I guess I am the first one to post here. I am suprised by the fact that no one posted here considering the number of Vietnamese math lovers we have here  :D . Anyway, I just want to meet people who like math including/but not limited to Vietnamese.

Name: Anh Nguyen (Nguyen Duy Anh in the Vietnamese way)
I lived in Vietnam until the age of 13. I am now living in Washington DC. I attended Trung Vuong Middle School - Math academy (you might have heard?). I would love to meet and talk with some people back home  ;)

Hopefully, I will see you in 2007  :lol:  (2007 IMO is in Vietnam)

Asians rule!  :P

You can use Vietnamese here (and it's encouraged actually) ;)

How long have u been in US ?? what high school u go to ?? ma` thi AMC vua roi` duoc bao nhieu diem??


induction
combinatorics proposed
combinatorics

Given an integer $ n\ge 6$. Denote S be the set of $ n$-digit numbers which is divisible by 3 and only contains digits 1,2,3,4,5. Determine |S|.

Let $ a_n$ the amount of $ n$-digits numbers such that the remainder of them mod $ 3$ is $ 0$ and contain those digits.
Similarly $ b_n$ with remainder $ 1$ and $ c_n$ with remainder $ 2$ then $ (|S|\equal{}a_n)$

$ a_1\equal{}1$, $ b_1\equal{}2$ and $ c_1\equal{}2$
with the following recursions:
$ a_{i\plus{}1}\equal{}a_i\plus{}2 b_i\plus{}2c_i$, $ b_{i\plus{}1}\equal{}2a_i\plus{}b_i\plus{}2c_i$ and $ c_{i\plus{}1}\equal{}2a_i\plus{}2b_i\plus{}c_i$
On the other hand: $ a_i\plus{}b_i\plus{}c_i\equal{}5^i$
thus: $ a_{i\plus{}1}\equal{}2 \cdot 5^i \minus{} a_i$

Then by induction $ a_i\equal{}\frac{5^i\plus{}2 \cdot (\minus{}1)^i}{3}$


number theory
greatest common divisor
LaTeX
algebra
polynomial
number theory unsolved

prove that the gcd of

$\frac{a^{p}+b^{p}}{a+b}$ and $a+b$ is $1$ or $p$ where $p$ is an odd prime number and  $\gcd(a,b)=1$.

can anybody write this in latex please

[mod.: $\LaTeX$ed]

I edited your post. But this problem was posted before.

I'm not very sure of the solution:
$ M=\frac{a^{p}+b^{p}}{a+b}=a^{p-1}-a^{p-2}b+a^{p-3}b^{2}-\dots+a^{2}b^{p-3}-ab^{p-2}+b^{p-1}$. Let $d=\gcd(M,a+b)$. Let $M=k(a+b)+r$, $0\leq r<a+b$. Then $d|r$, and consequently $d|r+l(a+b)$. Using $a\equiv(-b)$mod$(a+b)$, we get $M\equiv (-b)^{p-1}-(-b)^{p-2}b+(-b)^{p-3}b^{2}+\dots-(-b)b^{p-2}+b^{p-1}\equiv pb^{p-1}$mod$(a+b)$. We get $d|pb^{p-1}$. From $\gcd(a,b)=1$ results $(b,a+b)=1$, and since $d|(a+b)$, we obtain $\gcd(d,b)=1$, so $d|p$.

Well, that polynomial is irreducible over $\mathbb{Z}[x]$ by Eisenstein Criterion. Now, lets see if we can apply it to the problem.

This is related to the problem so I'll just post a link http://www.mathlinks.ro/Forum/viewtopic.php?t=113283.
Bye


function
algebra
polynomial
number theory

Hey, I've been reading about the number bases in AoPS Volume 1, and I know how to change between bases, but [i]why[/i] exactly?

This is a part of a larger problem I've been having... I actually understand a concept 95% and I can even explain it to others, but I keep telling myself I don't understand it fully.  So instead of accepting certain things as fact, I keep questioning myself.  Understanding something fully to me means understanding it intuitively and logically...I get the logical part, but it's not like a complete, satisfying answer in my head.  I think I need another explanation...

So I'd really appreciate a super detailed explanation, leaving even the obvious stuff in.  :)

yea i sort of have the same question

like... can anyone give an example question where you could probably solve it using other methods, but it'll take a while, vs using number theory or like number bases to solve it much easier and faster?

I know for a fact that a long time ago on the AHSMEs, once in a while they would ask questions about number bases.

[quote="Centy"]Pretty contrived example but:

The function $f : \mathbb{Z}_0 \mapsto \mathbb{Z}_0$ is defined as follows:

$f(1) = 1, f(3) = 3$ and then for every $n \geq 1$, $f(2n) = f(n)$, $f(4n + 1) = 2f(2n + 1) - f(n)$ and $f(4n + 3) = 3f(2n + 1) - 2f(n)$.

Find how numbers $n$ where $1 \leq n \leq 2006$ such that $f(n) = n$.

Try listing the values of $n$ and $f(n)$ is base 10 and then base 2 (binary).[/quote]

I don't really understand...

but how useful are they in problem solving. Like when would you think of using number bases to solve complicated problems unsolvable without using bases?

Some problems are easeir to solve using bases.  An example would be bijection.  ANother would be:
[i] Solve for $a,b,c\in\mathbb{N}$ such that $36a+6b+c=200$.[/i]

Masoud Zargar

Or let $n$ be a positive integer. Find the nmber of polynomials $P(x)$ with coefficients in $\{ 0, 1, 2, 3 \}$ such that $P(4) = n$.

[quote="chronoflare"][quote="brady00n"][quote="b-flat"]Number bases are useful for a lot of things (besides problems in competitions and on tests).  For example, computers are run using binary, and lots of programming languages use hexadecimal.

The point of learning new bases is to see how numbers are created, and how many things work, such as the rule for divisibility by 3.

I'm not really sure what you mean by a complete explanation.[/quote]

By a complete explanation, I mean detailed reasoning behind why the methods of changing bases work.[/quote]

Think of it like this... every number technically could be written as a tally which may become freakishly long. (Example 1024 = 1024 lines)  Thats what a number is.  A number is basically a way of counting something.  And you count something one at a time.  We developed base 10 out of convience.  It speeds up working with numbers by quite a bit... Without it the world would be impractical.
The way a base works is for number abc in base n
abc = a * n^2 + b * n + c.
Now for instance when you have a base 2 number, no longer can you use digits 2 and up. This number can be expressed still as a tally. For instance binary 10000000000 would be 1024 lines.  This is the definition of a base.  Now that tally is worthless, we want to convert it back to base 10.  So we take it and simply groups it within the powers of ten:
1024 - 1*1000 = 24 Number: 1xxx
24 - 0*100 = 24 Number: 10xx
24 - 2*10 = 4 Number: 102x
4 - 4*1 = 0 Number: 1024
Bases are just ways we developed to group the value behind digits.  Nothing more.[/quote]

That really helped.  Let me try to summarize it in order to determine if I completely understand it...

Ok, "numbers" in a strict sense are really entities either tangible or imaginary, which represent discrete quantities.  The $0, 1, 2, 3,...$ is just a visual representation of an entity that can be represented in any shape or form, only having to satisfy a given quantity.  Bases are used to group these entities into discrete "blocks," in order to allow numbers to be written down in a concise format.

For example, if the number base was 4, then only the digits 0-3 would be available for use.  As one counts up from 0 on base 4, one can proceed as: $0, 1, 2, 3$.  However, we must stop here, because only numbers 0-3 are usable.  Therefore, the place to the left of the numbers, which were previously zero, are now available for use.  This expands the original amount of numbers by a factor of 4, because instead of only one digit being able to be used, a second digit can be "stuck" to the left, being either $0, 1, 2, 3$

00  01  02  03
10  11  12  13
20  21  22  23
30  31  32  33

Why can a number in base 4 (or any base $n$), be written as the sum of the succesively larger powers (and their coefficients)?  Because any number can be broken down into discrete blocks.  A large chunk would be removed first, then smaller chunks to accurately get the exact numerical value.  In the numbers above, the first row can be represented as $n\times4^0$ and the entire block can be represented as $n\times4^1$.  Imagine that the process of increasing powers continues, as the number continues to grow.  The numbers directly below 30 would then be 100, 110, 120, 130, 200, 210, 220, 230, 300, 310, 320, 330.  Then this row would run out of numbers.  The total numbers are now 64, another power of 4.

Why does this work?  Because as a number allows itself to have more than one digit, the number of options expands.  In the block shown above, only 1 or 2 digits could be manipulated.  However, in the next set of powers, up to 3 digits can be represented, causing a greater possibility of numbers.

The block is shown once more

0     1     2    3
10  11  12  13
20  21  22  23
30  31  32  33

As the 3rd digit is added to these numbers, another number (that added digit), is free to change from 0-3.  The 1 and 2 digit numbers shown above could have 16 different combinations.  When the 3rd digit is added, it expands the selection to digits that have 1-3 as their first (as of now) digits, in addition to the 0 for the first digit it has above.  Therefore, the total number that can be expressed with an exponent of 2 is 4 times greater than the total number with the exponent one.

Why exactly four?  Because there are only four numerals.  The 1st digit can only go from 0-3, so there are four added options.  Not more, not less.  So every time this happens, a four is multiplied to accomodate the extra choices.  This corresponds to an increase in the exponent value.

Now back to actual numbers...  A number in a given base is basically a selection of what portions of certain blocks of numbers are necessary.  The coefficients in front of the base^exponent show this.

I have a feeling I totally went off on a tangent, but I want to know the fundamentals well.  :oops:


[quote="kunny"]In using $ab+bc+ca=1$, rewrite the given inequality,

$\frac{(a+bc)(b+ca)(c+ab)}{(1+ab)(1+bc)(1+ca)}\geq (\frac{\sqrt{3}+1}{4})^{3}$

$\Longleftrightarrow \frac{(abc-1)^{2}+abc(a+b+c-1)^{2}}{(abc)^{2}+abc(a+b+c)+2}\geq \left(\frac{\sqrt{3}+1}{4}\right)^{3}$.[/quote]

I don't understand! :oops:


Don't they usually have "A is to B as C is to D" type of problem?

Anyway "3@4 is to 4@3 as 7@98 is to...?"

um...i'm just wondering, is this another trick question, like the 1=5, 2=25 ... thing?

well, if just regularly, we get 98@7 but that is too easy, it multiple choice, since too many ways to look at this, like time 4^2/3^2 or just reverse or....

Well, made you think...

In other words can you say a function is symmetric if f(a)=b and f(b)=a?


There is a party going on. There are a number of people in the party. Some of them know each other, some others don't. But each person in the party knows at least one other person. Now a celebrity enters the party, everybody knows the celebrity but the celebrity knows no one in the party. After some time, you enter the party. You are allowed to ask only one question to each person in the party. The question should be of the form "Do you know this person?". "This person" could be anyone you choose from the party. How can you figure out who is the celebrity?  :wink:

[hide]To simplify things, consider asking person A if he knows person B.  If he does, then person A is not the celebrity.  If he doesn't, then person B is not the celebrity.  Therefore when we ask person A if he knows person B, essentially we find out that either A or B is not the celebrity.

At any point in the question-asking process, there are four types of people:

1)  Unknown identity, has not been asked a question
2)  Unknown identity, has been asked a question
3)  Known not to be the celebrity, has not been asked a question
4)  Known not to be the celebrity, has been asked a question

We can throw away all the people of type 4, since they are useless to us, so we only have to worry about types 1-3.  Now since at each step we ask one person a question and find out one person who is not the celebrity, the numbers of people of types 2 and 3 are equal, say they are both $ n$.  We claim that we can ask questions in such a way that $ n$ never exceeds 1.  Asking one question cannot increase $ n$ by more than 1, so when $ n \equal{} 0$ we can ask any question we want using two people of type 1.  Now suppose $ n \equal{} 1$.  Suppose person A is the person of type 2, and person B is the person of type 3.  We ask some person C of type 1 if he knows A.  If we find out that A is not the celebrity, then A becomes type 4 and C becomes type 2, so $ n$ remains at 1.  If we find out that C is not the celebrity, then A remains type 2 and C becomes type 4.  Here $ n$ again remains at 1.

We continue asking questions in this way until we are left with only two people not of type 4.  If we are left with two people X and Y of type 1, we ask X if he knows Y - if he does, Y is the celebrity, if not, X is the celebrity.  If we are left with people of types 2 and 3, we know the person of type 2 is the celebrity, and we're done.

Hope this made sense  :P .[/hide]

Nice explanation! I like your solution matt. Here is mine... a bit more wordy

[hide="Solution"]
Let all the people in the party be named A1, A2, A3 ... An.

