I would like to adopt this example , for a slightly modified Sierpinski triangle: The first step (the whole triangle is black) should be the same. Then in the second step, the three black triangles in the corners should only have a side length of 1/3 (instead of 1/2) of the side length of there "parent" triangle, and so on. Could someone help me with that, i've never used this "lindenmeyersystems" package...
22536
I have been trying to recreate the following image in LaTeX    and for me personally it has to be done with pdflatex. Although answers using pst-fractals From Jake I got a good start which led to the following code  \documentclass{standalone} \usepackage{tikz} \def\level{5} \usetikzlibrary{lindenmayersystems}  \begin{document} \begin{tikzpicture}[l-system={step=5pt, order=\level, angle=120},rotate=180]     \pgfdeclarelindenmayersystem{Sierpinski triangle}{     \symbol{X}{\pgflsystemdrawforward}     \symbol{Y}{\pgflsystemdrawforward}     \rule{X -&gt; X-Y+X+Y-X}     \rule{Y -&gt; YY}     }     \draw [black] (3,2) l-system     [l-system={Sierpinski triangle, axiom=X, anchor=north east},fill=white]; \end{tikzpicture}  \end{document}   There are however two problems with the code.   I layer a black triangle behind the lindenmayersystem? How do I keep the size of the triangle consistent with different number of recursion levels.
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To each vertex of a pentagon,we assign an integer $x_{i}$ with sum $s=\sum x_{i}&gt;0$. If x,y,z are the numbers assigned to 3 successive vertices and if $y&lt;0$,then we replace $(x,y,z)$ by $(x+y,-y,y+z)$.This step is repeated as long as there is a y&lt;0.Decide if the algorithm always stops.
123406
Prove:  If we color the plain with three different colours, then there will always be an equilateral triangle which has three vertices of the same colour.  I have proved it for two colours but I just can't solve this one.
99379
Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.   Continue this process ad infinitum.  What is the area of the limiting "polygon"?  My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.   Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance.
122903
given three lines $\ell_1,\ell_2, \ell_3 $ which intersect in one point $P$. How can one construct a triangle such that the given lines become its angle bisectors?  So far I tried to find conditions on how the three given lines have to intersect such that the desired triangle exists. There are altogether six rays $\omega_1^1, \omega_1^2; \omega_2^1,\omega_2^2;\omega_3^1,\omega_3^2$ - any line $\ell_j$ determines the two rays $\omega_j^1,\omega_j^2$- emanating from the common point $P$.  Any of the three vertices $V_1,V_2,V_3$ of the desired triangle has to lie on exactly one line, i.e.  $V_j\in \ell_j=\omega_j^1\cup \omega_j^2$. Hence we have either $V_j\in \omega_j^1$ or $V_j\in \omega_j^2$.  Therefore the triangle should exist if and only if it is possible to choose three rays $\tilde{\omega}_1\in \lbrace \omega_1^1, \omega_1^2\rbrace $, $\tilde{\omega}_2\in \lbrace \omega_1^2, \omega_1^2\rbrace$, $\tilde{\omega}_3\in \lbrace \omega_1^3, \omega_1^3\rbrace$ such that any three of the rays $\tilde{\omega_1}, \tilde{\omega_2}, \tilde{\omega_3}$ form an angle less than $\pi$.  But I do not know how to construct the triangle?!   I started to draw a circle around $P$ with any radius. Now I can choose a Point $V_1$ on the ray $\tilde{\omega_1}$ outside the circle. And now I can draw the tangents through $V_1$ to the circle. I think those tangents will meet the other rays at some points $V_2,V_3$ (Why?). Now we can draw the line $V_2,V_3$. But this line needs to be tangent to the circle and I'm not sure about that. Will this construction work, or is it done in a different way?  Best regards
115265
Problem:    How many triangles do $m$ lines form if   a) Every two lines intersect and no three lines intersect at one point.  b) There are $n$ lines among $m$ lines that are parallel to each other. No other line is parallel to these $n$ lines, and no other two lines are parallel to each other. Again no three lines intersect at one point.  Thank you.
112207
I'm not sure how to go about this. One of my thoughts was to take the number of edges in a complete graph, and subtract all of the edges that would be needed to make a smaller cycle.   If you have $C_5$, there is only one way to make a $5$-cycle on $5$ vertices, but there's $10$ ways to make a $3$-cycle, and...  I have no idea how to do this without counting things multiple times.
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How do draw a right triangular prism in latex in this point of view?   And moreover, how can I add the right angle label like this?   Finally, how to add lines in the same figure?   Thank you so much
36735
Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.
