geometry
3D geometry
tetrahedron
IMO Shortlist
linear algebra
algebra
Vectors

Let $a_{ij}$ $i=1,2,3$; $j=1,2,3$ be real numbers such that $a_{ij}$ is positive for $i=j$ and negative for $i\neq j$.

Prove the existence of positive real numbers $c_{1}$, $c_{2}$, $c_{3}$ such that the numbers \[a_{11}c_{1}+a_{12}c_{2}+a_{13}c_{3},\qquad a_{21}c_{1}+a_{22}c_{2}+a_{23}c_{3},\qquad a_{31}c_{1}+a_{32}c_{2}+a_{33}c_{3}\] are either all negative, all positive, or all zero.

[i]Proposed by Kiran Kedlaya, USA[/i]

I don't know if this works or not, a lot of it is intuitive, but I think there might be some good ideas in there.

First of all, if the vectors $\vec {u_i}=(a_{i1},\ a_{i2},\ a_{i3})$ are coplanar, then it's easy to show that all the three components of the non-zero solution of the system $a_{i1}c_i+a_{i2}c_2+a_{i3}c_3=0$ have the same sign. In this case we find positive $c_i$ s.t. all of those expressions are equal to $0$. We may thus assume that the three vectors are not coplanar (or, in other words, we may assume they're independent).

We have to find three positive numbers $t_i$ s.t. all the components of the solution of the system $a_{i1}c_1+a_{i2}c_2+a_{i3}c_3=t_i$ have the same sign. In other words, we have to find positive $t_i$ s.t. the three determinants obtained from $\left |\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{array}\right |$ by replacing a column with $\left (\begin{array}{c}t_1\\ t_2\\ t_3\end{array} \right )$ have the same sign.

This translates to the point $T(t_1,t_2,t_3)$ being on the interior of the trihedron determined by the lines $OP_1,OP_2,OP_3$, where $P_i(a_{1i},a_{2i},a_{3i}$. It really isn't hard to show that such a point $T$ exists (by "trihedron" I understand the figure formed by three lines in space, not three rays, as we usually understand). This is the part I'm not convinced about. It's related to the fact that the oriented volume of a tetrahedron which has $O$ as a vertex depends on the direction in which we consider the three vertices different from $O$ (clockwise/anticlockwise direction).

Opinions?

In the official solution (and maybe grobber's reply), it says that it suffices to show that if we set $ O(0,0,0)$, $ P(a_{11},a_{21},a_{31})$, $ Q(a_{21},a_{22},a_{23})$, $ R(a_{31},a_{32},a_{33})$ in 3d space then there exists a point in the interior of $ \triangle PQR$ all of whose coordinates are all positive, negative, or zero.
It does not give justification, though.
Can someone explain this? I am just a beginner  :) .

[b]1.[/b]

[quote="dgreenb801"]In the official solution (and maybe grobber's reply), it says that it suffices to show that if we set $ O(0,0,0)$, $ P(a_{11},a_{21},a_{31})$, $ Q(a_{21},a_{22},a_{23})$, $ R(a_{31},a_{32},a_{33})$ in 3d space then there exists a point in the interior of $ \triangle PQR$ all of whose coordinates are all positive, negative, or zero.[/quote]

That's because every point inside the triangle $ PQR$ has coordinates
$ \left( a_{11}c_{1}\plus{}a_{12}c_{2}\plus{}a_{13}c_{3},  a_{21}c_{1}\plus{}a_{22}c_{2}\plus{}a_{23}c_{3},  a_{31}c_{1}\plus{}a_{32}c_{2}\plus{}a_{33}c_{3} \right)$
for some positive reals $ c_1$, $ c_2$, $ c_3$. In fact, you can take $ c_1$, $ c_2$, $ c_3$ as the barycentric coordinates of the point with respect to triangle $ PQR$ (normalized to satisfy $ c_1\plus{}c_2\plus{}c_3\equal{}1$).

[b]2.[/b] I remember having solved this problem some years ago (alas, not on the TST) using some casebash and Helly. Now I know there is a solution on a few lines, alas using analysis:

Let $ A$ be the matrix $ \left(a_{ij}\right)_{1\leq i\leq 3,\ 1\leq j\leq 3}$. Let $ K$ be a real which is greater than $ \max\left\{a_{11},a_{22},a_{33}\right\}$. Then, all entries of the matrix $ K\cdot I_3\minus{}A$ are strictly positive. Hence, after the [url=http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem]Perron-Frobenius theorem[/url], there exists a right eigenvector $ v$ of $ A$ such that all three components of $ v$ are strictly positive. This vector $ v$ then satisfies $ \left(K\cdot I_3\minus{}A\right)v\equal{}\mu v$, where $ \mu$ is the corresponding eigenvalue. This rewrites as $ Av\equal{}\left(K-\mu\right)v$. Hence, the three components of $ Av$ are proportional to those of $ v$, so they are either all negative, or all zero, or all positive. Now, if we write $ v$ in the form $ v\equal{}\left(\begin{array}{c} c_1\\ c_2\\ c_3 \end{array}\right)$, then $ Av\equal{}\left(\begin{array}{c}  a_{11}c_{1}\plus{}a_{12}c_{2}\plus{}a_{13}c_{3} \\  a_{21}c_{1}\plus{}a_{22}c_{2}\plus{}a_{23}c_{3} \\  a_{31}c_{1}\plus{}a_{32}c_{2}\plus{}a_{33}c_{3} \end{array}\right)$, and we are done.

This generalizes to $ n$-vectors.

  darij

[quote=darij grinberg]Now I know there is a solution on a few lines, alas using analysis:

Let $ A$ be the matrix $ \left(a_{ij}\right)_{1\leq i\leq 3,\ 1\leq j\leq 3}$. Let $ K$ be a real which is greater than $ \max\left\{a_{11},a_{22},a_{33}\right\}$. Then, all entries of the matrix $ K\cdot I_3\minus{}A$ are strictly positive. Hence, after the [url=http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem]Perron-Frobenius theorem[/url], there exists a right eigenvector $ v$ of $ A$ such that all three components of $ v$ are strictly positive. This vector $ v$ then satisfies $ \left(K\cdot I_3\minus{}A\right)v\equal{}\mu v$, where $ \mu$ is the corresponding eigenvalue. This rewrites as $ Av\equal{}\left(K-\mu\right)v$. Hence, the three components of $ Av$ are proportional to those of $ v$, so they are either all negative, or all zero, or all positive. Now, if we write $ v$ in the form $ v\equal{}\left(\begin{array}{c} c_1\\ c_2\\ c_3 \end{array}\right)$, then $ Av\equal{}\left(\begin{array}{c}  a_{11}c_{1}\plus{}a_{12}c_{2}\plus{}a_{13}c_{3} \\  a_{21}c_{1}\plus{}a_{22}c_{2}\plus{}a_{23}c_{3} \\  a_{31}c_{1}\plus{}a_{32}c_{2}\plus{}a_{33}c_{3} \end{array}\right)$, and we are done.

This generalizes to $ n$-vectors.[/quote]

An analysis-free proof of this (replacing the use of Perron-Frobenius by some linear programming theory) can be found at [url]https://mathoverflow.net/questions/299318/constructive-proof-of-a-rational-version-of-perron-frobenius[/url] .

Clearly we can assume that, $a_{11} = 2$, $a_{12} = -1$, $a_{13} = -1$ and $c_1 = 1$. Let, $R_i = a_{i1}c_{1}+a_{i2}c_{2}+a_{i3}c_{3}$
\begin{tabular}{ |c|c|c|c| }
 \hline
 2 & -1 & -1 \\
 \hline
 - & + & - \\
 \hline
 - & - & + \\
 \hline
\end{tabular}
Now, let's work with $c_2+ c_3 = 2$. Then $R_1 = 0$ and $R_2$ will be minimum and $R_3$ will be maximum for $c_2 \to 0$ on the other hand $R_3$ will be minimum for $c_3 \to 0$
Clearly both minimums of $R_2, R_3$ are negative.
Now, if they don't reach $0$ then we can take any $c_2+ c_3 = 2$  and increase both them by $\epsilon > 0$ for small enough $\epsilon$ to get all of them negative.

Now if they can reach $0$, WLOG let $R_2$ becomes $0$ for some $c_2+c_3 = 2$.
If, $R_3 = 0$ then we are done.
If it is negative then we will increase $c_3$ by $\epsilon > 0$ to get all negative.
If it is positive then we will decrease $c_3$ by $\epsilon > 0$ to get all positive.  $\blacksquare$

Note that this is equivalent to showing that the span of \[ \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \] in ${\mathbb R}^3$ over positive input vectors intersecting either the $(+, +, +)$ or $(-, -, -)$ quadrant.
Let $A$, $B$, $C$ be the output vectors that the basis vectors are mapped too, which lie in the quadrants $(+, +, -)$, $(+, -, +)$, and $(-, +, +)$ in some order.
If the span contains $(0, 0, 0)$, the result follows. As such, if $A$, $B$ and $C$ are linearly dependent, by the sign of the output vectors, it holds.
If any of the lines $AB$, $BC$, or $CA$ cross through the $(+, +, +)$ quadrant the result follows.
Else, it follows through considering projections that for each basis vector, either it or its additive inverse are in the span.
As such, the span contains the entirety of one of the quadrants. The quadrant can't contain either $A$, $B$, $C$. It also can't contain their additive inverses by linear independence. Thus, the result follows.

Consider the vectors $\vec{a_1}=[a_{11},a_{12},a_{13}], \vec{a_2}=[a_{21},a_{22},a_{23}], \vec{a_3}=[a_{31},a_{32},a_{33}]$ and $\vec{c}=[c_1,c_2,c_3]$. Now, for any two vectors $\vec{c}$ and $\vec{a_i}$, $a_{i1}c_{1}+a_{i2}c_{2}+a_{i3}c_{3}$ is positive if and only if the angle between the vectors is acute, negative if and only if the angle is obtuse, and zero if and only if it is right.

For each vector $\vec{a_i}$, draw the plane through the origin normal to the vector. Color the side containing the head of $\vec{a_i}$ blue and the other side red. Note that any point that is on all three planes, on all three red regions, or on all three blue regions satisfies the following: if we make $\vec{c}$ have this point as its head, then the angles are all acute, all obtuse, or all zero, as desired.

Now, let's translate this onto the unit sphere. We project the entire space onto the surface of a unit sphere. Note that each plane is a great circle on this sphere, so for each $\vec{a_i}$, half of the sphere is colored red, and half colored blue. Note that all of $(+,-,-)$ is colored red by $\vec{a_1}$, $(-,+,-)$ is colored red by $\vec{a_2}$, and $(-,-,+)$ by $\vec{a_3}$. Clearly, our all-red region is a spherical triangle, and so it is connected, which means that since $(+,+,-)$, $(+,-,+)$, and $(-,+,+)$ all have blues, it must be a subset of the points with at least two negatives. If it were to be in two different octants, it must also be in the $(-,-,-)$ octant by connectedness. If it were only in one, say $(+,-,-)$, then there is a point in $(-,+,-)$, $A$, for which the distance from $A$ to any point in $(-,-,-)$ is longer than the distance from $A$ to any point in the all-red, clearly impossible. Thus, the all-red intersects with $(-,-,-)$. Since the all-blue is the antipode set of the all-red, it intersects with $(+,+,+)$ as desired.

oops?

By scaling, we can assume $a_{ii}=1$ for $1\leq i \leq 3$ and $c_3=1$ Let $a_{i,j}$ for $i\neq j$ instead be positive, we negate the term in the respective expression. It remains to show there exist $c_1,c_2$ such that \[c_{1}-a_{12}c_{2}-a_{13},\qquad -a_{21}c_{1}+c_{2}-a_{23},\qquad -a_{31}c_{1}-a_{32}c_{2}+1\]
are the same sign.

