Hello,

I wonder what is the general opinion in the forum about the good writing habits on Math Olympiads and in general in math (and why not, in this forum). By good writing practices I mean good orthography, grammatical concordance, concordance on the ideas, good use of the words (to use a word with the right meaning) and related habits.

As a teenager I always hated these matters, specially orthography (mainly because mine was really bad, it is A LOT better now, but still far from perfect). Nevertheless, by now I do consider almost rude and unpolite the poor writing, specially when you are writing a report or a homework... or the solution to the problem 2 of the IMO and you want those 7 points  :D .

Everybody knows that good writing habits are hard to get (It is hard enough to get them in your mother language, it is harder to get them on a second or third languate), my question to you is: How necessary is it for us, mathematicians and math competitors (and MathLinkers)? (I think a lot).

I could talk a lot about this topic, but I prefer to read other opinions first.

Good writing habits are VERY important, even (especially!) outside of mathematics. Writing is one of the main forms of communication that we have in this world, and thus it is very necessary to communicate efficiently.   :lol: There are many instances of miscommunications because of poor writing skills, some of them funny, some of them sad, and some of them simply cost-ineffective.

P.S. Since you bring it up, I was wondering if you would mind/prefer corrections on your posts? I know that English is not your first language, and you are still having difficulties with it. Would you find such corrections, by PM, to be useful in your study of the language?

Very important!
In a contest, if your answer sheet is clean and neat, you can give a good impression to the guy who give you the score. That can bring you lots of unexpected results.

Excellent oral skills are just as important as good writing skills, no what career you choose.  Even if you don't end up as a professor or teacher, your job duties will almost certainly include giving talks, collaborating with colleagues and mentoring younger coworkers.  The ability to express your thoughts clearly is very valuable.

[quote="bleumoose"]Since you bring it up, I was wondering if you would mind/prefer corrections on your posts? I know that English is not your first language, and you are still having difficulties with it. Would you find such corrections, by PM, to be useful in your study of the language?[/quote]

I wouldn't mind. I usually try to use an informal but correct English in my posts, but I am not perfect and as you say, English is not my first language. If you have any comment, you may PM me or post it in the same post, but ... in the same post it may be out of place.

Actually, I encourage to every person in the forum to give and receive such corrections (of course, in a respectful enviroment :D ) without feeling him/herlself attacked.

[quote="djimenez"]Hello,

I wonder what is the general opinion in the forum about the good writting habits on Math Olympiads and in general in math (and why not, in this forum). By good writting practices I mean good orthography, grammatical concordance, concordance on the ideas, good use of the words (to use a word with the right meaning) and related habits.

As a teenager I always hated these matters, specially orthography (mainly because mine was really bad, it is A LOT better now, but still far from perfect). Nevertheless, by now I do consider almost rude and unpolite the poor writting, specially when you are writting a report or a homework... or the solution to the problem 2 of the IMO and you want those 7 points  :D .

Everybody knows that good writting habits are hard to get (It is hard enough to get them in your mother language, it is harder to get them on a second or third languate), my question to you is: How necessary is it for us, mathematicians and math competitors (and MathLinkers)? (I think a lot).

I could talk a lot about this topic, but I prefer to read other opinions first.[/quote]

Spell "writing" correctly would help you a little bit.

[quote="henryyangrui"]Spell "writing" correctly would help you a little bit.[/quote]

Thank you for the remark. I will take it into consideration. Each day you learn something new :D  :D (The important part is not to forget two old things :P).

No, really, thanks for the remark. My English writing is far from perfect, but I am working to improve it.  ;)

I think that writing skills, especially persuasive writing skills, are essential to the study of mathematics.  Furthermore, I think that experience with logic and math often works to improve the writing skills.


If the lengths of a kite's diagonals are 6 cm and 12 cm, can one of the kite's sides be twice as long as another?  Prove your answer.

Same goes for this one. Proofs are not put in MATHCOUNTS (i haven't seen one yet.)

LOCK THIS TOPIC, its already in Intermediate Topics.

[url]http://www.mathlinks.ro/viewtopic.php?t=298711[/url]

Also, Peterhi needs to be educated about SPAMMING.


geometry
3D geometry
combinatorics unsolved
combinatorics

Is it possible to dissect a cube into 51 cubes?

Yes it is. In fact, it is possible to dissect a cube into $ n$ cubes, for any integers $ n\ge48$.

Read [url=http://mathworld.wolfram.com/CubeDissection.html]this[/url].


geometry
incenter
circumcircle
analytic geometry
inequalities
trigonometry
geometry proposed

Let $ ABC$ be a triangle with sides $ a, b, c$, centroid $ G$, incentre $ I$, and circumcenter $ O$. Prove that

$ [GIO]\geq\frac{(a\minus{}b)(c\minus{}b)(c\minus{}a)}{12R}$

where the square brackets denote the signed area of a triangle.

.

We use areal coordinates (normalized barycentric coordinates) with respect to $ \triangle ABC.$ Coordinates of the centroid, orthocenter and incenter are given by:

$ G \left(\frac {1}{3}: \frac {1}{3}: \frac {1}{3} \right ) \ , \ H(\cot B \cot C: \cot A \cot C: \cot A \cot B) \ , \ I \left ( \frac {a}{2p}: \frac {b}{2p}: \frac {c}{2p} \right )$

${\left(\begin{array}{ccc}\frac {1}{3} & \frac {1}{3} & \frac {1}{3} \\
 \\
\cot B \cot C & \cot A \cot C & \cot A \cot B \\
 \\
\frac {a}{2p} & \frac {b}{2p} & \frac {c}{2p} \end{array}\right) = \frac {[\triangle IGH]}{[\triangle ABC]} \ \Longrightarrow}$

$ 6p \cdot \frac {[\triangle IGH]}{[\triangle ABC]} = \sum_{\text{cyclic}} a(\cot A\cot C - \cot A\cot B)$

We use the well-known identities

$ \cot A = \frac {b^2 + c^2 - a^2}{4[ABC]} \ , \ \cot B = \frac {a^2 + c^2 - b^2}{4[ABC]} \ , \ \cot C = \frac {a^2 + b^2 - c^2}{4[ABC]}$

$ \frac {48[\triangle IHG] \cdot [ \triangle ABC]^2}{r} = \sum_{\text{cyclic}} a^4(c - b) +  a^3(c^2 - b^2)$

$ 48[\triangle IHG] \cdot [\triangle ABC]^2 = 2rp[a^3(c - b) + b^3(a - c) + c^3(b - a)]$

$ \Longrightarrow  [\triangle IHG] = \frac {a^3(c - b) + b^3(a - c) + c^3(b - a)}{24[\triangle ABC]}$

Since $ [\triangle IGO] = \frac {1}{2}[\triangle IHG]  \Longrightarrow \ [\triangle IGO] = \frac {a^3(c - b) + b^3(a - c) + c^3(b - a)}{48[\triangle ABC]}$

Simplifying the denominator and substituting $ [\triangle ABC] = rp$ gives

$ [\triangle IGO] = \frac {|(a - b)(b - c)(c - a)|}{24r} .$  Now, by Euler's inequality the result is obvious.

[quote="Ligouras"]Let $ ABC$ be a triangle with centroid $ G$, incentre $ I$ and circumcenter $ O$. Prove that $ [GIO]\geq\frac {|(a \minus{} b)(c \minus{} b)(c \minus{} a)|}{12R}$ .[/quote]

$ G\left(\frac 13,\frac 13,\frac 13\right)$ ; $ I\left(\frac {a}{2s},\frac {b}{2s},\frac {c}{2s}\right)$ ; $ O\left(\frac {aR}{2S}\cdot\cos A,\frac {bR}{2S}\cdot\cos B,\frac {cR}{2S}\cdot\cos C\right)$ ; $ S \equal{} [ABC]$ . Thus $ [GIO]$ is

$ S\cdot\mathrm{mod}\left|\begin{array}{ccc} \frac 13 & \frac 13 & \frac 13 \\
 \\
\frac {a}{2s} & \frac {b}{2s} & \frac {c}{2s} \\
 \\
\frac {R}{2S}\cdot a\cdot\cos A & \frac {R}{2S}\cdot b\cdot\cos B & \frac {R}{2S}\cdot c\cdot\cos C\end{array}\right| \equal{}$ $ \frac {R}{12s}\cdot\mathrm{mod}\left|\begin{array}{ccc} 1 & 1 & 1 \\
 \\
a & b & c \\
 \\
a\cdot\cos A & b\cdot\cos B & c\cdot\cos C\end{array}\right| \equal{}$

$ \frac {R}{12s}\cdot\frac {1}{8a^2b^2c^2}\cdot \mathrm{mod}\left|\begin{array}{ccc} 2bc & 2ca & 2ab \\
 \\
2abc & 2abc & 2abc \\
 \\
a\cdot\left(b^2 \plus{} c^2 \minus{} a^2\right) & b\cdot\left(c^2 \plus{} a^2 \minus{} b^2\right) & c\cdot\left(a^2 \plus{} b^2 \minus{} c^2\right)\end{array}\right| \equal{}$

$ \frac {R}{24sabc}\cdot\mathrm{mod}\left|\begin{array}{ccc} bc & ca & ab \\
 \\
1 & 1 & 1 \\
 \\
a\cdot\left(b^2 \plus{} c^2 \minus{} a^2\right) & b\cdot\left(c^2 \plus{} a^2 \minus{} b^2\right) & c\cdot\left(a^2 \plus{} b^2 \minus{} c^2\right)\end{array}\right| \equal{}$

$ \frac {1}{96s^2r}\cdot\mathrm{mod}\left|\begin{array}{ccc} bc & c(a \minus{} b) & b(a \minus{} c) \\
 \\
1 & 0 & 0 \\
 \\
a\cdot\left(b^2 \plus{} c^2 \minus{} a^2\right) & 4s(s \minus{} c)(a \minus{} b) & 4s(s \minus{} b)(a \minus{} c)\end{array}\right| \equal{}$

$ \frac {|(a \minus{} b)(a \minus{} c)|}{24sr}\cdot\mathrm{mod}\left|\begin{array}{cc} c & b \\
 \\
s \minus{} c & s \minus{} b\end{array}\right| \equal{}$ $ \frac {|(a \minus{} b)(b \minus{} c)(c \minus{} a)|}{24r}\ge$ $ \frac {|(a \minus{} b)(b \minus{} c)(c \minus{} a)|}{12R}$ .

[quote="Virgil Nicula"]With barycentric coordinates :

$ G\left(\frac 13\ ,\ \frac 13,\ \frac 13\right)$ , $ I\left(\frac {a}{2s}\ ;\ \frac {b}{2s}\ ,\ \frac {c}{2s}\right)$ ; $ O\left(\frac {R^2}{2S}\cdot\sin 2A\ ,\ \frac {R^2}{2S}\cdot\sin 2B\ ,\ \frac {R^2}{2S}\cdot\sin 2C\right)$ $ \implies$

$ [GIO] \equal{}$ $ \left|\begin{array}{ccc} \frac 13 & \frac 13 & \frac 13 \\
 \\
\frac {a}{2s} & \frac {b}{2s} & \frac {c}{2s} \\
 \\
\frac {R^2}{2S}\cdot\sin 2A & \frac {R^2}{2S}\cdot\sin 2B & \frac {R^2}{2S}\cdot\sin 2C\end{array}\right|\cdot [ABC] \equal{}$ $ \left|\begin{array}{ccc} 1 & 1 & 1 \\
 \\
a & b & c \\
 \\
\sin 2A & \sin 2B & \sin 2C\end{array}\right|\cdot\frac {R^2}{12s}$ a.s.o.[/quote]

nice nice nice!!!!!

dear friend [color=red][b]luisgeometra[/b][/color] your solution is similar to my construction.... thanks


Given $a,b,c >0$ such that $a+b+c=3.$

Prove that : $\sum_{cyc}\frac{2}{a^{2}+bc}\le \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$


This is an interesting problem, but I am not sure of how to solve it. I am thinking to use the Binomial Therorem, but I don't think that will work....any other ideas?

Compute the number of terms in the expansion of $(w+x+y+z) ^ {12}$

This is similar to the concepts taught in AoPS Volume II in chapter 15, but I still don't know how to do it.

[hide="Hint"]Think about the degree of each term in the expansion and then see if you can restate the problem.[/hide]

[hide]The answer is 455, can you work backwards to solve the problem?[/hide]

Sorry joml88, I still don't know what you are talking about, and fanaticsm, are you sure it is 455? I don't think thats the answer...

By degree I mean like the first term will be $w^{12}$ so its degree is 12.  Another term would be $w^3x^4yz^4$ so its degree would also be 12.  So the sum of the exponents of each term will always equal 12.

See if you can do anything with that.

I got 455 also.

:( Okay, how would you solve this (by this point, I have given up, and am looking for a solution) :(

Ok, I ll go one step farther but I wont solve it yet.

Each term in the expansion is of the form of $w^{e_1}x^{e_2}y^{e_3}z^{e_4}$ where $e_1+e_2+e_3+e_4=12$ and the $e_i$ are all greater than or equal to zero.

So try finding how many solutions there are to that last equation.

If you are still stumped I can go farther or if you want, AoPS Volume 2 Chapter 17.2 should help out a lot  ;) .

[hide="This is the solution"]let each term in the expansion be expressed as $w^ax^by^cz^d$, where $a$, $b$, $c$, and $d$ are non-negative integers.  The problem then becomes as follows: determine the number of solutions in non-negative integers there are in the following equation:
\[a+b+c+d=12\]
We will determine the number of solutions in the following manner: let n represent one unit, and let D be a divider.  I can't think of how to word that better, so I'll illustrate it with an example.  nDnnnDnnnnnDnnn would represent the solution $a=1$, $b=3$, $c=5$, $d=3$ (one n, then a D; three n's, then a D; five n's, then a D; and three n's).  Similarly, nnnnnnDDnnDnnnn would represent $a=6$, $b=0$, $c=2$, and $d=4$.  This further reduces the problem into determing the number of distinct ways to arrange the letters nnnnnnnnnnnnDDD.  This is just $\displaystyle{{15}\choose{3}}$, or $455$[/hide]


Please put a rose in front of your msn name to show your sympathy for the hundreds of dead children in russia who have died in a terrorist act, i hope you find it in ur heart to do this! paste this to all of your online contacts!

rose code=(f)

I'm just wondering, who did it? All I can do is show sympathy for the 155 children who died, but not anger and unhappiness at the perpetrators of the act. [size=9](Though that might be a good thing...)[/size]


Prove that

$ \sum_{k\equal{}m}^{n} {k \choose r} \equal{} {n\plus{}1 \choose r\plus{}1}\minus{}{m \choose r\plus{}1}$

Quick Question:  If we wanted to use induction for this, which variable would determine the base case?

