inequalities
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inequalities proposed

In this topic, I will present the my 13th proof for the famous inequality - Iran 1996 - which was first proposed by Ji Chen on Crux Mathematicorum. The inequality states:

[b]Problem 1.[/b] [i]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\right) \ge \frac{9}{4}.\][/i]

This inequality is really nice, interesting and hard. I think there are also many solvers trying to solve it in different ways (for example, my friend - Anas Bentaleb - with his 2 special proofs for this nice inequality). Of course, I am one of them. :) After working a lot on it, I have found for my own 13 proofs and in this topic, I will present one of them - the 13th proof. The proof is as follow:

[i][b]Proof 13. [/b][/i]We can assume (WLOG) that $ a \ge b \ge c$ and rewrite the inequality as
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{ab}{(a\plus{}b)^2} \plus{}c\left( \frac{a}{(a\plus{}c)^2}\plus{}\frac{b}{(b\plus{}c)^2}\right) \ge \frac{9}{4}.\]
Now, we take a look at the following interesting identity
\[ \frac{a}{(a\plus{}c)^2}\plus{}\frac{b}{(b\plus{}c)^2} \minus{}\frac{a^2\plus{}b^2\plus{}6ab}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)} \equal{}\frac{(a\minus{}b)^2(a\minus{}c)(b\minus{}c)}{2(a\plus{}b)(a\plus{}c)^2(b\plus{}c)^2} \ge 0,\]
which guides us to prove the sharper inequality
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{ab}{(a\plus{}b)^2} \plus{}\frac{c(a^2\plus{}b^2\plus{}6ab)}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{9}{4},\]
\[ \frac{a^2\plus{}b^2\plus{}c(a\plus{}b)}{(a\plus{}c)(b\plus{}c)} \plus{}\frac{c(a^2\plus{}b^2\plus{}6ab)}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)} \plus{}\frac{c}{a\plus{}b}\plus{}\frac{ab}{(a\plus{}b)^2} \ge \frac{9}{4}.\]
Now, the inequality being homogeneous, so we can assume $ a\plus{}b\equal{}1$ and put $ x\equal{}ab$, then $ \frac{1}{4} \ge x \ge c(1\minus{}c)$, and the above inequality becomes
\[ \frac{1\minus{}2x\plus{}c}{x\plus{}c\plus{}c^2}\plus{}\frac{c(1\plus{}4x)}{2(x\plus{}c\plus{}c^2)} \plus{}x\plus{}c \ge \frac{9}{4},\]
\[ \frac{2\plus{}3c\plus{}4(c\minus{}1)x}{2(x\plus{}c\plus{}c^2)} \ge \frac{9}{4} \minus{}x\minus{}c,\]
\[ 2(2\plus{}3c\plus{}4(c\minus{}1)x) \ge (9\minus{}4c\minus{}4x)(x\plus{}c\plus{}c^2),\]
\[ (4x^2\minus{}7x\plus{}4) \plus{}(4c^2\plus{}16c)x \plus{}4c^3\minus{}5c^2\minus{}3c \ge 0,\]
We have
\[ 4x^2\minus{}7x\plus{}4 \equal{}(4\minus{}x)(1\minus{}4x) \ge \left( 4\minus{}\frac{1}{4}\right) (1\minus{}4x) \equal{}\frac{15}{4} \minus{}15x.\]
It suffices to show that
\[ \frac{15}{4} \plus{}(4c^2\plus{}16c\minus{}15)x \plus{}4c^3\minus{}5c^2\minus{}3c \ge 0.\]
Now, with notice that $ 4c^2\plus{}16c \minus{}15 \le 0$, we obtain
\[ \frac{15}{4} \plus{}(4c^2\plus{}16c\minus{}15)x \plus{}4c^3 \minus{}5c^2 \minus{}3c \ge \frac{15}{4} \plus{}(4c^2\plus{}16c\minus{}15)\cdot \frac{1}{4} \plus{}4c^3\minus{}5c^2\minus{}3c \equal{}c(1\minus{}2c)^2 \ge 0.\]
This completes our proof. Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$ and its cyclic permutations.

[i][b]Comment. [/b][/i]This proof is not very nice but we can proud of it because we can prove a hard inequality without using any known method. :)

I hope that you all, after reading it, will tell me your ideas and your comments. Thank you very much.

Regards

Vo Quoc Ba Can

P/s: Very sorry for my bad English.

[quote="can_hang2007"]
[i][b]Proof 13. [/b][/i]
Now, we take a look at the following interesting identity
\[ \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2} \minus{} \frac {a^2 \plus{} b^2 \plus{} 6ab}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {(a \minus{} b)^2(a \minus{} c)(b \minus{} c)}{2(a \plus{} b)(a \plus{} c)^2(b \plus{} c)^2} \ge 0,
\]
which guides us to prove the sharper inequality
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{} \frac {c(a^2 \plus{} b^2 \plus{} 6ab)}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {9}{4},
\]

\[ \frac {a^2 \plus{} b^2 \plus{} c(a \plus{} b)}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {c(a^2 \plus{} b^2 \plus{} 6ab)}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \ge \frac {9}{4}.
\]
Now, the inequality being homogeneous, so we can assume $ a \plus{} b \equal{} 1$ and put $ x \equal{} ab$, then $ \frac {1}{4} \ge x \ge c(1 \minus{} c)$, and the above inequality becomes
\[ \frac {1 \minus{} 2x \plus{} c}{x \plus{} c \plus{} c^2} \plus{} \frac {c(1 \plus{} 4x)}{2(x \plus{} c \plus{} c^2)} \plus{} x \plus{} c \ge \frac {9}{4},
\]

\[ \frac {2 \plus{} 3c \plus{} 4(c \minus{} 1)x}{2(x \plus{} c \plus{} c^2)} \ge \frac {9}{4} \minus{} x \minus{} c,
\]

\[ 2(2 \plus{} 3c \plus{} 4(c \minus{} 1)x) \ge (9 \minus{} 4c \minus{} 4x)(x \plus{} c \plus{} c^2),
\]

\[ (4x^2 \minus{} 7x \plus{} 4) \plus{} (4c^2 \plus{} 16c)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge 0,
\]
We have
\[ 4x^2 \minus{} 7x \plus{} 4 \equal{} (4 \minus{} x)(1 \minus{} 4x) \ge \left( 4 \minus{} \frac {1}{4}\right) (1 \minus{} 4x) \equal{} \frac {15}{4} \minus{} 15x.
\]
It suffices to show that
\[ \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge 0.
\]
Now, with notice that $ 4c^2 \plus{} 16c \minus{} 15 \le 0$, we obtain
\[ \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)\cdot \frac {1}{4} \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \equal{} c(1 \minus{} 2c)^2 \ge 0.
\]
This completes our proof. Equality holds if and only if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ and its cyclic permutations.

[i][b]Comment. [/b][/i]This proof is not very nice but we can proud of it because we can prove a hard inequality without using any known method. :)[/quote]

Nice and very good. My dear brother. :)

Thanks for another proof for the nice and hard inequality. :)

yes , your idea iss very natural . ''Caychy bat doi'' is very usefull .
I thinks it very easy to understand this solutions .
Thanks Canhang .

Congratulations! Amazing Solution in my opinion. And it is very short... No brute force is used.
I think this proof is very hard to obtain, because as you said, no method or special technique is used.

[quote="can_hang2007"]
Now, we take a look at the following interesting identity
\[ \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2} \minus{} \frac {a^2 \plus{} b^2 \plus{} 6ab}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {(a \minus{} b)^2(a \minus{} c)(b \minus{} c)}{2(a \plus{} b)(a \plus{} c)^2(b \plus{} c)^2} \ge 0,
\]

[/quote]
I think this identity is "key" that helps to solve Iran 96.
Since I love identities, can you tell me how did you establish this, at least a hint?

Thanks for sharing!  :)

This is a possible way to motivate the identity...

Lets find a lower bound for $ \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2}\ge f(a,b,c)$. Of course this must be an identity for $ a \equal{} b$ so
$ f(a,a,c) \equal{} \frac {2a}{(a \plus{} c)(a \plus{} c)}$. Now lets try and find a function $ f$.

If we replace $ 2a$ with $ \frac {x(a\minus{}b)^2 \plus{} yab}{a \plus{} b}$, and $ (a \plus{} c)(a \plus{} c)$ with $ (a \plus{} c)(a \plus{} b)$ we get the a nice bound.

Note that a number of ideas suggest this. First, it gives us a symmetric denominator. Second, we want to replace $ a$ with a power mean. We just need to solve for the coefficients $ x,y$.

Comment: we can make the inequality sharper by replacing with different power means of $ a,b$.

[quote="shaaam"][quote="Altheman"]
If we replace $ 2a$ with $ \frac {x(a \minus{} b)^2 \plus{} yab}{a \plus{} b}$, and $ (a \plus{} c)(a \plus{} c)$ with $ \boxed{(a \plus{} c)(a \plus{} b)}$ we get the a nice bound.
[/quote]
Did you mean replace $ (a \plus{} c)(a \plus{} c)$ with $ (a \plus{} c)(b \plus{} c)$?
Thanks Altheman for your nice ideas. I don't know how to finish it.
Can you please explain the procedure of finding coeffiecients $ x$ and $ y$.
Thanks again!  :lol:[/quote]

Yes, thats correct.

The next step is to check that there is equality for a=b...now note that I chose the form carefully so $ x$ disappears and we find $ y$

$ 2a\equal{}\frac{y a a}{2a}$ so $ y\equal{}4$.

Now we factor the expression

[code]In[24]:= Factor[
 a/(a + c)^2 +
  b/(b + c)^2 - ( x (a - b)^2 + 4 a b)/((a + b) (b + c) (c + a))]

Out[24]= -((a - b)^2 (-a b - c^2 + a b x + a c x + b c x +
    c^2 x))/((a + b) (a + c)^2 (b + c)^2)[/code]

Now we want the expression $ c^2\plus{}ab\minus{}x(ab\plus{}ac\plus{}bc\plus{}c^2)$ to be positive.

We want the inequality to be as sharp as possible, so we want the largest possible value of $ x$. Setting $ a\equal{}b\equal{}c\equal{}1$, we get $ 2\minus{}4x\ge 0$ so $ x\le\frac{1}{2}$. If we plug that in, we get $ \frac{1}{2}(a\minus{}c)(b\minus{}c)$

It is natural to say WLOG $ c$ is the minimum because that gives us the sharpest bound.

[quote="can_hang2007"]In this topic, I will present the my 13th proof for the famous inequality - Iran 1996 - which was first proposed by Ji Chen on Crux Mathematicorum. The inequality states:

[b]Problem 1.[/b] [i]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ (ab \plus{} bc \plus{} ca)\left( \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right) \ge \frac {9}{4}.\]
[/i]

This inequality is really nice, interesting and hard. I think there are also many solvers trying to solve it in different ways (for example, my friend - Anas Bentaleb - with his 2 special proofs for this nice inequality). Of course, I am one of them. :) After working a lot on it, I have found for my own 13 proofs and in this topic, I will present one of them - the 13th proof. The proof is as follow:

[i][b]Proof 13. [/b][/i]We can assume (WLOG) that $ a \ge b \ge c$ and rewrite the inequality as
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{} c\left( \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2}\right) \ge \frac {9}{4}.\]
Now, we take a look at the following interesting identity
\[ \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2} \minus{} \frac {a^2 \plus{} b^2 \plus{} 6ab}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {(a \minus{} b)^2(a \minus{} c)(b \minus{} c)}{2(a \plus{} b)(a \plus{} c)^2(b \plus{} c)^2} \ge 0,\]
which guides us to prove the sharper inequality
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{} \frac {c(a^2 \plus{} b^2 \plus{} 6ab)}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {9}{4},\]

\[ \frac {a^2 \plus{} b^2 \plus{} c(a \plus{} b)}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {c(a^2 \plus{} b^2 \plus{} 6ab)}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \ge \frac {9}{4}.\]
Now, the inequality being homogeneous, so we can assume $ a \plus{} b \equal{} 1$ and put $ x \equal{} ab$, then $ \frac {1}{4} \ge x \ge c(1 \minus{} c)$, and the above inequality becomes
\[ \frac {1 \minus{} 2x \plus{} c}{x \plus{} c \plus{} c^2} \plus{} \frac {c(1 \plus{} 4x)}{2(x \plus{} c \plus{} c^2)} \plus{} x \plus{} c \ge \frac {9}{4},\]

\[ \frac {2 \plus{} 3c \plus{} 4(c \minus{} 1)x}{2(x \plus{} c \plus{} c^2)} \ge \frac {9}{4} \minus{} x \minus{} c,\]

\[ 2(2 \plus{} 3c \plus{} 4(c \minus{} 1)x) \ge (9 \minus{} 4c \minus{} 4x)(x \plus{} c \plus{} c^2),\]

\[ (4x^2 \minus{} 7x \plus{} 4) \plus{} (4c^2 \plus{} 16c)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge 0,\]
We have
\[ 4x^2 \minus{} 7x \plus{} 4 \equal{} (4 \minus{} x)(1 \minus{} 4x) \ge \left( 4 \minus{} \frac {1}{4}\right) (1 \minus{} 4x) \equal{} \frac {15}{4} \minus{} 15x.\]
It suffices to show that
\[ \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge 0.\]
Now, with notice that $ 4c^2 \plus{} 16c \minus{} 15 \le 0$, we obtain
\[ \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)\cdot \frac {1}{4} \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \equal{} c(1 \minus{} 2c)^2 \ge 0.\]
This completes our proof. Equality holds if and only if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ and its cyclic permutations.

[i][b]Comment. [/b][/i]This proof is not very nice but we can proud of it because we can prove a hard inequality without using any known method. :)

I hope that you all, after reading it, will tell me your ideas and your comments. Thank you very much.

