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In each cell of an 8x8 table we wrote either +1 or -1. In each step we switch the signs of all the cells in an 3x3 or 4x4 square. Is it possible to go from any initial configuration to any other?

There are 36 3x3 squares and 25 4x4 squares, so 61 squares in total.
The order squares are switched does not matter and switching the same square twice takes you back to the previous config,
so there are just $ 2^{61}$ configurations that can be reached from a given starting config.
But there are $ 2^{64}$ configurations in total, so some cannot be reached.

Would it be possible if it were a 9x9 table?

In fact it is not possible for any size of table when switching signs in two squares bigger than 1x1.

How interesting! I didn't know that counting still works! My original proof is based on coloring, this way, as you stated we can prove that for any two square bigger than one it is impossible to go from any state to any other. But your counting proof induced some ideas, I'll post some problems coming out of these ideas, for the time being, a simple consequence of this approach is that, there are 9 distinct configs which it impossible to go from any of them to another one.

The simple counting argument that I gave does not work for larger tables. However you can make a better counting argument by counting identities, i.e. different arrangements of squares that give the same result. For every 6x6 square there ia such an identity and all but one of these are independent. The identity is found by using every possible 3x3 and 4x4 sqaure that fits inside the 6x6 sqaure at once and showing that this switches each cell an even number of times. If you include these in the counting you can extend the argument to any size of table.

I looked for alternatives proofs, for example I wondered if there was a subset of cells for which there would always be an even number of them changed. I could only find such subsets for smaller tables e.g. 6 x 6. Is your colouring proof an extension of this to multiple subsets?

Edit: Actually I thought more carefully about this and I think that there are more dependencies between the identities than I previously realised. The number of independent identities is $ \frac{1}{2}(n\minus{}5)(n\minus{}4)$  so the transofrmations are not poissible if
$ n^2 \plus{} \frac{1}{2}(n\minus{}5)(n\minus{}4) > (n\minus{}2)^2 \plus{} (n\minus{}3)^2$
$ k \equal{} 12 \minus{} \frac{1}{2}(n\minus{}5)(n\minus{}6) > 0$

This is true for $ n$ up to 10 so I am not so sure about an 11 by 11 table.

For $ n \equal{} 8$ we have $ k \equal{} 9$ which agrees with the number of distinct configs found by pmvald.

I didn't understand the 6x6 square argument, but I was thinking about the same idea of counting identities. My idea was that if these two shapes (e.g. 3x3 and 4x4 square) both can cover (without overlapping) an mxm square for some $ m$, then we have $ (8\minus{}m\plus{}1)^2$ common configs that can be obtained by both shapes as the switching tool. I guess you are doing the same thing, is that right? Anyway, this idea leads to some other interesting problem. One of them is to find or estimate the value of $ m$. It is interesting even if we limit ourself to only one shape: Given a grid shape (e.g. 3x5 square, an L shape, etc), determine when this shape can cover a mxm square for some $ m$. In the next step one may ask, what is the min value of $ m$? I'll write more on this subject. As for an alternative proof, my proof is exactly based on coloring. highlight the first two rows with black and leave the third row uncolored. Then again, highlight the next two rows (4th and 5th) and leave the third (6th tow) uncolored and so on (this coloring is shown in Figure one in this article: http://pmvald.livejournal.com/1246.html). Then it is easy to see that any switching cell group, namely any 3x3 or 4x4, contains an even number of colored cells, thus the parity of number of colored +1 remains the same all through the switchings. So one cannot move from any config to any other. I wrote a short article about this coloring idea and how it leads to some other interesting problems, it can be found at http://pmvald.livejournal.com/1246.html

Your colouring argument is very nice. That's just the kind of solution that an olympiad problem should have. It extends to larger squares much more easily than the counting argument.

I was actually looking for just such a solution having found a suitable colouring for a 6x6 square as shown, but I was expecting something more complicated and missed the easy answer!

Using identities can be tricky because you have to work our how many indepedent identites you have. In general the problem is just linear equations over the field $ F_2$ so you can construct a rectangular matrix with all the equations and compute its rank numerically. That will tell you exactly how many independent identities there are for a given problem size. For general sizes you have to work quite hard to get a theoretical solution.

My search for identites was similar to yours exept that you do not have to limit yourself to non-overlapping coverings. The first one I spotted was a 7x7 identity. Place two 4x4 squares and two 3x3 squares in a 7x7 square so that they overlap only on the central square. You can rotate this through 90 degrees to get an identical covering and the equality of the two is an identity. I then noticed that a 6x6 identity is possible if you just place every possible 4x4 and 3x3 square inside the 6x6 square. These identities can be generalised for different sizes of squares. I got confused about which of these identities are indepedent. The 7x7 identity is not independent of the 6x6 identity, but I now think all the 6x6 identities are indpendent, in which case there are enough to solve all sizes of table.

I can see now how the colouring proof can be extended to any two square sizes. The counting argument generalises for the same cases. It would be nice to know what combinations of 3 or more square sizes can be used to switch to any configuration, if any.

You should write up everything you have on this in a paper.

I wrote a program to compute the rank. It confirms that all the 6x6 identities are indepedent.

For three squares I can compute the smallest table for which all configuration changes are possible:

(2x2,3x3,5x5)  is sufficient for 7x7 table and up.
(2x2,3x3,7x7)  is sufficient for 9x9 table and up.
(3x3,4x4,5x5)  is sufficient for 9x9 table and up.
(3x3,4x4,7x7)  is sufficient for 11x11 table and up.
(3x3,5x5,7x7)  is sufficient for 12x12 table and up.

What about higher dimensions? Perhaps four cubes are necessary in 3 dimensions, etc. I think a colouring argument confirms that, not sure.

I think I understand it a bit better now. The magic colourings are actaully the zero eignevectors of the matrix of equations over $ F_2$.
If it is not possible to go from every config to another then the rank of the matrix falls short of the number of squares and the eigenvectors must exist. They can be computed in polynomial time. It would be fun to modify my program and compute some magic colourings for some cases, especially when there is only one colouring. E.g for squares 3x3, 5x5, 7x7 in an 11x11 table. I dont have time to do it now but perhaps later.

Another coloring that solves the original problem :

0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1

Each $ 3 \times 3$ and $ 4 \times 4$ square cover even number of fields with number 1, and the statment follows in the same way as in one of the above arguments.

[quote="PhilG"]I got the program to work and now I can calculate magic colourings.
I did the case of an 11x11 table and squares of size 3x3, 5x5 and 7x7
In this case the colouring in unique.
You can readily check that any square of one of the three sizes will cover an even number of coloured squares.
In this case the pattern does not extend to bigger squares becuase the three sizes are then sufficient to generate any pattern.[/quote]

Good job! You did a valuable contribution. That's how math problems are born. I got your idea of identity this way: the mxm square is called an identity if a combinations of shapes (e.g. 3x3 and 4x4 squares) can cover any of its cells an even number of times (that's how 7x7 and 6x6 are defined to be identities), that's what you mean, isn't it? This concept of identity, in a way that is defined here, is not free from implications. To start with, one can come up with this question: given a set of shapes (e.g. 3x3 and 4x4 squares), what is the (minimum?) value of $ m$ in the previous definition? For the case of 3x3 and 4x4 square, you went as low as m=6, what about m=5? What about the same question for two squares with different dimensions? what about three squares .... These questions are comparable to my question of non-overlapping covering. You allow overlaps, but it should happen an even number of times for each cell. It is interesting to study these two problems parallel with each other and see how they lead to different results. Then you mentioned 'independent' identities, I didn't get that. How do you do define independent identities? But even without knowing its definition one can pose this problem: for a set of shapes, find the smallest m for which all mxm identities are independent (does it make sense?). You also proposed to study this problem through its linear algebraic model, i.e. by converting it to a set of linear equations in $ F_2$. That's a very nice idea, at least it gives an affirmative answer to Open Research Problem 6 (in my article found at: http://pmvald.livejournal.com/1246.html, you may check for yourself to make sure I'm not making any mistake here). Besides, the model itself is interesting too. For example each shape introduces a vector of size $ n^2$ ($ n$ is the size of the big table). I became interested in the structure of the vectors produced by square shapes. I'll write more one this kind of problems. You also came up with some sets of three squares for which we can go from any config to any other. Then, using the linear algebraic model (i.e. Open Research Problem 6), it is equivalent to saying that for these sets of three squares there is no magic coloring (is that right?). I actually posted this question in the section 'combinatorics open questions' of this forum (link: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=301111). Here is the original post (which received no answer until now!):

Find  all (or some!) finite sets $ D \subset N$ such that for any large enough $ n$ (in terms of members of $ D$), one can color some cells of the $ n$x$ n$ table $ T$ in black such that we have at least one black cell and for any $ d \in D$, any $ d$x$ d$ square in $ T$ has an even number of black squares.

Lets call such sets $ D$ good sets. So at lest we know that $ \{3,4\}$ is a good set. In fact using my coloring argument we can prove that any 2-element set $ \{a,b\}$ is good is at least one of its members is even. What about the case when both its members are odd? Then your program shows that these sets are not good: $ \{2,3,5\},\{2,3,7\},\cdots$. So one may guess that 3-element sets are not good. However we should note that any set $ D$ is good if the GCD of its members is larger than one (why?). So a trivial requirement for a set to be bad (=not good) is that the GCD of its members should be $ 1$. So, modifying my last statement about 3-element sets, I ask if all 3-element sets with GCD of its members equal to one are bad? You also computed the smallest table for which the magic coloring does not exists. It prompts the general problem of finding such a value for a bad set $ D$. I'll try to see if this value can be estimated in a good way or not. And finally your proposal about studying the higher dimensions is nice. As you said, maybe one can prove that any 4-element set (with GCD of all its members equal to one) is bad (3-dimension version) and $ 4$ is the smallest number for which this statement holds.

I think your understanding is very good, but to clarify a bit I'll summarise how the linear algebra works starting from your set description.

We have $ m$ sets $ A_1, .. , A_m$ of which are all subsets $ A_i \subset E \equal{} \{e_1, ..., e_k\}$ with $ |E| \equal{} k$.

Then we can represent each set $ A_i$ by a vector $ a_i$ from the vector space of dimension $ k$ over $ F_2$.
I.e. each vector $ a_i$ has $ k$ components $ a_{ij}$ such that $ a_{ij} \equal{} 1$ if $ e_j \in A_i$, otherwise $ a_{ij} \equal{} 0$

So the vectors form the columns of an $ m \times k$ matrix $ A$ over $ F_2$ with components $ a_{ij}$.

We start with a configuration of zeros and ones which is given by a column vector $ x$ with $ k$ components each equal to $ 0$ or $ 1$.
(These actually represent the $ 1$ and $ \minus{} 1$ numbers in the original problem.) $ x$ is transformed by adding any combination of
vectors $ a_i$ to get a new vector $ x$. This combination can be represented by a vector $ v$ with components $ v_j$ telling us how many
times the vector $ A_j$ is used in the sum, so

  $ y \equal{} x \plus{} \sum_{j \equal{} 1}^{m} a_j v_j \bmod{2}$

Since everything is carried out over $ F_2$ we can also consider $ v_j$ as a vector $ v$ over $ F_2$ and write this in terms of matrices as

 $ y \equal{} x \plus{} Av$

We want to know if we can transform from any given vector $ x$ to any other given vector $ y$. This is equivalent to asking whether we can solve this linear system of equations over $ F_2$. For matrices over any field the answer is that this is possible if the rank $ r$ of $ A$ is equal to $ k$
( see [url]http://en.wikipedia.org/wiki/Rank_(linear_algebra)[/url] )

In particular the rank $ r$ of $ A$ can be at most $ r \le \min(m,k)$ so not all transformations are possible if $ m < k$.

If $ r < m$ then there are $ h \equal{} m \minus{} r$ linearly independent vectors $ w_j$ such that $ A w_j \equal{} 0$. These are what we call independent identities.
The identities form a vector sub-space of dimension $ h$ so there are $ 2^h \minus{} 1$ identities excluding the trivial identity.

If $ r < k$ then there are $ g \equal{} k \minus{} r$ linearly independent vectors $ u_i$ such that $ u_i A \equal{} 0$. These are independent magic colourings.
The magic colourings form a vector sub-space of dimension $ g$ so there are $ 2^g \minus{} 1$ possible magic colourings in all (we exclude 0)

From this the conclusion follows that magic colurings exist iff not all transformations are possible.

We can also see how the counting argument works in general. We know that $ m < k$ is sufficient to show that not all transformations are possible.
We get a stronger counting test if we know the number of indepednent indentites $ h \equal{} m \minus{} r$, then $ g \equal{} k \minus{} r \equal{} k \plus{} h \minus{} m$,
so if we find enough independent identities to show that $ k \plus{} h > m$ then we know that not all transformations are possible.