We take two cases, one where they are all positive and one when they are all negative. If all are positive, then
\begin{align*}
c_1 > a_{12} c_2 + a_{13}\\
c_2 > a_{23} + a_{21}c_1 \\
1> a_{31} c_1 + a_{32} c_2
\end{align*}
Let $c_2 = a_{23} + a_{21}c_1 + \varepsilon$. Then the other two inequalities become
\begin{align*}
c_1 > a_{12} \cdot a_{23} + a_{12}a_{21} c_1 + a_{13} + \varepsilon\\
1 > a_{31}c_1 + a_{23}a_{32} + a_{32}a_{21}c_1 + \varepsilon
\end{align*}
(we loosely use $\varepsilon$, it does not have a fixed value)
If $a_{12}a_{21} \geq 1$ or $a_{23}a_{32}\geq 1$, it is easy to check the inequalities have no solutions. Otherwise, there are solutions when
\[
\frac{a_{12}a_{23} +a_{13}}{1-a_{12}{21}} \leq  \frac{1-a_{23}a_{32}}{a_{31}+a_{32}a_{21}}.
\]

Now repeat for the negative case. If $a_{12}a_{21} \geq 1$ or $a_{23}a_{32}\geq 1$, then making $c_1$ sufficiently large/small (depending on which one is true) suffices. Otherwise, it is easy to check there are solutions when
\[
\frac{a_{12}a_{23} +a_{13}}{1-a_{12}{21}} \geq  \frac{1-a_{23}a_{32}}{a_{31}+a_{32}a_{21}}.
\]

But one of these inequalities must be true, so we are done.


Find the number of 5-digit numbers that each contains the block '15' and is divisible by 15.

[hide]
ok, well first off, in order to be divisible by $15$, the number must be divisible by both $5$ and $3$
this means that the last digit must be a $5$ or a $0$, and the sum of the digits is $\equiv 0 mod 3$
for clarity sake, i will refer to the 5-digit number as $a$
given this, we have two cases:

case1 : the $15$ is at the end of $a$, so therefore $a$ is clearly divisible by $5$, and since $15 \equiv 0 mod 3$, then the remaining 3 digits must be $\equiv 0 mod 3$ as well. This is essentially equivalent to putting a 3 digit number divisible by $3$ in front of the $15$, so for this case, we simply need the number of 3 digit #s divisible by $3$ which $= \lfloor \frac{1000}{3} \rfloor - \lfloor \frac{100}{3} \rfloor = 300$

case2: the $15$ is somewhere in the middle of $a$. For this case, the last digit of $a$ must be a $5$ or a $0$. The $15$ must be somewhere in the first 4 digits of $a$. There are 3 possibilities for its positioning: $15ABC, A15BC, or AB15C$, where $A$ and $B$ represent arbitrary digits and $C$ represents $0$ or $5$. If $C = 0, AB \equiv 0mod3$ and if $C = 5, AB \equiv 1mod3$, but since the number of 2 digit numbers $AB \equiv 0mod3 =$ the number of 2 digit numbers $AB \equiv 1mod3 = 33$, the total number of possiblities for this case is $2*3*33$ ($2$ possible last digits $* 3$ possible positions for $15 * 33$ possibilities for $AB$) $= 198$

therefore, the answer = case 1 + case 2 = $\boxed{498}$[/hide]

hi rushil

i think the answer is 487

i have checked it many times.

The official answer is 479 ---- I think I got around 487 too in the exam~1

hi rushil

yes the ans is 479
i finally got it
i was certainly very wrong in giving the ans as 487 :oops:

i was not considering the case wen there r 2 blocks of 15
that gives exactly 8 cases which have to b sub frm 487

thnx

well, i see part of my mistake, but im still not getting the right answer, but what i know i did wrong was:
[hide]
i didnt notice there was overlap between case 1 and case2 so i double counted the case where we had $A1515$ and $15A15$ where A is an arbitrary digit, so we must subtract the overlap. for $a$ to be divisible by $15$, $A$ must $= 0,3,6,$ or $9$, but for $A1515, A$ cannot $= 0$ therefore, there are 7 overlaps b/t case1 and case2 bringing my answer to $\boxed{491}$
[/hide]
anybody see the rest of my mistakes?

I think in your case 2 you may have included a handful of 4-digit numbers by inadvertantly setting A to 0.  (For instance, in the AB150 case, you allowed AB to be the "two-digit" numbers 03, 06, 09.)

ahh yes youre right, i think i meant to say the number of 2 digit #s divisible by 3 = 30, so in that case, we can say the total for case 2 is actually 180, and so therefore:
case1 + case2 - overlap = 473... great now im too low... now what?

hi

u took the fol cases

ABC15 ---  300

AB15C----- wud give 30 cases n not 33 for C=0 or C=5

A15BC------  similarly gives  30

15ABC----- gives 33 for c=5
and gives 34 and not 33 for c=0  (one case that i think u might be missing is a=b=0)

thus total= 300+ (30*2)*2 + 33+34=487
 now my mistake was to consider upto this pt only

so we sub the cases in which there r more than 1 block of 15
ie. 8 cases
so ans=487-8=479 :)


function
real analysis
number theory theorems
number theory

The arithmetic function rad(n) is defined in the following link:

http://en.wikipedia.org/wiki/Radical_of_an_integer

The Dirichlet convolution inverse of this arithmetic function seems to agree with Euler's Totient for square-free natural numbers [b]and vanishes otherwise.[/b] Is this true? If so, has anybody proved it?

All mentioned functions are multiplicative, especially the inverse $f$ of $\mathrm{rad}$.
We we just want to show that $f(p)=p-1$ and $f(p^{k})=0$ if $p$ prime and $k \geq 2$. That's again rather easy checking.


How can I delete a blog category?  I looked at all the blog options but couldn't find it.

I don't think that's what he ment. In the right-side menu on your main blog page there are the categories listed, plus a button (link) below them saying "Edit categories". You click on that and an edit (pencil) and delete (x) buttons appear. Use those to manage your categories.

Sorry to revive, but I don't seem to be able to find this link anymore.  Can someone help me? Thanks.

Hmm, I don't think these instructions were quite correct.

Click on a category name.

Find the post you want to remove a category from

Edit that post

Click Tags (Oddly names, but categories are under there)

Click Remove next to the category name.

This isn't a well tested function. Please let us know if it works.

Hmm, I'm sorry to report that it doesn't seem to be working.... :(

It seems to just remove that post from that given category instead of deleting the category altogether.

Also, there is some REALLY odd behavior with page numbers in the categories.  For instance, in [url=http://www.artofproblemsolving.com/Forum/blog.php?u=53544&cat=836&start=0]my blog[/url], I have placed one post under the category 'Non-maths'.  But then, when clicking the link labeled "Non-maths [1]", the page count shows "22 blog entries Page 1 of 5", despite that there is only one entry!  And the links themselves labeled "1 2 3 4 5" take you to all 5 "pages" with exactly the same content... that one entry....

^ Should I put that under the Bug Reports thread?
EDIT: Never mind, I just noticed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1892782#p1892782]your post[/url] on that thing.

The only way to completely remove a category is to delete it from all blog entries that have that category.


There are 25 magnetic tapes on reels and 1 empty reel. One can rewind a tape from a full reel to an empty one, thus reversing its direction. Can one reverse the direction of every tape while leaving each on its original reel?

Pierre.

Hmm...
Enumerate all tapes with 0, 1, 2, ..., 25 and place them in row. Tape 0 is an "empty" tape.
Suppose it is possible to change direction of each non-zero tape.
We know that each transposition chages parity of number of inversions in the sequence of tapes. Each transposition involves 0 and non-zero number. Since all non-zero numbers were transposited odd number of times (in order to change direction), we conclude that we perfomed odd number of transpositions. It foolows that final number of inversions must be odd. Contradiction.

I don't see why each transposition involves 0.  :(
It involves an empty reel but not necessary the initial one.

Pierre.

Each time 0 means that it is an empty place :), so next transposition necessarily involves this 0.

So why is there a contradiction?

Pierre.

Finally we have the same permutation, so its parity of number of inversions is even, as it was in the beginning.


counting
distinguishability
combinatorics proposed
combinatorics

There are $ 11$ coins, which are indistinguishable by sight. Nevertheless, among them there are $ 10$ geniune coins ( of weight $ 20$ g each) and one counterfeit (of weight $ 21$ g). You have a two-pan scale which is blanced when the weight in the left-hand pan is twice as much as the weight in the right-hand one. Using this scale only, find the false coin by three weighings.

We put $ 4\minus{}2$ coins, $ 5$ remain. The worst case is when this is balanced and the counterfeit is among the remaining $ 5$.
We put $ 2\minus{}1$ coins, $ 2$ remain. So if we have the fake one among $ 2$ coins, we choose one from them and $ 2$ genuine coins, and we're done by at most $ 3$ weighings.


For the last couple of months I have been experimenting with mind maps for solving math problems.
I would like to describe some of its pros and cons.

What is mind mapping? (You may want to skip this if you are familiar with mind maps.)

Mind mapping is a special form of note-taking.
Here are some essential features:
- You take a (preferably large) sheet of paper in landscape format.
- You write the topic / the problem in the middle of the sheet and draw a frame around it.
- You write the main aspects and main ideas around that central topic and link them through lines to the center.
- You expand the ideas in these "main branches" into subbranches etc.
- Wherever appropriate, you should use figures, colours, arrows to link branches etc.

How can mind maps be used for solving math problems?

I will start with two principal uses:
- Using mind maps to examine a given problem.
- Using mind maps to organize problem solving tools.
Combining these two approaches could mean using two mindmaps at a time:
- a "problem map" for dealing with the problem itself and
- a "tool map" (or several of them) containing problem solving tools - from general ones (e.g. the ones presented in Polya's "How to Solve It") to highly specialized ones (e.g. for dealing with Poisson processes).

Using problem maps

Using mind mapping for solving a problem has a number of advantages:
- It's easy to keep track of several aspects or approaches, of aims and sub-aims etc.
- Due to the map's layout, spontaneous ideas can be placed easily and appropriately in the map.
- Generally speaking, due to the map's hierarchical structure, your ideas get rather well structured.
- Using words, mathematical terms and figures allows to exploit the advantages of each of these three representations.
- The ideas and chains of thought documented in the map can be scrutinized.
- Mind mapping is easy to learn and easy to use.
(Of course, some of these advantages apply to note-taking in general, but some are peculiar to mind mapping.)

The main disadvantage is that laying out the map can become tricky when lengthy computations are inevitable.

Using tool maps

The basic idea in using tool maps is to collect and structure tools for solving problems in mind maps.
These tool maps can be organized along several concepts, like:
- Stages of problem solving, e.g. Polya's scheme from "How to Solve It": "understanding the problem" - "devising a plan" - "carrying out the plan" - "looking back".
- Standard situations in problem solving, e.g. "looking for new approaches" - "overcoming frustration" - "need for information" - "my most frequent errors in problem solving".
- Mathematical objects involved, e.g. matrices, polynomes or inequalities.

Here comes a brief discussion of tool maps.
We start with its advantages:
- Most important: In constructing and improving your own tool maps, you learn a lot about problem solving and especially your personal problem solving behaviour.
- Tools maps act as reminders for techniques you might otherwise have overlooked.
- Tool maps can help novices with adopting new working heuristics.
- Tool maps are very flexible and can be adapted to all sorts of experience, needs and special fields.
- Due to their graphical representation and their structure, tool maps are easier to scan and to expand than conventional catalogues or lists.
- Tool maps may help to share problem solving techniques in a group by making "implicit" problem solving techniques "explicit".
Here are some disadvantages:
- Sometimes tool maps may become messy and overloaded and need redrawing.
- To use tool maps consistently, it's essential that the tool maps are easily accessible, e.g. as a poster at the working place.

I am most interested in experiences others have made with mind maps in problem solving.

Regards,
Thomas

Never tried this, but sounds really inefficient and I think it would impair creativity. Unless the problem isn't of that sort, in which case, why bother. Might be good for programming though.

[quote="thomasteepe"]
I am most interested in experiences others have made with mind maps in problem solving.
[/quote]

I've never tried physically mapping a problem out in math, but I think most of us do it at some level in our head. When you're stuck, you think about other techniques and think about what looks familiar or "nice" in a problem.