Regards

Vo Quoc Ba Can

P/s: Very sorry for my bad English.[/quote]
It is a nice solution.
By the way, I solved it without any lemma in the following way (My 2nd solution  :blush:  )
[b]Problem: (Iran 1996) Prove that $ \forall x,y,z\in\mathbb{R}^ \plus{} ;$
\[ (xy \plus{} yz \plus{} zx)\left(\sum \frac {1}{(x \plus{} y)^2}\right)\geq \frac {9}{4}\]
[/b]
[i]Solution 2[/i]

Let $ x \plus{} y \equal{} a, y \plus{} z \equal{} b, z \plus{} x \equal{} c$ hence
$ 2x \equal{} b \plus{} c \minus{} a\implies x \equal{} \frac {b \plus{} c \minus{} a}{2}$ and other similar relations.
Hence;
\[ \sum yz \equal{} \frac {1}{4}\sum (2bc \minus{} a^2)\]
Now assume $ a\geq b\geq c$ So
$ 2bc \minus{} a^2\leq 2ca \minus{} b^2\leq 2ab \minus{} c^2$
So these sequences are oppositely sorted. Now we apply Chebyshev's inequality:
\[ \frac {4}{9}\sum yz\sum \frac {1}{(x \plus{} y)^2} \equal{} \frac {1}{3}\sum (2bc \minus{} a^2)\cdot \frac {1}{3}\sum\frac {1}{a^2}\]

\[ \geq \frac {1}{3}\sum \left((2bc \minus{} a^2)\frac {1}{a^2}\right) \equal{} \frac {1}{3}\sum\frac {2bc}{a^2} \minus{} 1\]

\[ \geq 2\left(\prod_{cyc} \frac {bc}{a^2}\right)^{\frac {1}{3}} \minus{} 1 \equal{} 1\]
The last step follows from AM-GM inequality.
So $ \frac {4}{9}\sum yz\sum\frac {1}{(x \plus{} y)^2}\geq 1 \implies \sum xy\sum \frac {1}{(x \plus{} y)^2}\geq \frac {9}{4}$
Hence proved.
  :)
Do you like it? :blush:

Potla solution is wrong as Chebychev is applied in not correct way. :wink:

If sequences $ a_i$ and $ b_i$ are similarly sorted then $ \frac{1}{n}\sum a_ib_i\geq \frac{1}{n^2}\sum a_i\sum b_i$
If sequences $ a_i$ and $ b_i$ are [b]oppositely[/b] sorted then $ \sum a_ib_i\leq \frac{1}{n}\sum a_i \sum b_i$
I used  the second statement of Chebyshev..  :rotfl:
so i think it is correct... please correct me if I am wrong.  :blush:

$ 2bc\minus{}a^2$ and $ \frac{1}{a^2}$ sequences are similarly sorted. :wink:

Yes.  :blush:
Indeed these are not oppositely sorted .......

Really, I missed such a big flaw in my solution. :(

Sorry about that , dear friends - thanks to manlio for your notice.  :)

[quote="can_hang2007"]

[b]Problem 1.[/b] [i]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ (ab \plus{} bc \plus{} ca)\left( \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right) \ge \frac {9}{4}.\]
[/i]

[/quote]
For the proof enough to check only two cases:
1) $ b\equal{}1,$ $ c\equal{}0$;
2) $ b\equal{}c\equal{}1.$ :wink:

[quote="anas"]the following is stronger but  easier than iran96, :
let $ u,v,w$ positive reals such that : $ uw \plus{} uv \plus{} vw \plus{} 2uwv \equal{} 1$  then :
$ (1 \plus{} u)^2 \plus{} (1 \plus{} v)^2 \plus{} (1 \plus{} w)^2 \geq3( u \plus{} v \plus{} w \plus{} uv \plus{} wv \plus{} wu)$ it can be easily solved with pqr  :)  :wink:[/quote]
Dear friend,
We usually try to find a good solution for the nice problem, Iran 96. Many proofs has been proposed (SOS, Mixing Variables, ...). But I do believe that the simpliest solution is to use the trivial inequality $ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8abc$ (with only 15 lines in my bad writing). What do you think, friend?

[quote="can_hang2007"][quote="anas"]the following is stronger but  easier than iran96, :
let $ u,v,w$ positive reals such that : $ uw \plus{} uv \plus{} vw \plus{} 2uwv \equal{} 1$  then :
$ (1 \plus{} u)^2 \plus{} (1 \plus{} v)^2 \plus{} (1 \plus{} w)^2 \geq3( u \plus{} v \plus{} w \plus{} uv \plus{} wv \plus{} wu)$ it can be easily solved with pqr  :)  :wink:[/quote]
Dear friend,
We usually try to find a good solution for the nice problem, Iran 96. Many proofs has been proposed (SOS, Mixing Variables, ...). But I do believe that the simpliest solution is to use the trivial inequality $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8abc$ (with only 15 lines in my bad writing). What do you think, friend?[/quote]

simply WOW  :ninja: , [hide]I think , no I don't think I'm sure that you are one of the best and brilliant ineq solvers my dear friend , I will try to think of your solution[/hide]

[quote="can_hang2007"]In this topic, I will present the my 13th proof for the famous inequality - Iran 1996 - which was first proposed by Ji Chen on Crux Mathematicorum. The inequality states:

[b]Problem 1.[/b] [i]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Show that
\[ (ab \plus{} bc \plus{} ca)\left( \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right) \ge \frac {9}{4}.\]
[/i]

This inequality is really nice, interesting and hard. I think there are also many solvers trying to solve it in different ways (for example, my friend - Anas Bentaleb - with his 2 special proofs for this nice inequality). Of course, I am one of them. :) After working a lot on it, I have found for my own 13 proofs and in this topic, I will present one of them - the 13th proof. The proof is as follow:

[i][b]Proof 13. [/b][/i]We can assume (WLOG) that $ a \ge b \ge c$ and rewrite the inequality as
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{} c\left( \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2}\right) \ge \frac {9}{4}.\]
Now, we take a look at the following interesting identity
\[ \frac {a}{(a \plus{} c)^2} \plus{} \frac {b}{(b \plus{} c)^2} \minus{} \frac {a^2 \plus{} b^2 \plus{} 6ab}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {(a \minus{} b)^2(a \minus{} c)(b \minus{} c)}{2(a \plus{} b)(a \plus{} c)^2(b \plus{} c)^2} \ge 0,\]
which guides us to prove the sharper inequality
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{} \frac {c(a^2 \plus{} b^2 \plus{} 6ab)}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {9}{4},\]

\[ \frac {a^2 \plus{} b^2 \plus{} c(a \plus{} b)}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {c(a^2 \plus{} b^2 \plus{} 6ab)}{2(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \ge \frac {9}{4}.\]
Now, the inequality being homogeneous, so we can assume $ a \plus{} b \equal{} 1$ and put $ x \equal{} ab$, then $ \frac {1}{4} \ge x \ge c(1 \minus{} c)$, and the above inequality becomes
\[ \frac {1 \minus{} 2x \plus{} c}{x \plus{} c \plus{} c^2} \plus{} \frac {c(1 \plus{} 4x)}{2(x \plus{} c \plus{} c^2)} \plus{} x \plus{} c \ge \frac {9}{4},\]

\[ \frac {2 \plus{} 3c \plus{} 4(c \minus{} 1)x}{2(x \plus{} c \plus{} c^2)} \ge \frac {9}{4} \minus{} x \minus{} c,\]

\[ 2(2 \plus{} 3c \plus{} 4(c \minus{} 1)x) \ge (9 \minus{} 4c \minus{} 4x)(x \plus{} c \plus{} c^2),\]

\[ (4x^2 \minus{} 7x \plus{} 4) \plus{} (4c^2 \plus{} 16c)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge 0,\]
We have
\[ 4x^2 \minus{} 7x \plus{} 4 \equal{} (4 \minus{} x)(1 \minus{} 4x) \ge \left( 4 \minus{} \frac {1}{4}\right) (1 \minus{} 4x) \equal{} \frac {15}{4} \minus{} 15x.\]
It suffices to show that
\[ \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge 0.\]
Now, with notice that $ 4c^2 \plus{} 16c \minus{} 15 \le 0$, we obtain
\[ \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)x \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \ge \frac {15}{4} \plus{} (4c^2 \plus{} 16c \minus{} 15)\cdot \frac {1}{4} \plus{} 4c^3 \minus{} 5c^2 \minus{} 3c \equal{} c(1 \minus{} 2c)^2 \ge 0.\]
This completes our proof. Equality holds if and only if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ and its cyclic permutations.

[i][b]Comment. [/b][/i]This proof is not very nice but we can proud of it because we can prove a hard inequality without using any known method. :)

I hope that you all, after reading it, will tell me your ideas and your comments. Thank you very much.

Regards

Vo Quoc Ba Can

P/s: Very sorry for my bad English.[/quote]
Very nice proof,Can_hang2007!Thank you very much!
I also have one nice proof for IRAN 96 but I don't have a lot of time.
Sorry,I am very bad E! :blush:

[quote="can_hang2007"]I dedicate the following proof to my friend, manlio.

Without loss of generality, we may assume that $ a \ge b \ge c.$ Then, we will show that
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(a \plus{} c)^2} \plus{} \frac {1}{(b \plus{} c)^2} \ge \frac {1}{4ab} \plus{} \frac {2}{(a \plus{} c)(b \plus{} c)}.$
Indeed, this inequality is equivalent to
$ \frac {1}{(a \plus{} c)^2} \plus{} \frac {1}{(b \plus{} c)^2} \minus{} \frac {2}{(a \plus{} c)(b \plus{} c)} \ge \frac {1}{4ab} \minus{} \frac {1}{(a \plus{} b)^2},$
or
$ \frac {(a \minus{} b)^2}{(a \plus{} c)^2(b \plus{} c)^2} \ge \frac {(a \minus{} b)^2}{4ab(a \plus{} b)^2}.$
This is true because $ 4ab \ge 4b^2 \ge (b \plus{} c)^2,$ and $ (a \plus{} b)^2 \ge (a \plus{} c)^2.$

Now, using the above estimation, it is sufficient to prove that
$ (ab \plus{} bc \plus{} ca)\left[\frac {1}{4ab} \plus{} \frac {2}{(a \plus{} c)(b \plus{} c)}\right] \ge \frac {9}{4}.$
Since $ \frac {ab \plus{} bc \plus{} ca}{4ab} \equal{} \frac {1}{4} \plus{} \frac {c(a \plus{} b)}{4ab}$ and $ \frac {2(ab \plus{} bc \plus{} ca)}{(a \plus{} c)(b \plus{} c)} \equal{} 2 \minus{} \frac {2c^2}{(a \plus{} c)(b \plus{} c)},$ it is equivalent to
$ \frac {c(a \plus{} b)}{4ab} \ge \frac {2c^2}{(a \plus{} c)(b \plus{} c)},$
or
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8abc.$
The last one is true according to the AM-GM Inequality, so our proof is completed.[/quote]

 :lol:  nice

Thank you very much, my dear friend for this beautiful proof. It is really genial in its simplicity. :)

[quote="can_hang2007"]I dedicate the following proof to my friend, manlio.

Without loss of generality, we may assume that $ a \ge b \ge c.$ Then, we will show that
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(a \plus{} c)^2} \plus{} \frac {1}{(b \plus{} c)^2} \ge \frac {1}{4ab} \plus{} \frac {2}{(a \plus{} c)(b \plus{} c)}.$
Indeed, this inequality is equivalent to
$ \frac {1}{(a \plus{} c)^2} \plus{} \frac {1}{(b \plus{} c)^2} \minus{} \frac {2}{(a \plus{} c)(b \plus{} c)} \ge \frac {1}{4ab} \minus{} \frac {1}{(a \plus{} b)^2},$
or
$ \frac {(a \minus{} b)^2}{(a \plus{} c)^2(b \plus{} c)^2} \ge \frac {(a \minus{} b)^2}{4ab(a \plus{} b)^2}.$
This is true because $ 4ab \ge 4b^2 \ge (b \plus{} c)^2,$ and $ (a \plus{} b)^2 \ge (a \plus{} c)^2.$

Now, using the above estimation, it is sufficient to prove that
$ (ab \plus{} bc \plus{} ca)\left[\frac {1}{4ab} \plus{} \frac {2}{(a \plus{} c)(b \plus{} c)}\right] \ge \frac {9}{4}.$
Since $ \frac {ab \plus{} bc \plus{} ca}{4ab} \equal{} \frac {1}{4} \plus{} \frac {c(a \plus{} b)}{4ab}$ and $ \frac {2(ab \plus{} bc \plus{} ca)}{(a \plus{} c)(b \plus{} c)} \equal{} 2 \minus{} \frac {2c^2}{(a \plus{} c)(b \plus{} c)},$ it is equivalent to
$ \frac {c(a \plus{} b)}{4ab} \ge \frac {2c^2}{(a \plus{} c)(b \plus{} c)},$
or
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8abc.$
The last one is true according to the AM-GM Inequality, so our proof is completed.[/quote]


LOL I GOT MUTED AND LOOK WHAT HAPPEND TO MY RATING

TOO BAD 27R BEAT ME TO IT

Dude, you don't have to make a new topic just for that.

Its called spamming.

Where was I gonna post it?

It was about FTW

And it wasn't about the problems

And it wasn't questions or supports

[quote="billybobjoe95"]how does someone get muted? and why did it say none?[/quote]

 you can get muted for a lot of reasons like swearing, using caps, flooding, etc. pi's rating said none because he had no rating.... i think he was muted. i think levans fixed that problem though. although currently there is no limit to lowest rating and it is possible to go negative.

I bet I was muted b/c everyone hates me  :(  :(  :(

I think I should be more amiable like everyone else

Ok fine I was using caps, but I did put a lowercase letter in there, and RA was SPAMMING and putting UPPERCASE, and he still talks fine

SEE EVERYONE HATES ME

No one hates you, Pi. The only reason you're muted is because some of your posts are unproductive(a.k.a spam). We talked about this already.... :|

T_________________________T

T_________________________T

But I have productive posts too! (I mean, I have almost 400 posts, at least 40 have to productive right?)

lol i don't think it's the number of posts that count it's the percentage of spam u have. i think that's why levans was unable to give us an exact number of "productive" posts needed.

You could've posted it under Lowest Rated Player? if you really wanted to.  And to be quite frank with you, none of us care that you had no rating.

[quote="PI-Dimension"]Ok guyz we all know this is spam can we stop posting in it? Your just making me more humiliated.

 :([/quote]

'nuff said


complex analysis
function
complex analysis unsolved

Let $ f$ be an injective holomorphic function in the unit disc with $ f(0) \equal{} 0$ and $ f'(0) \equal{} 1$. Furthermore, write $ f(z) \equal{} z \plus{} a_2z^2 \plus{} a_3z^3 \plus{} \cdots$ and assume the $ a_n$ are real. Show that:
i) $ |a_2|\leq2$.
ii) $ |a_n| \leq n$.

I can give the hints in my textbook for part b) if it turns out to be too hard. :)

isn't b) just the bieberbach conjecture?