Let us start by assuming that someone, say Ak is a celebrity. We want to confirm that Ak is indeed a celebrity. One way of confirming this is by asking all the remaining (n-1) people the question "Do you know Ak" and receiving a "Yes" as an answer from all of them. However it is quite possible that the person Ak may not be the celebrity to start with. In that case we will have to change our "assumed celebrity" to be a different person.

Here is the sequence of steps to be followed to identify the celebrity:

Step 1: Ask A1 "Do you know A2?"

Case 1) the answer is "yes":
Cross out A1 from further consideration, because A1 cannot be the celebrity (celebrity knows no one)! Then ask any one of the remaining (n-2) people, say A3, if he/she knows A2. If the answer is “no” Cross out A2 from further consideration, because A2 cannot be the celebrity (every one knows the celebrity)! If the answer is "Yes" cross out A3 from further consideration, because A3 cannot be the celebrity (celebrity knows no one)! At this point we have (n-2) people with A2 as the "assumed celebrity". This is reduced to the original problem with (n-2) people in the party.
Continue this process of elimination of people untill you are left with 3 people in the list of consideration.

Case 2) the answer is "no":
Cross out A2 from further consideration, because A2 cannot be the celebrity (every one knows the celebrity)! Then ask any one of the remaining (n-2) people, say A3, if he/she knows A1 If the answer is “yes” cross out A3 from consideration (celebrity knows no one)! If the answer is "no" cross out A1 from further consideration, because A1 cannot be the celebrity (every one knows the celebrity)! At this point we assign our "assumed celebrity" to be “A3”. Now we have (n-2) people with A3 as the "assumed celebrity". This is reduced to the original problem with (n-2) people in the party.
Continue this process of elimination of people untill you are left with 3 people in the list of consideration.

Step 2: When you are left with 3 people say An-2, An-1 and An with An-2 as the "assumed celebrity", ask An-1 "Do you know An-2?"

Case a) the answer is "yes":
Cross out An-1 from consideration. Now ask An "Do you know An-2?", if the asnwer is yes An-2 is the Celebrity, otherwise An is the Celebrity.

Case b) the answer is "no":
Cross out An-1 from consideration. Now ask An "Do you know An-1?", if the answer is yes An-1 is the celebrity, otherwise An is the Celebrity.

[/hide]

Can anyone give a more concise explanation?


This was confusing me so any help would be appreciated:

Use a suitable test to determine whether the following series converges or diverges:

sigma starting at j=2 and going to infinity of:
1/(j^4 - j)

using limit comparison test and p-series test, $\lim_{n \to \infty}\frac{\frac{1}{n^{4}-n}}{\frac{1}{n^{4}}}= 1$, so $\sum_{n=2}^{\infty}\frac{1}{n^{4}-n}$ converges


let $f(x)$ be continue in$[0;1]$ and $f(0)=0$ and $f(1)=1$
  and satisfying $f(\frac{2x+y}{3}) =\frac{5f(x)+f(y)}{6}$  for all $x \geq y \in [0;1]$
 find the value of$f(\frac{8}{23})$

Posted before. But function, satisfied equation:
$f(ax+(1-a)y)=bf(x)+(1-b)f(y)$ can be continiosly if and only if
1. f(x)=const,
or
2. a=b.
You have $a=\frac{2}{3}\not =b=\frac{5}{6}$. Therefore continiosly function is const.

why ?
 i think that  f(x) =x is right .
    can you  say why f(x ) = const .
 thanks

[quote="tranthanhnam"]why ?
 i think that  f(x) =x is right .
[/quote]
I don't think that f(x)=x is right
LHS=$\frac{2x+y}{3}$,but
RHS=$\frac{5x+y}{6}$

Rust is right,f(x)=const

no, I think Rust don't say right.
BEcause, we prove that exist enique function f satisfied.
let g(x)=[ f(8/27+4x/27)-f(8/27)]/[f(4/9)-f(8/27)]=Af(Bx+C)+D
Thus,f=g
WE can caculate f(8/23)

It is posted before, were I prove, that for any y: $-\infty \le y\le+\infty$ exist sequence $x_{n}\to x_{0}=\frac{8}{23}$, suth that $f(x_{n}) \to y$.

I think that f is enique,g defined g(0)=0 g(1)=1 and $g(\frac{2x+y}{3}) =\frac{5g(x)+g(y)}{6}$  for all $x \geq y \in [0;1]$  thus f=g $x \geq y \in [0;1]$
=> caculate g(x)


Hello
$ 4^{\minus{}2x\minus{}8}\equal{}4^{4x\plus{}10}$
$ x\equal{}\minus{}3$
Thank u.

hello, alternatively we can write $ (2^{ \minus{} 2})^{2x \plus{} 8} \equal{} (2^4)^{2x \plus{} 5} \Leftrightarrow 2^{ \minus{} 2x \minus{} 16} \equal{} 2^{8x \plus{} 20} \Leftrightarrow \minus{} 4x \minus{} 16 \equal{} 8x \plus{} 20 \Leftrightarrow \minus{} 12x \equal{} 36$ , and hence $ \boxed{x \equal{} \minus{} 3}$.
Sonnhard.


calculus
integration
algebra
polynomial
LaTeX
calculus computations

Hey everybody, I'm new to the forum.

Can somebody please help me out with this problem? I've tried a few different 'u substitions' but I can't figure this one out.

Edit: I don't know how to use the code to make it look like an actual problem either.

integral of   [dx]/[(x^(1/3))+(x^(1/2))]

Use $ x\equal{}t^6$. Then the integral becomes $ \int\frac{6t^5}{t^3\plus{}t^2}\,dt$. Now, I hope you know how to integrate a rational function.

Thanks, I think I figured it out with your help.

I did polynomial long division and got

6 times the integral of [u^2 - u - 1 -((u^2)/(u^3 + u^2))]du

integrated then substituted x^(1/6) back in for u.

then my final answer is

2(x^(1/2)) - 3(x^(1/3)) - 12(x^(1/6)) - 6ln|x^(1/6)| + C

Almost. The last two terms should be $ \plus{}6x^{1/6}\minus{}6\ln|x^{1/6}\plus{}1|$.

[quote="whitesox23"]Edit: I don't know how to use the code to make it look like an actual problem either.[/quote]

That's [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX[/url]. The code
\int \frac{dx}{x^{1/3}+x^{1/2}}
produces
$ \int \frac{dx}{x^{1/3}+x^{1/2}}$
when placed between dollar signs. There are numerous tutorials on how to use $ \text{\LaTeX}$, and you can also see other people's code by hovering over the pictures or quoting the posts.


ABC is a triangle with [AB] = 7 and  [AC] = 9.

D is a point on AB such that [BD]=3.[DE] and a line segment from D to AC intersects AC at the point E.

Area of the tetragon BCDE is 5/7 of the triangle ABC... Find [CE]=?

isn't there anyone who can solve this problem?

Are you sure there is nothing missing?

[BD]=3.[DE] that is a product , or [BD]=3?

I think that he accidentally left a space out, and [BD]=3. If [BD]=3*[DE], then I can construct two triangles with differing lengths of [CE].
Now the problem isn't too hard.


Prove that [tex]15x^2-7y^2=9[/tex] has no integer solutions.

Clearly $y$ is a multiple of $3$. Thus, $15x^2$ is a multiple of $9$ so that $x$ is a multiple of $3$.
Let $x=3a,y=3b.$
The equation rewrite $15a^2-7b^2 = 1$ or, equivalentely $15a^2 = 1+7b^2.$
It follows that $1+7a^2 = 1+2a^2 = 0 \mod [5].$
But a square is $0$ or $1$ or $-1$ modulo $5$, which leads to an impossibility.

Pierre.

you could use modulo 5 directly:
\[ 15x^2 - 7y^2 = 9 \mod [5] \]
\[ 7y^2+9 = 0 \mod  [5] \]
\[ 2y^2 = 1 \mod  [5] \]
which is impossible as you stated.


geometry
perimeter
calculus
geometry proposed

Given a triangle $ABC$ with side length $BC=a,CA=b,AB=c\ (a>b>c)$.
Two points $P,Q$ lie on the perimeter of the triangle.
Find the minimum value of $\overline {PQ}$ such that the area of the triangle is bisected by the segment $PQ$.

The minimum is $\sqrt{\dfrac{c^2 - (a-b)^2}{2}}$

Let $P \in b, Q \in a$.

The two triangles $ABC,PQC$ have a common angle in $C$.

$(PQC) = \dfrac{1}{2} \cdot CP \cdot CQ \cdot sinC$

$(ABC) = \dfrac{1}{2} \cdot a \cdot b \cdot sinC$

$(ABC) = 2(PQC) \Rightarrow$

$a \cdot b \cdot sinC = 2 \cdot CP \cdot CQ \cdot sinC$

since $sinC \neq 0$ we have:

$ab = 2 \cdot CP \cdot CQ \ \ \ (1)$

So we have to minimize the distance $PQ$ when $(1)$ holds.

[hide="Doing a little calculus we find that"]

$PQ$ has the minimum length when $CP=CQ$.

Let's prove it:

$PQ^2 = CP^2 +CQ^2 -2 \cdot CP \cdot CQ \cdot cosC$

Let $x=CP$.

Then $CQ = \dfrac{ab}{2x} = \dfrac{k}{x}$, where $k = \dfrac{ab}{2}$

$PQ^2 = x^2 +(\dfrac{k}{x})^2 -2k \cdot cosC$

$PQ^2 = x^2 +(\dfrac{k}{x})^2 -2k +2k -2k \cdot cosC$

$PQ^2 = (x -\dfrac{k}{x})^2 +2k \cdot (1 - cosC) \ \ \ (2)$

Since $2k \cdot (1 - cosC)$ is constant, we have a minimum value for $PQ^2$ when $x^2 = k = \dfrac{ab}{2}$

$x^2 = k \Rightarrow x = \sqrt{k} \Rightarrow \dfrac{k}{x} = \sqrt{k} \Rightarrow x = \dfrac{k}{x} \Rightarrow CP=CQ$

So, we have the minimum length of $PQ$ when $CP = CQ$.

Let's find this minimum.

Applying $x = \dfrac{k}{x}$ in $(2)$ we get:

$PQ^2 = 2k(1 - cosC)$, where $k = \dfrac{ab}{2}$

So ${PQ^2}_{min} = ab(1-cosC)$

This is the minimum value of $PQ^2$ when $P,Q$ are on the sides $a$ and $b$.

Working in the same way, we can find that the minimum of $PQ^2$ is:

$bc(1-cosA)$, if $P,Q$ are on sides $b$ and $c$

$ca(1-cosB)$, if $P,Q$ are on sides $c$ and $a$

 :mad:
Unfortunately, we can't find the smallest value of $ab(1-cosC), \ bc(1-cosA), \ ca(1-cosB)$
This is because:

$c<b<a \Rightarrow C<B<A \Rightarrow cosC>cosB>cosA \Rightarrow$

$1-cosC < 1-cosB < 1-cosA$

and $ab>ca>bc$

(the product of the two other sides decreases and $1-cosX$ increases when we choose the angles $C,B,A$)

So we should express this minimums by a different way:

Let's see the first case, when $P,Q$ are on sides $a$ and $b$. I'll note this minimum as $min_C$.

$min_C = ab(1-cosC) = ab \dfrac{1-cos^2C}{1+cosC} = ab\dfrac{sin^2C}{1+cosC}$

Now, recall that $sinC = \dfrac{c}{2R} \Rightarrow sin^2C = \dfrac{c^2}{4R^2} \Rightarrow$

$min_C = \dfrac{abc^2}{4R^2(1+cosC)} = \dfrac{abc}{4R^2} \cdot \dfrac{c}{1+cosC} \Rightarrow$

$min_C = \dfrac{S}{R} \cdot \dfrac{c}{1+cosC}$ , where $S = \dfrac{abc}{4R}$ is the area of the triangle

Similarly, we have:

$min_A = \dfrac{S}{R} \cdot \dfrac{a}{1+cosA}$

$min_B = \dfrac{S}{R} \cdot \dfrac{b}{1+cosB}$

Now, it's easy to compare the minimum values in every case, since

$a>b>c$ and

$1+cosA < 1+cosB < 1+cosC$

So we find that $min_A > min_B > min_C$. Hence,[/hide]
the minimum value of $PQ^2$ is $ab(1-cosC)$, hence the minimum length of $PQ$ is $\sqrt{ab} \sqrt{1-cosC}$, and this happens when $P,Q \in a,b$

Since $cosC$ is not given, we should express it in terms of $a,b,c$.