I would like to reply to some topics mentioned by probability1.01

[b]Efficiency[/b]

- Mind mapping as used in the problem map is just a different way of note-taking. More precisely: It is first and foremost a special way of arranging your notes according to their context and relation. Most people I know use mind maps, and no one has reported difficulties or extra efforts in doing this.
- There are three main activities concerning tool maps:
a) Constructing: This is done once (or at least very rarely) for each tool map topic, and then the tool maps can be used in solving any number of problems.
b) Improving: This happens occasionally and can be done quickly.
Both a) and b) are not dull and mechanical activities - with a proper chance they may lead to real insight into problem solving.
c) Using: Of course, the tool maps need only be used when you're stuck or when you want to remind yourself of new tools you are unfamiliar with.

[b]Creativity[/b]

Problem maps and tool maps should perhaps best be viewed as modules that can be used separately (both of them) or in combination. Impairing creativity might occur when you start using mind maps in a dull routine. I think no one is advocating this. You may consider mind maps as just another tool in your box, even ignoring them altogether as long as your ideas flow freely and resort to maps when you get stuck.
If this sounds a bit random and arbitrary: I don't want to champion any fixed method, but develop and discuss its uses.
Another interesting question is on which level of expertise you can profit most by using mind maps (novice or expert; age etc.).

[b]"Unless the problem isn't of that sort..."[/b]
What is meant by this expression?

[b]Good for programming[/b]
I am curious about ideas on this!

Kind regards, Thomas

while i might go through some of these processes mentally, writing it out in a flow chart seems too structured and formal. though i'm sure it helps some people.


Why is the barycentric coordinate system (area ratios) equivalent to the mass-point system (center of mass method) with respect to a triangle?

OK I got it on my own just this afternoon. It's easy enough to see the equivalence for any point inside the triangle ; just show that each pair-of-area ratio is equal to the corresponding pair-of-mass ratio. I suppose it can't be too difficult for any point outside (just a little casework - haven't done it yet).
Thanks anyway.


inequalities
probability
combinatorics theorems
combinatorics

I had read a TITU's book and he is used Bonferonni's Inequaliy. But when I googled, I saw that Bonferonni's Inequalities is a inequality about probality, I don't know how can Titu apply to inclusion exclusion principle.
thanks  :)

I think you're referring to the fact that the partial sums of the inclusion-exclusion formula alternately overestimate and underestimate what the full sum counts. Bonferroni, though a probability inequality, says pretty much the same thing if you clear fractions.

yes  :lol:  but can you tell me more about that? (that why can we convert a probality inequality to Inclusion-Exclausion Inequality :D) thanks


[quote="AstroPhys"][hide="Question 1"]
81
[/hide]
[hide="Question 2"]
809
[/hide][/quote]

Eight billion comes before eighty one... how about eight billion one?

Yeah that works, did not think about 8 billion.

Yes, that would come before it.
Then the prime number should be wrong to, what are the prime numbers in the 8 billions?

[quote="AstroPhys"]Then the prime number should be wrong to, what are the prime numbers in the 8 billions?[/quote]

good luck finding it....

We should start by investigating 8 billion 9 and 8 billion 3, since 8...1 and 8...7 are clearly divisible by three.

[quote="236factorial"]eight billion five comes before eight billion one, doesn't it?[/quote]

how about eight billion eight million five?! And that is not the answer either ...

If that is a prime number then yes, it works  :) .

Is "eighty-five" one or two words? If two, we can have
8,088,088,085

[quote="xxxyyyy"]It is easy to write the integers in words - one, two, three, four, five, etc ....

We use the normal way of doing it, like one hundred seventy, one thousand twenty four, etc.

Suppose that you have done it for all numbers up to 10 billion.  Difficult? You can use this website: http://www.easysurf.cc/cnvert18.htm

Now suppose that you have arranged the results in ascending alphabetic order.

$Question 1$ What is the first odd number in this list?

$Question 2$ what is the first prime number in this list?

Edit1: When we arrange the words/phrases in alphabetical order, we follow the dictionary convention and not what MS-Excel uses (space being ascii 32 and so on)....So we ignore spaces, hyphens, etc in the sorting process.[/quote]

Answer 1
[hide]Eight billion eighteen million eighteen thousand eight hundred eighty five [/hide]
Answer2
[hide]Eight billion eighteen million eighteen thousand eight hundred fifty one [/hide]

:huh:  :rotfl: The solutions I posted were wrong be a [i]small[/i] difference.


topology
complex analysis
complex analysis unsolved

what is  the  basic  diffrence  between an  algebraic  structure  and  a  topological structure

That's kind of a vague question.  Structure of what?  In a nutshell, topology describes the open sets of a space.


Am I crazy, or on the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPS_Community_Awards]AoPS Community Awards Page[/url], the section for AMC 8 perfect scorers is no longer there?

I tried looking for it, and yes, it's no longer there.

What pianoforte said. I removed it myself.


I'm searching for the solution of the inequality
 \[\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}\ge \frac{3}{1+abc}\]
where \[a,b\text{ and }c\] are positive real numbers.

[quote="kam"]I'm searching for the solution of the inequality
\[ \frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}\ge \frac{3}{1+abc}\]
where
\[ a,b\text{ and }c \]
are positive real numbers.[/quote]
Maybe it:
http://www.mathlinks.ro/Forum/viewtopic.php?t=161059


conics
ellipse
AMC
USA(J)MO
USAMO
complex numbers

Given two complex numbers $ \alpha\equal{}6/7\plus{}11/7i$ and $ \beta\equal{}\minus{}4/7\plus{}1/7i$, and let the complex numbers $ \omega_1$ and $ \omega_2$ be two solutions to the equation $ |\omega\minus{}\alpha|\plus{}|\omega\minus{}\beta| \equal{}20/7$.
Find the maximum possible value of $ |\omega_1\minus{}\omega_2|$.

This problem is essentially saying that you have an ellipse with foci $ \alpha$ and $ \beta$. So the maximum value is the length of the major axis of the ellipse which is $ \frac{20}{7}$. This problem can also be done with the triangle inequality...

$ 2|\omega_1\minus{}\omega_2|\equal{}|(\omega_1\minus{}\alpha)\minus{}(\omega_2\minus{}\alpha)|\plus{}|(\omega_1\minus{}\beta)\minus{}(\omega_2\minus{}\eta)|$

$ \le |(\omega_1\minus{}\alpha)\plus{}|(\omega_2\minus{}\alpha)|\plus{}|(\omega_1\minus{}\beta)|\plus{}|(\omega_2\minus{}\beta)|\equal{}2\cdot\frac{20}{7}$.

...We can check that equality occurs.

[quote="Yongyi781"]Bonus: Find the source of this problem :D[/quote]
Haha, yeah...
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=270591

BTW for finding the max of $ |\omega|$ we could just use $ 1/2|\omega\minus{}\alpha\plus{}\alpha|\plus{}1/2|\omega\minus{}\beta\plus{}\beta|\le$
$ 22/7\plus{}1/2|\alpha|\plus{}1/2|\beta|$, but could we solve this geometrically?


Let $ a,b,c$ be positive numbers such that:$ a\plus{}b\plus{}c\equal{}1$.Prove that:

$ \sqrt{4a\plus{}(b\minus{}c)^2}\plus{}\sqrt{4b\plus{}(c\minus{}a)^2}\plus{}\sqrt{4c\plus{}(a\minus{}b)^2} \ge 3(\sqrt{a^2\plus{}abc}\plus{}\sqrt{b^2\plus{}abc}\plus{}\sqrt{c^2\plus{}abc})$

 :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ a \plus{} b \plus{} c \equal{} 1$.Prove that:

$ \sqrt {4a \plus{} (b \minus{} c)^2} \plus{} \sqrt {4b \plus{} (c \minus{} a)^2} \plus{} \sqrt {4c \plus{} (a \minus{} b)^2} \ge 3(\sqrt {a^2 \plus{} abc} \plus{} \sqrt {b^2 \plus{} abc} \plus{} \sqrt {c^2 \plus{} abc})$

 :)[/quote]
I dedicate the following proof to my friend-Tran Quang Hung and all mathlinks members. This maybe a long proof but the idea is natural and I hope you all enjoy it.

Regards

See here (the last post) for the geometric form of this inequality: http://www.mathlinks.ro/viewtopic.php?t=260387


find all square roots of
$1-x+\sqrt{22x-15-8x^2}$


Let $n$>$2$ be a positive integer and $S$ be a set containing $n$-elements. Find the smallest positive integer $k$ for which there exist subsets $A_{1},A_{2},...,A_{k}$ of the set $S$ with the following property: for any two different elements $a,b$ belong to $S$, there exist a number $j$ belong to ${1,2,...,k}$ ,such that $A_{j}$ ∩ {$a,b$} has exactly one element.

I'm sorry about my poor English!!!!

Isn't it just $k=n$?

No two elements can be in the same set, so each $A_{j}$ has exactly one element...


[quote="Fermat_gauss"]The problem is very hard...[/quote]

What is $AD$??

If you don't mind, write down the statement of the problem and the what we have to prove....

 :)


Just wondering...what happens if your mailbox is full and someone tries to send you a private messege?  Does the message remain in the other person's outbox until the reciever clears up his/her mailbox?  Or is the messege bounced back?  And is it possible to have a slightly larger mailbox?  Or no...never mind...I'll just delete some items.  Easy fix.

My guess is that the oldest PM in your inbox gets deleted.

The message never gets bounced back even if there was a problem. The inbox of the ricepient is independent from the sender's Sent/Outbox so even if one is deleted the other is still there. The only way for them to get deleted on their own is if your Outbox fills up also.

A simple solution would be to utilize the Savebox which can then serve as the secondary Inbox and thus multiply your storage power by two.

I do use my savebox.  It's just that it's filled up also.  I guess I'm too much of a packrat in that respect.

*looks at savebox* Wow...all of these messages are from the jojobird era.  Crazy.

You could also copy and paste them onto Word or something like that but I guess they aren't that valuable...

That's what I did.  I have two documents with Tare letters (one with random stuff and one about the Saga), one with rippledance letters, and one with Mock AMC responses.  Cool, no?

Plus, I now have plenty of blackmail material from rippledance stored on my computer.  Muahahahaha.... :twisted:


$ x>y>z>0$ and $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$. What is the minimum value of
\[ \frac{4}{z^2}\plus{}\frac{1}{(x\minus{}y)y}\plus{}\frac{1}{(y\minus{}z)z}.\]


here the line segments ($ BE, CF,AD$) from the vertices of the  triangle meet at a point.
$ FE$ is joined and extended to a point $ L$ where it meets the extended $ BC$.
prove that $ 2BD*BL \equal{} BC(BL \plus{} BD)$

SORRY FOR THE FREE HAND AND INACCURATE  FIGURE .

well it is my sincere request to the mods to transfer this problem to middle school.
THE PROBLEM IS SO STUPID AND EASY FOR THIS LEVEL


Let $ a;b;c>0$ and $ a\plus{}b\plus{}c\equal{}1$
$ Prove that (abc\plus{}1)^3 \ge (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2$

[quote="tranvanluan"]Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$
$ Prove that (abc \plus{} 1)^3 \ge (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2$[/quote]
But it is obivious;$ RHS < 1 < LHS$
I think you have some mistake in the post

[quote="tranvanluan"]Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$
$ Prove that (abc \plus{} 1)^3 \ge (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2$[/quote]
 it's very easy.
 We  have:
$ a\plus{}b  <1$ hence $ (a\plus{}b)^2  <1$
Similatry, $ (b\plus{}c)^2<1$ and $ (c\plus{}a)^2 <1$
Thus, $ RHS <1 <LHS$


Is there a perfect square $ x$ such that $ 2008\plus{}2\sqrt{x}$ is also a perfect square? If so, what is the smallest?

$ x\equal{}4k^2$, were $ k\equal{}m^2\minus{}502$ or $ x\equal{}4(m^2\minus{}502)^2$. Minimal for m=23.

$ x \equal{} a^2$

$ 2008 \plus{} 2a \equal{} b^2$

Clearly $ b$ must be even

The first even perfect square greater than $ 2008$ is $ 46^2 \equal{} 2116$.

So $ 2008 \plus{} 2a \equal{} 2116 \rightarrow a\equal{}\frac{2116\minus{}2008}{2} \equal{} \frac{108}{2} \equal{} 54$

In conlusion the minimum $ x$ is $ 54^2 \equal{} 2916$.


I have found this topic to be a useful icebreaker, so it seems to be as good a thread-starter as any:

To what movie events in the upcoming year are you looking forward?

