How many closed polygons will $n$ coplanar lines, all intersecting each other (no three concurrent), form on the plane?  Can we find a recursion formula for this?? Is there a way out?
17123
Show that $n$ lines separate the plane into  $\frac{n^2+n+2}{2}$ regions if no two of these lines are parallel and no three pass through a common point.   I know we start with the base case, where, if we call the above equation P(n), P(0), for 0 lines would be 0.  But I really have no idea how to begin the inductive step.  How do we know what k+1 we're supposed to arrive at?  Thanks!
114651
Does there exist $n$, and $r&lt;2n-2$, such that the $n\times n$ square grid can be connected with an unbroken path of $r$ straight lines?   Note: This has essentially already been asked - see  MSE thread.  I am posting this question because I wanted to explicitly focus on one of the unanswered parts.  Notice that $2n-1$ is the trivial solution.  This is obtained by either spiralling towards the center, or by zig-zagging up and down.  On the  MSE thread, I posted an answer showing $2n-2$ was possible for $n\geq 3$. (this is obtained by reducing to the $3\times 3$ case)  I felt $2n-2$ should be optimal, but I couldn't think of a proof.  (Also, it is not necessarily clear that for a 100 billion by 100 billion grid you can't use some kind of cleverness to eliminate one single line)  This has been bothering me since I saw it, and I would like to know if anyone has any ideas.  Thanks,  Edit:  This is regarding an question/answer I received:  Not only are 45 degree lines allowed, but lines in any direction.  Naturally 0,45,90 seem like the best since those angles will intersect the most lattice points, and we ask ourselves "Will any optimal solution consist only of lines at these angles?".  The answer is a definite no.  Consider the  to the $10\times 10$ grid which uses 18 lines (so $2n-2$, i.e. currently optimal).  Also there is  solution to the $8\times 8 $ grid using 14 lines.  Interestingly, this solution to the $10\times 10 $ generalizes to give another $2n-2$ line solution to the $n\times n$ grid when $n$ is even.  (I found this $8\times8$ and $10\times 10$ solution on a link posted by user3123 on the  question page.  Specifically it was  link.)
18569
I have come across numerous questions where I am given the picture such as the above one been asked "how many rectangles are there?". I have even come across some slightly different images that instead of rectangles you are supposed to find the number of triangles. Well, I was thinking whether there is any formula or strategy that is used to solve these problems without having to manually count every shape.  Help would be appreciated.  Thank you :)
1577
Support for N-Gons (polygons consisting of more than 4 vertices) has been in Blender for a while now, but many of us still may not know when to use them, and how to do so effectively.  In what situations do N-Gons really come in handy? When should they be just plain left out? Do they have any place in finished models?
69796
Let $X = \{A_1,A_2,... ,A_n\}$ be a set of distinct $3$-element subsets of $\{1, 2,... , 36\}$ such that  i) $A_i$ and $A_j$ have non-empty intersection for every $i,j$.  ii) The intersection of all the elements of $X$ is the empty set.  Show that $n ≤ 100$. How many such sets $X$ are there when $n = 100$?  Source:  BMO 2005 round 2 question.  Please help !! I am not even able to proceed with the question I tried to come up with a recurrence relation but its not working. I have only been able to figure out the number when 36 is replaced by 6, which is an easy thing to do
103967
Given a simple graph $G$ [no loops or parallel edges] on six vertices,  let $G^c$ denote its complement. It is known that either $G$ or $G^c$ must contain a triangle  $T$ in it. [An example of a Ramsey number I think.]   My question came up when I tried to find such a graph on six vertices  which had only one $T$ in it or its complement.  I believe if I did things right there are none of these. This I tried to check by drawing a copy of each graph $G$  on six vertices which has a $T$ in it,  and then looking at the complements to see if any had  no $T$'s in them. I found none.  One way to pose my question for a general number $n$ of vertices  is then as follows: Consider the set of all graphs $G$ on $n$ vertices,  and for each one construct $G^c.$  Count the number of $T$ in $G$ and add that to the number of $T$ in $G^c.$  Finally find the minimal such sum for the given $n$  and call that say $M(n).$ So my above six vertex conjecture  is that $M(6)=2.$   By considering a cycle graph on five vertices we find $M(5)=0.$   So going further one could ask if anything is known  about this minimum function $M(n)$ for larger $n$ [such as bounds, etc.] And does it have a name?  I would also like to see a simpler way to handle the six vertex case, one not involving drawing all the graphs.