An interesting observation is that we can generalise the transformation problem from $ F_2$ to any ring but we have used the fact that $ F_2$ is a field to draw some of the conclusions. Over rings in general the row rank and column rank fo a matrix can be different, but otherwise most of what I said would generalise.

I'll try and answer some of the extra questions you ask.
Your first question is "Is there an identity in a 5x5 table using 3x3 and 4x4 squares?"
To answer that we just need to solve the analogous one dimensional problem.
I.e. "given a one dimension strip of length 5, can we find an identity using strips of length 3 and 4?"

Its not immediately obvious that these questions are equivalent, but here is why.
Firstly, if we have an identity in the 5x5 square then the squares from the identity that intersect any row or column of the square will form an identity of the one dimensional problem in that strip, so if we have an identity in the 5x5 square then we have at least one identity of the strip of length 5.

Conversely, suppose we have an identity on the strip of length 5. The identity consists of a set of subset $ S_3$ of the strips of length 3 and a subset $ S_4$ of the strips of length 4. In algebraic terms the identity is the sum of two vectors $ w \equal{} s_3 \plus{} s_4$ such that $ Aw \equal{} 0$

The 5x5 problem is based on a vector space that is a tensor product of the 5 strip problem with itself and I can write an identity for this space as $ s_3 \otimes s_3 \plus{} s_4 \otimes s_4$. In other words a 3x3 square is in the identity for the 5x5 square if its projection onto horizontal and vertical axis is in the identity on the strip. I think I will just leave you to convince yourself that this works, rather than try to write it in algebraic terms.

In any case, to show that there are no 5x5 identities we just have to show that there are no identities on the strip of length 5. That is quite easy because there are only five shapes to use. In fact in this one dimensional case we have $ k \equal{} m \equal{} 5$ and the rank of the problem is also five because we can easily show that it is possible to go from any configuration to any other. Then the number of identites is $ h \equal{} m \minus{} r \equal{} 0$ so there are none.

Of course all this generalises and I think it generalises far enough to solve all 2 dimensional problems with two square sizes, if we can solve all the one dimensional cases, but I dont have a proof. What I think is that if you have $ h$ indepednent idenetities in a one dimensional problem then you have $ h^2$ independent identities in the 2 dimensional problem. It is not difficult to see how to construct those identities using the tensor product idea but proving that they are all independent and that there are none others is harder. If it's true then it can be used to solve any problem with two square sizes quite quickly.

For example, take the original 8x8 problem with 3x3 and 4x4 squares. For the one dimensional problem we have $ k_1 \equal{} 8$ and $ m_1 \equal{} 6\plus{}5 \equal{} 11$. In this problem you can go from any config to any other so there are no magic colourings $ g_1 \equal{} 0$ and $ h_1 \equal{} g_1 \plus{} m_1 \minus{} k_1 \equal{} 3$. Then for the two dimensional problem we must have $ h_2 \equal{} h_1^2 \equal{} 9$, but $ k_2 \equal{} 8\times8 \equal{} 64$ and $ m_2 \equal{} 6\times6 \plus{} 5\times5 \equal{} 61$ therefore $ g_2 \equal{} h_2 \plus{} k_2 \minus{} m_2 \equal{} 9\plus{} 64 \minus{} 61 \equal{} 12$ so there must be 12 independent colourings.

When there are three sizes of square this method does not seem to work. I dont know if it works for 3 dimensional problems with two sizes of cube.
Perhaps there are more complicated generalisations that can be made.

Here are some results that I get from my program without complete proof

For the case of two square sizes $ s_1$ and $ s_2$ the smallest square size with an identity is $ x \equal{} s_1\plus{} s_2 \minus{} \gcd(s_1,s_2)$
The number of independent identities in an $ n \times n$ table $ n \ge x$ is $ h \equal{} (n \plus{} 1 \minus{} x)^2$

For three square sizes $ s_1, s_2, s_3$ that are relatively prime in pairs $ \gcd(s_1,s_2) \equal{} \gcd(s_2,s_3) \equal{} \gcd(s_1,s_3) \equal{} 1$,
the largest table with a magic colouring is $ n \equal{} s_1\plus{}s_2\plus{}s_3 \minus{} 4$ (so its bad)
Any other combinations of three squares has a magic colouring for all table sizes (so its good)

I think I've spent enough time on it now, nice problem, thanks. :)


Let $ m_1,...,m__k$ be natural numbers and $ b_1,..,b_k \in \mathbb{Z}$ such that for all $ i,j \in \{1,...,k\}$, $ b_i \equiv b_j$ $ (mod$ $ {gcd(m_i,m_j)})$ .Prove that below system has
answer:
$ X \equiv b_1$  $ (mod$ $ {m_1})$
.
.
.
$ X \equiv b_k$  $ (mod$ $ {m_k})$

Fix some prime p which divides at least one of $ m_{i}$. Let  $ m_{k}$ be the number  that is divisible by the largest power of p . Let   $ p^{l}\parallel{}m_{k}$. Let $ X \equiv b_{k}(mod p^{l})$. By the  chinese remainder theorem we can find X satisfying this condition for each prime p. We shall prove this X satisfies the condition of the problem.
It is enough to prove that for all primes p if $ p^{l}\parallel{}m_{i}$ then $ X \equal{} b_{i}(mod p^{l})$. Let  $ m_{k}$ be the number  that is divisible by the largest power of p. We have
$ X \equal{} b_{k}\equal{}b_{i}(mod p^{l})$. Here we have used that $ p^{l}|gcd(m_{i}, m_{k)}$, and $ b_{k}\equiv b_{i} mod(gcd(m_{I}, m_{k})$.


[quote="mecrazywong"]a,b,c are the sidelengths of a triangle such that $a^2+b^2+c^2=9$. Prove that\[a+b+c\ge\sqrt{a^2b^2+b^2c^2+c^2a^2}\][/quote]
The inequality is quivalent to $(a+b+c)^2(a^2+b^2+c^2)\ge 9(a^2b^2+b^2c^2+c^2a^2)$
Put $c\longrightarrow 0,a=b=1$ to get $8\ge 9$

:oops_sign:
Sorry, it seems that I got some computational errors :(


Do you know the Jensen Inequality's geometric proof?

Does there exist something like geometric proof of Jensen ?  :? It is too hard to believe...

Why? Jensen's inequality only states that if $f$ is convex, the barycenter of the points images $f(x_i))$ is above the image by $f$ of the barycenter (with the same non-negative mass) of the $x_i$.

Pierre.

It is not a proof, in my opinion. It is just a "visualization". ;)

I didn't say it was a proof and I'm too lazy to try to turn it into a rigourous proof. :?  I just observed that Jensen's inequality has a really simple geometric interpretation so that it is not  completely stupid to ask for a geometric proof as Megus semmed to believe.

Pierre.


function
quadratics
algebra
quadratic formula

Solve the next system :

$ \{
\begin{aligned}
a^2 + b^2 + c^2 &= \frac {5}{3}
\\
abc &= 1
\end{aligned}
$

with $ a,b,c\in \mathbb{R}$

EDITED my poor algebra,,,

let $ a \equal{} \frac {1}{bc}$.  Then we have $ \frac {1}{b^2c^2} \plus{} b^2 \plus{} c^2 \equal{} 5/3$ so $ b^4c^2 \plus{} b^2c^4 \minus{} 5/3 b^2c^2 \equal{} \minus{} 1$ implies $ b^2c^2 (b^2 \plus{} c^2 \minus{} 5/3) \equal{} \minus{}1$ implies $ b^2\plus{}c^2 < 5/3$ so $ bc < 5/6$ by AM-GM... doesn't help

I think there are an infinite number of solutions

perhaps i should have continued, I think I can get some bounds

$ \minus{}1\equal{}b^2c^2(b^2\plus{}c^2\minus{}5/3) < 25/36(b^2\plus{}c^2\minus{}5/3)$
$ \minus{}36/25\plus{}5/3 \equal{} 17/75 < b^2 \plus{} c^2 < 5/3$

EDIT: nvm... I see it now.

$ 0\equal{}b^4c^2\plus{}b^2(c^4\minus{}5/3c^2)\plus{}1$

and solve by quadratic formula or completing the square.

Substitute a = 1/bc in the first equation to get that \[ \frac{5}{3} \equal{} a^2\plus{}b^2\plus{}c^2 \equal{} \frac{1}{b^2c^2}\plus{}(b\pm c)^2\mp 2bc \geq \frac{1}{b^2c^2}\mp bc \mp bc \geq 3.\]In the last equation we used AM-GM on $ \{1/b^2c^2,\mp bc, \mp bc\}$ (depending on the sign of $ bc$). No solutions!

[quote="alex2008"]Solve the next system :

$ \{ \begin{aligned} a^2 + b^2 + c^2 & = \frac {5}{3} \\
abc & = 1 \end{aligned} $

with $ a,b,c\in \mathbb{R}$[/quote]

Eh...by AM-GM $ \frac{5} {3}=a^2+b^2+c^2 \ge 3 \sqrt[3] {(abc)^2}=3$ contradiction.

Real numbers, not necessarily positive.

I knew that.


Wasn't this problem posted before? Here's my solution I posted:
[hide]We know that the last digit must end in $4$, because the units digit of $x^5$ is the unit digit of $x$, and $3+0+4+7= 14$, so the units digit is $4$.

Now here is the clever part:

$27$ is approximately $\frac{1}{5}*133$

$84$ is approximately $\frac{3}{5}*133$

$110$ is approximately $\frac{4}{5}*133$

Letting $x$ be $133$, we have:

$= x^5 + (\frac{4x}{5})^5 + (\frac{3x}{5})^5 + (\frac{x}{5})^5$

$= \frac{3125x^5}{3125} + \frac{1024x^5}{3125} + \frac{243x^5}{3125} + \frac{x^5}{3125}$

$= \frac{4393x^5}{3125}$

This is approximately $1.45x^5$

So this problem is reduced down to finding a number $y$ that has a units digit of $4$ such that $y^5= 1.45 x^5$

If $y$ is $154$ or approximately $1.15x$, it is too high because $(1.15x)^5$ > ${1.45x^3}$

Thus $y$ must be $144$, which makes sense because $(1.07)^5$ is approximately $1.45$

So the number is $144$[/hide]

[quote="1234567890"]We know that the last digit must end in $4$, because the units digit of $x^5$ is the unit digit of $x$, and... [/quote]
Just to clarify things, I think this is due to Fermat's little theorem which states $a^p\equiv a\mod p$. :)

Masoud Zargar

FWIW - It used to impress many people when I could "find out" fifth root of numbers without pen/paper/calculator and within a few seconds. (similar tirck was also used for seventh root, cube root etc.) .. ( Assume that the number is smaller than 15 digit and has integer root etc..)..

Any way if one assumes that indeed there is an integer  $k$, then finding it can be done almost right away without pen/paper/calculator..
It may take more time to write the method than actually do it but here is the idea. (Most of the work can be done in multitasking mode and obviousif you think the right way)

1.  $k>133$  [hide="because"] (because $k^5=133^5+something > 133^5$) [/hide]
2   $k <200$  (because [hide="because"]$k < \sqrt [5]{(133^5 \cdot 4)}$ (one can easly see $\sqrt [5] 4 < (3/2)$ because $(3/2)^5 = (9*27)/(8*4)>27/4>4$ and $133\cdot (3/2) < 400/2 = 200 )$[/hide]
3. unit digit is $4$ [hide="because"] (=unit digit of $3+0+4+7$  (use the fact that $a^5$ and $a$ has the same unit digit.)[/hide]
    (So the number is something like $1\_4$)
4.  $k$ is divisible by $3$ [hide="because "] (Mod 3 gives  $1^5 +(-1)^5+0+0 \equiv 0$[/hide]

This gives only choice - $144$ or  $174$

You check modulo 7, [hide](we get $0^5+5^5+0^5+(-1)^5 \equiv 5\cdot(25)^2-1 \equiv 5\cdot 4^2-1 = \equiv 10-1 \equiv 2$[/hide]Thus $174\equiv (-1)$ can be ruled out, (as ($(-1)^5 \equiv -1)$  and $144 \equiv 4$ works)

As I said, it takes much much longer to write/explain it much faster to do it....:)

(Actually you don't even need the step 4 (or mod 3) as $144$  is only number between $133$ and $200$ which is $4$ mod 7 and 10)

Have you ever read that book "Figuring" by some indian author? She shows you how to do pretty much everything in your head. On the cover it has her working out the 23rd root of some giant number in her head, it took her 53 seconds. The programed a computer to try find it, it took it 1:27. Pretty impressive!

its reeeaaalllllllllllyy old, so i would have been suprised if you did! My father, a maths teacher, gave it to me, and he had gotten it from his father, to give you some idea of how old it is...