[hide="But that is the easy part."]
$c^2 = a^2 +b^2 -2ab \cdot cosC$

$c^2 = a^2 +b^2 -2ab +2ab -2ab \cdot cosC$

$c^2 = (a-b)^2 +2ab(1-cosC)$

$c^2 - (a-b)^2 = 2ab(1-cosC)$

$ab(1-cosC) = \dfrac{c^2 - (a-b)^2}{2}$, which always positive since $c>|a-b|$[/hide]

So finally we have that the minimum length of $PQ$ is $\sqrt{\dfrac{c^2 - (a-b)^2}{2}}$

We can easily (using ruler and compass) find the positions of $P$ and $Q$ on the sides $a$ and $b$, since we know that

$CP^2 = CQ^2 = \dfrac{ab}{2} = b \cdot \dfrac{a}{2}$

That is, $CP$ is the geometric mean of $b$ and $\dfrac{a}{2}$.


Well TT, as Meghrabi said... Welcome.

This is how I would do it...
Height=$h$
Radius=$(h-7)$

Now the formula for the total surface area of a cyilinder is
$2\pi rh+2\pi r^2$ We will factor it a little bit to avoid having $\pi$ everywhere.

$2\pi(rh+r^2)$ Lets substitute

$2\pi[(h-7)h+(h-7)^2]=1248\pi$
$h^2-7h+h^2-14h+49=624$
$2h^2-21h-575=0$ Ok now just solve this equation to find $h$ and then find $r$.

hello!

for this problem, all you need to do is set up the equation correctly

maint thing you need to worry is plugging in (h-7) instead  ;)

[quote="Meghrabi"]Welcome

height: h
radius:  h-7 (7 feet less then the height)

The surface of a cilinder is (with r the radius of the base, and h the height) 2*r*pi*h=2*(h-7)*pi*h ;)[/quote]

I've forgotten to add  2*r²*pi  :blush:

[quote="LordoftheMorons"][quote="TT"]I'm new here so if this is in the wrong place my bad but w/e..I'm homeschooling in geometry, 10th grade, and it's just hard to teach myself everything from the book. This is probably really easy, but I'm kinda stuck on this problem:

The total area of a right circular cylinder is 1248 pi square feet. The radius of the base of the cylinder is 7 feet less than the height. Find the radius and the height of the cylinder.

Can someone walk me through it? Thanks...[/quote]

Buy the AoPS books. They're much more helpful than normal textbooks.[/quote]

They're more advanced as well. AoPS books are for those who want to do competitions and work with problems that take much more thought than a question from school would.

For school, a normal textbook would be fine. For everything else, theres AoPS.

Yes, maybe start with a book less advanced that AoPS and then you see...

I find Introduction to the Probabilities a little easier, but that one doesn't have any geometries

I didn't get the new geometry book yet :?


let $ f: l^2 \to R$ defined by $ f(x_1,x_2,...)\equal{}\sum_{n\equal{}1}^{\infty} \frac{x_n}{2^{n/2}}$ for $ (x_1,x_2,...)\epsilon l^2$ then what will be norm of $ f$

I think the norm is $ 1$.
We have,

$ |f(x)|\leq \sqrt{\sum|x_{n}|^{2}}\sqrt{\sum_{n\equal{}1}^{\infty}\frac{1}{2^{n}}}\equal{}\parallel{}x\parallel{}$, which implies that $ \parallel{}f\parallel{}\leq 1$.

On the other hand, consider the sequence $ x_{n}\equal{}\frac{1}{2^{n/2}}$.

A calculation shows that $ f(x)\equal{}1\equal{}\parallel{}x\parallel{}$ and hence $ \parallel{}f\parallel{}\geq 1,$ and the result follows.

I think didilica's solution is correct. I wanted to add I found it helpful since I'm trying to better understand linear functionals.

[quote="didilica"]I A calculation shows that $ f(x) \equal{} 1 \equal{} \parallel{}x\parallel{}$ and hence $ \parallel{}f\parallel{}\geq 1,$ and the result follows.[/quote]

What's the calculation which shows the above?

[quote="limsup"]
What's the calculation which shows the above?[/quote]

This is just a calculation of geometric series. What is the problem you met?

$ x_{n}\equal{}\frac{1}{2^{n/2}}$

Plug it back in.

[quote="limsup"]What's $ x$?[/quote]
$ x \equal{} (x_1,x_2,\cdots,)$, with $ x_n \equal{} \frac{1}{2^{n/2}}$.


Prove $ f\equal{} \sigma T^4$ σ= Stefan-Boltzmann , T=Temprature


Is log function log(x) taught in Middle school?
If yes, which grade will it be taught?

Depends on your school system and teacher. I taught logs in 8th grade Algebra 2. This doesn't mean all teachers of 8th grade Algebra 2 will do so. Logs are actually fairly simple. If you want to learn about them, I recommend chapter 1 of AoPS Volume 1.

Unfortunately, Algebra 2 in quite a few schools is at late as 9th, 10th, or in some horrendous cases, 11th grade.

Thanks for the info.
I have AOPS Volume 1 and Intro to Algebra books.
I tried to read as much as I can. But I don't have all the time I want.

My main goal for now is to prepare for mathcounts contests.
Is log function tested in Mathcounts?
If not, I will study it in next summer.
If yes, I will study it this year.

I haven't seen it in any MathCounts questions, and if it is, it is very basic.

eh i think thats algebra 2 for me thats not middle school

oh yeah this is for hide and go seek(ignore me if u dont know what im talking about)


search
quadratics
number theory unsolved
number theory

$ a \in N$, for all prime $ p$, $ a$ is a square mod $ p$. Show that $ a$ is a square.

see the last post in
http://www.mathlinks.ro/viewtopic.php?search_id=307244551&t=34935

Oh! I think there is a more elementary proof. With quadratic reciprocity, it is certainly done!

Isn't quadratic reciprocity the most elementary theorem that deals with squares $ \mod p$¿

Could you not just pick a prime $ p$ bigger than $ a$ and be done instantly?

Of course not.  $ 2$ is a square $ \bmod 7$, but that doesn't imply that $ 2$ is a square.

that's a nasty expression, even nastier than picking $ p$ bigger than $ a$.

See my second post in http://www.mathlinks.ro/Forum/viewtopic.php?t=64322 for a proof not needing Dirichlet's theorem.


Let $x,y,z$ be positive real numbers. Prove that:
$\displaystyle \frac {2x^2(y+z)}{(x+y)(x+z)}+\frac {2y^2(x+z)}{(y+x)(y+z)}+\frac {2z^2(x+y)}{(z+x)(z+y)}\leq x+y+z$

I've a soln, but it isn't nice.  :?
Can you post the soln??

i guess i have the same solution :D (multiply with (x+y)(x+z)(y+z), simplify, use $x^3y+xy^3\geq 2x^2y^2$).


[quote="hyderman"]hello

how to solve this

Combin(n,2)=28

thanx[/quote]

This doesn't really belong in Intermediate forum, I think  :wink:

[hide="solution"]$\binom{n}{2}=28$

$\frac{n!}{2!(n-2)!}=28$

$\frac{n(n-1)}{2}=28$

$n(n-1)=56$

$n^{2}-n-56=0$

$(n+7)(n-8)=0$

Since $n>0$, $n=\boxed{8}$.[/hide]
[hide="answer only"]8[/hide]


Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
I know if I wanted to express the sum of many terms, I could express it as,
$ {{\sum_{i \equal{} m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}} \plus{} {{x}_{m \plus{} 1}} \plus{} {{x}_{m \plus{} 2}} \plus{} ... \plus{} {{x}_{n \minus{} 2}} \plus{} {{x}_{n \minus{} 1}} \plus{} {{x}_{n}}$

In addition, I know if I wanted to express the product of many terms, I could express it as,
$ {{\prod_{i \equal{} m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}\cdot{x}_{m \plus{} 1}\cdot{x}_{m \plus{} 2}\cdot...\cdot{x}_{n \minus{} 2}\cdot{x}_{n \minus{} 1}\cdot{x}_{n}}$

However, how would I express many terms which are equal to each other? In other words, what is the correct expression for the following,
$ {\left({{\bigsqcup_{i \equal{} m}^{n}}{{x}_{i}}}\right)} \equal{} {\left({{x}_{m} \equal{} {x}_{m \plus{} 1} \equal{} {x}_{m \plus{} 2} \equal{} ... \equal{} {x}_{n \minus{} 2} \equal{} {x}_{n \minus{} 1} \equal{} {x}_{n}}\right)}$

I know: $ {\sum}$ is for adding and $ {\prod}$ is for multiplication; and in my example above I used $ {\bigsqcup}$, but what is the correct symbol for "equals"?

Also, is continuous subtraction of terms just a modification of the summation of terms?
In addition, is continuous division of terms just a modification of the product series of terms?

Lastly, besides the three types of ways of organizing terms as listed above (sum, product, and equals), are there any other ways of organizing terms?

[b]2. Relevant equations.[/b]
Not really calculus related, but I started covering this type of stuff in my calculus class.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
Yea, not really sure here.
[/hide]
Thanks,

-APSStudent

[quote="APSStudent"]how would I express many terms which are equal to each other? In other words, what is the correct expression for the following,
$ {\left({{\bigsqcup_{i \equal{} m}^{n}}{{x}_{i}}}\right)} \equal{} {\left({{x}_{m} \equal{} {x}_{m \plus{} 1} \equal{} {x}_{m \plus{} 2} \equal{} ... \equal{} {x}_{n \minus{} 2} \equal{} {x}_{n \minus{} 1} \equal{} {x}_{n}}\right)}$[/quote]
There is, as far as I know, no particularly universal notation for this, since it's not a very important concept.  It expresses a simple relationship rather than a function.  The notation $ x_1 \equal{} x_2 \equal{} ... \equal{} x_n$ is concise and unambiguous.

[quote="APSStudent"]Also, is continuous subtraction of terms just a modification of the summation of terms?
In addition, is continuous division of terms just a modification of the product series of terms?[/quote]
I'm not sure what you mean by either of these things.  Do you have examples?

[quote="APSStudent"]Lastly, besides the three types of ways of organizing terms as listed above (sum, product, and equals), are there any other ways of organizing terms? [/quote]
Again, I'm not sure what you're asking here.  Sum and product are your basic (binary) field operations.  Generically, you've always got union, intersection, and symmetric difference (the first two of which have symbols $ \bigcup$ and $ \bigcap$ respectively), which apply to sets of elements rather than elements.

Hey,
[quote="t0rajir0u"]
I'm not sure what you mean by either of these things.  Do you have examples?
[/quote]
Like how would you express the following,
$ {{x}_{m}}\minus{}{{x}_{m\plus{}1}}\minus{}{{x}_{m\plus{}2}}\minus{}...\minus{}{{x}_{n\minus{}2}}\minus{}{{x}_{n\minus{}1}}\minus{}{{x}_{n}} \equal{} {???}$
and
$ {{x}_{m}\div{x}_{m\plus{}1}\div{x}_{m\plus{}2}\div...\div{x}_{n\minus{}2}\div{x}_{n\minus{}1}\div{x}_{n}} \equal{} {???}$

[quote="t0rajir0u"]
Again, I'm not sure what you're asking here.  Sum and product are your basic (binary) field operations.  Generically, you've always got union, intersection, and symmetric difference (the first two of which have symbols $ \bigcup$ and $ \bigcap$ respectively), which apply to sets of elements rather than elements.
[/quote]
Yup that's all I am familiar with as well - the basic operations of: sum and product; and the operations of union and interesect with respect to sets.

Thanks,

-APSStudent

[quote="APSStudent"]Like how would you express the following,
$ {{x}_{m}} \minus{} {{x}_{m \plus{} 1}} \minus{} {{x}_{m \plus{} 2}} \minus{} ... \minus{} {{x}_{n \minus{} 2}} \minus{} {{x}_{n \minus{} 1}} \minus{} {{x}_{n}} \equal{} {???}$[/quote]
As $ x_m\minus{}\sum_{k\equal{}m\plus{}1}^nx_k.$

[quote]and
$ {{x}_{m}\div{x}_{m \plus{} 1}\div{x}_{m \plus{} 2}\div...\div{x}_{n \minus{} 2}\div{x}_{n \minus{} 1}\div{x}_{n}} \equal{} {???}$[/quote]
I wouldn't write that in the first place. Do you notice how seldom anyone who has taken an algebra course uses the symbol $ \div?$

I could of course do essentially the same as above and write $ x_m\prod_{k\equal{}m\plus{}1}^n\frac1{x_k}$

One reason sigma and pi notation is unambiguous is that addition and multiplication are both associative, which means it doesn't matter how you group the operations:  $ (x_1 \plus{} x_2) \plus{} x_3$ is the same thing as $ x_1 \plus{} (x_2 \plus{} x_3)$, and so forth.

Division and subtraction do not share this property; parentheses matter.  You will rarely find the need to use repeated subtractions and divisions in that manner.  Really, once you learn more algebra you'll come to realize that the basic arithmetic operations are [b]not[/b] addition, multiplication, subtraction, and division; they are addition, multiplication, additive inverse, and multiplicative inverse.  That shift in perspective will help.

