WINNERS ARE TWO THIS TIME.. RC-7th and Farazy since both of them each had one credit. They'll be send to the Championship Round with Jimmy.

Let us discuss the solution here.

For #1.

[quote]Question 1:

$z_1,z_2,....z_5$ are non-negative real numbers that holds the following equation true:

$z_1+z_2+z_3+z_4+z_5 = 200$.

The maximum value, which is called $\mu$, is defined by $z_1+z_2$,$z_2+z_3$,$z_3+z_4$, or$z_4+z_5$. If $\mu$ lies between $\alpha$ and $\beta$ such that $\beta - \alpha = 2$, find, with fully written proof, the value of $x+y$ if $\mu - \beta + \alpha = \frac {x}{y}$. NOTE: $\alpha$ is the smaller number that is closest to the real value of $\mu$, meaning $\mu - \alpha$ is less than one.
[/quote]

There are actually two possible answers.

We have to know that because of $z_1+z_2 = z_2+z_3 = z_3+z_4 = z_4 + z_5$, we get:

$z_1 = z_3 = z_5$ and:

$z_2 = z_4$.

We can either maximum the value of $z_1,z_3,z_5$ or $z_2,z_4$. The first one gives you $z_1 = z_3 = z_5 = \frac {200}{3}$ and second case gives you $z_2 = z_4 = 100$. The rest is 0.

For the first case, two intervals are 66 and 68. Therefore:

$\frac {200}{3} - 68 + 66 = \frac {194}{3}$ so $194+3 = \fbox {197}$.

For the second case, interval doesn't work well because 100 is an integer value. But if we have to pick them, it would be 99 and 101. So:

$100-101+99 = 98$. So, $\frac {98}{1} = 98 = 98+1 = \fbox {99}$.

Either answer was fine.

I'll soon post the contestant's answers as well.

Here is mine:

The problem was:

[quote]Question 2:

Given that $a,b,c$ represent the length of the side of the $\Delta {ABC}$ and $k$ is any positive integer. Prove that for all $a,b,c,k$, the following inequality holds true:

$\sqrt {\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c}} < \sqrt {2k^2}$[/quote]

Because of triangle inequality (I hope all of you know this so I don't have to prove it in here), we have:

$a<b+c\\
b<a+c\\
c<a+b$

By assuming $a=b+c$, we can have $2(b+c) = a+b+c$ and because this is NOT true, instead our real inequality is:

$2(b+c) > a+b+c$.

Similarly, we have:

$2(a+c) > a+b+c$
$2(a+b) > a+b+c$.

Since our $k$ can be removed by squaring the both sides and removing the $k^2$, we don't have to worry about it. By the way, $k^2$, was put in there to make it harder. :lol: It wasn't necessary on our proof.

So:

$\frac {a}{b+c} + \frac {b}{a+c} + \frac {c}{a+b} < 2$ is what we need to prove. Rather than describing it on the words, here is the visual:

$\frac {a}{b+c} + \frac {b}{a+c} + \frac {c}{a+b} = \frac {2a}{2(b+c)} + \frac {2b}{2(a+c)} + \frac {2c}{2(a+b)}\\
< \frac {2a}{a+b+c} + \frac {2b}{a+b+c} + \frac {2c}{a+b+c}\\
< \frac {2a+2b+2c}{a+b+c}\\
< 2$

Done. :)

Might want to check what you typed up for the last part of number 1...(second case)

100-101+99 is 98...not 300...

I wished I would have seen the algebraic method for #2...Mine was sloppier...it worked 3 cases prooving that it was true...but it was kind of vague... Good luck guys!

My solution for number 2 was algebraic, but much longer; I manipulated it to a negative number on the LHS and a positive number on the RHS, which definitley wasn't as elegant as Silverfalcon's.

[quote="Silverfalcon"]We have to know that because of z_1+z_2 = z_2+z_3 = z_3+z_4 = z_4 + z_5, we get: [/quote]

How come? Can you elaborate?

Its says that $\mu$ is defined by $z_1+z_2$,  $z_2+z_3$, $z_3+z_4$, $z_4 + z_5$

It took me a while to realize that it meant that they all must be equal...
I felt so dumb after I noticed it. :?

So...
$\mu=z_1+z_2 = z_2+z_3 = z_3+z_4 = z_4 + z_5$

RC-7th, I am just curious to see what your solution was for number 2.  Could you senc it it to me somehow?

RC-7th's answer:

Given that $a,b,c$ represent the length of the side of the $\Delta {ABC}$ and $k$ is any positive integer. Prove that for all $a,b,c,k$, the following inequality holds true:

$\sqrt {\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c}} < \sqrt {2k^2}$[/quote]

By the triangle inequality: \[a<b+c\Rightarrow a-b-c<0\], \[b<a+c\Rightarrow b-a-c<0\], and \[c<a+b\Rightarrow c-a-b<0\]

By definition, the product of three negative numbers is negative, so \[(a-b-c)(b-a-c)(c-a-b)\] is negative.

By definition, \[(a-b-c)(b-a-c)(c-a-b)<3abc\] because a negative number is less than a positive number.

By the distributive property: \[(a-b-c)(b-a-c)(c-a-b)=a^3+b^3+c^3-a^2b-a^2c-ab^2-ac^2-b^2c-bc^2+2abc\], hence \[(a-b-c)(b-a-c)(c-a-b)<3abc\Rightarrow a^3+b^3+c^3-a^2b-a^2c-ab^2-ac^2-b^2c-bc^2+2abc<3abc\]

We may freely add or subtract the same thing from both sides of an inequality, hence \[a^3+b^3+c^3-a^2b-a^2c-ab^2-ac^2-b^2c-bc^2+2abc<3abc \Rightarrow a^3+b^3+c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+3abc<2a^2b+2a^2c+2ab^2+2ac^2+2b^2c+2bc^2+4abc\]

By the distributive property, \[(a^3+a^2b+a^2c+abc)+(b^3+ab^2+b^2c+abc)+(c^3+ac^2+bc^2+abc)=a(a+b)(a+c)+b(a+b)(b+c)+c(a+b)(b+c)\] and \[2[(a^2b+a^2c+ab^2+abc)+(abc+ac^2+b^2c+bc^2)]=2(a+b)(b+c)(a+c)\], hence \[a^3+b^3+c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+3abc<2a^2b+2a^2c+2ab^2+2ac^2+2b^2c+2bc^2+4abc\Rightarrow a(a+b)(a+c)+b(a+b)(b+c)+c(a+b)(b+c)<2(a+b)(b+c)(a+c)\]

By definition, \[(a+b)(b+c)(a+c)\] is positive because it is the product of three positive numbers, each of which is the sum of two positive numbers, hence we may divide both sides of the inequality by it: \[a(a+b)(a+c)+b(a+b)(b+c)+c(a+b)(b+c)<2(a+b)(b+c)(a+c)\Rightarrow\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}<2\]

Both sides are positive, hence we may take the squareroots of them: \[\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}<2\Rightarrow\sqrt{\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}}<\sqrt{2}\]

k is non-negative, hence we may multiply both sides of the inequality by it : \[\sqrt{\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}}<\sqrt{2}\Rightarrow k\sqrt{\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}}<k\sqrt{2}\]

By definition, \[k\sqrt{\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}}<k\sqrt{2}\Rightarrow \sqrt{k^2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)}<\sqrt{2k^2}\]

By the distributive property, \[=\sqrt {k^2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)}=\sqrt {\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c}}\]

Farazy's answer:

I think this one was okay but I gave credit to RC-7th because personally in this type of problem, I prefer algebraic proof or inductive ones than case-by-case.

Plus, it was hard to see whether he had achieved all the points I was looking for. :?

Here it is:

The value of $\epsilon$ in this proof is defined as an infintesimal value.

We are given that: $a$,$b$, and $c$ are lengths of the sides of triangle ABC and $k$ is any positive integer.

We need to prove that:
$\sqrt {\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c}} < \sqrt {2k^2}$

Let us begin by simplifiying this inequality.

We can square both sides of our inequality:
$\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c} < 2k^2$

and divide by $k^2$:
$\frac {c}{a+b} + \frac {b}{a+c} + \frac {a}{b+c} < 2$

Now we have proven that $k$ can be any positive integer let alone any number.

We only have to deal with $a$,$b$, and $c$ now.  How much can we gather about these lengths?

A theorem states that any two sides of a triangle must be larger then the third side.
We can write this as three inequalities in terms of a, b, and c.
$a<b+c$
$b<a+c$
$c<a+b$

We can proove the inequality we were given true by developing the highest possible value of
$\frac {c}{a+b} + \frac {b}{a+c} + \frac {a}{b+c}$

How can we begin by developing the highest possible value?
We will begin by saying that for any values we select for $a$, $b$, and $c$, the following inequality is true.
$a\le b\le c$

We will work different cases that could possible develop the highest possible existing value.
The cases will be structured as follows.

1.$c$ has the highest possible value:
   a)while $a$ is almost equal to $b$
   b)$c$ has the highest possible value, while $b$ is almost equal to $c$
2.$c$ is equal to $b$ and is equal to $a$.

Case one part one:
The highest possible value of $c$ is always extremely close to $a+b$.
This can be expressed as $a+b=c+\epsilon$

This makes the value of $\frac{c}{a+b}$ an infintesimal away from one.  Since the value of $\epsilon$ is so small, we can completely ignore that it is there.

From this we can reduce our original inequality
$1+ \frac {b}{a+c} + \frac {a}{b+c} < 2 \Rightarrow  \frac {b}{a+c} + \frac {a}{b+c} < 1$

Now we can go into our two subcases.
In subcase a, $b$ is equal to $a$.  This is true while $a+b=c+\epsilon$.  So we can say that $a=b=\frac{c+\epsilon}{2}$ again we can ignore $\epsilon$, giving us $a=b=\frac{c}{2}$.

We can now subsitute these values into our inequality
$\frac {\frac{c}{2}}{\frac{c}{2}+c} + \frac {\frac{c}{2}}{\frac{c}{2}+c} < 1 \Rightarrow \frac {\frac{c}{2}}{\frac{3c}{2}} + \frac {\frac{c}{2}}{\frac{3c}{2}} < 1\Rightarrow \frac{1}{3}+\frac{1}{3}<1 \Rightarrow \frac{2}{3}<1$  which is true.   We can plug in our values  to simplify the case showing us  $1\frac{2}{3}<2$.  All of this is true showing us that the first case and subase (a) is true.

Now let us go into subcase b.
In this subcase, $c$ again is almost equal to $b+c$, giving us $a+b=c+\epsilon$.

we are also saying that b is to almost equal c.  So $c=b+\epsilon$.  Since $\epsilon$ can be ignored we can say that $c+b$.  This gives $a$ the value of 0, or more correctly $\epsilon$.

Using these values, we have:
$a+b=c+\epsilon$ and when $\epsilon$ is ignored $a+b=c$
$b+\epsilon=c$ and when $\epsilon$ is ignored $b=c$
$a=\epsilon$  and when $\epsilon$ is ignored $a=0$
Plug these values into our inequality.

$\frac {c}{c+0} + \frac {c}{0+c} + \frac {0}{c+c} < 2 \Rightarrow \frac {c}{c} + \frac {c}{c} + \frac {0}{2c} < 2 \Rightarrow 2<2$
That seems to have prooven everything wrong...but since we ignored $\epsilon$ our value is of by an infintesimal.  So lets reintroduce $\epsilon$ accordingly.

Now we have $2-\epsilon<2$ which is true, proving that subcase (b) is true.

Now for the final case, where $a=b=c$.

We can plug this into our original inequality:
$\frac {c}{c+c} + \frac {c}{c+c} + \frac {c}{c+c}<2 \Rightarrow \frac{1}{2}+\frac{1}{2}+\frac{1}{2}<2 \Rightarrow \frac{3}{2}<2$ which is true showing us that the final case is true.

What was the highest value?

In case 1a. we got something close to 1.6666666666666666...
In case 1b. we got something close to 1.9999999999999999...
In case 2 we got 1.5

As we can see, case 1b. was the highest possible case, which was true proving that for any triangle with sides $a$,$b$,$c$ the following is true:
$\frac {c}{a+b} + \frac {b}{a+c} + \frac {a}{b+c} < 2$
and we proved that $k$ can be any positive integer.

Proving that the inequality:
$\sqrt {\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c}} < \sqrt {2k^2}$
Is always true for the values given.

I didn't even try the problems, but if I did I could have gotten both right. :(  :(  :(  :(


Given two circle $M$ and $N$, we say that $M$ bisects $N$ if they intersect in two points and the common chord is a diameter of $N$. Consider two fixed non-concentric circles $C_1$ and $C_2$.
a) Show that there exists infinitely many circles $B$ such that $B$ bisects both $C_1$ and $C_2$.
b) Find the locus of the centres of such circles $B$.

See the 1st part of the solution of problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=57715]All such circles pass through two fixed points[/url].