Ok, whoever wrote this quiz was NOT paying very much attention or something...and it cost me points I tell you! Points! Okay, so maybe I'm overreacting, but there is a serious problem with one of the questions.

The question was like this:

Which of these are body parts or functions that help an animal survive?

(a) all of the above
(b) claws
(c) teeth
(d) running

The "correct" answer was (a). HOW CAN (a) BE RIGHT IF THERE ARE NO ANSWERS ABOVE IT?!  :huh:  :what?:
grrrr :roll:

:rotfl: Funniest thing I've heard all day.

There's a simple explanation. You know how experts claim that you should always choose C if you aren't sure of the answer? (some people think its B or wahtever)... well, FunTrivia.com wants to prevent people from "accidentally" making too many answers C so when you make a quiz, they ask you for the right answer and three wrong answers and put all of them in random order.

The questions from the AOPS Trivia Challenge are taken from FunTrivia.com.

So yes, you have a point that the test creator wasn't thinking... but keep in mind that all of these quizzes have to be OKed by editors... so I think the test creator wasn't the only one who wasn't thinking...

(... and couldn't this have merely gone into the regular thread?)

Yeah, the answers are put in randomly so the first answer could be all of the above.

Shouldn't this be in the AoPS Trivia Challenge topic?

[quote="ZennyK"]Yes, but I couldn't find it at the time and I wanted to release my frustration as quickly as possible.  :P[/quote]

I know how that is! :D

Nice, I once took a quiz where the answers were D: none of the above. E: all of the above.
yea a friend of mine made this stupid quiz.

I just took a math test were the answers were

A) 70% of total
B) 60% of total
C) 50% of total
D) 20% of total
E) 20% of total

And the answer was 20% of total.  And I, with my luck, managed to pick the wrong 20%.


calculus
integration
derivative
function
calculus computations

Determine the value of $a,b$ for which $\int_0^1 (\sqrt{1-x}-ax-b)^2 dx$ is minimized.

Maybe not an elegenat solution but solve the integral to get
[hide]$\frac{1}{2}+\frac{1}{3}a^2+ab+b^2-\frac{8}{15}a-\frac{4}{3}b$

Set $f(a,b)=\frac{1}{2}+\frac{1}{3}a^2+ab+b^2-\frac{8}{15}a-\frac{4}{3}b$. Setting the partial derivatives to 0 and solving yields

$a=-\frac{4}{5}$, $b=\frac{16}{15}$

The Hessian is given by
$\left[ \begin {array}{cc} 2/3&1\\\noalign{\medskip}1&2\end {array} \right] $, which has positive eigenvalues, and thus the given a b are a minimum of the function f[/hide]

We can observe that:
\[F(a,b)=\int_0^1(\sqrt{1-x}-ax-b)^2\,dx\]
is a continuos function, always > 0, and $F(a,b)\to +\infty\quad\text{when}\quad a^2+b^2\to +\infty$. Then F(a,b) has a minimum in the points where $\frac{\partial F}{\partial a}=0;\quad \frac{\partial F}{\partial b}=0.$
After is not necessary to render explicit the function since we can do the partial differentiation under integral sign and all is OK.


A basketball player runs down the court, following the path indicated by the vectors A, B, and C in the figure below.
The magnitudes of these three vectors are A = 11.6 m, B = 23.5 m, and C = 6.80 m. And as shown, q1 = 46.2 ° and q2 = 27.8 °. Calculate the magnitude of the net displacement of this player.
What is the direction of the displacement of this player? Give the angle counterclockwise relative to the right.

I will suggest you put the 3 vectors in their component forms, and add the components, then calculate the angle:

(left to right ) = i component
(back to front) = j component

then ~ A : A = (11.6m)i +0j
          B  :  B = -(23.5m)*sin(46,2)i + (23.5)cos(46,3)j
B = -(16.96m)i + (16.23m)j

C: C = sin(27,8)(6.8m)i + cos(27,8)(6.8m)j
C = (3.17m)i + (6.015m)j

A + B + C = (-2.17m)i + (22.23m)j

thus displacement = SQRT([2.17]^2 + [22.23]^2)
= 22.33m
Direction : 180- 84.42 = 95.58 degrees from the right


algebra
polynomial
vector
linear algebra

I'm asked to determine whether the set:

p1 = 1 - x
p2 = 5 + 3x - 2x^2 (x-squared)
p3 = 1 + 3x - x^2

form a linearly independent set in P sub 2. Up until now i've been using vectors, so i'm unsure how to approach. The example in the book jumps from a set of polynomials to a "thus it can be seen..." paragraph without a step in between the solution and the problem.

Any help is appreciated. Thanks in advance.

The vector space of polynomials with degree at most two is isomorphic to $\mathbb{R}^{3}$. The isomorphism comes from choosing a basis for the space of polynomials- $1,x,x^{2}$ is the standard choice. What does your set look like with respect to this basis?


function
algorithm
algebra
polynomial
number theory
prime numbers

$ \phi_m \equal{} 1 \plus{} \sum_{i \equal{} 1}^{m \minus{} 1} A_i$, where $ A_i \equal{} \begin{cases}1 & \quad a_i < a_m \\
0 & \quad a_i > a_m\end{cases}$

Find a closed form for this function without the sigma.

What is the sequence $ \{ a \}$? Also, what happens if $ a_i \equal{} a_m$?

Err...it is the number of terms less than a_m.

TZF please look at the second page of this post.http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228499.
i didnt undrestand it either.

.. where $ a_i$ represented the digits of $ \overline{a_1a_2\cdots a_n}$ .. It just says "$ \phi_m$ is the number of digits to the left of $ a_m$ which are smaller than $ a_m$" [as altheman already stated], and there's no point to attempt to close it.

please i need the closed form,if you can find me it will really help me.

someone atleast tell me if its possible to find a closed form.

I don't think it's possible to close it in a way you want since you need to compare what digit is greater and it can only be done by checking each digit separately. There is no algorithm to do so besides already mentioned "one by one" checking. And even if there were, it would be too complicated to be efficient.

i think this is new to maths, is that why it cant be closed?people havnt solved anything of this sort before?

Try to think about it. You have m-1 comparisons and there is no simpler way to compare 2 numbers but to look at them and see what is greater so what you have is an algorithm with m-1 iterations.
If you want to have computer calculate it, you would write program based on this simple algorithm.

My opinion is that the function is too unpredictable and that it cannot be closed.

What exactly are you expecting a "closed form" for this type of function to look like? It's impossible to have it in terms of $ n$ by itself since nothing about the [i]size[/i] of the set tells you how its elements compare to some other arbitrary value.

You have to look at each element individually. If you're looking at each element individually, you're not closing anything.

It is perfectly legitimate to have a function described without formulas. In this case, it was kind of dumb to try to express "$ \phi_m$ is the number of numbers in $ \{a_i\}$ less than $ m$" with a formula that dilutes the meaning and does not offer any more workability. The best you can do is probably something like $ \phi_m \equal{} \#\{x\in \{a_i\} | x < m\}$, but this isn't more useful or descriptive or enlightening than the word definition.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228499.
in that link if you see the post 24, its in another form but it can be turned to a polynomal.
You are sayin its impossible to find a formula for $ \phi_m$ but in a polynomial is it possible to have a coeficient which cannot be expressed in mathematical form ?
dont get me wrong please i am not arguing, i just want to know more.

It is expressed in a mathematical form already. If I understood correctly, you are claiming that you have some kind of formula that "always works" but where did you derive it from? What is it's usage?

And considering this particular problem of closing formula you are just not listening what everyone say. Please re-read everything. I'm sorry to say this but reading your posts only leads me to conclusion that you are missing some fundamentals of mathematics.

Returning to "the formula". What does it equal to? How could anyone express your function using that polynomial when the formula doesn't equal anything? What you are asking doesn't make any sense. Again, I believe that you are missing some fundamentals of mathematics or you are very poorly expressing your thoughts.

[quote]you are claiming that you have some kind of formula that "always works" but where did you derive it from? What is it's usage? [/quote]
this shows you havnt read that link yet? so u dont know what i am on about.
and you can make that formula a polynomial because its equal to 0

[quote="Flame"]And considering this particular problem of closing formula you are just not listening what everyone say. Please re-read everything. I'm sorry to say this but reading your posts only leads me to conclusion that you are missing some fundamentals of mathematics.[/quote]  I suggest you take a look through past threads by binomial_4eva -- s/he suffers from a strange notion that everything "has a formula," and s/he is resistent to any argument to the contrary.

Lol ,i have a point JBL. everything has a formula dont laugh at me lol
i can prove that everything hs a formula, if they didnt what would maths be for? i can even tell you this that the whole universe can be summarised with one formula. i mean the whole universe is one formula, nothing more.
this was off-toppicish sorry for that.

But seriously if that formula i have found can be turned in to a polynomial and the polynomial is equal to $ 0$ ? that means we will have a polynomial with a coifficient that cannot be written in mathematical form. have you heard of that before? its impossible so there must be a neat formula for that.
the reason why so many people disagree with me is because i dont just apply the maths i know from college and uni ,i question them to find new.

[quote="binomial_4eva"]i can prove that everything hs a formula, if they didnt what would maths be for?[/quote]Unfortunately, this is simply not true.  I would be very interested in seeing said proof.  :lol:

Try the $ n$th digit of $ \pi$ in base $ m$, without referencing $ \pi$ or anything that evaluates to it. Your answer must be in terms of $ m$ and $ n$.

For example, (assuming that $ m\geq 4$), for $ n \equal{} 0$, $ f(m,n) \equal{} 3$, and $ n\geq 1$, $ f(m,n) \equal{} 0$, and $ n < 0$ returns the digit $ \minus{} n$ values after the decimal point, in base $ m$.

For a simpler version, fix $ m \equal{} 2$ and express your answer solely in terms of $ n$.

But I sort of see what you're saying. However, saying that everything has a "formula" is not correct because the word "formula" is not the right word for what you're trying to say.

[quote="binomial_4eva"]by all means please disprove me ? and what thing do you want me to find a formula for ?
do u want to say things like prime numbers?[/quote]If you did have a formula for the $ n^{th}$ prime, you'd be a very rich man.

Those examples are fine,* but they allow the "just because you and I don't know a formula doesn't mean that one doesn't exist" cop-out.  But a simple cardinality argument suffices: a "formula" (a concept that binomial_4eva never defines, and doesn't appear to have a formal (as opposed to intuitive) grasp of) is some finite string of symbols.  The number of formulas is, therefore, countably infinite.  But there are many simple objects (infinite strings of 0s and 1s, say) of which we can prove there are uncountably many.  Thus, not even every 0-1 sequence can have a formula.  Similarly, one can't give a name to every real number, nor give an algorithm to compute every function from $ {\bf N}$ to $ {\bf N}$, nor do lots of other things that sound intuitively plausible.  One of the major conceptual results of (for example) computability theory is that not only can we prove that there are unsolvable problems of this sort, but also that some of those problems are problems we might actually want to solve (e.g., a universal debugger for C++).

* Actually, I don't really believe this: I think neither of those examples is very good.  In particular, it may be possible in both cases to give algorithms that complete the desired task (perhaps in very long amounts of time).  "Existence of an algorithm" is probably the closest formal analogue of the intuitive thing that binomial_4eva is reaching for (largely because we believe the Church-Turing thesis, which says that the notion of algorithm is well-defined and very robust one, while the notion of "formula" in the way b_4e says it is very poorly-defined and non-robust), and there is a very simple algorithm for miyomiyo's problem, and I certainly can't rule out the existence of an algorithm for Yongyi's problem (and neither can Yongyi).**  In other words, in both cases I think "just because we don't know the formula yet ..." really isn't a cop-out, but actually can be made precise and meaningful (though I don't know if b_4e has the mathematical depth to understand why the objection might be valid in these cases but still invalid in general).

** In fact, there is an algorithm (the Bailey-Borwein-Plouffe algorithm) that allows one to compute an arbitrary digit of $ \pi$ in base 16 (and so also in base 2).


I would be greatful if we can have a trigonometry forum on this site. Trigonometry is not the regular courses that is being served on every school but it is very important course to learn and I'm very enjoying it now. Also, I believe that it would help any students on AMC very much because some involves complicate trigonometric ideas.

Sincerely,

Silverfalcon

The Intermediate Forum is a great place to prompt trig discussions.

