.

http://upload.wikimedia.org/wikipedia/sr/6/67/Teziste-trougla.gif

Lenghts CP, BN, AM are x,y,z...

[quote="Tomekk"]The thing that bugs me is that I dont know how to say or pronounce what in triangle is "x", "y" or "z"....[/quote]

Unless I'm a complete idiot, I think you're talking about the [url=http://en.wikipedia.org/wiki/Median_(geometry)]medians[/url] of the triangle.

[quote]Unless I'm a complete idiot, I think you're talking about the [u][b]medians[/b][/u] of the triangle.[/quote]

Yep, exactly :). Thanks for translation.


I have 2 inequalities that I solved...but I want to make sure I did the right thing.... (the other is in another post)

1)$ \minus{}1\leq x \leq \frac{x}{x\minus{}2}$

First, I multiplied $ (x\minus{}2)^2$ to both sides because it's always positive
and the inequality sign won't change. then it becomes

  $ \minus{}(x\minus{}2)^2 \leq x(x\minus{}2)^2 \leq x(x\minus{}2)$. I divided the inequality into 2 parts, namely
$ \minus{}(x\minus{}2)^2 \leq x(x\minus{}2)^2$ and $ x(x\minus{}2)^2 \leq x(x\minus{}2)$.

The first inequality produced $ (x\plus{}1)(x\minus{}2)^2 \geq 0$ which means $ x > \minus{}1$
The second inequality produced $ x(x\minus{}2)(x\minus{}3) \leq 0$ which yields $ 2 \leq x \leq 3$

From here, there's no intersection, so i arrived with the answer No Solution. Is it right? thanks! :D

[quote="Jakey"]I have 2 inequalities that I solved...but I want to make sure I did the right thing.... (the other is in another post)

1)$ \minus{} 1\leq x \leq \frac {x}{x \minus{} 2}$

First, I multiplied $ (x \minus{} 2)^2$ to both sides because it's always positive
and the inequality sign won't change. then it becomes

  $ \minus{} (x \minus{} 2)^2 \leq x(x \minus{} 2)^2 \leq x(x \minus{} 2)$. I divided the inequality into 2 parts, namely
$ \minus{} (x \minus{} 2)^2 \leq x(x \minus{} 2)^2$ and $ x(x \minus{} 2)^2 \leq x(x \minus{} 2)$.

The first inequality produced $ (x \plus{} 1)(x \minus{} 2)^2 \geq 0$ which means $ x > \minus{} 1$
The second inequality produced $ x(x \minus{} 2)(x \minus{} 3) \leq 0$ which yields $ 2 \leq x \leq 3$

From here, there's no intersection, so i arrived with the answer No Solution. Is it right? thanks! :D[/quote]
try x=3 and x=0 you still get the correct solution  :)

err, so can you suggest how i should solve it?

firstly , you wrote $ x>\minus{}1$ and $ 2 \leq x \leq 3$ have no intersection , it is wrong :) ( 'cause $ 2 \leq x \leq 3$ then $ x >\minus{}1$  , the intersection is [2,3] :) )

How about go like this :
divide the first inequality in 2 parts $ x \geq \minus{} 1$ and $ x\leq \frac {x}{x \minus{} 2}$
solve the second one , $ x \minus{} \frac {x}{x \minus{} 2} \leq 0$ equivalent to $ \frac {x(x \minus{} 3)}{x \minus{} 2} \leq 0$ gives $ x(x \minus{} 3)(x \minus{} 2) \leq 0$ , we obtain $ x\leq 0$ or $ 2 < x \leq 3$
since $ x \geq \minus{} 1$ so we have the final solution is $ \minus{} 1\leq x \leq 0$ and $ 2\geq x \geq 3$

oh..how stupid of me to get that part wrong!:) otherwise, i'd gotten the same answer with you...but it should be $ 2\leq x\leq 3$ btw


What are the last three digits of $ 10!$

Please do not use congruence modulo.

Thank you in advance

Since its a low number we can quickly factorize it and easily.
$ 10! \equal{} 2^83^45^27$. Since each pair of factors of $ (2,5)$ give a trailing $ 0$, we get 2 trailing zeros. Thus we are left to find only the third digit.
This is found by looking at $ \frac {10!}{100} \equal{} 2^63^47$. I suppose implicitly I'm using congruency, but I am also familiar with the small numbers.
Considering only the last digit of the product we can say its the last digit of $ 4*1*7$ which is equal to $ 8$.
Thus the last three digits are $ 800$.


[quote="silouan"]Let a bicentric quadrilateral ABCD . Its incircle touches its sides
AB,BC,CD ,DA  at the points K,L,M,N . Prove that the lines KM, NL, AC,BD are concurrent .[/quote]
What do you mean by saying "bicentric",if it means that the qudrilateral can be both inscribed and circumscribed,then it isn't neccessery.This fact is true for all quadrilaterals which can be circumscribed around a circle,and it as a part of a well known fact,wich states that the major diogonals of a cirumscribed hexagon are concurent.  :)

Yes this is the meaning of bicentric .Could you post a solution of that you said ?Thank you.


Let us take the incenter I of the triangle ABC. Call T and T' the intersections of the perpendicular line to AB that pass throgh I with the incircle (T € AB). Call M the midpoint of AB. Show that T'C//MI

The point T is the point where the incircle of triangle ABC touches its side AB, and the point T' is the point diametrically opposite to T on the incircle of triangle ABC. Now, in post #3 of http://www.mathlinks.ro/Forum/viewtopic.php?t=5811 , I showed that (with changed notations), if Z' is the point where the C-excircle of triangle ABC touches its side AB, then the point T' lies on the line CZ', and the line MI is parallel to the line CZ'. Of course, since the point T' lies on the line CZ', the line CZ' coincides with the line T'C. Hence, we can rewrite MI || CZ' as MI || T'C. And the problem is solved.

  Darij


In a cannery peas of mass M come out of a pipe uniformly at a velocity V with a separation d. A fly of mass m sitting on the end of the pipe hops with  negligible speed on to a passing pea.

(1) How much energy must the fly's legs absorb in order to hang on to  the pea?
(2) How far does the fly travel before the pea behind catches up?

Just before the peas collide the fly lets go of the first pea with no change of velocity.The peas then collide without loss of energy,after which the fly catches hold of the second pea

(3 ) How fast is the first pea travelling after the collision?
(4) How far does the fly travel on the second pea before being caught by the third pea?

[hide]Using conservation of momentum, I'd say that the velocity of pea+fly is $\frac{M}{m+M}V$, therefore, the energy difference is $\frac{1}{2}\frac{m}{m+M}MV^2$. The pea+fly are $\frac{m}{m+M}V$ slower than the pea behind so the pea behind catches up after $\frac{M}{m}d$. When the two peas collide, it should be just an ordinary ellastic collision of two bodies of equal mass which means that they only exchange their velocities. The first pea will travel at velocity $V$ and the third pea will chatch up after $\frac{M}{m}d$.[/hide]


geometry
probability
MATHCOUNTS
number theory
\/closed

I'm in 7th grade and I'm finishing up Algebra 1... although I know quite a bit of Algebra 2. I completed Mastering Mathcounts/AMC8 a few months ago, (I think it's called Advanced Mathcounts/AMC8 now) that was a pretty good level for me... challenging but not too hard. Looking at the info that AoPS has online I have narrowed my next class choices down to:
          Intro to Number Theory
          Intro to Geometry
          Intermediate Counting and Probability
          Algebra 2 or Algebra 3
I wanted to take a class with Mr. Rusczyk, but would taking Algebra 2 or even Algebra 3 be pointless if I'm doing Algebra 2 in school next year? And which would be the best choice out of the other 3?
       One more thing... I have done mathcounts in 6th and 7th grade and will be doing it again in 8th grade, so can people also recommend the classes that are the best for preparing for mathcounts?

