About two matrices A,B

Prove AB=BA, such that A^2 + AB + B^2 = 0.

$ A(A^2 \plus{} AB \plus{} B^2) \minus{} (A^2 \plus{} AB \plus{} B^2)B \equal{} A^3 \minus{} B^3 \equal{} 0$

$ (A^2 \plus{} AB \plus{} B^2)A \minus{} B(A^2 \plus{} AB \plus{} B^2)B$
$ \equal{} (A^3 \minus{} B^3) \plus{} (AB \minus{} BA)A \plus{} B(BA \minus{} AB)$
$ \equal{} (AB \minus{} BA)A \minus{} B(AB \minus{} BA) \equal{} 0$

If, $ A \neq 0$ and $ B \neq 0$, $ AB \minus{} BA \equal{} 0 \iff AB \equal{} BA$.

$ A \equal{} 0$ or $ B \equal{} 0$ or $ A, B \equal{} 0$ is trivial.

[quote="Kouichi Nakagawa"] $ \equal{} (AB \minus{} BA)A \minus{} B(AB \minus{} BA) \equal{} 0$

If, $ A \neq 0$ and $ B \neq 0$, $ AB \minus{} BA \equal{} 0 \iff AB \equal{} BA$..[/quote]

How could you make this deduction ? In matrix world, $ UV\equal{}0$ does not imply $ U\equal{}0$ or $ V\equal{}0$ !

$ (A \plus{} B)(A \plus{} B) \equal{} A^2 \plus{} AB \plus{} BA \plus{} B^2$ and by hipotesis we got...$ A^2 \plus{} B^2\plus{} AB \equal{} 0$...then $ / \plus{} BA$ in  both sides in the hipotesis equation we got $ A^2 \plus{} B^2 \plus{} AB \plus{} BA \equal{} BA\Rightarrow (A \plus{} B)^2 \equal{} BA$

if  AB=BA ... so ....

$ (A \plus{} B)^2 \minus{} AB \equal{} 0$
$ A^2 \plus{} B^2 \plus{} BA \equal{} 0$

Then...

$ (A \plus{} B)^2 \minus{} AB \equal{} A^2 \plus{} B^2 \plus{} BA\Rightarrow BA \minus{} AB \equal{} A^2 \plus{} B^2 \plus{} BA$... then....$ / \plus{} \minus{} BA$ in both sides....$ \minus{} AB \equal{} A^2 \plus{} B^2\Rightarrow A^2 \plus{} B^2 \plus{} AB \equal{} 0$

uh guys I think that I solve it but my solution is to large and ugly  :(  :huh:

$ (A \plus{} B)^2 \equal{} BA/...then...B^{ \minus{} 1}*/....and.../*B$ we got...
$ B^{ \minus{} 1}(A \plus{} B)^2B \equal{} AB$....
$ (B^{ \minus{} 1}A \plus{} I)(AB \plus{} B^2) \equal{} AB$
$ (B^{ \minus{} 1}A \plus{} I)* \minus{} A^2 \equal{} AB$
$ \minus{} B^{ \minus{} 1}A^3 \plus{} \minus{} A^2 \equal{} AB/... \plus{} A^2$
$ \minus{} B^{ \minus{} 1}A^3 \equal{} \minus{} B^2/...* \minus{} 1/....B*$
$ A^3 \equal{} B^3$

$ B^4 \equal{} B^4$
$ ^\Rightarrow A^2*AB \equal{} BA*A^2$
$ ( \minus{} AB \minus{} B^2)*AB \equal{} BA( \minus{} AB \minus{} B^2)$
$ ABAB \plus{} B^2AB \equal{} BAAB \plus{} BAB^2$
$ AB^2 \plus{} B^2AB \equal{} BAAB \plus{} BAB^2$
$ (AB)^2 \equal{} BAB^2 \minus{} B^2AB \plus{} BA^2B$
$ (AB)^2 \equal{} B(AB \minus{} BA \plus{} A^2)B$
$ (AB)^2 \equal{} B(AB \minus{} BA \plus{} \minus{} (B^2 \plus{} AB))B$
$ (AB)^2 \equal{} B( \minus{} BA \minus{} B^2)B$
$ (AB)^2 \equal{} \minus{} B^2(A \plus{} B)B/...B*$
$ B(AB)^2 \equal{} B^3(A \plus{} B)B$
$ B(AB)^2 \equal{} \minus{} A^3(AB \plus{} B^2)$
$ B(AB)^2 \equal{} \minus{} A^3* \minus{} A^2$
$ (BA)^2B \equal{} A^5/...(BA)^{ \minus{} 1}$
$ (BA)B \equal{} (BA)^{ \minus{} 1}A^5$
$ BAB \equal{} \equal{} A^{ \minus{} 1}*B^{ \minus{} 1}A^5$
$ BAB \equal{} A^{ \minus{} 1}B^{ \minus{} 1}A^3A^2$
$ BAB \equal{} A^{ \minus{} 1}B^{ \minus{} 1}B^3A^2$
$ BAB \equal{} A^{ \minus{} 1}B^2A^2/...A*$
$ ABAB \equal{} B^2A^2$
$ (AB) \equal{} B^2*A^2$

$ (A \plus{} B)^2 \equal{} BA/...*(BA)^{ \minus{} 1}$
$ (A \plus{} B)^2(BA)^{ \minus{} 1} \equal{} I/...AB*$
$ AB(A \plus{} B)^2(BA)^{ \minus{} 1} \equal{} B(A \plus{} B)^2B^{ \minus{} 1}$.....*
$ AB(A \plus{} B)^2A^{ \minus{} 1}B^{ \minus{} 1} \equal{} B(A \plus{} B)^2B^{ \minus{} 1}/....*B$
$ AB(A \plus{} B)^2A^{ \minus{} 1} \equal{} BBA/.....*A$
$ ABBA \equal{} B^2*A^2$.....but $ (AB)^2 \equal{} B^2A^2$.....then....
$ ABBA \equal{} (AB)^2/....(AB)^{ \minus{} 1}*$
$ BA \equal{} AB$...$ AB \equal{} BA$

*$ (A \plus{} B)^2 \equal{} BA/...B^ \minus{} 1*/....*B\Rightarrow B^ \minus{} 1(A \plus{} B)^2B \equal{} AB$

well that`s all ,  thanks loup blanc

Hi Alberto,
what would you want to prove ?

Hi loup

well i`m try to prove that if $ A^2\plus{}B^2\plus{}AB\equal{}0\Rightarrow AB\equal{}BA$ , but my answer it`s to huge and ugly because :first i have to prove that $ A^3\equal{}B^3$,then prove $ (AB)^2\equal{}B^2*A^2$ and then that $ AB\equal{}BA$ starting with $ (A\plus{}B)^2\equal{}BA$ because $ A^2\plus{}B^2\plus{}AB\equal{}0/ then \plus{}BA in_both_sides...$ $ A^2\plus{}B^2\plus{}AB\plus{}BA\equal{}(A\plus{}B)^2\equal{}BA....so...(A\plus{}B)^2\equal{}BA$ . But if you  spot a miss please tell me, i`ll be very thankfull.

  :)

Have  a look on my first post. It's a counterexample : $ A^2\plus{}B^2\plus{}AB\equal{}0$ doesn't imply $ AB\equal{}BA$ !!

Facts: the result is false in $ \mathcal{M}_2(\mathbb{C})$ but is true in $ \mathcal{M}_2(\mathbb{R})$

After we have obtained $ (AB\minus{}BA)A\minus{}B(AB\minus{}BA) \equal{} 0$ one can easily prove that $ AB\minus{}BA$ is nilpotent on $ \mathbb{R}$ and $ \mathbb{C}$ by showing by induction that $ Tr((AB\minus{}BA)^k)\equal{}0$ for all $ k.$

Actually, if we denote by $ X\equal{}AB\minus{}BA$ there is a theory of solving $ AX\equal{}XB$ equations for any $ X$ and it depends on charesteristic roots of both $ A$ and $ B.$

[b]loup blanc, [/b] how to get your fact? Can one deduce it from what I wrote above?

Hi Rustam,
1) I think that $ (AB\minus{}BA)A\minus{}B(AB\minus{}BA)\equal{}0$ doesn't imply $ AB\minus{}BA$ is nilpotent.
2) In $ \mathcal{M}_2(\mathbb{C})$, take $ A\equal{}\begin{pmatrix}1&u\\0&j\end{pmatrix},B\equal{}\begin{pmatrix}j&v\\0&j^2\end{pmatrix}$ where $ j\equal{}exp(2i\pi/3)$.
3) In $ \mathcal{M}_2(\mathbb{R})$ necessarily $ A,B$ commute. See my post in
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=396184141&t=144008
where you change $ A$ into $ \minus{}A$.

[quote="loup blanc"]Hi Rustam,
1) I think that $ (AB \minus{} BA)A \minus{} B(AB \minus{} BA) \equal{} 0$ doesn't imply $ AB \minus{} BA$ is nilpotent.
[/quote]

Hi loup blanc! I found a mistake in my proof of my assumption.
thanx for the link


Given that $ x$ is a positive integer and $ y\equal{}5^x$, what is the greatest value for $ y$ for which $ y<$ 30,000?

[quote="ernie"]You can definitely use logs to your advantage here, but since this is a School Sprint, brute force reveals $ \boxed{6}$.[/quote]

What?? I think you mean x=6, making y=15625.


Find all polynomials $P(x)$ with integer coefficients such that $(x^2+6x+10)P(x)^2 -1 $ is a square of another polynomial $Q(x)$ with integer coefficients.


Hello,

I was wondering what topic I need to look up to solve the following:

Let $ x$ and $ y$ be positive integers such that $ 3x \plus{} 7y$ is divisible by $ 11$.  Which of the following must also be divisible by $ 11$?

(A) $ 4x \plus{} 6y$
(B) $ x \plus{} y \plus{} 5$
(C) $ 9x \plus{} 4y$
(D) $ 4x \minus{} 9y$
(E) $ x \plus{} y \minus{} 1$

I know the answer just by some guess and check, but I would like to learn theory behind it.  Thanks.

Source: GRE Math Subject Practice Test #25

One way to approach this is by modular division.  The given is equivalent to saying that $ 3x \equiv \minus{} 7y \bmod 11$ (the notation $ a \equiv b \bmod m$ means $ m$ divides $ a \minus{} b$).  Multiplying by four gives

$ 12x \equiv x \equiv \minus{} 28y \equiv 5y \bmod 11$

(in other words, $ \frac{\minus{}7}{3} \equiv 5 \bmod 11$, which is what I meant when I said we should use modular division.)  We can now check the validity of A) and D) by computing

$ 4x \equiv 20y \equiv 9y \bmod 11$

so the answer is D).


$a,b,c$ are non-negative real numbers such that :
$\frac{1}{a\sqrt{a}}+\frac{1}{b\sqrt{b}}+\frac{1}{c\sqrt{c}}=3$

Prove that

$\frac{a^{2}+c}{b}+\frac{b^{2}+a}{c}+\frac{c^{2}+b}{a}\geq 6$


function
algebra
functional equation
algebra unsolved

Determine all functions $ f: \mathbb{R}^\plus{} \mapsto \mathbb{R}^\plus{}$ which satisfy \[ f \left(\frac {f(x)}{yf(x) \plus{} 1}\right) \equal{} \frac {x}{xf(y)\plus{}1} \quad \forall x,y > 0\]

What was the motivation for creating $ h$?

Anyway... Exactly the same works for $ f$.

[quote=mathmanman]The only solution is : $ \forall x \in \mathbb{R}^\plus{}, \ \ f(x) \equal{} x.$

Here is a sketch of my proof :

at first, let $ h(x) \equal{} \frac 1{f \left( \frac 1x \right)},$ from $ \mathbb{R}^\plus{}$ to $ \mathbb{R}^\plus{}$ as well; then our functional equation becomes $ h(y\plus{}h(x)) \equal{} x \plus{} \frac 1{h \left( \frac 1y \right)}.$

Now, $ h$ is clearly injective (check it), but $ h$ is also surjective (same thing), hence bijective,
then we get that the inverse function $ g$ of $ h$ exists, and satisfies the same functional equation (again, check it).

Now, we'll prove that $ h(x) \equal{} x.$
If $ a > h(b),$ then $ h(a) \equal{} h((a\minus{}h(b)) \plus{} h(b)) \equal{} b \plus{} \frac 1{h \left( \frac 1 {a\minus{}h(b)} \right)} > b$ so $ a > h(b)$ implies that $ h(a) > b.$
Now since the inverse function $ g$ satisfies the same functional equation, we also have $ (c > g(d) \implies g(c) > d).$
Then, letting $ c \equal{} h(a)$ and $ d \equal{} h(b),$ we get $ (h(a) > b \implies a > h(b)).$
Therefore, we have that $ (a > h(b) \iff h(a) > b).$
Finally take $ a : \equal{} x$ and $ b : \equal{} x$ in the above to get the desired result.

Thus, eventually, $ f(x) \equal{} x.$[/quote]

Although it is easy to show $h$ is injective, how can we prove $h$ is surjective?
Also, it is not easy to see that the inverse function of $h$ (if exists) satisfies the same functional equation. Can you show it clearly?
Thank you for sharing.


My biggest problem is learning how to convert word problem data into a math equation leading to the answer by solving for a variable or variables.

This skill separates amateur math people from true mathematicians.  I believe the skill to create an equation from the given information in word problems is unique and wonderful.

Here is a sample question:

Ritter earns 9 dollars per hour weekdays and twice as much per hour during the weekend.  Last week he earned 378 dollars, including weekend hours.  Of the 35 hours Ritter worked for that week, how many hours did he work during the weekend?

MY WORK:

Monday through Friday = 9 dollars an hour.

Weekend = 18 dollars per hour.

Total earned = 378 dollars.

Total hours, including weekend hours = 35.

Let x = weekend hours.

I was not able to set up an equation using the given data.

I found the answer by playing with the information for about 30 minutes.

I know the answer.

The answer is 7 hours.

However, can someone here show me HOW TO set up the PROPER EQUATION to find the answer much faster?

dunno but

[hide]

systems of equations?