:rotfl:  :rotfl:  That "just" seems just perfect here! No, there's a small detail in the statement of the problem, that makes it very different from Bieberbach's conjecture: the words real coefficients. With this assumption, the result is due to Dieudonné, but the proof is much much easier than in the general case.

i actually looked up the statement of the bieberbach conjecture after posting and noticed that but i guess i forgot to edit.  in any event i'm not trying to assert that the bieberbach conjecture is easy, i meant 'just' in the sense of 'exactly' :)


solve for x.
sqrt(243^(x+1))  =  cubert(81^(2-x))


inequalities
geometry
perimeter
calculus
derivative
parameterization
triangle inequality

Prove that in a nondegenerate triangle ABC with perimeter 1, the following inequality holds:

$\displaystyle\ a^2 + b^2 + c^2 + 4abc < \frac {1}{2} $

That's essentially the WURZEL problem Kappa 13 by Sefket Arslanagic. Since my solution was an algebraic massacre (at that time I didn't know any smart way of solving such inequalities), I post the much more beautiful solution of Dr. Joachim Assion, in WURZEL 9+10/2003, page 225:

From the triangle inequality a + b > c and the fact that a + b + c = 1, it is clear that $a+b=\frac12+d$ and $c=\frac12-d$ for some positive d. Thus,

$a^2+b^2+c^2+4abc=a^2+b^2+\left(\frac12-d\right)^2+4ab\left(\frac12-d\right)$
$=a^2+b^2+\left(\frac12-d\right)^2+2ab-4abd$
$=\left(a^2+b^2+2ab\right)+\left(\frac12-d\right)^2-4abd$
$=\left(a+b\right)^2+\left(\frac12-d\right)^2-4abd$
$=\left(\frac12+d\right)^2+\left(\frac12-d\right)^2-4abd$
$=\frac12+2d^2-4abd=\frac14+2d\left(d-2ab\right)$.

Remains to show that d (d - 2ab) < 0, or, equivalently, that d - 2ab < 0. But this is easy to see, since $a<\frac12$ and $b<\frac12$ yield

$0<\left(a-\frac12\right)\left(b-\frac12\right) = ab-\frac12\left(a+b\right)+\frac14 = ab-\frac12\left(\frac12+d\right)+\frac14 = ab-\frac12d$,

so that $ab>\frac12d$ and 2ab > d, so that d - 2ab < 0.

  Darij

Does this work?  Wolog, let a>=b>=c (this is what I'm concerned about)

Now fix a = t, t E (1/3,1/2)

This gives: t^2 + b^2 + c^2 + 4tbc on the LHS with b+c = 1-t

We can complete the square to give:

t^2 + (1-2t)b^2 + (1-2t)c^2 + 2t(b+c)^2.  Since b+c is constant, we just need to maximize (1-2t)(b^2 + c^2).

In fact, by convexity, this is maximized when we maximize b.  Thus a=b, so we get a=b=t, c=1-2t.

Now expanding this, we get the cubic -8t^3 + 10t^2 - 4t + 1, which has first derivative -24t^2 + 20t - 4 which has critical points t=1/2 and t=1/3 convieniently.  It is increasing on this interval so t=1/2, which implies a=b=1/2, c=0, LHS < 1/2 (because the triangle is nondegenerate) as desired.

[quote="darij grinberg"][quote="blahblahblah"]Does this work?  Wolog, let a>=b>=c [b](this is what I'm concerned about)[/b][/quote]

This works, yes. For a symmetric inequality, we can always assume WLOG $a\geq b\geq c$.

As for the rest of the proof, I must say I don't understand it completely, but the above was correct.

  Darij[/quote]

Yup, the reason I asked is that I've seen, on occasion, inequalities involving triangles where the substitution a>=b>=c was not valid.  I guess those inequalities must have been cyclic and not symmetric.

Anyways, for the rest of my proof, I fixed the largest side, which made the inequality homogeneous, and then I used some very basic ideas from calculus to reduce it to one parameter t.

This inequality appears in ther book USSR Mathematical Olympiads 1989-1992 by A.M> Slinko.
Here is his proof :Let the area of the triangle be S.The triangle has semiperimeter 1/2
Herons formula gives S^2 =1/2(1/2-a)(!/2-b)(1/2-c). This together with the fact that a+b+c=1 gives
16S^2 =-1 +4(ab+bc+ca) -8abc

Squaring both sides of a+b+c =1 gives
2(ab+bc+ca) =1-(a^2 +b^2 +c^2) Therefore

16S^2 =1-2(a^2+b^2 +c^2)-8abc
          =2(1/2 -a^2 -b^2 -c^2 -4abc) from which we get the desired inequality

The easiest approach I think is to make the ineqaulity homogeneous,but the algebra is a little heavy .

This  nice ineq  was  a  problem  at the  23rd  ASU , 1989. Use  the  well  know  formulas
[tex] a^2+b^2+c^2=2(s^2-r^2-4Rr) [/tex]  and [tex] abc=4Rrs [/tex]  , put  [tex] s=\frac{1}{2} [/tex]  and its  trivial


let $ a, b, c$ be three sides of a triangle, $ n$ is a positive integer and $ n \geq 2$ find the smallest value of a positive real number $ k$ such that
$ \sqrt [n]{a^2 \plus{} b^2} \plus{} \sqrt [n]{b^2 \plus{} c^2} \plus{} \sqrt [n]{c^2 \plus{} a^2} < k\sqrt [n]{(a \plus{} b \plus{} c)^2}$

$ f(x) \equal{} \sqrt[n] {x}$ is concave.
Jensen:
$ \sqrt [n]{a^2 \plus{} b^2} \plus{} \sqrt [n]{b^2 \plus{} c^2} \plus{} \sqrt [n]{c^2 \plus{} a^2} \leq 3\sqrt [n]{\frac {1}{3}\times2(a^2 \plus{} b^2 \plus{} c^2)} < 3\sqrt [n]{\frac {1}{3}(a \plus{} b \plus{} c)^2}$
then $ k\geq 3^{\frac {n \minus{} 1}{n}}$


[quote="kennyb274"]I know that the summation from n=1 to infinity of (x^n)/n!=e^x [/quote]

This is incorrect. :)

[quote="kennyb274"]I know that the summation from n=1 to infinity of (x^n)/n!=e^x, so would the sum of (x^n)/(x*n!)=(e^x)/x[/quote]
the only problem with your thing(as kunny mentioned) is that:
$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$


geometry
rectangle
combinatorial geometry
area of a triangle
combinatorics unsolved
combinatorics

In the plane we are given $3 \cdot n$ points ($n>$1) no three collinear, and the distance between any two of them  is $\leq 1$. Prove that we can construct $n$ pairwise disjoint triangles such that: The vertex set of these triangles are exactly the given 3n points and the sum of the  area of these triangles $< 1/2$.

The following idea is very well-known, and the problem seems familiar too. I think it might have been discussed before.

Consider a line $d$ in the plane which has a direction different from all the directions of the lines determined by two points among the $3n$. Assume it's "far away" from our set (outside the convex hull of the set, for example), and start moving it s.t. it meets our points one by one. For all $k\in\overline{1,n}$, we consider the triangle formed by the $3k-2$'nd, the $3k-1$'st and the $3k$'th point that $d$ meets, and these are our $n$ triangles.

Consider now two lines $d_1,d_2\|d$ as close to each other as possible s.t. all the $3n$ points lie between them, and two other lines $e_1,e_2\perp d$ with the same property. These four lines form a rectangle with sides $\le 1$, and this rectangle can be divided by lines $\|d$ into smaller rectangles, each one containing one of the triangles (for each two consecutive triangles we consider a line $\|d$ which lies strictly between them). Since the area of a triangle is $<\frac 12$ times the area of such a small rectangle, and the areas of the small rectangles add up to the area of the large rectangle, which is $\le 1$, we are done.


Express the inradius and circumradius of a triangle in terms of its sides $a$, $b$, and $c$.

[quote="nateharman1234"][hide] Well we know that (a+b+c)/2 is the semiperimeter, so I will just leave that as s for simplicity.

By two different formulas for area we get:
Sqrt{s(s-a)(s-b)(s-c)} = rs    where r is the inradius

so..

r = Sqrt{s(s-a)(s-b)(s-c)} / s

and by a similar arguement we know

Sqrt{s(s-a)(s-b)(s-c)} = abc/4R  where R is the circumradius

so

 R = abc / 4Sqrt{s(s-a)(s-b)(s-c)}[/hide][/quote]

That's right, but you should try to use LaTeX.
[hide="Solution in LaTeX"]Well we know that $\frac{a+b+c}{2}$ is the semiperimeter, so I will just leave that as $s$ for simplicity.

By two different formulas for area we get:
$\sqrt{s(s-a)(s-b)(s-c)}=rs$ where $r$ is the inradius

so..

$r=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$

and by a similar arguement we know

$\sqrt{s(s-a)(s-b)(s-c)}=\frac{abc}{4R}$ where $R$ is the circumradius

so

$R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$[/hide]


请问一下数独有多少种？
注：（1）数独即为九个“九宫格”组成的9*9的大方格，要求每行，每列，以及每个“九宫格”中的数字都是1，2，3，4，5，6，7，8，9。
（2）对称或旋转n*$\frac{\pi}{2}$的都算一种。

No one interests in it？It is a popular game in many countries.

我在图书馆看过有关方面的书，不过没有细看， :blush: ，
等假期里，好好研究一下。

为什么数独总是跟图书馆扯上关系~~
因为别的地方更不可能看到数独，所以所有的数独（尽管不多）全在图书馆里，造成的错觉就是数独总是跟图书馆扯上关系。

wiki: Mathematics of Sudoku
http://en.wikipedia.org/wiki/Mathematics_of_Sudoku

上面很详细 :)


limit
real analysis
real analysis unsolved

If {s_n} is a complex sequence, define its arithmetic means t_n by t_n = (s_o + s_1 + ... + s_n) / (n + 1) where (n = 0, 1, 2, ...).

I already proved that if lim s_n = s, then lim t_n = s.

Prove that if {s_n}_n >= 0 is a real sequence such that lim s_n as n->oo = +oo, then lim t_n as n->oo = +oo.

let $R > 0$. since $s_{n}\rightarrow \infty$, there exists $M$ such that $s_{n}>R$ for $n>M$. let $s_{0}+...+s_{M}=k$. then $t_{M+N}>\frac{k+RN}{M+N+1}=\frac{k}{M+N+1}+R\frac{1}{1+\frac{M}{N}+\frac{1}{N}}$. now as $N\rightarrow \infty$, the RHS goes to $R$, so $\lim_{n\rightarrow\infty}t_{n}=\lim_{N\rightarrow \infty}t_{M+N}\geq R$. since $R$ was chosen arbitrarily, it follows that $\lim_{n\rightarrow\infty}t_{n}= \infty$


A block (mass m1 = 22.1 kg) is on a surface inclined at  = 33°. It is connected to a mass m2 = 11.2 kg by a massless cord passing over a pulley. The coefficients of friction are µk = 0.15 and µs = 0.25.

I really need a push to get started. THe problem asks for the magnitud of the force of friction.

[quote="=s"]what's figure 4-53?[/quote]

Sorry but i Need help

im not getting this!!!

[quote="jmerry"]That post really meant "You haven't told us enough for us to be able to help you."[/quote]

Yeah the problem is that im trying to find the magnitud of friction. But im stuck really bad, I don't know where to start. Can you help me.


Καθώς ο προκριματικός διαγωνισμος έρχεται, είναι καλό να αναρτήσουμε μερικά παλιά θέματα ώστε να βοηθήσουμε τους διαγωνιζόμενους.

Θέματα Μεσογειακού Μαθηματικου Διαγωνισμού 2007(πρέπει να ήταν ταυτόχρονα και προκριματικός)

Πρόβλημα 1ο
Έστω x,y,z πραγματικοί αριθμοί με
\[ x \leqslant y \leqslant z
\]
και
\[ xy \plus{} yz \plus{} zx \equal{} 1
\]
.
(α) Να αποδείξετε ότι
\[ xz < \frac {1} {2}
\]
(β) Είναι δυνατόν να βελτιώσουμε την τιμή της σταθεράς
\[ \frac {1} {2}
\]
Προβλημα 2ο'
Έστω ABCD ένα εγγραψιμο τετράπλευρο του οποίου οι διαγώνιοι AC και BD τέμνονται στο Ε. Δίνεται ότι AB=39,AE=45,AD=60 και BC=56
Να προσδιορίσετε το μήκος της πλευράς CD.