Algebra 2 or Algebra 3 will [b]not[/b] be pointless if you're taking Algebra 2 in school.  This year, I'm taking Geometry in school.  It's quite amusing how all the material covered in the first day in the AoPS class was covered in the first month in school.  Everything is studied more in-depth in the AoPS classes, so taking Algebra 2 or Algebra 3 definitely not be pointless.

I recommend starting with Intro NT and Intro Counting & Probability (in either order), and following that up with Intro Geometry.  It's a tough call on our Algebra 2 class - what you might want to do is get the Intro Algebra book and decide for yourself if you think you need a class for the latter half of the book.  You'll want to take the Algebra 3 class at some point, but I'd recommend the Intro classes first.

Whoa! What? You're skipping geometry? I'm in geometry right now, but I am going to pre-calc next year because I'm going to take the test to skip Algebra 2. Whhoa..... Geometry plays a big role in math dude. I know this kid that skipped it, but regretted it because he didn't know a lot of formulas and theorems.  :(

Sorry I didn't reply earlier but I wasn't able to come on...
       Anyway, no jonathonchou117, I'm not skipping geometry. I'm homeschooled and the curriculum we use has geometry spread out through Algebra and Pre-Calc... So I already know a lot of geometry. Good point, though.
     Thanks for all the feedback, guys!
         And if you were wondering my top choices now are Intro to Geo and Intro to Number Theory... although I'll consider Algebra 3 later. So Mr. Ruscyzk, you should teach Algebra 3, too!!  :lol: I suppose I'll just have to go to your math jams...

According to my friends, geometry is the most important for competitions.

For me, the real value in geometry is that geometry problems offer so many different lines of attack.  The key to being good at geometry is to be able to sift through all these different approaches and pick out the ones that will be most likely to succeed.  While I use very little geometry in my day-to-day life, I have to sift through many different facts and goals in my professional (and personal) life, and choose the most appropriate actions to meet the most important goals.  This is what geometry taught me how to handle.


advanced fields
advanced fields unsolved

Is it true that any bounded operator A:H-->H
(where H is a Hilbert space, and specifically Banach)
is also compact?

"Bounded" means continuous. The continuous image of a compact set is compact.

Certainly not! Compact means mapping bounded sets to relatively compact sets, completely continuous.

The prototype $ A$ is a diagonal operator, $ A(x_n) \equal{} (\alpha_n x_n)_{n\equal{}1\ldots\infty}$ where $ (x_n) \in \ell^2$. $ A$ is compact if and only if $ \alpha_n \to 0$

So, the standard counterexample here: the identity, in any infinite-dimensional space.


Happy New Year!

I was bored awhile ago and came up with this:

$ 2010\equal{}7^1\plus{}6^3\plus{}5^4\plus{}4^5\plus{}3^2\plus{}2^7\plus{}1^6$ - All of 1-7 used as bases and exponents.

$ 2010\equal{}2\cdot 3\cdot 5 \cdot (7\plus{}11\plus{}13\plus{}17\plus{}19)$.

First 8 primes.

Happy New Year everybody (especially the Canadians who are reading this)!

Cheers,

Hunter

Yeah happy new year, and good luck to everyone on the upcoming CMOR! :lol:


Find all integer $ (x,y,z,t)$, which $ x\plus{}y\plus{}z\plus{}t\equal{}0$ and $ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt$ is the perfect square.

I solved it :D  :D  :D
Prove this
$ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt\equal{}2(\displaystyle\sum(xy))^2$
($ \displaystyle\sum(xy)\equal{}xy\plus{}xz\plus{}xt\plus{}yz\plus{}yt\plus{}zt$)
We have
$ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt\equal{}x^2(y\plus{}z\plus{}t)^2\plus{}y^2(x\plus{}z\plus{}t)^2\plus{}z^2(x\plus{}y\plus{}t)\plus{}t^2(x\plus{}y\plus{}z)^2\plus{}4xyzt
\equal{}2\displaystyle\sum(x^2y^2)\plus{}2\displaystyle\sum(x^2yz)\plus{}4xyzt$
And
$ 2(\displaystyle\sum(xy))^2\equal{}2\displaystyle\sum(x^2y^2)\plus{}4\displaystyle\sum(x^2yz)\plus{}12xyzt$

$ 2\displaystyle\sum(x^2y^2)\plus{}2\displaystyle\sum(x^2yz)\plus{}4xyzt\equal{}2\displaystyle\sum(x^2y^2)\plus{}4\displaystyle\sum(x^2yz)\plus{}12xyzt$
$ \displaystyle\sum(x^2yz)\equal{}\minus{}4xyzt$
But
 $ 4xyzt\equal{}xyzt\plus{}xyzt\plus{}xyzt\plus{}xyzt\equal{}\minus{}(y\plus{}z\plus{}t)yzt\minus{}...\minus{}xyz(x\plus{}y\plus{}z)\equal{}\minus{}\displaystyle\sum(x^2yz)$
QED
WE have that
$ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt\equal{}2(\displaystyle\sum(xy))^2$
But $ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt$ is a perfect square and 2*(perfect square)
it's possible if $ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt\equal{}0$
$ |x^4\plus{}y^4\plus{}z^4\plus{}t^4|\equal{}|4xyzt|$
But by AM-GM $ |x^4\plus{}y^4\plus{}z^4\plus{}t^4|\ge|4xyzt|$ and $ |x^4\plus{}y^4\plus{}z^4\plus{}t^4|\equal{}|4xyzt|$ if $ |x\equal{}y\equal{}z\equal{}t\equal{}0|$.
We find all $ (x,y,z,t)$ - $ (0,0,0,0)$

P.S. In this solution $ \displaystyle\sum$ is cyclic sum.

And I have any solution this problem :lol:
$ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt\equal{}\frac{(x^2\plus{}y^2\plus{}z^2\plus{}t^2)^2}{2}$
We have that $ x^2\plus{}y^2\plus{}z^2\plus{}t^2\equal{}0$ and $ x\equal{}y\equal{}z\equal{}t\equal{}0$
But need prove
$ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}4xyzt\equal{}\frac{(x^2\plus{}y^2\plus{}z^2\plus{}t^2)^2}{2}\equal{}2(\sum xy)^2$
It's no hard, but I proved this 30 min


symmetry
ratio
geometry proposed
geometry

Let there be two circles. Find all points $M$ such that there exist two points, one on each circle such that $M$ is their midpoint.