[hide="hint"]Use Pascal's Identity repeatedly.[/hide]

I forgot to mention that we have prove this identity using double counting method.

[hide="Club-Forming"]
We will prove $ {m \choose r\plus{}1} \plus{} \sum_{k\equal{}m}^{n} {k \choose r} \equal{} {n\plus{}1 \choose r\plus{}1}$.
Say we want to form a club of $ r\plus{}1$ people from a group of $ n\plus{}1$ people.  The number of ways to do so is clearly counted by the RHS.
Let's say that the people in the group are all of different ages, and that for any club formed, the oldest member in the club is designated as the leader.
Say that a person $ P$ is the $ k\plus{}1$th youngest person in the group; that is, in the group, there are exactly $ k$ people younger than $ P$.  Then to form a club given that $ P$ is the leader, we must choose $ r$ other members from the $ k$ people younger than $ P$.  Hence,  $ \sum_{k\equal{}m}^{n} {k \choose r}$ counts the number of possible clubs across all leaders who are older than $ m$ people in the group.  Finally, to account for the clubs made entirely of the $ m$ youngest people in the group, we simply add $ \binom{m}{r\plus{}1}$.  Thus, the LHS = RHS.[/hide]


Let  $ F(x)=x^n+ n-1 \; . $ and consider  the statement:
 [i] ,,There are an infinite number of positive integres $ m $ such that   $ F(m) $ has at most ${\mathbf k}\; ,\;  {\mathbf k}={\mathbf k}(n)={\mathbf k}(F,n) ,$ prime factors." [/i]

[b]Question:[/b] [i] $ {\mathbf k}={\mathbf k}(n) = ?  $  Upper bounds,... ?  [/i]

Some known results :

[b]a)[/b] $ k(2)= 5 $ - B.V.Levin -[1960]

[b]b)[/b] $ k(3)= 4 $ - Y. Wang -  [1957]

Other informations: In case when $ G(x)=x(x+2) $ , A.Selberg-[1947]  has  proved that $ {\mathbf k}(G,n)=5. $

You can find the answer in Acta Arithmetica, several articles on this theme. A very hard fact!


Alternatively for #1:

Group the terms as follows:
\[ (1 \minus{} 2) \plus{} (3 \minus{} 4) \plus{} (5 \minus{} 6) \plus{} \dots \plus{} (2007 \minus{} 2008) \plus{} 2009\]
The terms inside each set of parentheses sum to $ \minus{} 1$, and there are $ \frac {2008 \minus{} 2}{2} \plus{} 1 \equal{} 1004$ of them.
\[ \underbrace{( \minus{} 1) \plus{} ( \minus{} 1) \plus{} \dots \plus{} ( \minus{} 1)}_{1004\ \minus{} 1\text{'s}} \plus{} 2009 \equal{} \minus{} 1004 \plus{} 2009 \equal{} \boxed{1005}\]

For #2, the easiest way to note that $ 15 \times 13 < 200$ is that $ 14^2\equal{}196$ and $ 15 \times 13 \equal{} (14\plus{}1)(14\minus{}1)\equal{}14^2\minus{}1\equal{}195$.  Clearly, now, $ 13 \times 16 > 200$.

For #3, I don't quite see how the sequence proceeds...could you post a few more terms? (if AndrewTom has the correct answer, disregard this part).


trigonometry
complex numbers
calculus
calculus computations

Compute the sum
\[ \sum_{k \equal{} 1}^{n} \binom{n}{k} \cos{kx}\] (edit)

Do you mean $ \sum_{k\equal{}1}^{n} {n \choose k} \cos kx$?  The sum you gave is just $ (2^n \minus{} 1) \cos nx$.

That's not a very interesting sum. As written, it's just $ (2^n \minus{} 1)\cos nx.$ I'm going to assume that you actually meant $ \cos kx$ instead of $ \cos nx.$

$ \sum_{k \equal{} 0}^{n}\binom{n}{k}\cos kx \equal{} \text{Re} \left(\sum_{k \equal{} 0}^{n}\binom{n}{k}e^{ikx}\right)$

$ \equal{} \text{Re}\left[\left(1 \plus{} e^{ix}\right)^n\right]$

Factor $ e^{ix/2}$ out of each factor to get

$ \left(1 \plus{} e^{ix}\right)^n \equal{} e^{inx/2}\left(e^{ \minus{} ix/2} \plus{} e^{ix/2}\right)^n$

$ \equal{} 2^ne^{inx/2}\cos^n\left(\frac {x}{2}\right)$

Take the real part of that to get

$ \sum_{k \equal{} 0}^{n}\binom{n}{k}\cos kx \equal{} 2^n\cos\left(\frac {nx}2\right)\cos^n\left(\frac {x}2\right).$

Or, if you insist on starting the sum at $ k \equal{} 1,$

$ \sum_{k \equal{} 1}^{n}\binom{n}{k}\cos kx \equal{} 2^n\cos\left(\frac {nx}2\right)\cos^n\left(\frac {x}2\right) \minus{} 1.$

[quote="Kent Merryfield"]That's not a very interesting sum. As written, it's just $ (2^n \minus{} 1)\cos nx.$ I'm going to assume that you actually meant $ \cos kx$ instead of $ \cos nx.$

$ \sum_{k \equal{} 0}^{n}\binom{n}{k}\cos kx \equal{} \text{Re} \left(\sum_{k \equal{} 0}^{n}\binom{n}{k}e^{ikx}\right)$

$ \equal{} \text{Re}\left[\left(1 \plus{} e^{ix}\right)^n\right]$

Factor $ e^{ix/2}$ out of each factor to get

$ \left(1 \plus{} e^{ix}\right)^n \equal{} e^{inx/2}\left(e^{ \minus{} ix/2} \plus{} e^{ix/2}\right)^n$

$ \equal{} 2e^{inx/2}\cos^n\left(\frac {x}{2}\right)$

Take the real part of that to get

$ \sum_{k \equal{} 0}^{n}\binom{n}{k}\cos kx \equal{} 2\cos\left(\frac {nx}2\right)\cos^n\left(\frac {x}2\right).$

Or, if you insist on starting the sum at $ k \equal{} 1,$

$ \sum_{k \equal{} 1}^{n}\binom{n}{k}\cos kx \equal{} 2\cos\left(\frac {nx}2\right)\cos^n\left(\frac {x}2\right) \minus{} 1.$[/quote]

Correct me if I'm wrong, but I think it should be

$ e^{inx/2}\left(e^{ \minus{} ix/2} \plus{} e^{ix/2}\right)^n$$ \equal{} 2^n e^{inx/2}\cos^n\left(\frac {x}{2}\right)$ instead.


Here is the code:

public class lesson11exercise5
{
  public static void main(String args[])
  {
    int iMus,b;
    for(iMus=0;iMus<10;iMus++)
    {
       b=19+iMus;
    }
    System.out.println(b);
  }
}

This is an exercise from the Blue Pelican Java textbook.
When I compiled this programme, the compiler said that 'variable b might not have been initialized'.

How came this?Plz help me.Thanx a lot! :?

The compile message really tells you everything. Even though in java saying 'int b' initialises b to 0 by default, the compiler still checks to make sure it actually gets initialised to something. So just adding b=0 to the start will make it compile..


function
inequalities
algebra open
algebra

Let see binary representation of $ cosh 1$: $ cosh (1) \equal{} 1.100010110000011...$.
I'm interesting in questions: how many consecutive digits "1" in this representation can be? Is this amount bounded?
And can you define function $ f(n)$ such that $ f(n) \minus{} th$ digit (for all $ n$, or for infinitely many $ n$) after "." is exactly "0"?

Any ideas? Give me your hints.
And I forget about: from what minimum position $ s(k)$ we wiil get $ k$ consecutive digits "1" in representation (for example, we get 7 consecutive digits "1" in more than 100-th position, so we can put $ s(7)\equal{}100$ etc) ?

OK, some another:
We have $ g(k) \equal{} \frac {1}{0!} \plus{} \frac {1}{2!} \plus{} ... \plus{} \frac {1}{(2k \minus{} 2)!}$.
How we can prove that in binary represantation of $ g(k)$ we cannot have $ k$ consecutive digits "1" in $ k \plus{} 1$, $ k \plus{} 2$,...,$ 2k$ position? It's clear for $ k \leq 255$, maybe it's true for all $ k$. How can we prove it? You hints?
Somebody, help me!

OK, I need last step:
Prove or disprove that for all integer $ n$ there isn't exist integer $ k$ such that inequality holds:
$ 0<2^n k (2n)! \minus{} 2^{2n}(1\plus{}2n(2n\minus{}1)\plus{}2n(2n\minus{}1)(2n\minus{}2)(2n\minus{}3)\plus{}...\plus{}(2n)!)<(2n)!$

[quote="Kirill"]
Prove or disprove that for all integer $ n$ there isn't exist integer $ k$ such that inequality holds:
$ 0 < 2^n k (2n)! \minus{} 2^{2n}(1 \plus{} 2n(2n \minus{} 1) \plus{} 2n(2n \minus{} 1)(2n \minus{} 2)(2n \minus{} 3) \plus{} ... \plus{} (2n)!) < (2n)!$[/quote]
Nobody? I tired to update this topic. Can somebody help me?


does anybody know how to solve factorial equations?
for example..:
solve for n
(n-2)! = 10...i don't know if this problem really works but i need help learning how to solve problems of this form...

With that kind of problem specifically you just have to guess and check, or know your factorials off the top of yourhead.

But some Mathcounts problems, such as the following, have nice solutions involving !'s.

Express $k$ as a common fraction in terms of $n$, where $n$ is an integer greater than 1.

$\frac{k(n-1)!}{(n+1)!}=\frac{(n-1)!}{(n-2)!}$

You can also try factoring the RHS to see if there is something resembling a factorial.

Many problems can be solved by cancellation, and factoring. For example: \[\frac{7!*8!* 9!}{7!+8!+9!}= \frac{8!* M!}{9}\] What is the value of $M$?

[quote="mihail911"]that's just it...i don't understand how to do problems like that...i found one from an old mathcounts test
(n! - (n-2)!) /23 =1...solve for n[/quote]

To solve,
\[\frac{7!*8!* 9!}{7!+8!+9!}= \frac{8!* M!}{9}\]
First cancel the $8!$s to get
\[\frac{7!*9!}{7!+8!+9!}= \frac{M!}{9}\]
Then, factor out the $7!$s from the left fraction
\[\frac{7!*9!}{7!(1+8+9*8)}= \frac{M!}{9}\]
\[\frac{7!*9!}{7!(81)}= \frac{M!}{9}\]
Cancelling out the $7!$s gives
\[\frac{9!}{81}= \frac{M!}{9}\]
Multiplying both sides by 9 gives
\[\frac{9!}{(9)}= M!\]
Now it should be easy to see $M=8$


Let $ a,b,c$ be distinct positive real numbers. Determine the minimum value of the following expression
\[ P\equal{}\frac{ab\plus{}bc\plus{}ca}{(a\minus{}c)^2}\cdot\left( \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{}3\right)\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be distinct positive real numbers. Determine the minimum value of the following expression
\[ P \equal{} \frac {ab \plus{} bc \plus{} ca}{(a \minus{} c)^2}\cdot\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right)
\]
:)[/quote]

Let $ k$ is the minimum value that we need to find, then rewrite our inequality
\[ \frac {{ab \plus{} bc \plus{} ca}}{{\left( {a \minus{} c} \right)^2 }}\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3} \right) \ge k
\]

\[ \Leftrightarrow \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {{k\left( {a \minus{} c} \right)^2 }}{{ab \plus{} bc \plus{} ca}}
\]
Let $ a \equal{} t^3 ,b \equal{} t,c \equal{} 1$ then we get
\[ k \le \frac {{t^8 \plus{} 2t^7 \minus{} 2t^6 \minus{} 2t^5 \plus{} t^4 \minus{} 2t^3 \plus{} t^2 \plus{} 1}}{{t^8 \minus{} 2t^5 \plus{} 1}}
\]
Now we let $ t \to \plus{} \infty$ then we get $ k \le 1$.