Πρόβλημα 3ο
Τριγώνου ABC δίνεται η πλευρά a=BC, η γωνία
\[ \hat A \equal{} \omega
\]
και ότι
\[ a \equal{} \sqrt {\rho R}
\]
όπου ρ η ακτίνα του εγγεγραμένου κύκλου και R είναι η ακτίνα του περιγεγραμμένου κύκλου του.Να προσδιορίσετε τα τρίγωνα που ικανοποιούν τη δεδομένη ισότητα, δηλαδή να υπολογίσετε τις πλευρεές τους b=Ca και c=AB

Πρόβλημα 4ο
Έστω x>1 μη ακέραιος πραγματικός αριθμός.Να αποδείξετε ότι:
\[ \left( {\frac {{x \plus{} \left\{ x \right\}}} {{\left[ x \right]}} \minus{} \frac {{\left[ x \right]}} {{x \plus{} \left\{ x \right\}}}} \right) \plus{} \left( {\frac {{x \plus{} \left[ x \right]}} {{\left\{ x \right\}}} \minus{} \frac {{\left\{ x \right\}}} {{x \plus{} \left[ x \right]}}} \right) > \frac {9} {2}
\]
όπου [x] και {x} παριστάνουν το ακέραιο και το κλασματικό μέρος αντίστοιχα του x

Μερικές βιαστικές υποδείξεις για τα 3 πρώτα θέματα
1 . Αν $ s\equal{}x\plus{}z ,p\equal{}xz$ τότε $ ys\plus{}p\equal{}1$ και $ (x\minus{}y)(z\minus{}y)\le 0 \Rightarrow y^2\minus{}sy\plus{}p\le 0$ τότε $ 2p\minus{}1\le \minus{}y^2\le 0$ αν $ y\equal{}0\Rightarrow x\le 0 \le z , xz\equal{}1$ αδύνατο άρα $ y \ne 0\Rightarrow p<\frac{1}{2}$ (*).η σταθερά $ \frac{1}{2}$ δεν μπορεί να βελτιωθεί σε κάποια $ c$ διότι ανάμεσα στο c και το 1/2 μπορεί να υπάρξει τιμή του y ώστε η (*) ν μην ισχύει

2 . Απο τα όμοια τρίγωνα EDA,ECB υπολογίζουμε το ΒΕ ενώ αν ονομάσουμε με λ τις αναλογίες των πλευρών των ΑΒΕ,CDE εκφράζουμε τα μήκη CE,CD,EB συναρτήσει του λ. Μετά με Πτολεμαίο  βρίσκουμε το λ και τα ζητούμενα

3. Αν Ι το έκκεντρο τότε  ΒΙΓ κατασκευάζεται από την α την ΒΙΓ=90+ω/2 και το ύψος του ρ αφού η R είναι γνωστή από τα α ,ω. άρα και ο κύκλος (Ι,ρ) κατασκευάζεται οπότε αν φέρω  τις εφαπτόμενες από τα Β,Γ βρίσκω και την Α άρα όλα τα στοιχεία του ΑΒΓ

Και μια λυση στο 4ο θεμα:

Ειναι εφαρμογη της ανισοτητας:
$ \sum{\frac {a \plus{} b}{c} \minus{} \frac {c}{a \plus{} b}} \geq \frac {9}{2}$, για $ a \equal{} x, b \equal{} \{x\}, c \equal{} [x], a \equal{} b \plus{} c$

Η τελευταια μπορει να δειχθει ετσι:
$ \frac {a \plus{} b}{c} \minus{} \frac {a}{b \plus{} c} \equal{} \frac {b}{c}\left(1 \plus{} \frac {a}{b \plus{} c}\right)$ και
$ \frac {a \plus{} b}{c} \minus{} \frac {b}{a \plus{} c} \equal{} \frac {b}{c}\left(1 \plus{} \frac {b}{a \plus{} c}\right)$
Οποτε ευκολα το αριστερο μελος ειναι:
$ \frac {1}{2}\left(2\sum{\frac {a}{b}} \plus{} \sum{\left(\frac {a}{b} \plus{} \frac {b}{a}\right)\left(\frac {c}{b \plus{} a}\right)}\right) \geq \frac {1}{2}\left(6 \plus{} 2\frac {3}{2}\right) \equal{} \frac {9}{2}$

Ισοτητα δεν μπορουμε να εχουμε εδω διοτι ειναι $ x > 0$.

[quote="Eukleidis"]Πρόβλημα 4ο
Έστω x>1 μη ακέραιος πραγματικός αριθμός.Να αποδείξετε ότι:
\[ \left( {\frac {{x \plus{} \left\{ x \right\}}} {{\left[ x \right]}} \minus{} \frac {{\left[ x \right]}} {{x \plus{} \left\{ x \right\}}}} \right) \plus{} \left( {\frac {{x \plus{} \left[ x \right]}} {{\left\{ x \right\}}} \minus{} \frac {{\left\{ x \right\}}} {{x \plus{} \left[ x \right]}}} \right) > \frac {9} {2}
\]
όπου [x] και {x} παριστάνουν το ακέραιο και το κλασματικό μέρος αντίστοιχα του x[/quote] επίσης [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=801778#p801778]εδώ[/url] κι [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=802280#p802280]εδώ[/url]


I think that each question should be weighted different.

The first 5 can be weight 1 point, the second five 2 points each and the third fivve 3 points each for example.

Not sure if everyone will agree. Just putting the thought of given different values to different questions out there. (Cuz I got #11 and #14 right while making dumb mistakes on #1,#2,#4 and #5)

I think getting the easy questions right is just as important as getting the hard questions right... if you to go the IMO and are the only person to get #6 but don't get anything else, you don't win. It seems almost like tradition in the AMC system to weight all questions on a test equally.

Another point is that they can't always judge the difficulty of a problem, and weighting problems just puts more pressure on them which is totally unnecessary. As for this AIME, I felt that #11 and #14 were both easier than, say, #4 (which I also got wrong).

Well, number 14 was kind of too guessable...

But I agree with Paladin. It's impossible to judge whether a number thirteen is really a thirteen or rather a five. You know? Then people will start complaining about: oh, well, I spent so much time on number seven, and meanwhile number fifteen was so trivial. God. This AIME sucks. *more complaining and finally more revision requests*

Bad Idea. This isn't HMMT. It works for actual short contests, not 3 hour examinations.

The emphasis is on getting questions correct, rather than getting just the hard questions correct. The idea is that you have to be able to master the easy questions in order to qualify, or be some ridiculously competent that you can get the last 8 and miss the first 7, or something of that nature. It is important to realize that you really need to get the beginning questions right, because they're generally easiest. You're not really "expected" to get all of the last ones on the AIME. While you feel like you've accomplished something for solving a number 13, its still worth just as much as number 1.

Lets imagine a rather average student takes the AIME. He gets 1-8 correct, and qualifies for the USAMO.

Now, lets imagine a very competent student takes the AIME. He gets 1,2, 5, 7, 12, and 13 right. This person definitely solved the more challenging problems, but didn't show his competency on all of the "easy" material, so therefore he shouldn't pass the examination. Its like knowing bits of algebraic number theory when you don't know Algebra II. Or something to that effect.

I don't like where this thread is going either. It seems to be going down the complaining road as well. I'm going to lock this topic, and ask that all format complaints be put in the concerning the AIME twice a year thing.


As long as there is no solution on the internet, like you said, it should be fine.

Please [b]do not[/b] post information about the problems while the round is in progress.


For each integer $ n$, $ 0 \le n < 2010$, let $ \theta (n)$ be the number of integers $ x$, $ 0 \le x < 2010$ such that $ x^2 \minus{} n$ divides 2010. Find $ \sum^{2009}_{k \equal{} 0} n \cdot \theta (n)$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1093491#1093491]Been posted here[/url]  :)


geometry
calculus
integration
analytic geometry
calculus computations

Here are two problems

I got up to drawing the graph and finding the point of intersection, but thats it.. can anybody help me finish them

1)   y= x^2  - 18    ,    y =  x - 6

and 2)     x = 0,  x = 4 ,  y = 1/(x+1)   ,   y = (x-1)/2

Thanks A lot

1)[url=http://www.mathlinks.ro/viewtopic.php?t=170850]See my last post[/url]

that really doesnt help much, i need the next step

Can you understand the area by Integration?

If you don't know integration, it's sort of difficult.  For 1, the graph of $ y\equal{}x\minus{}6$ is on top, so we set up our integral $ \int_{a}^{b}(x\minus{}6)\minus{}(x^{2}\minus{}18)\mbox{ d}x$ where $ a$ is the smaller $ x$-value of the intersection, and $ b$ is the larger one.  So we integrate and get $ \frac{x^{2}}{2}\minus{}12x\minus{}\frac{x^{3}}{3}$, which is then just evaluated from $ a$ to $ b$, or $ f(b)\minus{}f(a)$ is the region (of course, if part of the region is positive, and part is negative, there is more work to be done but I'm too lazy to do that).

If you don't know integration, then everything I just posted is useless anyway and you won't really be able to do these problems.

on the graph of    y = x^2   and   y = x^3,      what is the other point of intersection  besides 1,1    and what is the top and botton in this, ??

$ x^3\minus{}x^2\equal{}x^2(x\minus{}1)$ gives the $ x$ coordinates of the intersection points are $ x\equal{}0,\ 1$ and since $ x^2\geq 0$, for $ 0\leq x\leq 1$, we have $ x^2\geq x^3.$

i got   1/12    for m area between x^2 and x^3   is that right?

ok good, thank you so much for your help,, i know how to do these i think now but number 2 still as me stumped

1.Sketch the graphs.

2. Find the $ x$ coordinates of the intersection points.

3. Confirm that which graph is locatetd upper tha the other one.

4. If for $ y\equal{}f (x),\ g(x)$ such that $ f(x)\geq g(x)$ at $ a\leq x\leq b$, the area bounded by these graphs can be evaluated as $ \int_a^b \{f(x)\minus{}g(x)\}\ dx.$

Good luck. :)

right, but the grapph crosses and its wierd looking from 0,4 so i dont no how to get started

Let $ x \equal{} \alpha$ be the $ x$ coordinate of the intersection point, the desired area $ S$ can be calculated by splitting the region into two parts by the line $ x \equal{}\alpha$.

the intersection point is at x=1.732

so i would find the area from 0 - 1.732

and then

the area from 1.732 - 4???

Let $ f_1(x) \equal{} x^2 \minus{} 18$ and $ f_2(x) \equal{} x \minus{} 6$, we have:
$ f_1(x) \minus{} f_2(x) \equal{} 0$<=>$ x^2 \minus{} x \minus{} 12 \equal{} 0$<=>$ x \equal{} \minus{} 3$  or  $ x \equal{} 4$
The area we need to find is:
$ S \equal{} \int_{ \minus{} 3}^{4}\left|x^2 \minus{} x \minus{} 12\right|dx \equal{} \left| \left(\frac {x^3}{3} \minus{} \frac {x^2}{2} \minus{} 12x\right)|_{ \minus{} 3}^4 \right| \equal{} \left| \minus{} \frac {343}{6}\right| \equal{} \frac {343}{6}$

ppt, your answer is correct. See my first post above.

We have: $ \frac{1}{x\plus{}1}\equal{}\frac{x\minus{}1}{2}$
<=>$ x^2\minus{}1\equal{}2$<=>$ x\equal{}\sqrt{3}$ or $ x\equal{}\minus{}\sqrt{3}$.
We only get $ \sqrt{3}$.
The area we need to find is:
$ S\equal{}\int_{0}^{4}\left|\frac{1}{x\plus{}1}\minus{}\frac{x}{2}\plus{}\frac{1}{2}\right|dx$
$ \equal{}\left|\int_{0}^{\sqrt{3}}\left(\frac{1}{x\plus{}1}\minus{}\frac{x}{2}\plus{}\frac{1}{2}\right)dx\right| \plus{} \left|\int_{\sqrt{3}}^{4}\left(\frac{1}{x\plus{}1}\minus{}\frac{x}{2}\plus{}\frac{1}{2}\right)dx\right|$
$ \equal{}ln\frac{\left(\sqrt{3}\plus{}1\right)^2}{5}\plus{}\frac{1}{2}\plus{}\sqrt{3}$.


pigeonhole principle
calculus
integration
combinatorics unsolved
combinatorics

Solve:

Prove that some integral power of 2 has the decimal expansion which starts with the digits 2004:

2^n = 2004...

Using pigoenhole principle.

Well, since you have posted exactly the same question in the pre-olympiad forum, I think we can assume that the pigeon-hole principle will be used in this other forum. Here we will used some advanced tools (this is the name of the section  :D ).

It is a classical result that if $\alpha $ is irrational then the numbers of the type $n-k\alpha$ , where $n,k$ are non-negative integers, form a set which is dense in $\mathbb R$ (note that its proof may used the pigeon-hole principle, so it is ok :D ).

Use this for $\alpha = \frac {Ln(10) } {Ln(2)}$.
It follows that there exists a pair of nonnegative integers $(n,k)$ such that $\frac {Ln(2004)} {Ln(2)} < n-k \frac {Ln(10)}{Ln(2)} < \frac {Ln(2005)} {Ln(2)}$.
This leads to $2004\cdot 10^k < 2^n < 2005 \cdot 10^k$, and we are done.

Pierre.

Pierre, would it be possible for you to post a proof of that classical result?

Thanks.


Does the series $ \sum_{i\equal{}1}^{\infty}(\minus{}1)^{i\plus{}1}\frac{1}{i}$ diverge or converge?

Your series converges to 0. You have alternating positive and negative numbers, and $ |f{i}|>|f(i\plus{}1)|$.

That series is well known (it is the alternated harmonic series): it is convergent and its value is ln 2.

[quote="Temperal"][quote="Kyron"]Your series converges to 0. You have alternating positive and negative numbers, and $ |f{i}| > |f(i \plus{} 1)|$.[/quote]
Sorry, but I can't follow your logic. An alternating positive-negative sequence and $ |f(i)| > |f(i \plus{} 1)|$ doesn't imply convergence to $ 0$. Carcul is correct.[/quote]

I'm sorry, I didn't mean that $ \sum_{i\equal{}1}^{\infty}(\minus{}1)^{i\plus{}1}\frac{1}{i}$ [i]equals[/i] 0, I meant that the value of the individual,um, sum-number-thingies [i]approaches[/i] 0 as the value of $ i$ increases, meaning that there is [i]an[/i] answer, not that 0 [i]was the[/i] answer.

Do you mean because the terms go to 0, the sum converges?  Try \[ \sum_{n\equal{}1} ^{\infty} \frac{1}{n}\].

If the series converges, the terms go to 0, but it's not a necessary [b]and[/b] sufficient condition

This problem is beyond the scope of HSB.  And the [url=http://en.wikipedia.org/wiki/Alternating_series_test]alternating series test[/url] is what we're looking for.


topology
real analysis
real analysis unsolved

Let $ X$ be a topological space, $ Y$ a Hausdorff space, and let $ f: X\to Y$ and $ g: X\to Y$ be continuous. Show that $ {x\in X: f(x)\equal{}g(x)}$ is closed.

Hint:
Set $ F =${$ {x\in X|f(x) = g(x)}$}.
Consider $ X - F = X\bigcap \tilde F$.
Choose $ z\in X\bigcap \tilde F$. Then $ z\in \tilde F \implies f(z) \neq g(z)$.
Since $ Y$ is a Hausdorff space, there exist two open sets $ U_{1}$ and $ U_{2}$ with $ f(z)\in U_{1}$ & $ g(z)\in U_{2}: U_{1}\bigcap U_{2} = \emptyset$.
By the continuity of $ f$ and $ g$, $ f^{ - 1}(U_{1})$ and $ g^{ - 1}(U_{2})$ are open subsets of $ X$ and so they are are contained in $ \tilde F$.
We have,
$ f^{ - 1}(U_{1}) \bigcap g^{ - 1}(U_{2}) \in \tilde F$.

Now set $ O_{z} = f^{ - 1}(U_{1}) \bigcap g^{ - 1}(U_{2})$

And show that $ \tilde F = \bigcup O_{z}$

Edited: I realized that my first argument was not valid so I tried to correct it here.


give x,y in integer.Find the largest of
|7x^2+13xy-7y^2|;
that is end.
------Hikaru------

Hikaru, this paper is really strange.Are you sure it is VMO1978?
Let $ y=0 $ and $ f(x,y) =7x^2 $

i think he means that
find the min value of
$\mid 7x^2-13xy-7y^2\mid$
and of course $x^2+y^2 > 0$


Hi! One of the high school students' participents on the imo is a member of the moroccan delegation and we, as a member of his family, don't have any news about him. Can you tell us, please, the name of the hotel and the telephone number of this hotel if it is possible where the moroccan delegation is. And thank you very much.