Let $O_1 = (x_1, y_1)$, $O_2 =(x_2, y_2)$ , $r_1, r_2$ be the centres and radii of $C_1, C_2$ respectively. A circle of center $B$ and radius $R$ meets the criteria if and only if $$\overline{BO_1}^2+r_1^2=R^2=\overline{BO_2}^2+r_2^2$$ by the pythagorean theorem applied to the triangles connecting $B$, $O_i$ and the point of intersection of both circles (it is clear that $B$ lies on the perpendicular bisector of the common chord). This equation simplifies into a linear equation of the form $$2\Delta_xx+2\Delta_yy =\Delta_{x^2}+\Delta_{y^2}-\Delta_{r^2},$$ where we write we write $\Delta_z = z_2-z_1$ for short. We can observe that the locus described by this equation is a line perpendicular to the segment $\overline{O_1O_2}$, and that's a good place to stop.


Three farmers sell chickens at a market. One has 10 chickens, another has 16, and the third has 26. Each farmer sells at least one, but not all, of his chickens before noon, all farmers selling at the same price per chicken. Later in the day each sells his remaining chickens, all again selling at the same reduced price. If each farmer received a total of 35 dollars from the sale of
his chickens, what was the selling price before noon and the selling price after noon?

[hide="ok,"]The farmers sold $ x,y,z$ chickens before noon, respectively. Clearly $ x,y,z$ are positive integers

Let $ a=b+c$ be the price per chicken before noon and $ b$ the price after noon

The first farmer sold $ x$ chickens before noon, so he took $ x\cdot (b+c)+(10-x)\cdot b = \boxed{10\cdot b+x\cdot c}$ dollars

Similarly, the 2nd farmer took $ 16\cdot b+y\cdot c$ dollars and the 3rd took $ 26\cdot b+z\cdot c$ dollars

$ \begin{array}{l}10\cdot b+x\cdot c = 26\cdot b+z\cdot c \\ 16\cdot b+y\cdot c = 26\cdot b+z\cdot c \\ \end{array}\}\Rightarrow \begin{array}{l}(x-z)\cdot c = 16\cdot b \\ (y-z)\cdot c = 10\cdot b\\ \end{array}\}\Rightarrow \begin{array}{l}(x-z) = 8\cdot \frac{2b}{c}\\ \\ (y-z) = 5\cdot \frac{2b}{c}\\ \end{array}\}\Rightarrow$

$ \begin{array}{l}x-z = 8t \\ y-z = 5t\\ \end{array}\}\ \ (1)$, where $ t=\frac{2b}{c}>0$

From $ (1)$ we can see that $ 8t$ and $ 5t$ should be integer numbers. And since $ (8,5)=1$ we get that $ t$ is an integer (positive)

So we have $ (x,y,z) = (z+8t, \ z+5t, \ z)$

If we replace $ z=1$ and $ t=1$ at $ x=z+8t$ we'll get $ x=9$.
For any $ z>1$ or $ t>1$ we'll get $ x\geq 10$, impossible
$ \hline$
So the only solution is $ z=1$ and $ t=1$

$ \frac{2b}{c}=1\Rightarrow 2b=c$

$ 26\cdot b+z\cdot c=35 \Rightarrow$
$ 26\cdot b+1\cdot (2b)=35 \Rightarrow$
$ 28\cdot b=35 \Rightarrow \boxed{b = \frac{5}{4}}$

And $ a=b+c = 3b\Rightarrow \boxed{a=\frac{15}{4}}$

and $ (x,y,z)=(9,6,1)$
$ \hline$[/hide] but is there an elegant solution?


find all integer m and n s.t$m|n^2+1 ,n|m^3+1$

I remember posting this. I think it was never solved :) (I don't have a solution eihter; I posted it in the "Unsolved Problems" section).


Taken off Kalva site, which does not supply a solution:

Show that a square with side 1 cannot be covered by five squares with side less than 1/2.

Isn't this sort of easy?

The only way in which a corner can be covered is to have a small square share that corner with the large square, and this takes care of 4 small squares (there are 4 corners of the large square). Since the small squares have side <1/2, it means that the midpoints of the sides of the large square aren't covered by the first 4 small squares, so they all must be covered by the 5'th. This means that the 5'th small square must contain a segment of length 1 (the distance between 2 opposite midpoints of the sides of the large square), so its diagonal is >=1, so its side is >=1/\sqrt2 >1/2.

Yes, in that case the problem is easy.
I think it is maybe more difficult if the little squares are allowed to overlap regions outside of the big square(that is the big square is not assumed to cover each of the little squares).

Pierre.

Hmmm... I suppose you're right; nobody said that the big square has to cover the 5 small squares. I think the problem was meant in this more general way.

Well, we can say one thing: for each of the vertices of the big square and the center of the big square there is one and only one small square which covers it.

I'm not sure how we should continue. I had this idea about looking at the intersections between the center square and the corner squares, but it seems a bit boring because we have to consider some cases.

I think I tried this one before without success. But since this problem was proposed on Iberoamerican 2001 it should have a nice idea.

We had an earlier discussion of this problem at the link
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12615[/url].

rcv gave a complicated proof.  I didn't check the details.

My last post in that topic proposed an alternative approach, which I never completed.  In the large square, draw the two segments connecting the midpoints of opposite sides.  Consider any square of side length less than 1/2.  Look at the intersection of that small square with the two midsegments and the boundary of the large square.  My conjecture was that the measure of the intersection with the (boundary of the) large square, plus half the measure of the intersection with the two midsegments, is less than 1.  A proof of that conjecture would solve the problem.

It is not a new idea for me :D But why "half the measure of the intersection with the two midsegments"?

My original conjecture was without the word "half".  But that conjecture turned out to be false.  See my second post in that other topic.

Sorry, I still don't understand the meaning of this "half". Anyway, this aproach is computational, and we need a nice idea.


[quote="Dr Sonnhard Graubner"]hello, what do you mean with $ [ABC]$? the area of the given triangle?
Sonnhard.[/quote]

Yes my Friend [color=red][b]Sonnhard[/b][/color]


Given a line segment $AB$ in the plane, find all possible points $C$ such that in the triangle $ABC$, the altitude from $A$ and the median from $B$ have the same length.

The requested set of points $C$ is a sum of two circles passing through $B$ without point $B$.
Let $M$ be the midpoint of $AC$ and $N$ be the midpoint of $AB$. Let $O$ be the vertex of equilateral triangle $NBO$.
Denote the following circles:

$o_{1}=(N,NB)$

$o_{2}=(O,OB)$

$o_{3}=(B,BM)$

Let us take such point $C$ that $M$ lies on circle $o_{2}$. Let $P$ be the intersection of $o_{1}$ and $o_{3}$ on the other side of $AB$ than $C$. Obviously, $AP$ is tangent to $o_{3}$, therefore $BP\perp AP$. As $\triangle BON$ is equilateral, also $\triangle BMP$ is equilateral. So the altitude of $\triangle BMP$ from $M$ passes through $N$. Hence, $MN\perp BP$. This implies $AP\parallel BC$. So $BP$ is the length of the altitude of $\triangle ABC$ from $A$. But $BP=BM$, so our point $C$ satisfies the necessary conditions. For a given altitude from $A$ there is only one "good" point $C$ for each side of line $AB$. Therefore, the requested set of points $C$ is a circle $o$ homothetic to $o_{2}$ with respect to $A$ in scale $k=2$  and a circle symmetric to $o$ with respect to $AB$ (excluding point $B$).


calculus
integration
trigonometry
logarithms
calculus computations

Given $ f'(x)\equal{}\frac{\cos x}{x},f(\frac{\pi}{2})\equal{}a,f(\frac{3\pi}{2})\equal{}b$

Find $ \int^{\frac{3\pi}{2}}_\frac{\pi}{2} {f(x)}dx$ in terms of $ a,b$.

$ \{xf(x)\}'\equal{}f(x)\plus{}xf(x)$ Integrating both sides for $ \frac{\pi}{2}\leq x\leq \frac{3\pi}{2}$ gives the answer $ 2\minus{}\frac{\pi}{2}(a\minus{}3b)$.

[quote="kunny"]$ \{xf(x)\}' \equal{} f(x) \plus{} xf(x)$[/quote]

$ \{xf(x)\}' \equal{} f(x) \plus{} x\color{red}{f'(x)}$???

Yes, you are right, thank you!, but the answer is right, I think.

Ok, but I am little busy, please wait for a while.

This is a present for you, Rajiv.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=370060&search_id=739643759#370060[/url]

[quote="Rajiv"][quote="kunny"]Integrating both sides for $ \frac {\pi}{2}\leq x\leq \frac {3\pi}{2}$ gives the answer $ 2 \minus{} \frac {\pi}{2}(a \minus{} 3b)$.[/quote]

I couldn't appreciate your method very well. Could you please explain a little? :oops:[/quote]
hello
after integrating it will come:
$ \frac {3\pi }{2}f(\frac {3\pi}{2}) \minus{} \frac {\pi}{2}f(\frac {\pi}{2}) \equal{} \int^{\frac {3\pi}{2}}_{\frac {\pi}{2}}f(x)$+$ \int^{\frac {3\pi}{2}}_{\frac {\pi}{2}}[cosxdx]$
I think now u understood what kunny did.
Thank u

[quote="kunny"]

$ \{xf(x)\}' \equal{} f(x) \plus{} xf'(x)$

[/quote]
Yesssssssss! I get it! :wow:  :clap:

How stupid I was being :fool: , I thought there was some application of the above equation! So it was simple integration...

Thanks [b]kabi[/b]. Nice solution [b]kunny[/b]! :thumbup:

Thank you!  :lol:

For example $ \int x\sin x\ dx$, we imagin $ ( \minus{} x\cos x)' \equal{} \minus{} \cos x \plus{} x\sin x$ in our head, or $ \int x\sin x dx \equal{} \minus{} x\cos x \minus{} \int ( \minus{} \cos x)\ dx$ that is to say,

Subtract $ \int ( \minus{} \cos x)\ dx$ from $ \minus{} x\cos x$, which says Subtract by integrating unnecessary term from $ \minus{} x\cos x$. Another example:

$ \int x\ln x\ dx$

$ \left (\frac {x^2}{2}\ln x\right) ' \equal{} x\ln x \plus{} \frac {x}{2}$, thus we would subtract $ \int \frac {x}{2}\ dx$ from $ \frac {x^2}{2}\ ln x$.

[quote="kunny"]
$ \int x\ln x\ dx$

$ \left (\frac {x^2}{2}\ln x\right) ' \equal{} x\ln x \plus{} \frac {x}{2}$.
Thus we would subtract $ \int \frac {x^2}{2}\ dx$ from $ \frac {x^2}{2}\ ln x$.[/quote]
I'll remember that. thanks a ton!!! :D

[quote="kunny"]Sorry, I made a typo and have just edited. Please make a confirmation.[/quote]

Yet I got the method... :D

[quote="kunny"]Thank you for explaning that in place of me. :)[/quote][quote="Rajiv"]Thanks kabi. [/quote]
My pleasure guys.


trigonometry
geometry
circumcircle
incenter
inradius
analytic geometry
geometry theorems

In one of the posts there was an identity ($NI=3GI$) used to solve an ineq ($NI\cdot HI \leq OI^2$). Are the any other interesting identities between important points? These are all I know:

1) $OI^2=R^2-2Rr$
2) $AH=2R\cos A$
3) $OH^2=R^2\left(1-8\cos A\cos B\cos C\right)$
4) $IA=r/\sin\frac{A}{2}$
5) $OH=3OG$
6) $3GH=2OG$
7) $AI_a=p\cos\frac{A}{2}$
8) $OO_9=\frac{R}{2}\cdot\sqrt{1-8\cos A\cos B\cos C}$
9) $GI^2=\frac19\left(p^2+5r^2-16Rr\right)$
10) $O_9I=\frac{R}{2}-r$
11) $NI=3GI$

Hereby, ABC is an arbitrary triangle, O is its circumcenter, I is its incenter, $I_a$, $I_b$, $I_c$ are its excenters, H is its orthocenter, G is its centroid, N is its Nagel point, and $O_9$ is its nine-point center. Also, $p=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, and R and r are its circumradius and inradius, respectively.

I would love to see other identities.. thank you!
cheers!

[color=red][Moderator edit: Defined notations.][/color]

By the way..
can someone help me with 9).. a hint or a solution? :D

I think the best way to approach these are baricentric coordinates. Here's what I tried :

You can see easily that the baricentric coordinates of I are $\frac{a}{2p}$, $\frac{b}{2p}$ and $\frac{c}{2p}$, so you write $\overrightarrow{GI} = \frac{a}{2p}\cdot\overrightarrow{GA}+\frac{b}{2p}\cdot\overrightarrow{GB}+\frac{c}{2p}\cdot\overrightarrow{GC}$, using vectors. Then you square this sum (scalar product). It's just straightforward calculations from now on.


Prove AM-GM in the following way:

Let AM-GM-n denote AM-GM for n variables.