The inequality becomes
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {{\left( {a \minus{} c} \right)^2 }}{{ab \plus{} bc \plus{} ca}}
\]
\[ \Leftrightarrow \left( {ab \plus{} bc \plus{} ca} \right)\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right) \minus{} 3\left( {ab \plus{} bc \plus{} ca} \right) \ge \left( {a \minus{} c} \right)^2
\]
\[ \Leftrightarrow \left( {\sum {a^2 } \minus{} \sum {ab} } \right) \plus{} \left( {\sum {\frac {{ab^2 }}{c}} \minus{} \sum {ab} } \right) \ge \left( {a \minus{} c} \right)^2
\]

\[ \Leftrightarrow \frac {1}{2}\sum {\left( {a \minus{} b} \right)^2 } \plus{} \sum {\frac {{c\left( {a \minus{} b} \right)^2 }}{b}} \ge \left( {a \minus{} c} \right)^2
\]

\[ \Leftrightarrow \frac {{\left( {b \plus{} 2c} \right)\left( {a \minus{} b} \right)^2 }}{b} \plus{} \frac {{\left( {c \plus{} 2a} \right)\left( {b \minus{} c} \right)^2 }}{c} \ge \frac {{\left( {a \minus{} 2b} \right)\left( {a \minus{} c} \right)^2 }}{a}
\]
By Cauchy Schwarz inequality we have
\[ \frac {{\left( {b \plus{} 2c} \right)\left( {a \minus{} b} \right)^2 }}{b} \plus{} \frac {{\left( {c \plus{} 2a} \right)\left( {b \minus{} c} \right)^2 }}{c} \ge \frac {{\left( {b \plus{} 2c} \right)\left( {c \plus{} 2a} \right)\left( {a \minus{} c} \right)^2 }}{{2bc \plus{} 2ac \plus{} 2c^2 }}
\]
Thus it suffices to prove that
\[ \frac {{\left( {b \plus{} 2c} \right)\left( {c \plus{} 2a} \right)}}{{2bc \plus{} 2ac \plus{} 2c^2 }} \ge \frac {{\left( {a \minus{} 2b} \right)}}{a}
\]
which is true. We have done and the inequality holds if and only if $ a \equal{} b \equal{} c$.

great! but can U explain why you let a=t^3,b=t,c=1? why don't you let other ways for ex: let a=2,c=1,b=...?
from your ideal,show me,pls !thanks :D

[quote="RainL"][quote="can_hang2007"]Let $ a,b,c$ be distinct positive real numbers. Determine the minimum value of the following expression
\[ P \equal{} \frac {ab \plus{} bc \plus{} ca}{(a \minus{} c)^2}\cdot\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right)
\]
:)[/quote]

Let $ k$ is the minimum value that we need to find, then rewrite our inequality
\[ \frac {{ab \plus{} bc \plus{} ca}}{{\left( {a \minus{} c} \right)^2 }}\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3} \right) \ge k
\]

\[ \Leftrightarrow \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {{k\left( {a \minus{} c} \right)^2 }}{{ab \plus{} bc \plus{} ca}}
\]
Let $ a \equal{} t^3 ,b \equal{} t,c \equal{} 1$ then we get
\[ k \le \frac {{t^8 \plus{} 2t^7 \minus{} 2t^6 \minus{} 2t^5 \plus{} t^4 \minus{} 2t^3 \plus{} t^2 \plus{} 1}}{{t^8 \minus{} 2t^5 \plus{} 1}}
\]
Now we let $ t \to \plus{} \infty$ then we get $ k \le 1$.

The inequality becomes
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {{\left( {a \minus{} c} \right)^2 }}{{ab \plus{} bc \plus{} ca}}
\]

\[ \Leftrightarrow \left( {ab \plus{} bc \plus{} ca} \right)\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right) \minus{} 3\left( {ab \plus{} bc \plus{} ca} \right) \ge \left( {a \minus{} c} \right)^2
\]

\[ \Leftrightarrow \left( {\sum {a^2 } \minus{} \sum {ab} } \right) \plus{} \left( {\sum {\frac {{ab^2 }}{c}} \minus{} \sum {ab} } \right) \ge \left( {a \minus{} c} \right)^2
\]

\[ \Leftrightarrow \frac {1}{2}\sum {\left( {a \minus{} b} \right)^2 } \plus{} \sum {\frac {{c\left( {a \minus{} b} \right)^2 }}{b}} \ge \left( {a \minus{} c} \right)^2
\]

\[ \Leftrightarrow \frac {{\left( {b \plus{} 2c} \right)\left( {a \minus{} b} \right)^2 }}{b} \plus{} \frac {{\left( {c \plus{} 2a} \right)\left( {b \minus{} c} \right)^2 }}{c} \ge \frac {{\left( {a \minus{} 2b} \right)\left( {a \minus{} c} \right)^2 }}{a}
\]
By Cauchy Schwarz inequality we have
\[ \frac {{\left( {b \plus{} 2c} \right)\left( {a \minus{} b} \right)^2 }}{b} \plus{} \frac {{\left( {c \plus{} 2a} \right)\left( {b \minus{} c} \right)^2 }}{c} \ge \frac {{\left( {b \plus{} 2c} \right)\left( {c \plus{} 2a} \right)\left( {a \minus{} c} \right)^2 }}{{2bc \plus{} 2ac \plus{} 2c^2 }}
\]
Thus it suffices to prove that
\[ \frac {{\left( {b \plus{} 2c} \right)\left( {c \plus{} 2a} \right)}}{{2bc \plus{} 2ac \plus{} 2c^2 }} \ge \frac {{\left( {a \minus{} 2b} \right)}}{a}
\]
which is true. We have done and the inequality holds if and only if $ a \equal{} b \equal{} c$.[/quote]
Yeah, this is my solution. :)
Indeed, I have proposed this problem to math and youth magazine but they didn't choose this for the advanced problems, so I posted it here.

[quote="can_hang2007"]Let $ a,b,c$ be distinct positive real numbers. Determine the minimum value of the following expression
\[ P \equal{} \frac {ab \plus{} bc \plus{} ca}{(a \minus{} c)^2}\cdot\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right)
\]
:)[/quote]

Thank you for your nice problem, can_hang2007!

[quote="shfdfzhjj"][quote="can_hang2007"]Let $ a,b,c$ be distinct positive real numbers. Determine the minimum value of the following expression
\[ P \equal{} \frac {ab \plus{} bc \plus{} ca}{(a \minus{} c)^2}\cdot\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right)
\]
:)[/quote]

Sorry,I'm late.Nice problem!  :lol:
I think this is nice too. :)
I have two proof,one is ugly,the other is nice.[/quote]
Yes, really nice. :)
By the way, we also have some similar results
\[ \left( a \plus{} b \plus{} c\right) \left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \ge 9 \plus{} \frac {27}{2}\cdot \frac {(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}
\]

\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \ge \frac {3}{2} \plus{} \frac {27}{16} \cdot \frac {(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}
\]
All three inequalities can be proved by the same technique (AM-GM Inequality).
:)

[quote="can_hang2007"]Let $ a,b,c$ be distinct positive real numbers. Determine the minimum value of the following expression
\[ P \equal{} \frac {ab \plus{} bc \plus{} ca}{(a \minus{} c)^2}\cdot\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right)
\]
:)[/quote]
Here is my solution. :)


geometry
circumcircle
Euler
parallelogram
geometric transformation
reflection
trapezoid

Let triangle ABC inscribed in circumcircle (O), (I) is its incenter. Let D, E, F perspectively be the tagency point of (I) with BC, CA, AB. EF intersects (O) at X_ 1, X_2. Similar for Y_1, Y_2; Z_1, Z_2. Prove that the radical center of (DX_1X_2),  (EY_1Y_2), (FZ_1Z_2) lies on the Euler line of triangle DEF.  :D

Mashimaru has had the solution, :D , he will posted it later

[hide="Solution"]
Let $ X_1X_2$ intersect $ BC$ at $ X$ and let the midpoint of $ BC$ be $ M$.  Let the midpoints of $ AB$ and $ AC$ be $ P$ and $ N$ respectively.  Let the centers of $ (EY_1Y_2)$ and $ (FZ_1Z_2)$ be $ O_1$ and $ O_2$ respectively.  Let the parallel to $ BI$ through $ O$ intersect $ IE$ at $ P$ and the parallel to $ IC$ through $ O$ meet $ IF$ at $ Q$.

[b]Lemma:[/b] $ PQ\parallel EF$.
[i]
Proof:[/i] Let the feet of the perpendiculars from $ E$, $ O$, and $ P$ to $ BI$ be $ E'$, $ O'$, and $ P'$.  It is known that if $ H$ is the orthocenter of $ \triangle ABC$, then $ O$ and $ H$ are isogonal conjugates with respect to $ \triangle ABC$.  Thus, since $ IE\parallel BH$, we have that $ \angle EIE' \equal{} \angle OBO'$.  Furthermore, $ \angle OO'B \equal{} \angle EE'I$, so $ \triangle EE'I\sim \triangle OO'B$.  This means that \[ \frac {PI}{IE} \equal{} \frac {PP'}{EE'} \equal{} \frac {OO'}{EE'} \equal{} \frac {OB}{EI} \equal{} \frac {R}{r}\] where $ R$ is the circumradius of $ \triangle ABC$ and $ r$ is its inradius.  Similarly, $ \frac {IQ}{IF} \equal{} \frac {R}{r} \equal{} \frac {PI}{IE}$, so $ PQ\parallel FE$.

Now, we have that, by the complete quadrilateral $ XBCEAF$, $ (X, B, D, C) \equal{} \minus{} 1$.  This implies that $ X(B, D, M, C) \equal{} \minus{} 1$ since $ M$ is the midpoint of $ BC$ and hence $ XX_1\cdot XX_2 \equal{} XB\cdot XC \equal{} XD\cdot XM$, so $ X_1X_2MD$ is cyclic.  Similarly, the circumcircle of $ \triangle EY_2Y_1$ passes through $ N$ and that of $ \triangle FZ_1Z_2$ passes through $ P$.  This means that $ O_1$ is the intersection of the perpendicular bisector of $ EN$ and the perpendicular bisector of $ Y_1Y_2$.  Notice that the latter is the perpendicular to $ DF$ through $ O$.  Since $ IB\perp DF$, we have that it is the parallel to $ IB$ through $ O$.  Therefore, $ O_1$ is the intersection of the perpendicular through the midpoint of $ EN$ to $ AC$ and the parallel to $ BI$ through $ O$.  It follows that $ O_1$ is the midpoint of $ OP$.  Similarly, $ O_2$ is the midpoint of $ OP$, so $ O_1O_2\parallel PQ\parallel EF$.  This means that the radical axis of $ (EY_1Y_2)$ and $ (FZ_1Z_2)$ is perpendicular to $ O_1O_2$, so it is perpendicular to $ EF$.  Now, consider the circles $ (Y_1Z_2Y_1Z_1)$, $ (Y_1Y_2E)$, and $ (Z_1Z_2F)$.  By the Radical Axis Theorem, the radical axis of the latter two circles must pass through $ D$.  It follows that the radical axis of $ (Y_1Y_2E)$ and $ (Z_1Z_2F)$ must be the perpendicular through $ D$ to $ EF$.  Doing the same for the other radical axes, we see that they must concur at the orthocenter of $ \triangle DEF$, which clearly lies on the Euler Line of $ \triangle DEF$.  [/hide]

Let the angle bisectors $ AI, BI, CI$ cut the circumcircle $ (O)$ again at $ X, Y, Z.$ Let $ XX'$ be a diameter of $ (O).$ $ I$ is orthocenter of $ \triangle XYZ$ $ \Longrightarrow$ $ IYX'Z$ is a parallelogram, its diagonals $ IX', YZ$ cutting each other in half at $ K.$ Let $ (K)$ be a circle with center $ K$ and radius $ KD.$ $ IX'$ cuts $ (O)$ again at $ T,$ which is tangency point of the mixtilinear A-incircle $ (J_a)$ of the $ \triangle ABC$ with its circumcircle $ (O).$

Parallel to $ XT$ through $ D$ cuts $ (I)$ again at $ P.$ The angle $ \angle X'TX$ is right, $ IK \perp XT.$ Therefore, $ DP \perp IK$ is radical axis of $ (I), (K).$ Perpendicular to $ AI$ through $ I$ cuts $ CA, AB$ at $ E_a, F_a$; these are tangency points of $ (J_a)$ with $ CA, AB.$ The parallels $ EF \parallel E_aF_a$ are similar with center $ A$ and coefficient $ k \equal{} \frac {AI}{AJ_a}.$ $ XT$ cuts $ (J_a)$ again at $ D_a.$ $ T$ is the external similarity center of $ (O), (J_a)$ $ \Longrightarrow$ tangent of $ (J_a)$ at $ D_a$ is parallel to tangent of $ (O)$ at $ X,$ which is parallel to tangent of $ (I)$ at $ D.$ Since $ A$ is the external similarity center of $ (I), (J_a),$ it follows that $ A, D, D_a$ are collinear. But then the parallels $ DP \parallel XT$ are similar with the same center $ A$ and coefficient $ k \equal{} \frac {AI}{AJ_a}.$

$ (U), (V)$ are circles with diameters $ IA, IX$ and common tangent $ E_aF_a$ at $ I.$ $ (U)$ goes through $ E, F$ and $ (V)$ goes through $ T$ on account of the right angles at these points. $ (U)$ cuts $ (O)$ again at $ S.$ Pairwise radical axes $ AS, E_aF_a, XT$ of $ (O), (U), (V)$ meet at their radical center $ Q.$  Pairwise radical axes $ AS, EF$ of $ (O), (U)$ and $ (U), (I)$ meet at the radical center $ R$ of $ (O), (U), (I).$ Radical axis of $ (O), (I)$ goes through $ R.$ From similarity of the parallels $ EF \parallel E_aF_a$ and $ DP \parallel XT$ with the same center $ A$ and coefficient $ k,$ it follows that the radical axis $ DP$ of $ (I), (K)$ also goes through $ R.$ But then $ R$ is also radical center of $ (O), (I), (K)$ and the radical axis of $ (O), (K)$ also goes through $ R.$ But this radical axis is perpendicular to the center line $ OK,$ i.e., parallel to $ EF,$ hence it is identical with $ EF.$ As a result, $ (K)$ is the circle $ \odot(DX_1X_2).$ In exactly the same way, circles $ (L), (M)$ centered at the midpoints $ L, M$ of $ ZX, XY$ and with radii $ LE, MF$ are the circles $ \odot(EY_1Y_2), \odot(FZ_1Z_2).$

$ D$ is radical center of the circles $ (O), (L), (M).$ Radical axis of $ (L), (M)$ goes through $ D$ and is perpendicular to their center line $ LM \parallel YX \parallel EF,$ it is the D-altitude of the $ \triangle DEF.$ Similarly, radical axes of $ (M), (K)$ and $ (K), (L)$ are the E- and F-altitudes, they meet at the orthocenter of the $ \triangle DEF$ on its Euler line.