Okay.

Rep, where are transcrips available? I don't think they're accessible unless you took a class...

P.S. I don't have time to take classes neither do I have money to pay for it..


Solve the system of the equations:
 x+2(y+z)=1 (1)
 y+3(z+x)=2 (2)
 z+4(x+y)=-1 (3)
 where x,y,z are real numbers

[quote="Dr Sonnhard Graubner"]hello, the solution is $ x=-\frac{3}{23},y=-\frac{8}{23},z=\frac{21}{23}$.
Sonnhard.[/quote]
That is correct, but if you posted a solution it would be even better. :wink:

x+2(y+z)=1 (1)
y+3(z+x)=2 (2)
z+4(x+y)=-1 (3)

Multiply (2) with 2

x + 2y + 2z = 1
6x+ 2y + 6z = 4

Subtract...

-5x -4z = -3   or 5x + 4z = 3 --------(4)

again, multiply (2) 4

12x + 4y + 12z = 8
4x + 4y + z = - 1

subtract....

8x + 11z = 9 ----(5)

Multiply (4) with 8 and (5) with 5 then you get

40x + 32 z = 24
40x + 55 z = 45

subtract.....

- 23 z = - 21

z = 21/23

Substitute z = 21/23 into either (4) or (5) to get the value of X

After that you plug the values of x and z into (1), (2) or (3) to get the value of y

You are correct this is the the solution to the problem.Here is my solution:
                                                    Solution(by galois01)

                            x+2(y+z)=1 (1)
                            y+3(z+x)=2 (2)
                           z+4(x+y)=-1 (3)

                          Let x+y+z=k (4)
                          from (4) we get:
                           y+z=k-x
                           z+x=k-y
                           x+y=k-z
 So
 (1)=> x=2k-1
 (2)=> y=(3k-2)/2
 (3)=> z=(4k+1)/3
 with these (4) gives: k=10/23
 Finally we get:
 x=-3/23
 y=-8/23
 z=21/23
 which is the solution of the system

you could always use matrices and gaussian elimination..


1. In a plane , the  angles  of  a  regular  polygon   with  n  sides  add up  to  less  than  n^2  degrees. What  is  the  smallest  possible  value of  n?

2. The  zeros  of  the  polynomial  p(x)  =  x^3 - 6x^2 + ax  + b  are three consecutive  integers.  Find  a  and  b.

.[hide]1.  180(n-2)<n^2

well, looking at it, n must be really close to 180.  If we have n^2 be right between n-2, we will know that n^2>180(n-2), by idea of am-gm.  So by making n really close to 180 and making 180, n, n-2 equally spaced, we see 178 is the smallest value.

2.  y, y+1, y+2 are the 2 roots.  The sum of the roots is 6=y+y+1+y+2, so y is 1, and the 3 roots are 1, 2, 3.  -b will be the product of the roots, which is 6, so b is -6.   a will be the sum of the  product of the roots taken 2 at a time, which is (1*2) + (1*3) + (2*3) = 11, so a is 11.[/hide]


6 doesnt matter!

it itself is a multiple of 2!

i purposely did that..

i see u noticed that so u wouldnt subtract all the numbers divisible by 12..while adding all divisible by 6. because 6=2*3, i meant to do that

right...
thats cool, a loophole for the solver

[quote="usaha"]find the number of integers between 1 and 100,inclusive that are divisible by 2,5,or 6 but not all three[/quote]
[hide]We can ignore the multiples of 6 because all multiples of 2 are multiples of 6. Using PIE, we find that there are $50+20-10=60$ multiples of 2 and/or 5. Since we only want the numbers divisible by 2,5, or 6 but not all three, we must subtract the number of integers that are multiples of 30. So there are $60-3=\boxed{57}.$[/hide]

[hide]Using PIE, I get 57.  Nice trick for saying 6.  I almost did 12.[/hide]

[hide=" :stretcher: "]
use pie.
$60-3=57$
[/hide]

Nice question, I like the trick.
[hide]NOTE: All decimal numbers will be rounded DOWN to the nearest whole number in my work.
$(100/2)+(100/5)+(100/6)=86$. Then we do $(100/10)+(100/30)+(100+6)=29$. $86-29=57$.
ANSWER: $57$.
[/hide]

[quote="pascal_1623"][hide]
10 multiples of 5 are multiples of 2 as well (multiples of 10)
so 50 multiples of 2 + 20 multiples of 5 - 10 = 60.[/hide][/quote]
Don't forget that you have to subtract off multiples of all three, which means multiples of $\mbox{lcm}(2,5,6)=30$.


geometry
3D geometry
pyramid
tetrahedron

I'm trying to make a pyramid with a triangular base.Could someone tell me the details on the angles and stuff...? {It's a regular pyramid that I need to make..}

Do you mean you want to make a regular tetrahedron by folding a paper?

If that's what you meant, you simply need to cut out an equilateral triangle, and then take the midpoints of 3 sides and connect them. You should end up with dividing the original equilateral triangle into 4 small congruent equilateral triangles. Each small equilateral triangle is a face of the tetrahedron when it's folded.


calculus
integration
trigonometry
derivative
real analysis
real analysis unsolved

(1) Evaluate the integral from x = 0 to x = 1 of x^x dx.

(2) Find the indefinite integral of e^arcsin(x) dx (Without Mathematica!)

$I=\int e^{\arcsin x}dx$

With $x = \sin u$

$I = \int e^{u}\cos u \cdot du = \frac{1}{2}(\sin u+\cos u) e^{u}+C = \frac{1}{2}\left(x+\sqrt{1+x^{2}}\right)e^{\arcsin x}+C$

That's nice - the method I had in mind was a double integration by parts:

Write e^arcsin(x) = 1.e^arcsin(x).
Integrate by parts giving -xe^arcsin(x)/sqrt(1-x^2) in the integral term.
Recognise that -x/sqrt(1-x^2) is the exact derivative of sqrt(1-x^2).
Integrating by parts a second time leads to the original integral being retrieved as the sqrt(1-x^2) terms cancel perfectly.

Factorise to obtain the solution given above using the trig substitution.


Just curious, what is your favorite tv show?  I'm interested to know what AoPSers watch.

I watch nothing.  Maybe some Xbox 360 twice or trice a week but the rest is all laptoparized.

I watch:
The News, related news shows, ie (Nightline) once or twice a week
Simpsons if I get bored
The Tonight Show  :D Jay Leno is hilarious.

LOST is an awesome show. :D House is very interesting, too.

Those news things, like Primetime, 20/20 Dateline and stuff like that but the best show is....
House! Haha, Doctor House is awesome.

I almost never watch current TV (Sometimes the Simpsons and Family Guy, but not too often.)  However, I enjoy watching old episodes of Monty Python's Flying Circus and Dilbert.

No particular order:-

Friends

Simpsons

Becker

Just shoot me

Rockstar

Spongebob
Jimmy neutron
Kenannkel
Drakenjosh

Oprah

Desp. housewives

Lost(rarely though,never get the timings right)

The musters definitely! also, Lost in Space is good.

I am addicted to reality shows.

Big Brother and Amazing Race are my two staples. I usually try and watch Survivor, The Apprentice, the RR/RW one, Treasure Hunters is pretty new, America's Got Talent, I watch every reality show :D

Law & Order is pretty good, so is CSI and Numb3rs (only have seen it once)

While I download it because it's not on TV in the US, ANIME.

Too many other shows to list :(

Jeopardy at 5:00
World News at 5:30
Local News at 6:00

Still trying to figure out when Numb3rs is (only saw one episode, and that was on an airplane)

Jeopardy and The Simpsons.  I watch nothing else.

Well, I like 24 and "the Shield", the latter is a very rough and serious show about hard corrupt cop life in LA.

I really enjoy anime (like Rurouni Kenshin and Elfen lied) but Belgian TV doesn't broadcast much of that.

I liked the older Simpsons episodes, but season 8 and later are pretty worthless.

I hate reality TV : Big Brother, Queer eye for the straight guy, Hogan knows best, .... :mad:

[quote="akalra1"]comedy and reality shows

o yea![/quote]

Are you me?That's ALL I watch.

comedy central, monty phyton (cant spell), 3 stooges, the classics, etc


[quote="Beat"][i]Prove that if $a, b, c, d, e\ge 0$ then $\frac{a+b+c+d+e}{5}\ge \sqrt[5]{abcde}+\frac{q}{20}$, where $q=(\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+(\sqrt{c}-\sqrt{d})^2+(\sqrt{d}-\sqrt{e})^2$[/i][/quote]
Since $\frac{a+b+c+d+e}{5}\ge \sqrt[5]{abcde}+\frac{p}{20}$, where
$p=\sum_{sym}(\sqrt{a}-\sqrt{b})^2$ by Cauchy and $p\geq q.$ ;)

Bringing q/20 to the other side and simplifying,get LS:
(a + b + c + d + e + sqrt(ab) + sqrt(bc)+ sqrt(cd)+ sqrt(de)+ sqrt(ea))/10, which by AM-GM is
>=RS.nice result! :)

[quote="Beat"]You prove nothing[/quote]
What do you mean? $p\geq q$ or $\frac{a+b+c+d+e}{5}\ge \sqrt[5]{abcde}+\frac{p}{20}.$ What to explain more in detail? ;)

[quote="Detective"]The generalization is also true.

$\frac{a_1 + a_2 + a_3 + ... + a_n}{n} \geq \sqrt[5]{abcde} + \frac{(\sqrt{a_1} - \sqrt{a_2})^2 + (\sqrt{a_2} - \sqrt{a_3})^2 + (\sqrt{a_3} - \sqrt{a_4})^2 + ... + (\sqrt{a_n} - \sqrt{a_1})^2}{4n}$[/quote]
No, Detective. Your inequality is wrong. ;)
The generalization is a next known and very easy inequality:
If $a_1,a_2,...,a_n$ are non- negative real numbers and $n\in\mathbb N,n>1,$  then
$\frac{a_1+a_2+...+a_n}{n}-\sqrt[n]{a_1a_2...a_n}\geq\frac{1}{n(n-1)}\sum_{sym}(\sqrt{a_1}-\sqrt{a_2})^2,$
which true by Cauchy (AM - GM) only. ;)


Why  Ring test is not reliable in presence of Nitrites,Iodides and Bromides

a test to detect the presence of Nitrates in a mixture/compound.
it is done by mixing the Nitrate containing solution with excess $ \ce{FeSO4}$ and $ \ce{H2SO4}$
the the nitrates get reduced to $ \ce{NO}$ while ferrous is oxidised to ferric salt.
the Ferrous sulphate forms a Brown coloure complex arund the Ring of test Tube namely $ \ce{Fe[SO4 (NO)]}$

Nitrate in presence of bromide n iodide:
Brown ring test cannot be applied since the lib of free halogen wid conc. H2so4 will obscure the brown ring due to nitrate. The solution is therefore warmed with naoh to decompose nh4+ salts. Devarda’s alloy or al powder is then added and solution is warmed. The evolution of nh3 indicates presence of nitrate.
in fact Brown ring test fails in presence of chromates, chlorates,  sulfites, thiosulfites, iodates, cyanides, thiocyanates, hexacyanoferrate(II) and(III). These anions interfere either by complexing wid fe2+ or by redox reactions.they mebbe removed by adding ag2so4. Nitrites also give brown ring test. They are best removed by adding sulfamic acid


I forgot to bubble in the bubble asking if you'll take USAMO if eligible.

Does that mean I won't be able to take USAMO if eligible (pretty sure I qualified), or should I not worry? (I want to take USAMO.)

crap there was a space like that? i got pay attention more. and i think i got a 9 too. =(

I think they'll still let you take the test, even if you didn't bubble the thing in.

I think they may contact you via email or phone to ask you if you want to take the USAMO. So tell your parents or yourself to pay attention to the phone, and check your email in time.

[quote="dogseatcheese"]What if you said that you didn't want to take the USAMO but changed your mind after the test. Will they still ask if you want to take it?[/quote]You'll should contact them ASAP. That way, you should be able to take it for sure.

Okay, I sent them an e-mail, but haven't gotten a reply. I didn't indicate that I didn't want to take the USAMO, but I didn't indicate that I did either.

I can't give you guys the exact link (because my memory just isn't that good), but Mr. Dunbar (who's basically in charge of the AMC) said, on this board, that it didn't matter if you bubbled that you were willing to take the USAMO. If you said you would, they'd just go ahead and put your name on the list. Otherwise, they'll contact you and ask anyway.

I'd say there's no need to worry.