I think that they are fine, but I have no idea which hotel, but it's either Fiesta Americana or Conquistador. You can try calling at both ...  :?


[b]n > 2[/b]

Prove that [b]n(n+1)(n+2)...(3n-4)(3n-3)[/b] is divisible by a square of each natural number that is less or equal n

Any set of $k$ consecutive integers contains a multiple of $k$.
Thus, if $k \leq n-1$ there is a multiple of $k$ among the numbers $n,n+1, \cdots , 2n-2$ and there is another among the integers $2n-1, \cdots, 3n-3$. Therefore the given product is divisible by $k^2$.
If $k=n$ the result also holds since the product contains the factors $n$ and $2n$.

Pierre.


Let $ I$ be the incenter and $ \left(I\right)$ the incircle of a triangle $ ABC$.

Let the segments $ AI$, $ BI$, $ CI$ meet the circle $ \left(I\right)$ at points $ A_1$, $ B_1$, $ C_1$, respectively. Let $ A_2$, $ B_2$, $ C_2$ be the midpoints of the segments $ BC$, $ CA$, $ AB$, respectively. Prove that the lines $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ concur at one point.

[quote="pigfly"]suppose $I$ is incenter of triangle $ABC$ , $AI$ meet $(I)$ in $A_1$, such that $A_1$ in segment $AI$. $A_2$ is midpoint of $BC$, $B_1$, $B_2$, $C_1$, $C_2$ similar. prove that $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent.[/quote]

This has been asked on http://www.mathlinks.ro/Forum/viewtopic.php?t=20543 some time ago and I don't know a synthetic proof.

  Darij

Dear Mathlinkers,
two synthetic proofs of this known result originated from Clark Kimberling has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A_{i},B_{i},S_{i}$ ($i=1,2,3$) be invertible real $2\times 2$ matrices such that [list][*]not all $A_{i}$ have a common real eigenvector, [*]$A_{i}=S_{i}^{-1}B_{i}S_{i}$ for $i=1,2,3$, [*]$A_{1}A_{2}A_{3}=B_{1}B_{2}B_{3}=I$.[/list] Prove that there is an invertible $2\times 2$ matrix $S$ such that $A_{i}=S^{-1}B_{i}S$ for all $i=1,2,3$.

We can replace $A_{i}, B_{i}$ by $xA_{i}, yB_{i}$ with $x,y$ non-zero real numbers; we can also
replace $A_{i}, B_{i}, C_{i}$ by $X_{i}A_{i}X_{i}^{-1}, X_{i}B_{i}X_{i}^{-1}, X_{i}C_{i}X_{i}^{-1}$ with $X_{1},X_{2}\in GL_{2}(\mathbb{R}).$ We can thus assume $A_{1}=A_{2}.$ Of course, it is enough to find an $S$
such that $SA_{1}S^{-1}=A_{1}$ and $SB_{1}S^{-1}=B_{2}.$ We shall also assume that none of the matrices
$A_{i}, B_{i},C_{i}$ are scalars.

Suppose first that the characteristic polynomial of $A_{1}$ is reducible over $\mathbb{R}$.
Twisting $A_{1}$ by a scalar and conjugating appropriately, we need to consider the following
two cases:

i) $A_{1}=A_{2}=\begin{pmatrix}x & 0\\ 0& 1\end{pmatrix}$ with $x\neq 1.$ Set
$B_{i}=\begin{pmatrix}a_{i}& b_{i}\\ c_{i}& d_{i}\end{pmatrix}.$ The condition that
 $A_{1}B_{1}$ and $A_{1}B_{2}$ are conjugates forces $a_{1}=a_{2}, d_{1}=d_{2}, b_{1}c_{1}=b_{2}c_{2}.$ For an
appropriate diagonal matrix $S$ we get
$B_{2}=SB_{1}S^{-1}.$

ii) $A_{1}=A_{2}=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}.$ Again, let
$B_{i}=\begin{pmatrix}a_{i}& b_{i}\\ c_{i}& d_{i}\end{pmatrix}.$ Comparing traces of $A_{1}B_{1}, A_{1}B_{2}$ gives
$c_{1}=c_{2}.$ The condition on eigenvectors implies $c_{1}\neq 0.$ If we set
$S=\begin{pmatrix}1 & x\\ 0 & 1 \end{pmatrix}$ where $x$ is chosen so that
$SB_{1}S^{-1}=\begin{pmatrix}a_{2}& * \\ c_{1}& * \end{pmatrix},$ then $SA_{1}S^{-1}=A_{1}$ and
$SB_{1}S^{-1}=B_{2}.$

Now suppose that $A_{1}=A_{2}$ has irreducible (over $\mathbb{R}$) characteristic polynomial.
As in case i, we can find a $T\in GL_{2}(\mathbb{C})$ such that $TA_{1}T^{-1}=A_{1}$
and $TB_{1}T^{-1}=B_{2}=S_{2}B_{1}S_{2}^{-1}.$ Note the following: If $A\in GL_{2}(K)$ is a non-scalar
 ($K$ a field) and $MA=AM$ for some $2\times 2$-matrix over $K$, then $M=xA+yI$ with
$x,y\in K.$ (To see this, conjugate $A$ to a matrix
$\begin{pmatrix}0 & d\\ 1 & t\end{pmatrix}.$) We can thus find real numbers $x_{1},y_{1}, x_{2},y_{2}$ such that $S_{2}^{-1}(x_{1}A+y_{1}I)=x_{2}B_{1}+y_{2}I$ with $(x_{1},y_{1})\neq (0,0).$ (This can
be seen after writing $T$ as a complex linear combination of $A$ and $I$,  and then
separating real, imaginary parts.) Set $S=x_{1}A+y_{2}I.$ If this matrix $S$ is invertible, then we
are through. But this follows because the ring of matrices $\{ xA+yI: x,y\in \mathbb{R}\}$ is
isomorphic to $\mathbb{C}$ by the map which sends $A$ to one of its eigenvalues.


function
induction
advanced fields
advanced fields theorems

Prove that the set of [url=http://en.wikipedia.org/wiki/Computable_function]computable functions[/url] is countable.

The text by Oded Goldreich proves it in the following manner,

[quote]Since each computable function is computable by a machine that has a finite description, there is a 1-1 correspondence between the set of computable functions and the set of strings.[/quote]

I do not understand why a finite description results in a 1-1 correspondence. Or rather, I do not understand the difference between this "finite description of a function/machine" and the "finite description of a subset in the power set of the natural numbers".  The latter (ie power set of natural numbers) is known to be uncountable.

Sorry if this sounds confusing and if this post is in the wrong forum.

What is likely meant is that there is a 1-1 correspondence between the set of computable functions and some subset of the set of all strings of [i]finite[/i] length.  Since the set of finite strings is countable, all of its subsets are countable, and in particular so is the set of machines with finite descriptions.

There is a section of the [url=http://www.artofproblemsolving.com/Forum/index.php?f=331]forum[/url] specifically dedicated to computer science where this sort of question may be more appropriate.

If we have a finite alphabet {a,b,c,...,z} say, and we write programs as strings in this alphabet "ack(m,n) := ..." or whathaveyou since every string has finite length, We can see that the set of strings is countable because it simply corresponds to base 26 numbers. We have not constructed a bijection here (because not every string is a valid program) but since there is an injection both ways that's [url=http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem]good enough[/url].

If we had a countable alphabet, say natural numbers {0,1,2,..}. We could not consider strings as base-infinity numbers anymore.. but this is still in bijection with natural numbers the justification is slightly harder. Assuming that the product of two countable sets (for an explicit example NxN is in bijection with N witnessed by f(x,y) = (x^2+y^2+x+2xy+3y)/2) then (by induction) we can treat any set N^n as N. So for a length n string we can take f(n,<string>) where <string> is the string represented as an element of n as its encoding. This set of finite length strings of N is usually called N* and is distinct from the powerset.


i was going to have a poll but fire fox crashed on me and i dont have the time to make another one. plz keep an accurate ranking and when i can i will always keep it updated on this first post. but plz make your copies too so i can cut and paste! =P o and keep the ranking list hidden with the "click to reveal hidden content" thingymabob

[quote="kimmystar94"]1. Kirby

Kirby just owns.[/quote]

My response:

1. Kirby

Kirby just owns.

1. Snake
2. Marth
3. Meta Knight
4. Lucas

Pikachu is PWNAGE.  Why doens't anybody like it?

Might have somethign to do with him yelling "PIKA!" all the time...

Pika!  Was almost as annoying as kirby's taunt in brawl.  Better now, though.

[quote="erabel"]1. Toon Link
2. Ike
3. Meta Knight
4. Squirtle
5. Pit
6. Snake
7. Sonic
8. Lucario
9. All others
10. Ganondorf
11. Sheik
12. Falco=EPIC FAIL![/quote]
please explain why falco is epic fail. please explain the benefits of using ike (besides the fact that you can ko noobs quickly).

I like ike (rhyming power). I prefer power characters with a semi-decent recovery move that can also kill people. Excellent dair, and no-nonesense final smash.

hmm... for me all dairs are suicide moves (i suck at recovery) except for falco's which is a really fast spike
i like sort of fast characters that can land more hits. 1*10>5000*0
and just another thing falco's side smash is relatively strong

I hate Falco because my brother always beats me if he plays as Falco, unless I play as Toon Link, but then he starts to cry.
Ike is a beast. 'Nuff said.

hello????

[hide="i have cheat codes to brawl"]
[hide="righttt here...."]
[hide="almost there"]

PIKA!
[/hide][/hide][/hide]
Believe it or not, that is a valid cheat.

Not really. How come nobody realizes you can cancel it mid-smash by pressing b?

OMG

I can't believe nobody likes toon link. I could pwn you all with him. I actually can play all of the characters though.

[quote="007math"]My favorite character is probably SNAKE!!!  You can control his missile (side+b), thats how I kill most people.  And his anti-gravity thingy is awesome!!!  And he has good attacks!!and his final smash is awesome too![/quote]

I love facing Snake's that over-rely on Side-B. A dodge or jump causes the missle to miss, and in the time it takes to turn it around, you can easily own him with a smash attack.

Meh snakes final smash is pretty much ownage. unless you're really good at dodging....

My little brother always plays PIkachu and spams B-Down. Let me tell you, that leads to some very interesting matches (i.e. in a 20 life game me winning with 15 lives left).

Frankly, ZSS dair is beast, but if you go off the edge you have to be quick. Her regular special is paralyzing, but without any truly powerful moves, it's only useful when you are running or have a baseball bat :| Still, her Final Smash owns if they are like over 50% (on DR 1.0 of course, its OHKO on 2.0!) especially because people are always near you fighting over final smash. Still, she has no real ranged moves, and no real power. I still find her a very playable character.

Snake is just so awsome!! because you can like blast everyone with his missle. And his attacks are pretty strong.

wtf. don't revive old threads, spammer n00b.

Mine:

1. Zelda
2. Sheik
3. Marth
4. Kirby
5. Peach(hate her but she is strong)

I rock at Marth and Zelda...but I always die with Marth's KO-I am always on the edge for some reason.

I rock at Zelda though! And Link...kinda

Best
1. Captain Falcon
2. Toon Link
3. Ike
4. Marth
5. Meta Knight

Worst
1. Jigglypuff
2. Olimar
3. Mr Game and Watch
4. Lucas
5. King Dedede

Falco
Marth
Pit
Toon link.

PIt and toon link are just must easier to use, but I like falco and marth because they are used less commonly and have somewhat unique, quick but striking, fighting mechanics.  Unfortunately, Marth has no ranged attacks.

Kirby cause i suck at everyone else

well maybe yoshi with his egg troll move

Mr. Game and Watch. They gave him a HUGE buff. I also like Captain Falcon (Punch).

Best:
1. Snake(my fav)
2. Kirby (because you can just float around if they kick you off)
3. Mario & Luigi (very agile and pretty good moves)
4. Ike (very strong a combos)
5. Pikachu (nice recovery and moves)

Worst:
1. Jigglypuff (kicks and punches are hard to use)
2. Mr. Game and Watch ( slow)
3. Ganondorf (slow attacks and movement)
4. Ness ( weak moves)
5. King D ( slow attacks and moves)
[hide=" and P.S."] Why did everyone start reviving this thread after I did? I apologized already! [/hide]


geometry
3D geometry
tetrahedron
octahedron
pyramid

What is the volume of an tetrahedron with edge length of 1?

What is the volume of an octahedron with edge length of 1?

[hide]a tetrahedron is a pyramid with all faces congruent
a octahedron is a tetrahedron on another tetrahedron so twice the first answer
the base is 1 X (sqrt3).5
the height is sqrt.5
(.5(sqrt3) X sqrt.5)/3
[b](sqrt1.5)/6[/b]
[b](sqrt1.5)/3[/b][/hide]

[quote="apusey"]a octahedron is a tetrahedron on another tetrahedron so twice the first answer
[/quote]

sure about that (look at the bases)?

[hide="I think..."]
Tetrahedron: $\frac{\sqrt{2}}{12}$

Octahedron: $\frac{\sqrt{2}}{3}$[/hide]

I can prove my answer for tetrahedron, and I believe I can use a similar method for octahedron, too.

If no one else is posting a proof, I'll post mine later.
(Sorry, I don't have much time now.)

[quote="frt"]Tetrahedron: $\frac{\sqrt{2}}{12}$[/quote]
[hide="Proof"]
Let $A$, $B$, $C$, and $D$ be the vertices of the tetrahedron.
Draw a perpendicular line from $A$ to $\triangle BCD$ and let $P$ be the intersection. ($P$ is the centroid of $\triangle BCD$.)