[quote="t0rajir0u"]Really, once you learn more algebra you'll come to realize that the basic arithmetic operations are [b]not[/b] addition, multiplication, subtraction, and division; they are addition, multiplication, additive inverse, and multiplicative inverse.[/quote]
Thanks. That was essentially my point but you said it better.

Hey,
[quote="Kent Merryfield"]
As $ x_m \minus{} \sum_{k \equal{} m \plus{} 1}^nx_k.$
[/quote]
Thanks,
$ {{x}_{m} \minus{} {\sum_{i \equal{} m \plus{} 1}^{n}}{{x}_{i}}} \equal{} {{{x}_{m}} \minus{} {{x}_{m \plus{} 1}} \minus{} {{x}_{m \plus{} 2}} \minus{} ... \minus{} {{x}_{n \minus{} 2}} \minus{} {{x}_{n \minus{} 1}} \minus{} {{x}_{n}}}$

[quote="Kent Merryfield"]
I wouldn't write that in the first place. Do you notice how seldom anyone who has taken an algebra course uses the symbol $ \div?$
I could of course do essentially the same as above and write $ x_m\prod_{k \equal{} m \plus{} 1}^n\frac1{x_k}$
[/quote]
I agree, but I thought using,
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{1\div{x}_{i}}} \equal{} {{x}_{m}\div{x}_{m \plus{} 1}\div{x}_{m \plus{} 2}\div...\div{x}_{n \minus{} 2}\div{x}_{n \minus{} 1}\div{x}_{n}}$

was more clear/concise as opposed to,
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{1/{x}_{i}}} \equal{} {{x}_{m}/{x}_{m \plus{} 1}/{x}_{m \plus{} 2}/.../{x}_{n \minus{} 2}/{x}_{n \minus{} 1}/{x}_{n}}$
or
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{\frac {1}{{x}_{i}}}} \equal{} {{x}_{m}\cdot\frac {1}{{x}_{m \plus{} 1}}\cdot\frac {1}{{x}_{m \plus{} 2}}\cdot\frac {1}{{x}_{m \plus{} 3}}\cdot\ldots\cdot\frac {1}{{x}_{n \minus{} 2}}\cdot\frac {1}{{x}_{n \minus{} 1}}\cdot\frac {1}{{x}_{n}}}$

So, of the above three equations which would be the most conventional/rigorous way?

[quote="t0rajir0u"]
One reason sigma and pi notation is unambiguous is that addition and multiplication are both associative, which means it doesn't matter how you group the operations:  $ (x_1 \plus{} x_2) \plus{} x_3$ is the same thing as $ x_1 \plus{} (x_2 \plus{} x_3)$, and so forth.
Division and subtraction do not share this property; parentheses matter.
[/quote]
Ah, ok that makes sense for why there isn't a significant need for the equivalent of sigma notation and pi notation for: subtraction and division.

[quote="t0rajir0u"]
You will rarely find the need to use repeated subtractions and divisions in that manner. Really, once you learn more algebra you'll come to realize that the basic arithmetic operations are [b]not[/b] addition, multiplication, subtraction, and division; they are addition, multiplication, additive inverse, and multiplicative inverse.  That shift in perspective will help.
[/quote]
Thanks for the insight.

Thanks,

-APSStudent

[quote="APSStudent"]
was more clear/concise as opposed to,
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{1/{x}_{i}}} \equal{} {{x}_{m}/{x}_{m \plus{} 1}/{x}_{m \plus{} 2}/.../{x}_{n \minus{} 2}/{x}_{n \minus{} 1}/{x}_{n}}$
[/quote]
I [i]really[/i] wouldn't write that. It reeks of ambiguity and missing parentheses. Again, t0riraji0u's point holds: division is just not a fundamental operation.

On the other hand, [quote="APSStudent"]$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{\frac {1}{{x}_{i}}}} \equal{} {{x}_{m}\cdot\frac {1}{{x}_{m \plus{} 1}}\cdot\frac {1}{{x}_{m \plus{} 2}}\cdot\frac {1}{{x}_{m \plus{} 3}}\cdot\ldots\cdot\frac {1}{{x}_{n \minus{} 2}}\cdot\frac {1}{{x}_{n \minus{} 1}}\cdot\frac {1}{{x}_{n}}}$[/quote] is quite clear.  "Rigor" doesn't apply to notation, though: "consistency" and "clarity" are what one should strive for, with "agreement with standard conventions" frequently an important part of both.

[quote="Kent Merryfield"]
I [i]really[/i] wouldn't write that. It reeks of ambiguity and missing parentheses. Again, t0riraji0u's point holds: division is just not a fundamental operation.
[/quote]
I agree as t0rajir0u pointed out division does not follow the associative property, therefore without parentheses the following are ambiguous:
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{1\div{x}_{i}}} \equal{} {{x}_{m}\div{x}_{m \plus{} 1}\div{x}_{m \plus{} 2}\div...\div{x}_{n \minus{} 2}\div{x}_{n \minus{} 1}\div{x}_{n}}$
and
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{1/{x}_{i}}} \equal{} {{x}_{m}/{x}_{m \plus{} 1}/{x}_{m \plus{} 2}/.../{x}_{n \minus{} 2}/{x}_{n \minus{} 1}/{x}_{n}}$
[quote="JBL"]
"Rigor" doesn't apply to notation, though: "consistency" and "clarity" are what one should strive for, with "agreement with standard conventions" frequently an important part of both.
[/quote]
Ok, that make sense. However, since division is not associative, technically shouldn't the following have parentheses like,
$ {{x}_{m}\cdot\prod_{i \equal{} m \plus{} 1}^{n}\frac {1}{{x}_{i}}} \equal{} {{x}_{m}\cdot\left({\frac {1}{{x}_{m \plus{} 1}}}\right)\cdot\left({\frac {1}{{x}_{m \plus{} 2}}}\right)\cdot\left({\frac {1}{{x}_{m \plus{} 3}}}\right)\cdot\,\dots\,\cdot\left({\frac {1}{{x}_{n \minus{} 2}}}\right)\cdot\left({\frac {1}{{x}_{n \minus{} 1}}}\right)\cdot\left({\frac {1}{{x}_{n}}}\right)}$

Or is it understood that convention dictates we interpret the below as if it was above?
$ {{x}_{m}\cdot\prod_{i \equal{} m \plus{} 1}^{n}\frac {1}{{x}_{i}}} \equal{} {{x}_{m}\cdot\frac {1}{{x}_{m \plus{} 1}}\cdot\frac {1}{{x}_{m \plus{} 2}}\cdot\frac {1}{{x}_{m \plus{} 3}}\cdot\ldots\cdot\frac {1}{{x}_{n \minus{} 2}}\cdot\frac {1}{{x}_{n \minus{} 1}}\cdot\frac {1}{{x}_{n}}}$

Thanks,

-APSStudent

The expression you've written is a bunch of multiplications, and multiplication [i]is[/i] associative.  Just because what you're multiplying happens to be a set of multiplicative inverses affects nothing.


floor function
combinatorics unsolved
combinatorics

Let N be a positive integer.consider the set X of all triplets (a,b,c) of three positive integers which are distinct pairwise and
a+b+c=N .Consider K triplets  in X such that any two triplets have no common  elements.Let  maxK=K(N)
Prove that   N-6< 6K(N) <12N-9

The question is surely meant to read $2n-9$ on the right.

Consider the set of triples $\left\{(2i-1, 2i, N-4i+1) \, : \, 1 \leq i \leq \left\lfloor\frac{N}{6}\right\rfloor\right\}$.  $(2i-1)+2i+(N-4i+1) = N$ and $i \leq \left\lfloor\frac{N}{6}\right\rfloor$ implies $6i \leq N$ and so $2i < N-4i+1$, so all triples are of distinct integers.  By the nature of the triples, then, it is clear that no two triples have overlap.  Thus $K(N) \geq \left\lfloor\frac{N}{6}\right\rfloor > \frac{N}{6}-1$ and the lower bound is verified.

Now, there are $N-3$ possible numbers which can appear in a triple.  However, if $N-3$ appears in a triple it needs both 1 and 2.  If $N-4$ appears in a triple it needs both 1 and 3.  If $N-5$ appears in a triple, it needs either both 2 and 3 or both 1 and 4.  In either case, we can have at most one of the three numbers $\{N-3, N-4, N-5\}$, so there are actually at most $N-5$ possible numbers.  Thus $3K(N) \leq N-5$ so $6K(N) \leq 2N-10 < 2N-9$.


For some reason, the wiki seems to think that the page "permutations" redirects to the (nonexistent) page "Fundamental Counting Principle."  But it doesn't.  (It "thinks" this on the Broken Redirects page as well as on the Wanted Pages page.)

I deleted the permutations page and the problem seems to be solved.


function
induction
geometry
geometric transformation
rotation
symmetry
algebra proposed

Hard to write it out ... Let's try, however:
Denote $A=\{1,2,...,n\}^2$ and $S$ its image in the plane, i.e. something like the square of vertices $(1,1),\ (1,n,),\ (n,1),\ (n,n)$.
If $(a,b)$ is an arbitrary point in the plane with $a,b\in\mathbb{Z}$, let's call "its neighbours" the following points: $(a-1,b),\ (a+1,b),\ (a,b-1),\ (a,b+1)$. Condition $|a-x|+|b-y|=1$ means then: $(x,y)$ is a neighbour of $(a,b)$. It follows that f maps neighbours of any $(a,b)\in A$  to neighbours of $f(a,b)$.
It's easy to prove that f is injective, hence also surjective (since A is a finite set). An interior point of S has 4 neighbours in S, so its image has (at least, but in fact precisely) 4 neighbours in S (by injectivity). A point on a side of S , other than a vertex, has 3 neighbours, so its image has the same. It follows that f maps interior points to interior points, points on the sides other than vertices to similar points and finally, vertices to vertices.
Thus f(1,1) can be (1,1), (1,n), (n,1) or (n,n) and in every case f(1,2) can be chosen in two ways. Starting from vertex $(1,1)$, we obtain that f also preserves the order along sides, rows or columns. For instance, if P, Q, R are neighbours on the same row, then f(P), f(Q) and f(R) are also neighbours [b]on the same row or column[/b], , in the same orde (induction is needed to accomplish the proof).

In the end, we get 8 such functions: the identity function, 3 rotations around the center of the square ($\pi/2,\ \pi$ and $3\pi/2$) and 4 symmetries corresponding to the 4 symmetry axes of the square.

Why must $f$ be injective?

What about a function like
$f(a,b)=(1,1)$ if $a-b\equiv 0$ mod 2
$f(a,b)=(1,2)$ if $a-b\equiv 1$ mod 2

You're right, Agrippina, f must not be injective: there are lots of non-injective functions which satisfies the conditon in the statement ! (some of them quite irregular).

I think the injectivity of $f$ should have been stated from the beginning. Otherwise there would be too many different types of functions.

[quote="Raluca"]You're right, Agrippina, f must not be injective: there are lots of non-injective functions which satisfies the conditon in the statement ! (some of them quite irregular).
[/quote]
not really irregular. We generalize into m*n table, then by induction on m,n we can find out all the types. Let's do it!!
By the way, the solution is very interesting, it is like folding papers. :D :)


AMC
USA(J)MO
USAMO
probability
inequalities
pigeonhole principle

ABCD is a tetrahedron. The midpoint of AB is M and the midpoint of CD is N. Show that MN is perpendicular to AB and CD iff AC = BD and AD = BC.
it's easy with vectors. but how to solve it without it?

We have:
$\begin{cases}MN\perp AB\\ MN\perp CD\end{cases}\Longleftrightarrow\begin{cases}NA=NB\\ MC=MD\end{cases}\\ \Longleftrightarrow \begin{cases}BC^{2}+BD^{2}=AC^{2}+AD^{2}\\ AD^{2}+BD^{2}=AC^{2}+BC^{2}\end{cases}\Longleftrightarrow AD=BC$
Similar: $AC=BD$

[quote="srulikbd"]something strange 8th USAMO 1979: a1, a2, ... , an is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is a. Another member is picked at random, independently of the first. Its value is b. Then a third, value c. Show that the probability that a + b + c is divisible by 3 is at least 1/4.
isn't the probability is 1/3? the answer at kalva http://www.kalva.demon.co.uk/usa/usoln/usol793.html[/quote]
no, 1/4 is the minimum probability.
[hide="example"]
\[a_{n}=\left\{ \begin{array}{rl}1,&n\text{ odd}\\ 2,&n\text{ even}\\ \end{array}\right.\]
[/hide]
[hide="solution"]
Let $x,y,z$ be probabilities of getting 0,1,2 modulo 3 respectively. Then $x+y+z=1$. The chosen numbers in the sequence add to a multiple of 3 when all are congruent or all are different, so the probability is $x^{3}+y^{3}+z^{3}+6xyz$.