[quote="Phelpedo"]Which might mean that you'll find out if you qualified earlier than anyone else, because they'll have to contact you first.[/quote]

That's one way of looking at it. Or maybe, just to torture you, they'll print the qualifier list first and then contact you.

Here's the e-mail I received from the AMC:

[quote]Leaving this bubble blank would not exclude you from being invited to take
the USAMO if your index score is within the cutoff.[/quote]


On Friday, a snowboard originally priced at $ \$100$ was
discounted $ 50\%$. On Monday, that sale price was reduced by
$ 30\%$. In dollars, what is the price of the snowboard after the
Monday reduction?

Discounted by 50% we would multiply be .5 for the price to be 100(.5) or 50 on Friday.
Discounted again by 30% we would multiply by .7 for the price to be 50(.7) or [b]35 dollars[/b] on Monday.


[quote="mathwizarddude"]1. Find the number of natural numbers from 599 to 999 inclusively that do not contain the digit 7.

2. Find the total number of occurrences of the digit 7 in the previous range of numbers (of example, the number 677 has two occurences of 7, and 777 has 3, and their the total for these two numbers is 3).[/quote]

[b]1)[/b] $ 1\plus{}3\times 9 \times 9\equal{}244$ is the answer.
[b]2)[/b] Appear in hundreds digit : $ 100$ times
in tens digit :$ 4 \times 10\equal{}40$ times
in unit digit : $ 40$ times

So the answer is $ 100\plus{}40\plus{}40\equal{}180$ times.


AMC USA(J)MO USAMO AIME Harvard college MIT

YEs. it started last yr. Too bad i missed it... :(

I missed it as well....grr....if only I wasn't a sophomore :(

Wait, are they accepting the extra 30 freshmen again this year?

that started last year. I think they are intended to do it in the future. IT is encouraging and motivating younger students to try to challenge themselves on problem-solving.

[quote="ben12345_5"]Are all freshmen who qualify for USAMO invited to attend the Math Olympiad Summer Program? :?:[/quote]

Well, "Yes" is probably not quite an accurate answer. The statement was that approximately  30 freshmen qualifying for USAMO will be invited to attend MOP. Of course, last year there was only 25 or so and everyone was invited. Given that this year's AIME content was more accessible to students of lower grades who, on average, have less problem solving experience, it's possible that more than 30 freshmen will qualify. (That's my theory at least.) In that case, I'm not sure what they'd do. It would probably be done by USAMO score or something. And in that case, if you get 1 point you'll probably make MOP. So, my advice to you is submit any reasonable amount of work for any problem. Even disorganized thoughts, if hitting along the lines of the steps to an actual solution, may give you a couple of points.

[quote="DPopov"]:(  :(  :(  :(  :(  sophies have no chances....unless they are super smart and make top 12 or HM.[/quote]

Well, it's more like top 35ish if you consider the international USAMO takers and the senior non-winners. So...plan on solving half of the problems :-)

I think I got like 35th or something my freshman year on the USAMO and I made MOP 2003, which was a 30 person MOP. So you don't have to be Honorable Mention to make it.

Also your index matters (somewhat).

What was your score when you got 35th? Also, my index is a 256.5, although I don't think that really matters too much. From what I've heard 20+ is what you need to get into MOP, is this correct?

MOP qualifying score differs every year. Last year, it was a few points below 20.

And I think darktreb is right about USAMO index mattering. However, your actual USAMO score is a lot more important than your USAMO index.

Yes, your AMC and AIME scores can come back and hurt you.  The formula is AMC+10AIME+9USAMO.  For example, last year I got 22nd place on the USAMO but barely made MOP because of my bad AMC and AIME scores.

377, but that number does nothing to predict what is needed this year; the difficulty of the USAMO and, to a lesser extent, the AIME (which seemed very easy this year) will determine the cutoff.

Don't forget that the USAMO is out of 42.  Honestly, I think they shouldn't even consider the AMC and AIME after someone has made USAMO.  Those scores just don't seem very relevant anymore.

If I purposely fail all my classes this year, will I still be considered a freshman next year? btw, I'm a Freshman this year.

[quote="filletwho"]why's the AIME weighted heavier than the USAMO?[/quote]

It's not. Each USAMO question is worth 7*9 = 63 points while AIME questions are worth 10. So ths USAMO can get u 378 points while u maximize your AIME score at 150.

Basically, the AIME and AMC only help if you and someone else tie on the USAMO or are really close, say 1-2 points off.

What exactly qualifies a fresman for USAMO?

The AIME floor value or if they have a high enough index. Read about it on the AMC's website.

[quote="darktreb"]When considering AIME being "more heavily weighted" than USAMO, you also have to consider that ALL USAMO qualifiers have AIME scores in the range of 8 to 15 or something like that, so it's not really on a scale of 0 - 15, but rather on a scale of 0 to 7 or 8, since everybody got at least a certain number of problems so it really doesn't affect the system.

The same logic applies to the AMC. So in conclusion,

AMC variability: 100 to 150 (50)
AIME variability: 8 (hypothetically) to 15 (70)
USAMO variability: 0 to 42 (378)

That's the "true" MOP index distribution of scores.[/quote]

Yeah, that makes it seem a bit more convincing about USAMO affecting MOP qualification.

[quote="wow346"]Seniors have practically no chance at qualifying for MOP right?[/quote]

No? Seniors have chances that are just as good as sophomores and juniors. If you score in the top 30 or so on USAMO, you qualify for MOP.

This is true, seniors are only invited if they are in the top 12 aka "The Winners"

Why would they accept seniors to MOP if they didn't make top 12? What would be the point?

I have also heard a rumor that starting this year, if you make top 12 as a senior but fail to make the IMO team (which is determined within a week to ten days), you get sent home.

However, that really doesn't make that much sense to me. They won't save space because the number of housing floors used is still the same, and all they really save is the meager meal money for two weeks. I guess they don't want people being a distraction or whatever, but I still think it sounds like a pointless policy.

that would be a pretty harsh thing to do..... i mean what if it was a first-time senior coming to MOP? One-week of MOP COULD make a difference in someone's life.

that makes no sense to me either.

They don't care about making a difference in your life.

 MOP is not exactly like other math camps. First and foremost is the priority of producing an IMO team.

But yes, it would be harsh to send someone home.

[quote]Actually I believe they'll let you stay but demote you to Blue group or something. I know seniors had an option of leaving last year, but no one did.[/quote]

That is my experience as well, but what I'm saying is I heard a rumor that they were going to change the policy this year. I know they have never forced anyone to leave in the past.

At MOP, students are split into groups. Worry about that when you get there. ;-)

MOP is held at the University of Nebraska - Lincoln.

[quote]MOP is held at the University of Nebraska - Lincoln.[/quote]

The food is not bad but man could they please change the location ... Titu and Steve Dunbar etc... go somewhere else like that one year in 2001 or whatever it was.

[quote="darktreb"]The food is not bad but man could they please change the location ... Titu and Steve Dunbar etc... go somewhere else like that one year in 2001 or whatever it was.[/quote]

I'm sure you're aware, but after 3 weeks the food is quite despicable. I agree about relocating. I believe many universities would fight for the opportunity to host such a prestigous program. Personally, I wouldn't mind if Harvard or somewhere like that took on the task. Heck, it doesn't even have to be a university! They could have it at the Four Seasons in New York City! I hear it's quite a view. ^-^

I believe in 01 they had it in D.C. because the US was hosting IMO.

I find the rumor that senior non-winners will be sent home hard to believe.  I think it wouldn't save the AMC money to send them home early, because the AMC would have to book flights home on short notice, and it doesn't cost them much extra to keep them at MOP (they have the rooms already, they don't need to hire extra staff for a couple people, it might be hard to arrange with UNL so as to save money on the meal plans).  Besides, even though seniors who don't win the USAMO could theoretically slack off, my experience has been that they've generally continued to be engaged in the material and helpful and supportive to promising younger students.

[quote="henryyangrui"]All freshman participants in the nation go to one city?[/quote]

Yeah, but that's about 30 people...

[quote="seungjun"]What if you're an eighth grader? I haven't seen any 8th graders in the list of previous MOSP participants...[/quote]

Try the 2002 listing.

8th graders are not included in the "additional 30" selection. At least, they weren't least year.

Henry, you take the  USAMO where you live (with the exception of when MIT hosted). For MOP, all MOPpers (obviously) go to the same place for camp.


There were several problems that I didn't understand.

Compute $S = \frac{1}{5} + \frac{1}{25} + \frac{2}{125} + \frac{3}{625} + \frac{5}{3125} + \ldots$, where each numberator is the sum of the preceding numerators, and each denominator is 5 times the preceding one.

Can you use Phi for this? Or is there some easy method that I missed?

Let C denote the cylinder $x^2+y^2=4$, $0 \le x \le 6$, of radius 2 and height 6. What is the length of the shortest path on the cylinder from the point (2,0,6) to the point (2,0,0) which passes through the point (2,0,2) and passes twice around the cylinder? I got a REALLY weird answer with $\pi$ in the radical and everything.

An equilateral triangle is filled the the max with n rows of congruent circles. What is the limit as n approaches $\infty$ of the ratio of the area of the circles to the area of the triangle?

Simply not enough experience with limits to tackle this, but I'd like an explanation anyways.

Thanks for any help!

Edit: Spelling.

I'm a little foggy on what the problems are asking for #2 and #3 myself (typos?).  For #1, try something kinda similar to the way you calculate a geometric sum (multiply it by *something*).  That should be an at least somewhat useful hint.

Wouldn't number 3 be[hide]1?[/hide] I'm not very comfortable yet with limits, so I'm not sure though.

[quote="white_horse_king88"]Compute $S = \frac{1}{5} + \frac{1}{25} + \frac{2}{125} + \frac{3}{625} + \frac{5}{3125} + \ldots$, where each numberator is the sum of the preceding numerators, and each denominator is 5 times the preceding one.

Can you use Phi for this? Or is there some easy method that I missed?
[/quote]
[hide]Multiply $S$ by 5 to get these:
\[5S=1+\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{5}{5^4}+\cdots\]
\[25S=5+1+\frac{2}{5}+\frac{3}{5^2}+\frac{5}{5^2}+\cdots\]
\[125S=25+5+2+\frac{3}{5}+\frac{5}{5^2}+\cdots\]
Adding the first two, we get
\[30S=7+\frac{3}{5}+\frac{5}{5^2}+\cdots\]
Subtract this from the third to obtain
\[95S=25\Rightarrow S=\frac{5}{19}\][/hide]
This is a cool problem.  :)

AGH. 1 is so SIMPLE. *grimace*

Too bad I didn't get that during the competition.

You just find that in terms of $\frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \ldots$.

And you set that to n, and I got $\frac{1}{5}n(5+S) = S$.

It's late, but that's what I got after some experimenting.

Eh, and $x = \frac{5}{19}$.

[quote="litoxe2718281828"][quote="white_horse_king88"]
Let C denote the culinder $x^2+y^2=4$, $0 \le x \le 6$, of radius 2 and height 6. What is the length of the shortest path on the culinder from the point (2,0,6) to the point (2,0,0) which passes through the point (2,0,2) and passes twice around the cylinder? I got a REALLY weird answer with $\pi$ in the radical and everything.
[/quote]
I also get a nasty answer....
[hide]Roll out the cylinder (right?) onto a xy-plane, placing (2,0,0) on the origin, so that (2,0,6) corresponds to $(8\pi,6)$ (6 is height, $8\pi$ is two circumferences). We know that (2,0,2) has hight 2 (and therefore y-coordinate 2 in this xy-plane) and that it has an x-coordinate of either $0$, $4\pi$, or $8\pi$. We just draw lines now for each case:
Case I: (2,0,2) is $(0,2)$.
\[2+\sqrt{(8\pi)^2+(6-2)^2}=2+4\sqrt{1+4\pi^2}\]
Case II: (2,0,2) is $(8\pi,2)$
Same as above by symmetry.
Case III: (2,0,2) is $(4\pi,2)$
\[\sqrt{(4\pi)^2+2^2}+\sqrt{(4\pi)^2+(6-2)^2}=2\sqrt{4\pi^2+1}+4\sqrt{\pi^2+1}\]
It turns out that Case III is shorter...but this sounds a lot like what you got. (??)[/hide][/quote]

That's exactly what I did and I got that answer also.

But why would an answer like that be on a math competition? Sounds like a bad problem to me.


[quote="madiemae"]I am having a hard time establishing this identity

[u]1 + secθ   = sin²θ[/u]
Sec θ               1 - cosθ

[/u][/quote]

I'm going to go ahead and interpret that as $ \frac{1\plus{}\sec\theta}{\sec\theta}\equal{}\frac{\sin^2\theta}{1\minus{}\cos\theta}$.