$BP=\frac{2}{3}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}$

$AP=\sqrt{AB^2-BP^2}$ $=\sqrt{1-\left(\frac{\sqrt{3}}{3}\right)^2} =\frac{\sqrt{6}}{3}$

$\therefore V=\frac{1}{3}\cdot AP\cdot \triangle BCD$

$=\left(\frac{1}{3}\right)\left(\frac{\sqrt{6}}{3}\right)\left(\frac{\sqrt{3}}{4}\right)$

$=\boxed{\frac{\sqrt{2}}{12}}$

[/hide]

[hide]An octahedron is two pyramids connected at the rectangular base. The volume is then $\frac{2}{3} s^2 \cdot h$ where $h$ is the altitude to the common base. $h=\sqrt{1^2-\frac{1}{2}}=\frac{1}{\sqrt{2}}$. Thus $V=\frac{2}{3} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{3}$.[/hide]

[hide]Volume of a tetrahedron is $\frac{1}{3}ah$ where $a$ is area and $h$ is height.

The area of the base is $\frac{\sqrt{3}}{2}$

The height can be obtained by pythagorean's theorem, $c^2 - b^2 = h^2$
In this case, $c$ is $\sqrt{3}$ and $b$ is $\frac{\sqrt{3}}{2}$, thus we obtain $h = \sqrt{3 - \frac{3}{4}} = \frac{3}{2}$

Now $\frac{1}{3}ah = \frac{\sqrt{3}}{4}$

The volume of the octohedron is just twice the volume of the tetrahedron.

$V_2 = \frac{\sqrt3}{2}$[/hide]

[quote="breez"]The volume of the octohedron is just twice the volume of the tetrahedron.[/quote]

Sure about that? ;)

It is not? I see it just as 2 tetrahedrons joined on 1 side.

That would be a polyhedron with only 6 faces... :?

I guess there are some problems on your solution.

First, you said that the area of the base is $\frac{\sqrt{3}}{2}$, but I think that should be $1\cdot\frac{\sqrt{3}}{2} \cdot\frac{1}{2} =\frac{\sqrt{3}}{4}$. (The base is an equilateral triangle with side lengths $1$.)

Second, I don't know how you got the height of the tetrahedron.
Although the question wasn't clear about this, the question was asking the volume of a regular tetrahedron. So $c$ should be $1$ (=side length), not $\sqrt{3}$. Also, $b$ is $\frac{2}{3}\cdot \frac{\sqrt{3}}{2} =\frac{\sqrt{3}}{3}$. (Please refer to my previous post for further explanation.)

I see what you are saying, so the are of the base should be ${\sqrt{3}}{4}$

But how do you get the height? I see it as a triangle whose hypotenuse is $\sqrt{3}{2}$ and whose adjacent side is half that. So it should be $\frac{1}{2}$

yeah i get what frt got.

For the octahedron i get

[hide]sqrt(2)/3[/hide]

Sry about the lack of latex.


I need help with the following proofs:

Given triangle $ ABC$ with 3 acute angles. $ A,B,C$ denote the angles, $ a,b,c$ denote the lengths of the sides of the triangle. Prove:

1) $ \frac{sin(A)\plus{}sin(B)\plus{}sin(C)}{cos(A)\plus{}cos(B)\plus{}cos(C)}\le \frac{tan(A)tan(B)tan(C)}{3}$

2) $ tan^6(A) \plus{} tan^6(B) \plus{} tan^6(C) \ge 81$

3) $ \left(tan(A/2)\right)^{2\sqrt{2}}\plus{}\left(tan(B/2)\right)^{2\sqrt{2}}\plus{}\left(tan(C/2)\right)^{2\sqrt{2}}\ge 3^{1\minus{}\sqrt{2}}$

4) $ (sin(A))^{sin(B)}\plus{}(sin(B))^{sin(C)}\plus{}(sin(C))^{sin(A)}\ge 1.19$

5) $ (1\plus{}sin(A))^{1/{sin(B)}}(1\plus{}sin(B))^{1/{sin(C)}}(1\plus{}sin(C))^{1/{sin(A)}}>8$

6) $ 3(a\plus{}b\plus{}c)\equal{}\pi \left(\frac{a}{A}\plus{}\frac{b}{B}\plus{}\frac{c}{C}\right)$

7) $ tan^n(A) \plus{} tan^n(B) \plus{} tan^n(C) > 3\plus{}\frac{3n}{2}$ for $ n\in N$

8) $ sin^n(A) \plus{} sin^n(B) \plus{} sin^n(C) > \frac{n}{n\minus{}1}$ for $ n\in N$ and $ n\ne 1$

[hide="2"]By the Power Mean and Jensen inequalities,
$ \displaystyle\sqrt [6]{\dfrac{\tan^6 A \plus{} \tan^6 B \plus{} \tan^6 C}{3}} \ge \dfrac{\tan A \plus{} \tan B \plus{} \tan C}{3} \ge \tan \left(\dfrac{A \plus{} B \plus{} C}{3}\right)$
and $ \tan \left(\dfrac{A \plus{} B \plus{} C}{3}\right) \equal{} \tan \dfrac{180^\circ}{3} \equal{} \sqrt {3}$,
so
$ \displaystyle\sqrt [6]{\dfrac{\tan^6 A \plus{} \tan^6 B \plus{} \tan^6 C}{3}} \ge \sqrt {3}$.
Raising both sides to the sixth power, we get
$ \dfrac{\tan^6 A \plus{} \tan^6 B \plus{} \tan^6 C}{3} \ge 27$, which implies $ \tan^6 A \plus{} \tan^6 B \plus{} \tan^6 C \ge 81$, with equality when $ A \equal{} B \equal{} C \equal{} 60^\circ$.
[/hide]

for (2), use Jensen with f(x) = [tanx]^6

for (3) also, use Jensen with f(x) = [tanx]^(2*sqrt(2))  :)

No1.[hide]Without loss of generality let a $ \leq$ b $ \leq$ c
Then tanA $ \leq$ tanB $ \leq$ tanC and
cosA $ \geq$ cosB $ \geq$ cosC.By Chebychev's Inequality we get

$ \frac{1}{3}$.$ {\sum{\frac{sinA}{cosA}}.cosA}$ $ \leq$ $ \frac{1}{3}$.$ \sum{tanA}$.$ \frac{1}{3}$.$ \sum{cosA}$ which becomes the inequality in number after writing tanA.tanB.tanC = tanA +tanb +tanC(an identity true in every triangle[/hide])

No7 [hide]By the AM_GM inequality we have

($ \frac {tanA + tanB + tanC}{3})^{3}$ $ \geq$ tanA.tanB.tanC and using the identity tanA +tanB+tanC =tanA.tanB.tanC we deduce that tanAtanBtanC $ \geq$ $ \sqrt {27}$ when the angles A,B and C are acute.

Now use the following version of Bernouilli's Inequality:Let x be positive and not equal to 1.Then

$ x^p - 1$ > p(x-1) when p<0 or p>1 while

$ x^p - 1$  < p(x-1)  when 0 <p<1

Writing p=n and taking x=tanA,  TanB ,    tanC in succession and adding the three inequalities we get

$ \sum{tan^{n}A}$  -3 >n.$ \sum{tanA}$ -3n which becomes

$ \sum{tan^{n}A}$ >3 + n.$ \sum{tanA}$ -3n ,    and using $ \sum{tanA}$ = tanA.tanB.tanC $ \geq$ 3.$ \sqrt {3}$ we get

$ \sum{tan^{n}A}$ >   3 +3n($ \sqrt {3}$ -1) >  3+  $ \frac{3n}{2}$[/hide]

Any ideas on how to solve 4, 5, 8? Thanks.

No5 [hide]Bernoulli's inequality states:Let x be a positive number not equal to 1 then

$ x^p-1 >p(x-1)$    if     p<0    or     p>1.

Let x= 1+sinA and       p=$ \frac{1}{sinB}$        then applying the Bernoulli inequality we have

$ (1+sinA)^{\frac{1}{sinB}}$ -1 >          $ \frac{1}{sinB}$.(1+sinA -1)   ,     that is

$ (1+sinA)^{\frac{1}{sinB}}$ >       1 +$ \frac{sinA}{sinB}$ >        2.$ \sqrt{\frac{sinA}{sinB}}$  .......(1). Similarly

$ (1+sinB)^{\frac{1}{sinC}}$ >      2.$ \sqrt{\frac{sinB}{sinC}}$ ..........(2) ,and

$ (1+sinC)^{\frac{1}{sinA}}$ >      2.$ \sqrt{\frac{sinC}{sinA}}$ ..........(3).

Now multiply the inequalities 1,    2   and     3    together to get the required result.[/hide]

[quote="kcstevonne"]
8) $ sin^n(A) \plus{} sin^n(B) \plus{} sin^n(C) > \frac {n}{n \minus{} 1}$ for $ n\in N$ and $ n\ne 1$[/quote]

Is this true?
because if we try $ A\equal{}B\equal{}C\equal{}\frac{\pi}{3}$,then $ LHS\to 0$ on the other hand,$ RHS\to 1$ when $ n\to \infty$

Good catch, Anavel_Gato. The inequality in 8 is wrong.


How to prove that every real number has a decimal representation and it is unique unless we have some positive integer $ n$ such that for all $ i > n$ $ i$th decimal is $ 0$ or for all $ i > n$ $ i$th decimal is $ 9$.?

Here's an idea for proving existence.

Let $ x \in \mathbb{R}$, $ 0< x < 1$.  I want to show that there exist integers $ a_1, \ldots, a_n, \ldots$ such that $ 0 \leq a_n \leq 9$ and $ x \equal{} \frac{a_1}{10} \plus{} \frac{a_2}{10^2} \plus{} \ldots \plus{} \frac{a_n}{10^n} \plus{} \ldots$.

Okay, here goes.

Let $ a_1$ be the greatest integer such that $ \frac{a_1}{10} \leq x$.  Notice that $ 0 \leq a_1 \leq 9$.  Let $ r_1 \equal{} x \minus{} \frac{a_1}{10}$.  Notice that $ r_1 < \frac{1}{10}$.  (If it were not, then we could have selected a larger $ a_1$.)  And of course, $ r_1 \geq 0$.

Let $ a_2$ be the greatest integer such that $ \frac{a_2}{10^2} \leq r_1$.  Notice that $ 0 \leq a_2 \leq 9$.  (If    $ a_2$ were $ 10$ or more, that would contradict $ r_1 < \frac{1}{10}$.)  Let $ r_2 \equal{} r_1 \minus{} \frac{a_2}{10^2}$.  Notice that $ r_2 < \frac{1}{10^2}$.  (If it were not, then we could have selected a larger $ a_2$.)  And of course, $ r_2 \geq 0$.

Notice that $ x \minus{} \frac{a_1}{10} \minus{} \frac{a_2}{10^2} \equal{} r_1 \minus{} \frac{a_2}{10^2} \equal{} r_2 < \frac{1}{10^2}$.

Basically, continue to choose digits in this manner.  Make it rigorous using induction.

map x to its decimals as above.  showing uniqueness is only showing that

$ x \equal{} .a_1a_2a_3.... \equal{} .b_1b_2b_3 ...$

so $ 0 \equal{} x \minus{} x \equal{} \sum_{k \equal{} 1}^{\infty} \frac {a_k \minus{} b_k}{10^k}$

note that if $ |a_{k_o} \minus{} b_{k_o}| > 0$ for some minimal $ k_o$ then we always have

$ 1 \ge \sum_{k \equal{} k_o \plus{} 1}^{\infty} \frac {a_k \minus{} b_k}{10^{k \minus{} k_o}} \ge \minus{} 1$

with either limit being achieved only when $ a_k \minus{} b_k \equal{} \pm9$ for each k over $ k_o$

so in order to achieve zero, this can only happen when $ a_{k_o} \minus{} b_{k_o} \equal{} 1$ and $ a_k \minus{} b_k \equal{} \minus{} 9$ or $ a_{k_o} \minus{} b_{k_o} \equal{} \minus{} 1$ and $ a_k \minus{} b_k \equal{} 9$ when $ k>k_o$


algebra
function
domain
calculus
integration
complex analysis
complex analysis unsolved

let f be meromorphic in a domain G. f is analytic on the boundary of G and $ |f|<1$ there. show that the no of roots of $ f(z)\equal{}1$ inside G is equal to the number of poles of f inside G.

under the same conditions but suppose now that $ |f|>1$ on the boundary, then show that this time the no of roots of $ f(z)\equal{}1$ is equal to the no of zeros of f inside G. ( i think once we prove earlier part , this can be got by considering 1/f)

Apply the argument principle to $ f\minus{}1$.

yes i tried that.

we have the integral of $ \frac{f'}{f\minus{}1}\equal{}N\minus{}P$ .here no of poles is the same as the no of poles of f. but how do we show that the integral is 0?

i think i get it now. since $ |f| < 1$ on the boundary we can write the denominator of  $ \frac {f'}{f \minus{} 1}$ in powers of f with convergence being uniform. so we can interchange sum and integral and each of the terms in the integral is a primitive. since the curve is closed, integral is 0. right?

Easier: $ f\minus{}1$ does not wind around $ 0$ because it stays to the left of it.


induction
modular arithmetic
number theory
combinatorics proposed
combinatorics

Let $ A$ be the subset of the set of positive integers, having the following $ 2$ properties:

1) If $ a$ belong to $ A$,than all of the divisors of $ a$ also belong to $ A$;

2) If $ a$ and $ b$, $ 1 < a < b$, belong to $ A$, than $ 1 \plus{} ab$ is also in $ A$;

Prove that if $ A$ contains at least $ 3$ positive integers, than $ A$ contains all positive integers.

If $ A$ contains $ 1$ and $ a,b$, then it will contains with confidence $ 2$, too. Simply, if $ a$ and $ b$ aren't even, then $ ab \plus{} 1$ would be such one $ \Rightarrow 2\in A$.

By induction we can prove that the elements of $ A$ leap through all the remainders $ \mod 2^n$ for some nature $ n$ and using this fact we would be able to complete the solution very quickly.

1. Obviously for $ n \equal{} 1$, we will have some evens and some odds, hence we needn't prove anything.
2. Let's suppose we have proved that for $ n \equal{} k$ there always exists $ a\in A$ such that $ a\equiv r(\mod 2^k)$ for every $ r \equal{} 0,1,2\dots, 2^k \minus{} 1$.
Let's now look $ n \equal{} k \plus{} 1$:

If $ a \equal{} 2^kq \plus{} r\in A$ for $ 0\le r < 2^k$, $ \Rightarrow 2a \plus{} 1 \equal{} 2^{k \plus{} 1}q \plus{} 2r \plus{} 1 \equal{} c\in A$ for some odd $ 1\le 2r \plus{} 1 < 2^{k \plus{} 1}$.

