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stoutfellowBefore I say any more about isolated classes, I need to say a few words about area. The word has a slightly different meaning in the circles I move in, and I'd like to clarify that. I'll refer to the vernacular meaning as "absolute area"; "area", by itself, will refer to the technical meaning.
First off, an indication of the difficulties involved. Consider the quadrilateral with vertices A(0,0), B(1,1), C(1,0), D(0,1). This quadrilateral is shaped like a bow tie. What is its area? A natural answer might be the sum of the absolute areas of the two triangles it encloses, but that turns out to be an inconvenient choice mathematically.
The definition of area that I use rests on two declarations. First, the area of a triangle is the absolute area if the vertices are given in counterclockwise order; it's the negative of the absolute area if they are given in clockwise order. (This is related to the fact that a two-by-two matrix can have negative determinant, if you took and remember any linear algebra.) Second, if you have a polygon ABC..TU, its area may be computed by selecting a point Z and adding up the areas of the triangles ZAB, ZBC, ..., ZTU, ZUA. It turns out that the choice of Z makes no difference - the sum is the same, no matter where Z is.
For example, consider the triangle with vertices A(0,0), B(1,0), and C(0,1). The vertices ABC are given in counterclockwise order, so the area is the absolute area, which is 1/2. If you chose Z in the interior, all three of ZAB, ZBC, ZCA would also be counterclockwise, and obviously the area of ABC is the sum of the areas of those triangles. But what if you chose Z = (1,1)? Note that the triangles ZAB and ZCA are both counterclockwise, but ZBC is clockwise, so we get the sum of the absolute areas of ZAB (1/2) and ZCA (1/2), *minus* the absolute area of ZBC (1/2), so the area is 1/2 + 1/2 - 1/2 = 1/2. The same thing will happen for any choice of Z.
As for the bow-tie quadrilateral, the simplest choice of Z is at the intersection of AD and BC, (1/2, 1/2). Now ZAB is counterclockwise, but ZCD is clockwise, so the area is the absolute area of ZAB minus the absolute area of ZCD, or 1/2 - 1/2; the quadrilateral has area 0. (ZBC and ZDA are degenerate triangles, with area 0.)
If you want a challenge, work out the area of the hexagon whose vertices are A(0,0), B(1,0), C(1,1), D(1/3,1/3), E(2/3,1/3), and F(0,1).
First off, an indication of the difficulties involved. Consider the quadrilateral with vertices A(0,0), B(1,1), C(1,0), D(0,1). This quadrilateral is shaped like a bow tie. What is its area? A natural answer might be the sum of the absolute areas of the two triangles it encloses, but that turns out to be an inconvenient choice mathematically.
The definition of area that I use rests on two declarations. First, the area of a triangle is the absolute area if the vertices are given in counterclockwise order; it's the negative of the absolute area if they are given in clockwise order. (This is related to the fact that a two-by-two matrix can have negative determinant, if you took and remember any linear algebra.) Second, if you have a polygon ABC..TU, its area may be computed by selecting a point Z and adding up the areas of the triangles ZAB, ZBC, ..., ZTU, ZUA. It turns out that the choice of Z makes no difference - the sum is the same, no matter where Z is.
For example, consider the triangle with vertices A(0,0), B(1,0), and C(0,1). The vertices ABC are given in counterclockwise order, so the area is the absolute area, which is 1/2. If you chose Z in the interior, all three of ZAB, ZBC, ZCA would also be counterclockwise, and obviously the area of ABC is the sum of the areas of those triangles. But what if you chose Z = (1,1)? Note that the triangles ZAB and ZCA are both counterclockwise, but ZBC is clockwise, so we get the sum of the absolute areas of ZAB (1/2) and ZCA (1/2), *minus* the absolute area of ZBC (1/2), so the area is 1/2 + 1/2 - 1/2 = 1/2. The same thing will happen for any choice of Z.
As for the bow-tie quadrilateral, the simplest choice of Z is at the intersection of AD and BC, (1/2, 1/2). Now ZAB is counterclockwise, but ZCD is clockwise, so the area is the absolute area of ZAB minus the absolute area of ZCD, or 1/2 - 1/2; the quadrilateral has area 0. (ZBC and ZDA are degenerate triangles, with area 0.)
If you want a challenge, work out the area of the hexagon whose vertices are A(0,0), B(1,0), C(1,1), D(1/3,1/3), E(2/3,1/3), and F(0,1).