1. Prove that AM-GM-2 holds. (trivial)
2. Prove that AM-GM-n implies AM-GM-2n
3. Prove that AM-GM-n implies AM-GM-(n-1)

1.
Let A and B be our variables: $A,B \ge 0$
$\frac{A+B}{2}\ge \sqrt{AB}$
$(A+B) \ge 2 \cdot \sqrt{AB}$
$(A+B)^{2}\ge 4AB$
$A^{2}+2AB+B^{2}-4AB \ge 0$
$(a-b)^{2}\ge 0$
True by the trivial inequality.

2.
I'll post this when I get back from a party.

[quote="alex__ib"][quote]2. Prove that AM-GM-n implies AM-GM-2n
3. Prove that AM-GM-n implies AM-GM-(n-1)[/quote]

I don't get the text, when you will explain i will come with a solution  :)[/quote]
2. AM-GM for $n$ variables implies AM-GM for $2n$ variables.
3. AM-GM for $n$ variables implies AM-GM for $n-1$ variables.

13375P34K43V312, you spelled derivative wrong in your signature.  :wink:

Assume we've proven AM-GM for $n$ numbers. Let $a_{1},\dots,a_{2n}$ be $2n$ non-negative numbers. Then

$\frac{a_{1}+\dots+a_{2n}}{2n}=\frac{\frac{a_{1}+\dots+a_{n}}{n}+\frac{a_{n+1}+\dots+a_{2n}}{n}}{2}$

Now apply previously proven AM-GM-2 and AM-GM-n to that.

Part 3 coming soon :)

Good job, Kurt Gödel! :)

This is actually the only instance of such sort of "reverse induction" I know of, or jump-first-patch-it-up-later induction :)

I don't like to call this proof induction, technically induction is when you apply the principle of mathematical induction.

2006 Texas A&M Power Team Question 7:

Prove max>AM>GM>HM>min


Harvard college Princeton MIT calculus trigonometry AMC

Business and management requires a load of mathematical knowledge.

Now, how many years, what universities, and what courses do you have to take to have a "good" high-salary job in the business world?

What schools and colleges are recommended for doing business and management?

Cornell? Harvard? Princeton?

[quote="hwenterprise"]Now, how many years, what universities, and what courses do you have to take to have a "good" high-salary job in the business world?[/quote]
This depends on so many factors and thus it cannot be answered. What is "good" to you? What job? What field? "Business" is a bit too broad.
[quote="hwenterprise"]What schools and colleges are recommended for doing business and management?

Cornell? Harvard? Princeton?[/quote]I am not expert on business schools, but I think Wharton at UPenn and Sloan at MIT are among the best. I don't think you can go wrong if you attend any one of the Ivy-Leagues or "top tier" schools.

while it's true that in "business" in general, math is needed, you don't need to have an extensive knowledge of math in management, unless you're a CFO or something. The people in management typically leave the math work for their employees :) , and even if you're an employee, you don't typically need to know calculus or trig in business. at most, intermediate algebra might be needed for cost-benefit analysis or something....

[quote="Spoon"]from what I've observed in my 4 summers of interning, from a management standpoint, the more statistics you know, the more valuable you are to a corporation. USAMO type math goes out the window.[/quote]

Problem-solving skills do not go out the window.  For corporations that are aware, or that you can make aware, of the skill and selectivity level that a USAMO victory (or even USAMO qualification) represent, it will be considered highly relevant.   It doesn't matter whether Euclidean geometry puzzles are literally part of the job, but whether you are demonstrably analytically competent, and competitions are one of the few avenues of demonstration available outside of an employment history.  The competitions leading to the USAMO are the only ones in the US with tens of thousands of participants, and therefore establish a more meaningful level of selectivity.

furthermore, how on earth do you infer "problem solving skills" from "USAMO math". Honestly...

[quote="Spoon"]Of course problem solving skills are helpful in just about everything someone can do in life. But you can have excellent problem solving skills and not be able to answer a single USAMO question... I was talking specifically about the math.  [/quote]

You appeared to claim that knowledge of statistics is valuable (true) and that USAMO competence is irrelevant (superficially true but substantively very false).    There is virtually no job where USAMO participation is a prerequisite, but many many jobs where having reached that level of competence reflects valuable skills (and will be taken to indicate that by employers familiar with the competition or its selectivity level).

[quote] I would not agree with the claim that USAMO quals have better problem solving skills than those that do not qualify, necessarily, they merely have a more specialized set of problem solving skills.[/quote]

How do you suppose the specialized skills are acquired, and what sort of personal attributes might be involved?
Why do some people acquire them more than others, given similar (or lesser) resources?

There is no crystal ball that predicts specific, job-relevant problem solving skill short of having held an equivalent job.
However, as predictors go, USAMO qualification (not to mention USAMO top scoring) is far better than most other credentials I can imagine.   It indicates some mixture of traits such as intelligence, drive, ambition, work, and of course, problem solving skills over and above the specialized knowledge that you mention.

[quote="Spoon"]Of course problem solving skills are helpful in just about everything someone can do in life. But you can have excellent problem solving skills and not be able to answer a single USAMO question... I was talking specifically about the math. I would not agree with the claim that USAMO quals have better problem solving skills than those that do not qualify, necessarily, they merely have a more specialized set of problem solving skills.[/quote]

Sure it's not a for sure thing that USAMO qualifier has better problem solving skills than a non-qual, but all other things being equal, which would you hire?  It's an indicator, just like lots of other things.  It's not 100% accurate, but it's a much better measure than many other things which are commonly used to judge problem solving skills (and willingness to work hard towards a goal for an extended period of time).

[quote="Spoon"]from what I've observed in my 4 summers of interning, from a management standpoint, the more statistics you know, the more valuable you are to a corporation. USAMO type math goes out the window.[/quote]

This depends on what you are doing.  For a very clearly-defined algorithmic job (i.e. one that entry-level and interns are likely to have), this is likely true.  For higher-level work, it almost certainly isn't true that problem solving skills are less important than technical knowledge.  Basically, companies that hire USAMO people are making the sensible wager that the USAMO success is an indicator that they can learn the technical skills.  Put simply, a stat jockey with no problem solving skills is going to have a lot harder time expanding her/his use to a company than a problem solver who doesn't have a stat background.  The problem solver can learn the requisite stats very quickly.

As for inferring problem-solving skills from USAMO success, I'll note that success on the AIME/USAMO requires the very fundamental problem solving skills that define what 'problem solving' is - applying basic tools in combination to attain complex desired results.

Rrusszyc (sorry, butchered it),

I was wondering what companies actually know about the USAMO, and respect it. I know the companies sponsoring AoPS: Google, DE Shaw, Jane Street do, but what others?

Well, Business and Management type employers are hardly the type to look for brilliant, or even especially good mathematicians.  In light of that, I can see why stats skills would be desired.  Of course, in a job where you do things that are moderately sophisticated with numbers, skills like knowledge of stochastic processes and whatnot would become more useful than oversimplified statistical models.

However, one would wonder why you keep on bringing up USAMO type abilities.  Honestly, do you resent *those* people?  Why do you insist that, say, people who are able to solve USAMO problems do not have problem solving skills?  I'd like to think that I know a fair number of people who have excellent problem solving skills, whether in math, or physics, or computer science - but I can say all of the people I know who went to olympiads have excellent problem solving skills.  None of them got that way from mindlessly doing exercises - they are without exception very bright and very insightful.

[quote="rrusczyk"]This depends on what you are doing. [/quote]

The question was quite specific, "business and management", and I take that to mean more of the "business management" side rather than "technical job". I think "all other things being equal" isn't a very realistic situation. Management is a whole other world from the technical aspect of business, and it's basically how your people skills are, which, and sorry for being stereotypical, I think that the USAMO for the most part would not win as far as that is considered.

CLEARLY, if he was refering top a "very clearly-defined algorithmic job", my statement does not hold...

[quote="blahblahblah"]However, one would wonder why you keep on bringing up USAMO type abilities.  Honestly, do you resent *those* people?  Why do you insist that, say, people who are able to solve USAMO problems do not have problem solving skills?[/quote]

Well let me assure you, if I was judging solely by you, I would be fairly certain that USAMO quals do not possess any reading skills... because that is not what I said nor implied! You're such an idiot. Where do you come off making such an accusation?!

If you already think I'm such an idiot, then there's really not much point in responding, is there.  In any event, you have far too many opinions, formed with far too little life experience, and I fear that someday someone is going to read your posts without a grain of salt and be led into making a wrong decision.

*EDIT*
_______

I would like to personally apologize to blahblahblah for the religious comment in this post. In my haste, I mistook you for someone else. You're right, you agreed with me. I really am sorry for that.

*2nd Edit*: Thinking about more, I realized that I have been under this incorrent impression for quite some time, and I wrongly thought you were the one that got so personally upset with me. Looking at it now, I have no idea why I thought it was you :(, especially when you were one of the few who agreed with me. I feel really bad now :(, I hope you can forgive me.
_______

It is not harming anyone to hear what I've said about management. The day my manager tells me "Oh you should have referenced Pick's Theorem when describing that slide" is the day I quit.

Of course USAMO quals have EXCELLENT problem solving skills. That DOESNT MEAN that a non USAMO qual isn't just as good, especially in regards to a non mathematical subject. That is all I said, reread my post 6 times if that's what it takes you. It was fleeting_guest who originally made the completely ignorant (yes I said ignorant, it was indeed ignorant) change from "USAMO Math" to "Problem solving skills", and other seemed to follow. If anyone does not see how the two are different, go back to middle school, regardless of your age, because you quite obviously spent too much time on math and not enough on english.

I am very upset, as you can clearly tell. Why am I upset? Because people immediately attack me and my statement without bothering to actually read what I said. I even plainly stated "from a management standpoint" implying specifically what I meant. If you think that your day as a manager will revolve around math in way or form, you are sadly mistaken. If this forum wasn't more one sided, many more people would open to my opinion(s).

Spoon, I made two posts in the religion thread.  The only relevant one would be this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33938&postorder=asc&start=101

in which I specifically noted that I agree with you!  So I'm not sure what you're getting at.

As for fleeting_guest's comment, it seems to me that he is responding to your statement in post 4 (don't bother editing it, as fleeting_guest already quoted you) that USAMO-type math goes out the window.  In fact, his post seems to be quite relevant, and I imagine that most others would concur.

Next, you say that I've spent too much time on math and not enough on english.  That may be, but I'm not the one who came up with such gems as 'in many times', 'you blatently', 'to make me out to look bad', 'I will in turn not response to'.  I don't like to sink to the level of grammar nitpicking, but then again, I don't like people challenging my english skills.  I ran my high school newspaper way back when, I won essay contests - in fact, I firmly believe that I would be an English major if I wasn't a math major.  I will, however, concede that I misread one of your statements earlier - that is, I read infer as imply in post 7, and I thought that you were just misusing the word.

Fair enough - it's an easy mistake to make, and I certainly won't hold it against you.

I know that I have disagreed with you on a variety of topics recently, but it is nothing personal.  Rest assured that I don't follow you around looking for points on which to start conflict.

[quote="Spoon"]It is not harming anyone to hear what I've said about management. The day my manager tells me "Oh you should have referenced Pick's Theorem when describing that slide" is the day I quit. [/quote]

The following are direct quotations from postings in this thread:
---"It doesn't matter whether Euclidean geometry puzzles are literally part of the job"
---"There is virtually no job where USAMO participation is a prerequisite"

Apparently nobody is under the illusion that USAMO material is part of any job description.
The actual issue raised was that demonstrated mastery of such material (USAMO qualification or victory) is nevertheless meaningful (if not all-decisive) in the job market, especially for the first few years where there is not much prior employment history.

[quote]Of course USAMO quals have EXCELLENT problem solving skills. That DOESNT MEAN that a non USAMO qual isn't just as good, especially in regards to a non mathematical subject. [/quote]

Has anybody here actually claimed that a non-qualifier is automatically less good?  Is there any statement you can quote from this thread that would support such a ridiculous interpretation?

[quote="Spoon"]blahblahblah: Read Edits above[/quote]

Wow...that's spam, right?

Oooppps...this is too. :(

No, it was not spam... If a new post was not made, it is likely that blahblahblah would not have looked at the thread, thus he would not have seen my edit. Additionally, if he did look a thread that had no replies, it was not a guarantee that he would see my edit without me pointing it out.

[quote="rrusczyk"][quote="Spoon"]from what I've observed in my 4 summers of interning, from a management standpoint, the more statistics you know, the more valuable you are to a corporation. USAMO type math goes out the window.[/quote]

This depends on what you are doing.  For a very clearly-defined algorithmic job (i.e. one that entry-level and interns are likely to have), this is likely true.  For higher-level work, it almost certainly isn't true that problem solving skills are less important than technical knowledge.  Basically, companies that hire USAMO people are making the sensible wager that the USAMO success is an indicator that they can learn the technical skills.  Put simply, a stat jockey with no problem solving skills is going to have a lot harder time expanding her/his use to a company than a problem solver who doesn't have a stat background.  The problem solver can learn the requisite stats very quickly.[/quote]

"Stat jockeys" are problem solvers.  Careers in applied statistics require a substantial background in mathematics (usually at least a bachelors degree or equivalent) in addition to probability, inferential statistics, experimental/survey design and application knowledge.