This beautiful geometry problem can be stated in another way, this statement is a little more complicated than the one I had posted: "Prove that the radical center of  (DX_1X_2), (EY_1Y_2), (FZ_1Z_2) lies on OI." Actually, to prove this, it is not really hard :blush: . We just have to prove that O lies on the Euler line of triangle DEF by using the inversion with Pole and Polar. You can try :D  :lol:

This problem was proposed by [b]Cosmin Pohoata[/b] and [b]Darij Grinberg[/b] on [i]Mathematical Reflections[/i]. And here is my solution:

Let $ S$ be the intersection of $ EF$ with the sideline $ BC$, and $ M$ is the mid-point of $ BC$. We have the division $ (BDCS)$ is a harmonic, so that $ SD\cdot SM\equal{}SB\cdot SC\equal{}SX_1\cdot SX_2$. Thus, $ M$ lies on the circumcircle of triangle $ DX_1X_2$.

Let us call $ P$, $ Q$ the mid-points of arcs $ BC$ and $ AB$ of the circle $ \rho (O)$ (not containing $ A$, $ C$), respectively. Let $ N$ the mid-point of the arc $ BC$ containing $ A$ of $ \rho (O)$, $ O_1$ the mid-point of the segment $ IN$. We have $ OO_1$ is parallel to $ IP$, it means $ OO_1$ is perpendicular to $ X_1X_2$, and so it's also the perpendicular bisector of $ X_1X_2$. On the other hand, we have the quadrilateral $ IDMN$ is a trapezoid, so $ O_1$ lies on the perpendicular bisector of $ DM$. Now, we have the quadrilateral $ DMX_2X_1$ is cyclic and $ O_1$ is the intersection of the perpendicular bisectors of the segments $ DM$ and $ X_1X_2$, so that $ O_1$ is the center of the circumcircle of triangle $ DX_1X_2$.

We can easily to see that $ OO_1$ is parallel to $ IP$ and $ OO_1\equal{}\frac{1}{2}\cdot IP$. Similarly, $ OO_3$ is parallel to $ IQ$ and $ OO_3\equal{}\frac{1}{2}\cdot IQ$, where $ O_3$ is the center of the circumcircle of triangle $ FZ_1Z_2$. Hence, we obtain $ O_1O_3$ is parallel to $ PQ$. Now, let's take a look at the circumcircles of triangles $ DX_1X_2$ and $ FZ_1Z_2$, we have the radical exits of the circles is the line $ d_2$, which passes through $ E$ and perpendicular to $ O_1O_3$, i.e. $ d_2\perp PQ$. It follows $ d_2$ is the $ E$-altitude of triangle $ DEF$. Similarly, we have the radical axis $ d_3$ of the circumcircles of triangles $ DX_1X_2$ and $ EY_1Y_2$ is the $ F$-altitude of triangle $ DEF$. Hence, we deduce the radical center of the circles $ DX_1X_2$, $ EY_1Y_2$, and $ FZ_1Z_2$ is the orthocentre of triangle $ DEF$, which has known that lies on $ OI$. The solution is completed.

Hope that I didn't make any mistake!  :read:

When I proved this problem, I didn't use any midpoints of AB, BC, CA. I noticed that if we let C1, B1 respectively be the midpoint of arc AB, AC then the midpoint of B1C1 are the center of the circumcircle of triangle DX1X2 :D. To prove this I use the similarity that maps (I) to (O), and directed angle :oops:  :D
Then I also noticed that the radical center of (DX_1X_2), (EY_1Y_2), (FZ_1Z_2) are the orthocenter of triangle DEF, hence it lies on the Euler line of triangle DEF.
Proof is done then!!! :D

To April, Yetti, and The QuattoMaster 6000: Thanks for your all nice solution, you have widened my eyes :blush:
To April: May I ask your in Vietnamese???? :blush: I am just curious :oops:

Here is my solution for this problem
[b]Solution[/b]

Let $P$, $Q$, $R$, $J$ be the midpoints of $\stackrel\frown{BAC}$, $\stackrel\frown{ABC}$, $\stackrel\frown{ACB}$ and $\stackrel\frown{BC}$ which not contain $A$; $X'$, $Y'$, $Z'$, $M$ be the midpoints of $IP$, $IQ$, $IR$, $BC$; $G$ $\equiv$ $EF$ $\cap$ $BC$, $H$ be orthocenter of $\triangle$ $DEF$
It's easy to see that: $H$ lies on Euler line of $\triangle$ $DEF$
We have: ($BCDG$) = $-$ 1 so $\overline{GD}$ . $\overline{GM}$ = $\overline{GB}$ . $\overline{GC}$ = $\overline{GX_1}$ . $\overline{GX_2}$ or $M$ $\in$ ($DX_1X_2$)
But: $O$, $X'$ are midpoints of $PJ$, $IP$, respectively, then $OX'$ $\parallel$ $IJ$, so $OX'$ $\perp$ $X_1X_2$
It's easy to see that $X'$ line on perpendicular bisector of $DM$, combine with $D$, $M$, $X_1$, $X_2$ lie on a circle, hence $X'$ is center of ($DX_1X_2$)
Similarly: $Y'$, $Z'$ are centers of ($EY_1Y_2$), ($FZ_1Z_2$), respectively
But: $Y'Z'$ $\parallel$ $QR$ $\parallel$ $EF$, $DH$ $\perp$ $EF$ so $DH$ $\perp$ $Y'Z'$
We also have: $\overline{DY_1}$ . $\overline{DY_2}$ = $\overline{DZ_1}$ . $\overline{DZ_2}$, then $DH$ is radical axis of ($EY_1Y_2$) and ($FZ_1Z_2$)
Similarly: $EH$, $FH$ are radical axises of (($FZ_1Z_2$), ($DX_1X_2$)), (($DX_1X_2$), ($EY_1Y_2$))
Therefore: $H$ is radical center of ($DX_1X_2$), ($EY_1Y_2$) and ($FZ_1Z_2$)

No, I post the figure first and then I post the solution  :)


Trapezoid ABCD is circumscribed about circle P.  AB and CD are perpendicular to AD.  The length of AB is 13 and the length of CD is 10.  Find the perimeter of ABCD.

Source: Kentucky Governor's Cup Regionals 2003

[hide]Let the radius of the circle be $r$. Draw the radii of the circle that intersect the tangents. The shorter base has sectinos of $r$ and $10 - r$. Since tangents coming off of the same circle that meet outside of the circle are equal, the longer side also has a section with length $10 - r$.  The longer base has a section with length $13 - r$, and so the longer side has a section with length $13 - r$. The longer side is then equal to $13 - r + 10 - r = 23 - 2r$. The perimeter of the trapezoid is $23 - 2r + 2r + 13 + 10 = 46$[/hide]

Let the point of tangency of the circle to AB be E, the tangent point to AD be F, the tangent point to DC be G, and the tangent point to BC be H, then we have AE = AF = DF = DG=r, where r is the radius of the circle, then GC=CH=10-r, and EB=BH=13-r, so to find the perimeter we add these 4r+2(13-r)+2(10-r)=46.

Here's another way, but I'm not sure if this is also Jimmy's way too.

[hide]A to the point of tangency is f, B to the point of tangency is g, C to the point of tangency is h, and D to the point of tangency is j.  Note that the point of tangency doesn't matter.

Then, you can say that f+g = 13 and h+j = 10.  f+g+h+j = 23.  2(f+g+h+j) = perimeter of ABCD = 46.[/hide]

this is theorem.
For that trapozoid, there is always: $AB+CD=AD+BC$


linear algebra
matrix
geometry
geometric transformation
reflection

Here's a question from my FIITJEE exam today. I think its a nice and interesting one and everyone should give it a try.:
A source of sound emitting a frequency of $ 3000Hz$, moving towards a stationary reflecting wall with a speed $ 50 m/s$. there are four observers a shown in the figure. $ B$ aNd $ D$ are moving with $ 10m/s$ towards right. $ A$ and $ C$ are at rest.
What are the beat frequencies heard by the four?
Match the matrix:
$ \begin {align*} & A) Observer A \; \; \; \; \; \; \; \; \; \; \; \; p) 200Hz \\
& B) Observer B \; \; \; \; \; \; \; \; \; \; \; \; q) 0Hz \\
& C) Observer C \; \; \; \; \; \; \; \; \; \; \; \; r) 900Hz \\
& D) Observer D \; \; \; \; \; \; \; \; \; \; \; \; s) 875Hz$

State reasons for your answers.

Hello
Rajiv I have taken 340 m/s as velocity of sound thus getting  only approx answer.So I m not sure about answer.
I m getting this result:
$ A\minus{}\minus{}>0$
$ B\minus{}\minus{}>200$
$ C\minus{}\minus{}>875$
$ D\minus{}\minus{}>900$
Thank u.

we don't actually need any calculations here. just logic

[hide]
reflect the car across the wall and get rid of the wall. its the same thing.

now, observer A has two cars going towards him. they will get doppler shifted the same amount since doppler effect only varies with velocity. his beat freq. will be 0.
observer B has a car going at 40 towards him and 60 towards him.
observer C has a car going at 50 towards him and 50 away from him.
observer D has a car going at 60 towards him and 60 away from him.

consequently, we can expect B's beat freq to be low compared to that of C & D
we can also say that D's beat freq is likely to be slightly higher than that of C
so we assign B -> 200, C -> 875, and D->900.

yay multiple choice
[/hide]

[quote="kabi"]Hello
Rajiv I have taken 340 m/s as velocity of sound thus getting  only approx answer.So I m not sure about answer.
I m getting this result:
$ A - - > 0$
$ B - - > 200$
$ C - - > 875$
$ D - - > 900$
Thank u.[/quote]
Yes, you are right.
$ \begin{align*} & A \to q \\
& B \to p \\
& C \to s \\
& D \to r$
is the solution.

[hide="Hint:"]
$ f' = f_0 \left( \frac {v + v_o}{v + v_s}\right)$ where $ v_o$ is observer velocity ($ + ve$ if directed towards source), $ v_s$ is source velocity ($ + ve$ if directed towards observer).

Consider a particular frequency to be recieved by the wall and the same is transmitted by it which causes the beats.

[b]bowei's[/b] solution too is wonderful, quick thinking-saves time in the exam.
[/hide]


function
trigonometry
logarithms
limit
real analysis
real analysis unsolved

Is function $ f(x)\equal{}xe^{\sin x}$ uniformly continuous?

I think it isn't but I don't  know how to show it.

I assume that by "uniformly continuous" you meant "uniformly continuous on the entire real line". Then it isn't, indeed. Just take any small $ \delta>0$ and show that $ f(2\pi n\plus{}\delta)\minus{}f(2\pi n)\to\plus{}\infty$ as $ n\to\plus{}\infty$.

[quote="fedja"]I assume that by "uniformly continuous" you meant "uniformly continuous on the entire real line". Then it isn't, indeed. Just take any small $ \delta > 0$ and show that $ f(2\pi n \plus{} \delta) \minus{} f(2\pi n)\to \plus{} \infty$ as $ n\to \plus{} \infty$.[/quote]

But if I use a graph appliction to plot the graph, I get a smooth curve. :o

Yes, the function is infinitely smooth (even analytic) and, thereby, uniformly continuous on every compact interval. It is just like $ x^2$, which is also not uniformly continuous on the entire real line but still has a very nice graph.

Thx, and what about function $ f(x)\equal{}x\ln x$ in $ (0;1)$?

[quote="LideR_M"]Thx, and what about function $ f(x) \equal{} x\ln x$ in $ (0;1)$?[/quote]

Since $ \lim_{x\rightarrow 0^\plus{}} x \ln x \equal{} 0$, the following function $ g: [0,1]\rightarrow \mathbb{R}$:
\[ g(x)\equal{}x\ln x \text{ if }x\in (0,1], g(x)\equal{}0 \text{ if }x\equal{}0\]
is continuous. As $ [0,1]$ is a compact set, $ g$ is uniformly continuous. Since a restriction of uniformly continuous function is also uniformly continuous and $ g|_{(0,1)}\equal{}f$, $ f$ is uniformly continuous.


quadratics
number theory proposed
number theory

Prove that if $p=4k+1$ is prime then $\sum_{r=1,(r|p)=1}^{p}r = \frac{p(p-1)}{4}$

(r|p) is L symbol
______________________
Let me be love of your life!

[hide="hint"]If $a$ is a quadratic residues of $p$ less than $\frac{p}{2}$ then $p-a$ is a quadratic residues of $p$ greater than $\frac{p}{2}$. [/hide]

all right but can you post complete solution!
If not ,it will become sad topic! :)


Find a real number $ t$ and polynomials $ f(x)$ , $ g(x)$ and  $ h(x)$ with integer coefficients such that:

$ f(x)\equal{}\sqrt {2}$, $ g(x)\equal{}\sqrt {3}$ and $ h(x)\equal{}\sqrt {7}$

I assume you mean $ f(t), g(t), h(t)$ respectively.  Equivalently, we want to find $ t$ such that $ \mathbb{Q}[t] \equal{} \mathbb{Q}[ \sqrt{2}, \sqrt{3}, \sqrt{7} ]$.  The obvious choice is $ t \equal{} \sqrt{2} \plus{} \sqrt{3} \plus{} \sqrt{7}$ and the rest is linear algebra.


function
trigonometry
limit
calculus
calculus computations

Does this function $f: y = \frac{{1 + \sin \pi x}}{{x + 1}}$ has a limit when $x \to + \infty$? And what about the $\mathop {\lim }\limits_{n \to + \infty } y_n$ where $y_n=f(n)$?

both limits are $0$ the latter being a subsequence of the first

This fonction has limit $0$ when $x\to\infty$ because $1+\sin{\pix}$ is between $0$ and $2$ and because $x+1\to\infty$
It's the same for $n$ :roll:

Yes, this has a limit, and the limit is zero.

More generally: Suppose $f(x)=g(x)h(x),$ with the following conditions: $g(x)$ is bounded (That means that there is some constant $M$ such that $|g(x)|\le M$ for all $x$), and $\lim h(x)=0.$

Then $\lim f(x)=0.$

Once you have a formal definition ($\epsilon$-$\delta$ or $\epsilon$-$N$) of limit, the proof of this becomes a straightforward exercise in applying that definition.

You can also think of this result as a consequence of the "squeeze theorem." $f(x)$ is squeezed in between $-M|h(x)|$ and $Mh(x),$ both of which tend to zero, so $f(x)$ must tend to zero.