[hide="hint"]First answer this question:
For two circles $ w_{1}$, $ w_{2}$, find all points $ M$ such that the central symmetry of $ w_{1}$ is tangent to $ w_{2}$
[/hide]

Let $ P$ be a point fixed on one circle, then the locus of $ M$ is the other circle dialated by ratio 1/2. Repeat this for all points and you get like a ring.

let centers of circles be $ O_{1}$ and $ O_{2}$ and $ N$ be midpoint of $ O_{1}O_{2}$ and $ A$ be a point on $ (O_{1})$ and $ B$ be on $ (O_{2})$.
it's easy to prove that:$ \overrightarrow{NM}\equal{}\overrightarrow{O_{1}A}\plus{}\overrightarrow{O_{2}B}$ the maximum lenght of $ MN$ is $ r_{1}\plus{}r_{2}$ when $ O_{1}A\parallel O_{2}B$ and minimum of $ MN$ is $ |r_{1}\minus{}r_{2}|$ when $ \overrightarrow{O_{1}A}$ and $ \overrightarrow{O_{1}B}$ be in opposite direction.hence the locus of $ M$ is the pink part in the figure.


rotation
geometry
geometric transformation

The diagram below shows two bodies of masses 0.50 kg and 1.00kg connected by a light rigid rod of length 4.00 m and on a smooth surface .A body P of mass 0.50 kg which moves at velocities 4.00 m/s collides and sticks to the body of mass 0.50kg and the system rotates about the center of mass . Calculate the angular velocity of the system .

[img]http://tinypic.com/jfytg3.jpg[/img]

Can someone help ?

I think the center of mass isnt at the center of the rod since both the mass arent equal . Also can you please explain more detail but no just write down the numbers ...

I wonder how you got 2^2 in your first equation.

Ok ,I think I figure out how to do. Here is how I do but I dont know whether it is correct or not

First we find the center of mass of the rod which is $8/3$ cm from the 0.50kg body or $4/3$ cm from the 1.00 kg body .

Since angular momentum is conserved , the angular momentum before collision is

$mr^2(\frac{v}{r})$ where $m$ is the mass of $P$ and $v$ is its velocity while $r$ is distance from the 0.5 kg body to the center of mass ($=8/3$ cm) .

Hence we get $(0.5)(8/3)(4.00)=16/3$

Total angular momentum after collision is $I\omega$ , where $I=(0.5)(8/3)^2+(1.0)(4/3)^2=16/3$ .

Hence we get $16/3=16/3\cdot\omega$ $\implies \omega=1$

Are you saying that the moment of Inertia is now $I=1(2^2)+1(2^2)=8$ ? Also total momentum before collision is $(0.5)(2)(4)=4$ instead ?? so that would give us $\omega=0.5$ ? :wallbash:

[quote="bus"][quote="varun"]We can conserve the angular momentum about the centre of mass of the body(after the collision).
We get,
0.5*4*2=I*2^2 w=4I w
Where I is the moment of inertia about the centre of mass of the body, and w is the final angular velocity.
I=2^2 + 2^2 = 8 $Kg m^2$
Thus w = $1/8 rad/s$[/quote][quote="varun"]I have taken the centre of mass of the system after the collision.

Also $2^2$ is for $r^2$[/quote]You're saying that 0.5*4*2 is the angular momentum about the center of the rod before the collision, right? That seems to be all right. But why do you put it equal to I*r^2*w ? I've always thought that angular momentum = I*w...

Shyong: you're forgetting that after the collision the body P sticks to the 0.5 kg body. Therefore, I = (0.5[b] + 0.5[/b])(8/3)^2 + (1.0)(4/3)^2. Also, the point you've selected as the ceneter of mass won't be the center of mass anymore...[/quote]

Yes bus, you pointed out my error. Actually it was done in a bit of hurry. So I apologize for the error.
It should be I.w.

Also, shyong your solution was correct untill you took r= 8/3.
The point about which momentum is conserved is same. Thus, when you are taking later about the centre of rod. So, initially taking about the centre of the rod.
Understood.

varun , I dont quite get what you mean in your last post  :huh:

By the way , why we use r=2 before collision but not r=8/3 ? Isnt it before collision the center mass is still not at the center of the rod ? I still dont get this part  :(

You can take the angular momentum about any point - it doesn't matter but you have to take it about the same point before and after the collision. So you can do it about the center of the rod - that's precisely what varun did:

angular moemntum before the collision = 0.5 * 2 * 4 = 4

angular momentum after the collision = I * w = (1 * 2^2 + 1 * 2^2) * w = 8 * w

w = 1/2 rad/s,

or you can take the angular momentum about the point which is 8/3 m from the 0.5 kg body and 4/3 m form the 1 kg body like you did:

angular momentum before the collision = 0.5 * 8/3 * 4 = 16/3

angular momentum after the collision = I * w = (1 * (8/3)^2 + 1 * (4/3)^2) * w = 80/9 * w

w = 3/5 rad/s.

However, this second w is not the angular velocity you're looking for - it's "the rate of rotation" about the given point. When you talk about angular velocity, you usually want to know the rate of rotation about the center of gravity of the object. Nevertheless, angular momentum is still a valid physical quantity and the conservation of angular momentum holds about any point.

(Please correct me if I said something terribly wrong :).)

Well i didnt know that angular velocity means such thing  :roll: Thx bus :D

Something I'd like to add to "the rate of rotation" I've mentioned yesterday - notice that it won't be constant because the moment of inertia will change with time if we keep the point we take for center of rotation fixed. Therefore it usually only make sense to calculate $\omega$ about points about which the moment of inertia is constant.


calculus
integration
algebra
binomial theorem

Let $n=1667$. What is the first nonzero digit in the decimal expansion of $\sqrt{n^2+1}-n$?

with calculator
[hide]$2$[/hide]
Do you want this without calculator ?

Yeah... otherwise this problem becomes trivial :roll:

[hide]We're looking for $k$ such that

$\frac{k}{10^m}\leqslant \sqrt{n^2+1}-n<\frac{k+1}{10^m}$

for some $m\in\mathbb{N}$

As $\sqrt{n^2+1}-n=\frac{1}{\sqrt{n^2+1}+n}$, we have

$k\leqslant \frac{10^m}{\sqrt{n^2+1}+n}<k+1$

As $n=\frac{10002}{6}$, we have $\sqrt{n^2+1}+n=\frac{\sqrt{10002^2+6^2}+10002}{6}>\frac{2\cdot 10^4}{6}$

Therefore $\frac{10^m}{\sqrt{n^2+1}+n}<3\cdot 10^{m-4}$

Hence $m=4$ and it's very easy to prove that $\frac{10^4}{\sqrt{n^2+1}+n}>2$. Therefore, the first non-zero decimal is $2$, and it's fourth behind the point.[/hide]

Another way you can do it is by doing the binomial theorem with $\frac{1}{2}$ and obviously you can stop once you hit the first nonzero term.