It would be very difficult to launch a career in statistics without substantial formal training (Masters or PhD) in the field.  Many mathematics students will find these careers interesting - even those with theoretical interests.  Perhaps stat careers should be discussed in a separate subtopic.  If anyone is interested, I'd be happy to discuss them.


I can't seem to think of a more elegant solution to this problem. (One of those fancy geometric interpretations maybe?) I'd hope that there is one, as this problem seems somewhat interesting. Can someone lend me a hand? :D

Problem: Let $f(x)=(x+3)^{2}+9/4$ for $x\ge-3.$ Compute the shortest possible distance between a point on $f$ and a point on $f^{-1}.$

My Brute Force Methodology: The minimum distance from $f$ to $f^{-1}$ should be twice the minimum distance from $f$ to the line $x=y.$

To find this, we use the point-to-line distance formula. Let the coordinates of the nearest point on $f$ be $(x_{0},y_{0}).$ It follows that the desired distance is (I hope?)
\begin{eqnarray*}2\times \frac{|x_{0}-y_{0}|}{\sqrt{2}}&=& \sqrt{2}|x_{0}-(x_{0}+3)^{2}-9/4| \\ &=& \sqrt{2}|(x_{0}+5/2)^{2}+5| \\ &=& 5\sqrt{2}. \end{eqnarray*}

uhhh...you just want the 2 times the distance from y=x and the tangent to f(x) with a slope of 1


Hey, Miren esto!!!

[url]http://www.awesomemath.org[/url]

pascual me lo dijo leo, pero de  nada vale ir para mi porque es depsues de la imo...
cierto?

o crees que valeria la pena ir solo para aprender y para putnam?

Pascual2005 sé que vas a participar de este evento... podrías comentar algo de este evento a la comunidad ?

$Tipe$

Claro:

Este es un campo de verano que se llevara a cabo en Austin Texas, dirigido por Titu Andreescu y Zuming Feng. Esta basado creo en el MOSP (Mathematical Olympiad summer program) que es la preparacion de EEUU para la IMO, con la diferencia de que este campo esta dirigido a jovenes de todo el mundo que deseen y puedan asistir.

Resulta muy recomendable ya que Titu y Zuming son de los dos mayores pioneros de  olimpiadas matematicas en el mundo, ademas asistiran como mentores y asistentes 15 jovenes de todo el mundo con buenas participaciones a nivel internacional, incluyendo a Leo y a mi.

Entonces espero se animen a aplicar, les gustara!


geometry
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USAMO
Ross Mathematics Program

I am trying to prepare for the 05' USAMO and I really stink at geometry.  Does anyone happen to know of any good books or other resources that would be helpful in preparing for USAMO geometry problems? Thanks.

Thanks, I'll see if I can get ahold of those books.

I would also suggest [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883856395/qid=1092737678/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-6043065-0271943?v=glance&s=books&n=507846]Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry (New Mathematical Library)[/i], Washington 1995[/url]. Also, there are lots of nice internet ressources, like [url=http://www.cut-the-knot.org/geometry.shtml]Alexander Bogomolny's "Cut the Knot"[/url] and some other sites [url=http://de.geocities.com/darij_grinberg/Links.html]linked from my page[/url].

  Darij

[quote="zscool"]I would recommend [i]Geometry Revisited[/i] by Coxeter and Greitzer

and [i]Challenging Problems in Geometry[/i] by somebody both books should provide more than enough material to tackle usamo geometry problems.[/quote]

Challenging Problems in Geometry is by Alfred S. Posamentier, I believe.


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[quote="RacerX"][url]http://mathworld.wolfram.com/PolygonArea.html[/url]

This formula is very interesting but I still have some questions:

1. What's the "name" of this formula?
2. Does it only work for convex polygons?
3. Where can I find its proof?

Post some links if you can.[/quote]

I believe this is what everyone calls "shoestring" because of the way you multiply the numbers.

True, my coach called it the "Shoestring Theorem" or "Shoelace Theorem". I don't know if it can be used with a figure though or where to find the proof. Try a Google Search  :P .

It works for any figure in the plane, if you take the points in clockwise or counterclockwise order. The general method of proof is to first divide the figure into convex polygons, then smaller triangles, and then apply the general formula for triangles which is easily found by the cross product (involves vectors), which is of course not MATHCOUNTS level. It's mainly just tedious; there is also a more MATHCOUNTS level technique to find the area of the triangles, but that is very messy.


What is the units digit of the sum $ 1! \plus{} 2! \plus{} 3! \plus{} 4! \plus{} 5! \plus{} \cdots \plus{} 1000!$?

1!=1 units digit=1
2!=2 units digit=3
3!=6 units digit=9
4!=24 units digit=3
5!=120 units digit=3
6!=720 units digit=3
7!=5040 units digit=3
...
...

If you take a close look, starting with 5 factorial, each one ends with a zero, thus, the question may as well be asking 1!+2!+3!+4!, giving a units digit of
[b]3[/b].

NOTE: The units digits are obtained from adding up all the previous terms.

We know that 5!=120 since there is a 2 and a 5. That means we only have to look at 4! and less for the units digit. So 4! ends in 4 3! ends in 6 2! ends in 2 and finally 1!  ends in 1. adding these together we find that the units digit is 3.


Let $ a_i$, for all $ i\in\left\{1,2,...,n\right\}$, be positive reals. Prove that:

$ a_1\plus{}\left(a_1a_2\right)^{1/2}\plus{}\left(a_1a_2a_3\right)^{1/3}\plus{}...\plus{}\left(a_1a_2...a_n\right)^{1/n} \leq \left(1\plus{}\frac{1}{n}\right)^n \cdot \left(a_1\plus{}a_2\plus{}...\plus{}a_n\right)$.

Sorry orl,but what 'a_1_a_2a_3' means??? :?

Ok, I got it. I assume that the a_i are positive  if not there would be some trouble with the n-th roots...
Now, the desired inequality follows easily from the two lemmas above (lemma 2 gives a stronger result than the desired one, since from lemma 1 we deduce that c_i/i is increasing) :

Lemma 1 :
The sequence U_n = (1 + 1/n) n  is increasing.

Proof :
From AM/GM :
((1 + 1/(n-1) ) n-1  ) n  < [(n-1)(1 + 1/(n-1) ) + 1]/n = 1 + 1/n, hence result.

Lemma 2 :
Let b_i = (a_1*...*a_i) 1/i  and c_i = i*(1 + 1/i) i .
Then, for each n :
 \sum {1 \leq i \leq n} b_i  <  \sum {1 \leq i \leq n} (c_i*a_i)/i.

Proof :
For each n, from AM/GM we have :
b_n  \leq (c_1*a_1 + ... + c_n*a_n)/(n*P_n) , where P_n = (c_1*...*c_n) 1/n .
But our choice of c_i leads to P_n = n+1. Thus :
b_n  \leq  (c_1*a_1 + ... + c_n*a_n)/[n(n+1)].
Since 1/[n(n+1)] = 1/n - 1/(n+1), we deduce that :
 \sum {1 \leq j \leq n} b_j  \leq  \sum {1 \leq j \leq n} ( \sum {1 \leq i \leq j} (c_i*a_i/j - c_i*a_i/(j+1) ) )
= ( \sum {1 \leq j \leq n} c_i*a_i/i ) -  \sum {1 \leq j \leq n} c_j*a_j/(n+1)
<   \sum {1 \leq j \leq n} c_i*a_i/i , as desired.

Pierre.

In good soviet book "Foreign mathematic olympiads" (1987) the source of this problem is indicated as "France, 1982". The proof in this book is the same as pbornsztein's one.
When I was 3-year student I proved in my term work general inequality. It contains a sum of averages of p-degree, i.e. (((a1)p+...+(ak)p)/k)1/p. The inequality mentioned above is special case for p=0.
Maybe it is well known generalization. Does anybody know?

???? Have you more informations about what 'France 1982' means?

Pierre.

Moreover, I do not understand what you mean by 'a 3 years student' (I cannot beleive that you were 3-years old :D ).
And I don not understand either what is your generalization : What is p? Why p = 0? Do you compare with what?

Pierre.

I have no supposition. Only remark on this problem is "Jury(or committee), France 1982".

Actually, "3-year student" is third year of education at university (as I see it).

Inequality I proved is
\sum {k=1,2,...} ((a1p+...+akp)/k)1/p \leq Cp * \sum {k=1,2,...} ak, where p<1, Cp=(1-p)-1/p and Cp can't be replaced by smaller value.

If p->0 then ((a1p+...+akp)/k)1/p -> (a1...ak)1/k.

Ok... So, I think your inequality is only valid for p < 1.
Thus, it is equivalent to theorem 326, which is attributed to Hardy, in :

G. Hardy, J.E. Littlewood, G. Polya, 'Inequalities', Second edition, Cambridge Mathematical Library.

Pierre.

Thank you, Pierre.
I will ask my friends to look through it in russian edition.

two other sources:

[1] T. Carleman: Sur les fonctions quasi-analytiques, Conferences faites au cinquieme congres des mathematiciens scandinaves, (Helsingfors, 1923), 181-196

[2] Herbert S. Wilf: Finite Sections of Some Classical Inequalities, (Berline, Heidelberg, New York, 1970), 70-79


y= ax + bz^2 (a and b are constants)

Find the fractional error in y(specifically through the logarithm method).

Fractional error in $ y$ is $ \frac {\Delta y}{y} \approx \frac {dy}{y}$
$ y \equal{} ax \plus{} bz^2$
$ dy \equal{} adx \plus{} 2bzdz$
$ \frac {dy}{y} \equal{} \frac {a\Delta x \plus{} 2bz\Delta z}{ax \plus{} bz^2}$

I had said to do by the [b]logarithm[/b] method specifically


Let $ n \in \mathbb{N}, n \ge 2$. Prove that if the numbers $ n, n^2\plus{}n\plus{}1$ and $ n^3\plus{}n^2\plus{}n\plus{}4$ are primes, then the numbers $ n^4\minus{}2$ and $ n^5\minus{}2$ are primes, too.

If $ n\equal{}1\mod 3$, then $ 3|n^2\plus{}n\plus{}1$,
if $ n\equal{}2\mod 3$, then $ 3|n^3\plus{}n^2\plus{}n\plus{}4$.
Therefore from $ n^2\plus{}n\plus{}1,n^3\plus{}n^2\plus{}n\plus{}4,n>1$ primes we get $ 3|n$ or $ n\equal{}3$.


$f(x): R--> R        ; g(x):  N --> N$
find all g(x) that the funtional equation have the proof
$f(f(x)+g(x))=g(x)$
I have tried for a long time but...........

[quote="Killer"]Function g is defined sur N ?[/quote]
oh i am sorry both f and g are from N to N

[quote="Peter VDD"]we need to find all couples f,g? or a g that verifies this for every f?[/quote]
Dear Peter
We "only" want to find g that verifies this for every f :D


[quote="pianoforte"]source: KCATM 2004

It took 3 hours to row a boat $18$ km against the current. The return trip with the current took $1 \frac{1}{2}$ hours. Find the speed of the rowboat in still water.[/quote]
[hide]
Speed against current: $\frac{18}{3}=6$
Speed with current: $\frac{18}{\frac{3}{2}}=12$
Still water: $\boxed{8 \text{km\h}}$?[/hide]

I would say 9 as you know the current is equally big all the time. And you know when he is sailing against the current it is slowing him down and when it is with the current it is speeding him up. So therefore.

$y - x = 6$
and
$y + x = 12$

Where y is the speed.
then you can solve that and it gives.
$y = 9$

Wait, so then when you have the against current and with current speed, it's just the average of the two for the still water speed. I thought the average might not be linear, but I guess i'm wrong.

i just did what the guy with the long name that begins with c  did and took the average of the two numbers

[hide]The speed against the current is 6km/h, with the current, 12km/h.

Let x be the boat speed in still water, y be the current.

x+y=12
x-y=6

Add and divide by 2, x=9. [/hide]

So it is just the average of the speeds (if the problem had asked for the average speed the boat if it had went up stream 18km and then went back down stream, then the harmonic mean would be correct).

[hide]$\frac{\frac{18}{3}+\frac{18}{1.5}}{2}$ is the speed with out the current.  Simplify to $\frac{6+12}{2}=9$[/hide]


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linear algebra

A quadrilateral has vertices at $ A(0, 0)$, $ B(2, 4)$, $ C(8, 5)$ and $ D(6, 1)$. What is the area of quadrilateral $ ABCD$?

hello, the area is after a well-known formula $ A = |\vec{AD} \times \vec{AB}|$
with $ \vec{AD} = \left(\begin{array}{c}{a}6\
1\end{array}\right)$ and $ \vec{AB} = \left(\begin{array}{c}{b}2 \
4\end{array}\right)$,
so we have
$ A = \left|\left(\begin{array}{c}{a}6\
1 \
0\end{array}\right)\times\left(\begin{array}{a}2 \
4 \
0\end{array}\right)\right| = 22$.
Sonnhard.