In your example, $(1+\sin\pi x)$ is bounded and $\frac1{x+1}$ tends to zero.

The answer is the same for the sampled sequence as for the continuous function, and for the same reasons.


Assuming that the real line has the least upper bound property.

Show the sets:

$ [0.1]\equal{}${$ x|0 \leq x \leq 1$}
$ [0.1)\equal{}${$ x|0  \leq x < 1$}

have the least upper bound property

[b]The least upper property[/b] of a partially ordered $ S$ means, that every non-empty subset $ S$ has a least upper bound.

So how to show that every non-empty subset of your two sets $ [0,1]$ and $ [0,1)$ has a least upper bound?

We have to use the so called [b]least upper bound axiom[/b] http://www.artofproblemsolving.com/Forum/posting.php?mode=reply&t=256822, which states, that if a non-empty set of real numbers has an upper bound, then it has a least upper bound. Since $ [0,1]$ and $ [0,1)$ are bounded, the conclusion follows.


trigonometry
geometry
incenter
geometric transformation
reflection
circumcircle
perpendicular bisector

triangle
Point P lies inside triangleABC, and is such that <PAC = 18°, <PCA = 57°, <PAB = 24°, and <PBA = 27°.  Show that ABC is isosceles.

$\angle BAC = \angle BAP+\angle PAC = 24^\circ+18^\circ = 42^\circ,$ $\angle ABC' > \angle ABP = 27^\circ.$ Let $\triangle ABC'$ be isosceles with the base angles $\angle ABC' = \angle BAC' = \angle BAC = 42^\circ,$ so that is C somewhere on the line AC'. The remaining angle of this isosceles triangle is $\angle BC'A = 180^\circ-2\ \angle ABC' = 96^\circ.$ The angles from P are $\angle PAC' = \angle PAC = 18^\circ,$ $\angle PAB = 24^\circ,$ $\angle PBA = 27^\circ,$ same as in $\triangle ABC,$ and in addition, $\angle PBC' = \angle ABC'-\angle PBA = 15^\circ.$ Let the perpendicular bisector of AC' meet PA at Q. $\triangle AC'Q$ is isosceles with the base angles $\angle AC'Q = \angle C'AQ \equiv \angle C'AP = 18^\circ.$ We have $AC' = 2\ AQ\ \cos 18^\circ$ and

$AB = 2\ AC'\ \cos 42^\circ = 4\ AQ\ \cos 18^\circ\ \cos 42^\circ =$

$= 2\ AQ\ (\cos 60^\circ+\cos 24^\circ) = AQ+2\ AQ\ \cos 24^\circ$

Let $R \in AB$ be a point, such that $\triangle ARQ$ is isosceles with base angles $\angle ARQ = \angle RAQ \equiv \angle BAP = 24^\circ.$ Then RQ = AQ and $AR = 2\ AQ\ \cos 24^\circ,$ so that BR = AQ = RQ, which means that $\triangle BQR$ is also isosceles with base angles $\angle BQR = \angle QBR = \frac{\angle QRA}{2}= 12^\circ.$ It follows that $\angle QBP = \angle ABP-\angle QBA = 15^\circ = \angle PBC'$ and consequently, BP bisects $\angle QBC' = 30^\circ.$ From the isosceles $\triangle AC'Q,$ $\angle PQC' = 2\ \angle C'AQ = 2 \cdot 18^\circ = 36^\circ.$ On the other hand, from the isosceles $\triangle ARQ,\ \triangle BQR,$ $\angle PQR = 2\ \angle RAQ-\angle BQR = 2 \cdot 24^\circ-12^\circ = 36^\circ,$ so that QP bisects $\angle BQC' = 72^\circ.$ As a result, the angle bisectors BP, QP meet in the incenter P of $\triangle BQC'.$ Hence C'P bisects the remaining $\angle BC'Q = 180^\circ-(\angle QBC'+\angle BQC') = 180^\circ-(30^\circ+72^\circ) = 78^\circ$ and $\angle PC'Q = \frac{\angle BC'Q}{2}= 39^\circ.$ But then $\angle PC'A = \angle PC'Q+\angle QC'A = 39^\circ+18^\circ = 57^\circ = \angle PCA,$ which implies that $C' \equiv C$ are identical and the isosceles $\triangle ABC' \equiv \triangle ABC$ also identical.

Another solution, with no trigonometry at all, starting as the previous one:

[quote="yetti"]$\angle BAC = \angle BAP+\angle PAC = 24^\circ+18^\circ = 42^\circ,$ $\angle ABC' > \angle ABP = 27^\circ.$ Let $\triangle ABC'$ be isosceles with the base angles $\angle ABC' = \angle BAC' = \angle BAC = 42^\circ,$ so that is C somewhere on the line AC'. The remaining angle of this isosceles triangle is $\angle BC'A = 180^\circ-2\ \angle ABC' = 96^\circ.$ The angles from P are $\angle PAC' = \angle PAC = 18^\circ,$ $\angle PAB = 24^\circ,$ $\angle PBA = 27^\circ,$ same as in $\triangle ABC,$ and in addition, $\angle PBC' = \angle ABC'-\angle PBA = 15^\circ.$[/quote]

Let the reflection of BC' in BP meet AP at Q, BP bisects $\angle C'BQ = 2\ \angle C'BP = 30^\circ.$ Let the perpendicular bisector of BQ meet AB at R and let a circle (R) passing through B, Q meet BC' at S. Since $\angle SBQ \equiv \angle C'BQ = 30^\circ,$ $\angle SRQ = 60^\circ,$ $\triangle QRS$ is equilateral. From the isosceles $\triangle BRS$ with the base angles $\angle RSB = \angle RBS \equiv \angle ABC' = 42^\circ$ and from the equilateral $\triangle QRS,$ $\angle QRA = \angle SRA-\angle SRQ = 2\ \angle RBS-\angle SRQ = 2 \cdot 42^\circ-60^\circ = 24^\circ.$ It follows that $\triangle ARQ$ is isosceles with the base angles $\angle QRA = \angle QAR \equiv \angle PAB = 24^\circ$ and AQ = RQ. Quadrilateral ARSC' is cyclic, because the isosceles $\triangle SBR \sim \triangle ABC'$ are similar, having equal base angles, hence SR is antiparallel of AC' with respect to $\angle ABC'.$ Since AQ = RQ = SQ, Q is the circumcircle of $\triangle ARS$ and of quadrilateral ARSC', yielding AQ = C'Q. Thus $\triangle AC'Q$ is isosceles with the base angles $\angle AC'Q = \angle C'AQ \equiv \angle C'AP = 18^\circ$ and $\angle PQC' = 2\ \angle C'AQ = 36^\circ.$ On the other hand, from the non-isosceles $\triangle ABQ$ with angles $\angle BAQ \equiv \angle BAP = 24^\circ$ and $\angle ABQ = \angle ABC'-\angle QBC' = 42^\circ-30^\circ = 12^\circ,$ we get $\angle PQB = \angle ABQ+\angle BAQ = 36^\circ.$ It follows that QP bisects $\angle BQC' = 72^\circ.$ The rest is the same as before:

[quote="yetti"]As a result, the angle bisectors BP, QP meet in the incenter P of $\triangle BQC'.$ Hence C'P bisects the remaining $\angle BC'Q = 180^\circ-(\angle QBC'+\angle BQC') = 180^\circ-(30^\circ+72^\circ) = 78^\circ$ and $\angle PC'Q = \frac{\angle BC'Q}{2}= 39^\circ.$ But then $\angle PC'A = \angle PC'Q+\angle QC'A = 39^\circ+18^\circ = 57^\circ = \angle PCA,$ which implies that $C' \equiv C$ are identical and the isosceles $\triangle ABC' \equiv \triangle ABC$ also identical.[/quote]


function
algebra
domain
geometry
calculus
derivative
calculus computations

I really appreciate, if somebody could help me with these tasks, I am beginner and really need fast help. Also, as my mothertonque is not English, some mathematical terms may be wrong, hope you understand.

1. Function y=ln (x+10/x-8)+80

a)find domain, b)asymptotes equations, c)Extremums, d)grow and decline areas, e)curvature and concava areas, f)point of inflections

2.Function:  z=xy+40x, a) find Extremums in condition: x^2+y^2=256

b) control sufficiency, c) designate type, d) calculate the value of function

Thank you!

Most of this stuff comes down to analyzing the derivative, so I'll get you started with those:
$ \frac{dy}{dx}\equal{}\frac{1}{x\plus{}\frac{10}{x}\minus{}8}\cdot (1\minus{}\frac{10}{x^{2}})$.  Did I interpret the denominator correctly?


In my solution, I typed the wrong round number for the template in LaTeX. I just left the 1 from last round.

What should I do?

Don't worry about it.  It'll be obvious from your solutions what round it is.


Prove that 1/a+1/b-1/c <1/abc if a^2+b^2+c^2=5/3, a,b,c positive.

Quite easy exercise, we can even replace <1/abc by <= 5/(6*abc). Try.


A class of $ 31$ students was polled about which soft drink was their favorite. Seventeen students liked Fizz-Up, eighteen liked Fizz-Down, and six students didn't like either. How many students liked both?

The number of people who liked either was 31-6=25. 17+18=35, which exceeds 31 by 4. The overcounted also happens to be one times the people who liked both (venn diagrams are good), so the answer is 4.

[quote]The number of people who liked either was 31-6=25. 17+18=[b][color=red]35, which exceeds 31 by 4[/b][/color]. The overcounted also happens to be one times the people who liked both (venn diagrams are good), so the answer is 4.[/quote]

By this logic, shouldn't we instead find how much 35 exceeds 25 (the number of who like either)? This would give 10 as an answer. As a Venn Diagram, the numbers in the circle would be 7 (only Fizz-Up), 10 (both), 8 (Fizz-Down), 6 neither.

The Archer is right.....

$ 25$ people voted, and $ 35$ people like Fizz-Up or Fizz-Down.... So $ \boxed{10}$ people like both.


geometry
geometric transformation
reflection
LaTeX

A particle moves inside the square with vertices A=(0, 0), B=(1, 0), C=(1, 1), and D(0, 1).  It begins at the point (2/3, 0), travels to some point on BC, then travels to some point on CD, then travles to some point on DA, and then travels to the point (1/6, 0).  What is the minimum distance that the particle could have travelled?

[hide]I drew a bunch of boxes which represent the reflections of the unit square. I drew a straight line from start to end point, making sure it goes through each of the required side.

Via the distance formula:

$\sqrt{2^{2}+(1+\frac16+\frac13)^{2}}$
$\sqrt{4+\frac{9}{4}}$
$\sqrt{\frac{16+9}{4}}$
$\boxed{\frac52}$[/hide]
Am i right?

[hide="hint"]

In order to find the shortest root we create a mirror for the lines, reflecting it sequentially from last to first and then we draw the shortest path from the original point to the final line.  Then where it hits a side we reflect it obeying the law of reflection, when it hits the next side we reflect it again and so on until it comes to the final line.  The distance is equal to that to the final mirrored line.  And since the shortest distance is a straight line following the reflecting line is the shortest path since it is the shortest path to the mirrored line.

[/hide]

[hide="my answer"]

using 3 refections i got a minimum distance of: $\frac{\sqrt{17}}{2}$

[/hide]

yeah, just like demji and ShoeFactory said, just reflect the box in a series of ways so that you get a straight line and then finding the length of that line is not very hard. My answer was $5/2$

could someone please post a picture of the reflection? thanks.

ok here is a picture:

Its not quite to scale and some proportions are off but ti gets across the general idea

I don't really get his diagram but its probably the same thing i did.

I never thought transformations could be useful in problems like this. I should have payed more attention in Geometry.

We never learned transformation/rotation/reflections in Geometry. :maybe:

My reflections kind of looked like this:
$\ \qquad \Box\Box$
$\Box\Box$
EDIT: OK, the top 2 boxes are supposed to be shifted over to the right, but I can't figure out how to do it in $\LaTeX$.

[quote="ShoeFactory"]That is a pro drawing indeed![/quote]

If that is truly what you mean, then you are crazy ( :D ) because those are very cheap paint skills I have, with a mouse that cannot move without jerky movements.

[quote="pkerichang"][quote="ShoeFactory"]I never did this type of thing in geometry. I encountered one of these by reading a book, which had lots of little problem/puzzles.[/quote]

Well...That's why you should take geometry at Troy...we did these kinds of stuff in class (ACoPS mention this technique too).[/quote]

Eric you fool.

I learned it on my own so its ok.  I just have to hope that there are less than 5 geometry on the AIME. And hopefully they are 12,13,14,15 so I have a chance to get the rest. :D


trigonometry
calculus
calculus computations

can someone help me replace the polar equation with an equivalent cartesian equation: tanµ=3

$ \tan\theta \equal{}3\Rightarrow \frac{\sin\theta}{\cos\theta}\equal{}3\Rightarrow r\sin\theta\equal{}3r\cos\theta\Rightarrow x\equal{}3y$


real analysis
function
integration
calculus
abstract algebra
group theory
real analysis unsolved

If $ E_1$ and $ E_2$ are measurable subsets of $ \mathbb R^1$ with $ |E_1| > 0$ and $ |E_2| > 0$. Show that the set $ \{ x \in \mathbb R^1 : x \equal{} x_1 \minus{} x_2, x_1 \in E_1, x_2 \in E_2\}$ contains an open interval with positive length.

Any hints?

You're asking about the set $ E_1\minus{}E_2.$ It wouldn't change the problem to flip $ E_2$ around and then ask about the set $ E_1\plus{}E_2,$ so that's what I'll do.

Assume WLOG that $ |E_1|<\infty$ and $ |E_2|<\infty.$ If not, we simply intersect each with some interval to make them smaller; if the conclusion holds for the smaller sets then it would hold for the larger sets.