I personally am eagerly awaiting a) an opportunity to see Around the World in 80 Days (just because Jackie Chan is my hero [see avatar]), b) the theatrical release of Alien Vs. Predator on August 13 (from what I've seen, it appears extraordinarily promising) and c) the DVD release of the original Star Wars Trilogy in September (my dad and I have been waiting for the DVD release for quite some time now).

What about you?

i'm not looking forward to much. i guess i'll look forward to when dodgeball and farenheit 9/11 comes out on dvd. btw, around the world in 80 days is doing really badly in the box offices.

Phooey.  Oh, well.  I just love martial arts action scenes.  Plot is SLIGHTLY overrated in my movie-going experiences (as far as action films go, at least).

hildalgo on dvd
star wars trilogy on dvd
when i finally get time to see the terminal
lotr3 on extended dvd

[quote="RobinHood3000"]I personally am eagerly awaiting a) an opportunity to see Around the World in 80 Days (just because Jackie Chan is my hero [see avatar]), b) the theatrical release of Alien Vs. Predator on August 13 (from what I've seen, it appears extraordinarily promising)
What about you?[/quote]
Around the World in 80 Days is nice, but c'mon in all the movies with Jackie Chan, everything rounds up to something the Chinese Mafia did or will do if he doesn't stop them ... :)

AVP sucked right from the begining. It has a 30 mins introduction of the whole team, and everybody except a couple or so are eaten in less than 15 minutes (or so it seems). You don't actually see the Aliens lurking around as we did in previous Alien movies, and the end is very, very tipical: humans and predators join each others' forces to fight the evil Aliens, and of course, to have a start for the AVP2, the little alien is brought aboard the Predator ship ... just yuk.

I am waiting for the last part of the StarWars 6-logy, in which we can finally see how does Anakin becoming Darth Vader.

[quote="Valentin Vornicu"]AVP sucked right from the begining. It has a 30 mins introduction of the whole team, and everybody except a couple or so are eaten in less than 15 minutes (or so it seems). You don't actually see the Aliens lurking around as we did in previous Alien movies, and the end is very, very tipical: humans and predators join each others' forces to fight the evil Aliens, and of course, to have a start for the AVP2, the little alien is brought aboard the Predator ship ... just yuk.
[/quote]

Bah, humbug.

As a former major poster on the PlanetAvP forums, I must thank you for bringing back a bit of nostalgia for me; it is greatly appreciated.  Now, to the rebuttal.

I'll start with your last point and work backwards: there is no way, earthly, Yautja, or Xenomorphic way that the PredAlien chestburster will possibly be the kickoff to AvP2.  The reason is because it is a wittle baby Alien on a ship full of Elder Predators.  To clarify, Elder Predators = the Best of the Best of the species.  The PredAlien chestburster stands no shot at survival whatsoever unless it survives long enough to molt, and even then it's far too likely to be discovered in the long journey (presumably) to the Yautja planet, where it would be exterminated in a matter of minutes, or some distant other hunting grounds, where it would be the only Alien on the planet because, as a PredAlien drone, it is not only horrendously ugly but also incapable of impregnating (presumably; there is some supposition that plain ol' Drone can impregnate exactly once, but those who think so are strictly in the minority) others.  Either way, the PredAlien's final destination isn't going to be a human populated planet, which basically eliminates a shot a it having even a cameo in AvP2.  More likely, the Queen that sinks into the Antarctic at the end will be the source of inspiration, if AvP2 continues the first timeline at all.

The end is typical because it's an homage to an AvP Comic, in which a human woman teams up with a Predator and is eventually accepted into his clan.  Admittedly, most fans of the franchise feel that the execution was not as well-done in the movie, but that hardly makes it typical.  This is especially true considering that the Lead Predator (known as Scar, incidentally) ended up dying, and Lex comes out of the whole thing with a spear and a coupon for 10 years of psychiatric therapy.

As for the Aliens not lurking around, they do indeed lurk in a number of segments in the movie; when Gill (the first Pred to die) is killed and when Scar semi-decapitates a Drone.  Not to mention the fact that, in [i]Aliens,[/i] the Aliens hardly "lurked" at all.  The Aliens, when in a group, specialize in swarm tactics, and consequently would have no cause to lurk in this particular situation.

As for the minimum of time used for introduction, there isn't much call to particularly endear the characters to the audience when they'll all die anyway.  I, for one, am grateful they made an attempt at all.  And as for the short time in which the first two Predators died, that can be attributed to studio need/greed for short time and maximum possible showings.  I, for one, extol the movie as an excellent popcorn flick, and immensely enjoyed watching the Predators (my favorite) and the Aliens (boo!!) duke it out on the big screen.

Okay, thanks for listening.

A small note: I wasn't complaining about the intro being too short, but rather too long! :) The action itself is "blood-less", it should have been an R-rated movie ... I can bet you that AvP2 will be connected to the Alien on the vessel that took off ...

Oh, my mistake.  Hmm...I've seen movies with longer (or at least it feels that way) intros, and half-an-hour probably is pretty long relative to the showing time; about one third of the film, if memory serves.

And just out of curiosity, WHY should it have been an R-rated movie?

[quote="LynnelleYe"]Harry Potter and the Goblet of Fire. Can you tell I'm obsessed?[/quote]
yep

[quote="RobinHood3000"]My favorite of the four books; assuming they don't leave as many plot holes as in the third movie, the fourth one should be terrific.[/quote]

agree totally. soo excited bout the 4th movie! the first official pictures were released a few days ago at [quote]www.mugglenet.net[/quote] ...my fave HP site. it's gonna be awesome.

There was plenty of action between the Aliens and the Predators.  The reason it wasn't R-rated was because the gore and blood being spattered was mostly acidic or fluorescent rather than red, and from the lack of language.

And if it makes you feel any better, I hear the sequel's gonna have all-out war.

[quote="Valentin Vornicu"][quote="RobinHood3000"]And just out of curiosity, WHY should it have been an R-rated movie?[/quote]I wanted more action, more battle. That's what my expectations were about this movie (a full-contact between Aliens and Predators). Wasn't so :([/quote]

Violent, aren't we, Valentin?  ;)  :P
Jk, jk.

[quote="RobinHood3000"]There was plenty of action between the Aliens and the Predators.[/quote]There was more action between the walls of the pyramide and the humans than between the two alien species :)

Well, I for one liked the one-on-one wrestling match between Celtic and Grid...and Scar was awesome taking on the Queen.  That spin-flip-kick-spear-throw move near the end was [b]excellent[/b].

Dude, Orville Reddenbacher wouldn't have enough popcorn to last through all of the bone-crunching, blood-spilling, evil-defeating glory that is the Lord of the Rings Trilogy.

LOL. haha. yea, 11 hours 6 minutes. I'll be stocking up on food ^_~ (no I'll try not to eat TOO much...maybe...TRY..)  :D

[quote=joyofpi]LOTR marathon: to watch the whole extended trilogy of Lord of the Rings with ice cream and popcorn and food and friends. all 666 minutes ^_~
pure bliss.  :)[/quote]

LOTR is pretty good!  Not as good as the books, though.


calculus
integration
trigonometry
complex analysis
calculus computations

how can I calculate the following integral:
$\int_{0}^{\infty}\frac{x^{2}\cos{2x}}{x^{4}+6x^{2}+9}dx$

Use a calculator?  :P
Mathematica says the indefinite integral is
\[\(-\frac{x \cos (2 x)}{2 \left(x^{2}+3\right)}-\\ \frac{1}{24}i e^{-2 \sqrt{3}}\left(\left(6+\sqrt{3}\right) e^{4 \sqrt{3}}\text{Ei}\left(-2 \left(i x+\sqrt{3}\right)\right)-\right.\\ \left.\left(-6+\sqrt{3}\right) \text{Ei}\left(2 \sqrt{3}-2 i x\right)\right)+\\ \frac{1}{24}i e^{-2 \sqrt{3}}\left(\left(6+\sqrt{3}\right) e^{4 \sqrt{3}}\text{Ei}\left(2 i x-2 \sqrt{3}\right)-\right.\\ \left.\left(-6+\sqrt{3}\right) \text{Ei}\left(2 i x+2 \sqrt{3}\right)\right)\)\]
so I don't really see how you would do that without a calculator, but I could be missing something.
The improper integral is equal to $\frac{1}{12}(-6+\sqrt{3})e^{-2\sqrt{3}}$

First, this is really complex analysis, not real analysis.

$\int_{0}^{\infty}\frac{x^{2}\cos{2x}}{x^{4}+6x^{2}+9}dx = \int_{0}^{\infty}\frac{x^{2}\cos{2x}}{(3+x^{2})^{2}}dx$, which has poles at $\sqrt{3}i$ and $-\sqrt{3}i$.  Use the residue theorem, while considering the upper semicircular contour as R goes to infinity.  Note that over curved part of the circle, the integral will be zero since the x^4 term dominates, and that from negative infinity to 0 the integral is equal to that of 0 to infinity since the cosines and squares are even.

This means that the integral is 1/2 of the residue of order two at $\sqrt{3}i$, which is a relatively elementary calculation.


geometry
analytic geometry
graphing lines
slope
trigonometry
geometric transformation
reflection

$\triangle ABC$ is isoceles with $\angle A=20^{\circ}, AB=AC$,  let points $M$ and $N$ be on AB and AC respectively such that: $\angle BCM=50^{\circ}, \angle CBN=60^{\circ}$

find $\angle MNB$

http://www.mathlinks.ro/Forum/viewtopic.php?t=64298

By the way, this is definitely not an intermediat problem.

i think an easy solution would be to put this in a coorinate system...it might be somewhat messy, but it is really straight forward

extend MN to meet BC at Q, we get MQB=30 (after the cooridinate geometry), then it is trivial, and the answer is 130

i am not sure how to show this with out such a mess, but one thing is that finding that slope lies on having exact values of tan 10 and tan 60, so the solution is not general, but this is a case that works out well

For this problem: Constructions and transformations are much easier than trig methods, trig methods are much easier than coordinate methods.

[quote="Altheman"]i am not sure how to show this with out such a mess, [/quote]
Did you click my link?

oh okay...but i don't know how you saw that...this does not really seem to be at a reasonable difficulty for intermediate...the real thing i don't get is that how would you know that reflecting that point would lead to a solution because many of the steps after that were not completely obvious, and just worked out because the numbers of the problem were special

Expirences tell me that this kind of problems always have a constuction method and the final answer must be very "whole". Also, since one angle is $30^o$, so reflect will give a $60^o$ angle thus there will be an equilateral triangle, then the angle chasing job would be much easier.


[size=150][b]ΠΩΣ ΓΡΑΦΟΥΜΕ ΕΛΛΗΝΙΚΑ ΣΕ $\LaTeX$[/b][/size]

Μιας και είχε τεθεί το ερώτημα από τον κύριο Στεργίου, εν συντομία:

Αν θέλουμε να γράφουμε ελληνικά σε $\LaTeX$ κατεβάζουμε το πακέτο greek-tex , από την ιστοσελίδα του κυρίου Μοσχοβάκη Ιωάννη στο ucla, δηλαδή από την εξής διεύθυνση: http://www.math.ucla.edu/~ynm/greektex/

Στην ουσία δηλαδή, [i]den gr'afoume 'etsi[/i], αλλά [i][b]έτσι[/b][/i]. Με άλλα λόγια, υπάρχει η υποστήριξη για ελληνικούς χαρακτήρες. Απλώς, θα πρέπει να ακολουθηθούν οι λειτουργίες κατά την εγκατάσταση του πακέτου κατά γράμμα, έτσι να γίνει επιτυχώς και η εγκατάσταση του greek-hyphenation-system .

[b]Δεν[/b] θα πρέπει να ξεχνάμε ότι απαραίτητη προϋπόθεση για να γράφουμε αρχεία σε $\LaTeX$ στον προσωπικό υπολογιστή μας, είναι η ύπαρξη του προγράμματος σε αυτόν. Στα unix, μπορείτε να το βρείτε στο /usr/ports, στα windows, αφού μπείτε στη σελίδα http://www.latex-project.org , ψάξτε για το αρχείο [b]Miktex[/b] και κάντε το εγκατάσταση.