Therefore, the problem is equivalent to proving $4(x^{3}+y^{3}+z^{3}+6xyz)\geq 1$ given $x,y,z$ nonnegative with a sum of 1.

WLOG let $x\geq y\geq z$. The left hand side is equal to
\[4(x^{3}+y^{3}+z^{3}+6xyz)\geq 4(x^{3}+y^{3}+7z^{3})\geq (x+y)^{3}+28z^{3}=(1-z)^{3}+28z^{3}\]

Then note that
\[(1-z)^{3}+28z^{3}-1=3z(9z^{2}-z+1)=3z((3z-1)^{2}+5z)\geq 0\]
so the inequality is true.
[/hide]

[quote="srulikbd"]do you mean that the sequence should be periodic?
also, I don't get this one 5.  X has n members. Given n+1 subsets of X, each with 3 members, show that we can always find two which have just one member in common. when they just them mean exactley one? cause if {1,2,3,4} then every subset have 2 members in common..[/quote]

- it doesn't say that the sequence has to be periodic, but i showed that there was a sequence where the probability is equal to 1/4. That means the lower bound can't be raised to 1/3.

- You can't have $n=4$ at all because there aren't 5 subsets of $\{ 1,2,3,4\}$ with 3 elements.

[hide]
Suppose there existed $n+1$ distinct 3-element subsets of $X$ that all have two elements in common. By the Pigeonhole Principle, some element appears in at least 4 of the subsets. Let $a$ be this element, $S_{1}$ through $S_{4}$ be four of the subsets that contain $a$, and $T_{1}$ through $T_{4}$ be the two-element subsets such that $T_{k}\cup \{ a\}=S_{k}$ for each $k$ (that is, remove $a$ from $S_{k}$).

Each of the four $T_{k}$ have two elements each and exactly one element in common. If no three had a common element, then PIE would give
\[|T_{1}\cup T_{2}\cup T_{3}\cup T_{4}|=4\cdot 2-6\cdot 1=2\]
which is ridiculous (this could only be true of the union of four 2-element sets if they were all the same). So some element, say $b$, appears in at least 3 of the subsets.

Now some [i]pair[/i] of elements appears in at least three sets. Suppose such sets are $\{ a,b,c\},\ \{a,b,d\},\ \{ a,b,e\}$. Suppose $A$ is any of the 3-element subsets. Since $A$ shares a pair with each subset, it must have $a$ or $b$. WLOG let it have $a$; if it did not have $b$, then it would need to have $a,c,d,e$ which isn't possible. So [i]all[/i] of the $n+1$ subsets have a common pair. But now each set must consist of the common pair and one of the other $n-2$ elements, so there are at most $n-2$ possible subsets, a contradiction. It follows that two of the $n+1$ subsets have at most 1 element in common.
[/hide]


In a group of $n$ people, every weekend someone organizes a party in which he invites all of his acquaintances. Those who meet at a party become acquainted. After each of the $n$ people has organized a party, there still are two people not knowing each other. Show that these two will never get to know each other at such a party.

It's not Swiss! It has appeared earlier in St. Petersburg's Mathematical Olympiad (year 1994-1999, I do not remember precisely).


algebra
polynomial
calculus
calculus computations

How to solve the differential equation $y^{\prime \prime} = ky$?

Look up how to solve homogenous differential equations (characteristic polynomials and stuff).

Yes, WLOG we can write $y(x) = z(x)e^{r_1x}$ where $r_1$ is a root to the characteristic polynomal. Plugging this into the equation and using some relationships between coefficients of polynomials and the values of the roots the equation reduces to a first order equation in $z$ which we know how to solve for all solutions.

[quote="perfect_radio"]Is there any (easy) way to prove that there aren't other solutions? (besides the ones we obtain with the characteristic polynomial)[/quote]
The existence-uniqueness theorem. If you have enough solutions to cover all initial conditions, you have everything. For a linear homogenous equation of degree $n$, $n$ linearly independent solutions generate everything.

Of course we can do that. But I was looking for a way to avoid that theorem. (which seems a little too complicated for my current knowledge)


LaTeX
trigonometry
inequalities unsolved
inequalities

x ,y,z are three real numbers,find the maximum and minimum of
 cos(x)sin^ 2(y)+cos(y)sin^2(z)+cos(z)sin^2(x)
   the answer is 5/4 and -5/4, now I have a solution, but it is really ugly,
who can give a better solution?[/hide][/list][/b][/i]

i remenber it was appered in russion‘s MO a few years ago.

[quote="suweijie"]which year? I will appreciate it if you write the solution.大家都是中国人,我来自浙江宁波,你呢?[/quote]
my hometown is ningbo,too.
but now I'm in wuhan :)
can I ask which school are you in?

I come from Yuyao high school.In Chinese: 余姚中学

my middle school is in 鄞洲
my approach for the problem .
for the maximum.
let  assume $0 \leq \cos x \leq \cos y \leq cos z$,we can assume $\frac{\pi}{2} \geq x \geq y \geq z \geq 0$
$f(x,y,z)=cos(x)sin^ 2(y)+cos(y)sin^2(z)+cos(z)sin^2(x)$
can we get one of $f(\frac{\pi}{2},y,z)-f(x,y,z) \geq 0$,$f(x,y,0)-f(x,y,z) \geq 0$ must be true?
I haven't tried it. :)

[quote="suweijie"]Could you tell me how to type mathematics with LaTeX in the forum?
Excuse me for my poor English.[/quote]
sorry,I don't know how express how to use here:(

1993全俄数学奥林匹克一道题的加强，《中学数学月刊》2004.第4期有讨论。
( I don`t know how to translate it into english)
BTW:i come from heilongjiang.


No, i didn't write anything like that. And i don't have to.
You can try to prove in the case where x,y,z are positive, if you feel it's easier. But you have to think about the other cases too.  ;)

I just hope you are completely sure about your problem. It looks kind of weird.

And we have way too many false inequalities on the forum already *sigh*

I've just edit my post. Now i'm sure about it.

Indeed, you are right.
I misunderstood something in my proof (which similar to those used in #16) and consider the negative cases too.
Anyway, i didn't know that there was such a post.
I didn't even know that this was "Vasc's" inequality. I read the "starting" inequality in Crux posted by someone named Simon Marshall.

[quote="darij grinberg"][quote="Anto"]Let $x,y,z$ be any reals s.t the inequality is well-defined.
Prove that :
$\frac{1}{x}+ \frac{1}{y}+ \frac{1}{z} + 4\cdot (\frac{1}{x+y}+ \frac{1}{y+z}+ \frac{1}{z+x}) \geq 12\cdot ( \frac{1}{3x+y}+ \frac{1}{3y+z}+ \frac{1}{3z+x})$.[/quote]

If x, y, z are positive, then see http://www.mathlinks.ro/Forum/viewtopic.php?t=22364 post #16.

If x, y, z are allowed to be negative, your inequality is [b]wrong[/b]. Just look at what happens when y approaches -3x. Then, the fraction $\frac{1}{3x+y}$ on the right hand sides flies away to $+\infty$ and comes back from $-\infty$ (or conversely). Meanwhile, all the other terms on the right hand side and the whole left hand side behave smoothly. It's clear that the inequality such an inequality cannot be true.

  Darij[/quote]
Another to verify why negative numbers do not satisfy the inequality is to take negative on both sides in the original inequality.


Go ahead, post ideas.  Of any sort.  And feel free to evaluate/implement previously posted ideas.

(Thanks to BanishedTraitor for thinking up the idea for this sticky.)

Idea 1: Get rid of Introductions/Chem Book During Summer stickies.

yes get rid of the introduction part atleast  :P

Another thing is that you probably have noticed that most of our topics are 'jumbled' up. We don't have subforums for General Chemistry, Organics Chemistry, Physical Chemistry etc.

This is probably due to the fact we are merely a small extension of a much larger site and also due to our small community.

I suggest we should try and organize, though I am unsure of how yet.

[quote="Carcul"]
That would be a good idea, the problem is that this forum does not probably have enough users for that separation to be worth. However, if Gregx007 can make that happen, it would be very nice, since organization would be more easy (for those who post and for those how browse the forum).
[/quote]

I'm not really sure what the standard procedure is for making new subforums.  I don't think that I have privileges to do so; it would probably have to go through an administrator.  That being said, I think that to make a real case for expanding our board, we would need evidence that there is sufficient interest (although again, I don't really know what the standard procedure is; it may not be that complicated).

[quote="Carcul"][quote="BanishedTraitor"]We don't have subforums for General Chemistry, Organics Chemistry, Physical Chemistry etc.[/quote]

That would be a good idea, the problem is that this forum does not probably have enough users for that separation to be worth. However, if Gregx007 can make that happen, it would be very nice, since organization would be more easy (for those who post and for those how browse the forum). Here are some possible sections:

1) High school chemistry.
2) General chemistry.
3) Organic chemistry.
4) Physical chemistry (which includes Thermodynamics, Quantum Chemistry, Kinetics, and Statistical Mechanics).
5) Inorganic and Analytical chemistry.
6) Problems from Chemistry Olympiads and similar sources(?).

Another thing: this semester I am giving a Physical Chemistry course, so in the next months I was thinking (if I have time) in posting here a collection of very nice problems. So, a Physical chemistry section would be helpfull, otherwise making a "Collection of Physical Chemistry Problems" post can result in a mess.

Another section that would be very helpfull: "Chemistry books", where we could make relevant comments (something like what is done in Amazon) about usefull books for students (and where students could find usefull information).

Let's make this a reference Chemistry forum![/quote]
This is quite a nice input. Although, the '1) High school chemistry.' can be inserted into '2) General chemistry.' so we have one less subforum.

About the collection of problems, can I suggest the way our user [b]kunny[/b] does with his series of 'Today's Calculation of Integral ###'? This is particularly helpful not only for searching but the numbering makes it organized.

If a petition is needed for subforums, count me in :)

[quote="10000th User"]About the collection of problems, can I suggest the way our user kunny does with his series of 'Today's Calculation of Integral ###'?[/quote]

I already do that with the "Organic Synthesis, Problem ##" series. Even so, having all those topics in a single place would be easier.

Ok.  Well, I don't know the policy for forum creation (does someone here?).  Isn't there a suggestions forum somewhere?  Also, the structure drawing package should be enabled in a few weeks (delayed due to holidays).

Unlikely because none of us are moderators. I think you should ask the Administrators. But I would understand if they don't agree because after all this is a 'Mathematics' forum. But the Chemistry forums on other sites don't have $ \text{\LaTeX}$ typesetting  :( .

Actually there is no need for subforums. Subsections is enough, where a subsection could be a sticky post (and so readily visible on the forum) whose organization were to be maintained by the moderator.

Done.  But now I feel there are too many stickies.  Any objections to removing the Chem Book/Marathon stickies?

[quote="gregx007"]Ok, so everyone please post in the metathreads!  (Just links to actual discussions, preferably with a description or summary; I don't think they are the place for actual discussion to occur.)[/quote]

Great now we are more organized!

P.S. Is it possible to get rid of the other moderator? Just wondering, not trying to be hostile towards anyone.

greg: can you move posts around within the forum? Because I think Pardesi posted a discussion in the metathread which should probably be made into a topic and then a link posted in the metathread.

Removal of moderator can only be done by an admin, or so I believe.  So if we requested it, I'm sure it could be done, but I don't see any need for it for the moment.

Pardesi's discussion has been made into a separate thread.  I am updating the welcome message to tell people not to post discussions in the metathreads.

I think, not sure though, that Carcul may have the most knowledge of Chemistry here. So I was wondering what level of Chemistry is Carcul capable of? Is Carcul comfortable with things like Quantum Chemistry, Schrodinger Equations etc. Because I might need help with some of these later.

Also anyone who can help please tell me.

If you know anyone who had an olympiad experience with chemistry on aops it would be cool to hear from them about their experience. Two people per school in the chem competition seems really limiting...

I was just wondering if anyone has good problems involving the chemistry of main group elements (e.g. s blocks, p blocks, first row transition metals, halogens etc.....). I think we should post some of those. There are some good ones on this year's IChO prepatory problems!

Is it possible to have a forum for olympiad? I think it'd be useful more useful than topics, if so we can have different problems, books and softwares available

Hey can the moderators start a marathon for chemistry olympiad.It will be very useful

There are tests online for NCHO (Chem Olympiad) but no solutions. Maybe we could have something similar to the AMC/AIME wiki?


superior algebra
superior algebra solved

Let (G,*) be a group such that there is a in G\{e} and p in N,p\geq 2 such that x^p=axa , for all x in G. Prove that if p=4k-1 then x^(p-1)=e, for all x in G.