Let $ f_1\equal{}\mathbf{1}_{E_1}$ and $ f_2\equal{}\mathbf{1}_{E_2}$ and let $ g\equal{}f_1*f_2$ where $ *$ denotes convolution. By following the definition of convolution, we can see that $ \{x: g(x)\ne 0\}\subseteq E_1\plus{}E_2.$ But we can show that the result of the convolution of two $ L^2$ functions belongs to $ C_0$ - the set of continuous functions that tend to zero at $ \infty.$ (There are several ways besides $ L^2$ to make that argument.) On the other hand, $ \int_{\mathbb{R}}g\equal{}\int_{\mathbb{R}}f_1\cdot\int_{\mathbb{R}}f_2\equal{}|E_1|\cdot|E_2|>0.$

A nonnegative continuous function with a positive integral must be positive at some point, and since it is continuous, it must be positive on some interval, which is what we needed to show.

This result can be generalized beyond $ \mathbb{R}$ to any locally compact abelian group (with Haar measure).

Thanks Kent Merryfield, your approach is really interesting :)


Each of the ten letters in $ ``MATHCOUNTS''$ is written on its own
square tile and placed in a bag. What is the probability that a tile
randomly selected from the bag will have a letter on it that is in the
word $ ``TEACH''$? Express your answer as a common fraction.

Letters: $ M,A,T,H,C,O,U,N,T,S$.
Letters who should be chosen: $ T,A,C,H$

Probability: $ \frac{4}{10}\equal{}\frac{2}{5}$

Well there are two T's, so it would be $ \frac{5}{10}\equal{}\boxed{\frac{2}{5}}$.

1) It says that [b]every[/b] letter in the word has [b]its own[/b] square tile
2) $ \frac{5}{10}\equal{}\frac{1}{2}\neq \frac{2}{5}$

There are 2"T"s so total of 5 letters. So the answer is $ \frac{5}{10}=\boxed{\frac{1}{2}} $

So AIME15, your Official answer is wrong-.- how is 5/10 equal to 1/2?? XD


If $ ABC$ is a triangle with incentre I and D is the point where the incircle touches $ BC$, the line connecting I with the midpoint of $ BC$ bisects $ AD$. :wink:


integration
calculus
geometry
rectangle
calculus computations

Is $i=\sqrt{-1}$ in this problem, kunny? If so, I can simplify the answer.

[hide="Answer"]$\displaystyle\int_{i-1}^i \left(x^4+2x^3+x^2-\frac{1}{30}\right)\, dx=\left[\frac{x^5}{5}+\frac{x^4}{2}+\frac{x^3}{3}-\frac{x}{30}\right]_{i-1}^i$

$=\displaystyle\left(\frac{i^5}{5}+\frac{i^4}{2}+\frac{i^3}{3}-\frac{i}{30}\right)-\left(\frac{(i-1)^5}{5}+\frac{(i-1)^4}{2}+\frac{(i-1)^3}{3}-\frac{(i-1)}{30}\right)$[/hide]

No, $i$ is not $\sqrt{-1}$ here, JesusFreak197 :)

You can simplify your result.

kunny

That's right,  JesusFreak197. :)

Then Calculate

\[\sum_{i=1}^n \int_{i-1}^i \left(x^4+2x^3+x^2-\frac{1}{30}\right)dx\]

Kunny: I think that $\Sigma$ and $\int$ have a lot in common. Some say if we pull the two ends of $\Sigma$, we get $\int$, is that true?

The pulling the two ends thing is just some humor, I believe.

And yes, the integral is often times interpreted as the limit of a sum.  Remember that the integral is the area under a curve.  To find this area we take the Sum of a bunch of rectangles that approximate this area under the curve.  As we make the widths of the rectangles smaller (limit as they go to 0) our approximate area becomes the actual area and the sum becomes an integral.

that is how i understand it. I think that integral is more powerful than sigma, well, at least in physics...

ok, sorry to interrupt all of you BUT i have to say this:

You are posting CALCULUS topics here. DO NOT POST calculus in Intermediate Topics

want proof? please see what Mr. DPatrick has to say:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26438]http://www.mathlinks.ro/Forum/viewtopic.php?t=26438[/url]

make sure you guys fully understand it because it is doing an injustice (not to me) but to the forum. thank you and sorry to bother you all

[quote="kunny"]$\sum_{i=1}^n \int_{i-1}^i \left(x^4+2x^3+x^2-\frac{1}{30}\right)dx$

$=\int_0^n \left(x^4+2x^3+x^2-\frac{1}{30}\right)dx$[/quote]

For those wondering - what Kunny has essentialy done is to break the integral up into unit sized steps, and then sum these up.


induction
number theory open
number theory

It's a one player number game! You start with any (positive or negative) integer $a_{0}$. On turn $i$, you can either multiply $a_{i}$ by a whole (positive or negative) power of $1/2$ to generate $a_{i+1}$, or multiply $a_{i}+1$ by a whole (positive or negative) power of $3/2$ to generate $a_{i+1}+1$. If you can make $a_{n}= 1$, you win!!!

For example, this is a valid sequence of moves: $a_{0}=-1, a_{1}=-64, a_{2}=-1/4, a_{3}= 1/8, a_{4}= 1$ (I used the second rule when I went from $a_{2}+1 = 3/4$ to $a_{3}+1 = 9/8$).

When can you win? If the Collatz Conjecture holds, then you can definitely always win, but as the example shows, the converse is not true. (In fact, the Collatz Conjecture ought to be the same thing, just with everything restricted to positive numbers). In other words, this is easier than the Collatz conjecture. And it's a GAME :D !!!!!

It's a joke isn't?
I take $a_{0}=1$ and win.   :lol:

The steps are inversible and inverse steps have the same form so both players in their turn can return previous  position. Hence there is no other winning number to start exept stupid partner.

Sorry, you can't choose the number you start out with. :roll:

Nothing understand. How to start game? Who choose starting number? What prevents to my opponent to keep the following scenario?

$\rightarrow a_{0}\rightarrow b \rightarrow c_{0}\rightarrow b \rightarrow c_{1}\rightarrow b \rightarrow c_{2}\rightarrow b \rightarrow c_{3}\rightarrow b \ ... \ \rightarrow c_{n}\rightarrow b \rightarrow \ ...$

The only thing your opponent does is pick the starting number. You do the rest.

Oooops! My opponent give me any integer and then leave me to do what I want.  Can my opponent win, who win if he choose $a_{0}=1$?  :lol:
I find a strange to call that a game.

Huge possibilities appear. I think it is not game - it is a toy when compared with Collatz Conjecture.
If I understood correctly it occurs to be easy and

[hide="I did my rest"]If $a_{0}\le 0$ then I take $a_{1}=\frac{a_{0}}{2^{n}}$ with sufficiently large $n$ to produce $a_{2}=(a_{1}+1)\cdot \frac{2}{3}-1$ be positive.
Now induction step for positive $a_{0}$:
$a_{0}=3m-1 \rightarrow \frac{3m-1}{4}\rightarrow \frac{3m+3}{4}\cdot\frac{2}{3}-1=\frac{m-1}{2}\rightarrow m-1$ (if $a_{0}\ne 2$ of course  :lol:  )
$a_{0}=3m+1 \rightarrow \frac{3m+1}{2}\rightarrow \frac{3m+3}{2}\cdot\frac{2}{3}-1=m$ (for $a_{0}\ne 1, 4$  :lol: )
$a_{0}=6m \rightarrow 3m$
$a_{0}=6m-3 \rightarrow (6m-2)\cdot\frac{3}{2}-1 = 9m-4 \rightarrow \frac{9m}{4}-1\rightarrow \frac{9m}{4}\cdot \frac{4}{9}-1=m-1$
$a_{0}=3 \rightarrow (3+1)\cdot\frac{3}{2}-1=5 \rightarrow (5+1)\cdot\frac{3}{2}-1=8 \rightarrow 1$[/hide]

[quote="me"]I find a strange to call that a game.[/quote]
Up to a giraffe came the idea at last:
[quote="zeb"]It's a [b]one player[/b] number game!
[/quote]

Nice! I didn't expect it to be solved so fast. (Oh, and the game thing was just thrown in their to make the problem more appealing :P )

Now to generalize it! This seems to be the best generalization so far:

You start with any (positive or negative) integer $a_{0}$. On turn $i$, you can either multiply $a_{i}$ by a whole (positive or negative) power of $(m-1)/m$ to generate $a_{i+1}$, or multiply $a_{i}+1$ by a whole (positive or negative) power of $m/(m+1)$ to generate $a_{i+1}+1$. When can you eventually reach $m-1$? (so the original problem was the case $m=2$)

Can you always reach $m-1$ starting from $1$?

Hmm, the case $m=1$ is easy! $a_{0}\to  a_{0}*0 = 0$ :lol:

For the case $m=3$, we have the sequence $1 \to  1/2 \to  3/4 \to  4/3 \to  2$. Also, $0 \to 1/3 \to  1/2 \to  ... \to  2$.


Show that, for every positive integer n, $121^{n}-25^{n}+1900^{n}-(-4)^{n}$ is divisible by 2000.


An integer between 1000 and 9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits.  How many balanced integers are there?

[hide]
call balanced integer $abcd$, where $a,b,c,d$ are the digits of the number. $1\leq a+b \leq 18$. To give sum 1 for $a,b$ we have 1 possibility:
$(1,0)$. For $c+d$ we have 2 possibilities, because we have to consider also $(0,1)$. To have sum 2 we have 2 possibilities for $(a,b)$ and 3 for $(c,d)$ and so on. So the number of balanced integers is $1*2+2*3+3*4+4*5+...+18*19=2280$[/hide]

That answer is way too big. Make sure your pattern works (it doesn't).

[hide="Hint"]What if $a+b=11$?  Are there still $11$ possibilities for $(a,b)$?  Are there still $12$ possibilities for $(c,d)$?  Remember that $0\le a,b,c,d\le 9$.[/hide]

[quote="robinhe"][hide]It's just like hydro said up to $9\times 10$ but for digits that sum up to 10, there are only $9$ possibilities (not $10$) and none of them have a 0, so there is no need to subtract one before multiplying.

Thus the answer is $1\cdot 2+2\cdot 3+...+8\cdot 9+9^2+8^2+7^2+...+1^2=\boxed{615}$[/hide][/quote]

You mentioned "It's just like hydro said up to $9\times 10$" in your solution, but you didn't include $9\cdot 10$ when you added everything up.  Otherwise, very nice solution!  :)

Right, I was a little confused.

I did the problem first, then wrote the solution, instead of doing the problem as I wrote the solution.  Sorry :oops:

Why does this AIME only have 10 questions?

Because I forgot to add them.   :P   You can find all 15 problems now.

[/hide]
first work with the two digits.  find how many ways u can add the two digits to 1,2,3,.18.  you find there are 2 ways u can add up to 1 (0,1 and 1,0), 3 ways to add up to 2, 4 ways to add up to 3 and so on.. until u get to 9 (10 ways).  then u notice there are only 9 ways to add up to 10, 8 ways to add up to 11, and so on until theres 1 way to add up to 18.  then u find how many terms there are.  if both sides add up to 1, there are 1*2 ways (as the first digit can not be 0), to add up to 2, there are 2*3 ways, all the way up to 9*10 ways for 9.  then if both sides add up to 10, there are 9^2 ways (as the first digit will not be 0), and so on until 1^2 ways for 18.  then if u add them all together, (1*2 +2*3+3*4...9*10+9^2+...1^2), the answer will be 615.

Is there some quick way to add them all up, or do you have to do it all by hand?

please don't bump 11-year-old threads. Thanks


a subject long ignored since the death of Einstein, but a friend of mine has completely re-worked the subject, working from Schroedinger, Plank and so on to develop equations to unify classical forces, here's some links

http://pages.globetrotter.net/srp/geomax2a.htm

http://pages.globetrotter.net/srp/unifying_classical_force_equations.pdf

http://pages.globetrotter.net/srp/discrete_electromagnetic_fields.pdf

[quote="durt"][url]http://www.timecube.com/[/url][/quote]

what is that ?

[quote="durt"][url]http://www.timecube.com/[/url][/quote]

see, now this is trolling i can get behind.


many people do. I don't but i'm pretty sure all the chapters are good for competitions (AIME and USAMO)

can someone who has actually read the book please post about it?

thanks

None of the chapters are geared to contests. The first four, maybe, are relevant.

[quote="Smartguy"]hey,
does anyone have this book?

If you did read it, can you tell me which chapters are most helpful for preparation for the math contests?

Thanks[/quote]

Good books spaek for themselves, I have been benefited and I am sure so will you if you read the whole book!


$ A$ is a set and has an upper bound. $ A$ is not empty set. $ A$ is a set with non-negative numbers.

Let:   $ A^{2}\equal{}\lbrace y\in R: \ y\equal{}x_{1}x_{2} \ \ \ \ \ \ \ x_{1}, x_{2}\in A\rbrace$

Show that $ sup \ A^{2}\equal{}(sup \ A)^{2}$

Let $ x,y \in A$, then

\[ x \cdot y \leq (\sup A)^2 \Rightarrow \sup (A^2) \leq (\sup A)^2\]

On the other hand,

\[ x^2 \leq \sup (A^2) \Rightarrow x \leq \sqrt{\sup(A^2)} \Rightarrow \sup A \leq \sqrt{\sup(A^2)}.\]

Thus, $ (\sup A)^2 \leq {\sup(A^2)}$ (since the elements in $ A$ are nonnegative)

Therefore, $ (\sup A)^2 \equal{} {\sup(A^2)}$


This thread is for serious discussion of the Magic: The Gathering card game.

Let's start with a question. Does anyone here (besides me) actually play?

I've never even heard of the above terms. I played back in 8th edition to 10th.

[quote="eggylv999"][quote="starkiller2"][quote="eggylv999"]ME! I use Esper. Conflux sucks. Alot.[/quote]
Conflux is OK, at least for us who opened Nicol Bolas at the prerelease...
Anyways, with Esper aren't u vulnerable to artifact hate?[/quote]

Yes. (which sucks for me)

Nicol Bolas is cheap.[/quote]
He is, but he is the incarnation of win.