[u]ΠΡΟΣΟΧΗ[/u]: Πρέπει να αλλαχθούν και οι environmental variables από το control panel των windows !

Για περισσότερες πληροφορίες για το $\LaTeX$, απευθυνθείτε εδώ:

http://www.mathimatiko.net/latex

Η ιστοσελίδα είναι υπό την υπόβλεψη του φίλου Θάνου Τσουάνα  :D

[i]Για ό,τι απορίες έχετε, θα χαρώ να βοηθήσω, αν βέβαια μπορώ  :) [/i]

παιδια το ειπα και στον αμβροσιο. οποιος θελει να προτεινει ασκηση και δεν το κανει λογω λατεξ μπορει να απευθυνθει σε εμενα και να την γραψω εγω για αυτον , :)


Nice problem! :lol:

[hide="Hint for a fast solution"]
$x^n+x^{n-1}+x^{n-2}+...+x+1=\frac{x^{n+1}-1}{x-1}$
[/hide]

Frt, that's only for successive exponents though, isn't it? The exponents go in multiples of 5 and 3 (there's only one successive term, which surge took care of with an alternate method).

Well, if you let $y=x^3$ etc... ;)

Or think about the sum of geometric serites.

Oh, you're right, my bad. I hadn't thought about it that way. Nice hint.

surge, you don't need to do that!
frt has the correct idea :)
[hide="solution"]

it simplifies to:
$\frac{(\frac{x^6-1}{x-1})(\frac{x^{30}-1}{x^3-1})}{(x^3+1)(\frac{x^{30}-1}{x^5-1})}=\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$[/hide]


algebra
polynomial
analytic geometry
number theory proposed
number theory

Find all polynomials $p$ with real coefficients that if for a real $a$,$p(a)$ is integer then $a$ is integer.

this one is very easy: suppose deg(p(x))>=2. Suppose leading term is positive. Eventually the first term will 'overwhelm' the other terms, and similarly with the derivative. So we can find some integer q, s.t. p(x) is strictly increasing and p'(x) is also increasing for all x>q. But we have a contradiction in this because: draw lines corresponding to y=p(q)+1,y=p(q)+2,etc. The x-coordinates corresponding to these points are integers but they become closer and closer together, contradiction. a similar contradiction is reached if the leading term is negative. So it remains to consider when p(x) is linear or constant  :D .

The solution above has main idea, but here is a better-written version.

First, constant polynomials work, thus suppose $P(\cdot)$ is non-constant. Furthermore, notice we can take wlog the leading coefficient of $P$ to be positive. Now, as above, $P'$ is eventually positive, thus $P$ is eventually strictly increasing. Now, let $(N,\infty)$ be the interval on which $P$ is strictly increasing, and let $n\in(N,\infty)$ be an integer. Then, there is an $x_n$ such that $P(x_n)=n$ holds. The conditions of the problem imply $x_n\in\mathbb{Z}$. Now, there is another integer $x_{n+1}$, such that $P(x_{n+1})=n+1$. This implies, $x_{n+1}-x_n \mid P(x_{n+1})-P(x_n)=1$, and since we focus on the region on which $P$ is increasing, we get $x_{n+1}=x_n+1$. Similarly, get an $x_{n+2}$ such that $P(x_{n+2})=n+2$, and obtain that $x_{n+2}=x_n+2$. It therefore holds that, $P(x_n+\ell)=n+\ell = (n-x_n)+x_n+\ell$ for every non negative integer $\ell$, that is, $P(r)=r+C$ holds (for $C=n-x_n$ constant) infinitely often. However, this implies $P(r)-C-r$ has infinitely many roots, and thereby must itself be the zero polynomial.

This yields, $P(x)=x+C$ where $C\in\mathbb{Z}$, in addition to constant solutions.


algebra
polynomial
analytic geometry
linear algebra
linear algebra unsolved

Is the set of complex/ real cyclic ( that is, characteristic polynomial equals the minimal one) matrices connected? If not, find its connected components.

Over $\mathbb{C}$, absolutely.
The set of matrices in companion form is path connected (isomorphic to $\mathbb{C}^n$), and the set of invertible matrices is path connected; Conjugate the first by the second, and we can easily construct paths by moving in one coordinate or another.

Over $\mathbb{R}$, we can tweak this argument to work for odd dimension; we only need to conjugate by matrices of positive determinant in that case.
For even dimension, we need something else.


Prove that there exist infinitely many positive integers m such that

n4 + m

is not prime for any integer n.

Chinaboy wrote:[hide]I saw this question on AOPS forum before...
ok... m = 4k^4
n^4 + 4k^2n^2 - 4k^2n^2 + 4k^2

(n^2 + 2k^2) - (2nk)^2
= (n^2 + 2k^2 - 2nk)(n^2 +2k^2 + 2nk)
which is composite.[/hide]

This is commonly known as the Sophie Germain Identity. Very useful for solving these problems.

[hide=" another solution"]
For example

If $ n^4\equal{}1$, $ m\equal{}4k\minus{}1$ works

If $ n^4\equal{}0$, $ m\equal{}4k$ it's ok.

If $ n^4>1$, $ m\equal{}kn^4$ satisfy the condition.

[/hide]

 :)

[quote="ElChapin"]
If $ n^4 > 1$, $ m \equal{} kn^4$ satisfy the condition.
[/quote]

I think the question statement indicates that $ m$ is independent of $ n$.  :wink:

:oops:  sorry...

[quote="Arne"]is not prime for any integer n.[/quote]

 :)


for the first one, start off with the ozonolysis products, gorup together MeCHO and (Me)2CHCHO to get one of te alkenes, and other two to get other alkene, as both alkenes have six carbons.
3. the benzene ring attached to O and not CO is brominated, as CO is a deactivating group
4.it is a friedel crafts reaction.
and abt the seconf one, that statement is weird, what it seems to mean is that all carbons are chiral, whihc is not possible, but i think what the problem setter  means is that the groups attached to CO are diff.

Can u give the mechanism of the 4th reaction?
the solution set says the major product is Ph-CH2-Ph

yes, that is rite.
 the mech is:
Ch2Cl2 + FeCl3-> CH2Cl+(carbocation) + FeCl4-
CH2Cl+ + C6H6->C6H5-CH2Cl + H+
C6H5-CH2Cl + FeCl3-> C6H5-CH2+(carbocation) + FeCl4-
C6H5-CH2+ + C6H6-> C6H5-CH2-C6H5 + H+.
as i said this is a normal friedel crafts reaction

thank u very much....the problems seem so trivial now.....i shud have got it on my own...but neways :fool:


inequalities
trigonometry
inequalities unsolved

Let $ f(x)\equal{}\frac{\sin x\plus{}\cos x}{\sin x\plus{}\tan x}\plus{}\frac{\tan x\plus{}\cot x}{\cos x\plus{}\tan x}\plus{}\frac{\sin x\plus{}\cos x}{\cos x\plus{}\cot x}\plus{}\frac{\tan x\plus{}\cot x}{\sin x\plus{}\cot x}$
Then find the minimum when $ x\in (0,\frac{\pi}{2})$

[quote="tobeno_1"]Let $ f(x) \equal{} \frac {\sin x \plus{} \cos x}{\sin x \plus{} \tan x} \plus{} \frac {\tan x \plus{} \cot x}{\cos x \plus{} \tan x} \plus{} \frac {\sin x \plus{} \cos x}{\cos x \plus{} \cot x} \plus{} \frac {\tan x \plus{} \cot x}{\sin x \plus{} \cot x}$
Then find the minimum when $ x\in (0,\frac {\pi}{2})$[/quote]
$ f(x) \equal{} \frac {\sin x \plus{} \cos x}{\sin x \plus{} \tan x} \plus{} \frac {\tan x \plus{} \cot x}{\cos x \plus{} \tan x} \plus{} \frac {\sin x \plus{} \cos x}{\cos x \plus{} \cot x} \plus{} \frac {\tan x \plus{} \cot x}{\sin x \plus{} \cot x}$
$ \equal{}(sinx\plus{}cosx)(\frac{1}{sinx\plus{}tanx}\plus{}\frac{1}{cosx\plus{}cotx})\plus{}(tanx\plus{}cotx)(\frac{1}{cosx\plus{}tanx}\plus{}\frac{1}{sinx\plus{}cotx})$
$ \geq \frac{4(sinx\plus{}cosx)}{sinx\plus{}cosx\plus{}tanx\plus{}cotx}\plus{}\frac{4(tanx\plus{}cotx)}{sinx\plus{}cosx\plus{}tanx\plus{}cotx}\equal{}4$

[quote="chien than"][quote="tobeno_1"]Let $ f(x) \equal{} \frac {\sin x \plus{} \cos x}{\sin x \plus{} \tan x} \plus{} \frac {\tan x \plus{} \cot x}{\cos x \plus{} \tan x} \plus{} \frac {\sin x \plus{} \cos x}{\cos x \plus{} \cot x} \plus{} \frac {\tan x \plus{} \cot x}{\sin x \plus{} \cot x}$
Then find the minimum when $ x\in (0,\frac {\pi}{2})$[/quote]
$ f(x) \equal{} \frac {\sin x \plus{} \cos x}{\sin x \plus{} \tan x} \plus{} \frac {\tan x \plus{} \cot x}{\cos x \plus{} \tan x} \plus{} \frac {\sin x \plus{} \cos x}{\cos x \plus{} \cot x} \plus{} \frac {\tan x \plus{} \cot x}{\sin x \plus{} \cot x}$
$ \equal{} (sinx \plus{} cosx)(\frac {1}{sinx \plus{} tanx} \plus{} \frac {1}{cosx \plus{} cotx}) \plus{} (tanx \plus{} cotx)(\frac {1}{cosx \plus{} tanx} \plus{} \frac {1}{sinx \plus{} cotx})$
$ \geq \frac {4(sinx \plus{} cosx)}{sinx \plus{} cosx \plus{} tanx \plus{} cotx} \plus{} \frac {4(tanx \plus{} cotx)}{sinx \plus{} cosx \plus{} tanx \plus{} cotx} \equal{} 4$[/quote]
Very nice! Thank you!


Given a  △ABC , △ABC’△BCA’△ACB’ are isoceles triangles outside △ABC
∠A’+∠B’+∠C’=360
Prove:2∠A’B’C’=∠AB’C

[quote="April"]I guess triangles $ BCA'$, $ CAB'$, $ ABC'$ are isosceles at $ A'$, $ B'$, $ C'$ respectively.

Construct a point $ K$ such that $ \triangle BCA'\sim\triangle KBC'$. We can easily to see that $ \triangle CAB'\sim\triangle KAC'$, therefore $ \triangle AB'C'\sim\triangle ACK$. It implies $ \angle AKC \equal{} \angle AC'B'$. Similarly, $ \angle BKC \equal{} \angle BC'A'$. Hence, $ \angle AC'B' \plus{} \angle BX'A' \equal{} \angle AKB \equal{} \frac {1}{2}\angle AC'B \equal{} \frac {1}{2}\left(\angle AKB \plus{} \angle B'C'A'\right)$. The result follows.[/quote]
yes，the same way to mine.but I denote D instead of your K

See [url=http://www.mathlinks.ro/viewtopic.php?t=167759&sid=2624edc17dd9dd03e814f7d9ad308bc0]here[/url] for a general theorem which "solves" this problem...


Um, yes, but if you have such a high rating, you should be winning more often.

dude, calm down

i wud be happy to trade ratings with u

[quote="Jongao"]I believe that people with higher ratings are put at a disadvantage.
A loss could mean a drop of 40 points, and a win adds 0 or 1 point.[/quote]

You might be right.  If you want to do some research on rating systems to find something more appropriate (and how to code it up), let me know.  Right now we are using the ELO rating system (modified to let it have a bit more impact on ratings), which is what the official Chess rating system is and the rating system used on most web based multi user games (e.g., Yahoo Games, Comedy Central, etc.)  Just like here, if you were 1500 and playing a 1200, your rating wouldn't go up much, if at all, because you would be definitely expected to win.

Yeah, I play chess offically, and I'm 1815.

If I was playing some 1200, I would be expected to win. If s/he wins, then it

would mean a rating boost. If I win, then I might gain 1 point, more likely 0.