Could you explain that with Mathcounts-level explanation?

Hi!  The previous poster has the right tool, but made an error.

Search for "cross product" at wikipedia.  On the article under "geometric meaning"  You'll find that the cross product of two vectors in 3-D space is the area of the parallelogram you can build out of two of each of them (times 1 or -1... since areas can't be negative, remove the minus).

*Long story short*  For 2-D vectors (a,b) ,(c,d)  the cross product (a,b)X(c,d)=ad-bc.  (That's the determinant of the matrix formed by those two vectors!).

Unfortunately the last poster chose (8,5) as one of his vectors, which is a mistake since that's the diagonal of this parallelogram.  Using (2,4) and (6,1) you get 2-24=-22... the absolute value of that is 22, that's the area.

PS  IMPORTANT:  This only works for parallelograms.  I notice they put "quadrilateral", which isn't very helpful of them :)  So, remember you can't use this for just any quadrilateral, just parallelograms.

hello, thank you for your hint, i will correct it.
Sonnhard.


vector
Functional Analysis
real analysis
real analysis unsolved

Let $ H$ be a Hilbert space and $ T: H\to H$ be an isometry that is not unitary (i.e. not surjective). Then the spectrum $ \sigma(T) \equal{} \overline{\mathbb{D}}$.

We had a solution outlined in an exercise by evaluating conformal disc automorphisms at $ T$. I felt this proof was not particularly intuitive,  so if someone knows a different solution I'd like to see it.

Assume that $ x\in H$, $ x\perp TH$, $ x\ne 0$. Consider the vector $ v\equal{}\sum_{k\equal{}0}^\infty \lambda^k T^k x$ where $ |\lambda|<1$. Now, $ \langle v, Ty\rangle\equal{} \sum_{k\equal{}1}^\infty \lambda^k \langle T^kx,Ty\rangle\equal{}\lambda \sum_{k\equal{}0}^\infty \lambda^k \langle T^kx,y\rangle\equal{}\lambda \langle v,y\rangle$ for all $ y\in H$. Thus, $ T^*v\equal{}\lambda v$ and, clearly, $ v\ne 0$, so $ \lambda\in\sigma(T^*)$. The rest is easy.

That is very nice. Just want to say thanks for being tremendous help for me this semester  :)

There is also another approach that just came into my head (it is also completely standard but I just forgot it for some reason).

[b]Claim.[/b] If $ T(s)$ ($ s\in [0,1]$)  is a continuous family of bounded operators in a Banach space $ X$ and $ \|T(s)x\|\ge c\|x\|$ for all $ x\in X$ with some $ c>0$, then all operators in the family are invertible or not invertible simultaneously.

I guess your FA textbook has it somewhere; if not, try to prove it yourself :).

Now just apply this claim to the family $ T\minus{}s\lambda I$ with $ |\lambda|<1$.


inequalities
LaTeX
algebra
polynomial
inequalities proposed

[b]For any positive numbers a, b, c, d there is the following inequality:[/b]
\[\sum a\cdot\sum a^2\cdot\sum \frac{1}{a}\ge 3\cdot\left(\sum a\right)^2+\sum a^3\cdot\sum\frac{1}{a}.\ \ \ (1)\]
[b]Generalization. For [/b] $n\in N,\ n>3$ [b]and any positive numbers[/b] $a_k,\ k\in \overline{1,n}$,[b] there is the following inequality:[/b]
\[s_1\cdot s_2\cdot s_3\ge \frac{3n}{2(n-2)}\cdot s^2_3+\frac{n-1}{n-3}\cdot s^2_1\cdot s_4.\ \ \ (2)\] [b]where[/b] $s_k,\ k\in \overline{1,n}$ [b]are the symetric basic[/b] [b]forms of order k between[/b] $a_k,\  k\in \overline{1,n},$ [b] i.e.:[/b]

$s_1=a_1+a_2+\dots +a_n;$
$ s_2=a_1a_2+a_1a_3+\dots+a_1a_n+a_2a_3+a_2a_4+\dots+a_2a_n+\dots +a_{n-1}a_n;$
[b]----------------------------[/b]
$s_n=a_1a_2\dots a_n.$

[b]The inequality (1) is a easy processing of the inequality (2) for n=4. [/b]

Oh,Levi cool inequalities.Although nobody gave a solution,please do not develop your idea,not a bit....
I just saw these two and i find them very interresting.I have what to do tomorrow.Now it's 3.00 am,and i'm very sleepy. :sleeping:

i haven't tried the generalized one, but the first one is not difficult.
It is equivalent to
$(\sum a^2b) \sum \frac{1}{a} \ge 3(\sum a)^2$
which is simply cauchy.
Write $A=ab^2+ac^2+ad^2+ba^2+bc^2+bd^2+ca^2+cb^2+cd^2+da^2+db^2+dc^2$
Then, $A(\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+...+\frac{1}{d}) \ge (3(a+b+c+d))^2$
the result then follows.

For the generalized one, i have a question. Do you mean $s_1 = \frac{...}{n}$? $s_2= \frac{...}{nC2}$ as well?

\[<\bi >\underline{\overline{|LEVI\ (VIRGIL)|}}<\bi >\Longrightarrow\prod_{k=1}^n (X+a_k)=\sum_{k=1}^n s_k\cdot X^{n-k}\Longleftarrow <\bi > \underline{\overline{|SIUHOCHUNG|}}<\bi >\]     :!: The Mac-Laurin's and Newton's inequalities are less used in this section of Forum Math. I try (and I do one's best) to learn Latex with some funy images !
 :spider:  :starwars:  :starwars: :gathering:  :starwars:   :starwars:  :spider:

Nobody, after approx. two weeks !
In the preceding message I already gave a direction for approach of the second problem.
After a week I will present the my solution, i.e. $7^{th}\ august$.

To be sincere, immediately after you posted the problem I knew the approach since all these sharp inequalities can be solved in the same way (just look through a classical proof of Mac-Laurin or Newton inequalities for polynomials to see I'm right), but I never had the patience to make the computations to see I'm right. Well, this evening I made them and yes, I was right: it is direct consequence of the $n=4$ case. For this just look at the polynomial $P$ whose roots are $a_i$ and by Rolle's theorem the polynomial $P^{n-4}$ has 4 positive roots. Apply the case $n=4$ for these roots and make the computations. Surprisingly, you'll find exactly the general case.

[b]I Will use the Newton's inequality:[/b] $ P_k=\frac{s_k}{\binom nk},\ k\in \overline {o,n}\Longrightarrow P_{k-1}\cdot P_{k+1}\le P^2_k.$

$k:=1\Longrightarrow P_0\cdot P_2\le P^2_1\Longrightarrow 2n\cdot s_2\le (n-1)\cdot s^2_1\ \ (1)$

$k:=2\Longrightarrow P_1\cdot P_3\le P^2_2\Longrightarrow 3(n-1)\cdot s_1\cdot s_3\le 2(n-2)\cdot s^2_2\ \ (2)$

$k:=3\Longrightarrow P_2\cdot P_4\le P^2_3\Longrightarrow 4(n-2)\cdot s_2\cdot s_4\le 3(n-3)\cdot s^2_3\ \ (3)$
[b]
We multiply the ineq. (1) with the ineq. (2) and the ineq. (2) with the ineq. (3). We[/b] [b]obtain:[/b]

$(4)\ \ s_1\cdot s_2\ge \frac {3n}{n-2}\cdot s_3\ \ \ \ \ (5)\ \  s_2\cdot s_3\ge \frac{2(n-1)}{n-3}\cdot s_1\cdot s_4$

[b]We multiply the ineq. (4) with[/b] $s_3$ [b]and add to the the ineq. (5) multiplied with[/b] $s_1$. [b]We obtain:[/b]

$2\cdot s_1\cdot s_2\cdot s_3\ge \frac{3n}{n-2}\cdot s^2_3+\frac{2(n-1)}{n-3}\cdot s^2_1\cdot s_4\Longrightarrow$ [b]proposed inequality ![/b]

[b]Remark 1. [/b] This inequality (2) belongs to Serban Nacu who generalized my inequality (1).
 Where is my dear Serban Nacu now ?

[b]Remark 2.[/b] I have (and I will offer soon) a generalization of the inequality (2) which is binding
 the any four consecutive terms of the sequence ${s_k,\ k\in \overline 0,n}.$

Very nice,Levi.I think from this beautiful method we can deduce lots of inequalities. ;)  :coolspeak:


[quote="math10"]Let $ f$: $ Q.Q$ to $ Q^{ \plus{} }$ such that :
$ f(xy,z) \equal{} f(x,z).f(y,z)$ ,$ f(z,xy) \equal{} f(z,x).f(z,y)$,$ f(x,1 \minus{} x) \equal{} 1$ for all rational number $ x,y,z$. Prove that: $ f(x,x) \equal{} f(x, \minus{} x) \equal{} 1$ and
$ f(x,y).f(y,x) \equal{} 1$[/quote]

1) If $ \mathbb Q^ \plus{}$ is the set of all positive rational numbers, then :
Using $ y \equal{} 0$ in $ f(xy,z) \equal{} f(x,z).f(y,z)$, we get $ f(0,z) \equal{} f(x,z)f(0,z)$ and so, since $ f(u,v) > 0$ $ \forall u,v$ : $ f(x,z) \equal{} 1$ $ \forall x,z$
So $ f(x,y)$ is the constant $ 1$ function and the result is immediate.

2) If $ \mathbb Q^ \plus{}$ is the set of all non negative rational numbers, then :
Let $ g(x,y)$ such that $ g(0,0) \equal{} 0$ and $ g(x,y) \equal{} 1$ $ \forall (x,y)\ne(0,0)$
It's easy to see that $ g(x,y)$ is a solution but $ g(x,x) \equal{} 1$ is wrong for $ x \equal{} 0$

So the problem is wrong since there exist solutions $ f(x,y)$ for which $ f(x,x)\ne 1$ for some $ x$


The sequence goes like this....

$1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 14, 15, 16, 17, 18, 19$

Find the next number....

Here is the solution:
[hide]$a)$ There are no more missing numbers,
$b)$ Thus, answer = $20$

How many possible explanations can you give?[/hide]

one of the probable explainanatins could be that -

all such trio of numbers is removed where sum of squares of two prime nos is another prime no.

We consider that fact that any prime number greater than 3 can be written as 6p + 1 or 6p - 1.

Consider two such prime nos. ie
6a + 1 and 6b + 1

Sum of squares of these two would be
$36a^2 + 12a + 1 + 36b^2 + 12b + 1$
= $36a^2 + 36b^2 + 12a + 12b + 2$
Clearly this shall always be divisible by 2 and hence not prime.

The only one case would be of prime nos 2 and 3. Once square of these numbers and the sum of their squares is removed no other combination would yield a prime no. as given above.

Hope its one of the correct explaination.

[quote="JBL"]This sort of question has traditionally been posted in the Puzzles forum.  This sort of question also traditionally draws some scorn.  After all, if you give me any number there, I can justify it as the next term of the sequence by taking a polynomial of sufficiently large degree.[/quote]

I understand that, after posting it I realized it that it may not be the right palce to post it, though i was not aware that it traditionally invites scorn.

[quote="numbergeek"]hey but lets discuss the probable explainations..... :?[/quote]

sure, the thread is still open.


[b]Εύχομαι σε όλους τους μαθητές :  [/b]

                           [size=150][color=red] Καλή Επιτυχία στις εξετάσεις ![/color][/size]

[b] Μπάμπης [/b]

Καλή επιτυχία  :)

Εύχομαι και εγώ[color=red] ΚΑΛΗ ΕΠΙΤΥΧΙΑ[/color] :)

[color=red]Καλή επιτυχία[/color] σε όλους (ακόμα και σε αυτούς που έχουν πάρει το άνευ :wink: )

εύχομαι κι εγώ καλή επιτυχία ΣΕ ΟΛΑ ΤΑ ΠΑΙΔΙΑ  :starwars:  :first:

Κανείς από εδώ δε θα τη χρειαστεί στα μαθηματικά. :wink:  Ελάχιστα ίσως για το τέταρτο θέμα.
 :P Καλή επιτυχία πάντως σε όλα τα άλλα. :D

Καλή επιτυχία  :lol:

Kali epitixia kai apo emena stis eksetaseis gia olous tous ipopsifious!

Alexandros

Καλή επίτυχία σ αυτούς που δίνουν πανελλήνιες,
και καλό κουράγιο σ όλους εμας τους υπόλοιπους,
μεχρι να περάσουν οι φετινές εξετάσεις και να δώσουμε του χρόνου πανελλήνιες!!!!! :ninja:

Είδατε τα θέματα στα Μαθηματικά Γενικής Παιδείας στις Πανελλήνιες :?:

Οχι δεν τα είδα...μα σήμερα δεν γράφουν :?: έχουν βγεί ήδη? κι αν ναι πού μπορώ να τα βρώ?