$x+y=35$
$9x+18y=378$

[/hide]

solve for y...i think that's it...im not sure  :D


Let $n$ points $A_1, A_2, \ldots, A_n$, ($n>2$), be considered in the space, where no four points are coplanar. Each pair of points $A_i, A_j$ are connected by an edge. Find the maximal value of $n$ for which we can paint all edges by two colors – blue and red  such that the following three conditions hold:

[b]I.[/b] Each edge is painted by exactly one color.
[b]II.[/b] For $i = 1, 2, \ldots, n$, the number of blue edges with one end $A_i$ does not exceed 4.
[b]III.[/b] For every red edge $A_iA_j$, we can find at least one point $A_k$ ($k \neq i, j$) such that the edges $A_iA_k$ and $A_jA_k$ are blue.

i put a solution yesterday but then realized that there was an error in what i had earlier. here is a correct solution.

in the first place if we regard the graph formed only by the blue edges on $n$ vertices, then the graph has $\Delta\leq 4$ and has diameter =$2$.
firstly note that if some vertex has degree $3$ or lesser then there are atmost $1+3+3^2=13$ vertices(each term counts teh number of vertices at distance $0,1$ or $2$ from the chosen vrtex of degree atmost $3$). so suppose that we have a regular graph of degree $4$ and diamter $2$. as above we get a bound of $1+4+4\cdot 3=17$ vertices. if the graph has $17$ vertices then it is a Moore graph but then the only possible Moore graphs occur with degree $2,3,7$ or possibly $57$(this is because they become what are called strongly regular graphs and so have to satisfy a so-called Integrality condition), so $n=17$ is ruled out.
 if the graph has $16$ vertices, then note that it cannot have any triangles(else, atmost 15 vertices are at a distance $\leq 2$ from one of the vertices of the triangle-this is easy to work out so i won't give the details). hence if we count the number of colored triangles with $2$ blue and $1$ red edge(or in this graph, configurations of the form $(u,v,w) u\leftrightarrow v,u\leftrightarrow w$) then since each vertex contributes six such configurations, there are $96$ such configurations and $32$ edges. Hence some non-adjacent pairs have more than one vertex adjacent to both the vertices of the pair-in other words, quadrilaterals(there are $8$ of them).
if $u\not\leftrightarrow v$ and $x,y,z$ are vertices adjacent to both $u$ and $v$ then it is easy to see that $u$ is at a distance of atmost $2$ from atmost $15$ vertices and that is a contradiction.
if $u,v,x,y$ form a quadrilateral and there is another quadrilateral sharing the side(say) $(u,x)$ then again one can see that $u$ is at a distance $\leq 2$ with atmost $14$ vertices and so that is a contradiction again.
finally if $u$ belongs to two disjoint quadrilaterals then again one can see that $u$ is at a distance of atmost 2 from atmost 15 vertices and so that is a contradiction again.
the upshot is that any two such quadrilaterals are disjoint(vertex-wise). as we have counted here, there are $8$ such quadrilaterals but only $16$ vertices and these quadrilaterals are disjoint-that is again a contradiction.so $n\leq 15$.

now, for $n=15$ such a graph is possible(this is rather well-known) and i'll just attach a file for verification.

sorry, don't quite see why number of quadrilaterals is 8. i only came up with 4


trigonometry
inequalities
function
inequalities proposed

Q: Let x,y, and z be acute angles. Show that:
(sinx + cosx)(siny + cosy)(sinz + cosz)(sinxsinysinz + cosxcosycosz) is
less than or equal to 2.
I would be delighted to hear from you.
Sincerely,
Can.

Well, it is definitely true. By using AM - GM and the following inequality : $cosx+cosy+cosz \leq \frac{3}{2}$ and ${sinx + siny+sinz \leq \frac{3\sqrt{3}}{2}}  $.We may replace 2 by $\frac{(3\sqrt{3}+1)(\sqrt{3}+1)^3}{64}$

Treegoner's solution is true if x + y + z = 180 degrees. Here the sum need not be 180 degrees. So AM-GM doesn't work.
Can. :)  :)  :)

The key of the problem in general is the Jensen 's inequality by considering the two function $f(x) = sinx$ and $f(x) =ln(sinx)$ in $[0,\pi]$. :D to make a one variable inequality.  :D

Hi,
The problem I asked is best attacked by making the substitution m = tanx, and n = tany and p = tanz. We are then back to the non-homogenous case. Now with the help of a solved problem of APMO 04' ( please check out the site of John Scholes if you are interested ) I think we are a lot closer to getting the answer. Further discussion is certainly welcomed.
Can. :)


LaTeX
induction
algebra
binomial theorem

Find a formula, using the Newton Binomial, to find the value of:

$ 1^4 \plus{} 2^4 \plus{}3^4 \plus{} 4^4 \plus{} 5^4...........n^4$  where $ n$ is a natural number.

I have tried it by many ways!!

I started rewriting the addition like that:

$ (1)^4 \plus{} (1\plus{}1)^4 \plus{} (1\plus{}2)^4 \plus{} (1\plus{}3)^4.........$ but the problem is that at the end of the expression I get a four power number again!!!!

Please help me, and explain it step-by-step

Thanks! :lol:

Hi Chalao.

One way of doing what you want is:

Begin with

$ (r+1)^{5} - r^{5} = 5r^{4} + 10r^{3} + 10r^{2} + 5r +1$

Substitute $ r=1, r=2, ..., r=n$ and add these.

This assumes you already know $ \sum_{1}^{n} r = \frac{n(n+1)}{2}, \sum_{1}^{n} n^{2} = \frac{n(n+1)(2n+1)}{6}$ and $ \sum_{1}^{n} n^{3} = (\frac {n(n+1)}{2})^{2}$.

It is quicker, however, to use

$ (2r+1)^{5} - (2r-1)^{5} = 2(80r^{4} + 40r^{2} +1)$ and then to do the same thing, as this only requires knowing $ \sum_{1}^{n} n^{2} = \frac{n(n+1)(n+2)}{6}$.

I can probably write this out in more detail if you want as my CTRL key seems to be working OK this morning so I can type \ and use $ \text{\LaTeX}$.

Yes AndrewTom, that will work, but he seems to want a proof using Binomial Coefficients.

Yes, the binomial $ (r\plus{}1)^{5}$ is expanded using the binomial theorem which, in this case, involves the binomial cofficients $ \binom{5}{i}, i\equal{} 0, 1, 2, 3, 4, 5$ as

$ \sum_{i\equal{}0}^{5} \binom{5}{i} r^{5\minus{}i} \equal{} \binom{5}{0}r^{5} \plus{} \binom{5}{1}r^{4} \plus{} \binom{5}{2}r^{3} \plus{} \binom{5}{4}r^{2} \plus{} \binom{5}{1}r \plus{} \binom{5}{5}$

$ \equal{} r^{5} \plus{} 5r^{4} \plus{} 10r^{3} \plus{} 10r^{2} \plus{} 5r \plus{}1$

[quote="Chalao"]Find a formula, using the Newton Binomial, to find the value of:

$ 1^4 \plus{} 2^4 \plus{} 3^4 \plus{} 4^4 \plus{} 5^4...........n^4$  where $ n$ is a natural number.

I have tried it by many ways!!

I started rewriting the addition like that:

$ (1)^4 \plus{} (1 \plus{} 1)^4 \plus{} (1 \plus{} 2)^4 \plus{} (1 \plus{} 3)^4.........$ but the problem is that at the end of the expression I get a four power number again!!!!

Please help me, and explain it step-by-step

Thanks! :lol:[/quote]

[hide="solution?"]
suppose that $ 1^4 \plus{} 2^4 \plus{} 3^4 \plus{} 4^4 \plus{} 5^4...........n^4\equal{}an^5\plus{}bn^4\plus{}cn^3\plus{}dn^2\plus{}en$
then $ 1^4 \plus{} 2^4 \plus{} 3^4 \plus{} 4^4 \plus{} 5^4...........n^4\plus{}(n\plus{}1)^4\equal{}a(n\plus{}1)^5\plus{}b(n\plus{}1)^4\plus{}c(n\plus{}1)^3\plus{}d(n\plus{}1)^2\plus{}e(n\plus{}1)$
$ (n\plus{}1)^4\equal{}a[(n\plus{}1)^5\minus{}n^5]\plus{}b[(n\plus{}1)^4\minus{}n^4]\plus{}c[(n\plus{}1)^3\minus{}n^3]\plus{}d[(n\plus{}1)^2\minus{}n^2]\plus{}e[(n\plus{}1)\minus{}n]$
$ n^4\plus{}4n^3\plus{}6n^2\plus{}4n\plus{}1\equal{}a(5n^4\plus{}10n^3\plus{}10n^2\plus{}5n\plus{}1)\plus{}b(4n^3\plus{}6n^2\plus{}4n\plus{}1)\plus{}c(3n^2\plus{}3n\plus{}1)\plus{}d(2n\plus{}1)\plus{}e$
$ 1\equal{}5a$
$ 4\equal{}10a\plus{}4b$
$ 6\equal{}10a\plus{}6b\plus{}3c$
$ 4\equal{}5a\plus{}4b\plus{}3c\plus{}2d$
$ 1\equal{}a\plus{}b\plus{}c\plus{}d\plus{}e$
easily we get $ a\equal{}\frac15,b\equal{}\frac12,c\equal{}\frac13,d\equal{}0,e\equal{}\minus{}\frac{1}{30}$
then $ 1^4 \plus{} 2^4 \plus{} 3^4 \plus{} 4^4 \plus{} 5^4...........n^4\equal{}\frac15n^5\plus{}\frac12n^4\plus{}\frac13n^3\minus{}\frac{1}{30}n$
        $ \equal{}\frac{1}{30}n(n\plus{}1)(2n\plus{}1)(3n^2\plus{}3n\minus{}1)$
[/hide]

Yet another, more complicated way, is as follows:

We can prove by mathematical induction that

$ \sum_{1}^{n} 1 \equal{} n$, $ \sum_{1}^{n} r \equal{} \frac{1}{2}n(n\plus{}1)$, $ \sum_{1}^{n} r(r\plus{}1) \equal{} \frac{1}{3} n(n\plus{}1)(n\plus{}2)$, $ \sum_{1}^{n} r(r\plus{}1)(r\plus{}2) \equal{} \frac{1}{4}n(n\plus{}1)(n\plus{}2)(n\plus{}3)$, $ \sum_{1}^{n} r(r\plus{}1)(r\plus{}2)(r\plus{}3) \equal{} \frac{1}{5}n(n\plus{}1)(n\plus{}2)(n\plus{}3)(n\plus{}4)$.

Then we can use $ r^{2} \equal{} r(r\plus{}1) \minus{} r$ to get $ \sum_{1}^{n} r^{2} \equal{} \sum_{1}^{n} r(r\plus{}1) \minus{} \sum_{1}^{n} r \equal{} \frac{1}{3}n(n\plus{}1)(n\plus{}2) \minus{} \frac{1}{2}n(n\plus{}1) \equal{} \frac{1}{6}n(n\plus{}1)(2n\plus{}1)$.

Similarly, $ r^{3} \equal{} r(r\plus{}1)(r\plus{}2) \minus{}3r(r\plus{}1) \plus{}r$ gives $ \sum_{1}^{n} r^{3} \equal{} \frac{1}{4}n^{2}(n\plus{}1)^{2}$, and finally

$ r^{4} \equal{} r(r\plus{}1)(r\plus{}2)(r\plus{}3) \minus{}6r(r\plus{}1)(r\plus{}2) \plus{}7r(r\plus{}1) \minus{}r$ gives $ \sum_{1}^{n} r^{4} \equal{} \frac{1}{30}n(n\plus{}1)(2n\plus{}1)(3n^{2} \plus{}3n\minus{}1)$.


Let $ a,b,c$ be positive reals. Prove that

$ 4(a^2\minus{}a\plus{}1)(b^2\minus{}b\plus{}1)(c^2\minus{}c\plus{}1)\geq1\plus{}2abc\plus{}(abc)^2$

It is enough to show $ 2(a^{2}\minus{}a\plus{}1)^{\frac{3}{2}} \geq a^{3}\plus{}1$
(because of cauchy schwartz inequality)
but it is equivalent to $ 3(a\minus{}1)^{2} \geq 0$
so we are done.


If $ a$ is a nonzero integer and $ b$ is a positive number such that $ ab^2\equal{}\log{10}{b}$, what is the median of the set {$ 0,1,a,b,1/b$}?

I don't understand how they came up with the solution to this problem[hide]  the solution packet says, "if $ b>\equal{}1$, then $ 0<\log{10}{b}/b^2<1$, so $ a$ cannot be an integer.  I understand their logic, but how do you know to try $ b>\equal{}1$ in the first place?  why not $ b>\equal{}2$ ? [/hide]  what section in the AOPS books should I review to get a grasp on these types of problems?

You don't need to read a chapter of AoPS to get a grasp on "these types of problems."  Just try different values of $ b$ or just explicitly solve for $ a$ in terms of $ b$ and figure out when it's possible for $ a$ to be a nonzero integer.  It's a matter of not being intimidated and being organized more than anything else.


I've recently come across four Miquel's Theorems, all different and I'm not sure if they are actually known as Miquel's Theorem and I just want to clarify what I currently perceive the theorems to be.

Anyway:

[b][u]Miquel's (Pivot) Theorem[/u][/b]
Given a triangle $ABC$ and three points $A', B', C'$ on $BC, CA, AB$ respectively. Then the circumcircles of $\triangle AB'C', \triangle BC'A', \triangle CB'A'$ meet at a point.

[b][u]Miquel's Theorem[/u][/b]
Given four concylic points $A, B, C, D$, four circles through $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$ will have four new points of intersection. Then these new points are concylic.

[b][u]Miquel's (Pentagram) Theorem[/u][/b]
Given a convex pentagon, extend the sides to a pentagram. Externally to the pentagon, there are five triangles. Construct the five circumcircles. Each pair of adjacent circles intersect at a vertex of the pentagon and a second point. Then these five points are concyclic.