It's easy to see that a^2=e and then :
x^(p^2) = a x^p a = x => x^(p^2-1) = e , for all x
x^(p+1) = (ax)^2 => x^(p+1)(p-1) = (ax)^2(p-1) = e => (ax)^(2p-2)=e for all x in G => x^(2p-2) = e for all x in G.

What next ?? I cannot continue from here and the problem isn't solved at the end of the book.

For all x we have x p=axa. [b](1)[/b]

Iandrei showed:
a 2 =e.
x p+1=(ax) 2 =(xa) 2 [b](2)[/b] and
x 2p-2=e, for all x

Suppose we have x such that x 4=e.

Then p[u]=[/u]3 (mod 4), so in (1) we have x 3=axa. By multiplying to the left and to the right by x, we get x 4=(ax) 2 =(xa) 2 , so (ax) 2 =(xa) 2=e.
Take in (2) x->xa and get (xa) p+1=(xaa) 2 =x 2. But (xa) 2 =e and 2|p+1, so e=(xa) p+1=x 2 .

So I have proved that x 4=e =>x 2 =e. [b](3)[/b]
We have for all x, x 2p-2=e, so x 4*(2p-2)/4=e, so y 4=e, where y=x (2p-2)/4. So y 2=e, so x p-1=e.

------------------------------------------------------------------------------------
This posts humby begs you, Iandrei, to spare it's life and brings the solution to this problem as tribute. He knows his predecessor was not as worthy, but it made you edit your message. :D


Let $a_1,a_2,...,a_n$ be a posstive real sequence that is non-increase or non-decrease $(n>2)$ and $a_1\neq a_n$.
$p,q$ are 2 possitive real numbers such that : $\frac{p}{q}\ge \frac{a_1-a_2}{a_1-a_n}$.
Prove that $\sum_{i=1}^n \frac{a_i}{a_{i+1}p+a_{i+2}q}\ge \frac{n}{p+q}$ $(a_{n+1}=a_1;a_{n+2}=a_2)$
I’m sorry if it has already been post

 :P


linear algebra
matrix
function
algorithm
calculus
derivative
real analysis

Suppose $A(t)$ is a smooth family of $n \times n$ real matrices with $A(0)=I$.

(a)  For each t, write $A(t)=Q(t)R(t)$ where $Q$ is an orthogonal matrix and $R$ is upper triangular.  Suppose that $Q(0)=I$ and $Q(t)$ is a continuous function of t.  Show that $Q(t)$ and $R(t)$ are smooth functions of t if $\det(A(t))\neq 0$ for all t.

(b)  Find an example with $det(A(t))=0$ for some t showing that $Q(t)$ need not be unique or a smooth function of t.

(c)  Find an algorithm (formula) that involves $O(n^{2})$ operations that computes $\frac{dQ(0)}{dt}$ and $\frac{dR(0)}{dt}$ from $\frac{dA(0)}{dt}$.  (where these derivatives of course are being evaluated at the point zero).

Here are my thoughts on the solution (for what it's worth):

(a)  If $A$ is smooth in its entries as a function of $t$, then so is the $det(A)$ since it is an expanded sum of multiplications of its entries.  Therefore we can write:

$0 \neq det(A(t))=det(Q(t)R(t))=det(Q(t))det(R(t))= det(R(t))$
($Q(t)$ is orthogonal means its determinant is $\pm 1$, but we know $Q(0)=I$, therefore $det(Q(t))=1$).

Now since $det(A)$ is smooth, we must have $det(R(t))$ smooth by the equality, which then implies $R(t)$ is smooth.

Now since $Q(t)=A(t)R^{-1}(t)$ in the case of $det(A(t))\neq 0$, and since $A$ and $R$ (and therefore $R^{-1}$) are smooth, we have $Q$ smooth as well.

(b)  Let $A(t)=\left[ \begin{array}{cc}t& \pm \left( t-\frac{1}{2}\right) \sqrt{1-{t}^{2}}\\\noalign{\medskip}\mp \sqrt{1-{t}^{2}}&{t}^{2}-\frac{1}{2}t\end{array}\right]$.  Then $A(t)$ can be factored as $A(t)=Q(t)R(t)$ where:

$Q(t)=\left[ \begin{array}{cc}t& \pm \sqrt{1-{t}^{2}}\\\noalign{\medskip}\mp \sqrt{1-{t}^{2}}&t\end{array}\right]$

and

$R(t)=\left[ \begin{array}{cc}1&0\\\noalign{\medskip}0&t-\frac{1}{2}\end{array}\right]$.

Clearly, $det(A(\frac{1}{2}))=0$ and $Q(t)$ is neither smooth ($t=1$) nor unique.

(c)  Not sure about this part.  Clearly we have two equations which can be differentiated:

$A=QR \rightarrow A'=Q'R+QR'$ and $Q^{T}Q=I \rightarrow Q^{T}'Q+Q^{T}Q'=0$ where the prime indicates differentiation and not transpose.  If we allow $Q(0)=I$ as in (a), we just get that $Q$ is skew-symmetric which is not too helpful.  Any thoughts on how to finish up this problem?

Is there a better way to do part (a)?  Thanks.

you have to notice that if you impose that if every element on the diagonal of $R$ is positive, then the decomposition is unique. There are even some formula (I think) to compute all these matrices, since it is just Gram-Schmidt process in disguise. From here, everything follows..

Yes, it is true that the QR factorization is unique provided $det(A) \neq 0$ and also provided the sign of the diagonal elements is appropriately dealt with.  But that alone is not sufficient to prove part (a) for example.  For a given $t$, the factorization is unique, but an infinitesimal change in $t$ will give rise to a new (unique) factorization and one can not assume -- a priori -- that the change between these factorizations is smooth; this must be shown.  Part (b) then shows that the rank of $A$ makes a difference for the continuous factorization.  As for part (c), an explicit factorization in the abstract of $A$ (based on Gram-Schmidt) is not helpful to solve this part of the problem, so this is still open.

for a)  Remark: The Edoo's solution doesn't work.

$A^{T}{A}=R^{T}{R}$ is the Cholesky decomposition of the positive symmetric matrix $A^{T}{A}$. (Recall that $R_{i,i}>0$.) The construction of $R$ is done by recurrence and each step of this recurrence is a smooth function of the $A_{i,j}$. Thus $Q=AR^{-1}$ is also a smooth function of the $A_{i,j}$.
for c) The derivatives are considered in $0$.
$A'^{T}+A'=R'^{T}+R'$ is known; thus we know $2R'_{i,i}$ and $R'_{i,j}$ with $i\not={j}$; Moreover $A'=Q'+R'$ is known; thus we know $Q'$.
The operations are the calculation of $A'^{T}+A'$, $n$ divisions by 2 and the calculation of $A'-R'$ that is to say $O(n^{2})$ operations.

[quote="edoo"]Suppose $A(t)$ is a smooth family of $n \times n$ real matrices with $A(0)=I$.
[...]
(c)  Find an algorithm (formula) that involves $O(n^{2})$ operations that computes $\frac{dQ(0)}{dt}$ and $\frac{dR(0)}{dt}$ from $\frac{dA(0)}{dt}$.  (where these derivatives of course are being evaluated at the point zero).[/quote]
Note that if $A(0) = I$, then $Q(0) = R(0) = I$ by uniqueness of the QR-factorization. Also, since $Q(t)$ is orthogonal for all $t$, we have $Q'^{T}(t)Q(t)+Q^{T}(t)Q'(t) = 0$ for all $t$ and at $t = 0$ especially $Q'^{T}(0)+Q'(0) = 0$, i.e. $Q'(0)$ is skew symmetric. So it suffices to compute the lower triangular portion of $Q'(0)$.  Furthermore of course $R'(t)$ is upper triangular for all $t$.

Now let $A'(0) = (\alpha_{i,j})_{i,j}, \, Q'(0) = (\beta_{i,j})_{i,j}, \, R'(0) = (\gamma_{i,j})_{i,j}$. Then $A'(t) = Q'(t) R(t)+Q(t)R'(t)$ and for $t = 0$
\[A'(0) = Q'(0)+R'(0) \, .\]
So the formula in question is
\[\beta_{i,j}= \alpha_{i,j}\quad (i>j), \, \beta_{i,i}= 0, \quad \beta_{i,j}=-\alpha_{j,i}\quad (i < j) \]
\[\gamma_{i,j}= 0 \quad (i > j), \, \gamma_{i,i}= \alpha_{i,i}, \, \gamma_{i,j}= \alpha_{i,j}+\alpha_{j,i}\quad (i < j) \, .\]
Clearly $O(n^{2})$ operations. :lol:


He has a magical radio that lets him hear the broadcasts a day in the future, of course :rotfl:

[hide]The score of a game before it starts is always 0-0[/hide]

[hide="That was easy"]The score of a football game before it starts is always 0-0[/hide]

lol funcia...

and i guess the other two posts are good too

[quote="Ihatepie"]Bob says that he can tell the score of any football game before it even starts. Many think he is psychic and posseses supernatural powers. How is it possible that he can be accurate 100% of the time?[/quote]

By saying that he can tell the score it implies either at the beginning of the game or the end...

If he predicts the beginning score, people won't think that he is psychic because the beginning score is obvious.  However, there is no other score that must happen during the game (unless you count the score $x,y$ for some $x$ and some $y$, but then people will still not think he is psychic).

Yes I will argue semantics, semantics is everything! :P  :roll:

hmmmm

[hide]Bob is from the future and can predict scores  :rotfl:  :rotfl:  :rotfl: [/hide]

[quote="Ihatepie"]Bob says that he can tell the score of any football game before it even starts. Many think he is psychic and posseses supernatural powers. How is it possible that he can be accurate 100% of the time?[/quote]

The football game is on TV, and it isn't live. He's already seen the game and memorised it.        (Other peoples 0-0 theories make more sense..)

[quote="Ihatepie"]Bob says that he can tell the score of any football game before it even starts. [b]Many think he is psychic and posseses supernatural powers.[/b] How is it possible that he can be accurate 100% of the time?[/quote]Emphasis mine.

Personally I think Nonie's theory is more likely to be correct than 0-0 for the reason that I have already mentioned, i.e., no one would think him psychic if he predicted the beginning score.

this is like my unfair math final which asked me a question involving a "diamond" in a baseball field
since i didn't get it i got a 97
it would've been a hundred had i known that "diamonds" are squares

why must everyone understand these sports?

He is SUPER GOOD (and I mean it!!!) at it and can control the score easily because... everybody else... is downright awful.

No...

He's the judge or referee (whatever you call it)

Other than that, I don't know.

[hide]Nonono, it says before a game, so it's always 0-0[/hide]

[quote="kstan013"][quote="Ihatepie"]Bob says that he can tell the score of [b]any[/b] football game before it even starts. Many think he is psychic and posseses supernatural powers. How is it possible that he can be accurate 100% of the time?[/quote]Yes. Note the word [i]any[/i].[/quote]

In other words, no matter what score he predicts, it will be the score for some game in the future.  Presumably Bob is smart enough to not pick a score like 1024-512. :P

[quote="D3m0n Shad0w"]or Bob is SUPER rich and tells the two teams how many field goals and touchdowns they can score[/quote]
If he takes that tactic, he won't be "SUPER rich" for very long. :harhar:  :yankchain:


[b]Consider the system of equations:
$ a_1\plus{}8a_2\plus{}27a_3\plus{}64a_4\equal{}1\\
8a_1\plus{}27a_2\plus{}64a_3\plus{}125a_4\equal{}27\\
27a_1\plus{}64a_2\plus{}125a_3\plus{}216a_4\equal{}125\\
64a_1\plus{}125a_2\plus{}216a_3\plus{}343a_4\equal{}343$
These four equations determine $ a_1,a_2,a_3,a_4$. Show that
$ a_1\cdot (x\plus{}1)^3\plus{}a_2\cdot (x\plus{}2)^3 \plus{}a_3\cdot (x\plus{}3)^3\plus{}a_4\cdot(x\plus{}4)^3 \equiv (2x\plus{}1)^3$
i.e. these two polynomials are identically the same. Use this to show that $ a_1\plus{}a_2\plus{}a_3\plus{}a_4\equal{}8$ and that $ 64\cdot a_1\plus{}27\cdot a_2\plus{}8\cdot a_3\plus{}a_4\equal{}729$[/b]

[quote="123456789"][hide="Solution"]Firstly, to show that two cubic polynomials (LHS & RHS) are equivalent to each other, we need only to show that they are equal at four points, because four points determine a cubic ($ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$). The system of equations tells us that the two polynomials are equal at $ x \equal{} 0,1,2,3$.

So, we have shown that the two polynomials are equal. Now, if we equate the coefficients of the $ x^{3}$ term in both polynomials, we see that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \equal{} 2^{3} \equal{} 8$.