Ναι σήμερα γράφουν...Στο ιντερνετ δεν τα έχουν βγάλει ακόμη αλλά πιστεύω σύντομα θα τα αναρτήσουν....Στην έκθεση 11.30 στο in.gr τα είχαν βγάλει.

Καλή επιτυχία κι από εμένα σε όλους!  :)

Λουκάς

Παιδιά, τα θέματα έχουν ήδη δημοσιευθεί στο http://www.in.gr. Όποιος εβδιαφέρεται μπορεί να τα δει από εκει!

Υ.Γ. Πιστεύω ότι όπως ήταν αναμενόμενα και βατά!

Για όποιον θέλει τα θέματα έχουν δημοσιευτεί και στο http://www.forthnet.gr/templates/specials.aspx?c=10009728

Τα θέματα στα μαθηματικά γενικής ήταν δυσκολότερα σε σχέση με τα τελευταία χρόνια, κατα την γνώμη μου, αλλά κάποιος καλά διαβασμένος μαθητής δεν νομίζω πως αντιμετώπισε κανένα πρόβλημα.... Αύριο να δούμε τι θα πέσει...

Kali epitixia kai panw apo ola..OXI AGXOS!

[quote="silouan"][quote="Xaris N."]Τα θέματα στα μαθηματικά γενικής ήταν δυσκολότερα σε σχέση με τα τελευταία χρόνια[/quote]

Τα περσινά ήταν δυσκολότερα νομίζω με διαφορά .
Αν και τώρα κακά τα ψέματα μόνο από μεγάλη χαζομάρα(όχι ότι δεν γίνονται από όλους) μπορείς να χάσεις το 100 .[/quote]

silouane min eisai apolytos!!giati o marios o papamixalis (kai o fanis) i o stefanos o aretakis pou den egrapsan 100 stin panellinies??

mallon itan poly eukola kai berdeutikan!! :lol:

Θα λεγα πως επειδή είχαν εξασφαλίσει εισαγωγή είχαν λιγότερο άγχος... :wink:

Niko perisi den edosa panellinies... den eimai kai toso mikros... kai panellinies Mathimatika egrapsa 100, otan egrafa. (Mprabo mou!!!)....

Πως σας φάνηκαν τα σημερινα θεματα Μαθ. Κατ? Εγώ δυστυχώς έχασα το 20 για ένα Σ-Λ, συγκεκριμένα το (γ) το οποίο πιστευω οτι ηταν μεγάλη παγίδα, αφου αν δεν το διαβάσεις προσεκτικά σου φαίνεται ολοσωστο.  :mad:

 Παντως το 4ο θεμα ηταν δυσκολουτσικο, το οποίο βέβαια το έβγαλα ολο  :winner_first:

Τα θέματα τα έβαλα στο pdf

Μου άρεσε το 4ο !  :)

Και 'γω το  γ σωστό θα το 'βαζα  :maybe:

Πολύ ωραία τα θέματα. Δυστυχώς όμως στο 4β έγραψα μόνο κάτι περίεργες εμπνεύσεις... Εν ολίγοις το έχασα. Στεναχωρήθηκα πολύ, αλλά τουλάχιστον έπιασα όλα τα άλλα. Τέλος πάντων πάμε για Βιολογία τώρα.

[quote="cretanman"]...Omws einai mia apli kataskeuastiki anisotita... [/quote]

Cretanman ευχαριστώ...Μου έφτιαξες τη διάθεση   :)

[quote="cretanman"]Na deite omws posoi mathites tha pigan na lisoun to 4o thema me tin voitheia paragwgwn (pou fisika den einai swsto afou h $f(x)$ den leei oti einai paragwgisimi sinartisi kai malista den mporoume na to diasfalisoume me kapoio tropo).

Omws einai mia apli kataskeuastiki anisotita pou stirizetai sto oti an

$f(x)\geq 0$ sto diastima $[a,b]$ tote ean h sinartisi $f$ den einai pantou miden sto $[a,b]$, tote isxiei $\int_{a}^{b}f(t)dt>0$ (kai oxi apla $\geq 0$).

Alexandros[/quote]

Mia mikri iposimiosi...Gia na isxiei afto pou eipe o Aleksandros tha prepei na kseroume oti i f(x) einai sinexis se olo to diastima [a,b](an kai sto likio theoroume dedomeno oti milame gia sinexis oloklirosimes sinartiseis)

Iparxoun kapou , episimes liseis twn thematwn?Apo tous thematodotes? Giati toulaxiston apo to http://www.in.gr pou kitaksa ligo tis liseis exei kapoies paralipseis , kate me , sovares , pou dimiourgoun provlima...Px sto 4d tha prepei prwta na dikseis afto pou edikse o kirios stergiu edw http://www.mathlinks.ro/Forum/viewtopic.php?t=144475&start=40 , meta apo mia sizitisi pou eixame , giati apo ta theorimata tou vivliou den diasfalizete oti mporeis na xrisimopoiiseis DelHospital 'h sto 4b theorei to t<x enw to to t anikei sto [0,x]

Sas stelnw sto episinaptomeno pdf tis liseis mou stis Panellinies tou 2007.

Nomizw oti einai arketa analitikes (Elpizw na min exw ksexasei kati).

An iparxei kapoio provlima peite mou. Tha itan kalitero na tis grapsw se LateX kai na sas tis steilw alla den eixa to xrono giati eixa ligi douleia simera.

Alexandros

[quote="KAZA"]Ούτε εγώ εχω βει κάπου επίσημες λύσεις των θεμάτων...Γενικά, δεν θα χαρακτηρίζατε τα σημερινά θέματα  ως πολύ δύσκολα?[/quote]

Φίλε ΚΑΖΑ, η δυσκολία τών θεμάτων είναι πάντοτε σχετική. Για έναν μέσο μαθητή πιστευώ ότι δεν ήταν και πολύ εύκολο να πιάσει πάνω από 15. Από την άλλη, για ένα άτομο το οποίο είναι εξοικειωμένο με τα μαθηματικά δεν νομίζω ότι θα έγραφε κάτω από 18, αφού τα συγκεκριμένα θέματα δεν έκρυβαν κάποια ιδιαίτερη παγίδα.

[quote="DenPoNaEi"]kalhmera se olous . ego prepei na egrapsa 94/100 giati den ekana to 4b( eipa esto f(x) einai paragogisimh kai elusa mono ama isxuei auto). Den katalava omos giati sto oloklhroma t>x h t=x.
an mporousa na mou to eksigisei kapios plz.  :lol:[/quote]

[b]Στο ορισμένο ολοκλήρωμα η μεταβλητή ολοκλήρωσης (το $t$ δηλαδή) , διατρέχει το διάστημα που ορίζουν τα άκρα του ολοκληρώματος.[/b]
[color=red]    Συγχαρητήρια για το άριστα και καλά αποτελέσματα  ![/color]

 Μπάμπης


[hide="Solution"]Let the side length of the original square be $x$.

From the given information, we can write an equation: $x^{2}-36=(x-2)^{2}$

Expanding the RHS gives $x^{2}-36=x^{2}-4x+4$.

The squares cancel, leaving $4x=40$.

Finally, $x=10$, so the area of the original square was $100\text{ cm}^{2}$.[/hide]

[hide="Alternate Solution"]This problem involves sums of squares, and coincidentally, $36=6^{2}$.

Checking some small Pythagorean triples reveals the answer: $6^{2}+8^{2}=10^{2}$.

Since $10-8=2$, the original square had a side length of $10$, and area $100\text{ cm}^{2}$.[/hide]

Maybe this will help $x^{2}$ is the area of a square with side length $x$.  :maybe:

If the side length is x, then the area of the square is x^{2}.
x^2 = 36 + (x - 2)^2
solve the equation using the formula, (a-b)^2 = a^2 - 2ab +b^2.
x^2 = 36 + x^2 - 4x + 4
40 = 4x
x = 10.

Just to clarify some things and summarize i_like_pie's ideas. :)

[quote="send_it_all"]I'm not a math wiz by any stretch, but  friend called me with  a homework problem (I'm long out of school and he's taking courses to advance in his career). The problem is this:

If you take a square and reduce each side by 2 cm, you have reduced the total area by 36 cm, how big was the original square?....I come up with 38.85....ish, but am not sure how....could someone give me the answer and more importantly the formula for getting it?....Thanks[/quote]

[hide=" :welcomeani: "]
We know that $(x-2)^{2}=x^{2}-36$.  Therefore, $x^{2}-4x+4=x^{2}-36$, and thus $-4x=-40$.  Therefore, [b]x=10[/b][/hide]

I think :
we call original side is x cm,(x>0)
so x-2 is measurement after we reduce each side
( x-2)^2 = 36
x^2 -4x +4 -36=0
x^2-4x-32=0
x^2+4x-8x-32=0
x(x+4)-8(x+4)=0
(x+4)*(x-8)=0
x=-4(disagree)
or x=8(agree)
the original square is 64 cm^2

[quote="Babylo"]I think :
we call original side is x cm,(x>0)
so x-2 is measurement after we reduce each side
[b]( x-2)^2 = 36[/b]
x^2 -4x +4 -36=0
x^2-4x-32=0
x^2+4x-8x-32=0
x(x+4)-8(x+4)=0
(x+4)*(x-8)=0
x=-4(disagree)
or x=8(agree)
the original square is 64 cm^2[/quote]
Babylo, The square's area is [i]reduced[/i] by 36, so that line should be
$ (x-2)^{2}=x^{2}-36$.

[quote="AstroPhys"]Maybe this will help $ x^{2}$ is the area of a square with side length $ x$.  :maybe:[/quote]
right that means you get $ x^{2}-4x+4=x^{2}-36$ subtract $ x^{2}$ from both sides, then subtract 4 from both sides and get $ 4x=40$ so $ x=10$
[hide="sort of an idea"]is there a way to do this by factoring?$ (x-2)(x-2)=(x+6)(x-6)$ is that solvable?[/hide]

[quote="abacadaea"]is there a way to do this by factoring?$ (x-2)(x-2)=(x+6)(x-6)$ is that solvable?[/quote]
If $ (x-2)^{2}$ can be represented as the product of two other numbers, both of them must also be squares. They are $ 12$ apart, and it is easy to see that they are $ 4$ and $ 16$. Therefore $ x=10$.

But I prefer the standard factoring way as this might not always work.

[quote="send_it_all"]I'm not a math wiz by any stretch, but  friend called me with  a homework problem (I'm long out of school and he's taking courses to advance in his career). The problem is this:

If you take a square and reduce each side by 2 cm, you have reduced the total area by 36 cm, how big was the original square?....I come up with 38.85....ish, but am not sure how....could someone give me the answer and more importantly the formula for getting it?....Thanks[/quote]

[hide]$ \text{Call the side length x. Then }(x\minus{}2)^{2}\equal{}36$
$ x^{2}\minus{}4x\plus{}4\equal{}x^{2}\minus{}36$
$ \minus{}4x\equal{}\minus{}40$
$ x\equal{}10$[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

(a) For each element of $ \mathbb{F}_8$, determine its irreducible polynomial over $ \mathbb{F}_2$.
(b) Factor the polynomial $ x^{16}\minus{}x$ in the fields $ \mathbb{F}_4$ and $ \mathbb{F}_8$.

I know that $ \mathbb{F}_4\cong\mathbb{F}_2[x]/(x^2\plus{}x\plus{}1)$ and $ \mathbb{F}_8\cong\mathbb{F}_2[x]/(x^3\plus{}x\plus{}1)$, but I do not know how this would simplify the irreducible polynomials or $ x^{16}\minus{}x$, which factors as $ x(1\plus{}x)(1\plus{}x\plus{}x^2)(1\plus{}x\plus{}x^4)(1\plus{}x^3\plus{}x^4)(1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4)$ modulo 2.

Every degree $ k$ field over $ F_p$ splits completely over $ F_{p^k}$. Over the field with 16 elements, $ x^{16}\minus{}x$ is the product of all monic linear polynomials.
Over $ F_8$, it doesn't split any farther than you already have it, since the degrees are relatively prime.


group theory
abstract algebra
superior algebra
superior algebra unsolved

1.  IF N is a normal subgroup of a group G and if x^2 is an element of N for ever x in G, prove that every nonidentity element of the quotient group G/N has order 2.

2.  Let G be a group of order pq, with p and q (not necessarily distinct) primes.  Prove that the center Z(G) is either <e> or G.

1. this is immediate from definition.

2. In more general manner, show that if $ G/Z(G)$ is cyclic, then $ G$ is abelian.


Welcome to X-Man.

Very similar to the Mole, but this game has twists.

First of all, I am your host, alekim. If you know what The Mole is, then you think you have clear idea of what's about to happen. If you don't know, you still have to read on.