[b][u]Miquel's (Five Circles) Theorem[/u][/b]
Let five circles with concyclic centers be drawn such that each intersects its neighbors in two points, with one of these intersections lying itself on the circle of centers. By joining adjacent pairs of the intersection points which do not lie on the circle of center, an (irregular) pentagram  is obtained each of whose five vertices lies on one of the circles with concyclic centers.

I've always known the first theorem you mentioned as "Miquel's theorem".

I remember proving the third one myself, I didn't really know it was a theorem. It's a nice exercise, try it.

I've also heard about "Miquel points" quite often :)

Miquel Points are associated with the frst Miquel's Pivot Theorem. I was just getting concerned because like you I've always associated his theorem with the first one I mentioned.

But then last month I read an article about a different Miquel's Theorem. The proof is really trivial but nevertheless it appears to be the Miquel's Theorem according to Cambridge University mathematics glossary.

Miquel's theorem (for 2 tupe-4 points)
the circles (Ci), i=1,2,3,4 such that (Ci) and (Ci+1) intersects at Ai and Bi. If A1,A2,A3,A4 are concyclic then B1,B2,B3,B4 are concyclic


conics
exterior angle
geometry
angle bisector

Given and ellepse with focus $F_1$ and $F_2$. $P$ is a mobile point on the ellipse. Through $F_1$ construct a perpendicular line to the exterior angle bisector of $\angle{F_1PF_2}$. Find the locus of the projection.

Let $H$ be the projection of $F_{1}$ onto the exterior angle bisector of $\angle F_{1}PF_{2}$
Let $PD$ be the internal angle bisector of $\angle F_{1}PF_{2}$.Then of course $F_{1}H \parallel PD$

Let $Z \in F_{1}H \cap F_{2}P$.
Then it is known that the triangle $PZF_{1}$ is isosceles (because $\angle PF_{1}Z = \angle PZF_{1}= \angle F_{1}PD = \angle F_{2}PD$)

Hence $PZ=PF_{1}\Rightarrow F_{2}Z = F_{2}P+F_{1}P = 2a = constant$

$H$ is the midpoint of $F_{1}Z$, so the parallel line to $F_{2}Z$ through $H$ meets $F_{1}F_{2}$ on its midpoint $O$.

So $OH = \frac{F_{2}Z}{2}= a\Rightarrow H$ lies on the circle $w=(O,a)$

Inversely, for any point $H \in w$, we can see (using the same steps) that $F_{1}P+F_{2}P=2a$

So, the locus of $H$ is the circle $w$


analytic geometry
graphing lines
slope
real analysis
real analysis solved

3.Let (a_n)n>0  R*. For each n  N*\{1}, we consider e_n : a_nx+a_(n-1)y=0.
We denote with (b_n)n respectively (c_n)n those (a_n)n for wich e_n and e_(n+1) are perpendicular respectively parallel.
Study the convergence of (b_n)n and (c_n)n.

what did u do with this Kuba?

oh.. this is very easy
u only put the condition for pantes(panta in romanian)...
me_n*me_(n+1) = -1.
it's only routine... but if u want i will put the solution

[quote="Kuba"]pantes(panta in romanian)...
[/quote]
actually it's [u]slope[/u] in English :)


For real constant $ k$, consider the cubic equation : $ |x^3\minus{}x|\equal{}x\plus{}k$.

(1) Find the values of $ k\equal{}k_1,\ k_2\ (k_1<k_2)$ such that the cubic equation has exactly 3 distinct real roots.

(2) For $ k\equal{}k_2$, find 3 real roots $ x_1,\ x_2,\ x_3\ (x_1<x_2<x_3)$ of the cubic equation.

(3) For $ k_1<k<k_2$, how many real roots does the cubic equation have?

1) $ k_1\equal{}1$ and $ k_2\equal{}\frac{4\sqrt6}9$.

2) i'm too lazy to do it. :P

3) 4

Correct. I'm looking forward to answer to (2).

hello kunny, answer to (2)
$ x_1\equal{}\minus{}\sqrt{\frac{2}{3}}$
$ x_2\equal{}2\sqrt{\frac{2}{3}}$
$ x_3\equal{}\minus{}\frac{2^{\frac{5}{6}}}{\sqrt{3}}$
Sonnhard.

Your answer is incorrect, Dr Sonnhard Graubner.


The area of a circle is increasing at a rate of $ 250$ square metres per hour. Find the rate of increase of the radius when the area is $ 1900$ square metres.

The area of a circle is $ A\equal{}\pi r^{2}$.  We are told that $ \frac{dA}{dt}\equal{}2\pi r(t)\frac{dr}{dt}\equal{}250$.  When the area is 1900, we get that the radius is $ r\equal{}\sqrt{\frac{1900}{\pi}}$, so $ \frac{dr}{dt}\equal{}\frac{125}{\pi r}\equal{}\frac{dr}{dt}\equal{}\frac{1250\sqrt{19}}{\sqrt{\pi}}$ meters per hour.


trigonometry
calculus
calculus computations

How do you prove that \[ \cos x=\frac{e^{ix}+e^{-ix}}{2} \]

You can't put an apostrophe inside your alternate hide text. There's no problem with multiple hides, but your second hide is broken code.

[quote="FMako"][hide="you could just know that..."]
$e^{ix} = \cos{x} + \i \sin {x}$
Then,  $\frac{e^{ix}+e^{-ix}}{2} = \frac{2\cos{x}}{2} = \cos{x}$
[/hide]
OR
[hide="if I am allowed to use taylor series ..."]
Then,
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \rightarrow e^{ix} = \sum_{n=0}^{\infty} \frac{{ix}^n}{n!}$
Also,
$\cos{x} = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^2n}{(2n)!}$
$\sin{x} = \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n-1)!}$
Then, $e^{ix} = \cos{x} + \i \sin {x}$
[/hide]

How come I can't have double hides?
Edit: problem solved.[/quote]

ooh... man...i can't believe i didnt see that... heh  :blush:


search
inequalities unsolved
inequalities


IMO Shortlist
combinatorics proposed
combinatorics

Prove that if in a graph every  pair of 3-cliques has at least one vertice in common, and there are no 5-cliques, then there are 2 or fewer vertices such that , if you remove them from the graph, then there will be no 3-clique reamaining.

obs: a n-clique is a graph with n points where every point has an edge with every other point (in other words , every 2 points have an edge that has endpoints at these 2 points) .

Many IMO shortlist problems have been posted on the forum before.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17455 for this one.


$k\geq 2;1<n_1<n_2<...<n_k$ is positive integers.$a,b\in\mathbb{N}^*$ satisfying:
 $\prod\limits_{i=1}^{k}(1-\frac{1}{n_i})\leq \frac{a}{b}<\prod\limits_{i=1}^{k-1}(1-\frac{1}{n_i})$.Prove that $\prod\limits_{i=1}^{k}n_i\geq (4a)^{2^k-1}$.


In a game of chance, there is a $ \frac{1}{8}$ chance of winning a car, a $ \frac{1}{4}$ chance of winning a sofa, and a $ \frac{1}{2}$ chance of winning a goat. What is the probability that a player will not win anything? Express your answer as a common fraction.

The probability of winning SOMETHING is $ \frac{1}{8}\plus{}\frac{1}{4}\plus{}\frac{1}{2}\equal{}\frac{7}{8}$.  Then the probability that a player won't win anything is $ 1\minus{}\frac{7}{8}\equal{}\boxed{\frac{1}{8}}$.

Unless you interpret it as having idependent probability, in which case its 7/8x3/4x1/2=21/64

It seems like Aime15's solution is better.
Also, I got this question on FTW and 1/8 is correct.


geometry
inequalities
3D geometry
prism
tetrahedron
algebra
polynomial

[size=150]Austrian Polish 2001, Single Competition, Day 1[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=486903]Problem 1:[/url]
Determine the number of positive integer $a$ for which there exist nonnegative integers $x_{0},x_{1},\cdots,x_{2001}$ satisying \[a^{x_{0}}=\sum_{k=1}^{2001}a^{x_{k}}\] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633010]Problem 2:[/url]
Let $n$ be a positive integer greater than $2$. Solve in nonnegative real numbers the following system of equations \[x_{k}+x_{k+1}=x_{k+2}^{2}\quad , \quad k=1,2,\cdots,n\] where $x_{n+1}=x_{1}$ and $x_{n+2}=x_{2}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73396]Problem 3:[/url]
Real numbers $a,b,c$ are lengths of the sides of a triangle. Prove that \[2< \frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}-\frac{a^{3}+b^{3}+c^{3}}{abc}\leq 3\] [size=150]
Austrian Polish 2001, Single Competition, Day 2[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7673]Problem 4:[/url]
Prove that if $a,b,c,d$ are lengths of the successive sides of a quadrangle (not necessarily convex) with the area equal to $S$, then the following inequality holds \[S \leq \frac{1}{2}(ac+bd).\] For which quadrangles does the inequality become equality?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633020]Problem 5:[/url]
The fields of the $8\times 8$ chessboard are numbered from $1$ to $64$ in the following manner: For $i=1,2,\cdots,63$ the field numbered by $i+1$ can be reached from the field numbered by $i$ by one move of the knight. Let us choose positive real numbers $x_{1},x_{2},\cdots,x_{64}$. For each white field numbered by $i$ define the number $y_{i}=1+x_{i}^{2}-\sqrt[3]{x_{i-1}^{2}x_{i+1}}$ and for each black field numbered by $j$ define the number $y_{j}=1+x_{j}^{2}-\sqrt[3]{x_{j-1}x_{j+1}^{2}}$ where $x_{0}=x_{64}$ and $x_{1}=x_{65}$. Prove that \[\sum_{i=1}^{64}y_{i}\geq 48\] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=230479]Problem 6:[/url]
Let $k$ be a fixed positive integer. Consider the sequence definited by \[a_{0}=1 \;\; , a_{n+1}=a_{n}+\left[\root k \of{a_{n}}\right] \;\; , n=0,1,\cdots\] where $[x]$ denotes the greatest integer less than or equal to $x$. For each $k$ find the set $A_{k}$ containing all integer values of the sequence $(\sqrt[k]{a_{n}})_{n\geq 0}$.

[size=150]Austrian Polish 2001, Team Competition[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111372]Problem 7:[/url]
Consider the set $A$ containing all positive integers whose decimal expansion contains no $0$, and whose sum $S(N)$ of the digits divides $N$.

(a) Prove that there exist infinitely many elements in $A$ whose decimal expansion contains each digit the same number of times as each other digit.

(b) Explain that for each positive integer $k$ there exist an element in $A$ having exactly $k$ digits.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111375]Problem 8:[/url]
The prism with the regular octagonal base and with all edges of the length equal to $1$ is given. The points $M_{1},M_{2},\cdots,M_{10}$ are the midpoints of all the faces of the prism. For the point $P$ from the inside of the prism denote by $P_{i}$ the intersection point (not equal to $M_{i}$) of the line $M_{i}P$ with the surface of the prism. Assume that the point $P$ is so chosen that all associated with $P$ points $P_{i}$ do not belong to any edge of the prism and on each face lies exactly one point $P_{i}$. Prove that \[\sum_{i=1}^{10}\frac{M_{i}P}{M_{i}P_{i}}=5\] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=179620]Problem 9:[/url]
Let $n>10$ be a positive integer and let $A$ be a set containing $2n$ elements. The family $\left\{A_{i}: i=1,2,\cdots,m\right\}$ of subsets of the set $A$ is called suitable if:

(1) for each $i=1,2,\cdots,m$ the set $A_{i}$ contains exactly $n$ elements,

(2) for all $i \neq j \neq k \neq i$ the set $A_{i}\cap A_{j}\cap A_{k}$ contains at most one element.

For each $n$ determine the length of a maximal suitable family.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111366]Problem 10:[/url]
The sequence $a_{1},a_{2},\cdots,a_{2010}$ has the following properties:

(1) each sum of the 20 successive values of the sequence is nonnegative,

(2) $|a_{i}a_{i+1}| \leq 1$ for $i=1,2,\cdots,2009$.

Determine the maximal value of the expression $\sum_{i=1}^{2010}a_{i}$.

[size=150]Austrian Polish 2002, Single Competition, Day 1[/size]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25161]Problem 1:[/url]
Find all triples $(a,b,c)$ of nonnegative integers such that the number $2^{c}-1$ divides the number $2^{a}+2^{b}+1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111902]Problem 2:[/url]
Let $P_{1}P_{2}\dots P_{2n}$ be a convex polygon with an even number of corners. Prove that there exists a diagonal $P_{i}P_{j}$ which is not parallel to any side of the polygon.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111903]Problem 3:[/url]
Let $ABCD$ be a tetrahedron and let $S$ be its center of gravity. A line through $S$ intersects the surface of $ABCD$ in the points $K$ and $L$. Prove that \[\frac{1}{3}\leq \frac{KS}{LS}\leq 3\] [size=150]
Austrian Polish 2002, Single Competition, Day 2[/size]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111910]Problem 4:[/url]
For each positive integer $n$ find the largest subset $M(n)$ of real numbers possessing the property: \[n+\sum_{i=1}^{n}x_{i}^{n+1}\geq n \prod_{i=1}^{n}x_{i}+\sum_{i=1}^{n}x_{i}\quad \text{for all}\; x_{1},x_{2},\cdots,x_{n}\in M(n)\] When does the inequality become an equality ?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111906]Problem 5:[/url]
Let $A$ be the set $\{2,7,11,13\}$. A polynomial $f$ with integer coefficients possesses the following property: for each integer $n$ there exists $p \in A$ such that $p|f(n)$. Prove that there exists $p \in A$ such that $p|f(n)$ for all integers $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111904]Problem 6:[/url]
The diagonals of a convex quadrilateral $ABCD$ intersect in the point $E$. Let $U$ be the circumcenter of the triangle $ABE$ and $H$ be its orthocenter. Similarly, let $V$ be the circumcenter of the triangle $CDE$ and $K$ be its orthocenter. Prove that $E$ lies on the line $UK$ if and only if it lies on the line $VH$.