......My friend and i once made up this game called mag-i-oh, magic and yugioh together......it was weird......but fun! :lol:

i got owned by this 8 year old in magic the gathering.  Then again, nobody would be surprised.


function
topology
real analysis
real analysis unsolved

If $X$ be an complete metric space,$f$ is a upper semicontinuous function:$X \to \mathbb{R}$,Prove the set $E=\{x\in X: w_{f}(x)<\epsilon\}$ is open and dense in $X$ for any $\epsilon>0$.

Where $w_{f}(x)=\inf_{\sigma>0}\sup_{|y-x|<\sigma,|y'-x|<\sigma}|f(y)-f(y')|$
And it is easy to prove $E$ is open without using $f$ is a upper semicontinuous.
But how to prove  $E$ is dense?Thanks for help

Assume the contrary. Then for some $\varepsilon > 0$ there exists a ball $B$ s.t. $B\subset X\setminus E$.
Pick a point $x_{1}\in B$. Since $f(x)$ is upper semicontinuous, there exists $\delta$ s.t. for all $y\in B(x_{1},\delta)$, $f(y) \leq f(x_{1})+\varepsilon/2$. Now, by the definition of $E$ there are two points $y_{1}$ and $y_{2}$ in $B\cap B(x_{1}, \min(1/2, \delta))$ s.t. $f(y_{2})-f(y_{1}) \geq \varepsilon$. We have \[f(y_{1}) \leq f(y_{2})-\varepsilon \leq (f(x_{1})+\varepsilon/2)-\varepsilon = f(x_{1})-\varepsilon/2.\] Denote $x_{2}= y_{1}$. Similarly we find $x_{3}$ in $B\cap B(x_{2}, 1/4)$ s.t. \[f(x_{3}) \leq f(x_{2})-\frac{\varepsilon}2\leq f(x_{1})-2 \cdot \frac\varepsilon2.\] Then we find $x_{3}, x_{4}, \dots$: \[f(x_{k}) \leq f(x_{1})+(k-1) \cdot \frac\varepsilon2,\] Note that \[\liminf_{k\to\infty}f(x_{k}) \leq f(x_{1})+\liminf_{k\to\infty}-(k-1) \cdot \frac\varepsilon2=-\infty.\] We proved that the function $f(x)$ is unbounded from below in $B$. Moreover, the same proof shows that $f(x)$ is unbounded from below in every ball in $B$.
Observe that the set $A_{k}= \{x: f(x) > k\}$ is nowhere dense: in every ball $D$ there is a point $\hat x$ s.t. $f(\hat x) < k-1$. In a sufficiently small neighborhood of $\hat x$, $f(x)$ is less than $k-1/2$. On the other hand \[X = \bigcup_{k\in \mathbb Z}A_{k}.\] I.e. $X$ is a countable union of nowhere dense sets. This contradicts to the fact that every complete metric space is a Baire space (and, therefore, is not a countable union of nowhere dense sets).

Update: I corrected the proof.

Hi,Thank you for your solution.I also consider in this way but I only get a sequence $\{x_{k}\}$ such that $f(x_{k}) \to-\infty$,and It  doesn't leads to a contradiction :blush: .

But can you explain the following:
[quote="Yury"] \[f(y_{1}) \geq f(y_{2})+\varepsilon \geq (f(x_{1})-\varepsilon/2)+\varepsilon = f(x_{1})+\varepsilon/2.\] .[/quote]
How do you get:$f(y_{2})\ge f(x_{1})-\varepsilon/2$
In the condition,It should be $f(y_{2})\le f(x_{1})+\varepsilon/2$ which can't get that :(

Sorry my original proof was incorrect  :( . Now I corrected it (in my reply above).

I believe this time must be right.Because the problem is which I derive a problem in the Baire's Theorem section of Royden's book :lol:
And thanks for your nice solution.
[quote="Yury"] $A_{k}= \{x: f(x) > k\}$ .[/quote]
Anyway,I think we should set $A_{k}= \{x: f(x) \ge k\}$,because when you apply the Baire's theorem,you must need the set is closed :)

[quote="zhaobin"]
Anyway,I think we should set $A_{k}= \{x: f(x) \ge k\}$,because when you apply the Baire's theorem,you must need the set is closed :)[/quote]
No, not really. A union of countably many nowhere dense subsets is not equal to the entire space no matter whether the sets are closed or not.

Yes,you are right. :blush: But if when it is close,then it is easy to check it is nowhere dense.


floor function
logarithms
limit
function
number theory unsolved
number theory

Hi Mathematica; what is the value of e_1?
  Hi Hawk Tiger; if e_n=(e_(n-1))^2-e_(n-1)+1 and if e_0=1 then for all n: a_n=1.
  All that appears obscure to me.

I think $e_{1}=1$ ,And I think mathematica means the same.

Hi  Hawk Tiger.
That does not function with e_1=1.
(we find in this case E~1.1243..).
In fact the formula is e_n=e_0....e_(n-1) +1 (and e_1=2) ;
thus e_n=(e_(n-1))^2-e_(n-1)+1 is valid for n>=2.


calculus
linear algebra
linear algebra unsolved

Several people have asked me for these so I am attaching my lecture notes for calculus, multivariable calculus, and linear algebra.

This set is linear algebra, based off of the Honors Linear Algebra course, which uses Friedberg, Insel, and Spence's "Linear Algebra", 3E.

If you spot typos or errors, please tell me.

I was only able to give it a quick glance and so far just spotted:

* p. 1 of Lecture 3: "each constant is 0" in the proof of Thm 1.1 should probably be "each coefficient is 0".

* p. 1 of Lecture 13: Equation (1) has $ (\minus{}1)^{1\plus{}j}$ instead of $ (\minus{}1)^{i\plus{}j}$.

* p. 1 of Lecture 15: "If $ \{v_1,\ldots,v_n\}$ are the eigenvectors of $ A$" - it's not quite clear what you mean by "the" here. [i]All [/i]the eigenvectors? A complete set of eigenvectors belonging to different eigenvalues? Similarly, "if there are $ n$ distinct eigenvectors of A, we can write down a basis that diagonalizes A" doesn't make much sense. Over an infinite field there are infinitely many distinct eigenvectors associated with any eigenvalue; of course that doesn't mean you can diagonalize. This becomes clearer in the next Lecture but that paragraph is quite confusing.

* p.2 of Lecture 16: Theorem 1.7, "Let $ \lambda_1,\ldots,\lambda_k$ be distinct eigenvalues of $ T$ (maybe with multiplicity)". The phrase "distinct... maybe with multiplicity" is quite confusing, especially since you're allowing the $ \lambda_i$ to be any set of eigenvalues, not necessarily [i]the [/i]complete set of eigenvalues. I think what you mean to say is "Let $ \lambda_1,\ldots,\lambda_k$ be all the distinct eigenvalues of $ T$, where each $ \lambda_i$ possibly occurs with algebraic multiplicity $ >1$".


linear algebra
matrix
induction
linear algebra unsolved

Let $A$ be fixed matrix with real entries. For arbitrary $n \times n$ matrix $X_0$ define sequence by
\[ X_{n + 1} = X_n A - A X_n \]
Prove that if $X_{n^2} = A$, then $A = 0$. Also prove that if $X_{n^2 - n + 1} = A$, then $A = 0$.

alexilic

lets try some terms of it
$X_0=X_0$
$X_1=X_0A-AX_0$
$X_2=X_1A-AX_1=(X_0A-AX_0)A-A(X_0A-AX_0)= \\ =X_0A^2-AX_0A-AX_0A-A^2X_0= \\ =X_0A^2-2\cdot AX_0A -A^2X_0\\ X_3=(X_0A^2-2\cdot AX_0A -A^2X_0)A - A(X_0A^2-2\cdot AX_0A -A^2X_0)= \\ =X_0A^3-2\cdot AX_0A^2 -A^2X_0A - AX_0A^2-2\cdot A^2X_0A -A^3X_0=\\ =X_0A^3-3\cdot AX_0A^2 -3\cdot A^2X_0A -A^3X_0$
it seems to follow certain pattern...
we'll go one more time
$X_4=(X_0A^3-3\cdot AX_0A^2 -3\cdot A^2X_0A -A^3X_0)A-A(X_0A^3-3\cdot AX_0A^2 -3\cdot A^2X_0A -A^3X_0) =\\ =X_0A^4 - 3\cdot AX_0A^3 -3\cdot A^2X_0A^2 -A^3X_0A -AX_0A^3 -3A^2X_0A^2 -3\cdot A^3X_0A -A^4X_0=\\ =X_0A^4 -4\cdot AX_0A^3 -6A^2X_0A^2 -4\cdot A^3X_0A -A^4X_0$

we can try to make a map of the coefficients...those are important
$1,1$
$1,-2,-1$
$1,-3,-3,-1$
$1,-4,-6,-4,-1$
$....$
so we can see that every if we consider this a matrix of the form
$\left( \begin{array}{cccccc} a_{1,1} & a_{1,2} & ... & a_{1,n} \\ a_{2,1} & a_{2,2} & ... & a_{2,n} \\ . \\ . \\ . \\ a_{n,1} & a_{n,2} & ... & a_{n,n} \end{array} \right)$

we have the following rule for constructing this matrix
$a_{n,m}=a_{n-1,m-1}+a_{n-1,m}$
but it is visible that $a_{i,j}=0$ for $i<j$

all of this could be showed thgrough induction now
we can say the general form for
$X_n=\sum_{i=0}^{n}a_{n,i+1}A^iX_0A^{n-i}$

i made a mistake...
the conclusion was to find out that $A=0$ so if $A=0_{n^2}$
there can be no $A^{-1}$ because $det(A)=0$


Hello!!! I have noticed that my original game is coming to an end, so I am going to start my second game. This is going to be a very fast game. Sign-ups will last a week. Once it has been a week after this post has been made then the game begins. Anybody that has signed up before then will be in the game. Here is how the game will go. The object is to be the last remaining person in the game. You lose if another player requests that you be taken out of the game!!! Once that happens, it is over for you, and you have nothing else to do with the game. However the winner isn't going to be the first person to post a request that everybody else becomes eliminated. This is because no person can be responsible for two consecutive eliminations. At the final three the first person to get rid of somebody else will have to wait ten hours before they can make the winning request. If there are any questions please say so, but the game begins now!!!

sign up

dont quite get the game, but o wel......

the name is ok, i guess

sign up

hmm, I don't get the game either??????

sign up.
 I wish I will not be the first one to lose :D

I think 2lnx is a duplicate, eggylv999.

Sure I will sign up!  Sounds fun to me!

I do have to admit Speed is the Key!!! is kinda... well... drab.. and.... kinda corny mebbe not sure.  Oh well :D.

When the gmae begins it will be a fast game. It will probably only last like three or four days to tell you the truth. Basically everything has its ups and downs. Activity does because if you are not active than you might not even log on before somebody has decided to get rid of you. However, being too active might make you seem like a threat. Basically anybody can choose to kill somebody and eliminate them from the game. However, you can't do it twice in a row. Here is an example of a game played with game.guy, eggy, Joe, and ernie. ernie is the first person to post and he gets rid of eggy. Then game.guy is the next person to post and I get rid of ernie. However, Joe fails to log on before ten hours after my post, so I seize the opportunity and get rid of him, and then I win of course. Still any questions?

Contestants;
Joe10112
e,i,pi
yaofan
eggylv999
ak909090
2lnx
mousy
1995127

can we start? if not, then dont worry about my post; if we did, then look at my choice.

[hide="Choice"]rid ak

srry

random.org[/hide]

Umm, has this game started yet or not? I'll join... If it has I choose
     [hide]2lnx[/hide]

Contestants;
Joe10112
e,i,pi
yaofan
eggylv999
ak909090
2lnx
mousy
1995127
kookamunga
1=2

This game begins one week after the very first post on this thread.

You realize that there are three, and I suspect, four, multiple accounts in this game?

BTW I don't join, I just wish to mention.

Joe10112
yaofan
eggylv999
ak909090
mousy
kookamunga
1=2

what it should be

[quote="1=2"]It should be 2lnx. Who agrees with me?[/quote]

is that your alt.?

Umm well it's been eight days since the first post... so the game has officially started, right?
[b]ak909090[/b]

1=2

u made fun of me in catacombs....:( :mad:

[b]mousy[/b]- for lack of being here+random.org

the players list

brightzhu
Joe10112
yaofan
eggylv999
kookamunga

Three people remain;

kookamunga
yaofan
Joe10112

ZOMGZORD W8 FOR ME!!!! aww it's too late.... eggylv999 voted me out.... I forgot to tune in on this game.

This game was not planned out well. With three people left nobody is gonna post.

lol I pick [b]yaofan[/b]... hmm yeah this game doesn't really work in the end.

Well, yaofan could have gotten rid of you, but he had to wait ten hours after eliminating Joe10112.

Congratulations kookamunga you are the winner of Speed is the Key!!!


linear algebra
matrix
algebra
polynomial

Is there $ N \in M_{2009} (\mathbb{C}) \ni N^{2009} \equal{} E_{1,2009}$
note: $ E_{i,j}$ is a matrix which its (i,j)-th entries are 1, besides are 0

No: $ N$ is necessarily nilpotent, hence the characteristic polynomial of $ N$ is $ \minus{}X^{2009}$. By Cayley-Hamilton, we get $ N^{2009}\equal{}0$.


calculus
derivative
calculus computations

Help me on the nth derivative problem, please.
Find the nth derivative   $ {{\left( \frac {{{e}^{{{x}^{2}}}}}{x} \right)}^{\left( n \right)}}$

What is the derivative by $ n$?  Do you mean the $ n$th derivative?

hello, i think he meant the n-th derivative with respect to $ x$ of $ \frac{e^{x^2}}{x}$.
Sonnhard.


I'm interested in installing some sort of latex compiler on my forum, and I was wondering how I could go about this. Any help would be much appreciated.

That depends what access you have to your server:
1) If you have root access and can install (or have installed) LaTeX, GhostScript, ImageMagick then use [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] as this site does.
2) If you don't have such access then use [url=http://www.forkosh.com/mimetex.html]mimeTeX[/url]. You could use the better quality images of [url=http://www.forkosh.com/mathtex.html]mathTeX[/url] but unless you have LaTeX and dvipng on your server and can install CGI programs, you will get randomly occurring adverts.


