Geometry question

[ladyjlh]

How can I divide a triangular region into 22 parts by using only 6 lines?

 

[Loren]

Are you assuming they have to be straight lines?

 

[Deleted member 4993]

How can I divide a triangular region into 22 parts by using only 6 lines?

Do not forget to count the "bigger" triangles and the rectangular regions. Also make sure "no more than 2" lines go through a point - inside the given triangular region.

 

[TchrWill]

How can I divide a triangular region into 22 parts by using only 6 lines?

You do not state the groundrules for the location of the lines.Do they all radiate from one or more of the vertices? Are they parallel to the sides? Are they not parallel to the sides?

For instance, 6 lines drawn parallel to one side yields 7 regions. If three are parallel to each of two of the sides, none intersecting any of the others, you get 13 regions. Three radiating from one side and three radiating from another side yields 16 regio9ns. Two radiating from each vertex, no more than two intersecing an intersection, yields 19 regions.

The following is one way of location the 6 lines. Note that there are many other ways of defining the locations of the lines.

Given any triangle. Draw "n" lines from each vertex to "n" points on the opposite sides such that no three of the lines intersect at the same point. How many regions do the 3n lines divide the interior of the triangle into?

Lets try a small value of n = 1 and see what happens when we deal with just two sides of the triangle. Draw one line to one point on the opposite side. We immediatley see that we have divided the triangle into regions. Draw a second line and we now have 4 regions. Add the 3rd line and we have 7 regions.
Try n = 2. Draw two lines to the opposite side and we obtain 3 regions. Draw two more lines and we now have 9 regions. Draw the last two lines and we now have 19 regions.
Try n = 3. Draw three lines to the opposite side and we obtain 4 regions. Draw three more lines and we now have 16 regions. Draw the last three lines and we now have 37 regions.
Lets see what this data tells us so far.
Pts..................Lines/Regions
.1..............1/2.........2/4.........3/7
.2..............2/3.........4/9........6/19
.3..............3/4........6/16.......9/37

Notice anything about these numbers? Clearly, the first line, or lines, divides the triangle up into (n + 1) regions. Just as obvious, the second line, or lines, divides the triangle up into n^2 regions. If you take careful note of the difference between the 2nd and 3rd region numbsrs, you might recognize that the differences between pairs within each set of points are every other triangular number. For 1 point, 7 - 4 = 3. For 2 points, 19 - 9 = 10. For 3 points, 37 - 16 = 21, or every other triangular number. On this basis, my educated guess tells me that the next few sets of data follows the following pattern:
Pts..................Lines/Regions
.1..............1/2.........2/4.........3/7
.2..............2/3.........4/9........6/19
.3..............3/4........6/16.......9/37
.4..............4/5........8/25......12/61
.5..............5/6.......10/36.....15/91
.6..............6/7.......12/49.....18/127 and so on.

If we just enumerate the region numbers we get
Points...............1.....2.....3.....4.....5.....6................n
Regions.............7....19...37...61...91...127.............N
1st difference........12...18...24...30...36
2nd difference...........6.....6.....6.....6

Thus, we have a finite difference series where the second differences are constant, which leads to the expression for the nth term being of the second order, i.e., N = an^2 + bn + c.

Using the data points (n1, N1), (n2,N2), (n3,N3), etc., we substitute them into N = an^2 + bn + c as follows:
(n1,N1) = (1,7) produces a(1^2) + b(1) + c = 7 or a + b + c = 7
(n2,N2) = (2,19) produces a(2^2) + b(2) + c = 19 or 4a + 2b + c = 19
(n3,N3) = (3,37) produces a(3^2) + b(3) + c = 37 or 9a + 3b + c = 37
Solving, a = 3, b = 3, and c = 1 making the nth term of this regional series N = 3n^2 + 3n + 1.

Therefore, the number of regions developed within a triangle resulting from the creation of "n" lines being drawn from each vertex to "n" points on the opposite sides derives from N = 3n^2 + 3n + 1.

If you have some other definition of line locations, please repost your specific problem.