[size=150]Austrian Polish 2002, Team Competition[/size]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111908]Problem 7:[/url]
Find all real functions $f$ definited on positive integers and satisying:

(a) $f(x+22)=f(x)$,

(b) $f\left(x^{2}y\right)=\left(f(x)\right)^{2}f(y)$

for all positive integers $x$ and $y$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111912]Problem 8:[/url]
Determine the number of real solutions of the system \[\left\{ \begin{aligned}\cos x_{1}&= x_{2}\\ &\cdots \\ \cos x_{n-1}&= x_{n}\\ \cos x_{n}&= x_{1}\\ \end{aligned}\right.\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111915]Problem 9:[/url]
A set $P$ of 2002 persons is given. The family of subsets of $P$ containing exactly 1001 persons has the property: the number of acquistance pairs in each such subset is the same. (It is assumed that the acquaintance relation is symmetric). Find the best lower estimation of the acquaintance pairs in the set $P$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111916]Problem 10:[/url]
For all real number $x$ consider the family $F(x)$ of all sequences $(a_{n})_{n\geq 0}$ satisfying the equation \[a_{n+1}=x-\frac{1}{a_{n}}\quad (n\geq 0).\] A positive integer $p$ is called a [i]minimal period[/i] of the family $F(x)$ if

(a) each sequence $\left(a_{n}\right)\in F(x)$ is periodic with the period $p$,

(b) for each $0<q<p$ there exists $\left(a_{n}\right)\in F(x)$ such that $q$ is not a period of $\left(a_{n}\right)$.

Prove or disprove that for each positive integer $P$ there exists a real number $x=x(P)$ such that the family $F(x)$ has the minimal period $p>P$.

Bump. I am guessing the reason that the Austrian-Polish does not appear in the resources section is because this was posted so long ago. Can it be added or is there another reason it does not appear there?

Links to posts in stead of topics:
Austrian-Polish 2002
Day 1
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=158499#p158499]Problem 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635702#p635702]Problem 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635704#p635704]Problem 3[/url]
Day 2
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635730#p635730]Problem 4[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635722#p635722]Problem 5[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635715#p635715]Problem 6[/url]
Team Contest
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635726#p635726]Problem 7[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635732#p635732]Problem 8[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635735#p635735]Problem 9[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635737#p635737]Problem 10[/url]

Oh, I see now that the links are to topics and not to the specific post. Let me give those:
Austrian-Polish Contest 2001
Day 1
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=486903#p486903]Problem 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633010#p633010]Problem 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=424153#p424153]Problem 3[/url]
Day 2
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=34993#p34993]Problem 4[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633020#p633020]Problem 5[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=230479#p230479]Problem 6[/url]
Team Contest
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=632708#p632708]Problem 7[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=632715#p632715]Problem 8[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=179620#p179620]Problem 9[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=632693#p632693]Problem 10[/url]

PS: I did the same [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3116977#p3116977]here[/url].

Thanks for posting!  Sorry for the delay, but here they are: http://www.artofproblemsolving.com/Forum/resources.php?c=11&cid=258&year=2001

Thanks for posting! Here you are: http://www.artofproblemsolving.com/Forum/resources.php?mode=edit_problems&c=11&cid=258&year=2002
Again, sorry for the delay.


Is there any way to play Starcraft without the cd in the cd drive?

Yes. It's probably not legal, though. See [url=http://en.wikipedia.org/wiki/Software_cracking]here[/url].

Since I have a copy, here's a relevant excerpt of the [hide="license"]1. Limited Use License. Blizzard Entertainment (“Blizzard”) hereby grants, and by installing the Program you thereby accept, a limited, non-exclusive license and right to install and use one (1) copy of the Program for your use on either a home or portable computer. In addition, the Program has a multi-player capability that allows up to eight players per registered version of the Program to play concurrently. These additional copies of the Program are known as and referred to hereafter as “Spawned Versions.” You may install Spawned Versions of the Program on an unlimited number of computers. However, Spawned Versions of the Program must be played in conjunction with the registered version of the Program from which they were spawned. All of the terms and conditions of the License Agreement shall also apply to the Spawned Version of the Program, with the exception that the Spawned Version Program may be installed on as many computers as you wish. The Program also contains a Campaign Editor (the “Editor”) that allows you to create custom levels or other materials for your personal use in connection with the Program (“New Materials”). All use of the Editor or any New Materials is subject to this License Agreement. The Program is licensed, not sold. Your license confers no title or ownership in the Program.[/hide]
The "Spawned Versions" referred to don't require the CD, but can only be used in multiplayer games with the original version.

There's no specific language about defeating the "must have CD in drive" protection, and that's a legal gray area if you still have the CD. Exploiting that to install many copies off the same CD is a clear violation, and is enforced by checking CD keys in online play.

Yes. As you can read in the article, in general software cracking for personal use is still a largely uncharted area in the legal arena. Even if it is actually illegal, it is highly unlikely that anyone will ever catch you or care enough to prosecute you, as long as you really only want to do this for personal use (and you legitimately bought the game).

Change to Warcraft 3, the newest official patch removes the protection legally  :D

It's also at the borders of law, but I don't see any rational reason not to be allowed to play without the CD, as long as you are the owner of one. Note that "rational reason" is not what judges or lawayers always have  :wink:

Generally, an important thing to remember is that laws regarding activities like this vary from country to country. I know that Germany has recently tightened its laws to insane lengths in this area.

there still have those "no-cd-cheat" over internet...
make a couple search and paste it into the folder...

The latest patch of starcraft actually has a no-cd patch. Check the blizzard site for more info, but you just have to copy something on the CD to your install folder.

My friends have Starcraft (or Warcraft, or whatever) on their flash drives, and it runs without CDs. Of course, I have no idea whether or not it's legal.

It's legal as long as you have a legal copy (to run with crack and w/o CD, CD is annoying like crap) I think..
Even if it weren't legal you won't get prosecuted because the person prosecuting you would be wasting their time.

Blizzard made a patch awhile back, now all you have to do is copy a file into your program files and you can play SC.

Eh? I should update it then. Thanks a lot though.

Meh I should do it too for my W3 also.

take the install.exe file from your starcraft CD folder and copy it into your starcraft folder that was made on your hard disk.  Rename it as BroodWar.mpq to finish.  If you have only the original, copy install.exe and rename it is StarCraft.mpq instead.

It's on blizzard's website you know.


linear algebra
matrix
vector
geometry
abstract algebra
linear algebra unsolved

Let $A \in \mbox{GL}_{2}(\mathbb{Z})$ s.t. $A \not=I_{2}$ and $A^{3}=I_{2}$.
Show that there exists $P\in \mbox{GL}_{2}(\mathbb{Z})$ and $A=P \begin{pmatrix}0 &-1\\ 1 &-1 \end{pmatrix}P^{-1}$

Nice problem! Of course, it follows that $A^{2}+A+I=0$ and so $A$ can be written as $(a,b)$ and $(c,-a-1)$ (by lines). We need to prove the existence of a vector $v\in Z^{2}$ such that $(v,Av)$ is a basis in $Z^{2}$. One may assume that $c>0$ (follow the argument and you will see why). It is enough to find $(x,y)$ in $Z^{2}$ with $\det(v,Av)=1$, which comes down to the following: if $-bc=a^{2}+a+1$ then there are integers $x,y$ with $cx^{2}-(2a+1)xy+(-b)y^{2}=1$. This is standard application of Minkowski's theorem from geometry of numbers.
   We discussed in class another type of approach: put $(a+b\epsilon).v=av+bAv$ for $\epsilon$ root of $X^{2}+X+1$ and $a,b$ integers. If $Z[\epsilon]$ is principal (I think it's the case here), then $Z^{2}$ becomes a $Z[\epsilon]$ free $Z$ module for this new operation (it is free because it is finitely generated and torsion free) and by taking a basis into this module one can create a $Z$ basis in which the matrix has the desired form.

Very good problem for oral examination of ENS ULM


analytic geometry
vector
linear algebra
linear algebra unsolved

1. Let A, J, and P  Mn(R) with P nonsingular and J = P^-1AP (J not necc. Jordan Canonical Form of A). Let ek be the kth coordinate vector in Rn. Prove that if the kth column of J is b ek, where x  belongs  R, then  b is an eigenvalue of A and the kth column of P is a corresponding eigenvector.

similar matrices have same eigenvalues. So it suffices to prove that b is an eigenvalue of J.
It is well known that Ae_i represents the i-th column of A. So Ae_k = be_k, showing that b is an eigenvalue and e_k is an eigenvector.


quadratics
modular arithmetic
number theory unsolved
number theory

Prove that:
a) 2 is a primitive root of the prime $p=4q + 1$ if $q$ is an odd prime.
b) 2 is a primitive root of $p = 2q + 1$ if $q$ is a prime congruent to 1 mod 4.
b) -2 is a primitive root of $p = 2q + 1$ if $q$ is a prime congruent to -1 mod 4.

Peng Shi

contradictions:

a) it is easy to see we just need to considere that the order of $2$ is $q$,$2q$. but then $2$ is a square modulo p, but $p==5 (mod8)$, contradiction #1.

b) Again we just have to considere when the order of $2$ is $q$, it follows that $2$ is an square mod p, but $p==3(mod8)$, contradiction#2.

c)Supose the order of $-2$ is $q$, then $-2$ is a square modp, it implies $p==1$, or $3 (mod 8)$ but $p==7(mod8)$ contradicion#3

another nice one: if $p=2^q+1$ and p is a prime, show $3$ is a primitive root modp.

It's of the form $2^{2^n}+1$, of course, and all we need to show is that $3$ is not a quadratic residue (since all the divisors of $p-1$ are, in fact, $\frac{p-1}{\hbox{some power of two}}$). We can do this by quadratic reciprocity: $\left(\frac p3\right)\cdot\left(\frac 3p\right)=(-1)^{2^{2^n-1}}=1$. Since $p\equiv 2\pmod 3$ is not a quadratic residue of $3$, $3$ is not a quadratic residue of $p$, and we are done.


geometry
rectangle
geometric transformation
reflection
circumcircle
geometry solved

What's BMO?

Let $E$ be the reflection of $C$ in $B$. $CQE$ is a right triangle with hypothenuse equal to $CE=2BC$ and circumcenter equal to $B$ (because the circumcenter of a right triangle is the midpoint of its hypothenuse), so $BQ=BC$, as we wanted.

[quote="paul_mathematics"]ABCD is a rectangle, P is the midpoint of AB, and Q is the point on PD such that CQ is perpendicular to PD. Prove that the triangle BQC is isosceles.[/quote]

Another [i]solution:[/i] We will use directed angles modulo 180°.

Since the point P is the midpoint of the side AB of the rectangle ABCD, we have < BPC = - < APD by symmetry. In other words, < BPC = < QPB. But since ABCD is a rectangle, $AB\perp BC$, so that < PBC = 90°, and since $CQ\perp PD$, we have < PQC = 90°. Thus, the points B and Q lie on the circle with diameter PC; thus, < BPC = < BQC and < QPB = < QCB. Hence, the equation < BPC = < QPB becomes < BQC = < QCB. Hence, the triangle BQC is isosceles with BC = BQ.

  darij


MATHCOUNTS
geometry
perimeter
Pythagorean Theorem
area of a triangle
Heron\u0027s formula

Find the areas of the following triangles:
(a) The triangle with sides 4, 4, and 4
(b) The triangle with sides 3, 4, and 5
(c) The triangle with sides 2, 5, and 5

Does an equilateral triangle always have a larger area than all other triangles with the same perimeter?  Why or why not?

[quote="MCrawford"]Find the areas of the following triangles:
(a) The triangle with sides 4, 4, and 4
(b) The triangle with sides 3, 4, and 5
(c) The triangle with sides 2, 5, and 5[/quote]

(a)   4:rt3:
(b)   6
(c)   2 :sqrt: 6

now for the solutions.

(a) the area of an equilateral tringle (yes, tringle) is always (s :^2:  :rt3: )/4, this is so because if you split the triangle in half with the altitude, you have two 30-60-90- tringles. so the known leg of each triangle is 2, so the height is 2 :rt3:. so now the area of a tringle is 1/2 x bh=1/2 x (4)(2 :rt3:)=4 :rt3:
one could also use herons theorem, but its faster to use (s :^2:  :rt3: )/4 with equilateral tringles.

(b) ugh, no questions here

(c) there are 2 ways you could do this that i know of (easily at least, i hope grr) you could use herons theorem  :sqrt: (s)(s-a)(s-b)(s-c)
where s is the semiperimeter=(a+b+c)/2, and a,b, and c are the legs
so plugh the values in to get  :sqrt: (6)(1)(1)(4)= :sqrt: 24=2 :sqrt: 6

the other way is to draw the altitude from the vertex to make two right tringles. draw it out for yeself. so the height is :sqrt: 24=2 :sqrt: 6. cause of the pythagorean theorem is a :^2: +b :^2:=c :^2:.
so 1 :^2: +b :^2: =5 :^2:. manipulating that b :^2: =24. so b= :sqrt: 24=2 :sqrt: 6, then using the 1/2 x bh area formula thingie
1/2 x (2)(2 :sqrt: 6)=2 :sqrt: 6

hope i helped, peace

Again, I agree with Syntax, but again, my physics teachr is going to come get me soon. So cya all later.

[quote="Barnacle"]"What about the conjecture that equilateral triangles always have a larger area than any other triangles with the same perimeter?"

2 years later. im interested in knowing the answer!! i can't prove it  :?:  :?:  :?:[/quote]

I bet that if you take that problem to the Intermediate High School forum you're likely to get a good conversation out of it.

Yes it does, using heron's formula (that isn't above this level, is it?) we get the area is $\sqrt{s(s-a)(s-b)(s-c)}$where s is constant (semiperimeter) and a, b, and c are the side lengths. We expand $(s-a)(s-b)(s-c)$ to get $s^3-s^2c-s^2b-s^2a+abs+acs+bcs-abc$ Regrouping gives: $s^3-s^2(a+b+c)+s(ab+ac+bc)-abc$ because s is the semiperimeter, a+b+c = 2s THus we get:

$s^3-2s^3+s(ab+ac+bc)-abc$ To maximize this we ignore the constants (and the last term for now) and maximize $ab+ac+bc$ because $a+b+c$ is constant, if we increase one product, we lower another. As we will lower it more than we raise it because each number is multiplied by each other number, we want $a=b=c$ and thus, the equilateral triangle has the greatest area. It's probably faulty, as I had no idea how to actually maximize $ab+ac+bc$.