[size=24]Rules:[/size]
1. There will be two teams, the Noobs and the Spammers. Each team will consist of same amount of players with 1 leader. (however, teams won't matter at the end)

2. Your basic goal is to win several mini games and quizes. Your team will win $10,000 (PRETEND since I'm poor) if your team wins the round.

3. There is one person assigned as X-Man among everyone who's participating. Their main goal is to make their team LOSE the round by making stupid mistakes.

4. Discussions are allowed outside of the thread, but only within the team.

5. Since there are two teams, I will switch up the teams once. HOWEVER, after you switch, you are not allowed to discuss anything among the team outside of the thread. (you will know why)

Additional Rules:
1. Most rounds will consist of questions or random tasks that has to be performed online. However, you MAY NOT look up answers to questions online. Only tasks are allowed.

2. After each round, I will announce the winning team. The leaders from each team will select a person from their team (could be themselves). This will be crucial in the next step.

3. I will private message the team members the answers of the OTHER team's members. For example, the Red team will receive Blue team's answers and vice versa. The names won't be listed on the answers.

HOWEVER

If you have been chosen by the leader (and leader cannot discuss this with other team members about their choice) then your name WILL BE revealed to the other team's database next to your answer. This is crucial because if X-Man puts down a stupid answer and if he is suspected by the leader, his identity can be reveal to the other team.

Leaders: Use it as a strategy. If you think someone in your team is slacking off, then use it as an advantage to reveal him to the other team. If you need to prove that someone's innocent because of their stupidity in the forums, this might prove that they aren't the X-Man.

This system is the reason why the teams cannot discuss among themselves outside of the thread AFTER the team switch. It will basically reveal the X-Man right away if random.org does not comply with me.

At the end of x amount of rounds, everyone will vote on a person who they think is the X-Men. (remember he/she is also allowed to vote). Discussion is not allowed during the last segment.

If the group gets the X-Man, the teams will get their earnings while the X-Man walks home with nothing.

If the X-Man isn't voted, then he will win.

Basically, its the X-Man versus the entire group in the end.

This was created under 5 minutes. If you have suggestion to the rules, PM me and I'll see what I can do. Also, keep in mind that if I need to clarify something, then just ask me right away. I will answer ASAP and fix it among the rules.

Remember: This is my first attempt at AoPS game, so I might need little help with clarifying the rules.

Sign-ups Open. PM or post.

1. AIME15
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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14.


$ \left\{\begin{array}{c}
\triangle\ ABC\ \ : \ \ B\equal{}2A\ ,\ A<45^{\circ}\\\
\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\\\
M\in (AB)\ ,\ m(\widehat {ACM}) \equal{}A\\\
N\in (AC)\ ,\ m(\widehat {ABN}) \equal{}\frac A2\end{array}\right\|\implies m(\widehat {ANM}) \equal{} 90^{\circ}\plus{}\frac A2\ .$

[quote="Virgil Nicula"]$ \left\{\begin{array}{c} \triangle\ ABC\ \ : \ \ B \equal{} 2A\ ,\ A < 45^{\circ} \\
\ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\
\ M\in (AB)\ ,\ m(\widehat {ACM}) \equal{} A \\
\ N\in (AC)\ ,\ m(\widehat {ABN}) \equal{} \frac A2\end{array}\right\|\implies m(\widehat {ANM}) \equal{} 90^{\circ} \plus{} \frac A2\ .$[/quote]

$ m(\widehat {CMB}) \equal{} A \plus{} m(\widehat {ACM}) \equal{} A \plus{} A \equal{} 2A \Rightarrow BC \equal{} CM$
$ m(\widehat {CBN}) \equal{} 2A \minus{} \frac {A}{2} \equal{} \frac {3A}{2}$
$ m(\widehat {CNB}) \equal{} A \plus{} m(\widehat {ABN}) \equal{} A \plus{} \frac {A}{2} \equal{} \frac {3A}{2} \Rightarrow BC \equal{} CN$
$ CM \equal{} CN \Rightarrow m(\widehat {CMN}) \equal{} m(\widehat {CNM}) \equal{} 90 \minus{} \frac {A}{2} \Rightarrow m(\widehat {ANM}) \equal{} 90 \plus{} \frac A2$


Let P(x)=(x-1)(x-2)(x-3) .  For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 such that P(Q(x))=P(x)R(x)?

I'm not sure how to narrow down what polynomials are possible.

My first thought was to divide both sides by P(x) since the roots x=1,2,3 appear on both sides.  I thought there might be some sort of synthetic division involved, but I'm just not sure

Q(x) is obviously a quadratic polynomial.

Any help would be appreciated.  This is my first post here, so sorry if I didnt get the syntax with writing functions correct.  I'm also sorry if I posted this in the wrong forum.  I thought that this was a little more difficult than the beginner's problems so I decided to put it here.

[quote="pkerichang"]well... I believe all AMC problems are actually intermediate level

[hide]
As he mentioned, $Q(x)$ has degree 2.  plug in 1, 2, and 3 for x, we see that $Q(x)$ maps 1, 2, and 3 to 1, 2, and 3.  There are $3*2*1 = 6$ possiblities for one to one mapping, and $3*3*2 = 18$ two to one mapping.  but note that 1-1, 2-2, 3-3 and 3-1, 2-2, 1-3 are linear equations, therefore can't be right.  So there are $6 + 18 - 2 = 22$ Q(x) that satisfies the condition.
[/hide]

Am I right?[/quote]

Well, that is one of the choices, and I dont have a solutions manual, so I think you're probably right.

Could you explain one to one mapping and two to one mapping?  I'm not exactly sure what you mean by that, and how you're getting these numbers.

[quote="pkerichang"][quote="Petoria"][quote="pkerichang"]well the one to one just means that for $x = 1,2,3$ that $Q(x)$ has different values for each.  However, it is possible that there exist distinct $x$ and $y$ such that $Q(x) = Q(y)$.  for example, $Q(1) = 1$, $Q(2) = 3$, and $Q(3) = 3$.  Therefore, we also have to count these cases, which has $3*3*2 = 18$ ways of doing so.[/quote]

Ok, that makes a lot of sense.  I was definitely trying to do way too much work on that problem.

Why do we discount the ones that are linear equations?[/quote]

well, as you said that $Q(x)$ is quadratic, therefore it's not linear.  If you have three collinear points, then you will have something like $0x^2 + ax + b$, which does not work at all.[/quote]

Good point.

 :)

Thanks again.


invariant
linear algebra
matrix
algebra
polynomial

I have 2 Queations.please help
Q1
let A be a nxn matrix
how to show that:
dim{span(I, A,A^2,A^3,......)} <= n?
Q2
Let T and U be a linear operator on V, and
Suppose V is a T-cyclic subspace of itself
Show that UT=TU iff U=g(T) for some polynomial g(t)

Regarding the first question it's a simple deduction from Cayley Hamilton
for P=Sum ak*X^k (k^<=n) which is the carachteristic polynome of A we have :  P(A)=0(Mn(K( the field this time ;)))) Thus for i>n the expression A^(i-n)*P(A)=0 enables us to have the expression of A^i in terms of the set {A^k,0<=k<=n} (a simple recursion).
This way we conclude that Vect(I,A,A²,...................)=Vect(I,A,........,A^n) ,we keep in mind that dim(Vect(I,A,.........,A^n))<=n and now it's done :).
for the second question i didn't really get what you meant as i've studied math in french i find some difficulties to figure out english vocab sometimes ,would you better explain your question plz?

oh i typed something worng....
Let T and U be a linear operator on V, and
Suppose V is a T-[b]cyclic[/b] subspace of itself
(that mean suppose we can suppose V is generated by v)
Show that UT=TU iff U=g(T) for some polynomial g(t)


One of my peers misbubbled their name on the answer sheet, and came back with a 144 on AMC 10A.  When he took the 12b, he got 103.5,(bubbling his name correctly) and wants to take the AIME with his 10A score.  However, because it's under two different names, how will he do this?  We are both going to take the AIME at the school we took the B exam in.

Ha, I did this last year...Just email the AMC people, and give the necessary info. (I know you need name, school, CEEB (?) code, etc.) They'll change it for you.

I still wonder how you misbubble a name?

Well, I hope everything goes smoothly for you and your friend, AIME15

He put last-first instead of first-last or something, then erased and rebubbled, and it came out as something really funky :)

[quote="Smartguy"]I still wonder how you misbubble a name?[/quote]
Oh, it always happens. I get the full California listing of AIME qualifiers in each year, and I'm looking at last year's. I see several first names that start after several blank spaces. I see first names like "R MOND." I see a (short) last name repeated twice with several spaces in between. I see an entry in which the first name is given as "illegible" and the last name (which starts with numbers) is obviously a street address. The grade level column is blank is a significant number of cases. I see a first name missing altogether, although there is a last name.

Things like that always happen.

I almost did that this year, did last and first o n the same line, then erased


conics
hyperbola
geometry
circumcircle
analytic geometry
algebra
polynomial

Let $ ABC$ be a triangle and $ g$ be the line passing through the circumcenter of triangle ABC.  Prvoe that isogonal conjugate of $ g$ is rectangular hyperbola.

Let us consider barycentric coordinates WRT $\triangle ABC.$ $px+qy+rz=0$ is the equation of a line $\ell.$ Isogonal conjugation $\mathcal{I}:$ $(x:y:z) \mapsto \left( \frac{_{a^2}}{^{x}}:\frac{_{b^2}}{^{y}}:\frac{_{c^2}}{^{z}} \right)$ takes $\ell$ into a curve $\mathcal{L}$ with equation $\mathcal{L} \equiv pa^2yz+qb^2zx+rc^2xy=0,$ i.e the equation of a homogeneous polynomial of secong degree.

Thus, $\mathcal{L}$ represents a conic passing through the vertices of $\triangle ABC.$ Since the isogonal conjugate of the circumcircle $(O)$ of $\triangle ABC$ is the line at infinity, we conclude that $\mathcal{L}$ is either  hyperbola, parabola or ellipse if $\ell$  has either two, one or any common point with $(O).$ However, I don't know how to prove that $O \in \ell$ $\Longleftrightarrow$ $\mathcal{L}$ is a rectangular hyperbola.

a short proof: the isogonal of a line is an gyperbola from A,B,C so we have to prove that an hyperbola from A,B,C,H i rectangular.  Let the conics from A,B,C,H, on this bunch there exist 2 equilateral hyperbola than all the conics of the bunch are equilateral hyperbola.

Gabriel, you did not even show that isogonal conjugate of a line is a conic, not to mention a hyperbola.

Anyway, I put it down at [url]http://www.mathlinks.ro/viewtopic.php?t=129699[/url], click on "Show details".

Thank you for your help. Could you explain easily about lemma in Show details? What is parallel projection?

Parallel projection is a central projection from a point $ P^*$ at infinity. Just like central projection, parallel projection takes points to points, lines to lines, and conics to conics. Suppose it carries points $ A, B, ...$ to points $ A', B', ...$ Since all rays $ P^*AA', P^*BB', ...$ concur at a point $ P^*$ at infinity, they are all parallel.  It has the same invariant as central projection (it preserves cross ratios) and additional invariants: It preserves ratios of line segments on the same line and ratios of areas. Therefore, all points at infinity remain at infinity and parallel lines go to parallel lines. Parallel projection can also be defined as an affine transformation, i.e., the composition of a linear transformation and a translation.


A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac{m}{n}$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$.

I read the answers for the problem but there was one part that confused me. Could someone clarify this part for me?

[hide]
So there's $\frac{12!}{3!4!5!}$ total ways to arrange all 12 trees.
And then you consider that the birch trees can go in to any of the blank spaces: _M_M_M_O_O_O_O_
Thus, 8 spaces for 5 birch trees to go into.

Here's my question, why is the number of ways for the birch trees to be arranged $\binom{8}{5}$ instead of $8 nPr 5$? Doesn't the order matter in this case?

Similarily, why is the number of ways the maple + oak trees can be arranged = $\binom{7}{4}$ instead of $7 nPr 7$?

Thanks!
[/hide]

[quote="ark_xm"]
So there's $\frac{12!}{3!4!5!}$ total ways to arrange all 12 trees.
And then you consider that the birch trees can go in to any of the blank spaces: _M_M_M_O_O_O_O_
Thus, 8 spaces for 5 birch trees to go into.

Here's my question, why is the number of ways for the birch trees to be arranged $\binom{8}{5}$ instead of $8 nPr 5$? Doesn't the order matter in this case?

Similarily, why is the number of ways the maple + oak trees can be arranged = $\binom{7}{4}$ instead of $7 nPr 7$?

Thanks!
[/quote]
Say the birch trees are b1, b2, b3,....  Well, isn't b1,m,b2,m,b3,m,o,b4,etc the same as b3,m,b1,m,b4,m,o,b2,etc? Yes, because the birch trees are not distict (we can't tell b1 apart from b2).  If we used permutations instead of combinations, these two cases would be counded separately.  But we want them to be considered the same, so we use combinations.