I thought this was a cute problem.  If $ X_1,\ldots,X_n$ are independent random variables with $ X_j\to X$ almost surely, then $ X$ is almost surely constant.

Consider the sequence $ (X_n, X_{n\plus{}1})$: on the one hand, if $ (X,Y)\equal{}a.s. \lim (X_n,X_{n\plus{}1})$, then $ X,Y$ are independent. On the other hand, they are equal, hence the only possibility is that they are constants.


[quote="arqady"][quote="bbbbbbbbbb"]
\[ %Error. "eqalign" is a bad command.
{ \&{\rm{Let x}}{\rm{, y}}{\rm{, z > 0 and x+y+z = 1}}{\rm{.}}\cr \&{\rm{Prove that}}\cr \&{{\sqrt{z-xy}}\over x}+{{\sqrt{x-yz}}\over y}+{{\sqrt{y-zx}}\over z}\ge 3\sqrt 2 \cr}\]
[/quote]
Try $ x=y=\frac{5}{12}$ and $ z=\frac{1}{6}.$ :wink:[/quote]
Are u sure ???
We must have z-xy >=0 :)


Αν και είχα σκοπό να τη γράψω στα ελληνικά , δεν ξέρω πώς ξεχάστηκα και  την έγραψα σύντομα στα αγγλικά. Όμως σίγουρα θα σας αρέσει:

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

In the isosceles triangle ABC we have $ AB \equal{} AC , D$ is a point on $ BC , DE$ is perpendicular to $ AB , DZ$ is perpendicular to $ AC , H$ is the point of intersestion of  $ BZ , CE$  and  $ K$  the is the intersection point of  $ AH , EZ$.
 Prove that  $ DK$  bisects $ \angle EDZ$

 [b]ΣXOΛΙΟ[/b]

Έβαζα και το σχήμα αλλά μου βγάζει ότι έφτασα στο Quota limit 10 MB. Ξέρει κανένας σας τι είναι αυτό  και τι πρέπει να κάνω;

[quote="stergiu"][color=red][b]ΑΣΚΗΣΗ[/b][/color]

In the isosceles triangle ABC we have $ AB \equal{} AC , D$ is a point on $ BC , DE$ is perpendicular to $ AB , DZ$ is perpendicular to $ AC , H$ is the point of intersestion of  $ BZ , CE$  and  $ K$  the is the intersection point of  $ AH , EZ$.
Prove that  $ DK$  bisects $ \angle EDZ$[/quote]

[quote="stergiu"]
In the isosceles triangle ABC we have $ AB \equal{} AC , D$ is a point on $ BC , DE$ is perpendicular to $ AB , DZ$ is perpendicular to $ AC , H$ is the point of intersestion of  $ BZ , CE$  and  $ K$  the is the intersection point of  $ AH , EZ$.
 Prove that  $ DK$  bisects $ \angle EDZ$
[/quote]

Ωραίο πρόβλημα κ. Στεργίου. Παραθέτω τη λύση μου.

Κατ' αρχάς πολύ εύκολα $ \angle{EDB}\equal{}90\minus{}B\equal{}90\minus{}C\equal{}\angle{ZDC}$

Από αυτό εύκολα συνάγουμε ότι η $ BC$ είναι η εξωτερική διχοτόμος της γωνίας $ EDZ$. Εστω ότι η $ EZ$ τέμνει την $ BC$ στο $ M$. Τότε λοιπόν το $ M$ θα είναι το ίχνος της εξωτερικής διχοτόμου. Αρα αρκεί να αποδείξουμε ότι $ K$ είναι αρμονικό συζηγές του $ M$ ως προς $ E, Z$.

Πράγματι όμως αυτό ισχύει απ' το θεώρημα που λέει ότι "κάθε διαγώνιος ενός πλήρους τετραπλέυρου τέμνεται αρμονικώς υπό των άλλων 2". Ετσι στο πλήρες τετράπλευρο με διαγώνιους τις $ BC$, $ EZ$, $ AH$  η  $ EZ$ τέμνεται αρμονικώς από την $ AH$ (στο $ K$) κι από την $ BC$ (στο $ M$).

[quote="stergiu"][quote="bilot4"][quote="stergiu"]
In the isosceles triangle ABC we have $ AB \equal{} AC , D$ is a point on $ BC , DE$ is perpendicular to $ AB , DZ$ is perpendicular to $ AC , H$ is the point of intersestion of  $ BZ , CE$  and  $ K$  the is the intersection point of  $ AH , EZ$.
 Prove that  $ DK$  bisects $ \angle EDZ$
[/quote]

Ωραίο πρόβλημα κ. Στεργίου. Παραθέτω τη λύση μου.

Κατ' αρχάς πολύ εύκολα $ \angle{EDB} \equal{} 90 \minus{} B \equal{} 90 \minus{} C \equal{} \angle{ZDC}$

Από αυτό εύκολα συνάγουμε ότι η $ BC$ είναι η εξωτερική διχοτόμος της γωνίας $ EDZ$. Εστω ότι η $ EZ$ τέμνει την $ BC$ στο $ M$. Τότε λοιπόν το $ M$ θα είναι το ίχνος της εξωτερικής διχοτόμου. Αρα αρκεί να αποδείξουμε ότι $ K$ είναι αρμονικό συζηγές του $ M$ ως προς $ E, Z$.

Πράγματι όμως αυτό ισχύει απ' το θεώρημα που λέει ότι "κάθε διαγώνιος ενός πλήρους τετραπλέυρου τέμνεται αρμονικώς υπό των άλλων 2". Ετσι στο πλήρες τετράπλευρο με διαγώνιους τις $ BC$, $ EZ$, $ AH$  η  $ EZ$ τέμνεται αρμονικώς από την $ AH$ (στο $ K$) κι από την $ BC$ (στο $ M$).[/quote]

[b]Δημήτρη , [/b]

 δεν είναι μόνο το πρόβλημα ωραίο , αλλά [color=red][b]πολύ καλή είναι επίσης η λύση σου. :) [/b][/color]

 ΑΝ έχεις χρόνο ,

γράψε μια ακόμα  λύση (αξίζει να την έχουμε στο αρχείο μας) με το θεώρημα CEVA στο τρίγωνο $ AEZ$ με σεβιανές τις $ AK,EC,ZB$ και στη συνέχεια από την ομοιότητα των δύο μικρών ορθογώνιων τριγώνων θα καταλήξεις αμέσως στο αντίστροφο θεώρημα  της εσωτερικής διχοτόμου στο τρίγωνο $ DEZ$.

 Καλό Σαββατοκύριακο !

 Μπάμπης[/quote]

[color=red][b]Ο Δημήτρης έγγραψε και την παρακάτω λύση , η οποία είναι αρκετά διδακτική. Τον ευχαριστώ πολύ .[/b][/color]

[quote="bilot4"] [quote="stergiu"]
[b]Δημήτρη , [/b]

 δεν είναι μόνο το πρόβλημα ωραίο , αλλά [color=red][b]πολύ καλή είναι επίσης η λύση σου. :) [/b][/color]

 ΑΝ έχεις χρόνο ,

γράψε μια ακόμα  λύση (αξίζει να την έχουμε στο αρχείο μας) με το θεώρημα CEVA στο τρίγωνο $ AEZ$ με σεβιανές τις $ AK,EC,ZB$ και στη συνέχεια από την ομοιότητα των δύο μικρών ορθογώνιων τριγώνων θα καταλήξεις αμέσως στο αντίστροφο θεώρημα  της εσωτερικής διχοτόμου στο τρίγωνο $ DEZ$.

 Καλό Σαββατοκύριακο !

 Μπάμπης[/quote]

Πολύ καλή η λύση που προτείνετε κ. Στεργίου (κι αρκετά πιο απλή απ' τη δική μου).

Σύμφωνα λοιπόν με το θεώρημα CEVA στο τρίγωνο $ AEZ$ με σεβιανές τις $ AK,EC,ZB$ παίρνουμε:

$ \frac{EK}{KZ}\cdot\frac{CZ}{CA}\cdot\frac{BA}{BE}\equal{}1\iff$

$ \frac{EK}{KZ}\equal{}\frac{BE}{CZ}$ (1) αφού $ AB\equal{}AC$.

Ομως τα τρίγωνα $ BED, CZD$ είναι όμοια αφού είναι ορθογώνια και $ \angle{B}\equal{}C$.

Αρα $ \frac{BE}{CZ}\equal{}\frac{DE}{DZ}\iff$ (από την (1))

$ \frac{EK}{KZ}\equal{}\frac{DE}{DZ}\iff$ $ DK$ η διχοτόμος της $ \angle{EDZ}$.

Qed
[/quote]


AMC
USA(J)MO
USAMO
inequalities
function
Olimpiada de matematicas

Aki les paso un problema, k a mi pesar no puede llamarse asi, fue de un examen k nos pusieron, y la solucion es horrenda.

Encuentra todos los numeros naturales n, para los cuales existen a1, a2, ..., ak numeros racionales positivos (k>1) que cumplen:
a1+a2+...+ak = (a1)(a2)...(ak) = n

Espero sus soluciones, o al menos sus comentarios sobre este problema.

 :blush:

Pues no me pareció tan horrible la solución así que creería que estoy mal. Pero veamos:
$(a_{1}a_{2}. ... a_{n})/n=(a_{1}+a_{2}+...+a_{n})/n=1$
Luego, por desigualdad media aritmética-geométrica:
$1=(a_{1}+a_{2}+...+a_{n})/n \geq (a_{1}a_{2}. ... a_{n})^\frac1n$
Luego elevando a la n, y sustituyendo
$(a_{1}a_{2}. ... a_{n})/n=1$ tenemos que
$((a_{1}a_{2}. ... a_{n})/n \geq (a_{1}a_{2}. ... a_{n})$
Lo cual implica
$1 \geq n$ Luego n=1

Estas equivocado darth vader por lo que veo estas dejando mal a los siths .

K y n son distintos pues ademas la solucion no esta en el campo del algebra sino en el campo de la teoria de numeros  .

Haber si puedes corregir tu error .

Mi estimado Darth Vader, dejame decirte que tu solucion esta mal, es facil de ver el xq, pues observa que $2+2=(2)(2)=4$ y eso cumple la condicion!!!!
Hace poco me entere que fue USAMO 2006 problema 1 (o 4 nome acuerdo bien)!!!
Sigan intentando, para que vean los verdaderamente horrible que es!!!

jaja   :P  :P

Para $n=pq$ con $p,q>1$ se puede tomar $k=pq-p-q+2=(p-1)(q-1)+1>1$,  $a_{1}=p,\;\;a_{2}=q\;\;a_{i}=1 \;\;\forall i/3\leq i\leq k$.
Para $n=p\geq 11$, siendo $p$ primo, $k=\frac{p-1}{2}-1$, $a_{1}=\frac{p}{2},\;\;a_{2}=\frac{1}{2}\;\;a_{3}=4,\;\;a_{i}=1 \;\;\forall i/4\leq i\leq k$.

Con $1$ y $2$ es fácil ver que no se puede; con $3$ el asunto también parece bastante asequible; con $5$ y $7$  (si es que de verdad no se puede con ellos).... pues sí, la cosa se pone fea; es verdad, tiene toda la pinta de tener que acabar haciendo montones de obscenas acotaciones o quien sabe que otras bajezas fruto de la más absoluta y total depravación, de una mente enferma afectada por un grave complejo de Muirhead debido a toda una larga lista de deseos reprimidos.

Igual mañana me pongo con ello (aunque no sé si en lo de antes habrá algún error).

De todas formas igual otros encuentran una forma bonita de resolverlo; nunca se debe subestimar el poder del lado oscuro.

Muy weno el comentario, y dejame decirte mimoide, que verdaderamente debio de ser d algun enfermo traumado, xq el caso 7 si tiene solucion, y da un montooooooooooooooooon de trabajo encontrar la configuracion!!!
Todo lo demas es weno, la respuesta final es que se puede con
$n=4,6,7,8,9,.....$
Del $1$ al $5$ no se puede(con excepcion del 4) y se demuestra por $MA-MG$.

 :D  :D

[quote="conejita"]
Encuentra todos los numeros naturales $n$, para los cuales existen $a1, a2, ..., ak$ numeros racionales positivos $(k>1)$ que cumplen:
$a_{1}+a_{2}+...+a_{k}= (a_{1})(a_{2})...(a_{k}) = n$

Espero sus soluciones, o al menos sus comentarios sobre este problema.

 :blush:[/quote]
bueno, mm estas seguro que te piden calcular todos los $n$, por que todo depende de $k$, uno puede sacar el valor de $n$ en funcion de $k$, pero de varias formas, lo que si se puede probar es para todo numero $k=par$,  que bien recuerde, de la forma de expresar los $a_{i}$ cono solo numeros $1$ ó $2$.


Baza $A_{1}A_{2}..A_{n}$ a  piramidei $VA_{1}A_{2}..A_{n}$ este un poligon regulat. Daca $m(VA_{1}A_{2})=m(VA_{2}A_{3})=m(VA_{3}A_{4})=...=m(VA_{n-1}A_{n})$, atunci piramida este regulata.

[hide="hint"]Ducem din varf perpendiculara pe fiecare latura a bazei.[/hide]

[quote="perfect_radio"]Se pare că ai mancât [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31926]din ipoteză[/url]. Oricum, aştept şi alte idei.

PS: Am luat 0p la problema asta, ca de altfel mai toti concurenţii (în topicul ăla Valentin menţionează nişte [i]nume grele[/i] care n-au făcut-o).
PPS: Soluţia oficială e aia care apare şi-n Engel.[/quote]

stiam asta, ca "nume grele" nu au facut'o. nu stiu solutia din Engel, hintul e de la solutia mea (la care m'am chinuit ceva).
imi pare rau ca am mai sarit din ipoteza. am scris enuntul din memorie.

Este intr-o gazeta un articol al profesorului Mihai Piticari in care rezolva problema cu analiza matematica.