See -- this conversation has immediately moved to an Intermediate or even Pre-Olympiad level.


a corporate executive officer is attempting to arrange a meeting of this 3 vice presidents for tomorrow morning. He believes that each of these three busy individuals, independantky of the others, has a 60% chance of being able to attend the meeting

A) find the probability distribution of X, the number of vice presidents who can attend the meeting

B) What is the probability than none of the three vice presidents can attend the meating?

C) If the meeting will be held tomorrow morning only if everyone can attend, what is the probability that the meeting will take place at the same time?

D) How many of the vice presidents should be CEO expect for tomorrow morning's meating?

THNX A TON!!

actually i figured out b and C.....now i all i need is a and D...please!

[hide="As answer"]A [url=http://en.wikipedia.org/wiki/Binomial_distribution]binomial distribution[/url] with $ n \equal{} 3$, $ p \equal{} 0.6$[/hide]

I don't really understand D.

[quote="shybox"]D) How many of the vice presidents should be CEO expect for tomorrow morning's meating?[/quote]

That depends ... if they will be meating one of the vice presidents, I imagine they'll want to stay away.  If they are meating the CEO and I were a vice president, that would probably be something I'd want to watch.  (I'm assuming here that the process of being meated is as unpleasant as it sounds.)

More on-topic, I suggest that you go find your textbook, read the relevant section and answer your questions yourself.  (D), for example, is instantaneous if you know what expected value means -- just apply the definition.


Find all pair's of integers (x,y), such that

$x^3+x^2y+xy^2+y^3=8(x^2+xy+y^2+1)$

Davron

Thank you very much Jin but i have one question for you:

Let $f(x)=8(x^2+xy+y^2+1)$, then why do you say that the minimal value of $f(x)$ is $4(x^2+y^2)+8$? Why not this one$8 (x^2+y^2)+8$ ?

Davron

$x=-y$.In the problem,$x,y$ canbe negative.

Yes sure it is true for integers, but not only for postitive once.

Davron


ARML
geometry
probability
analytic geometry
graphing lines
slope
trigonometry

Let's post detailed solutions so that Sean can understand the problems, shall we?

1) Find the last two digits of $6^{710}$.

2) Find the area of a convex pentagon, inscribed in a circle, with sides 16, 16, 16, 16, 6.

3) If $x^2+y^2-30x-40y+24^2=0$, find the greatest value of $\frac{y}{x}$.

4) A computer is programmed to randomly generate a string of six symbols using only the letters A,B,C. What is the probability that the string will not contain three consecutive A's?

More to come...

In #3 just complete the square and get the equation of the circle.

Math Jam? Weren't these the problems on AIM yesterday?

Anyways....

[hide="1"]
We look at $6^{710}$ $\mod{100}$ but we have to split it up to $\mod{25}$ and $\mod{4}$ since $6$ and $710$ are not relatively prime. So $\phi{25}=20$, so $6^{710} \equiv 6^{10} \equiv 1 \mod{25}$ and then we know that $6$ to any power greater than $1$  is divisible by $4$ (since we'll always have atleast two $2s$, and so we get $6^{710} \equiv 0\mod{4}$, therefore we have to find a number between $00$ and $99$ which satisfies that, and we find that  $\boxed{76}$ satisfies it, so that's our answer.
[/hide]

[hide="2"]
I first labeled the points of the pentagon $ABCDE$, where $DC=6$ and everything else is $16$. From here I decided to split the pentagon into triangles and realized that there were only two types of triangles the pentagon can be split up into (that we know atleast two sides of), triangles with two sides of $16$ and triangles with sides of $6$ and $16$, so therefore $AD=EB=AC$ and $CE=DB$. So then I called each of the first three segments $x$ and each of the second two segments $y$ and then I decided to split the pentagon into quadrilaterals so I can use Ptolemy's Theorem. First I had quadrilateral $ABDE$ and using Ptolemy's I had: $16^2+16y=x^2$ and then I then made the other possible quadrilateral which was $EBCD$ and used Ptolemy's to get $6x+16^2=y^2$ and the last possible quadrilateral $ABCD$ and got $yx=16x+96$
Okay so after simplifying for $x$ and getting the equation $x^3-512x-1536$, I used rational root theorem and using trial and error I found that only $x=24$ worked for this equation and so $y=20$ and so after using Heron's formula on the three triangles and adding their areas up, we have the area of the Pentagon is $\boxed{123\sqrt{7}}$[/hide]

[hide="3"]After completing the square, we get the equation $(x-15)^2+(y-20)^2=7^2$, which is a circle with radius $7$ and centered at $(15,20)$. From here, we know that the tangent line from the origin gives us the largest value of the slope ($\frac{y}{x}$), so we draw in the tangent line and then use $\tan$, so if we call the origin, point $O$ and call the tangent point $X$, and call the center of the circle $P$, we have that $XP=7$, and using pythagorus $PO=25$, thus $XO=24$ follows. So then $\frac{y}{x}$, can be seen using $\tan{a+b} \Longrightarrow \frac{\frac{4}{3}+\frac{7}{24}}{1-\frac{4}{3} \cdot \frac{7}{24}}=\boxed{\frac{117}{44}}$[/hide]

I'll do number 4 later.

EDIT: i can't do number 4 :(

[quote="dark_sunburst"]Is #4
[hide]4/27?[/hide]
If it's not, I wouldn't be surprised...my logic was a bit fuzzy, now that I think about it.[/quote]

dark_sunburst, i think you made a mistake :( , it would be very *unlikely* that three consecutive As would happen, if i were to guess, i would say it would be very close to $\frac{8}{9}$ or $\frac{9}{10}$ or something.

:blush: ...Yeah, I just realized that answer was what I got for the probability that the string [i]will[/i] contain three consecutive A's. So, um, that was backwards.

Still not sure if I was right, though.

can u please post your solution dark_sunburst :? ...maybe I do need that aops counting/probability book after all ;)

Bella's answer seems correct (dark_sunburst [i]is[/i] bella, right? lol).

[hide="Solution"]
Let us find the probability that the computer [i]does[/i] print 3 consecutive A's. Let us treat the sequence of letters $AAA$ as a single subunit, and let us call this subunit $X$. Let us represent a letter the computer outputs as $Y$. Then, if the computer is to print three consecutive A's, the string in question will look something like this: $XY_1Y_2Y_3$. This string can be rearranged in $\frac{4!}{3!}=4$ ways. Because each $Y_j$ can have $3$ different choices (A, B, or C), the total number of six letter strings containing $AAA$ is equivalent to $4\cdot3^3$. The total number of strings is easy to determine: 6 letters, each of which have three possibilities, yield a total of $3^6$ total strings.

The probability for there to be 3 consecutive $A$s is therefore $\frac{4\cdot3^3}{3^6}=\frac{4}{27}$. Because we are asked for the probability this event does NOT occur, the answer we are looking for is $\boxed{\frac{23}{27}}$.
[/hide]

This is a bit easier if you've read the counting book.  Complementary counting is helpful here...


trigonometry
integration
calculus
calculus computations

By $\tan x=\frac{\sin x}{\cos x}$,

$\int \frac{\tan x}{\sin^{2}x \sqrt{\cos^{2}x-16}}\ dx$

$=\int \frac{\sin x}{\cos x (1-\cos^{2}x)\sqrt{\cos^{2}x-16}}\ dx$

Let $\cos x =t$,....


The diagonals of a convex quadrilateral are dividing it in 4 triangles with integer areas.Prove that the product of this areas is a perfect square

The triangles formed have areas $ \frac{1}{2}ab\sin A,\frac{1}{2}bc\sin(180\minus{}A),\frac{1}{2}ad\sin B,\frac{1}{2}cd\sin(180\minus{}B)$
Multiplying all of them, we have $ (\frac{1}{4}abcd\sin A\sin B)^2$, where the term in bracket is also an integer


There were a certain number of people on a bus. Each person had a seat. At the next bus stop, 1/4 of the total number of people left, and 21 got on. Now there are 6 people who did not have seats. How many people are currently on the bus?

[hide]Let's say there were originally $ x$ people on the bus. Now there are $ x \plus{} 6.$ Therefore, $ 21 \minus{} \frac{1}{4}x \equal{} 6.$ Solving, $ x \equal{} 60.$ So $ x \plus{} 6 \equal{} \boxed{66}!$[/hide]

[hide]K. So say there x is the number of people on the buss.....1/4 of x left, so 3/4 x people are remaining.
3/4x+21 people have now got on the bus, so that is the total number...
Then, You do 21-1/4x=6 and you get the answer as 60[/hide]


Given $ a\equal{}27$ and $ b\equal{}33$, find $ 25a\plus{}32b\plus{}75a\plus{}68b$.

hello, we have
$ 25a\plus{}75a\plus{}68b\plus{}32b\equal{}100(a\plus{}b)\equal{}100\cdot60\equal{}6000$
Sonnhard.

[quote=huh._.]SwanGoose said I memmed this.[/quote]

good job


function
calculus
integration
calculus computations

Hi everyone,

Could you please tell me if this is correct?

Find the average value of the function on the given interval

g(x)=x^2sq.(1+x^3) [0,2]

1/2int. [1/3sq. u]du=
u^3/2/(9)=

.314

Thank you very much

I must be honest and say that I cannot understand you're notation. If you have the function $ g(x)\equal{}x^2\sqrt{1\plus{}x^3}$ and want to calculate it's average value on a certain interval, shouldn't the same function be under the integral? You have:

$ \frac{1}{2\minus{}0}\int^{2}_{0}{x^2\sqrt{1\plus{}x^3}}dx$

use the substitution $ x^3\equal{}t$ and get:

$ dx\equal{}\frac{dt}{3x^2}$

and

$ \frac{1}{2}\int^{8}_{0}{x^2\sqrt{1\plus{}t}}\frac{dt}{3x^2}\equal{}$

$ \frac{1}{2}\frac{1}{3}\int^{8}_{0}{\sqrt{1\plus{}t}}dt\equal{}$

$ \frac{1}{6}\int^{8}_{0}(1\plus{}t)^{\frac{1}{2}}dt\equal{}$

$ \frac{1}{6}\frac{2}{3}(1\plus{}t)^{\frac{3}{2}}\mid^{8}_{0}\equal{}$

$ \frac{1}{9}(1\plus{}8)^{\frac{3}{2}}\minus{}\frac{1}{9}(1\plus{}0)^{\frac{3}{2}}\equal{}\frac{26}{9}$

Thank you very much

The integral is going from 0 to 2, so wouldn't the answer be 3? You could set u=1+x^3, right?

so you would have the integral

1/2 int 1/3sq. u from 0 to 2=
2(1+x^3)^3/2/18=
3

Thank you very much

[quote="Milin"]use the substitution $ x^3 \equal{} t$ and get:[/quote]

There is no need for a substitution: since (1+x^3)' = 3x^2, we can write

$ \int x^2 \sqrt{1\plus{}x^3}\,dx \equal{} \frac{1}{3} \int 3x^2 \sqrt{1\plus{}x^3}\,dx \equal{} \frac{2(1\plus{}x^3)^{3/2}}{9} \plus{} C$.


[quote="now a ranger"]The sum of three numbers is 20. The first is 4 times the sum of the other two. The second is seven times the third. What is the product of all three?[/quote]
[hide]
$a+b+c=20$
$a=4(b+c)$
$b=7c$
$a=4(8c)$
$a=32c$
$40c=20$
$c=\frac{1}{2}$
$a=16$
$b=\frac{7}{2}$
$abc=28$

[/hide]

[quote="Quevvy"][quote="Rafaelloaa"][hide]28[/hide][/quote]
I know that you are new, but you should post how you got your answer, not just the answer. For all I know, you could have looked at davidli's post and just put the answer in. Just telling you for next time...[/quote]

trust me, you don't want people thinking u do that (guess who's gotten in trouble with this around davidli before :P )

jorian

[hide]
I converted all the terms in terms of z

so

32z+7z+z=20
z=1/2
16*1/2*7/2=[hide][u][i][b]28[/b][/i][/u][/hide][/hide]


vector
geometry
topology
abstract algebra
analytic geometry
advanced fields
advanced fields unsolved

Let $ \mathcal{X}(M)$ be the set if all $ C^\infty$ vector fields defined on the $ C^\infty$ manifold $ M$. Then show that $ \mathcal{X}(M)$ is infinite-dimensional over $ R$ but locally finitely generated over $ C^\infty(M)$ , that is each $ p\in M$ has a neighborhood $ V$ on which there is a finite set of vector fields which generate $ \mathcal{X}(M)$ as a $ C^\infty(V)$ module.

This is a problem from "[i]An Introduction to DIfferentiable Manifold and Riemannian Geometry[/i]" by [b]William Boothby[/b]. I can understand the first part of the problem, but I am not sure about the second part. Any light into that would be really appreciated.

Thanks
Ajoye

Smooth vector fields on $ \mathbb R^n$ are finitely generated (as a $ C^{\infty}(\mathbb R^n)$-module) by $ \frac {\partial}{\partial x_1}, \dots,\frac {\partial}{\partial x_n}$. Use coordinate charts to push vector fields between $ M$ and $ \mathbb R^n$.

Thanks.. I think I understand it now.  That was really a stupid question, isnt it? :(


function
complex analysis
complex numbers
complex analysis unsolved

This is my own, it has a real quick solution to it, hope you like it.

Show that there [b]doesnt[/b] exists a function $ f(z)$ which is holomorphic on the disk $ \{|z|<\alpha : z\in \mathbb{C}\}$ for any $ \alpha >0$ such that $ f(x) = e^{-1/x^{2}}$ for all $ -\alpha<x<\alpha , \ x\not = 0$.

This follows immediately from the Identity theorem.   :P

Im not familar with the Identity theorem  :oops: , I was thinking of something else.