Yeah it would get annoying, but its the fairest system around, if the system was

x points for winning, -x points for losing, then some 1400 would play some 1100

a bunch of times, and ratings would be pretty meaningless

[quote="BOGTRO"]Yeah, I play chess offically, and I'm 1815.

[/quote]

 [size=25][color=white]alex yr such a show-off[/color][/size] :P

im like 1400

but im underrated

but i mean yeah, it makes sense. unless u uber-mem, your rating would be higher b/c yr probably more adept at math

so dont complain, and just win the easy games!

[quote="firecricket91"]Or you can just purposely deprove your rating to 1000 and win a 10 player game and it jumps up to 1450.

Its better than having it stay at 1350 because even at 1350 and you win a 10 player game your rating only goes up to like 1400.[/quote]

are you serious? That works lol, and I was trying to keep my rating up... (currently something like 1281 or something)


Does reasonable solution ?
Solve:$ x^2\plus{}x\plus{}1\equal{}0$      (1)
Realize $ x\equal{}0$ not satisfied (1)
When $ x\neq 0$, multiplication 2 sides of (1) we have:$ x^3\plus{}x^2\plus{}x\equal{}0$   (2)
From (1) we have :$ x^2\plus{}x\equal{}\minus{}1$ change into (2) we are: $ x^3\minus{}1\equal{}0$ equivalent $ x^3\equal{}1$$ \equal{}>x\equal{}1$
SO,solution of an equation is $ x\equal{}1$
What you have comments?Where error

[quote="mathz"]Does reasonable solution ?
Solve:$ x^2 \plus{} x \plus{} 1 \equal{} 0$      (1)
Realize $ x \equal{} 0$ not satisfied (1)
When $ x\neq 0$, multiplication 2 sides of (1) we have:$ x^3 \plus{} x^2 \plus{} x \equal{} 0$   (2)
From (1) we have :$ x^2 \plus{} x \equal{} \minus{} 1$ change into (2) we are: $ x^3 \minus{} 1 \equal{} 0$ equivalent $ x^3 \equal{} 1$$ \equal{} > x \equal{} 1$
SO,solution of an equation is $ x \equal{} 1$
What you have comments?Where error[/quote]
You just used $ \implies$ and not $ \iff$ :

$ x^2 \plus{} x \plus{} 1 \equal{} 0$

$ \iff$ $ x\ne 0$ and $ x^3\plus{}x^2\plus{}x\equal{}0$

$ \iff$ $ x\ne 0$ and $ x^2\plus{}x\plus{}1\equal{}0$ and $ x^3\minus{}1\equal{}0$

and so $ 1$ is not solution (while $ j$ and $ j^2$, the two other cubic roots of unity are)


geometry
circumcircle
inequalities
geometry proposed

A ray emanating from the vertex $A$ of the triangle $ABC$ intersects the side $BC$ at $X$ and the circumcircle of triangle $ABC$ at $Y$. Prove that $\frac{1}{AX}+\frac{1}{XY}\geq \frac{4}{BC}$.

We have to prove $\frac{1}{AX}+\frac{1}{XY}\geq \frac{4}{BC}$. After multiplication with BC, this inequality takes the form $\frac{BC}{AX}+\frac{BC}{XY}\geq 4$. But BC = BX + CX; thus, we can rewrite this inequality as $\frac{BX}{AX}+\frac{CX}{AX}+\frac{BX}{XY}+\frac{CX}{XY} \geq 4$. But this is clear, since $BX\cdot CX=AX\cdot XY$ by the Intersecting Chords Theorem, so that $\frac{BX\cdot CX}{AX\cdot XY}=1$, and thus AM-GM yields

$\frac{BX}{AX}+\frac{CX}{AX}+\frac{BX}{XY}+\frac{CX}{XY} \geq 4\sqrt[4]{\frac{BX}{AX}\cdot\frac{CX}{AX}\cdot\frac{BX}{XY}\cdot\frac{CX}{XY}}$
$=4\sqrt[4]{\left(\frac{BX\cdot CX}{AX\cdot XY}\right)^2}=4\sqrt[4]{1^2}=4$.

[quote="Megus"]where can I find proof of Intersecting Chords Theorem, because to be honest, this is the first time I've come across it.  :maybe:[/quote]

You may have heard of it under the name Power of a Point.  It's in AoPS vol. 1.

Anyway, the proof for a point inside a circle is easy.  Consider a point X inside a circle, and chords AC and BD passing through X.  $\angle AXB$ and $\angle CXD$ are equal, due to the fact that they are vertical angles.  Also, $\angle DCA$ and $\angle DBA$ are equal, since they are inscribed in the same arc.  Thus, ABX and CDX are similar triangles.  Therefore, we have $\frac{CX}{DX}=\frac{BX}{AX}$, or $CX\times AX=BX\times DX$, and we are done.

Let $AC,BD$ be intersecting chords, and let $X$ be their intersection point. Then, from the fact that $ABCD$ is cyclic, we get $\angle XAB=\angle XDC,\ \angle XBA=\angle XCD$, and we thus find the triangles $XAB,XDC$ to be similar. This means that $\frac{XA}{XB}=\frac{XD}{XC}$, which, in turn, proves $XA\cdot XC=XD\cdot XB$.

Holy ****  :blush: After all I remarked what it is indeed - I've known this theorem well since few years...  :blush:  At least I know what does the chord [I cannot find it in a dictionary so far] mean eventually  :D

Hm.  I was just editing my post to include the proof... guess you beat me to it.


calculus
integration
function
trigonometry
real analysis
real analysis solved

Compute the indefinite integration ( primitive function):
	$ \int \frac{t}{1+cos(t)}d(t)$

Questions about elementary ("closed form") antiderivatives don't usually make for interesting topics on this forum. For one thing, elementary functions without elementary antiderivatives are abundant.

But in this case, there is an elementary antiderivative. I'll give you the first step:

$\frac t{1+\cos t}=\frac{t(1-\cos t)}{(1+\cos t)(1-\cos t)}=$
$= \frac{t(1-\cos t)}{\sin^2t} =t(\csc^2t-\csc t\cot t)$

Now go finish it yourself, integrating by parts where it makes sense to do so.

Or you can perform a trignometric substitution : $\cos t = \frac{1-x^2}{1+x^2}$ where $x = \tan \frac{t}{2}$, thus $t = 2\arctan x$ and $dt = \frac{2}{1+x^2} dx$

$\int \frac{t}{1+\cos t}dt = \int \frac{\frac{4 \arctan x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}}dx = \int 2\arctan x dx$

However, this does not seem to give a neat primitive...


Let $F_{n}$ denote the $n$th Fibonacci number (with $F_{0}=F_{1}=1$). What is the minimum value of the sum

$|x-F_{F_{0}}|+|x-F_{F_{1}}|+|x-F_{F_{2}}|+\ldots|x-F_{F_{6}}|$

[quote="Chocoboom"][hide="i think"]Listing out $F_{0}$ to $F_{6}$, we get $1, 1, 2, 3, 5, 8, 13$
Here we see that it goes up to $F_{13}$, so lets list out a couple more terms. Starting with with $F_{7}$, we have $21, 34, 55, 89, 144, 233, 377$
Our equation then becomes:
$|x-1|+|x-1|+|x-2|+|x-3|+|x-8|+|x-34|+|x-377|$
So basically the sum of the differences between x and the fibonacci term.
To minimize it, we pick x as the middle term which is $3$.
$|3-1|+|3-1|+|3-2|+|3-3|+|3-8|+|3-34|+|3-377|=415$[/hide][/hide][/quote]

Well, graphing it, we see that you're right. But how did you know that picking the middle term would give you the minimum?

Umm.. I suck with proofs, but I'll try to explain it with logic. If the x gets bigger and bigger from the middle term, then for each unit the x increases, then all the lower terms, in this case 4 of them, their difference increases 4 times (1 for each term). However in the above terms, the difference decreases only by 3 times (1 for each term bigger then x). Same goes if x is decreased. So we have an increase in difference.

you should work on your proof abilities...

this one is not so bad...hint:

take x in intervals, namely, 0<x<F_F_1, or F_F_1<x<F_F_2, etc

also, whats up with the awk numbers?

Use triangle inequality to show that's the minimum

in particular,

$|a|+|b| \geq |a+b|$ and $|a|+|b|\geq |a-b|$ should make the solution obvious...


[quote="Nameless"][quote="pvthuan"]$\text{Prove that if} \; a,b,c\geq 0,\; \text{then}$
\[ \frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ca}+\frac{a+b}{c^2+ab}\geq\frac8{a+b+c}.  \][/quote]
I think it must be
\[ \frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ca}+\frac{a+b}{c^2+ab}\geq\frac9{a+b+c}.  \][/quote]
no, 1,1,0.

Note that equality holds when one of the numbers is $0$, other two equal, say $a=b=1$, $c=0$.


trigonometry
function
integration
calculus
derivative
calculus computations

find value of cosine series of function $\displaystyle F(x)=\sin{x}$ on $\displaystyle [0,\pi]$ at $\displaystyle x= 2005$.

for this i started by first finding the fourier series
for this i got $\displaystyle a_0 = \frac{4}{\pi}$
and for coefficient of cosine series i am stuck at
$\displaystyle \frac{2}{\pi}\int_0^\pi \sin{x}\cos{nx}dx $
how do i go about integrating this? (or is there a shorter different approach to this problem??)
thanks,

There are several different approaches, including product to sum trigonometric identities and just expressing everything in terms of complex exponentials. Here's another approach: integrate by parts:

$\int_0^{\pi}\sin x\cos nx\,dx=-\cos x\cos nx|_0^{\pi}- n\int_0^{\pi}\cos x\sin nx\,dx$

$=-\cos x\cos nx|_0^{\pi}- n\sin x\sin nx|_0^{\pi} +n^2\int_0^{\pi}\sin x\cos nx\,dx$

$\int_0^{\pi}\sin x\cos nx\,dx=-((-1)(-1)^n-1)+0-0+n^2\int_0^{\pi}\sin x\cos nx\,dx$

$(1-n^2)\int_0^{\pi}\sin x\cos nx\,dx=-((-1)^{n+1}-1)$

$\frac2{\pi}\int_0^{\pi}\sin x\cos nx\,dx=-\frac4{(n^2-1)\pi}$ if $n$ is even and zero is $n$ is odd.

The function $f(x)=\sin x$ on $[0,\pi]$, considered as an even function periodic of period $2\pi$ (which actually makes it periodic of period $\pi$) is continuous but not $C^1$. Its first derivative has a jump discontinuity. Such a function should have Fourier coefficients that are $O(\frac1{n^2}).$ Also the fact that it is actually periodic of period $\pi$ means we should be able to express it as a series in $\cos2kx.$ Both of these predictions are satisfied by this calculation.

thanks and for finding value at 2005 should i divide it by $\pi$ to find the remainder.(as 2005 is not in the interval and it is a periodic extension)??

Actually, you never needed to find the series at all. The value you are looking for is simply $|\sin2005|,$ since the even extension of this function is $|\sin x|.$


I'm a high school senior looking for a greater review of geometry than AOPS V1 can offer. I've ventured to obtain the book "Geometry" by McDougal Littell. I have just one question as of yet.

*39 Write a paragram proof: The sum of the lengths of the segments drawn from any point in the base of an isosceles triangle perpendicular to the legs is equal to the lengths of the alititude drawn to one leg.

**Before further consideration, please only use :
A)Parallel lines and Planes
B)Congruent Triangles
C)Quadrilaterals

I understand this is probably super easy to prove, but I don't have much time. Please don't use pyth. , similar triangles, LOC ect, which also make it very easy as I've realized.

Someone please, I don't see a conventional way of proving it. The wording is quoted with 100% accuracy.