[hide]If so, you could simply cross multiply to obtain $ \sin^2\theta\sec\theta\equal{}1\plus{}\sec\theta\minus{}\cos\theta\minus{}1\equal{}\sec\theta\minus{}\cos\theta$.  Dividing both sides by $ \sec\theta$ gives $ \sin^2\theta\equal{}1\minus{}\frac{\cos\theta}{\sec\theta}\equal{}1\minus{}\cos^2\theta$.  This is obviously true as a rearrangement of $ \sin^2\theta\plus{}\cos^2\theta\equal{}1$.[/hide]

[quote="madiemae"]thank you, but how would i get one side to look like the other without cross multipling?[/quote]

[hide]$ \frac {\sin^2\theta}{1 \minus{} \cos\theta} \equal{} \frac {1 \minus{} \cos^2\theta}{1 \minus{} \cos\theta} \equal{} \frac {(1 \minus{} \cos\theta)(1 \plus{} \cos\theta)}{1 \minus{} \cos\theta} \equal{} 1 \plus{} \cos\theta$
$ \equal{} (1 \plus{} \cos\theta)\left(\frac {\sec\theta}{\sec\theta}\right) \equal{} \frac {\sec\theta \plus{} 1}{\sec\theta}$[/hide]


Hi, I'm Richard and new to this forum and competition.
I went to the Rice tournament yesterday and talked to some friends (from LBJ) about the Cody Bowl.
I was wondering if it was too late to join or if anybody knows who I need to contact to see about this.  My friends told me that I might be able to go on free agency and make a biography or something like that if it wasn't too late.

Thanks

For more information about the league, you would need to talk to Cody, mathjoker both on this forum and on AOL Instant Messenger. You'll probably then go into the free agency and be able to participate by Week 10 (since Week 9 rosters have already been sent).

i think im on sunnyvale now, so that's really cool.
thanks for bidding so high on me  :D

anyone else from sunnyvale have aim, so i can ask them some questions about the general format?

Hello from Sunnyvale! My AIM is the same as my user name here.

Hello and welcome to Sunnyvale. My AIM is the same as what you see here as well.

Hi Richard!

My name is Keone, and I'm on Sunnyvale as well.  You can e-mail me at $keone.hon@gmail.com$.

You'll start in week 10, which will hopefully occur in about two weeks.  Sorry to keep you waiting so long - it was unlucky timing, I guess.

Depends on whether there's an expansion or not, perhaps, which I expect for NOML V.  I'm fairly sure that at the least you always keep your core 4.

Yes, my plan is that each team will be allowed to keep four players.  The rest will go in a draft, like last time.  Hopefully I'll have better safeguards in place to keep people out of the draft who aren't going to actually play... but it's always a lot easier for people to say "yeah, I want to play!" than to actually take the few hours per two weeks to do it.

[quote="sheepwarrior"]Wow dallas has a lot of players now....
welcome to the dallas team, young jeong (xxreddevilxx right?) and phillip hu.

and a question for cody or any1 else who knows.. every season, do the teams from the preevious season carry over, or is there a new draft??[/quote]

Am I on? how do I know? and uh... what do I do? I got the individual round sent to me, but I havent' gotten to it now since I'm so busy this week.

Awesome! I'm looking forward to playing Dallas now.

If you were recently drafted, you will NOT be playing in round 9, since the round was in progress already at the time.  Get in touch with your team captain by e-mail and he should be able to help you more.  If you're on Dallas, that's Sameer Deshpande (shake9991@yahoo.com).

hi to the new Dallas members, my name is Jim and i am your teammate. my email adress is professorchesssss@hotmail.com
once again, because week 9 is already in progress, you will start in week 10. The team captain, Sameer, will be sending out a roster before week 10 start to indicate whether you will take A test or B test.

did dallas just add like 3 ppl to their team or what?

yea, we got someone from some other team and signed two free agents...

we'll be dropping like 3 or 4 others: Joshua, Nick, and Zach definitely.


This feels like a stupid question, but does a divide b mean $ a|b$ or $ b|a$?

[quote="diophantient"][quote="NapoleonXIV"]$ a|b$ means $ b$ is multiple of $ a$, or $ b \equal{} k*a$ where $ a,b,k \in N$, or $ b \equiv 0$ $ mod$ $ a$
Example:
$ 2|10$ because $ 2*5 \equal{} 10$[/quote]
That's not what I'm asking, what I'm asking is which does the statement "a divides b" imply?[/quote]
a divides b means "a goes into b evenly: or "b is a multiple of a".  For example, 13 divides 39.

a divides b implies that if you have b cookies, and $ a$ children, then each child will get an equal, integral number of cookies. in other words, a times some integer is b.

so 2|10 means that if you have 10 cookies and 2 children, then each child will get an integral amount of cookies(5 in this case)

also, it means that 2( which is a) times an integer(5 in this case) is 10(b)


Let $ a_1,a_2$ and $ a_3$ be a positive real numbers such that $ a_1a_2a_3\equal{}1$ Prove that :

$ 2(a_1^2\plus{}a_2^2\plus{}a_3^2)\plus{}a_1\plus{}a_2\plus{}a_3 \ge 6\plus{}a_1a_2\plus{}a_2a_3\plus{}a_3a_1$

[quote="Abdek"]Let $ a_1,a_2$ and $ a_3$ be a positive real numbers such that $ a_1a_2a_3 \equal{} 1$ Prove that :

$ 2(a_1^2 \plus{} a_2^2 \plus{} a_3^2) \plus{} a_1 \plus{} a_2 \plus{} a_3 \ge 6 \plus{} a_1a_2 \plus{} a_2a_3 \plus{} a_3a_1$[/quote]
ooo men very very easy indeed.

Just $ \sum a_1^2 \ge \sum a_1a_2$

And $ a_1^2 \plus{} a_2^2 \plus{} a_3^2 \plus{} a_1 \plus{} a_2 \plus{} a_3 \ge 6 \sqrt [6]{a_1^3a_2^3a_3^3}\equal{}6$

done

Dimitris


induction
combinatorics proposed
combinatorics

There is a group of $2n$ people.
Each has no more than $n-1$ enemies.
Prove that they can be sat at the round table that any two neighbours are friends.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=knights&t=21903]Here you go[/url] :).

Dirac's theorem is really helpful ;) !
But can you find more elementary proof (for example using induction) ?
I think Myth was thinking about it...  :)

quite interesting... is there a proof by [b]induction[/b]  ? :roll:


I don't know exactly where to put this, so mods you can move it if you please.

8 queens puzzle: Try to fit 8 queens on an 8 by 8 chess board so that none attack each other

I recently wrote a program that finds all the solutions and it worked perfectly, and I want to get your insight?

How many solutions do you think/know there are?

What are some of those solutions?

I know of one possibility Ill surely send it soon

It seems like when I get to the seventh row, everything falls apart.

Notice when you have two queens as close as possible, if they can not take each other, they must be in the corner of a 2 by 3 rectangle. (or in other words, if either queen is converted to knight, it can immediately take the other one. ) So imagine starting at top right, use a knight to move down and left. the result is:

The only problem with that is the two queens in the corners.

good logic, the only moves the queen cant make are those of knights ....

Yes, there are some more.  Maybe a little more than some

can you compute how many combinatorially? i tried but I dont think so... the fact that each gets its own diagonal as well as row/column makes it hard


geometry
circumcircle
incenter
geometry unsolved

Let circle O be the circumcircle of triangle ABC. Circle O1 touches the sides AB,AC, touches the circle O internally at D. Circle O2 touches the extensions of sides AB,AC, touches the circle O externally at E. the bisector of ∠BAC intersects again circle O at M. MD intersects again circle O1 at P. ME intersects again circle O2 at Q. Prove that (1)  line PQ is common tangent of circle O1 and circle O2. (2) PQ∥BC

$ D$ is external similarity center of the circumcircle $ (O)$ and A-mixtilinear incircle $ (O_1)$ $ \Longrightarrow$ tangent to $ (O_1)$ at $ P$ is parallel to $ BC.$ Similarly, $ E$ is the internal similarity center of $ (O)$ and the A-mixtilinear excircle $ (O_2)$ $ \Longrightarrow$ tangent to $ (O_2)$ at $ Q$ is also parallel to $ BC.$ If $ X_1, Y_1$ are tangency points of $ (O_1)$ with $ AB, AC$ and $ X_2, Y_2$ tangency points of $ (O_2)$ with $ AB, AC,$ then the incenter $ I$ and A-excenter $ I_a$ of $ \triangle ABC$ are midpoints of $ X_1Y_2, X_2Y_2.$ Therefore, the circle pair $ (O_1), (O_2)$ is centrally similar to the circle pair $ (I), (I_a)$ with similarity center $ A.$  Let $ T, T_a$ be tangency points of $ (I), (I_a)$ with $ BC.$ It follows that $ A,T,P$ are collinear, $ A,T_a,Q$ are collinear, and $ PQ \parallel BC$ is the common internal tangent of $ (O_1), (O_2).$


geometry
circumcircle
geometric transformation
rotation
symmetry
homothety
function

time for me to propose a geometry problem  :D
I discovered this while playing around the fermats point

construct equilateral triangles A'BC,B'AC, C'AB on any triangle ABC
extend BB' to meet circumcircle of A'B'C' at B''
show the rotation of B'' 60 degrees about C' will have the image point B''' on AA'.

and obviously by symmetry we can do it for all the other points

:D  :D  :D

You've found a particular thing. A much more general result is true:

If we take a point B" on BB' and rotate it 60 deg about C' we get a point B''' which is on AA':

Points A,B''' are obtained from points B,B'' respectively by a similitude (rotation composed with a homothety) of angle 60 about center C'. This means that the angle between the lines BB'' and AB''' must be 60 deg. But it's well-known that the angle (the one <=Pi/2) between BB' and AA' is 60 deg, so B''' must be on AA'. So you see, we don't need B'' to be on the circumcircle of A'B'C'.

You could propose it for a contest or something. It looks kind of scary (I don't know any problems or nice properties which involve the circumcircle of A'B'C' in this configuration) :D

another thing, rotate B'' at center C' 60 degrees, C'' at center A' 60 degrees and A'' at center B' 60 degrees to get points B''', C''', A'''.  show its equilateral  can you prove that?   :D  :D

Now this might be more interesting... :D

the lengths AA'', BB'', CC'' can be expressed as some symmetrical function in terms of FA, FB, FC.  hence the same goes for UC', VA', WB' as the same symmetrical functions in terms of FA, FB, FC
however, since F is not the middle of the chord AA', BB', CC', it follows that FU, FV, FW are different from FA, FB, FC.  although UC', VA', WB' is equilvalent to AA'', BB'', CC'' respectively

It's easy to see that triangle A'''B'''C''' equilateral iff one of the relations FA'''=FB'''+FC''' (*) or the analogous ones is satisfied. Let's assume it's (*) (these relations come from ptolemy for the quadrilateral FC'''B'''A'''). If we write
FB'''+FB''=FC'
FA'''+FB'=FA''
FC'''+FA'=FC'' (obtaind just like (*))we can see that what we want is equivalent to FA'+FB'+FC'=FA''+FB''+FC'' (**). Bugzpodder sent me a PM saying that the relation FA+FB+FC=AA''+BB''+CC'' is known. It's easy to see that this is equivalent to (**) by using FA+FB=FC' and the other similar ones. I have a proof for this (because it wasn't enough for me just taking what bugz said for granted  :D ), but I'll post it some other time (no time right now).

I have my own proof for this  :D
I will post this proof when i get time do do it

grobber, i have to ask you to hold your proof until end of the week because of a due date.  okay?  thanks!