[hide="My solution"]
One of my favorite results from complex analysis is that holomorphic functions a analytic (abusing formalities). Now if $ f(z)$ was holomorphic on $ \{z\in \mathbb{C}: |z|<\alpha\}$ and fit the criteria then that would mean $ f(z)$ was analytic. But then the real part of $ f(z)$ i.e. the real function part $ f(x) = e^{-1/x^{2}}$ would have a Taylor series expansian centered at $ 0$ which is impossible. (It is a famous fact that $ e^{-1/x^{2}}$ does have a Taylor series).
[/hide]


trigonometry
calculus
derivative
function
complex analysis
algebra unsolved
algebra

Prove that if $ a_1,a_2,a_3\in \mathbb{R} $ and $ \forall x\in \mathbb{R} , a_1\cos a_1x+a_2\cos a_2x+a_3\cos a_3x\geq 0 $ then $ a_1a_2a_3=0 $

Nice!!! This follows from a problem that you posted some minutes ago and which Grobber has already solved, the one with the sum of periodical functions. For this one it suffices to take $ f_1 (x)=sin (a_1 x), f_2(x)=sin(a_2 x), f_3(x)=sin(a_3 x)$ and to observe that the derivative of their sum is nonnegative, thus the sum has a limit at infinity and thus it is constant. From here it is very simple, since you have that $ sin(a_1x)+sin(a_2 x)+sin(a_3 x)=0$ for all $x$. Develop this as an entire function and identify coefficients. You'll get $ a_1^k+a_2^k+a_3^k=0$ for all odd $k$. After that is nothing.


for positive integers $ a,b$ satisfying the equation $ a^2 \plus{} b^2 \minus{} ab \equal{} c^2$
prove that  $ (a \minus{} c)(b \minus{} c)\le\ 0$

You need to prove that $ ab\plus{}c^2\le ac\plus{}bc\Leftrightarrow a^2\plus{}b^2\le \sqrt{(a^2\plus{}b^2\minus{}ab)}(a\plus{}b)$

$ \Leftrightarrow a^4\plus{}b^4\plus{}2a^2b^2\le a^4\plus{}b^4\plus{}a^2b^2\plus{}a^2b^2\plus{}2a^3b\plus{}2ab^3\minus{}a^3b\minus{}ab^3\minus{}2a^2b^2$

$ \Leftrightarrow 2a^2b^2\le a^3b\plus{}ab^3$ , which is MA-MG

well ..... i have one doubt...in my proof........i assumed  $ a\ge\ b$ and proved that for this assumption $ a\ge\ c$ and $ c\ge\ b$....... and for the other case that is for $ b\ge\ a$
i can prove $ c\ge\ a$ and $ b\ge\ c$  ...................is it correct? :blush:

Yes , if your proof has no mistake , then is correct . Can you post it ?

from the given condition,
$ a^3 \plus{}b^3 \equal{} (a\plus{}b)c^2$
transposing suitably,
$ a^3 \minus{} a(c^2) \equal{} b(c^2)\minus{} b^3$,
or,
$ a(a\minus{}c)(a\plus{}c) \equal{} b(c\minus{}b)(c\plus{}b)$
we can thus conclude that,
$ (a\minus{}c)$ and $ (c\minus{}b)$ have the same sign(+ve or -ve),
which means,
$ (a\minus{}c)(b\minus{}c) \le 0$
$ Q.E.D$

write the given equation as
$ a^2 \minus{} c^2 \equal{} b^2 \minus{} ab$
$ (a \plus{} c)(a \minus{} c) \equal{} b(a \minus{} b)$
from my first assumption  $ a\ge\ c$
and we can have $ (b \plus{} c)(c \minus{} b) \equal{} a(a \minus{} b)$ again by my assumption i have $ c\ge\ b$
similarly i think i can prove for my second assumption also

This a notorius inequality (OMM 1996)

Other solution :

$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos\ 60^{o}$

So $ a,b,c$ verify the law of cosinus $ \Rightarrow \ a,b,c$ are the sidelenghts of a triangle with $ \alpha \equal{} 60^{o}$

If $ \beta\le 60^{o}$ and $ \gamma\ge 60^{o}$ or $ \beta \ge 60^{o}$ and $ \gamma \le 60^{o} \Rightarrow a\ge c\ge b$ or $ a\le c\le b$ .

Equality occurs when the triangle is equilateral .


inequalities
trigonometry
geometry
circumcircle
inradius
geometry solved

Prove that in any triangle ABC we have \[\left(1 - \frac{2a}{b + c}\right)^2 + \frac{2r}{R} \leq 1\] and \[\frac{2r}{R} \leq \sqrt{\cos{\frac{A - B}{2}}\cos{\frac{B - C}{2}}\cos{\frac{C - A}{2}}}.\]

[color=red][Moderator edit: Hereby, R is the circumradius and r is the inradius of triangle ABC. Of course, these are well-known notations, but it is still better to define them explicitely.][/color]

[quote="Arne"]Prove that in any triangle ABC we have \[\left(1 - \frac{2a}{b + c}\right)^2 + \frac{2r}{R} \leq 1\] and \[\frac{2r}{R} \leq \sqrt{\cos{\frac{A - B}{2}}\cos{\frac{B - C}{2}}\cos{\frac{C - A}{2}}}.\][/quote]

I haven't got the solution for the first inequality yet but here's my solution for the second:
$D= \sqrt{\cos{\frac{A - B}{2}}\cos{\frac{B - C}{2}}\cos{\frac{C - A}{2}}}\ = \sqrt{\frac{b+c}{a}\sin{\frac{A}{2}}\cdot\frac{a+b}{c}\sin{\frac{C}{2}}\cdot\frac{a+c}{b}\sin{\frac{B}{2}}}$
$= \sqrt{\frac{r}{4R}\frac{a+b}{c}\frac{a+c}{b}\frac{b+c}{a}}$
 From $R\geq 2r$ and $\frac{b+c}{a}\frac{a+b}{c}\frac{a+c}{b}\geq 8$ it derives $D\geq\frac{2r}{R}$
Artiona

I have another solution for the second inequality but the first one is still not solved. Anyone?

We have 4a^2bc >= (b+c)^2(a+b-c)(a+c-b) because the difference is (b-c)^2 (a+b+c)(b+c-a)
Multiplying by (b+c-a)/(abc)/(b+c)^2, we get
1-(1-2a/(b+c))^2 >= 2r/R.
Note that equality occurs if and only if b=c.
Kind regards

Some remarks on this topic:

[quote="Arne"]Prove that in any triangle ABC we have \[\left(1 - \frac{2a}{b + c}\right)^2 + \frac{2r}{R} \leq 1\] and \[\frac{2r}{R} \leq \sqrt{\cos{\frac{A - B}{2}}\cos{\frac{B - C}{2}}\cos{\frac{C - A}{2}}}.\][/quote]

The second of these two inequalities was proven before on http://www.mathlinks.ro/Forum/viewtopic.php?t=6038 and on http://www.mathlinks.ro/Forum/viewtopic.php?t=19253 (note that $r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$, so that $\frac{2r}{R}=8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$). The first one was posted on http://www.mathlinks.ro/Forum/viewtopic.php?t=6032 , but no proof was given. My proof of the first inequality is more or less equivalent to JPE's one, but a bit less algebraic, so I post it here:

We have to prove the inequality

$\left(1-\frac{2a}{b+c}\right)^2+\frac{2r}{R}\leq 1$.

We can rewrite it by subtracting $\left(1-\frac{2a}{b+c}\right)^2$ on both sides:

$\frac{2r}{R}\leq 1-\left(1-\frac{2a}{b+c}\right)^2$, what is equivalent to
$\frac{2r}{R}\leq 1-\left(1-2\cdot\frac{2a}{b+c}+\left(\frac{2a}{b+c}\right)^2\right)$, what is equivalent to
$\frac{2r}{R}\leq 2\cdot\frac{2a}{b+c}-\left(\frac{2a}{b+c}\right)^2$, what is equivalent to
$\frac{2r}{R}\leq 4\cdot\frac{a}{b+c}-4\left(\frac{a}{b+c}\right)^2$, what is equivalent to
$\frac{r}{2R}\leq \frac{a}{b+c}-\left(\frac{a}{b+c}\right)^2$, what is equivalent to
$\frac{r}{2R}\leq \frac{a}{b+c}\cdot\left(1-\frac{a}{b+c}\right)$, what is equivalent to
$\frac{r}{2R}\leq \frac{a}{b+c}\cdot\frac{b+c-a}{b+c}$.

Introducing the notation $s=\frac{a+b+c}{2}$ for the semiperimeter of triangle ABC, we have b + c - a = 2 (s - a) on the one hand, while a well-known formula states

$\frac{r}{R}=\frac{4\left(s-a\right)\left(s-b\right)\left(s-c\right)}{abc}$,

so that

$\frac{r}{2R}=\frac{2\left(s-a\right)\left(s-b\right)\left(s-c\right)}{abc}$.

Thus, the inequality in question,

$\frac{r}{2R}\leq \frac{a}{b+c}\cdot\frac{b+c-a}{b+c}$,

rewrites as

$\frac{2\left(s-a\right)\left(s-b\right)\left(s-c\right)}{abc}\leq \frac{a}{b+c}\cdot\frac{2\left(s-a\right)}{b+c}$, what is equivalent to
$\frac{\left(s-b\right)\left(s-c\right)}{abc}\leq \frac{a}{b+c}\cdot\frac{1}{b+c}$, what is equivalent to
$\frac{\left(s-b\right)\left(s-c\right)}{bc}\leq \frac{a^2}{\left(b+c\right)^2}$, what is equivalent to
$\left(\frac{b+c}{a}\right)^2\cdot\frac{\left(s-b\right)\left(s-c\right)}{bc}\leq 1$, what is equivalent to
$\frac{b+c}{a}\cdot \sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}}\leq 1$.

Now, there are two ways to prove this inequality: The first one is to note that the half-angle theorem shows that

$\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}}=\sin\frac{A}{2}$,

while the Mollweide formula yields

$\frac{b+c}{a}=\frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}}$,

so that we have to prove $\frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}}\cdot\sin\frac{A}{2}\leq 1$, or, equivalently, $\cos\frac{B-C}{2}\leq 1$, what doesn't need any further explanation ;) . The second way is less elegant, but I show it here since it shows a tighter inequality: We rewrite the inequality in question,

$\frac{b+c}{a}\cdot \sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}}\leq 1$,

as

$\frac{b+c}{a}\cdot \sqrt{\frac{s-c}{b}\cdot\frac{s-b}{c}}\leq 1$.

Now, by the AM-GM inequality, applied to the numbers $\frac{s-c}{b}$ and $\frac{s-b}{c}$, we have

$\sqrt{\frac{s-c}{b}\cdot\frac{s-b}{c}}\leq \frac{\frac{s-c}{b}+\frac{s-b}{c}}{2}$.

Thus, it will be enough to show the inequality

$\frac{b+c}{a}\cdot \frac{\frac{s-c}{b}+\frac{s-b}{c}}{2}\leq 1$.

In order to show this inequality, we multiply it by 2a and get

$\left(b+c\right)\cdot\left(\frac{s-c}{b}+\frac{s-b}{c}\right)\leq 2a$.

But

$\left(b+c\right)\cdot\left(\frac{s-c}{b}+\frac{s-b}{c}\right)=\left(b+c\right)\frac{s-c}{b}+\left(b+c\right)\frac{s-b}{c}$
$=\frac{b+c}{b}\left(s-c\right)+\frac{b+c}{c}\left(s-b\right)=\left(2-\frac{b-c}{b}\right)\left(s-c\right)+\left(2+\frac{b-c}{c}\right)\left(s-b\right)$
$=2\left(s-c\right)-\frac{b-c}{b}\left(s-c\right)+2\left(s-b\right)+\frac{b-c}{c}\left(s-b\right)$
$=2\left(s-c\right)-\left(b-c\right)\frac{s-c}{b}+2\left(s-b\right)+\left(b-c\right)\frac{s-b}{c}$
$=2\left(\left(s-b\right)+\left(s-c\right)\right)+\left(b-c\right)\left(\frac{s-b}{c}-\frac{s-c}{b}\right)$.

But (s - b) + (s - c) = a, so that

$\left(b+c\right)\cdot\left(\frac{s-c}{b}+\frac{s-b}{c}\right)=2a+\left(b-c\right)\left(\frac{s-b}{c}-\frac{s-c}{b}\right)$.

And the inequality in question,

$\left(b+c\right)\cdot\left(\frac{s-c}{b}+\frac{s-b}{c}\right)\leq 2a$,

transforms into

$2a+\left(b-c\right)\left(\frac{s-b}{c}-\frac{s-c}{b}\right)\leq 2a$,

i. e. into

$\left(b-c\right)\left(\frac{s-b}{c}-\frac{s-c}{b}\right)\leq 0$.

But this is immediate, since some algebra shows that

$\left(b-c\right)\left(\frac{s-b}{c}-\frac{s-c}{b}\right)=-\frac{\left(b-c\right)^2\left(s-a\right)}{bc}$,

what is clearly $\leq 0$.

This completes the proof of our inequality. It is easy to see from this proof that equality holds if and only if b = c.

  Darij


AMC USA(J)MO USAMO articles blogs trigonometry vector

I need some advice...I was thinking of purchasing the Arbelos volumes from the AMC, but I was wondering (from those of you who might have these volumes), how useful are they for preparing for the USAMO contest? Do they just contain short articles and problems, or are there solutions as well? Basically, would it be worth buying these volumes, or would it be best just to practice by solving old problems?

Thanks in advance!

Mr. Crawford made an entry in his blog about it a while ago. http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=237

I just got a few old issues today. Volume 2 issue 1 discusses:

>Spherical trigonometry (Big article)
>Problems from the Eotvos/Kurshak competition
>Vectors (Extensive article, Problems: ex: Prove that in any parallogram, the sum of the squares of the sides is equal to the sum of the squares of the diagonals)
>Parity problems (ex: Solve x^y+1=z, Find all integers a^2+b^2+c^2=a^2*b^2, ..., some with solutions)
>Assorted IMO rejected problems

Correct me if I'm wrong, but I think that the majority of problems in the Arbelos don't have solutions though.