[hide="hint"]Try drawing an auxiliary line to create a rectangle.[/hide]

[hide="solution"]Let the vertices of the triangle be $ A$,$ B$, and $ C$. With $ AB \equal{} CD$
Let $ Y$ be a point on $ AC$ and let $ X$ and $ Z$ be the intersections of the perpendiculars to side $ AB$ and $ CB$ respectively. Let $ M$ be the point of intersection of the altitude from $ A$ to $ BC$. Let $ L$ be the intersection of $ XY$ and $ AM$

We can see $ m\angle YZM \plus{} m\angle AMZ \equal{} 180$ So $ YZ \parallel AM$
Now draw an auxiliary line that is perpendicular to $ AM$ from $ Y$ through $ AM$ and let $ P$ be the intersection of the lines, and $ L$ be the intersection with $ AB$.
We can see that $ YZ \cong PM$ all we need to do now is show that $ LY \parallel BC$ so we can see that $ \Delta ALY$ is Isosceles and that $ \angle A \cong \angle LYA$ and show that $ \Delta XAY \cong \Delta PYA$, so that $ PA \cong XY$. We can see that $ m\angle ZMP \plus{} m\angle YPM \equal{} 180$ so $ LY \parallel BC$. And since $ AP \plus{} PM \equal{} AM$ and $ PM \equal{} YZ$, $ AP \equal{} XY$ Then $ YZ \plus{} XY \equal{} AM$ [/hide]

Thanks very much with the help. Due to my lack of effort in geometry 2 years ago, I feel like a rookie, thanks. I forget sometimes how flexible geometry is.


Show that $ \pi_{1}(\mathbb{R}^{2}\minus{}\mathbb{Q}^{2})$ is uncountable ,by using Van Kampen's theorem (there are other ways to prove this as well)

I just know that$ \pi_{1}(\mathbb{R}^{2}\minus{}\mathbb{Q}^{2})$ is uncountable since $ \mathbb{R}^2\minus{}\mathbb{Q}^2$ is uncountable
Could you show me how to use Van Kampen's theorem to prove this? I am really curious ! thanks

[quote="1234567a"]I just know that$ \pi_{1}(\mathbb{R}^{2} \minus{} \mathbb{Q}^{2})$ is uncountable since $ \mathbb{R}^2 \minus{} \mathbb{Q}^2$ is uncountable[/quote]
That has nothing to do with each other. There are uncountable spaces that have finite fundamental group, there are some that have countable fundamental group and there are some that have uncountable fundamental group.


Here is a problem I was working on, couldn't find a real slick solution to it maybe someone else has some insight.

Find all natural numbers equal to the sum of the factorials of their digits.

GL

http://mathworld.wolfram.com/Factorion.html

These are called factorions.

1
2
145
40585

Is there any good way to prove those 4 numbers are factorions (well 1 and 2 are obvious  :P ) without simply trial and error? What about proving they are the only ones (other then just trial and error and using the upper bound). Just wondering b/c the question was in a text.

They found that the upper bound must not exceed $ 2540160$, so they just brute-forced, I assume.

[quote="t0rajir0u"][quote="Spacecow"]Is there any good way to prove those 4 numbers are factorions (well 1 and 2 are obvious  :P ) without simply trial and error?[/quote]
The definition of a factorion isn't very deep.  What's wrong with calculating the sum of the factorials of the digits?[/quote]

What I meant was is there some reason you would expect those numbers to be factorions, if you didn't know ahead of time. Other then 1 and 2 of course.

Not particularly, no.  You can reason about the range such a number must take based on (the factorial of) its largest digit, but other than that this is a very computational and "accidental" problem (in my opinion).


Pablo copied from the blackboard the problem:
[list]Consider all the sequences of $2004$ real numbers $(x_0,x_1,x_2,\dots, x_{2003})$ such that: $x_0=1, 0\le x_1\le 2x_0,0\le x_2\le 2x_1\ldots ,0\le x_{2003}\le 2x_{2002}$. From all these sequences, determine the sequence which minimizes $S=\cdots$[/list]

As Pablo was copying the expression, it was erased from the board. The only thing that he could remember was that $S$ was of the form $S=\pm x_1\pm x_2\pm\cdots\pm x_{2002}+x_{2003}$. Show that, even when Pablo does not have the complete statement, he can determine the solution of the problem.

I am not very sure about this. Never used the condition $0\leq{x_i}$.
We can prove the general case, for any integer $n$,
$S=\pm{}x_1\pm{}x_2\cdots\pm{}x_{n-1}+x_n$ is maximized when $x_i=2^i$.
Use induction on $n$. If $n=1$, then the case is obvious. For $n=k$:
If the sign before $x_{n-1}$ is positive, then $S-x_n$ is maximized by assumption, so will be $x_n$ if $x_i=2^i$.
If the sign before $x_{n-1}$ is negative, then $-x_{n-1}+x_n\leq-x_{n-1}+2x_{n-1}=x_{n-1}$, with equality if $x_n=2x_{n-1}$. So $S\leq\pm{}x_1\pm{}x_2\cdots\pm{}x_{n-2}+x_{n-1}$, which is maximized if $x_i+2^i$. Altogether, $S$ is maximized if $x_i=2^i$ for $1\leq{i}\leq{n-1}$ and $x_n=2x_{n-1}$.


search
probability
number theory unsolved
number theory

let Pi denotes i th prime is (P1)(P2)...(Pm)+1 always squarefree?for all natural m.

I think it is not true. You can tested it by computer.

Search it yourself after you got a cluster of PCs.

i just want to make sure whether its already done or is rust fake.

I checked that $p_{1}\cdot\ldots\cdot p_{n}+1$ is squarefree for $n\leqslant 10^{5}$.

I was wrong  :wallbash_red:  :wallbash_red:  :wallbash_red: .  I evaluated in Mathematica 4.0 a small program:
[code]i = 1; p = 2; While[i <= 100000, If[Not[SquareFreeQ[p + 1]], Print[i]];
  p = p*Prime[i + 1]; i++][/code]
and didn't get any output. I thought that it meant that all numbers $p_{1}\cdot\ldots\cdot p_{i}+1$ for $1\leqslant i\leqslant 10^{5}$ were squarefree, but some error with previously evaluated
[code]<< NumberTheory`NumberTheoryFunctions`[/code]
(a command loading a package containg command SquareFreeQ among other things) caused that SquareFreeQ[p+1] wasn't evaluated and Not[SquareFreeQ[p+1]] didn't return value True, and this was the reason that none of numbers $i$ was printed at output. In fact, it was checked only for $i\leqslant 33$ (note that $p_{1}\cdot\ldots\cdot p_{33}+1= 72047817630210000485677936198920432067383702541010311$).
I wonder how did [b]Rust[/b] get his result.

Algoritm. We have 6542 primes less than 65536. Let p[i] i-th prime.
Construct massive
for (i=0,i<6542,i++)a[i]=0,b[i]=1,
were M=b(i)+p(i)a(i) (mod p(i)).

Then for(j=0;j<6542,j++){
for(i=j,i<6542;i++{
b(i)*=p(j), x=b(i)/p(i), b(i)-=(x*p(i),
a(i)*=p(j),a(i)+=x;a(i)%=p(i),
if((b(i)==p(i)-1)and (a(i)==p(i)-1)) cout<< p(j)<< p(i),
}}
Calculate 1 sec. For check primes p>65536 we have p^2>2^{32},
therefore it need modificied for naturals>2^{32}.

This seems to be an open question, so good luck for solving this!

Let \[M_{n}=1+\prod_{i=1}^{n}p_{i}.\] I check, that $q^{2}\not |M_{n}\ \forall n\in N, \forall q\le 2^{24}$.
Let $q=p_{k}$ prime, then $q_{k}\not |M_{n}, n\ge k$, therefore probabity that exist i, suth that $p_{k}|M_{i}$ is $\frac{k-1}{p_{k}}$ and probability, suth that exist i: $p_{k}^{2}|M_{i}$ is $\frac{k-1}{p_{k}^{2}}$, therefore probability suth that all $M-i$ are squarefree is \[P=\prod_{k}(1-\frac{k-1}{p_{k}^{2}})\] . I sheck that $k>1077871$, therefore $P>0.93$.


Given a right triangle whose side lengths are integral multiples of 7, how many units are in the smallest possible perimeter?

The smallest pythagorean triplet is 3-4-5, so the smallest possible triangle is 21-28-35. The perimeter is 84.

What dragon96 is doing is taking the smallest pythagorean triple and scaling it up 7.  This will always work.

[quote="markoboy"][hide="this is my answer ... i think"]the smallest triplet is

3-4-5

so we want to find the one with the multiples of 7 so we multiply and get

21-28-35=84

and their perimeter is 84 so the answer is 84 [/hide]

[b][i][u]please rate my post thanks ![/u][/i][/b][/quote]

It's the exact same solution the people above you wrote...

-1


[quote="mathmanman"]All positive integers are distributed among two disjoint sets $N_{1}$ and $N_{2}$ such that no difference of two numbers belonging to the same set is a prime greater than 100.

Find all such distributions.

[i]Proposed by N. Sedrakyan[/i][/quote]
If $n$ is in $N_{1}$ then $n+101$ must be in $N_{2}$ and $n+103$ must be in $N_{2}$. But then $n+2$ must be in $N_{1}$.
If $m$ is in $N_{2}$ then $m+101$ must be in $N_{1}$ and $m+103$ must be in $N_{1}$. But then $m+2$ must be in $N_{2}$.

So, if $1$ is in one of the to subsets (and it must be), then all the odd positive integers are in the same subset.
But if $2$ is in the same subset too than all the numbers are in only one subset $\to$ contradiction.
So, $2$ (and all the evens) are in the other subset.

[b]The answer is[/b]: We have only $2$ possibilities:
[b]1)[/b] All the odds are in $N_{1}$ when all the evens are in $N_{2}$
[b]2)[/b] All the odds are in $N_{2}$ when all the evens are in $N_{1}$

Obviously, this distibution garantees that the difference between two numbers in the same subset will be even, so it cannot be prime greater than 100.

I hope I was clear.
Sorry for my bad English!


$ a,b,c>0$ prove:
$ \frac{a^{2}}{b}\plus{}\frac{b^{2}}{c}\plus{}\frac{c^{2}}{a}\geq\frac{bc}{a}\plus{}\frac{ac}{b}\plus{}\frac{ab}{c}$

[quote="zaya_yc"]$ a,b,c > 0$ prove:
$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\geq\frac {bc}{a} \plus{} \frac {ac}{b} \plus{} \frac {ab}{c}$[/quote]
$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\geq\frac {bc}{a} \plus{} \frac {ac}{b} \plus{} \frac {ab}{c}$[/quote]
This inequality does not hold for all $ a,b,c>0$.  For example, the inequality is equivalent to
\[ \sum ab^3 \ge \sum a^2b^2\]
Now, let $ c \rightarrow 0$ and $ a\equal{}2,b\equal{}1$.
By the way, it holds for $ a,b,c$ are the side lengths of a triangle, this is IMO 1983 problem. :)

IMO 1983 inequality is equivalent to
\[ \sum\limits_{cyc}(a \minus{} b)(a \minus{} c)(b \plus{} c)(a \plus{} c \minus{} b)\geq 0
\]
we can prove it by Vornicu Schur Inequality. Or we can prove it by S.O.S method
in following form
\[ \sum\limits_{cyc}(a\minus{}b)^2(a^2\plus{}ab\minus{}b^2)\geq0\]


Show that the area under the curve f(x) = xx between x = 0 and x = 1 is equal to [tex] \displaystyle {\frac {1}{1^1} - \frac {1}{2^2} + \frac {1}{3^3} - \frac {1}{4^4} + ...} [/tex]

I assume we are allowed to use calculus

[hide]

x^x = e^(xln(x)) =  :Sigma: x^n ln(x)^n/n!

Now xln(x) is bounded so we are allowed to swap int and sum.

int(0..1) x^n*ln(x)^n = (1)^n * n! / (n+1)^(n+1) (by parts).[/hide]


Source: Neopets (<-- :rotfl:)

Source jokes aside, this problem is quite interesting.

How many circles of diameter 1 can be packed into a 10 x 10 square so that no two circles overlap?