$\sqrt{1667^2+1}=(1667^2+1)^{\frac{1}{2}}$ and so $\binom{\frac{1}{2}}{0}1667+\binom{\frac{1}{2}}{1}\frac{1}{1667}+\binom{\frac{1}{2}}{2}\frac{1}{1667^2}...$

you'll clearly see that the first nonzero digit is $\boxed{2}$, although it is VERY close to $3$

[quote="breez"]Um...how do you "choose 1" from $\frac{1}{2}$?[/quote]

It is called the "extended" binomial, wher instead of $\frac{n!}{k!(n-k)!}$, you change it into $\frac{n*(n-1)*(n-2)*\dots*(n-k+1)}{k!}$, so $n$ doesn't have to be positive integer.  It comes out when you apply this advance form to non-integral powers, the binomial theorem still holds.

idea: you could rationalize the numerator!

Another fun way (if you happen to notice that n is about (10000/6) ..  and know your continued fractions etc )  to see that the expression is  nothing but
${\cfrac 1 {2n + \cfrac 1 { 2n + \cfrac 1 \dots }}}$
Obviously it is  ([i]only slightly[/i]) less than $\frac 1 {2n} = \frac 3 {6n} = \frac 3 {10002} < 0.0003$ so answer must be 2
(Obviously denominator (2n)  is ([i]a lot)[/i])  less than 5000 so expression  must be > 0.0002)
(By the way the expression and the value 3/10002  are equal to 8   significant figures)
Added later: The method may be similar to one  hidden above ..


How many non-overlapping 2 by 2 squares will fit into a circle with radius 8?

Because it is somewhat relevant right now and I can't seem to find the solutions to the Purple Comet contests before 2007, here's stuff (the proof for this problem which is in one file does not seem to actually prove anything except that 40 is achievable):


geometry
geometric transformation
rotation
geometry unsolved

It seems that, constructing the triangle BTC with BT = TC, angle BTC=angle A' and A and T on each side of BC, AT will intersect BC at A', then parallel lines drawn through A' to BT and CT will intersect AB and AC respectively at C' and B'.
Similar constructions for CA and AB, provided the intersection point (A' for BC is on the segment BC), therefore A' cannot be any given point on the border of the triangle ABC, but maximum three points.

Best regards,
sunken rock

[b]Solution 1.[/b] Perform a rotation with center $ A'$ and angle equal to the given angle $\widehat{B'A'C'}.$ Such a rotation takes the line $c \equiv  AB$ into a line $ c',$ thus  $B' \equiv{} AC \cap c'.$ Now the construction of the isosceles $\triangle A'B'C'$ is straightforward.

[b]Solution 2. [/b] Circles $\odot(AB'C'),$ $\odot(BC'A'),$ $\odot(CA'B')$ concur at Miquel point $P.$ Angle chasing using the cyclic quadrilaterals $PA'BC'$ and $PA'CB'$ gives $\widehat{BPC}=\widehat{BAC}+\widehat{B'A'C'}$ and similarly $\widehat{APC}=90^{\circ}-\tfrac{1}{2}\widehat{B'A'C'}+\widehat{ABC}.$ Thus knowing the measure of $\widehat{B'A'C'},$ we can construct the point $P$ intersecting the circular arcs that see $BC,CA$ under the referred angles. The construction of $B'$ and $C'$ follows.


A book's pages are numbered from 1 to 100, and you know that two pages are torn out. How many different combinations of torn pages are there if the sum of their page numbers is 126?

what do you mean by "page?" because books are put together so that each 'page' actually contains four page numbers, and this sum is constant (2(100+1)=202).

I meant one half of a page, if we're using your definition of a page. Two consecutive page numbers.


Hello I'm from Poland and I have such a problem to solve:
[quote]In how many ways can we choose k books from n books standing on the shelf, if between 2 chosen books there must be at least 2 not chosen books[/quote]
Sorry for my not perfect English  ;)  and thanks in advance for your help.

Answer: $C_{n-2k+2}^k$ (binomial coefficient, $\frac{(n-2k+2)!}{k!(n-3k+2)!}$). Consider any sequence of $k$ 1-s and $n-3k+2$ 0-s. Add two 0-s between any two consecutive $1$-s. You get a sequence of $k$ 1-s and $n-k$ 0-s, in which there are at least 2 0-s between any two 1-s. And it is a bijection. Now replace 1-s to chosen books and 0-s two not chosen books.


(i) Prove that  $\lim_{n \to \infty}a_{n}= 2\/$ where $a_{n}=\sum_{r=0}^{n}{n \choose r}^{-m}\/$,  $m >0$.

(ii) Let  $k$ be a positive integer and $ S_{n}=\sum_{r=0}^{n}{kn \choose kr}\/$. Then $ \lim_{n \to \infty}(S_{n})^{\frac{1}{kn}}=2\/$.


να δειξετε οτι σε οποιαδηποτε αριθμητικη προοδο μπορούμε να βρουμε αυθαιρετο πληθος διαδοχικων ορων οι οποιοι να ειναιν ολοι συνθετοι...

dn katalava kati....
ti enois legontas af8ereto pli8os diadoxikon klp klp....?
enois....osodipote megalo enois??i ....tulaxiston dio diadoxikus orus....?
kai episis mipos 8a prepi o $a_{1}$ kai i diafora $w$ na ine akerei...?

 :D

αυθαιρετο εννοω οσοι θελουμε εμεις ,και  προφανως μιλαω για ακεραιους ορους  :)

diladi se ka8e akolu8ia mporume na vrume diaforetiko to pli8os diadoxikus orus me ayti tin idiotita,sosta?
gia tus akereus,ontos itan yiloprofanes ala epidi dn to dieukrinizes gi ayto to rotisa.. :)

 :D

edit:diladi arki an apodikso oti se ka8e akolu8ia iparxun dio diadoxiki ori pu na ine sin8eti  sosta??(afu ine af8ereto pli8os...)

Υπεθεσε μια ακολουθεια t_λ=aλ+b.Aν υποθεσουμε οτι n>=(k-1)a+b τοτε  οι ακεραιοι n!+b,n!+a+b,n!+2a+b,.....και n!+(k-1)a+b ειναι κ το πληθος διαδοχικοι οροι της ακολουθειας.

[quote="Αρχιμήδης 6"]τοτε  οι ακεραιοι n!+b,n!+a+b,n!+2a+b,.....και n!+(k-1)a+b ειναι κ το πληθος διαδοχικοι οροι της ακολουθειας.[/quote]
pew dyo logia perissotero, kai orise ti einei ta a,b,n,k.fisikoi akeraioi ?
 προς Ραλλη---> πρεπει να το αποδειξεις για καθε φυσικο αριθμο
καλο το αποτελεσμα ?

ΟΙ α,b akeraioi kai t,λ θετικοι.Υπεθεσε την ακολουθεια t_λ=5λ+3.Τοτε αν π.χ θελω να βρω 5 διαδ ορους της που να ειναι συνθετοι επειδη t_5=23 oi αριθμοι 23!+3,23!+8,23!+13,23!+18 και 23!+23 τιρουν τις απαιτησεις του προβληματος.οκ?

Αληθεια που την βρηκες?

βιβλιο του Sierpinski  με τιτλο 250 θεματα θ.α  1970.μεταφραση στα ελληνικα  2004 ( επιστημονικη επιμελεια Μ.Λαμπρου κτλ) :)

ουαου το ειχα δει......εκπληκτικο βιβλιο για να σε οδηγησει στον σωστο δρομο σκεψης..........