Ok, Gotcha! I wasn't going to post it b4 the weekend anyway.

consider A'''B'''C''' midpoints of A'A'', B'B'', C'C''.  they form a right triangles with diameter OF, where O is the center of the circle and F is the fermat point.  hence they lie on a cycle.  then we can show that the midpoints form an equilateral triangle, which by ptolemy, A'''F=B'''F+C'''F (or some permutations of it)  and hence we have A'A''+B'B''+C'C''=2(A'F+B'F+C'F) and everything readily follows.

i do not know how to prove this:

suppose we pick U,V,W such that AA''=VA', BB''=WB', CC''=UC'
then if you construct equilateral triangles outward on the sides of UVW, they will always coincide with A'', B'', C''
conversely
there exists an unique U, such that U lies CC', such that V,W can be obtained by rotating U 60 degrees about center B'' and A''. furtherefore, V, W are on the line AA', BB' respectively, and the rotation of V about center C' 60 degrees coincides with W.

this transformation has the property that AA''=VA', BB''=WB', CC''=UC'

I think the first thig you said (not the converse) can be proved like this:

We can prove the affirmation for C'' and the triangle constructed on VW. The others are proven similarly. I considered the case W between F and B' and V between F and A' (I have noidea what happens if that's not the case, I don't even know if it's possible for it to be otherwise;  :D I'm kind of in a hurry):

Since FC'=FA+FB we get FA'+FB'+FC'=FA'+FB'+FA+FB=>FC''+FA''-FA+FB''-FB=FA'+FB' so FA'+FB'=FC''+BB''+AA'' so FB'-BB''+FA'-AA''=FC'' so FC''=FB'-WB'+FA'-VA' so FC''=FV+FW. Since WFC''=C"FV=60 and FC''=FV+FW it means that VWC'' is equilateral. This is what you're trying to prove, if I'm not mistaken. The same for UWA''and UVB''. I think it can be proved similarly for all other situations (when V isn't between F and A', so FV isn't FA'-VA' anymore), but I'm not sure. Anyway, this is the main idea.

I just wanted to add this so it's not too confusing: In the previous post I used the fact that FA'+FB'+FC'=FA''+FB''+FC'', although I haven't posted my proof for that yet. It's not ugly, it involves some constructions but it would take some time to type it (I'm a slow typist :D ) and I didn't have the necessary time during the past weekend...

i was also in a rush i should make some amendments here:
indeed consider again the midpoints of A'A'', ...,..., to be X,Y,Z , (so no conflicts) we have shown that XF=YF+ZF (from the fact that XYZ is equilateral)
and A'A''+B'B''+C'C''=2(A'X+B'Y+C'Z)=2(A'X+B'Y+C'Z+(XF-YF-ZF))=2A'F+2B'F+2C'F

A'A''+B'B''+C'C''=2(A''X+B''Y+C''Z)=2(A''X+B''Y+C''Z-(XF-YF-ZF))=2A''F+2B''F+2C''F
which yields A'F+B'F+C'F=A''F+B''F+C''F


Concursul " La scoala cu ceas" este preconizat a se desfasura in perioada 27-29.03.2009.


Solve in natural numbers:

$5(xy+yz+zx)=4xyz$

If $z=1$, we get $(x+5)(y+5)=25$, which has no solutions in natural numbers. So none of the numbers can be $1$.

But then $x,y,z\ge2\Rightarrow0\le\frac{1}{x},\frac{1}{y},\frac{1}{z}\le\frac{1}{2}\Rightarrow 4\ge4-\frac{1}{x}-\frac{1}{y}\ge3$.  But $z=\frac{5}{4-\frac{1}{x}-\frac{1}{y}}\Rightarrow \frac{5}{4}\le z \le \frac{5}{3}$, which has no solutions in natural numbers.

This solution is wrong somewhere, the solutions are $(2,4,20)$ and $(2,5,10)$.

Oops.  :blush:

Yeah, I left out a five.  It should be $z=\frac{5}{4-5(\frac{1}{x}+\frac{1}{y})}$.

I think I can still salvage some shred of my dignity though... since $z\ge2$, we get $\frac{5}{2}\ge4-5(\frac{x+y}{xy}$ $\Rightarrow 3xy-10(x+y)\le0$ or $(3x-10)(3y-10)\le100$.

So at least one of the factors (say $x$) must be at most $10$, so $x\le\frac{20}{3}\Rightarrow x\le6$.

With that bound, you can just consider the different cases of the equation $((4x-5)y-5x)((4x-5)z-5x)=25x^2$.

It's not quite as elegant as my first one would have been... but I hope this one has the advantage of being correct.


function
calculus
algebra unsolved
algebra

hi everybody,
i have a problem which i want to discuss with all memberlist in our forum.
here the problem:
Let f be a function from N to N (N:nature number) satify:
$ (f(2n)\plus{}f(2n\plus{}1)\plus{}1)(f(2n\plus{}1)\minus{}f(2n)\minus{}1)\equal{}3(1\plus{}2f(n)), f(2n) >\equal{}f(n) , \forall n \in N$
denote set: $ M: \equal{}\{m\in f(N): m<\equal{}2007\}$
question: you can find all elements of M?
thanks to help me  solve this problem!

Not calculus. Moved to a more appropriate forum.

hi,
you can help me, give me link to forum approximate calculus

[quote="vuthailuan"]hi everybody,
i have a problem which i want to discuss with all memberlist in our forum.
here the problem:
Let f be a function from N to N (N:nature number) satify:
$ (f(2n) \plus{} f(2n \plus{} 1) \plus{} 1)(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1) \equal{} 3(1 \plus{} 2f(n)), f(2n) > \equal{} f(n) , \forall n \in N$
denote set: $ M: \equal{} \{m\in f(N): m < \equal{} 2007\}$
question: you can find all elements of M?
thanks to help me  solve this problem![/quote]

(I suppose $ 0\notin\mathbb{N}$ and so that $ f(0)$ is meaningless)

Then, according to me, it is impossible to find all elements of M since a lot of different functions are solutions :

$ (f(2n) \plus{} f(2n \plus{} 1) \plus{} 1)(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1) \equal{} 3(1 \plus{} 2f(n))$ implies $ f(2n\plus{}1)\minus{}f(2n)\geq 2$
But, if $ f(2n\plus{}1)\minus{}f(2n)\geq 4$, then LHS $ \geq (2f(2n)\plus{}5)3 \geq 6f(n)\plus{}15 >6f(n)\plus{}3\equal{}RHS$, which is impossible.

So $ 4>f(2n\plus{}1)\minus{}f(2n)\geq 2$
Then :

If $ f(2n\plus{}1)\minus{}f(2n)\equal{}2$, we have $ (2f(2n) \plus{}  3)(2 \minus{} 1) \equal{} 3(1 \plus{} 2f(n))$ and so $ f(2n)\equal{}3f(n)$ and $ f(2n\plus{}1)\equal{}3f(n)\plus{}2$

If $ f(2n\plus{}1)\minus{}f(2n)\equal{}3$, we have $ (2f(2n) \plus{}  4)(3 \minus{} 1) \equal{} 3(1 \plus{} 2f(n))$ which is impossible since LHS is even and RHS is odd.

And so we can arbitrarily build a lot of solutions :

$ 1)$ Choose $ f(1)\equal{}$ any positive integer
$ 2)$ Use rules $ f(2n)\equal{}3f(n)$ and $ f(2n\plus{}1)\equal{}3f(n)\plus{}2$ in order to build all other values.

It's easy to see that $ f(n)$ is obtained by taking the binary representation of $ n$, replacing all ones but the leftmost by twos, replacing the leftmost one by the ternary (base_3) representation of $ f(1)$ and considering the result as the base_3 representation of $ f(n)$.

Example :
$ f(1)\equal{}17$ (and the base_3 representation of 17 is $ B_3122$
$ f(2)\equal{}51$
$ f(3)\equal{}53$
$ f(4)\equal{}153$
$ f(5)\equal{}155$
...
$ f(13)\equal{}f(B_21101)\equal{}B_3122202\equal{}479$

So,
if $ f(1)\equal{}3000$, M is emptyset
if $ f(1)\equal{}2000$, M=$ \{2000\}$
if $ f(1)\equal{}1000$, M=$ \{1000,2000,2002\}$
and so on

thanks pco very much!
but in this problem i denote N is nature number so 0 \in N and it's easy see $ f(0)\equal{}0$ from $ f(2n)\equal{}3f(n)$ then $ f(1)\equal{}2$
and i can prove that the sequence $ f(n)$ is strict increase, so since $ f(107)\equal{}2006,f(108)>2007$ we can find all elements of M following
$ M\equal{}\{f(0),f(1),....,f(107)\}$
but i think your suppose with condition 0 is not in M well.
Luan.


case 1.Consider a circuit in which there is a 2 ohms resistor connected to a 12v voltage source .

by looking at the circuit we can say the potential drop across 2 ohms resistor is 12v

case 2.Now consider a circuit in which 2 ohms resistor and another 4 ohms resistor which are connected in series, and in turn

they are connected to  the same voltage source

                                         ---2 ohms----4 ohms----
                                         |                                                              |
                                         |                                                              |
                                         --------(+)12v(-)---------

                   (hehe dont laugh at my diag.) :D

ok,now if we solve , the the potential drop across the 2ohms resistor is 4v and across 4 ohms resistor is 8v.

potential drop is  defined as the work done in moving the electrons over the resistor[e.g:-for 2 ohms resistor for current to

pass  through it requires 4v equivalent work(am i right?)].

as the current goes from + to -ve ,that means electrons move from -ve to +ve ) that means the electrons first move through

the 4 ohms resistor and then  towards 2ohms,ok?,so now my question is why cant it use total 12 v work equivalent for 4 ohms

instead of using 8 ohms,i mean as the electrons flow from 4 ohms resistor,they still dont know that there exists a 2ohms

resistor(because they didnt travel yet)and instead  why dont they consider it as a case 1 and use total 12v for 4 ohms itself,

i dont understand how and what will happen after the  resitors are joined in series and then after they're connected to the

source(please some one explain the mechnism involved ,is there any good book  which explains the internal theory of a

circuit) ,my question seems to sound ridiculous  :huh: ,but please some  explain,as i am stuck with it for a long time.

hi dude
I hope u have gone through Quantum Physics.
there was a famous experiment some years ago. Scientists had send a single electron in a two way junction .
They found the presence of that single electron in both the path at the same time.
More surprising thing happened when a detector was placed in any path ,it was observe that wave probability function was in minimum there.
That's why ur comment that  electron do not know that there is a 2 ohm resistor after 4 ohm resistor  is actually wrong.
Experiments like Schrodinger's cat,photons passing through an infinitely long wall etc will hopefully tell u nature of electrons and about wave probability function.

Tell me if still u have some doubt.
Pardon me if u actually know this.

I don't know about that. But i have read something similar wherein an electron gun is used as an electron source in a sort of [url=http://en.wikipedia.org/wiki/Young%27s_double_slit]Young's double slit[/url] setup and the probability wave changes when a detector is placed near one of the slits.

Is it the same one which Feynman talks about?

very mauch thanks kabi, i actually dont know about that experiment,it's very good to understand ,can u suggest any book

which explains more about two way path,can you give me a link or elaborate abou photons travelling in an infinite wall

and the detector and wave probability function,by the way which country u are from and what field did you study? :)

i am very much interested in quantum mechanics

Hello
I m glad that u r interested in quantum mechanics.It is always imp.for physics to generate interest.
Well I had read about these things two years ago.That time I was totally crazy about it.I remembered when Newton read about Schrodinger equation and all other development done in this field ,he said that if a person is not shocked after reading it means that he didn't understand any thing.
Even the father of quantum Schrodinger(according to me) said that he was upset that he had done something in this field as he too was doubtful about this.
U have asked for some links::
*http://en.wikipedia.org/wiki/Schrodinger%27s_cat
*some videos:
 http://video.google.com/videosearch?q=experiments+in+quantum+mechanics&ie=utf-8&oe=utf-8&rls=org.mozilla:en-US:official&client=firefox-a&um=1&sa=X&oi=video_result_group&resnum=5&ct=title#
(remember do it in browser FIREFOX.)
*

I would suggest u clear ur doubts about dual nature of electrons.

right now I m busy in finding whats going in the field of nanotech .

well here is one experiment which I had seen few months ago.
http://www.fourmilab.ch/autofile/www/section2_84_14.html

if u are interested in nano too have a look on experiments done by IBM.
 there Scientist r just brilliant.
if u want more info pm me.I know some physics site which are having much more info.

And ya as asked something about me.I m sorry I don't like to write about me but I could tell u that I m much smaller than what u r thinking.
SAYONARA

hehe :) ok no problem,is newton younger or born after schrodinger?that's nice

IBM,i thought it only deals with computing and only with computers,it does phsics also!! cool

how come these guys  develop them itself in 86 and now i am reading it ,lol,really great.

whenever i remember any cat , i remember a experiment where cat is stored in a box and high Em waves are involved

by which the cat disappears inside the box , same as philadelphia erxperiment done by tesla and einstein?

Pardon me.
Newton died before Schrodinger birth.
but as my memory is not like that of goldfish I m pretty sure that he said that.
May he had given this statement as work on quantum stated on late 17th century.

Yes IBM is all about IT sector but u would also admit that how nanotech has come in this field as nano is on priority .