Actualy for this problem, if you used combination and permutaion, the answer will not change:

Combination- (8C5)/(12C5)=7/99-they are not dstinct
Permutation-(8P5)/(12P5)=7/99-they are distinct
*Note-we sperated the cases into 5 birch trees and 7 NON-BIRCH trees-not birch trees, maple trees, and oak trees (however you dont have to do that, as you'll still get the same answer)

You just have to keep them consistent


function
advanced fields
advanced fields unsolved

If $A$ and $B$ are two subsets of $\mathbb{R}^n$  that have fix point property then what about $A+B$ where $+$ is minkowski's sum.Has $A+B$ the fix point property?

A set in $\mathbb{R}^n$ have fix point property if every continious function from this set to itself have fix point.

If $A$ and $B$ are two subsets of $\mathbb{R}^n$ then $A+B=\{x+y: x\in A,y\in B\}$ .

Saman Gharib

I think It's nice problem to think about it ..... :!:

This result is false. My example is in $\mathbb{R}^2.$

Let $A=\{(x,y): x^2+y^2=1\}\setminus B(1;0.1).$ That is, $A$ is the unit circle with one small open chunk taken out. It is topologically equivalent to a closed bounded interval on the line, and hence has the fixed point property.

Let $B=\overline{B(0;0.2)}.$ The closed ball with a suitably small radius; it has the fixed point property by the two dimensional Brouwer fixed-point theorem.

Note that $A+B$ consists of all points whose distance from any point of $A$ is no greater than 0.2.

But $A+B$ is topologically equivalent to an annulus, and an annulus does not have the fixed point property.

You did it. Is it taste nice :?:

Saman Gharib


Εστω οτι ισχυει $ 5sin{a}\minus{}12cos{a}\equal{}13ln(\frac{e^b}{b})$.
Να βρεθει ο θετικος αριθμος $ b$.

[hide]$ 5^2 \plus{} 12^2 \equal{} 13^2$.
[hide]$ \max \text{LHS} \equal{} \min \text{RHS}$ οπότε και προκύπτει $ b \equal{} 1$.[/hide][/hide]

Cheerio,

Durandal 1707

Πλακα με κανεις :!: .
Κιολας?????
Μπραβο!
Εγω την βρηκα αρκετα δυσκολη και περιμενα οτι δεν θα εμπενε λυση τοσο γρηγορα.

Vrika oti b=1.As elpisoume pws eimai swstos.
I eksisosi grafetai:
           $ 5sina \minus{} 12cosa \equal{} 13blne \minus{} 13lnb \equal{} 13b \minus{} 13lnb$ or
           $ \frac {5}{13}sina \minus{} \frac {12}{13}cosa \equal{} b \minus{} lnb$  (1)

Epeidi $ 13^2 \equal{} 12^2 \plus{} 5^2$ yparxei gwnia c pou anikei sto 0,2p tetoia wste $ cosc \equal{} \frac {12}{13} kai sinc \equal{} \frac {5}{13}$.

Epomenws i (1) grafetai twra:
           $ cosacosc \minus{} sinasinc \equal{} lnb \minus{} b$ or
           $ cos(a \plus{} c) \equal{} lnb \minus{} b$

Opote              $ \minus{} 1 \equal{} < lnb \minus{} b < \equal{} 1$. Alla isxyei oti $ lnx < \equal{} x \minus{} 1 gia x > 0.$

Opote anagkastika prepei na exoume $ lnb \minus{} b \equal{} \minus{} 1 or lnb \equal{} b \minus{} 1$ pou simainei oti b=1 mias kai i monadiki timi tou b gia na isxyei i isotita
einai b=1.

ps Den tin paleyo me tin Latex LOL

Sakis thx, αλλά απλώς έτυχε να είμαι online.  Πάντως, η ταυτότητα $ A\sin \theta \plus{} B\cos \theta\equal{}\sqrt {A^2 \plus{} B^2} \sin(\theta \plus{} \phi)$ είναι πάντα χρήσιμη :)

Cheerio,

Durandal 1707


inequalities
induction
combinatorics proposed
combinatorics

There are exactly $ 2n \minus{} 1$ vertices in graph.It is known that if we delete any vertex then it is possible to find complete subgraph with $ n$ vertices.
Prove that initial graph contains complete subgraph with $ n \plus{} 1$ vertices.

Assume the contrary.
We will construct a sequence of $ K_n$'s $ A_1,A_2,\ldots ,A_j$; such that.
1) Take an arbitrary $ K_n$ as $ A_1$.
2) For $ i > 1$, if the set $ (A_1\cap A_2\cap\ldots\cap A_{i \minus{} 1})$ is empty, select $ i \minus{} 1 \equal{} j$ and stop the construction. Otherwise delete an arbitrary vertice from $ (A_1\cap A_2\cap\ldots\cap A_{i \minus{} 1})$. There must be a $ K_n$ in the remaining graph. It is obvious that this $ K_n$ is different from $ A_1,\ldots A_{i \minus{} 1}$, since it doesn't contain the deleted vertice. Take it as $ A_i$
For every $ i \equal{} 1,\ldots j$, say $ B_i$ is the set of vertices that none of them is an element of an $ A_i$.
Claim: For every $ i \equal{} 1,\ldots j$, $ \mid (A_1\cap A_2\cap\ldots\cap A_{i})\mid > \mid B_i\mid$.
Proof goes by induction. For $ i \equal{} 1$, it is $ n > n \minus{} 1$ which is true. Assume that it is true for $ i \equal{} k \minus{} 1$. Since $ \mid (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})\mid > \mid B_{k \minus{} 1}\mid$, $ (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})$ isn't empty. Now let $ A_k\cap B_{k \minus{} 1} \equal{} a$, $ A_k\cap (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1}) \equal{} b$ and $ A_k\cap ((A_1\cup A_2\cup\ldots\cup A_{k \minus{} 1}) \minus{} (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})) \equal{} c$. We must prove that $ b > \mid B_{k \minus{} 1}\mid \minus{} a$
Claim': $ c\le n \minus{} \mid (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})\mid > \mid B_{k \minus{} 1}\mid$
If it is not; take this $ c$ element and all elements of $ (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})$, they will be a $ K_{n \plus{} 1}$. Contradiction! From here we can see that every $ A_i$ has an element from $ B_{i\minus{}1}$.
Hence $ n \equal{} a \plus{} b \plus{} c\le n \minus{} \mid (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})\mid \plus{} a \plus{} b\le n \minus{} \mid B_{k \minus{} 1}\mid \plus{} a \plus{} b$ and the result follows.
So when there is no vertice that is not an element of some $ A_i$'s, there must be a vertice such that it is an element of all $ A_i$'s. Delete this vertice; there must be a $ K_n$ and also with the deleted vertice, a $ K_{n \plus{} 1}$. Contradiction!

Hey Umut,
Could you elaborate on the following part?

[quote]
For every $ i \equal{} 1,\ldots j$, say $ B_i$ is the set of vertices that none of them is an element of an $ A_i$.
Claim: For every $ i \equal{} 1,\ldots j$,$ |(A_1\cup A_2\cup\ldots\cup A_i)| > |B_i|$
[/quote]

As I understand, $ A_i$ are complete subgraphs with $ n$ vertices each, and $ B_i \equal{} V\setminus\{A_1,\ldots,A_i\}$, right?
On the other hand, following your construction we could have, say, for $ n \equal{} 4$
$ A_1 \equal{} \{v_1,v_2,v_3,v_4\}$, $ A_2 \equal{} \{v_2,v_3,v_4,v_5\}$, $ A_3 \equal{} \{v_1,v_4,v_5,v_6\}$
Here, $ A_1\cap A_2\cap A_3 \equal{} \{v_4\}$ and $ A_1\cup A_2\cup A_3 \equal{} \{v_1,v_2,v_3,v_4,v_5,v_6\}$, $ B_3 \equal{} \{7\}$.
So that we have $ |A_1\cap A_2\cap A_3| \equal{} 1 \equal{} |B_3|$ and a sharp inequality does not hold.

This part of your argument is confusing to me:
[quote]Hence $ n \equal{} a \plus{} b \plus{} c\le n \minus{} \mid (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})\mid \plus{} a \plus{} b\le n \minus{} \mid B_{k \minus{} 1}\mid \plus{} a \plus{} b$ and the result follows.[/quote]

I might have misunderstood you, but how do we know that $ n \minus{} \mid (A_1\cap A_2\cap\ldots\cap A_{k \minus{} 1})\mid \plus{} a \plus{} b\le n \minus{} \mid B_{k \minus{} 1}\mid$?
Aren't you just trying to use the inequality from the (yet unproven) first claim here?  From what I understood, we're still trying to prove it.

Wojtek

1) I am making a construction. My claim is for my construction of $ A_i$'s, not for any set of $ A_i$'s.
2) It is from induction step. I assumed $ k \minus{} 1$ and proved for $ k$, as usual.
Sorry for writing the solution confusing. I had got a short time to use a PC so i was able to to write it just careless and quick.

Hey Umut,
Thanks for your reply!

[quote]2) It is from induction step. [...][/quote]
Right!  My bad, poor reading comprehension.

[quote]1) I am making a construction. My claim is for my construction of $ A_i$'s, not for any set of $ A_i$'s. [/quote]

This is still unclear to me.  I was trying to say exactly this, but perhaps I misunderstood your construction.
If you could give me a spare few minutes read and pinpoint my mistake or misunderstanding in the following paragraph, I would be really grateful.
Let my try to go through it step by step as clearly as possible.

Suppose $ n \equal{} 4$, the set of vertices is $ V \equal{} \{v_1,\ldots,v_7\}$, and that the set of complete subgraphs of size $ n$ contains sets $ C_1 \equal{} \{v_1,v_2,v_3,v_4\},C_2 \equal{} \{v_2,v_3,v_4,v_5\},C_3 \equal{} \{v_1,v_4,v_5,v_6\}$ among some others.
Since we may take $ A_1$ arbitrary subgraph of size $ n$, let's take $ A_1 \equal{} C_1 \equal{} \{v_1,v_2,v_3,v_4\}$.
In the second step of your construction we need to take an arbitrary vertex from $ A_1$, so let's say we pick $ v_1$.  Now we look for $ A_2$ which doesn't contain this vertex ($ v_1$), so let's say we pick $ A_2 \equal{} C_2 \equal{} \{v_2,v_3,v_4,v_5\}$.
Now we're in the third step, since $ A_1\cap A_2 \equal{} \{v_2,v_3,v_4\}$ is still not empty, we pick from the intersection for example $ v_3$ and find $ A_3$ that doesn't contain $ v_3$... since $ C_3$ is good, we may take $ A_3 \equal{} C_3 \equal{} \{v_1,v_3,v_5,v_6\}$.

Here is a question for you:
Is it in accordance with the procedure you suggested?  I didn't "delete" $ v_1$ permanently, only for construction of $ A_2$.  I mean, we're not given in advance that by removing any two vertices we can find a $ K_n$ outside, so I assume that we may look for $ A_3$ anywhere in $ V\setminus \{v_3\}$.?

So far we have $ B_1 \equal{} \{v_5,v_6,v_7\}$, $ B_2 \equal{} \{v_6,v_7\}$, $ B_3 \equal{} \{v_7\}$
and $ A_1 \equal{} \{v_1,v_2,v_3,v_4\}$, $ A_1\cap A_2 \equal{} \{v_2,v_3,v_4\}$, $ A_1\cap A_2\cap A_3 \equal{} \{v_4\}$.
The procedure goes on at least one more step (hence $ j\ge 3$) so that according to your claim the sharp inequality $ |A_1\cap A_2\cap A_3| > |B_3|$ needs to hold for these three sets, but $ |A_1\cap A_2\cap A_3| \equal{} 1 \equal{} |B_3|$.

The problem is about my Claim'. When you take $ C_3 \equal{} (v_1,v_4,v_5,v_6)$ it makes $ (v_1,v_2,v_3,v_4,v_5)$ a $ K_{n \plus{} 1}$. I am considering it in Claim'.
Edit: I read my solution one more time and there is a typo. In Claim' there mustn't be $ >B_{k\minus{}1}$. I don't know why it is there but sorry :blush:

Hey Umut,

Finally it clicked to me :-)
Thank you for your comments, they were helpful a lot!
(This problem was puzzling me for some time.)

best regards,
Wojtek


If $ f(x)\equal{}x^2\plus{}x$, prove that the equation $ 4f(a)\equal{}f(b)$ has no solutions in positive integers $ a,b$.

Write the equation as a quadratic equation on $ a$: $ a^2\plus{}a\minus{}\frac{b^2\plus{}b}{4}\equal{}0$, with solutions $ a\equal{}\frac{\minus{}1\pm\sqrt{b^2\plus{}b\plus{}1}}{2}$.  If $ a$ is integer, then so is $ \sqrt{b^2\plus{}b\plus{}1}$.  But $ (b\plus{}1)^2\equal{}b^2\plus{}2b\plus{}1>b^2\plus{}b\plus{}1>b^2$, and $ b^2\plus{}b\plus{}1$ cannot be a perfect square, contradiction hence $ a$ is not integer.