λυνεται με τουλαχιστον αλλους δυο τροπους ,ο  ενας ειναι μια ασκηση απο το ιδιο βιβλιο (παλι με ακολουθιες ) και ο αλλος [hide]με κινεζικο θεωρημα υπολοιπων[/hide] αν μπορει καποιος ας δοκιμασει τον β τροπο

kosta...+apo tin apodiksi tu arximidi 6 ine fanero oti ayto pu les...
< αυθαιρετο πληθος διαδοχικων ορων...>den ine ayto pu ipes..<οσοι θελουμε εμεις>...
ala osodipote megalo pli8os diadoxikon oron pu na ine oli sin8eti....
gt otan les..<οσοι θελουμε εμεις>ine arketo na diksume oti iparxun panta dio diadoxiki ori opiasdipote akolu8ias pu na ine sin8eti..(gegonos pu apodikniete poli eykola..)
eno to osopipote megalo pli8os diadoxikon oron me ayti tin idiotita ine entelos diaforetiko...

 :D

στο παραδειγμα ο Αρχιμηδης 6 θελει να βρει 5 διαδοχικους ορους και τους   βρισκει. ομοιως μπορει να βρει m διαδοχικους ορους .καταλαβες ?

e...ayto lei kai ego...oti epidi mpori na vri m diadoxikus orus opu to m den eksartate apo to n...ine isodinamo me to na pis oti vriski osodipote megalo pli8os diadoxikon oron..(afu to m mpori na epilexti osodipote megalo 8elume..)+oxi me to na pis oti vriski osus 8eli...(gt poli apla ego mporo na po oti 8elo na vro dio diadoxikus orus...dn mu to apoklii i ekfonisi<osus 8elis>eno mu to apoklii i ekfonisi<osodipote megalo pli8os diadoxikon oron..>)

 :D

Προς Κωστα και RAKAR:
[hide]Τελικα το αυστηρο στα μαθηματικα ειναι και το απλο (+το σωστο) :lol:  :wink: [/hide]


Unknown source:

Solve the system:
\[ x^4 \plus{} y^4 \equal{} 82
\]
and
\[ x \minus{} y \equal{} 2
\]

[hide="Solution"]Let $ z$ be the arithmetic mean of $ x$ and $ y$, that is, $ z \equal{} x \minus{} 1 \equal{} y \plus{} 1$. Then the first equation is equivalent to $ (z \plus{} 1)^4 \plus{} (z \minus{} 1)^4 \equal{} 82 \iff 2z^4 \plus{} 12z^2 \plus{} 2 \equal{} 82 \iff z^4 \plus{} 6z^2 \minus{} 40 \equal{} (z^2 \plus{} 10)(z^2 \minus{} 4) \equal{} 0 \Rightarrow z \equal{} \pm \sqrt {10}i, \pm2$.
So the solutions are $ (x,y) \equal{} (1\pm \sqrt {10}i, \minus{} 1\pm \sqrt {10}i), (3,1), ( \minus{} 1, \minus{} 3)$[/hide]
Edit: If this is an unknown source, why is it in the AMC forum?


I stumbled across these websites, they have some very good links to free online literature in various mathematical subjects

http://us.geocities.com/alex_stef/mylist.html
http://gotmath.com/notes.html

before the kalva website, i used http://www.gotmath.com/aime.pdf for AIME problems... they had a whole bunch of the tests there.


Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$ and let $ t\in [1,5]$ . Show that :

$ \frac{a}{a^2\plus{}t}\plus{}\frac{b}{b^2\plus{}t}\plus{}\frac{c}{c^2\plus{}t}\le \frac{3}{t\plus{}1}$

I can prove it with the condition $ a\plus{}b\plus{}c\equal{}3$, but not with $ abc\equal{}1$ :(


modular arithmetic
probability
expected value

This is an example of a real life math problem that I ran into (although it may not look that practical, believe me, it is  :lol: ) . I am just looking for some insight on it. There may not be an answer, and this may be in the wrong section, but I am really unsure.

Let $S$ be the sum of $n$ random integers from 0-5, inclusive, where $n$ is a positive integer. You do not know what $S$ is, but you know that it is the sum of $n$ random numbers 0-5 and you know what $n$ is. You can pick an integer $p$ from $0-n$, inclusive, and an integer $q$ from 0-5, inclusive. You win if 	$S+q\equiv p \pmod{n+1}$.

My question is sort of vauge, but what is the best strategy for winning? Some more specific guided questions may be:

For specific values values of $n$, are there strategies for winning? (i.e. $n=3$)

Are there strategies for winning with sufficiently high values of $n$?

Thanks for your help.

If you choose $N \ge 5$, the expected value of $\frac{S}{N}$ is should be $\frac{0+1+2+3+4+5}{6}$ or $\frac{15}{6}$

I don't know if this helps at all, but I would look at expected value for all $N : N \ge5$


Let $ S$ be the set of all arrays $ (a_1,a_2,...,a_n)$ whose elements are from the set $ \{0,1,2\}$. Let a transformation of the array $ (a_1,a_2,...,a_n)$ be the replacement of one element $ a_j$ with $ b_j\in \{0,1,2\}$ such that no one of the numbers $ a_j,b_j$ appears in the array $ (a_1,a_2,...,a_{j\minus{}1})$. Prove that each array from $ S$ can be obtained from $ (0,0,...,0)$ with finitely many transformations.


prove that if p is a prime, then p^p-1 has a prime factor that is congruent to 1 modulo p  .........

Let prime q is divisor $q|p^{p}-1$. We have $p^{q-1}=1(mod \ q)$, and $p\not =1(mod \ q)$. Therefore $p|(q-1)\Longrightarrow q=1(mod \ p)$.

excuse me but why p\not=1 (mod \q) ??????  I solved it except prooving this already

If p=2 it give q=1. If p is odd prime, then it is possible, for example q=2.
In this case we can consider divisor of second multiple.$p^{p}-1=(p-1)(p^{p-1}+p^{p-2}+...+1)$.


probability
Duke
college
expected value
geometric sequence

You are stuck in a cave with 3 tunnels. The first tunnel takes you back to your starting point in 5 hours and the second does this in 7 hours. The third tunnel takes you out of the cave in 4 hours. Assuming that each time you randomly choose a tunnel, what is the expected number of hours you will be stuck in the cave?

[quote="Altheman"]isn't it just
$\frac{1}{3}*4+\frac{1}{3}*5+\frac{1}{3}*7=\frac{16}{3}$[/quote]

Can I please know why the calculation works like this?

Masoud Zargar

because the expected value is $E(X)=\sum XP(X)$

[quote="Altheman"]isn't it just
$\frac{1}{3}*4+\frac{1}{3}*5+\frac{1}{3}*7=\frac{16}{3}$[/quote]

Not really ... this will be the expected value,[b] if [/b]  the first two tunnels    took you[b] out[/b] instead of bringing you back to the starting poiint. ... ..Think of this way, if you change, all the timings   (of all tunnels) to 4 hours, do you really expect that the expected value to get out would be only   (4/3+4/3+4/3 = 4 ) hours when two of the trree tunnels brings you back to the starting point?