Did you order them from the AMC? I am going to order the first two volumes of Arbelos but I was wondering does a volume contain many issues, because you said "Volume 2 issue 1". How difficult do you think the books are?

Every volume contains 5 issues, for about 120 pages a book. I would say they are Harder AIME/Olympiad level mainly, but I'm not sure. I ordered them from the AMC's, but they can also be ordered from the MAA's bookstore.


LaTeX
number theory proposed
number theory

[b]   Let's take p a prime number and m,n natural numbers whith m,n>1.If n devides (m^p(n-1)) -1, prove that (m^n-1, n)>1.[/b] :)

[quote="ZetaX"]You can learn $\LaTeX$ here: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/quote]
   [b]Thank you![/b] :thumbup:

[quote="Chen241290"][quote="stancioiu sorin"][quote="spanferkel"]Please use Latex. Do you mean
[b]If n divides $m^{p(n-1)} -1$, prove that $(m^n-1, n)>1$. [/b] :?:
Or do you even mean $(m^p-1, n)>1$ which seems more logical :?:
anyway, if I get it right, it's all false for $m=2,n=3,p=5$, even with $m^p(n-1) -1$ instead. :?: ????[/quote]
   [b] Oh, I'm sorry. :blush: The conclusion is ([m^(n-1)]-1, n)>1.I don't know to use LATEX[/b].[/quote]
m=2, n=77, p=3 or any prime number?????????????[/quote]
     You didn't understand the conclusion.Please read it again!

[quote="stancioiu sorin"][/quote][quote="Chen241290"][quote="stancioiu sorin"][quote="spanferkel"]Please use Latex. Do you mean
[b]If n divides $m^{p(n-1)} -1$, prove that $(m^n-1, n)>1$. [/b] :?:
Or do you even mean $(m^p-1, n)>1$ which seems more logical :?:
anyway, if I get it right, it's all false for $m=2,n=3,p=5$, even with $m^p(n-1) -1$ instead. :?: ????[/quote]
   [b] Oh, I'm sorry. :blush: The conclusion is ([m^(n-1)]-1, n)>1.I don't know to use LATEX[/b].[/quote]
m=2, n=77, p=3 or any prime number?????????????[/quote][quote="stancioiu sorin"]
     You didn't understand the conclusion.Please read it again![/quote]
i am sorry, i was mistaken, this is the solution:
$n-1=p^a.b$; (b,p)=1
let q be a prime divisor of n
let g be the order of m: $m^g\equiv 1(mod q)$
q divides n $\Rightarrow q \ divides \ m^{p(n-1)} -1, \Rightarrow g \ divides \ p(n-1)$
$g \ divides \ p^{a+1}.b, \Rightarrow p^{a+1}.b=gh$
$If \ (h,p)=1 \forall q$. Therefore $p^{a+1} \ divides \ g; \forall q$ ;q:prime divisor of n
Then $p^{a+1} \ divides \ q-1$, since g divides q-1
$q\equiv 1(mod p^{a+1}); \forall q$
Therefore $n\equiv 1(mod p^{a+1})$ (contradiction)
$If \ p \ divides \ h$, for some q. Therefore g divides $p^a.b=n-1$
Therefore $q \ divides \ m^{n-1}-1$
Hence: $(m^{n-1}-1, n)>1$

[quote="Chen241290"][/quote][quote="stancioiu sorin"][/quote][quote="Chen241290"][quote="stancioiu sorin"][quote="spanferkel"]Please use Latex. Do you mean
[b]If n divides $m^{p(n-1)} -1$, prove that $(m^n-1, n)>1$. [/b] :?:
Or do you even mean $(m^p-1, n)>1$ which seems more logical :?:
anyway, if I get it right, it's all false for $m=2,n=3,p=5$, even with $m^p(n-1) -1$ instead. :?: ????[/quote]
   [b] Oh, I'm sorry. :blush: The conclusion is ([m^(n-1)]-1, n)>1.I don't know to use LATEX[/b].[/quote]
m=2, n=77, p=3 or any prime number?????????????[/quote][quote="stancioiu sorin"]
     You didn't understand the conclusion.Please read it again![/quote][quote="Chen241290"]
i am sorry, i was mistaken, this is the solution:
$n-1=p^a.b$; (b,p)=1
let q be a prime divisor of n
let g be the order of m: $m^g\equiv 1(mod q)$
q divides n $\Rightarrow q \ divides \ m^{p(n-1)} -1, \Rightarrow g \ divides \ p(n-1)$
$g \ divides \ p^{a+1}.b, \Rightarrow p^{a+1}.b=gh$
$If \ (h,p)=1 \forall q$. Therefore $p^{a+1} \ divides \ g; \forall q$ ;q:prime divisor of n
Then $p^{a+1} \ divides \ q-1$, since g divides q-1
$q\equiv 1(mod p^{a+1}); \forall q$
Therefore $n\equiv 1(mod p^{a+1})$ (contradiction)
$If \ p \ divides \ h$, for some q. Therefore g divides $p^a.b=n-1$
Therefore $q \ divides \ m^{n-1}-1$
Hence: $(m^{n-1}-1, n)>1$[/quote]

    Wonderful solution! :!: Keep going!Thank you!

Can some moderator please move it to Number Theory¿

Done.

stancioiu sorin, many of your posts are quite unnecessary. Try to limit the number of "posts without a content" you make.

OK.I've understood. :fool: Sorry.But if I have a solution at a problem, can I write it?

Of course you can!! Just don't make *too many* posts only saying "Thank you!" or something like that.


AIME 1983 problem 3:
Find the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$
Would you solve this equation with the quartic equation (which I do not know) or is there another way?

[hide="answer"]
if you set $s = x^2+18x+30$  then you can solve $s = 2\sqrt{s+15}$ for s which will give $s = 10$ as only solution. Now you can solve
$10 = x^2+18x+30$ for x and you will get $-9 +/- \sqrt{61}$, their product is 20. [/hide]

edit: latexed

yes I agree with you,
[hide="hint1"]
Real no.s mean POSITIVE no.s not neccesarily rational
[/hide]

[hide="hint2"]
Let $x^2+18x+30=a$....
[/hide]

In general solving trying to actually use the quartic equation is nigh impossible.


limit
analytic geometry
graphing lines
slope
function
topology
real analysis

1. Let $ M$ be a compact metric space and suppose that $ f: M\rightarrow M$ satisfies $ d(f(x),f(y))<d(x,y)$ whenever $ x\neq y$.  Show that $ f$ has a fixed point.

I know this means that $ g(x)\equal{}d(x,f(x))\equal{}0$ for some $ x\in M$ but I can't show that this must be true.  I tried considering the intermediate value theorem but I couldn't get there.

2. If $ f: (0,1)\rightarrow\mathbb{R}$ is uniformly continuous, show that $ \lim_{x\rightarrow 0^\plus{}}f(x)$ exists.  Conclude that $ f$ is bounded on $ (0,1)$.

If it's uniformly continuous on $ (0,1)$, then it's continuous on $ (0,1)$.  It seems that this implies that the limit exists but I'm not sure how to be completely rigorous.

3. Suppose that $ f: \mathbb{R}\rightarrow\mathbb{R}$ is continuous and that $ f(x)\rightarrow 0$ as $ x\rightarrow\pm\infty$.  Prove that $ f$ is uniformly continuous.

If $ f$ goes to $ 0$ as $ x$ gets bigger/smaller and $ f$ is continuous, then $ f$ has to be bounded on $ \mathbb{R}$.  By definition, if $ f\rightarrow 0$ as $ x\rightarrow\infty$, then $ \left|f(x)\right|<\varepsilon$ for $ x$ sufficiently large.  This seems to mean that the magnitude of the slope of $ f$ must not exceed some value for sufficiently large $ x$, which seems to imply uniformly continuous, but I'm not sure.

1. Take any $ y \in M$, and show that the sequence of iterates $ f^n(y)$ is a cauchy sequence, hence converges. Show that the limit is a fixed point. This is known as Banach's fixed point theorem, I think.

2. Take a sequence converging to 0. Show that f applied to the sequence is a cauchy sequence. Show then that if you take any other sequence converging to 0 you get the same result.

3. If $ |f(x)| < \epsilon$ for $ |x| > N$, $ f$ will be uniformly continuous on $ [N,N]$ by compactness. Put this together to get uniform continuity elsewhere. Another viewpoint is that $ f$ is continuous on the one-point compactification of $ \mathbb{R}$ (which is metrizable), hence uniformly continuous.

For 3, do you mean $ [ \minus{} N,N]$ since what you wrote is a one point set.  Or $ [N,\infty)$?  Could we take the limit as $ N\rightarrow\infty$?

For 2, suppose $ (x_{n})\in (0,1)$ is a sequence such that $ x_{n}\rightarrow 0$.  Because it is convergent, it is Cauchy.  Because $ f$ is uniformly continuous, it maps Cauchy sequences to Cauchy sequences, so $ f(x_{n})\rightarrow f(0)$, so $ (f(x_{n}))$ is Cauchy.

Does this get me anywhere?

[quote="JRav"]it maps Cauchy sequences to Cauchy sequences[/quote] was good; [quote="JRav"]$ f(x_{n})\rightarrow f(0)$, so $ (f(x_{n}))$ is Cauchy.[/quote] is bad, since $ f(0)$ doesn't exist and you're also reproving something you claimed to prove in the first sentence I quoted.

So what would be the next step?

And for the third one, could I use an argument similar to what I stated before?  Is there a way to make it rigorous?

[quote="Kalle"]1. Take any $ y \in M$, and show that the sequence of iterates $ f^n(y)$ is a cauchy sequence, hence converges. Show that the limit is a fixed point. This is known as Banach's fixed point theorem, I think.[/quote]

We cannot so easily tell if $ f^n (x)$ is Cauchy or not. If compactness of $ M$ is omitted, we even have a counter-example. For example, take $ M \equal{} [0, \infty)$ and define $ f(x) \equal{} \sqrt {x^2 \plus{} 1}$. Then with standard metric $ d(x, y) \equal{} |x \minus{} y|$, Mean Value Theorem shows that $ |f(x) \minus{} f(y)) \equal{} |x \minus{} y| |f'(c)|$ for some $ c$ between $ x$ and $ y$. Since $ f'(c) \equal{} \frac {c}{\sqrt {1 \plus{} c^2}} < 1$, $ M$ and $ f$ satisfies the condition except for compactness.

Now let's return to the original problem. Define $ g(x) \equal{} d(f(x), x)$. Since $ M$ is compact, $ f(M)$ is compact, so $ f$ must have a minimum value, say $ m$. Then $ m \equal{} f(x_0)$ for some $ x_0 \in M$. Now assume $ m > 0$. Then $ g(f(x_0)) \equal{} d(f(f(x_0)), f(x_0)) < d(f(x_0), x_0) \equal{} m$, a contradiction! Therefore $ m \equal{} 0$ and $ x_0$ is a fixed point. (In fact, it is 'the' fixed point.)

Yes, I just figured that one out so thank you anyway.  I'm stuck on the last two.

[quote="sos440"]
We cannot so easily tell if $ f^n (x)$ is Cauchy or not.[/quote]
I was thinking that $ (x,y) \to d(f(x), f(y))/d(x,y)$ was a continuous function on $ M \times M$ and hence assumed a maximum strictly less than 1. However, I suppose this argument is not true. It seems to me like it can be made into an upper semicontinuous function with values strictly less than 1, but maybe it is not the case. Is there an example of this type of lipschitz map on a compact space that does not actually have lipschitz constant < 1?

My apologies for being misleading. I replied very quickly just trying to outline what do.

Let $ M \equal{} \left[ 0, \frac {1}{2} \right]$ and let $ f(x) \equal{} x^2$.  One has $ |x^2 \minus{} y^2| < |x \minus{} y|$ for $ x \neq y$ but $ f' \left( \frac {1}{2} \right) \equal{} 1$.  Of course, the issue here is that $ M \times M \minus{} \{ (x, x) : x \in M \}$ isn't compact.


vector
algebra
polynomial
linear algebra

Let $ A$ be a linnear operator on an $ n$-dimensional vector space , $ n>1$ such that $ N^n\equal{}O,N^{n\minus{}1}\neq O$
Prove there is no operator $ X$ such that $ X^2\equal{}N$

$ X^{2n} \equal{} O$ implies that $ X^n \equal{} O$ since $ X$ has a characteristic polynomial of degree $ n$, but then $ X^{2n\minus{}2} \equal{} 0$; contradiction.


Suppose we had a $8\times8$ chessboard [b]but[/b] the top right square and the lower left square are missing.  We want to tile the entire board with dominoes.  Each domino tiles exactly two adjacent squares ( diagonals are not allowed).  Is it possible to tile the entire chessboard with no domino overlapping?

[hide="Solution"]
Color the squares of the checkerboard alternating with black and white, in the usual manner. The two removed squares are of the same color. Since each domino must cover one square of each color, the board cannot be covered completely. [/hide]

[quote="AstroPhys"]Was color a restriction in the initial problem?
Even so, it is still impossible to cover all the squares in manner suggested by the problem.[/quote]

The point of coloring the board is to [b]prove[/b] that the situation is impossible.


vector
abstract algebra
linear algebra
matrix
linear algebra unsolved

Let $ \mathbf{n}$ be a unit vector in $ \mathbb{R}^3$. Identify the image and kernel (null space) of each of the following linear maps $ \mathbb{R}^3 \rightarrow \mathbb{R}^3$:
\[ \textup{(a)} \, \, \mathcal{T} \colon \mathbf{x} \mapsto \mathbf{x'} \equal{} \mathbf{x \minus{} (x \cdot n)n}, \, \, \, \, \, \, \textup{(b)} \, \, \mathcal{Q} \colon \mathbf{x} \mapsto \mathbf{x'} \equal{} \mathbf{n \times x}\] Show that $ \mathcal{T}^2 \equal{} \mathcal{T}$ and interpret the map $ \mathcal{T}$ geometrically. Interpret the maps $ \mathcal{Q}^2$ and $ \mathcal{Q}^3 \plus{} \mathcal{Q}$ and show that $ \mathcal{Q}^4 \equal{} \mathcal{T}$.