Hint: The answer is not 100  :D

[hide="Interesting hypothesis"] Do you think we could approximate the digits of $\pi$ this way?[/hide]

[quote="cincodemayo5590"][hide="Solution"] First, the most space efficient arrangement of the circles is honeycomb-style, with repeating rows of 10 and 9 circles, 10 in the bottom row. Connecting radii of three triangles in the bottom left corner, we see that the vertical distance between radii is $2\sin{60}=\sqrt{3}$. The vertical distance from the topmost point of the bottom circle and the radius of the upper circle is thus $\sqrt{3}-1$, meaning that the vertical distance between the tops of the rows of circles is $1+(1-\sqrt{3})=2-\sqrt{3}$. This means that the total height of the honeycomb is $1+n(2-\sqrt{3})$, where n is the number of rows. Thus, $1+n(2-\sqrt{3}) \leq 10 \implies n=33$. Counting circles from the bottom, we have $17(10)+16(9)=\boxed{314}$.
I think that the answer being the first 3 digits of $\pi$ is really cool  :D [/hide][/quote]

I think you miscounted somewhere (probably number of rows), because $314$ circles of diameter $1$ have area $314 \cdot \frac{\pi}{4} > 100$...

[quote="paladin8"][quote="cincodemayo5590"][hide="Solution"] First, the most space efficient arrangement of the circles is honeycomb-style, with repeating rows of 10 and 9 circles, 10 in the bottom row. Connecting radii of three triangles in the bottom left corner, we see that the vertical distance between radii is $2\sin{60}=\sqrt{3}$. The vertical distance from the topmost point of the bottom circle and the radius of the upper circle is thus $\sqrt{3}-1$, meaning that the vertical distance between the tops of the rows of circles is $1+(1-\sqrt{3})=2-\sqrt{3}$. This means that the total height of the honeycomb is $1+n(2-\sqrt{3})$, where n is the number of rows. Thus, $1+n(2-\sqrt{3}) \leq 10 \implies n=33$. Counting circles from the bottom, we have $9(5)+8(4)=\boxed{314}$.
I think that the answer being the first 3 digits of $\pi$ is really cool  :D [/hide][/quote]

I think you miscounted somewhere (probably number of rows), because $314$ circles of diameter $1$ have area $314 \cdot \frac{\pi}{4} > 100$...[/quote]
[hide]
yea, i messed up alot... in the morning i'll get rid of it and retype it if i have time. i messed up with subtraction (yay! i can't do basic math!) and then i flipped the signs...

[hide]
Fill in the first row of $10$ circles, with a height of $1$. Adding the row of $9$ and another row of $10$ to form the honeycomb pattern adds a height of $\sqrt{3}$  (a 30-120-30 triangle connects the vertices of the three circles). The height of the honeycomb will be $1$ added to some multiple of $\sqrt{3}$. Since it has to be less than or equal to 10, we find that that multiple is $5$. The remaining height is not enough to add even just a row of $9$ circles which would add. The total area is then $(10+5\times 19)(\frac{\pi}{4})=\frac{105\pi}{4}$.

[/hide]

[hide]If you place the circles honeycomb-style, you can take two circles from the bottom row and the circle from the next row up (a row of 9) and connect the centers to form an equilateral triangle.  The height is $\frac{\sqrt3}{2}$, and the radius of a circle is $\frac{1}{2}$, so the center of a circle in the row of 9 is $\frac{\sqrt3-1}{2}$ above the top of a circle in the bottom row.  Multiplying this by two, we find that the distance between the first two rows of 10 is $\sqrt3-1$, and we can use a value of $\sqrt 3$ for the height of 2 rows, because this factors in the overlap between the row of 9 and the next row of 10.  We can include 5 groups like this into a 10x10 square, so that is $5*19=95$ circles, plus we can fit in 1 more row of 10, giving us a total of $\boxed{105}$.[/hide]

:blush: Oops, I didn't read your post carefully enough, Adunakhor...  I pretty much repeated your solution...

All answers listed are incorrect  :D

[hide="hint"]one of the answers given is only $\sin^2 x + \cos^2 x$ away from the correct answer[/hide]

So are you saying it's somehow 106?  I don't think that works.  It probably would areawise, but the shape doesn't work...

How can the answer be either $313$ or $315$ circles? Unless it's one off from $105$ (DanK's answer and mine were the same, I just included the area to show that it was less than $100$).

It's definitely not in the 300s, you're right there...  I'm thinking someone just did it with areas and forgot that shape matters too...?

[hide="tricky problem"]It is all correct until the last three rows. In the last three rows, you could have one set of 19 circles (2 lines) and 1 final line of 10 circles, but you could do something else...[/hide]

I just got 106. I'm pretty sure you can fit 2 consecutive rows of 10 in, with all of the other rows alternating between 9 and 10. When checking the height, I computed $8(\sqrt{3})\2 +2(1/2) +2(1) = 9.93$

So you mean that extra little bit of height is enough to make the last 3 rows all rows of 10, which is why it would be 106?

Yeah, in the 105 circle setup, there was $10-(1+5\sqrt{3})$ left over for the height. Changing a row of 9 circles to a row of 10 only adds $2-\sqrt{3}$ which is less than the excess so it fits.


Given that $ x, y, z$ are non-negative reals such that $ x \plus{} y \plus{} z \equal{} 1$, prove:

$ \dfrac{3}{2} < \dfrac{1 \minus{} x}{1 \plus{} x} \plus{} \dfrac{1 \minus{} y}{1 \plus{} y} \plus{} \dfrac{1 \minus{} z}{1 \plus{} z} \le 1 \plus{} \dfrac{x^2}{1 \minus{} x} \plus{} \dfrac{y^2}{1 \minus{} y} \plus{} \dfrac{z^2}{1 \minus{} z}$

[edited]

see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1187437#1187437[/url]

it is not hard

also the right inequality should be nonstrict


Let $k$ be number of the positive terms and $m$ be number of the negative terms, $x$ be arithmetic average of the positive terms and $y$ be arithmetic average of the negative terms.
It is clear $k+m=n$ and $a_n-a_1\geq x-y$. So we need to prove $x-y\geq\frac{2S}{n}$.
We can write $kx-my=S$ and $kx+my=0$, so $x=\frac{S}{2k}$ and $y=-\frac{S}{2m}$. Therefore,
\[x-y=\frac{S}{2}\left(\frac{1}{k}+\frac{1}{m}\right)\geq\frac{S}{2}\cdot\frac{4}{k+m}=\frac{2S}{n}.\]


<sarcasm>
Good job on answering the above question.
</sarcasm>

I may or may not be going depending on how robotics conflicts.

WISC is on that day, so Henry won't be there, I probably won't be there, Abubakar might be, Edward will be (too bad), and Claus may or may not be (in no particular order). There's not anyone else in Science Club and in Math Team who might get in the top, and we'll have the usual sparse scattering of Walton students who don't have anything else to do.

by the way, this is allen using edward's account.

omg hi billy :harhar:

Northviews coming. Heng can't come, because he has some ultimate frisbee tournament thing. Haha, he can't defend his title.

There is also an honor orchestra thing that day (auditions?). So between WISC and orchestra, Walton will be missing a few folks. I might even have to compete. That spells bad news for Walton!  :D

Don't worry Mr. Fulton; with you as our cheerleader, quite a few people will come running back (or away?)

Mr. Fulton, are you referring to All-State?  Cuz I'm pretty sure auditions are the week before, so I'll be able to go this time.  Then again, I might not cuz Heng will totally abuse the TST and make it all calc.

But everyone knows that calc is bad manners. ;\

Norcross will be attending.  This will be the first time in...4 years.

lawlz, nvmd, I kinda overestimated the amount of calc Heng would put on the TST.  It turns out the most calc he decided to put was like 2 or 3 questions on limits.  Too bad I didn't get a perfect, though. :\

I don't know exactly which (auditions, performance, practice) is occurring for (band, orchestra, chorus (I'm actually pretty sure it's not chorus)) but I'm pretty sure that something musical is happening that day that will keep some of my folks away.

Sitan, which test are you talking about that Heng made?

Hi Mr. Fulton!  I was just referring to our team selection test.  Anyways, I just found out that All-State is still on the 25th, so I can't go this year (again). :|

Let's just hope that I'll be exempt from the first round of auditions next year by making it into All-State this year, provided that I don't mess up horribly on sight reading again.

And this may be too early, but I'm probably not going to Rockdale either, for musical reasons as well.  I have a southern-states-wide piano competition that very same day, so that's two major tournaments I'll be missing this year........ :(

When is Rockdale?  I'm hoping that Norcross can attend a lot more math tourneys this year (although we missed Fulton).

I'm pretty sure it's on the 1st of November.

Thanks Sitan.

Have you seen the excerpts for 11-12?  Those octaves are ridiculously hard.

Hey,

Does Cobb County Invitational allow on site registration? If so, how much would the costs be for an individual/team?

Thanks,
B. Wang

Yes, it's all-state orchestra auditions.  I have missed Cobb every year until now for that reason.  But I [i]am[/i] exempt from all-state, so only auditioning for honor orchestra, so my audition isn't until afternoon. :)

Re azn :  Excerpt?!  As in for the second audition?  Where are they posted?  What pieces are we playing?  Do you mean étude?

Hi Miles!

To your confusion? (you never get confused, so question?) about the excerpt:

Psh, I wish the excerpts were out.  I'm pretty sure that he means etude.

I hope I make All-State this year so I can be exempt next year, but I might not. :(

I know it's rather late... but does anyone know when the tournament starts? I have to arrange the transportation without that little tidbit. Also, Mr. Stone doesn't seem to be picking up his cell phone.

Tournament starts at 9am. Be there earlier. Kennesaw Mountain High School

Also, any idea when the competition ends? I will probably be leaving for robotics right afterward.

I'm rather disappointed about my #6, in contrast to Nathan's #5, Miles's #2 and Santosh's #1.

Congratulations to Miles and Santosh for owning this tournament (again).

I placed in a real tournament!

Um...all I remember is

1- Santosh
2- Miles
3- Edward
4- Oliver
6- Ben
7- Jerry
8- Billy

hey...who's Nathan?  Is he good?
And Miles, I meant the All-state etudes.  Sheesh, so picky.

I don't think Nathan's that good, but he has placed in a few competitions over the last two years.

Edward is so beastly... :o
If feasible and allowed, could anyone PM me the questions? Thanks.

Grar...  I messed up on All-State sight reading, so I'll probably not make the 2nd round.  That means another year of non-Cobbing....

What was the highest score?  What was the total possible score?

5 Nathan Li(think)

[b]9[/b] Billy

Any of you guys have the video of my performance? I should've sung some other song, but I hadn't prepared.

I have video of some of your performance, but I dare not post it for fear of being smitten by an angry user/users of AoPS.

i didnt stay for the awards. what schools won?


to MOD: you can move this post to any other subforum if necessary.

How to find the roots of polynomials if it cannot be factored.

Ex: How to find the roots of $x^{3}+x^{2}+1=0$

http://planetmath.org/encyclopedia/CubicFormula.html

http://planetmath.org/encyclopedia/QuarticFormula.html

I believe that there isn't a formula for 5th-degree polynomial equations such as $x^{5}-4x^{4}-17x^{3}+2x^{2}+11x-2$.

Such a general formula cannot exist for degree five or above, according to the Abel–Ruffini theorem. See [url]http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem[/url]

In practice, the cubic and quartic formulas are staggeringly cumbersome, so approximate methods are usually used instead (say, by your calculator):

http://mathworld.wolfram.com/NewtonsMethod.html

As long as were on this topic, I would like to ask:

Does factoring count if we don't do it over the rationals?
I mean, are these examples of factoring:

$x^{2}+3 x+1 = \left(x-\frac{-3+\sqrt{5}}{2}\right) \left(x+\frac{3+\sqrt{5}}{2}\right)$

$x^{2}+x+1 = \left(x-\frac{-1+i \sqrt{3}}{2}\right) \left(x+\frac{1+i \sqrt{3}}{2}\right)$

It depends on the context.  "Factoring" can mean factoring over $\mathbb{Q}$, over $\mathbb{R}$, or over $\mathbb{C}$, or even over more exotic fields.

But generally, you'd factor over $\mathbb{Q}$, because $\left(x-\left(4+\sqrt{17}\right)\right)\left(x-\left(4-\sqrt{17}\right)\right)$ looks a LOT uglier, and probably more difficult to deal with, than $x^{2}-8x-1$, even though they are equivalent.

And don't even get me started with cubics or quartics.  For those who have seen the cubic and quartic formula, you'll find that if you factored one of those out, and the roots aren't pretty, that'd be a real pain.