Hope this helps.

ahh, i did not read the problem correctly

we have  $\frac{1}{3}$ chance for each of the times to occur so we add those, and $\frac{2}{3}$ chance of having to repeat the choices, and each of those adds on the time, then if he is stuck for a second time, that probability is $\left(\frac{2}{3}\right)^2$, etc,

$\frac{1}{3}*4+\frac{1}{3}*5+\frac{1}{3}*7+$
$\left(\frac{2}{3}\right)\left(\frac{1}{3}*4+\frac{1}{3}*5+\frac{1}{3}*7\right)+$
$\left(\frac{2}{3}\right)^2\left(\frac{1}{3}*4+\frac{1}{3}*5+\frac{1}{3}*7\right)+...+$

this is just an infinite geometric sequence
$\left(\frac{16}{3}\right)\frac{1}{1-\frac{2}{3}}=\boxed{16\,\text{hours}}$

The answer(16 hours)  may be correct but I am not sure,  if the above method  is very clear. Okay, let me change the problem a little, suppose two tunnles (4 hours and 5 hours ) takes you out but the third tunnle (7 hours) brings you back to where you started. What will be your answer?

Added a little later (Okay I see, and understand  what you did )

Also,  I find  the  problem a little funny, I mean, even if I did not have my dog with me  (a dog can sniff and find the right tunnle the first time) it is not at  all likely that   I will keep on choosing the tunnle at random..I mean .. If I took a tunnle and 7 hours later, I am back to where I started .. the  prob, of taking that tunnle [b] again [/b] will be zero and not 1/3  :) ... (But I will not critisize more, as I have a hunch who is the author of that problem in that  Duke meet..)

here is another nice way to solve it:
let $E$ be the expected value. then after returning to the starting point the expected value is still $E$. so
$E= \frac13 (5+E)+\frac13 (7+E)+\frac13 \cdot 4 \implies E=16$
or for your problem Gyan:
$E= \frac13 \cdot 4 + \frac13 \cdot 5+\frac13(7+E) \implies E= 8$


[quote="arkmastermind"]you could also try to use newtons sums to find a cubic polynomial satisfying those conditions and then finding the sum of the roots to the fourth power.[/quote]

Of course, I know. :lol:

since i dont feel like learning latex, ill use A = alpha, B = beta, C = delta
then

$x^4 + y^4 + z^4 = \frac{\alpha^4 - 6{\alpha^2}\beta + 8\alpha\delta + 3\beta^2}{6}$

// edited to use latex

[quote="arkmastermind"]since i dont feel like learning latex[/quote]

Aw, Asif, you really ought to learn it!  It's not hard and doesn't really take that much time.  And, of course, you don't need to learn it all at once.  Just start off with the basics, like putting dollar signs around your math.  You'll gradually pick it up.

It's nice seeing you post ;)


If $ x \plus{} y \equal{} 10$ and $ x^2 \minus{} y^2 \equal{} 40$, what is the value of $ 2x \plus{} 3y$?

Dividing the 2nd equation by the first, we get x-y=4

x+y=10

x-y=4

2y=6

y=3, x=7

14+9=23


solve in the set of positive integers the equation $ x^2\minus{}y^3\equal{}7$

Posted before...

Assume tht some solution exists.
$ \mod 4$ gives $ x$ even, $ y$ odd.
Now the equation is equivalent to $ x^2\plus{}1 \equal{} (y\plus{}2)(y^2\minus{}2y\plus{}4)$, and by $ y$ odd we get $ y^2\minus{}2y\plus{}4 \equiv 3 \mod 4$. Thus there is a prime $ p \equiv 3 \mod 4$ dividing $ y^2\minus{}2y\plus{}4$, thus dividing $ x^2\plus{}1$. This means that $ x^2 \equiv \minus{}1 \mod p$ giving $ \minus{}1 \equal{} (\minus{}1)^\frac{p\minus{}1}{2} \equiv (x^2)^\frac{p\minus{}1}{2} \equal{} x^{p\minus{}1} \equiv 1 \mod p$, the last step by Fermats theorem. But this gives the contradiction $ 1 \equiv \minus{}1 \mod p$, thus there are no solutions.


Hum... about the third problem, I don't understand the "official solution" but I arrived at different answer completely different. Take a look at my proof:

----------------------------------------------
I shall proof that there's no maximum value for the sum of absolute values for a, b and c.

Let's assume that x=1 and c=100. therefore we got "abs(a+b+100) is less or equal to 100".

Now, what if a=n, and b=-n, for some positive real number n?? Assuming so, we got "abs(n-n+100) is less or equal to 100" ---> "abs(100) is less or equal to 100", which is true.

Next, we can calculate the "abs(a)+abs(b)+abs(c)"--->"abs(n)+abs(-n)+abs(100)"-->2n+100.

Since n can be anything, there's no maximum values.

------------------------------------------

Tell me what you think.

It seems you've misunderstood the problem: the inequality $ |ax^2 \plus{} bx \plus{} c| \leq 100$ needs to be satisfied for [i]all[/i] $ x$ where $ \minus{} 1 \leq x \leq 1$. Thus, when $ a \equal{} n$, $ b \equal{} \minus{} n$, and $ c \equal{} 100$, by plugging in $ x \equal{} \minus{} 1$ we get $ |n \plus{} n \plus{} 100| \leq 100$, so $ n$ can't be positive.

EDIT: About the "for all real numbers $ x$ with $ |x| \leq 1$" phrase - more specifically, this means we need to find three numbers $ a$, $ b$, and $ c$ such that if $ x$ is a real number and $ |x| \leq 1$, then $ |ax^2 \plus{} bx \plus{} c| \leq 100$. So if there is any value of $ x$ between $ \minus{} 1$ and $ 1$ where $ |ax^2 \plus{} bx \plus{} c| > 100$, then $ a$, $ b$, and $ c$ do not form a valid triple.

By the way, the USAMTS is directed by some very experienced people, so when you think there is an error in the official solution, it's generally safe to assume you've made a mistake like misunderstanding the problem.

(Don't worry about it too much though - this is something that happens to everybody. :) )

hum... well, if x=1, there's no maximum values... but I still don't understand what it is asking... :( "for all real number abs(x)<=1" part...

We must find the maximum value of |a|+|b|+|c| AMONG all parabolas that satisfy the condition if $ |x|\le 1$ then $ |f(x)|\le 1$, or equivalently, the parabolas such that the part of the parabola where $ \minus{}1\le x \le 1$ is such that $ \minus{}100 \le f(x) \le 100$.

Your example does not satisfy $ \minus{}100\le f(x)\le 100$ for all $ \minus{}1\le x\le 1$.

[geogebra]b1aa106449bf5908ba3e47083a572cc284ed1f7f[/geogebra]

OH. Got it... well, I think I had a brain-freeze over the last month about this problem. thanks a lot for y'all's help ("help from you all" for non-southerners...) :) .

By the way, how did you do that applet? it's amazing!

above the box where you type your message, theres tools like "bold", italic underline, quote, code...the last one is geogebra.  click it.