Also, $ 64\cdot a_1 \plus{} 27\cdot a_2 \plus{} 8\cdot a_3 \plus{} a_4 \equal{} 729$ is true when $ x \equal{} \minus{} 5$.[/hide][/quote]
wait, how do you know that the two polynomials are equal at $ x\equal{}0,1,2,3$?
also, what $ x^3$ coefficients are you equating?

Let $ f(x)\equal{}a_1\cdot (x \plus{} 1)^3 \plus{} a_2\cdot (x \plus{} 2)^3 \plus{} a_3\cdot (x \plus{} 3)^3 \plus{} a_4\cdot(x \plus{} 4)^3$ and $ g(x)\equal{}(2x \plus{} 1)^3$.

Then, the system of equations you have is basically a list showing that:
$ f(0)\equal{}g(0)$
$ f(1)\equal{}g(1)$
$ f(2)\equal{}g(2)$
$ f(3)\equal{}g(3)$

Do you see the pattern?

Now, for the $ x^{3}$ coefficients...
If you fully expand the polynomials on both sides of the equivalence:

$ a_1\cdot (x \plus{} 1)^3 \plus{} a_2\cdot (x \plus{} 2)^3 \plus{} a_3\cdot (x \plus{} 3)^3 \plus{} a_4\cdot(x \plus{} 4)^3\equiv (2x \plus{} 1)^3$

you'll see that that the coefficient of the $ x^{3}$ term on the left hand side is $ a_1\plus{}a_2\plus{}a_3\plus{}a_4$ and that the coefficient of the $ x^{3}$ term on the right hand side is $ 2^{3}\equal{}8$. Since the two polynomials are the same, we can equate the coefficients, and that shows $ a_1\plus{}a_2\plus{}a_3\plus{}a_4\equal{}8$.


Without even looking at it, I can tell you're posting loose change for the 100th time, and that it's been debunked many times over.

True, but I am posting it at a better time than anyone else in the past one year  :P

[quote="K81o7"][quote="mustafa"]Is this going to be one of those ridiculous videos saying that 9/11 was fake? Those are just ridiculous. Can you explain the thousands of eyewitnesses/dead people/ absence of the twin towers?[/quote]

The lack of the twin towers, dead people, and eyewitnesses don't at all contradict conspiracy theory arguments.[/quote]
Which one are we talking about?

[quote="funcia"]why are we having a national holiday? to rile people up to fight iran.[/quote]
No.

There were enough threads about this already. We need not to discuss everything again (nonetheless a good strategy of those that believe in this, it's like an DoS attack).

@Penguin: all, or are there some serious ones that say the towers didn't collapse (but got stolen by aliens or whatever)¿

And the  life goes on ...Please take a look:
[url=http://www.boston.com/news/globe/city_region/breaking_news/2006/09/mit_hacks_mark_1.html]here[/url]

[quote="bubka"]I am in no mood for rambling about the twin towers and how 'unfortunate' the incident was, but I happen to have a nice link handy:

[url=http://video.google.com/videoplay?docid=-5946593973848835726&q=Loose+Change&hl=enorhttp://video.google.com/videoplay?docid=-7218920724339766288&q=Loose+Change&hl=enorhttp://video.google.com/videoplay?docid=7866929448192753501&q=Loose+Change&hl=enorhttp://video.google.com/videoplay?docid=7167151489146455453&q=Loose+Change&hl=en]The truth about the twin towers - 9/11 cover up[/url]

If anybody can download the full video, please send it to me. I have seen most of it, but not the whole. Enjoy it![/quote]

i've heard of it... but is it true?


A unit square has an equilateral triangle inscribed inside, with one vertex of the triangle shared by the square. The other two vertices of the triangle are on sides of the square, not on a vertex. Find the area of the triangle.

I tried solving two different ways and got two different answers... can anyone point out my obvious algebraic mistake?
[hide="answers"]
Note that the smaller right triangles with one leg 1 have the same hypotenuse, making them congruent.  Therefore the third right triangle is 45-45-90.  I solved it using x as different parts of the triangle and got different answers.

1) x is the side length of the equilateral triangle.
The smaller right triangles have legs $1$ and $\sqrt{x^2-1}$.  So the legs of the isosceles right triangle have length $1 - \sqrt{x^2-1}$.  Now we can say
$\sqrt 2 (1 - \sqrt{x^2-1}) = x \rightarrow x^2 = 2\sqrt{x^2-1} \rightarrow x^4 - 4x^2 + 4 = 0$
$x^2 = 2 \rightarrow x = \sqrt 2 \rightarrow A = \frac{s^2 \sqrt 3 }{4} = \frac{ \sqrt 3 } 2$

2) x is the other leg of the smaller right triangles with first leg 1.
The length of the equilateral triangle is $\sqrt{x^2+1}$, and the legs of the isosceles right triangle have length $1 - x$, so
$\sqrt 2 (1-x) = \sqrt{x^2 +1} \rightarrow x^2 - 4x + 1 = 0$
$x = 2 - \sqrt 3 \rightarrow A = \frac{(x^2 + 1)\sqrt 3 }{4} = \frac{(8 - 4\sqrt 3 )\sqrt 3 }{4} = 2\sqrt 3 - 3$

Clearly the answers aren't the same, but they should be... I redid my work and got the same things, so I need someone else to tell me what I'm doing wrong.
[/hide]

[quote]1) x is the side length of the equilateral triangle.
The smaller right triangles have legs $1$ and $\sqrt{x^2-1}$.  So the legs of the isosceles right triangle have length $1 - \sqrt{x^2-1}$.  Now we can say

$\sqrt 2 (1 - \sqrt{x^2-1}) = x \rightarrow$[/quote]

$2 (1 - \sqrt{x^2-1})^2 = x^2$

$2 (1 -2\sqrt{x^2-1} +x^2-1) = x^2$

$2 ( -2\sqrt{x^2-1} + x^2) = x^2$

$2 x^2 -4\sqrt{x^2-1} = x^2$

$x^2 = 4\sqrt{x^2-1}$

[color=red](Here is your mistake! You found $x^2 = 2\sqrt{x^2-1}$)[/color]

$x^4 = 16(x^2-1)$

$x^4 = 16x^2-16$

$x^4 -16x^2+16=0$

we get $x^2 = 8-4\sqrt 3$ which is the same as in (2)

$Area = \frac{(8-4\sqrt 3)\sqrt 3}{4}$

Another soln

Let $x$ the side length of the triangle.

Then its altitude is $x\frac{\sqrt 3}{2}$

The isosceles right triangle has the same base $x$ so its altitude is $\frac{x}{2}$

The diagonal of the square is equal to $\sqrt 2$

So $\frac{\sqrt 3}{2}x+\frac{1}{2}x = \sqrt 2\Rightarrow$

$x = \frac{2 \sqrt 2}{\sqrt 3 +1}$

[hide="or else"]$x = \frac{2 \sqrt 2}{\sqrt 3 +1} \cdot \frac{\sqrt 3 -1}{\sqrt 3 -1}$

$x = \frac{2 }{(\sqrt 3)^2-1^2}\sqrt 2 (\sqrt 3 -1)$

$x = \sqrt 2 (\sqrt 3 -1)$

and $x^2 = 2 (\sqrt 3 -1)^2 = 2(3-2\sqrt 3 +1) = 2(4-2 \sqrt 3) = 8 - 4 \sqrt 3$[/hide]

[hide]Let $s$ be the side of the square and let the side of the triangle be $t$, thus the height of the triangle is $\frac{\sqrt{3}}{2}t$, now seeing the $45,45,90$ triangle (sharing the hypotenuse as a sidelength of the triangle), we have it's altitude to be $\frac{t}{2}$ (since the altitude of a $45,45,90$ triangle is $\frac{1}{2}$ of it's hypotenuse. So then we have $\frac{t}{2}+\frac{\sqrt{3}}{2}t=s\sqrt{2}$, from there we see that $t=s\sqrt{6}-s\sqrt{2}$, so the area of the triangle is $\boxed{2s^2\sqrt{3}-3s^2}$[/hide]


Sometimes the LaTeX-formulas won't be shown correctly. But when I move mouse over the code, it looks good. Firefox 3.0.13 on Ubuntu, Xfce.

This has also been happening to me recently (Firefox 3.0.13 on Vista, and also Firefox on Linux) -- it started relatively recently (past couple of weeks, maybe?)  It also happens to other images on the site (e.g., the "new posts" icons on fora, or the "challenge" and the like icons on individual posts).

What exactly happens? This seems like a local issue, rather than a server issue ...

When I open some discussion, Firefox starts to render the $ \text{\LaTeX}$ code to images. Then sometimes it won't render everything, as in the screenshot.

This has happened to me too, (I'm using Firefox on Linux, and Windows), but the images always appear after I wait like a minute...

Same here. The problem is that Firefox is sometimes too slow. I notice it more with alcumus, but it happens on the forum too. If it doesn't work, just reload the page.

Sometimes I see that temporarily, but the page finishes loading and the images are properly rendered. I haven't had it stick at that level.

(Firefox on Mac.)


AMC AIME USA(J)MO USAMO MIT college Harvard

I had a great experience with this USAMO. I really learned a lot from actually forcing myself to take the test timed. The problems were great. They forced me to really think. There was not much that was particularly contrived (except the factorization for #5).

I really liked problem #2 because it forces the solver to take an intuitive idea and prove it rigirously. For some reason, I don't feel like I have really had that experience with a problem before. While my solution fell short, I learned the lesson of how carefully you have to justify your solutions. I made the assumption that when you have 3 circles, and you move them closer, the circle externally tangent to all 3 [and inside] decreases in radius. I was further unclear about how you chose the first 3 circles.

I learned about writing solutions in a timed environment. I wrote my solution for #1 and after reading it over, I realized how unclear it was. I decided that I had to rewrite it. I never really noticed how difficult it is to write clear solutions in the time pressure situation. I feel that I still need to develop that technique; perhaps some sort of outline would be a good idea for future solutions.

Along with a few little facts that I picked up, like <OAI=B+A/2-90 can be written symmetrically as 1/2(B-C), I learned about writing rigirous proofs to seemingly intuitive propositions. I really enjoyed the test this year. It did not really matter that I didn't qualify, I could still take the test on my own.  I am pretty sure that I would have gotten 21 to 23 based on the feedback I have gotten on this forum. While I am somewhat bitter because I probably could have made blue MOP this year, I still am happy with the experience. I guess I will just have to pay for a math camp this year.

--Alex

Yes, number three was particularly nice.

Yeah the USAMO has really helped peak my interest in math particulary problem solving/olympiad type problems. I will be sure to practice more next year.

I thought they were all good problems except for #4 which was slightly ambigious and led to me making a misinterpretation. The actual proof of what they wanted is still fine, but I proved a lot of stuff which is now obviously false first. Hope they don't hit me for that :(

I thought all six problems were very nice, problems 4 and 6 were particularly interesting.

This was actually the first time I had done this using all the time, it was much shorter than I thought, and I found working on the problems much more fun that I would have expected!

Like others, this has inspired to work on more olympiad problems.

Heh, I agree with the last comment. I found that all the problems were approachable/interesting and seemed fun right from the moment I read them, but they were definitely hard and good.

I thought they were all really good, although 1 and 4 were easier than usual, and the rest seemed a bit harder.

Yeah, looking at this year's USAMO has definitely motivated me a lot to try to make USAMO next year, and do more olympiad problems. They were all really good problems, and I actually got somewhere on #1 and #4 by myself...  :P

Even though a lot of people complained about #2, I think it is the most natural in terms of a question that might actually appear on real mathematical research...in fact I wondered about it before after reading some article about disc covers, but didn't put in very much thought.

[quote="Valentin Vornicu"]Wow I never thought I would see this thread coming. I was expecting complaints as usual :)[/quote]

What is there to complain about in an olympiad?

"Oh no I made stupid mistakes, I accidentally proved the converse of what we were supposed to prove!"

I guess if a solution is particularly contrived, maybe? 5 is in danger of that I guess, though I am surprised that so few people solved it (not that I did, but I would think that at least one of Alex and Brian and Arnav would).

Yeah...lol...you can't really make stupid mistakes; usually you know if you made some mistake during the course of your proof, i.e. it doesn't work.

I really liked the problems individually as well. I just didn't like how the test was strung together, with no problems of difficulty level 2/5 (except maybe #4). But all of the problems were very interesting and have creative and interesting, if not reasonable ;), solutions.