[quote="bkunchanapalli"]very mauch thanks kabi, i actually dont know about that experiment,it's very good to understand ,can u suggest any book

which explains more about two way path,can you give me a link or elaborate abou photons travelling in an infinite wall

and the detector and wave probability function,by the way which country u are from and what field did you study? Smile

i am very much interested in quantum mechanics
[/quote]
[b][size=150]I Hate Quantum Physics.[/size][/b]
$ \mathfrak{ \minus{} Albert \; Einstein}$,
$ \mathfrak{ \minus{} Me \; too.}$

Einstein was opposed to the Dual Particle Nature theory (the Heisenberg's stuff) as well as to the claims that the Universe is Probabilistic. He was a strong advocate of the Universe being Deterministic, ie: he believed everything is pre-determined. He put forth three challenges to Quantum Mechanics and the Heisenberg's Principle, the most popular being the EPR ([url=http://en.wikipedia.org/wiki/EPR_paradox]Einstein-Polodosky-Rosen[/url]) Paradox (if anyone's interested in the paradox, a thought experiment unsolved as yet, click on the link given in the previous bracket. Its pretty interesting!).

Another reason why Einstein was opposed to the Dual Particle Nature theory might be that (atleast I feel so) Heisenberg was Hitler's supporter and was actually helping the Nazis make the [url=http://www.stormfront.org/forum/showthread.php?t=189925]Atomic Bomb[/url] for use against the allies.

[hide="Disclaimer:"]What I've written here may be wrong. The past is more shrouded that the future![/hide]

Once you connect a circuit like that with resistors to the battery, it is definitely causing an imbalance right away.
There is electron movement already at the positive terminal of the battery. This happens to bring the balance the charges.
So,  if you imagine the circuit linearly,
|+[-+][-+][-+]^^^^[-+][-+][-+]^^^^[-+][-+][-+]-|

[] = part of the wire or a finite element of the wire.
Of course, both ends or terminals +||-are part of the battery and are connected.
So, there are differences in potential all along the circuit, and those differences are accounted by the established potentials at those points.
Now, if you want to move an electron over that path, equivalent work comes into account.
Is this a little clear ?


Let $ a$ and $ b$ be real numbers such that $ a\plus{}b\equal{}\minus{}1$.  Compute

$ a^2\minus{}b^2\plus{}a\minus{}b$.

$ a^2\minus{}b^2\plus{}a\minus{}b \equal{} (a\minus{}b)(a\plus{}b\plus{}1) \equal{} \boxed{0}$.


Solve this equation in $ R$:

$ \left (\sqrt{5x^3\minus{}4}\plus{}2\sqrt[3]{7x^2\minus{}1} \right)(x\plus{}1)\equal{}6(x\plus{}4)$

$ \sqrt{5x^3\minus{}4}\plus{}2\sqrt[3]{7x^2\minus{}1}\minus{}\frac{6(x\plus{}4)}{x\plus{}1}\equal{}0$
$ LHS$ is a monotone function hence  $ x\equal{}2$ is unique root of this equation


trigonometry
calculus
derivative
function
integration
calculus computations

Solve $ \frac{d^2y}{dx^2}\minus{}2\frac{dy}{dx}\plus{}5y\equal{}e^x\cos 2x$.

Let $ y(x)\equal{}e^{x}(A\cos{2x}\plus{}B\sin{2x})$.  Then $ y'(x)\equal{}e^{x}(\minus{}2A\sin{2x}\plus{}2B\cos{x})\plus{}e^{x}(A\cos{2x}\plus{}B\sin{2x})$
$ \equal{}e^{x}((A\plus{}2B)\cos{2x}\plus{}(B\minus{}2A)\sin{2x})$
and
$ y''(x)\equal{}e^{x}((\minus{}4A\minus{}3B)\sin{2x}\plus{}(4B\minus{}3A)\cos{2x})$

Then just plug in an match coefficients, assuming I got the derivatives correct.

First, solve the corresponding homogenous equation
$ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \minus{}2\dfrac{\mathrm{d}y}{\mathrm{d}x}\plus{}5y\equal{}0$
The auxiliary equation is
$ \lambda^2 \minus{}2\lambda\plus{}5\equal{}0$
with roots $ \lambda \equal{} 1\pm 2i$. Thus, the general solution for the homogenous equation is
$ y \equal{} e^x(C_1 e^{2ix}\plus{}C_2 e^{\minus{}2ix})\equal{}C_1y_1\plus{}C_2y_2\qquad(1)$
with $ y_1 \equal{} e^{x\plus{}2ix}$ and $ y_2\equal{}e^{x\minus{}2ix}$.
To obtain the general equation of the non-homogenous equation, we use the method of variation of the arbitrary constants. For this purpose, we suppose $ C_1$ and $ C_2$ to be functions of $ x$, and select them so as to satisfy the following two equations
$ C_1^\prime y_1 \plus{} C_2^\prime y_2 \equal{}0,\qquad C_1^\prime y_1^\prime \plus{} C_2^\prime y_2^\prime \equal{} e^x\cos2x$
We have $ y_1^\prime \equal{} (1\plus{}2i)e^{x\plus{}2ix}$, $ y_2^\prime \equal{} (1\minus{}2i)e^{x\minus{}2ix}$. Thus, the above set of equation becomes
$ C_1^\prime e^{2ix} \plus{} C_2^\prime e^{\minus{}2ix}\equal{}0,\qquad (1\plus{}2i)C_1^\prime e^{2ix} \plus{} (1\minus{}2i)C_2^\prime e^{\minus{}2ix}\equal{}\cos2x$
Solving these two equations for $ C_1^\prime$ and $ C_2^\prime$, we get
$ C_1^\prime \equal{} \minus{}\dfrac{1}{4}\,ie^{\minus{}2ix}\cos2x$ and $ C_2^\prime \equal{} \dfrac{1}{4}\,ie^{2ix}\cos2x$
from where after integration, we get
$ C_1 \equal{} \minus{}\dfrac{i}{8}\left(\dfrac{i}{4}e^{\minus{}4ix}\plus{}x\right)\plus{}A_1$, and $ C_2 \equal{} \dfrac{i}{8}\left(\minus{}\dfrac{i}{4}e^{4ix}\plus{}x\right)\plus{}A_2$
Plugging in these expressions in $ (1)$ and simplifying, we obtain
$ y\equal{} e^x ( A_2\cos2x \plus{}A_1\sin 2x) \plus{} \dfrac{e^x}{16}(\cos2x\cos4x\plus{}\sin2x\sin4x\plus{}4x\sin2x)$

[quote="JRav"]Let $ y(x) \equal{} e^{x}(A\cos{2x} \plus{} B\sin{2x})$.  Then $ y'(x) \equal{} e^{x}( \minus{} 2A\sin{2x} \plus{} 2B\cos{x}) \plus{} e^{x}(A\cos{2x} \plus{} B\sin{2x})$
$ \equal{} e^{x}((A \plus{} 2B)\cos{2x} \plus{} (B \minus{} 2A)\sin{2x})$
and
$ y''(x) \equal{} e^{x}(( \minus{} 4A \minus{} 3B)\sin{2x} \plus{} (4B \minus{} 3A)\cos{2x})$

Then just plug in an match coefficients, assuming I got the derivatives correct.[/quote]

[quote="kabi"]Hi kunny I have tried this problem there is one problem in JRav hint
The roots of the characteristic equation are 1+2i and 1-2i which r complex.
y can be imagined to be of form e^x(A sin(2x) + B cos(2x) ).
when we are putting the value of y y' y'' the coefficient of cos(2x) and sin (2x) are itself becoming zero.
Dude kindly try it and tell where I m doing mistake?[/quote]

In this case, the right hand of the equation has the same as the solution of the homogeneous form, so  $ y(x) \equal{} e^{x}(A\cos{2x} \plus{} B\sin{2x})$ is not a particular solution. Judging from getting my solution, the particular solution seems to be $ y(x) \equal{} x e^{x}(A\cos{2x} \plus{} B\sin{2x})$, although I can't expalinl you the reason.

The situation where the particular and homogeneous solutions coincide and one must multiply by $ x$ is referred to in physical contexts as critical damping.  For a mathematical explanation I would prefer to direct you to the [url=http://en.wikipedia.org/wiki/Laplace_transform]Laplace transform[/url], which does away with all this nonsense about varying parameters.

[quote="kaymant"]First, solve the corresponding homogenous equation
$ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \minus{} 2\dfrac{\mathrm{d}y}{\mathrm{d}x} \plus{} 5y \equal{} 0$
The auxiliary equation is
$ \lambda^2 \minus{} 2\lambda \plus{} 5 \equal{} 0$
with roots $ \lambda \equal{} 1\pm 2i$. Thus, the general solution for the homogenous equation is
$ y \equal{} e^x(C_1 e^{2ix} \plus{} C_2 e^{ \minus{} 2ix}) \equal{} C_1y_1 \plus{} C_2y_2\qquad(1)$
with $ y_1 \equal{} e^{x \plus{} 2ix}$ and $ y_2 \equal{} e^{x \minus{} 2ix}$.
To obtain the general equation of the non-homogenous equation, we use the method of variation of the arbitrary constants. For this purpose, we suppose $ C_1$ and $ C_2$ to be functions of $ x$, and select them so as to satisfy the following two equations
$ C_1^\prime y_1 \plus{} C_2^\prime y_2 \equal{} 0,\qquad C_1^\prime y_1^\prime \plus{} C_2^\prime y_2^\prime \equal{} e^x\cos2x$
We have $ y_1^\prime \equal{} (1 \plus{} 2i)e^{x \plus{} 2ix}$, $ y_2^\prime \equal{} (1 \minus{} 2i)e^{x \minus{} 2ix}$. Thus, the above set of equation becomes
$ C_1^\prime e^{2ix} \plus{} C_2^\prime e^{ \minus{} 2ix} \equal{} 0,\qquad (1 \plus{} 2i)C_1^\prime e^{2ix} \plus{} (1 \minus{} 2i)C_2^\prime e^{ \minus{} 2ix} \equal{} \cos2x$
Solving these two equations for $ C_1^\prime$ and $ C_2^\prime$, we get
$ C_1^\prime \equal{} \minus{} \dfrac{1}{4}\,ie^{ \minus{} 2ix}\cos2x$ and $ C_2^\prime \equal{} \dfrac{1}{4}\,ie^{2ix}\cos2x$
from where after integration, we get
$ C_1 \equal{} \minus{} \dfrac{i}{8}\left(\dfrac{i}{4}e^{ \minus{} 4ix} \plus{} x\right) \plus{} A_1$, and $ C_2 \equal{} \dfrac{i}{8}\left( \minus{} \dfrac{i}{4}e^{4ix} \plus{} x\right) \plus{} A_2$
Plugging in these expressions in $ (1)$ and simplifying, we obtain
$ y \equal{} e^x ( A_2\cos2x \plus{} A_1\sin 2x) \plus{} \dfrac{e^x}{16}(\cos2x\cos4x \plus{} \sin2x\sin4x \plus{} 4x\sin2x)$[/quote]

That's the correct answer. :)

Since you can summarize $ \cos2x\cos4x \plus{} \sin2x\sin4x \equal{}\cos 2x$, so the solution can be written as

$ y\equal{}\frac{1}{4}xe^x\sin 2x\plus{}C_1e^x\cos 2x\plus{}C_2e^x\sin 2x$, where $ C_1\equal{}A_2\plus{}\frac{1}{16},\ C_2\equal{}A_1$.

Of course, the end result is nicer in form now. And yes, Laplace transformations would have solved it readily.

$ (D^2 \minus{} 2D \plus{} 5)y \equal{} e^x\cos 2x\Longrightarrow y_p \equal{} Re {\frac {1}{\{D \minus{} (1 \minus{} 2i)\}\{D \minus{} (1 \plus{} 2i)\}}e^{(1 \plus{} 2i)x}}$

$ \equal{} Re \frac {1}{D \minus{} 1 \minus{} 2i}\cdot \frac {e^{(1 \plus{} 2i)x}}{D \minus{} 1 \plus{} 2i} \equal{} Re \frac {1}{D \minus{} 1 \minus{} 2i}\cdot \frac {e^{(1 \plus{} 2i)x}}{1\plus{}2i\minus{}(1\minus{}2i)}$

$ \equal{} Re \frac {1}{4i}\cdot xe^{(1 \plus{} 2i)x} \equal{} \frac {1}{4}xe^x\sin 2x$.

$ \therefore y \equal{} \boxed{e^x(C_1\cos 2x \plus{} C_2\sin 2x) \plus{} \frac {1}{4}xe^x\sin 2x}$

hello, i have found the following solution
$ y(x)\equal{}e^x(C_1\cos(2x)\plus{}C_2\sin(2x))\plus{}\frac{1}{8}(\cos(2x)\plus{}2x\sin(2x))$.
Sonnhard.