How can I prove this inequality using mathematical induction? What I find hard is that the RHS is a constant and not a term depending on n:

$ 1/1^2 \plus{} 1/2^2 \plus{} 1/3^2 \plus{} ... \plus{} 1/n^2 < 1.674$

ah ,thanks , I forgot a 1 ...

This problem was in a math contest but it was to prove $ < 2$  .   :maybe:

endbox u need to prove it is less than 1.674 not 2


geometry
3D geometry
sphere
calculus
integration

Our instructor showed us in class that by applying brute force it is possible to give a value to the gravitational intensity on the surface of a thin spherical shell.

My question is whether it is ethical to do this.After all,the limits at r=R should not exist because,inside the sphere it is 0 and outside the sphere it has got a finite value.Therefore the LHL and RHL do not match.Please confirm.

Of course 'on' the surface of the spherical shell there is no meaning of gravitational intensity now i assume he showed u via integration then you instead of integrating the limits from $x$ to $R$ integrate it to $x$ to $R-\epsilon$ where $\epsilon$ is arbitrarily small then the 'error' is so small that it's negligible or in other words immaterial.

well what i said above follows from the basic definition of gravitational field for bodies over which mass is distributed continously.

[quote="kolahalb"]What should be the conclusion?
I do not find any justification of this limiting value.[/quote]
can u elaborate his method .i don't get anything when u say limiting value

It was a limit (epsilon)--->0 on the integral that determines the value of intensity.

The limits involved epsilon.Lower limit epilon to upper limit R-epsilon
Ordinarily 0 to R limits give infinity.

well i still couldn't exactly understand what ur instructor has told u .But if we proceed as i say like u integrate it from $x$ the distance from the centre whetre u need to find the intensity to a distance arbitrarily close to the outer shell say $R+\epsilon$ then u also need not limit (i don't understand why ur instructor neede to do so )


Find the maximum of $ 3(xy\plus{}yz\plus{}zx)\minus{}xyz$

where $ x,y,z$ are positive reals such that $ x^3\plus{}y^3\plus{}z^3\equal{}3$.

hello, the maximum is 8 will be obtained at the point $ x\equal{}1,y\equal{}1,z\equal{}1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the maximum is 8 will be obtained at the point $ x \equal{} 1,y \equal{} 1,z \equal{} 1$.
Sonnhard.[/quote]

Please prove it  :)

[quote="outback"]Find the maximum of $ 3(xy \plus{} yz \plus{} zx) \minus{} xyz$

where $ x,y,z$ are positive reals such that $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$.[/quote]
Yes, for positive $ x,$ $ y$ and $ z$ we obtain:
1) $ \max_{x^3\plus{}y^3\plus{}z^3\equal{}3}(3(xy \plus{} yz \plus{} zx) \minus{} xyz)\equal{}8.$

2)By the way, $ \max_{x^3\plus{}y^3\plus{}z^3\equal{}3}(3(xy \plus{} yz \plus{} zx) \minus{} 4xyz)\equal{}5$
3)but $ \max_{x^3\plus{}y^3\plus{}z^3\equal{}3}(3(xy \plus{} yz \plus{} zx) \minus{} 5xyz)>4.$ :wink:

Normalize the inequality,we should to prove:
$ \frac {8}{3}(x^3 \plus{} y^3 \plus{} z^3) \plus{} xyz \ge (xy \plus{} yz \plus{} zx)\sqrt [3]{6(x^3 \plus{} y^3 \plus{} z^3)}$

Use the fact Schur's inequality and othere well-knowns,we have
$ \frac {1}{3}(x^3 \plus{} y^3 \plus{} z^3) \plus{} xyz \ge \sum_{sym}x^2y \ge \frac {2}{3}(xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z) \ge 2(\frac{xy \plus{} yz \plus{} zx}{3})^{3/2}$

The rest we should to prove that:
$ 7\frac {x^3 \plus{} y^3 \plus{} z^3}{3} \plus{} 2(\frac{xy \plus{} yz \plus{} zx}{3})^{3/2} \ge (xy \plus{} yz \plus{} zx)\sqrt [3]{6(x^3 \plus{} y^3 \plus{} z^3)}$

Use $ AM \minus{} GM$,you will kill it. :)

$ \sum x^3\plus{}3xyz\geq\sum(x^2y\plus{}xy^2)\geq 2\sum xy$
$ \Rightarrow \sum x^3\geq 2\sum xy\minus{}3xyz\equal{}3P\minus{}7\sum xy$
that $ \sum(x^3\plus{}y^3\plus{}1)\geq 3\sum xy\Rightarrow \sum xy\leq 3\Rightarrow P\leq 8$

We prove the stronger:\[ 8 \geq 3(xy+yz+zx)-xyz\qquad x+y+z=3\]
[hide="Proof"]
Suppose $ x\geq y\geq z$, and let $ w=\frac{y+z}{2}$.
Then
\begin{eqnarray*}3(xy+yz+zx)-xyz&=&3x(y+z)+(3-x)yz \\ &=&(3x+yz)(y+z) \\ &\leq& (3x+w^2)(2w) \\ &=&2w(3-w)^2\leq 8\end{eqnarray*}
Where the last inequality follows by $ AM-GM$:\[ 2=\frac{2w+(3-w)+(3-w)}{3}\geq\sqrt[3]{2w(3-w)^2}\]
[/hide]


[quote="ZzZzZzZzZzZz"]On number 5, if we want to use a fact about regular icosahedra, do we have to prove the fact, or can we just use it without proving it? (since it's a fact)[/quote]

This is not a question we can answer while a round is in progress.


MATHCOUNTS
function
AMC
USA(J)MO
USAMO
AIME
National Assessment

Thought it'd be nice to take a break from solving math-problems and just talk about MATHCOUNTS.

I'd like to make the nationals, and I'm guessing I'll make it *snickers from the audience* :) ...is the abbreviation for "MATHCOUNTS National Competition" "nats"? And why is "abbreviation" such a long word? :lol:

you're right rep...competitions start in like 4 months...we have our school competition like in the beginning of December and that's when we cut all 8th graders that don't make the top 8. I'm like 99% positive that I'll make it on the team unless someone unexpected gets smart overnight...anywas, back to the original topic, I'm pretty sure we'll make it to state cuz we've won Chapter for the past several years by quite a big margin, and I REALLY hope to make it to "nats," yes Tare, it's nats, this year, but I highly doubt it will happen  :(

Oh, OK. Just wanted to check since National Assessment & Testing also looked like nats :)

Another question: If you register as a team and you do really well as an individual but your team sucked, would you be able to move on?

So if you want to do Team Round yet still want to move on to the Nationals even if the team does bad, then you have to register as both a team member and as an individual?
Also if you start out as a member of a team, yet you do really good on everything, then is there any hope of making the nationals?