31830
I am trying to make in LaTeX this probability tree:    I tried using the tikz package but I keep having some issues. This is the best I got so far (link: ):      \documentclass{article}      \usepackage{tikz}     \usetikzlibrary{trees}      \begin{document}     \pagestyle{empty}      \tikzstyle{level 1}=[level distance=3cm, sibling distance=3cm]     \tikzstyle{level 2}=[level distance=3.5cm, sibling distance=2cm]      \tikzstyle{bag} = [circle, minimum width=10pt, inner sep=2pt]     \tikzstyle{end} = [circle, minimum width=3pt, fill, inner sep=0pt]      \begin{tikzpicture}[grow=right, sloped]      \node[bag]{}     child {             node[bag] {O}                         child {                     node[end, label=right:{O}] {}                     edge from parent                     node[above] {}                     node[below]  {$1-\lambda$}                     child {                     node[end, label=right: {O ...}] {}                     edge from parent                     node[above] {}                     node[below]  {$1-\lambda$}                 }                 child {                     node[end, label=right: {S ...}] {}                     edge from parent                     node[above] {$\lambda$}                     node[below]  {}                 }                 }                 child {                     node[end, label=right: {S}] {}                     edge from parent                     node[above] {$\lambda$}                     node[below]  {}                 }                 edge from parent                  node[above] {}                 node[below]  {$1-\lambda$}         }         child {             node[bag] {S}                     child {                  edge from parent                     node[above] {1}                     node[below]  {}                     node[bag] {O'}                                child {                                     edge from parent                                     node[above] {$1$}                                     node[below] {$$}                                     edge from parent                     }                     }             edge from parent                          node[above] {$\lambda$}                 node[below]  {}         };       \end{tikzpicture}     \end{document}   As you can see, they are quite different, but I can't seem to be able to progress any further, would you be able to help me? I can't seem to realize what are the errors.  Best.
20344
In a book of word problems by V.I Arnold, the following appears:   The hypotenuse of a right-angled  triangle  (in  a  standard  American  examination) is 10 inches, the altitude dropped onto it is 6 inches. Find the area of the triangle.  American school students had been coping successfully with this problem for over a decade. But then Russian school students arrived from Moscow, and none of them was able to solve it as had their American peers (giving 30 square inches as the answer). Why?  's the book. I assume the answer is some joke at the expense of the Americans, but I don't get it. Possibly a joke about inches? Anyone?
97653
To what value does the sum of the areas of the dark triangles converge if the area of ​​the square is 1? I already calculated the first to layers of triangles and got that one is $\frac{1}{2}$ and the other is $\frac{1}{2^5}$, but I don't see a sequence between those areas.
48006
Is it possible to get some help on this. I'm trying to help my friend solve this but it had been over 14 years since I did this type. All I have to work with is this.   Sorry if it is vague. Any help would be greatly appreciated!  Cosecant = cscx=1/sinx.
73673
You are given four points (on a Euclidian plane) that make up the corners of a square. You may change the positions of the points by a sequence of moves. Each move changes the position of one point, say p, to a new location, say p', by "skipping over" one of the other 3 points. More precisely, p skips over a point q if it is moved to the diametrically opposed side of q. In other words, a move from p to p' is allowed if there exists a point q such that $q = (p + p') / 2$.  Find a sequence of moves that results in a larger square. Or, if no such sequence is possible, give a proof of why it isn't possible. (The new square need not be oriented the same way as the original square. For example, the larger square may be turned 45 degrees from the original, and the larger square may have the points in a different order from in the original square.)  I've tried many different combinations and still haven't gotten anything more than 3 points in the right position and one point completely off. Is this possible?
3687
I have a mesh and in the region around each triangle, I want to compute an estimate of the principal curvature directions. I have never done this sort of thing before and  does not help a lot. Can you describe or point me to a simple algorithm that can help me compute this estimate?  Assume that I know positions and normals of all vertices.
76637
I saw this question today, it asks how many triangles are in this picture.    I don't know how to solve this (without counting directly), though I guess it has something to do with some recurrence.  How can count the number of all triangles in the picture ?
12824
I've been having a problem with these problems for a week now so I thought I'd post it on MSE. Read the text below. . I've also typed it up:  Annabel made a shape by placing identical square tiles in a frame as shown in the diagram above. The tiles are arranged in columns. Each column touches the base but no columns touch the side or top. There are no empty gaps between columns. The frame can be enlarged as needed.  These are the questions. I've also typed them up.    A. Show that it is possible to arrange $7$ tiles so that the ant walks $8,9,10,11,12,13,14,15$ tile edges.   B. Show 6 ways of arranging 7 tiles so that the ant walks a total of 9 tile edges.  C. Show it is possible to arrange 49 tiles so that the ant walks a total of less than 21 tile edges.  D. Show four arrangements of 137 tiles, each arrangement with a different maximum height, so that the ant walks a total of 34 tile edges.  For question $a$, is there a formula relating the number of tile edges exposed to the number of tiles. These are my answers for $a$    For the above picture, it's the left hand side which has the $a$ answers.  However, I'm having a bit of trouble with $b$. I've already got four answers down but where are the other two. My four current answers are in the right hand side of the last image. Also, I'll also appreciate a formula to show the relationship.  I'm also having trouble with $c$. I found out that you get 21 tile edges exposes when you make a $7*7$ square but the answer is asking for less than 21. Can anyone help me. Also, as in the previous questions, a formula to show the relationships would be great.  And for $d$, I am totally lost. I have no idea where to start, and I certainly can't find out four arrangements. So can anyone help me? And as always, a formula is nice  P.S. Please explain this in an understandable formula, as I'm only a Year 7 with the ability to understand linear algebra and parts of quadratics and trigonometry. Also, sorry about the poor imaging.