Now let us get $ c_1\equiv 2r \minus{} 1$ and $ c_2\equiv 1$ $ (\mod 2^{k \plus{} 1})$, $ \Rightarrow c_1c_2 \plus{} 1\equiv 2r(\mod 2^{k \plus{} 1})$ and we are ready.

After we have this, it remains to prove that every even belongs to $ A$ and the rest is elementary.

To get $ \pmod m$, use the command \pmod{m}.

I have a different solution to this one.
Easy to prove that numbers $ 1$ and $ 2$ belong to $ A$.Let's show that $ 3$ is also in set $ A$. Assume that the next positive integer is $ p$, where $ p$ is a prime number. (If $ p$ is composite and not of the form $ 2^{n}$, than it must have the prime divisor more than $ 2$, but if it's $ 2^{n}$, than $ 2^{n \plus{} 1} \plus{} 1$ is also contained in $ A$ and it undoubtedly has prime divisor). If $ p \equal{} 3k \plus{} 1$, for some positive $ k$, than $ 2(3k \plus{} 1) \plus{} 1 \equal{} 3(2k \plus{} 1)$ is in $ A$, implying that $ 3$ is in $ A$ as well. Assume $ p \equal{} 3k \plus{} 2$. Obviously, all the numbers of the form $ 2^{n}(p \minus{} 1) \plus{} 1$ are contained in $ A$. It means that if $ 3$ is not in $ A$, than all of the divisors of $ 2^{n}(p \minus{} 1) \plus{} 1$ must be of the form $ 3k \plus{} 2$. But if it has two divisors of this form, or has a divisor in greater degree than $ 2$, it will obviously have one with form $ 3k \plus{} 1$. So, remaining case is when all of the numbers of the form $ 2^{n}(p \minus{} 1) \plus{} 1$ are prime, but if $ n \equal{} p \minus{} 1$, we have $ p|2^{n}(p \minus{} 1) \plus{} 1$(By Fermat's little theorem), where $ 2^{n}(p \minus{} 1) \plus{} 1 > p$, implying that $ 2^{n}(p \minus{} 1) \plus{} 1$ is composite, q.e.d.
   So $ 3$ is in $ A$. We can easily get that $ 4$ and $ 5$ are also elements of the set $ A$. Let's use induction to prove the following claim: If $ n > 5$ and the set $ S \equal{} {1,2,3,...,n \minus{} 1}$ is subset of $ A$, than $ A$ contains $ n$. Proof: If $ n$ is odd, $ 2\frac {n \minus{} 1}2 \plus{} 1 \equal{} n$ is in $ A$, q.e.d. If $ n$ is even, $ 2\frac {n}{2} \plus{} 1 \equal{} n \plus{} 1$ is in $ A$. Hence, $ (n \plus{} 1)(n \minus{} 1) \plus{} 1 \equal{} n^{2}$ is in $ A$, giving $ n$ is in $ A$, q.e.d.
This ends our proof. :) Lasha Lakirbaia.

[quote="lasha"] Obviously, all the numbers of the form $ 2^{n}(p \minus{} 1) \plus{} 1$ are contained in $ A$. [/quote]
How to get this?  :oops:

It is trivial to prove that numbers $ 1,2 $ belong to $ A $.It is not that hard for $ 3 $ neither.$ 4 $ goes easily too.Just suposse oposite and examine cases which easily lead to contradiction.Now suposse otherwise,that there exist some numbers that don't belong to $ A$.Let $ k $ be the smallest one of them.Now because $ k>4 $ it holds $ \frac{k}{2} $ is not equal to $ 2 $.So if we take $ a=2 $ and $ b=\frac{k}{2} $ we easily obtain that $ A $ has $ ab+1=k+1 $ in itself.Now taking $ a=k-1 $ and $ b=k+1 $ we obtain $ A $ has $ ab+1=(k-1)(k+1)+1=k^2-1+1=k^2 $ so $ A $ contains $ k $ too,so we have that $ A $ contains all positive integers for $ 2|k $.Now if that isn't the case we will have to prove that $5$ belongs to $A$ and then we would have that $ \frac{k-1}{2}>2 $ so by taking $ a=2 $ and $b=\frac{k-1}{2}$ we obtain that $k$ belongs to $A$ which is a contradiction so the task is proved.

Actualy,i will write the part for $ 3 $ as well:
Suposse there are numbers $ 1,2,p $ in $ A $ such that $ p $ is prime (it is easy because there can't only be number $2$).Now if $ p \equiv 1 \pmod {3} $ it is trivial.Now consider case $ p \equiv 2 \pmod {3} $.Because of this $ A $ will have in itself also number $ 2p+1 $ which implies that it will also contain number $ 2p^2 + p + 1 $. It is clear that this number is not prime because $ p $ isn't equal to $ 2 $.So it has more than one divisor.Now it is trivial by just considering possible divisors.

An easy problem.We easy obtain that the sequence has 2(if all are odd then 1+ab will be even),and it has 3(if we have two integrs such that they are not congruent to each other mod3,we are finished,now,suppose that all are congruent to each other mod 3,now it is trivial that if all are 1 mod 3,so let all of them be 2 mod3,now if we have an even number bigger then 2 congruent to 2 mod3,than that number has a divisor bigger than 1 congruent to 1,so if all are odd and 2 mod 3,than 1+ab is even and 2 mod3,so we are finished),now from 2 and 3 we obtain 7 and then 5 and use that if a and a+2 belong to A,then a+1 belongs to A,and now we have 2,3,4,5,6 and 7,and now it is easy to see we are finished(induction)

First , we'll prove that $1$ (trivial) , $2,3,4$ belong to the set .
-For $2$ :Suppose one of the elements is even , then $2$ blongs to the set since it divides that element . Otherwise , if all the elements are odd then $ab+1$ must be even for some $a,b$ and so $2$ belongs eitherways.
-For $3$: If one of the elements is divisible by $3$ we're done . Otherwise suppose out of the three elements , two are not congruent to eachother mod 3 , then their product is congruent to 2 mod 3 , increased by $1$ is divisible by $3$  . Now suppose a=b=c=/=0[3] , then ab+1=bc+1=ac+1=2[3] , and so  considering the new element $ab+1$ , we treat two cases , either a=1[3] and so a(ab+1) +1 is divisible by 3 . Or a=2[3] in wich case all of the current elements of the set are 2 mod 3  , now take an element from the set , wich has more than two divisors (obviously exists ) , either this element has a divisor congruent to 1 mod 3 , or it is a product of even number of numbers congruent to 2 mod 3 , in the latter case we simply take away one factor congruent to 2 mod 3 and we get a divisor call it $p$ (so it belongs to the set since it divides an element from teh set)  and p has an odd number of divisors 2 mod 3 and so its 1 mod 3 . and now we can just take the number pa+1 wich is divisible by 3 , and so 3 belongs to the set .
And now easy to prove that 4,5,6,7 .. belong by just plugging in .
We generalize by induction  .
Suppose we proved stuff for numbers less than $X$ . If $X$ is odd then $\frac{X-1}{2}$ belongs and so $2.\frac{X-1}{2}+1=X$ also belongs .
If $X$ is even then $\frac{X}{2}.2+1=X+1$ belongs and so $(X-1)(X+1) +1=X^2=0[X]$ belongs to the set and so does $X$ .

If $n$ is in the set then $n-1$ is in the set. Note $n\equiv 1\pmod {n-1}$ . Take $nk+1$ for any $k$, this new term(let it be $k_1$) will be $\equiv k\pmod {n-1}$. Take $k_1$ with $n$ and we get a term $\equiv k+1\pmod {n-1}$. Repeating we will eventually get a term $\equiv 0 \pmod{n-1}$, hence $n-1$ is in the set too.

A very beautiful solution:-

A must contain 1(obvious) and it contains other 2 element x,y (say) so it must contain x+1( property 2) and similarly all the elements greater than x
So now it is suffice to show that it contains 2,3,4....(x-1)
Observe that 2x,3x,4x......(x-1)x are in the set so is it's factors ie 2,3,4.... are also in A

QED


When the following five numbers are written in order from least to greatest, what is the product of the second and fourth values: $ \minus{}1.4$, $ \frac{4}{3}$, $ \frac{11}{8}$, $ \frac{\minus{}3}{2}$, $ \minus{}2 \frac{2}{3}$?

In order, the numbers are $ \minus{}\frac83,\minus{}\frac32,\minus{}\frac75,\frac43,\frac{11}8$, so the answer is $ \minus{}\frac32\cdot\frac43\equal{}\boxed{\minus{}2}$.


logarithms
trigonometry
algebra unsolved
algebra

[hide]Is this a trick question? :)

$\log\tan 45^\circ=0$ ;)

(I assume we're dealing with degrees.)[/hide]

oh,I didn't notice this point.:)
thanks farenhajt.

and what is your idea about this?
$log tg1+log tg2+...+log tg89$
 :)

Well... :)

$\tan k^\circ=\cot (90-k)^\circ$, hence $\tan k^\circ\tan(90-k)^\circ=1$

And the sum of logarithms is the logarithm of the product...

I think you can finish it now :)


What is the probability that at least 1 head will be obtained after flipping 3 coins?

Just a more complete explanation on kunny's calculations. There is only one possibility that no heads will show up in three flips of the coin. Because there is a $\displaystyle\frac{1}{2}$ chance that on any flip, tails will show up, the probability of three tails showing up in a row is $(\frac{1}{2})^3$. Thus, the probability of having at least one head in three flips follows from kunny's calculations.

[hide]$\frac{7}{8}$[/hide]

Ok, so what about:

What is the probability that at least 2 heads will occur after flipping 3 coins?

[hide="Solution"]
The number of ways to get 2 heads is the number of arrangements of $HHT$, which is $\frac{3!}{2!}=3$ ways.

The number of ways to get 3 heads is the number of arrangements of $HHH$, which is 1.

The probability is then $\frac{3+1}{2^3}=\frac{1}{2}$.
[/hide]

Ah ok I see. So its outcome satisfying situation divided by total possible outcomes.


linear algebra
matrix
vector
inequalities
linear algebra unsolved

4. [/b] Let $E_m$ be $m\times m$ identity matrix, $A\in \mathbb{R}^{m\times n}$ and let $B$ be symmetric $n\times n$ matrix such that block matrix $\left(\begin{array}{cc}E_m & A \\ A^T & B\end{array} \right)$ is positively defined. Prove that "matrix determinant" $B-A^TA$ is also positively defined. (Symmetric matrix $M$ is said to be positively defined if for every vector-column $x\ne0$ the inequality $x^TMx>0$  holds.)
(A. Kukush)

the given matrix by bloc have determinant equal to $det(B-A^{t}A)$

use if a real symetric matrix is positively defined then its determinant
is >0

Grobber solved this problem here http://www.mathlinks.ro/Forum/viewtopic.php?t=98170.


Let $n$ be positive integers. Two persons play a game in which they are calling a integer $m\ (1\leq m\leq n)$ alternately. Note that you may not call the number which have already said. The game is over when no one can call numbers, if the sum of the numbers that the lead have said  is divisible by 3, then the lead wins, otherwise the the second move wins. Find $n$ for which there exists the way of forestalling.

if  $n=9$ ,when the first person pick $9$ , he wins automatically .....

I think that it should be this way

(i) $n\equiv 0\mod 3$ : first person will win if he pick $n$

(ii)$n\equiv 1,2\mod 3$ : second person will win .
this is because all the numbers in mod 3 are
$1,2,0,1,2,0\cdots 1,2$ or $1,2,0,1,2,0\cdots 1$
If the first person pick a number $a$ where $a\equiv 0\mod 3$ then second person will win by picking $n$ .
Else if first person pick $a\equiv 1,2\mod 3$ , second person will pick $b\equiv 2,1\mod 3$ to make it $0\mod 3$ in each (1,2,0) . So finally the first person have to pick a number from $1,2\mod 3$ and lose the game .

[quote="shasha"]This is pretty standard. If n=0,4,5 (mod 6) then first wins, else second wins.
All the cases are similar. For example:
n=3 (mod 6) Then we have 2k groups of 1 2 0 and another one group 1 2 0.
Second player can play in such a way that he takes from the first
6k numbers the same numbers (mod 3) as first player (by playing symmetric),
and from rest 1 2 0 he can take one number different from 0.[/quote]

Your answer is correct, but you need to explain in detail. :)

[quote="barasawala"][quote="shasha"]
All the cases are similar. For example:
n=3 (mod 6) Then we have 2k groups of 1 2 0 and another one group 1 2 0.
Second player can play in such a way that he takes from the first
6k numbers the same numbers (mod 3) as first player (by playing symmetric), [/quote]

You are wrong. The second player can't play symmetric. There are $2k+1$ of each number. So if first player always pick $0$, then when he picks his last $0$, the second player can't pick $0$ anymore...[/quote]

No, Im not wrong, just read more carefully. The second plays symmetric between
the first 6k numbers, the last triple is excluded.

Strategy for n=5(mod6)
We break the set ${1,2,....,6k+5}$ into
\[T_0={3,6,....,6k+3}\]
\[T_1={1,4,....,6k+4}\]
\[T_2={2,5,....,6k+5}\]

where $T_i$ is the set consisting of all members which are congruent to i mod 3.
[u][b]Strategy[/b][/u]: First player picks the number 3. Then there are an even number of elements in each set.In every step, the first player takes a number from $T_i$ only when second player had taken a number from $T_i$ (in the previous step)
[u][b]Inference[/b][/u]:In this process the first player picks up k+1 numbers from each set and hence the sum of the numbers is congruent to (k+1)(0+1+2)=0(mod 3)

Can anyone give a correct solution for the other residues modulo 6.


The meson pi has a half-life of $ 2.6\times10^{\minus{}8}\ \text{s}$. In the Chacaltaya Laboratory of Cosmical Physics (Bolivia), it is observed that many mesons created 20 km above the laboratory are able to reach it. Find the (estimated) speed of these particles.