(a) I will write $ T: T(x) \equal{} x \minus{} \langle x,n \rangle n, \,\,\, \forall x \in \mathbb{R}^3$. We will show that $ T^2 \equal{} T$ in three different ways.

I. Observe that the map T corresponds in fact to the first two steps of Gram-Schmidt orthogonalization process applied to the vectors $ n$ and $ x$, since $ \langle x,n \rangle n \equal{} \frac{ \langle x,n \rangle}{ \langle n,n \rangle} n \equal{} \mathrm{proj}_n(x)$. Thus, for the fixed unit vector $ n$, given an arbitrary vector $ x$ T finds an orthogonal vector to $ n$ by substrating from $ x$ its orthogonal projection on $ n$. Therefore, since $ T(x)$ is already orthogonal to $ n$, its projection on $ n$ is the zero vector, and so $ T(T(x)) \equal{} T(x)$.

II. $ T^2(x) \equal{} T(T(x)) \equal{} T(x \minus{} \langle x,n \rangle n) \equal{} x \minus{} \langle x,n \rangle n \minus{} \langle x \minus{} \langle x,n \rangle n,n \rangle n$

$ \equal{} x \minus{} \langle x,n \rangle n \minus{} \langle x,n \rangle n \plus{} \langle x,n \rangle\langle n,n \rangle n$

$ \equal{} x \minus{} \langle x,n \rangle n \minus{} \langle x,n \rangle n \plus{} \langle x,n \rangle n$

$ \equal{} x \minus{} \langle x,n \rangle n$

$ \equal{} T(x)$.

III. Let us find a matrix representation for $ T$ relative to the canonical basis of $ \mathbb{R}^3$. If $ n \equal{} (n_1,n_2,n_3)$, then we have $ T(1,0,0) \equal{} (1\minus{}n_1^2, \minus{}n_1n_2, \minus{}n_1n_3)$, $ T(0,1,0) \equal{} (\minus{}n_1n_2, 1\minus{}n_2^2, \minus{}n_2n_3)$, $ T(0,0,1) \equal{} (\minus{}n_1n_3, \minus{}n_2n_3, 1\minus{}n_3^2)$. Then the matrix

$ A \equal{} \begin{bmatrix} 1\minus{}n_1^2 & \minus{}n_1n_2 & \minus{}n_1n_3 \\ \minus{}n_1n_2 & 1\minus{}n_2^2 & \minus{}n_2n_3 \\ \minus{}n_1n_3 & \minus{}n_2n_3 & 1\minus{}n_3^2 \end{bmatrix}$

is the sought-after representation. But it is easy to see that $ A \equal{} I \minus{} nn^T$, and since $ nn^T$ is idempotent, for $ (nn^T)^2 \equal{} nn^Tnn^T \equal{} (n^Tn) nn^T \equal{} (1)nn^T \equal{} nn^T$, then $ I \minus{} nn^T \equal{} A$ is also idempotent. So, $ A^2 \equal{} A$, which implies $ T^2 \equal{} T$. $ \Diamond$

Now, since A is idempotent, then $ rank(A) \equal{} tr(A) \equal{} 3 \minus{} (n_1^2 \plus{} n_2^2 \plus{} n_3^2) \equal{} 3 \minus{} 1 \equal{} 2$, and so $ \dim Ker(T) \equal{} 1$ and $ \dim Im(T) \equal{} 2$. It is easy to see that any vector parallel to $ n$ is mapped onto the zero vector. Therefore, $ Ker(T)$ is the set of all vectors parallel to $ n$, or $ Ker(T) \equal{} \langle n \rangle$. It is also easy to see that $ Im(T)$ is the plane orthogonal to $ n$ that intersects the origin.

(b) We have $ Q(x) \equal{} n \times x, \forall x \in \mathbb{R}^3$, where $ \times$ is intended to represent cross product. Then,

$ Q^2(x) \equal{} n \times (n \times x) \equal{} \langle n,x \rangle n \minus{} \langle n,n \rangle x \equal{} \langle n,x \rangle n \minus{} x \equal{} \minus{} T(x)$.

As a consequence of the idempotency of $ T$, we see immediatly that $ Q^4 \equal{} (Q^2)^2 \equal{} (\minus{}T)^2 \equal{} T^2 \equal{} T$.

Finally we have

$ Q^3(x) \equal{} n \times (\langle n,x \rangle n \minus{} x) \equal{} \langle n,x \rangle (n \times n) \minus{} (n \times x) \equal{} 0 \minus{} (n \times x) \equal{} \minus{} Q(x)$

and so $ Q^3(x) \plus{} Q(x) \equal{} 0$.

This is comprehensive - for (a) I have the same solution as your second one (II). Thanks!

For the transformation (b), we can see that $ Ker(Q) \equal{} \langle n \rangle$, and $ Im(Q)$ is the plane perpendicular to $ n$, that is, $ Im(Q) \equal{} \{ x \in \mathbb{R}^3: \langle n,x \rangle \equal{} 0 \}$.


$s = (x,y,z)$

$x = r. \sin( \theta) . \sin(\phi)$
$y = r. \cos( \theta) . \sin(\phi)$
$z = r. \cos(\phi)$

$\theta = \frac{2 . \pi}{n}. \alpha$
$\phi = \frac{ \pi}{m}. \beta$

$\alpha,\beta,m,n$ are integers

The question is:

There is some $(m,n)$ such that $s$ are the points of plato solids?!

What are those $(m,n)$?

The plato solids are symmetrical, so, it can be circuncribed.
So the question Hold. If not, can we add some Phase to it to correct?

thanks

I need this for a icosaedron, but I want to know if it can be generalized for others plato solids.

PS: We make $\alpha,\beta$  integers variables ;)

After a Hard try i Found for icosaedron $n=5, m=3$

and for the others? dodecadron?! etc.

help....

there is a easy way.... must be


function
limit
algebra
functional equation
algebra unsolved

Find $ f: \Set{R} \rightarrow \set{R}$: $ f(x+y)=f(x).f(y).f(xy)$

I'm not sure whether or not a nontrivial solution exists, but if it does then it is necessarily not continuously differentiable in any neighborhood of $ 0$, so we shouldn't expect it to be any nice combination of elementary functions.
[hide="proof"]If $ f(y) \equal{} 0$ for any $ y$ then $ f(x) \equal{} 0$ for all $ x$.  Thus, if we're interested in nontrivial solutions, let us assume without loss of generality $ f(x) > 0$ for all $ x$.  If we look in the space of functions that are continuously differentiable in a neighborhood $ B$ of $ 0$ then the functional equation implies
\[ \lim_{y\to 0}\frac {f(x \plus{} y) \minus{} f(x)}{y} \equal{} \lim_{y\to 0}\frac {f(x)\left(f(y)f(xy) \minus{} 1\right)}{y}
\]
exists for $ x\in B$.  For this limit to exist we must have $ f(0) \equal{} 1$ and
\[ \lim_{y\to 0}f(x)\left[f'(y)f(xy) \plus{} xf(y)f'(xy)\right] \equal{} f'(x) \\
\Rightarrow f'(0)(x \plus{} 1)f(x) \equal{} f'(x)
\]
for which the solution is
\[ f(x) \equal{} e^{x \plus{} \frac {x^2}{2}}
\]
in some neighborhood of $ 0$.  But this does not satisfy the functional equation, so it does not admit solutions differentiable in any neighborhood of $ 0$.[/hide]

I think ur solution have some problem. There're exactly 3 funtion $ f(x)=1 \forall x \in \set{R}$, $ f(x)=-1 \forall x \in \set{R}$, $ f(x)=0 \forall x \in \set{R}$ satisfy.

Well, I did say any [u]nontrivial[/u] solution is not differentiable near $ 0$, though you're right I should have more specifically said non-constant instead, which is assumed in this step:[quote="JoeBlow"]
\[ \Rightarrow f'(0)(x \plus{} 1)f(x) \equal{} f'(x)
\]
for which the solution is
\[ f(x) \equal{} e^{x \plus{} \frac {x^2}{2}}
\]
in some neighborhood of $ 0$.[/quote]
Of course $ y \equal{} c$ is another solution to the ODE, which is where the three constant solutions come from.  For non-constant solutions, I don't see any errors in the argument.  And this was by no means intended to be a solution, rather an observation about the properties of nontrivial solutions [i]if[/i] they exist.

Here's a solution that doesn't assume anything about continuity of $ f$.

[hide="Solution"]
Obviously $ f(x) \equiv 0$ is one solution, so from now on we assume that $ f$ is not identically zero. If for some $ x_0$ we have $ f(x_0)\equal{}0$ then $ f(x_0\plus{}y)\equal{}0$ for all $ y$ which makes $ f \equiv 0$, so in fact we can assume that $ f$ is never zero.

In particular, $ f(0) \neq 0$. Define $ a\equal{}f(0)$ and from the functional equation with $ x\equal{}y\equal{}0$ we get  $ a\equal{}a^3$ so $ a\equal{}\pm 1$.

Now let $ b \equal{} f(1)$. Then $ f(2) \equal{} f(1\plus{}1) \equal{} b^3$. Now $ f(4)\equal{}f(2\plus{}2)\equal{}b^3b^3f(4)$ and since $ f(4) \neq 0$ we get $ b^6\equal{}1$ or $ b \equal{} \pm 1$ since $ b$ is real.

Set $ x\equal{}1$, $ y\equal{}\minus{}1$ in the functional equation: $ f(0) \equal{} f(1)f(\minus{}1)f(\minus{}1)$ or $ a \equal{} b f(\minus{}1)^2$. This implies that $ a$ and $ b$ have the same sign and since they're both $ \pm 1$ we have in fact $ b\equal{}a$.

To summarize so far: $ f(0)\equal{}a\equal{}\pm 1$ and $ f(1)\equal{}f(2)\equal{}a$ as well.

Now take an arbitrary $ x$ and set $ f(x)\equal{}z$. Then

$ f(x\plus{}1) \equal{} f(x)f(1)f(x) \equal{} az^2$

$ f(x\plus{}2) \equal{} f((x\plus{}1)\plus{}1) \equal{} f(x\plus{}1)f(1)f(x\plus{}1) \equal{} az^4$

$ f(x\plus{}3) \equal{} f((x\plus{}2)\plus{}1) \equal{} f(x\plus{}2)f(1)f(x\plus{}2) \equal{} az^8$

but we also have

$ f(x\plus{}2) \equal{} f(x)f(2)f(2x)$ so $ az^4 \equal{} zaf(2x)$ which implies

$ f(2x) \equal{} z^3$.

Now we can calculate $ f(x\plus{}3)$ another way, as $ f((x\plus{}1)\plus{}2)$. We get

$ az^8 \equal{} f(x\plus{}3) \equal{} f((x\plus{}1)\plus{}2) \equal{} f(x\plus{}1)f(2)f(2x\plus{}2) \equal{} az^2\cdot a \cdot f(2x\plus{}2)$, namely

(*) $ f(2x\plus{}2) \equal{} az^6$.

However, we just proved that if $ f(x)\equal{}z$ then $ f(x\plus{}2)\equal{}az^4$. Using that for $ 2x$ instead of $ x$ and remembering that $ f(2x)\equal{}z^3$ we find $ f(2x\plus{}2) \equal{} az^{12}$. Comparing this with (*) we get $ az^6 \equal{} az^{12}$ which proves that $ z \equal{} \pm 1$. But in fact, from $ f(x\plus{}1) \equal{} az^2$ we now get $ f(x\plus{}1) \equal{} a$ for all $ x$, so we proved that either $ f(x) \equiv 1$ or $ f(x) \equiv \minus{}1$. Together with the option $ f(x) \equiv 0$ this yields only 3 possible solutions.

[/hide]


Let $mn+1$ different real numbers be given. Prove that there is either an increasing sequence with at least $n+1$ members, or a decreasing sequence with at least $m+1$ members

All replies should go to this thread:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153730[/url]

It's the same problem.


Determine the number of paths from the origin to the point $(9,9)$ which never pass through $(5,5)$, in paths consisting only of steps going $1$ unit north and $1$ unit east.

[hide]
I first determine the number of paths to the point (9,9)
There need to be 18 moves in total, 9 of which go to the right, so there are $\binom{18}{9}$ paths.

Now I determine the number of paths to the point (9,9) that pass through the point (5,5)
Using the logic above, there are $\binom{10}{5}$ paths to (5,5) and from there $\binom{8}{4}$ paths to (9,9). This makes $\binom{10}{5}\binom{8}{4}$

So the answer is $\binom{18}{9}-\binom{10}{5}\binom{8}{4}=48620-(252)(70)=\boxed{30980}$
[/hide]

[quote="Kev-Eile"]Well that's a much better way than trying to count out the number of individual paths....[/quote]

:roll:

Look, it worked, okay? :P

As an aside to Kev: tjhance's solution is the same as Derek's.

That's proof enough for me that he IS Derek!

Is there any way, other than with the combinatorics (which is what I thought of also), to do this problem? (quickly)

[quote="JBL"]Yes: block walking.  Put a 1 at point $(0, 0)$.  For each point, you sum the numbers below it and to its left.  That counts paths.[/quote]

That's cool. I got the same answer, [hide]30980[/hide] from the block walking method, and it took maybe 5 minutes. I'll have to remember this the next time i've got an hour left on the AIME and feel like checking my work on the problems that I actually knew how to do. Thanks  :)

I got the answer using block walking as well, but only after doing it three times due to addition mistakes :blush:

[quote="breez"]I think block walking is more useful for cases in which multiple points are forbidden as opposed to cases of a single forsaken point such as this one, which can be done in less than a minute using combinatorics.[/quote]

Right, but block walking gets too messy and large for a bigger path. E.g.:

Find the number of paths of shortest distance from the origin to the point $(30,30)$ which do not pass through the lattice square with vertices $(10,12), \ (13, 12), \ (13,15), \ (10,15)$.


hello i m new i want just to try if i can participat (with this message)

There's a Test forum for this, if you just want to try things out and see how the forum works.
http://www.artofproblemsolving.com/Forum/index.php?f=224