Making nationals is entirely a function of your individual performance.  You can make it to state because your team does well or because you do well (or both).  When you register as part of a team, you are automatically registered as an individual as well.  MATHCOUNTS has their bases pretty well covered with regard to this now, I think.

no. top 4 teams and top 4 individuals go to states from each regional (i can't remember whether it's top 4 or someat mebbe 3) but anyways that's the general method. when i say top 4 individuals, i mean top 4 individuals who aren't already going to state because of their team performance.

now nats is different

just top 4 individuals go to nats from each state

Whew, that's good to here. Thanks for all your help everyone, thought I won't make it because of my school's educational level :lol:

...I wonder who voted for the first and the last...

uh-oh...we've already got 2 peeps for champ. that's not possible...

well, last year, there were at least 15 peeps who thought they would be champ, so yeah...

...wait, so someone did the same thing last year!? I thought I was being original :lol:

Umm, although I agree, this poll is [b]not[/b] about how you [b]want[/b] to do but how you [b]could[/b] do...
Just seeing how strong my competition is :)

no, tare, i don't think anyone actually made a poll, but i'm fairly sure just by talking to my teammates and others that at least 15 people aspired to be the champ (most of the top 10 at least, and then some)

well im perdy sure  ill go to nats. i would say "guaranteed" but i know something bad like last year prbly is going to happen, or has a chance to happen. but its all gravy to me. good luck to you all!

its odd how this thread was so active in the early morning. i just woke up anad its 340ish here

Yeah, I guess you could say that.
[quote="Syntax Error"]its odd how this thread was so active in the early morning. i just woke up anad its 340ish here[/quote]
Today people are going all over the place...only one of my friends are actually staying home all day today, so I guess if this applies to the rest of the US people won't be home about now, only in the mornings.

We need to have some options like : "the team im coaching will win states" or "i will be the nats team winner coach" for the older people here...

then hes guaranteed not to make chapter. which he said

he'll probably take the USAMO. that's sort of higher than the school comp.

*sigh* As I said before [b]don't vote if you're not eligible![/b] Geez, no wonder there's 2 votes for the first one...

2 votes for the first one?? isn't it 6??

oh, i see, 2 for not farther than school, my bad.

Fine, we can have a "how far do you think you'll go?" in the AMC forum, too.

Oops, I lied, I didn't vote.  I was only speaking figuratively.

i've already gone to nationals so here's how i went:

7th grade: 5th in chapter, 11th in state
8th grade: 1st in chapter, 3rd in state, 56th in nationals (on the team of national champion, washington :D )

i guess you could rate yourself according to this, but i guess my 7th grade chapter was really bad compared to my state :?

7th grade
2nd in chapter (lost to a kid who got 8th in nationals)......38th in state

8th grade
1st in chapter...4th in state...59th in nationals (I did horrible)

[quote="WarpedKlown1335"]Jeez...it sucked when I didn't even get to take the AMCs or AIMEs in 8th grade because I had FCAT standardized testing... :evil:  Instead, the only AMCs and AIMEs I took were for practice.  Oh well, I hope I make USAMO this year.[/quote]
I thought you were in Japan :?

7th grade: didn't even know MATHCOUNTS existed :)
8th grade: ?

j/w, but chinaboy told me that to break the ties in nationals (outside of top 10) they did it by your last name. but why wouldnt they just use the same formula thing they use for the top 10?

i think they shouldnt even break the ties outside of the top 10, just leave it as it was.

"Whaaaaaaaat? Whtcha talkwin bout?" :lol:
That's obsurd, an urban legend at the most. That's segregating people with last names starting with Zs (which includes many Russian people thus it qualifies as discrimination)

j/w...mmm, looks like w/e but still...er, just wondering?

yeah, i'm fairly sure j/w means just wondering.

anywayyyyysss.....

6th grade: school didn't let me go
7th grade: scheduling conflict
8th grade: 1st at chapter, 1st in state, 6th at nats :cry:


I guess you mean $\arcsin(x) = \sin^{-1}(x)$, etc. Then this equality just follows from $\arcsin(x)+\arccos(x) = \frac{\pi}{2}= \arctan(x)+\textrm{arccot}(x)$.


[hide]We have number of $1's$ equal number of $-1's$ in each sets $\{b_{i}\},\{a_{i}\},\{a_{i}b_{i}\}$ therefore we have q.e.d![/hide]

very long :D (both : Rust and aviateurpilot)
[quote="aviateurpilot"] number of $1's$=the nombre of $-1's$
$n=2k$
[/quote]

We have $\prod a_{i}=(-1)^{k},\prod b_{i}=(-1)^{k}, \prod a_{i}b_{i}=(-1)^{k}$ therefore $k$ is even.

[quote="cuopbientcd"]SOLUTION'S [i]Cuopbientcd[/i]
Very simply I have $n= \sum_{1}^{n}(a_{i}+1)(b_{i}+1)$
 but $a_{i}+1 and b_{i}+1 =0 or 2$ then $4|n$ :)[/quote]
Wow, very nice!  :wink:


Does not belong in [i]Pre-[/i]Olympiad. Moved

I have no idea what you are talking about  :maybe:

[quote="inom"]Is it ok if i use ABC method( abstract concreteness) without proof in international olympiads?[/quote]
ABD method,I think its authour is Nguyen Anh Cuong,a stdent in my country

ABC method was profiled in "Diamonds in mathematical inequalities", a book proposal which was donated for free to IMO students at last year's IMO, which was held in Vietnam. It covers strange stuff which seems to work well. (I was loaned it by a former contestant)

ABC method with non-negative reals a, b,c : very simply, if a+b+c and ab+bc+ca are made to have fixed values, then the maximum value of abc occurs when (a-b)(b-c)(c-a)=0 or abc=0. (I think!).

So the easy way to attack some problems is to rewrite them in terms of the elementary symmetric functions (a+b+c, ab+bc+ca, abc) and then use the ABC method, eg by proving monotonicity of the resulting function (ie treating a+b+c and ab+bc+ca as constants, for instance).


what is simon's factoring trick? i have not seen them in books? can  you give several examples of factoring that uses it? and who is simon? is he a mathematician?

Quick search:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1649850143&t=145007

I think/thought Simon is ComplexZeta. Is he?

The AoPSWiki article is [[Simon's Favorite Factoring Trick]].

They, unfortunately, never made a third one as stated in the second one.

[quote]I think/thought Simon is ComplexZeta. Is he?[/quote]

Yeah.

The Introduction to Algebra book by Richard Rusczyk also talks about simon's trick.

Yeah, it's a useful thing to know for MC...


inequalities
algebra
polynomial
complex analysis
complex analysis unsolved

Let $A(z)=z^{n}+\sum_{k=0}^{n-1}a_{k}z^{k}$ be a polynomial with complex coefficients and zeros $z_{1},\ldots,z_{n}$. Prove that
$\bigl|z_{1}z_{2}\ldots z_{n}\bigr|^{2/n}<1+\max_{1\lek\le n}|a_{n-k}|^{2}.$

Sorry, there is a typo in the topic posted previously. Here is the correct one:

Let $A(z)=z^{n}+\sum_{k=0}^{n-1}a_{k}z^{k}$ be a polynomial
with complex coefficients and zeros $z_{1},\ldots,z_{n}$. Prove that
$\Bigl|z_{1}z_{2}\ldots z_{n}\Bigr|^{2/n}<1+\max_{1\le k\le n}|a_{n-k}|^{2}.$