Thanks in advance for your help. :)

[hide="My Answer"]$ v>.999999924c$?
You find $ v$ such that $ vt>h$, where $ t\equal{}\frac{t_{0}}{\sqrt{1\minus{}\frac{v^{2}}{c^{2}}}}$[/hide]

well historically speaking it is similar o the famous $ \pi\text{\minus{}mesons}$ paradox

[quote="shreyas_patankar"]$ v > .999999924c$?
You find $ v$ such that $ vt > h$, where $ t \equal{}\frac{t_{0}}{\sqrt{1\minus{}\frac{v^{2}}{c^{2}}}}$[/quote]

Indeed, I realized the problem is actually related to relativity. Using the time dilatation formula I got the same answer than you, Shreyas. :) Thank you.


trigonometry algebra function domain inequalities

solve the equation: :lol:
$ \sqrt{\sin x\minus{} \sqrt{\sin x \plus{} \cos x}}\equal{}\cos x$

[hide="hint"]Just check for domain in $ [ \minus{} \pi,\pi]$ only region left will be $ (\pi /2,3 \pi/4]$. In which  lhs is positive and rhs is negative
hence no solution[/hide]

[hide="Answer"]$ \cos x\equal{}0,\ \sin x\equal{}1$ is the only solution .[/hide]

hello, by squaring the given equation two times  we get
$ \sin(x)^2\plus{}\cos(x)^4\minus{}2\sin(x)\cos(x)^2\minus{}\sin(x)\minus{}\cos(x)\equal{}0$
After factoring we have
$ (\minus{}\sin(x)\plus{}\cos(x)^2\minus{}\cos(x))(\minus{}\sin(x)\plus{}\cos(x)^2\plus{}\cos(x)\plus{}1)\equal{}0$.
Sonnhard.

[color=darkblue]Hi Dr Sonnhard Graubner. You are true, but we ought note conditional of it when square LHS and RHS.[/color]

[quote="mathson"]solve the equation: :lol:
$ \sqrt {\sin x \minus{} \sqrt {\sin x \plus{} \cos x}} \equal{} \cos x$[/quote]

$ \sqrt {\sin x \minus{} \sqrt {\sin x \plus{} \cos x}} \equal{} \cos x$

$ \sin x \minus{} \sqrt {\sin x \plus{} \cos x} \equal{} \cos^2 x$ and $ \cos x \ge 0$

If $ \cos x > 0$, then

$ \sin x > \sqrt {\sin x \plus{} \cos x}$

$ \sin^2 x \minus{} \sin x > \cos x > 0$

So $ \sin x < 0$ and then $ \cos^2 x \equal{} \sin x \minus{} \sqrt {\sin x \plus{} \cos x} < 0$ ,

that's contradiction.

So $ \cos x \equal{} 0$ and $ \sin x \equal{} 1$ .

[i]A slightly cleaner approach to essentially the same argument as the above:[/i]

$ \blacktriangleright \sin x \minus{} \sqrt {\sin x \plus{} \cos x} \geq 0 \implies \sin x \geq \sqrt {\sin x \plus{} \cos x}$, because we take the squareroot of it

$ \blacktriangleright \cos x \geq 0$, because it is the squareroot of another number

$ \blacktriangleright \sin x \leq \sqrt {\sin x} \leq \sqrt {\sin x \plus{} \cos x}$
- The first part is true because $ \sin \leq 1$, equality iff. $ \sin x \equal{} 1$
- The second part is true because $ \cos \geq 0$, achieves equality iff. $ \cos x \equal{} 0$

We can only satisfy all inequalities if each is satisfied in the equality case; that is, $ \sin \equal{} 1$, $ \cos \equal{} 0$. We verify that this is possible, and conclude that $ x \equal{} \frac {\pi}{2} \plus{} 2k \pi$


linear algebra
matrix
superior algebra
superior algebra unsolved

Let G be a finite group. Let g be an element of G. Can there be a representation

T:G--> Gl (3,C)

such that

T(g)=

 1  0  2
 0  1  0
 0  0  1

?

No because the order of the mentioned matrix is $ \infty$.


A cone with a glass surface is very thin 30cm
time and is initially with the vertex down. In
position, the liquid contained inside is level with the
20cm apex,

Reversing the position of the cone so that its base is
on a horizontal plane, the distance from the apex of the cone
surface of the liquid is x = 10 times the cube root of 3 n, where n is equal to:

(A) 19.
(B) 20.
(C) 21.
(D) 24.
(E) 26.

Portuguese question:

Um cone com superfície de vidro muito fino tem 30cm de
altura e está inicialmente com o vértice para baixo. Nessa
posição, o líquido contido em seu interior está com nível a
20cm do vértice,

Invertendo a posição do cone de forma que sua base fique
sobre um plano horizontal, a distância do vértice do cone à
superfície do líquido é x =10 vezes a raiz cúbica de 3 n , onde n é igual a:
(A) 19.
(B) 20.
(C) 21.
(D) 24.
(E) 26.

Hi luiseduardo,

I've read the Portuguese version and translate the first paragraph (more or less word for word) as

A cone with a very thin glass surface is 30 cm high and is initially (placed) with its vertex downwards. In this position the liquid contained in its interior is at a level 20cm from the vertex.

Unless I've made a mistake, I think the answer is 10 times the cube root of 19.

Please explain me how do you do. I'm sorry, I translated by Google Translator.

Bom dia luiseduardo,

In the initial position, the height of the cone of water is $ \frac {2}{3}$ of $ 30$, so, if the radius of the base of the whole cone is $ R$, then the volume of water is $ (\frac {2}{3})^{3}\frac {30\pi R^{2}}{3} \equal{} \frac {80}{27}\pi R^{2}$.

When the cone is inverted, there is a smaller cone of air (above the water) of  height $ h$ and rarius $ r$. By similar triangles $ \frac {h}{r} \equal{} \frac {30}{R}$. So $ r \equal{} \frac {Rh}{30}$.

Then $ \frac {80 \pi R^{2}}{27} \equal{} \frac {30 \pi R^{2}}{3} \minus{} \frac {\pi r^{2} h}{3}$.

Therefore, multiplying by $ 3$ and substituting, we get

$ 30R^{2} \minus{} \frac {80R^{2}}{9} \equal{} \frac {R^{2}h^{3}}{900}$

$ \frac {190}{9} \equal{} \frac {h^{3}}{900}$

$ h^{3} \equal{} 19000$

$ h \equal{} 10 \times (19)^{\frac {1}{3}}$.

I hope this is OK.

P.S. I'm not sure how good Google Translator is; if you're not sure about the English in a question, you could write it in Portuguese as well as English and I'll try to translate it for you.

I think so.
Thanks for your help Andrew ;)


IF $ a\plus{}b\plus{}c\plus{}d\equal{}4$ for a,b,c,d positive reals.prove that $ \frac{a}{b^2\plus{}b} \plus{} \frac{b}{c^2\plus{}c} \plus{} \frac{c}{d^2\plus{}d} \plus{} \frac{d}{a^2\plus{}a} >\equal{}\frac{8}{(a\plus{}c)(b\plus{}d)}$

See http://www.mathlinks.ro/viewtopic.php?t=299899&start=260 post #266 and #262
Also see here http://www.mathlinks.ro/viewtopic.php?t=154913


Express each product as a sum containing ONLY sines and cosines.

(1) cos(4x) cos(2x)

(2) sin(x/2) cos(5x/2)

[quote="ZzZzZzZzZzZz"]Do you mean sines and cosines of x?  Because the expression already contains only sines and cosines?[/quote]

Express as a [b]sum[/b].

$ \cos a \cos b \equal{} \frac {e^{ia} \plus{} e^{ \minus{} ia}}{2} \cdot \frac {e^{ib} \plus{} e^{ \minus{} ib}}{2} \equal{} \frac {e^{i(a \plus{} b)} \plus{} e^{i(a \minus{} b)} \plus{} e^{ \minus{} i(a \minus{} b)} \plus{} e^{ \minus{} i(a \plus{} b)}}{2 \cdot 2}$

$ \equal{} \frac {e^{i(a \plus{} b)} \plus{} e^{ \minus{} i(a \plus{} b)}}{4} \plus{} \frac {e^{i(a \minus{} b)} \plus{} e^{ \minus{} i(a \minus{} b)}}{4} \equal{} \frac {1}{2}[\cos(a \plus{} b) \plus{} \cos(a \minus{} b)]$

$ \sin a \cos b \equal{} \frac{e^{ia} \minus{} e^{\minus{}ia}}{2i} \cdot \frac{e^{ib} \plus{} e^{\minus{}ib}}{2} \equal{} \frac{e^{i(a \plus{} b)} \plus{} e^{i(a \minus{} b)} \minus{} e^{\minus{}i(a \minus{} b)} \minus{} e^{\minus{}i(a \plus{} b)}}{4i}$

$ \equal{} \frac{e^{i(a \plus{} b)} \minus{} e^{\minus{}i(a \plus{} b)}}{4i} \plus{} \frac{e^{i(a \minus{} b)} \minus{} e^{\minus{}i(a \minus{} b)}}{4i} \equal{} \frac{1}{2}[\sin(a\plus{}b) \plus{} \sin(a\minus{}b)]$


Convert 86.4444444... (base 10) into base 8.

I don't know if it is right...
[hide] $\displaystyle86.\overline{4}=86+\frac 49=126_8 + 0.\overline{34}_8=126.\overline{34}_8$ [/hide]

How did you convert $0.\overline{44}_{10}$ to $0.\overline{34}_8$ :?

Hm I'm getting $.\overline{36}$

$\frac{4}{9}$ is $\frac{4}{11}$ in base 8.. I think?

4/7 is wrong. bornformath, you are using the approach to go from base 8 to base 10, which is not we want. i'm not exactly sure how to go to base 8, but i haven't tried to hard so...

I did a little research on base change and I think I'm right, at least about the $\frac{4}{11}$ part.. I'm not good with division in base 8, so yeah.

[hide]
The 86 part is easy to convert: $86_{10}\to 126_8.$  Now for the $.4444\ldots=\frac 49$ part.  Think about the place value system, the first decimal is worth $8^{-1}$, the second $8^{-2}.$  So let's find the first few decimals and look for a pattern.  The first digit is 3 since we can take away $\frac 18$ twice from $\frac 49$.  So we have $.3\ldots$ so far.  After taking 3 $\frac 18$'s away we have $\frac 49-\frac 38=\frac 5{72}$ left over.  So we see how many times $\frac 1{64}$ goes into this.  Continuing in this manner we get $.3434\ldots.$ One would now guess that the decimal part equals $.\overline{34}.$

The best way to confirm this would be to convert back to base 10.  \[.\overline{34}_8=\frac 38 + \frac{4}{8^2}+\frac{3}{8^3}+\frac{4}{8^4}+\cdots=\left(\frac 3{8}+\frac 3{8^3}+\cdots\right)+\left(\frac 4{8^2}+\frac 4{8^4}+\cdots\right)=\frac 8{21}+\frac 4{63}=\frac 49.\]

Thus the answer is $126.\overline{34}.$ [/hide]

Whatever, I'll wait till high school  :lol:  still one year!


this same sequence was on one of paladin8's mock aime's

[quote="bui_haiha2000"]For a given positive integer $a_{0}$ define the sequence ${a_{n}}$ by :
           $a_{i+1}$ $=$ $\frac{a_{i}}{2}$ if $a_{i}$ is even

           $a_{i+1}$ $=$ $3a_{i}$ $-$ $1$ if $a_{i}$ is odd
                                                                    (for $i$ $=$ $0$,$1$,$2$,..)

Prove that if n is a natural number such that $a_{n}$ $=$ $a_{0}$ , then $2^{n}$ $>$ $a_{0}$[/quote]
Also Polish 57th MO 1st round 3th series problem 12. It is very hard.

[quote="me@home"]Is this the actual problem you mentioned from the 57th Polish MO? Then you should ask someone to post it here:
http://www.mathlinks.ro/Forum/resources.php?c=1&cid=22
because no one has put up any of those problems yet...[/quote]
It is a problem from [u]national[/u] Polish MO (not Austrian-Polish) from the beginning of this school year. BTW, how can I post problems in [b]Contest Resourcess[/b]? Just PM a moderator?


Hi, I am faced with a situation where I'd like to sort some things, but a comparison takes much longer than anything else in the sort. So, which (comparison) sort algorithm can be easily be implemented and takes the fewest comparisons? My list is probably going to be about 10-20 items (possibly up to 100 - no more), so it's not exactly asymptotic. Clearly the different O(NlogN) algorithms have differing constants, which will make a big difference, so my question is which algorithm would be the best in this situation?

Binary insertion sort has relatively few comparisons.  It's like regular insertion sort except that we use binary search to determine where to place each new number.  There is a lot of movement associated with this algorithm (to make room for each new number), but your problem description seems to say that comparisons matter more than movement.

The number of comparisons is asymptotic to $ N \log_2 N$ (with no multiplicative constant).

And since you'll never be able to devise an algorithm which would use less comparisons than this, I would definitely go for it. However, maybe even a better idea would be to somehow optimize the comparison routine so that it won't take so much time - find a potential function, use hashing, or something like that, it really depend on what you're sorting :).

I did a test with 7 different sorting algorithms, 6 popular ones and the binary insertion algorithm described by Ravi B.

I used a compare function which is relatively slow compared to all other operations. Data set was 40 items

Max&Mins of 5 runs:
Binary insertion sort: 209..219 comparisons
Quick sort: 243..336 comparisons
Heap sort: 339..352 comparisons
Linear insertion sort: 416..536 comparisons
Shell sort: 446..511 comparisons
Bubble sort: 733..897 comparisons
Exchange sort: 903 comparisons always

Is 5 tests sufficiently much data though to draw a conclusion from?


Find the least positive integer $ N$ such that the sum of its digits is 100 and the sum of the digits of $ 2N$ is 110.

Minimal number, when it had minimal number od digits or maximum digits equal 9.  Because $ S(2N)\minus{}S(N)\equal{}10$, the sum of digits <5 is 10 and 100=10+10*9. Minimal case when $ 2449...9 (ten \ 9)$

Call the number $ \overline{abcdefg.....}$
Consider multiplying a number by 2 by hand. For a digit less than 5, there won't be a carry, so the resulting digit will be twice the original. Otherwise, you will carry the one, so the resulting digit will be twice it minus 10, but because of the carry, you will add one to the next number. So for each digit n, the amount contributed to the sum of the digits of twice the number is either 2n or 2n-9. So if we subtract these from the original digit to get the difference in the sums, each digit contributes either n or n-9 and the difference in the digit sums is 10. So we have a+b+c+d...=100 and {a or a-9}+{b or b-9}+{c or c-9}...=10. Hence, 10 of the digits must contribute n-9, so there must be 10 carries, so since the digits add up to 100, the least is 2449999999999.

we only have to consider how much times the carries occur,and can easily get the answer $249999999999$.
(if there is no carry,then the digit sum of $2N$ will be 200;since the digit sum will decrease by $9$ after we increase a craay,hence $10$ carries must occurs).