Divide The Whole Grid Into Smaller Geometric Shapes By Drawing Straight Lines Either Following The

Mathematics: How can I teach a 5 year old how to divide, using a simple technique?

If it's one thing I learned while teaching special education, it is breaking things down into small steps.I am THE MASTER of small steps.Please notice what is REALLY going on on the page shown. All this is is counting by 5's, 2's, and 10's.Apparently, they must be doing this. This page is just having them count by those groups and either drawing or separating items into those groups.If your child does not seem to know how to count by these numbers, start practicing with him, as a game, every day, in the bath, while driving, before bed, count by 5's,10's,and 2's, like it's a way of life.Then grab anything near and put them into those sets. It can be all your eating spoons tossed on a table and put into groups of 2's, or better yet shoes in your home put into groups of twos and then counting by twos while touching the sets. It can be pennies put into groups of 5's, it can be cans of food put into groups of 10 and then counting by 5's and 10's while touching those sets.Do these as if it's a normal way of life, and then next add 3's, and periodically add another such as counting by 4's. Keep practicing the 2's, 5's, and 10's.Make it so he sees these sets in his sleep. By the time he gets to multiplication and actual problem division, it will be so easy he will wonder what the big deal is.

Can you draw a circle without a compass and only with a ruler and a pen?

It is possible to draw a circle with a ruler and a set square.Draw a square having side 10 centimeters .Divide each side of the square into 10 equal parts.Join the points as shown in the figure.For getting a more perfect circle , you can divide each side into more number of parts ( like 20 or 40 )

What are the algorithms for determining if a point is inside an arbitrary closed shape or not?

Testing if a point is inside a polygon is pretty hard for a human if the polygon is a bit more complex.(picture from http://www.ics.uci.edu/~eppstein...)O(n) algorithms:Ray casting: shoot a ray from your point and see how many sides of the polygon it intersects. If the number is even, your point is outside the polygon. If it's odd your point is inside.Winding number: go around the sides of the polygon and sum up the sum of signed angles the points on the sides make with your current point. The result should be 2π if the point is inside and 0 if the point is outsideConvex polygon/Starred polygon O(log n) algorithm:Choose a vertex of the convex polygon. Shoot n-1 rays from this vertex to every other vertex. Use binary search to find where your query point lies between two consecutive rays using their angle. Then you just need to test if the point is within a triangle.Grid solution:For a fixed precision split the plane into squares and compute if each square is inside or outside the polygon.Quadtree solution O(log U)?:Build a quadtree to represent the plane for some fixed precision. Each square in the quadtree is either totally inside, totally outside the polygon or intersects the polygon. If it intersects the polygon you either split it in 4 sub squares, or stop because the level of precision is small enough.Vertical slabs (O(n^2) preprocessing O(log n) query):Draw a vertical line through each vertex of the polygon. This splits the plane in n vertical slabs. For each slab compute the segments that cross the slab and sort them by the y of their middle point.For a given query point first use binary search to find the slab your point falls in. Within the slab you can binary search for the lowest segment that crosses the slab and is above our query point. Now you can count how many segments are intersected by a vertical ray that starts from the query point and know if you are inside the polygon or not.This approach can be improved to O(n) pre processing and O(log n) query time.There's another clever approach that splits at each level one triangle into 3 smaller ones, but I don't remember the details exactly.