Even stupid mistakes are not something to complain about...because that has nothing to do with the test writers

[quote="rem"]Thank you very much for number 6. It was one of the most beautiful inequalities that I have encountered.[/quote]

#6 was approximately as beautiful as my dog's vomit

[quote="Altheman"]ok i guess but you are just being unnecessarily negative...for #2...it is a good thing to encourage rigor..."it can favor those with more formal training" is a bizarre argument...an incomplete proof is incomplete no matter what...[/quote]

Uhh, no. Those with more formal training are more likely to be able to put together a rigorous proof. I don't have a problem with #2 from a purely competitive perspective. A problem like that is going to favor me since
a) I will solve it fast
b) Many other people will lose points for their writeups

The problem is, personally I think #1, 2, 4, and to some extent 5 should be used as separators for determining who should go to MOP. These problems are crucial in the development of the US team. If you think that "IMO quality people don't need help" you are pretty much flat out wrong in a majority of cases. You want 1, 2, and 4 to reward young talent that may not have the formal training but does have the ingenuity and perseverance to solve Olympiad level problems without training. These are the people you want to train for the future.

#2 may somewhat reward ingenuity, but it also punishes lack of training moreso than most #2's. There's a reason why you don't see inequalities as 1, 2, or 4 ever (or at least, not traditional ineqs) ... it's because you simply can't do some inequalities if you don't know some basic things (such as AM-GM, or Cauchy, or whatever), and it's not prudent to "filter out" some 8th or 9th grade kid from going to MOP simply because he/she didn't know an inequality.

I remember when I took the 2002 USAMO as an 8th grader. I got a total of 5 points, all for #6 seeing as how I submitted nothing of any discernible value for the other 5 problems. Ironically that was a test where #6 was the one problem that required very little actual training to do. Yes, #1 only required induction (which I didn't know), #2 only required basic algebra and knowing what a cotangent was (which I also didn't know), #4 was a functional equation which is hard to approach rigorously without at least some formal training. #5 probably also didn't require much training but I just didn't get it. The point is, even people who know almost nothing (like me in 8th grade) can get points on hard problems if they don't require too much training (and yes, #6 was certainly harder than most of the rest of the test, if not all of it). You want to give those people a chance.

[quote="calc rulz"][quote="rem"]Thank you very much for number 6. It was one of the most beautiful inequalities that I have encountered.[/quote]

So how was it beautiful? It seemed that the way to solve it was to bash through algebra/inversion to find a formula for $P_{a}Q_{a}$ and then use well-known inequalities to finish it off.[/quote]

Well, assuming that my solution is correct, you don't need to bash very much. But in any case I don't think #6 was particularly beautiful.

Also assuming probability1.01 is right, it doesn't take very long, either.

It's beautiful in a sense that something random and seemingly impossible reduces to something trivial.
I dunno about u guys, but I liked it.

[quote="calc rulz"][quote="rem"]Thank you very much for number 6. It was one of the most beautiful inequalities that I have encountered.[/quote]

So how was it beautiful? It seemed that the way to solve it was to bash through algebra/inversion to find a formula for $P_{a}Q_{a}$ and then use well-known inequalities to finish it off.[/quote]
You shouldn't confuse a result with its proof.

By that logic, Fermat's Last Theorem is about as ugly as it gets. I am inclined to disagree with that conclusion. :P

Though I didn't like that problem that much either. The less that needs to be defined in the problem, the better; there were way too many circles in that problem. Also add that geometry is probably my worst subject. :lol:

thats the other way around lol...something seemingly trivial ends up incredibly complicated...i think what rem meant is that the inequality reduced to $2r\le R$

But that's only if you did it one way.  The other way (dilation + law of cosines) has very few creative steps.  You honestly just compute and get AM-GM.

First of all, I'd like to thank the USAMO writers for another good test. All of the problems were approachable, understandable, and clear. That said, I personally disliked this test for many of the same reasons as darktreb. It seemed to me that there were relatively few moderately easy problems. The way I see it is that each problem has two parts, the question and the write-up. You can solve the question (which is more than just finding the answer, it's understanding why), but still not be able to communicate this in the write-up. (At MOP, that just means you're the person assigned to write up the solution for the book at the end.) To score well (6/7), you need to solve the question [i]and[/i] write it up well.

Back to this year. From my observations, the questions went this way: 1 was very easy, 2 and 4 were easy, 6 was moderately hard, 3 was hard, and 5 was very hard. For write-ups, assuming you got the questions, the order was reversed with the exception of #1 still being easy. If you got the factorization on #5, you had better get a 7, but if you understand the concept of picking a square and considering the sizes of things bordering it, you still might not do well on #4.

I personally think the emphasis should be on solving the question, as I like to call it. Multi-step questions, as opposed to simply multi-step write-ups, should be more common. Along these lines, problems like the whole 2005 Day 2 were some of the best. Of course, this is also tied to my own success; I made blue MOP in 2005 but haven't made it back since. But my point still stands: Success or failure (MOP or no MOP) should not depend on one's write-up, rather, one's problem-solving skills. Write-ups can easily be improved at MOP.

I could go back and analyze each of those 05 Day 2 problems, but I don't know if anyone would read that. Anyway, this post is long enough. The more tired I get, the more I think of, I guess. Yeah, definitely. I think I've learned more about what I've done wrong in the past by thinking about what I thought was wrong with this test. I think I'll start another post (if I ever get around to it) tomorrow.

I think the reason this issue (I don't have much opinion about whether it's bad or not) exists is because USAMO tries not to use too much stuff that you can really train for.  This leads to lots of arguments that really need to be hashed out in the writeup (i.e 2 / 4).  A problem like #6 can be written up almost mindlessly; it's easier to write up than any problem except for #5.  But it rewards training and in particular computation a lot more, which isn't really what the USAMO is primarily about.

On the other hand, what you call "solving" and "writing up" are more correctly, in my opinion, referred to as "gaining intuition" and "solving the problem". In most problems, you had better have some intuition about the problem before you solve it. However, this is not really solving it, as your intuition could very well be wrong, and at any rate you do not yet have a rigorous argument. Hopefully without too much arrogance I can call myself a fairly experienced olympiad taker -- number four this year took me very little time (less than ten minutes, probably) to gain the necessary intuition about, but easily over an hour, possibly going into two to two and a half hours, to actually come up with a decent rigorous method of proof. Consequently, I ended up by just generalizing the whole result and inducting, which didn't necessarily have that much to do with my original intuition other than to give me a result to generalize.

I guess my point is that, while intuition is very important, it is not going to solve the problem for you very often. If your intuition is good and nontrivial, and you put it down on paper, you will most likely get two to four points out of seven, which I would consider pretty good partial credit. The key is that you have to actually communicate it on paper, as opposed to trying to turn it into a handwavy proof (unless the hanwavy proof itself communicates the intuition). Also, there is a big emphasis on non-trivial. I think that most people who did number 2 eventually got to the three tangent circles (admittedly that took me probably over an hour, but I have never claimed geo to be my strong suit), and most people who did number 4 eventually got to the correct configuration, and most likely had some pretty good reasons why, intuitively, there could be no better configuration. With so many people making these steps, it does not make sense to award very many points of partial credit.

EDIT: Reading your post a second time, you may have meant something completely different than what I had first thought. I'll leave this here anyways because (a) maybe you did mean what I first thought, and (b) hopefully this will help new olympiad takers somewhat understand the grading process / what constitutes a rigorous proof.

To Yi: Stop posting here and go back to working on our project

To other people: We can argue all day about differences between intuition and solving and writing and whatever. Still, it is certainly possible to write high quality problems of a wide-variety of difficulty levels without having them require iffiness in a write-up. That was not really true on #2 and maybe not #4 either.

Put it this way, since I've already used my 8th grade self as the "can figure stuff out but has no idea how to write anything" standard, I'm pretty sure I would have figured out #2 my 8th grade year, and I'm positive I wouldn't have been able to write it up well and that I would have lost points for it. Now, for young students, from the training perspective you definitely care that they were able to figure out that you need to shift/expand the circles and get to a point where you have 3 tangent circles. You don't really care as much if they were able to write it rigorously, since with training that will surely come.

I was fortunate in getting to go to MOP 2002 in 8th grade because it was the one year they let everyone in. The next year I made it back to MOP only it was only 30 people this time (this is also the very last time MOP has been only 30 people). Would there have been ANY chance of me making MOP 2003 without the training I had the summer before? No way. Training makes a big difference, and an even bigger difference for people who have intuition, creativity, work ethic, and so on. It's sad that there are probably some promising kids who got 2 on #2 instead of 6 or 7 because of the lack of formal training, and won't make MOP as a consequence. I guess the 30 ninth graders thing mitigates this somewhat now, but still, it's food for thought.


Given a triangle with base $ a$ and perimeter $ P \equal{} 2p$. Find 2 sides of triangle, so that the area $ S$  gets the maximum value

The triangle will be isosceles with sides (P-a)/2,a,(P-a)/2


So what is the toughest problem that you guys have ever seen appear on an olympiad?

Fierytycoon

I don't think this is the toughest problem ever seen on an olympiad, but here's a notoriously difficult problem from a MOP 2002 test:

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

(a/(b+c))^(2/3)+(b/(c+a))^(2/3)+(c/(a+b))^(2/3)>=3/4^(1/3).

Enjoy!

Is it 3/4^(1/3) or (3/4)^(1/3) at the end?
(Just making sure)


Gauss
geometry
geometric transformation
reflection
analytic geometry
linear algebra
matrix

let $ABC$ be a triangle . For a point M inside the triangle , let E be a point of intersection of AM and BC , let F be the point of intersection of CM and AB .Let  N be the reflection of B in the midpoimt  of EF .
prove that  the line MN passes through a fixed point when M move inside triangle ABC.

The fixed point is the reflection (symmetrical) of the point $A$
w.r.t. (with respect to) the middlepoint of the side $[BC]$. Is it enough ?

Take a skew system of coordinates such that $A(0,1)$, $B(0,0)$, $C(1,0)$, $E(a,0)$, $F(0,b)$. Then we have
$N(a,b)$ and $M(\frac{ab-a}{ab-1},\frac{ab-b}{ab-1})$. Then the line $MN$ passes through the (fixed) point of coordinates $(1,1)$, and for seeing that it suffices to compute
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & a & b \\ 1 & \frac{ab-a}{ab-1} & \frac{ab-b}{ab-1} \end{vmatrix}=0  \]

You prove only a right-angled and isosceles triangle ! the triangle $ABC$ is anybody.

No. You solve the problem for any triangle because I said that you take a skew system of coordinates, I refer to a system of coordinates with the axes along the sides $BA$ and $BC$ of any triangle, and of course the units of measure along this axes are not the same.

Let $D$ be the reflection of $B$ in the midpoint of $AC$, and let $P,Q$ be the intersections of the parallels through $E,F$ to $AB,BC$ with $AD,CD$ respectively.

By applying Desargues to the two triangles $EFN,ACD$, proving that $DN,AE,CF$ are concurrent becomes equivalent to showing that $PQ,EF,AC$ are concurrent. The projective version would be this:

Let $ABCD$ be a quadrilateral. Also let $S=AB\cap CD,T=AD\cap BC$, and take $N$ to be any point in the plane. We define $E=SN\cap BC,F=TN\cap AB,P=SN\cap AD,Q=TN\cap CD$. Prove that $EF,PQ$ intersect on $AC$.

This would follow if we prove that $(FA,FC;FE,FQ)=(QA,QC;QP,QF)$. We have $(FA,FC;FE,FQ)=(B,C;E,T)$ and $(QA,QC;QP,QF)=(A,D;P,T)$. It's clear that $(B,C;E,T)=(A,D;P,T)$ because the two quadruples are perspective from $S$, and we're done.

The problem is pretty trivial once one has identified the fixed point:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle and M a point in its plane. Let the lines AM and BC meet at a point E, and let the lines CM and AB meet at a point F. Let N be the reflection of the point B in the midpoint of the segment EF, and let D be the reflection of the point B in the midpoint of the segment CA. Then, the points M, N and D are collinear.[/color]

(This theorem answers the problem, since the point D is our required fixed point.)

[i]First proof of Theorem 1.[/i] Consider the complete quadrilateral formed by the lines AB, BC, AM and CM. The segments BM, EF and CA are the diagonals of this complete quadrilateral. Since the midpoints of the diagonals of a complete quadrilateral are collinear (they lie on the Gauss line of the complete quadrilateral), it thus follows that the midpoints of the segments BM, EF and CA are collinear. Now, undertake a homothety with center B and factor 2. This homothety takes every point X to the reflection of the point B in the point X. Hence, this homothety takes

- the midpoint of the segment BM to the reflection of the point B in the midpoint of the segment BM, i. e. to the point M;
- the midpoint of the segment EF to the reflection of the point B in the midpoint of the segment EF, i. e. to the point N;
- the midpoint of the segment CA to the reflection of the point B in the midpoint of the segment CA, i. e. to the point D.

Since the midpoints of the segments BM, EF and CA are collinear, their images under our homothety must also be collinear; but since these images are the points M, N and D, we thus see that the points M, N and D are collinear, and Theorem 1 is proven.

A [i]second proof of Theorem 1[/i] would use the Pappos-Pascal theorem. Try finding the right points on yourself (two of them are infinite points).

  darij