780
I'm writing a computer graphics library, and I'd like to compute the plane that passes through three arbitrary points, $p$, $q$ and $r$. I'm defining the plane in the form of $Ax + By + Cz + D = 0$.  My algebra is rather rusty, but I suspect this is a job for some simultaneous equations I'll need to solve a system of equations:  $$0 = \begin{cases} Ax_p + By_p + Cz_p + D \\ Ax_q + By_q + Cz_q+ D \\ Ax_r + By_r + Cz_r + D \end{cases}$$  How would I efficiently find the values of $A$, $B$, $C$ and $D$ with respect to $p$, $q$ and $r$?
44947
If there are 5 points on the surface of a sphere, then there is a closed half sphere, containing at least 4 of them.  It's in a pigeonhole list of problems. But, I think I have to use rotations in more than 1 dimension.  Regards
17074
Pentagon $ABCDE$ is inscribed in a circle. For any edge of $ABCDE$, we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent.  We just finished a unit on vectors in my Pre-Calculus class but I have no idea on how to do this final problem. Any help would be greatly appreciated!
13657
Find the general form equation of a plane that contains the point $(2, -2, 1)$ and passes through a line $(x, y, z) = (1, 2, -3) + t(2, -3, 2)$.  I managed to use the fact that it contains a given point to write the equation of the plane as $A(x-2) + B(y+2) + C(z-1) = 0$ but got stuck there.
16445
If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).  I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?
59652
Why is it impossible to partition the edges in $K_n$ into $\lfloor \frac n6 \rfloor$ planar subgraphs for $n \ge 6$? I'm just stuck at the beginning and can't figure out how to go about this problem.     Thanks in advance for the help!
96057
What I'm looking to do is create a 10-sided die, specifically in the shape of a 'pentagonal trapezohedron', i.e., a polyhedron that has 10 kite shaped sides.  With the following link I was able to produce a d4, d6, d8, d12, d20, &amp; d30:   What I'm looking for is something similar for a 10-sided die.  Thanks
28539
In QGIS, is there a way to count the number of vertices of polygons and lines in a given layer, and create a separate field containing these values?   I suspect it's a function somewhere in the field calculator but I can't find it.
103809
How am I to "[u]se induction to show that $n$ straight lines in the plane divide the plane into $\frac{n^2+n+2}{2}$ regions"? It is assumed here that no two lines are parallel and that no three lines have a common point. Further, this is not a non-Euclidean problem, but I wouldn't mind a discussion on the non-Euclidean nature of the problem.  I was thinking it might be easier to show using the unit sphere.    A GOOD REFERENCE I FOUND THAT WILL BE USEFUL TO ANYONE COMING ACROSS THIS PROBLEM IS Concrete Mathematics, Graham, Knuth, Patashnik pp. 4–8.
66698
As the title suggests, I’m simply looking for a method to compute the number of Eulerian Paths in the complete graph $K_5$, where all paths start from the same node. Since there are 5 nodes, all Eulerian paths must be Eulerian circuits as well.  Thanks.
61581
I believe that this is a pretty simple question in terms of combinatorics.  Assuming we have the following set $\{ {1,2,3...n} \}$  In how many ways can we choose 3 subsets (A,B,C) such that the intersection between all them is empty?   $ A \cap B \cap C = \emptyset $  Well, what I have so far is the following:  For A we have $ 2^n $ options (where n is the size of the set) Then, for B, the remaining options are exactly $ 2^{n-|A|} $ since we can choose only from subsets who does not have elements in common with A. And finaly, for C, since we already have an empty intersection, we can choose whatever subset we want and get the expected result $ \emptyset $.  But for some reason it feels like the numbers are too big. If we take for example the set {1,2}, then, according to my "formula", the number of options should be:  $ \binom{4}{1} * \binom{2}{1} * \binom{2}{1} = \frac{4!}{3!} * \frac{2!}{1!} *  \frac{2!}{1!} = 16 $  Does it make sense?