8 major rewrite

I think I can prove this. I start by getting three consecutive points of the pentagon ABC equal to (0, 1),(0, 0) (1, 0) and the remainder with coordinates (x_1,y_1) and (x_2,y_2) both with sum of coordinates greater than one and both with both coordinates positive using area preserving transformations as mentioned in a comment to another post.

I start by choosing as three consecutive points the three consecutive points which form the greatest area, if there is I tie, I pick one.

I expand one dimension and contract an orthoganal orthogonal direction by the same amount. Doing this I first get the angle to be right then after this I use the same type of transformation contracting along one ray of the now right angle and expanding along the other ray. Doing this I can any ratio I want. Then I translate the middle point to the origin and rotate till I have the three desired points in the desired position and the other conditions on the remaining points come from the convexity of the polygon which is preserved by the transformations. I think I also have to note that the following proof will not be changed by an expansion of both coordinates by the same amount because while I can get an equilateral right trianangle triangle by area preserving transformations The area of the original triangle may not be 1/2.

Then I use the fact that the area in the plane of the triangle formed by any two points with coordinates (x_1,y_1) and (x_2,y_2) and the origin is the one half of the determinant of the matrix containing the two points. I think you have to be careful the points are going in clockwise order or else you will get a negative number. The points being in clockwise order means that y_1/x_1 > y_2/x_2. I apply this to the pentagon and get the area the 1/4 1/2 of determinant of (x_1,y_1) and (x_2,y_2) plus 1/4(x_11/2(x_1) + 1/4(y_1)1/2(y_2). I sweep clockwise from the origin and count the areas of the five coordinates formed by the midpoints this overestimates by 1/8 which I will have to subtract. I look at the determinants. For the second and fourth of these triangles it becomes clearer if I subtract one row from another. In any case the result of all this is the area of the midpoint polygon is 1/4 of the determinant of (x_1,y_1) and (x_2,y_2) plus 1/4(x_21/4(x_1) + 1/4(y_11/4(y_2) +1/8(x_1)+1/8(y_2)-1/8. Now the fact noted above that y_1/x_1 > y_2/x_2 means either x_1 is less than x_2 or y_2 To get this is less than y_1 in either case we can add 3/4 of the x_1 term to area of the x_2 term getting 3/8 or polygon I need
1/8 the y_2 term to determinant of (x_1,y_1) and (x_2,y_2) + 1/8(x_1) + 1/8(y_2) + 1/8 is greater than1/8(x_2)+1/8(y_1) or1/8 the y_1 term. This will raise one determinant of 1/4(x_2(x_1,y_1) and (x_2,y_2) + 1/8(x_1) + 1/4(y_11/8(y_2) to 3/8. Now + 1/8-1/8(x_2)-1/8(y_1) +1/8

The above expression is 1/4 of the area of the triangle with vertices (x_1,y_1),(1,1),(x_2.y_2) if the other term point (1,1) is also less on the same side of line connecting (x_1,y_1) and (x_2.y_2) plus 1/8.If this holds then the expression is positive and we are donebecause we will subtract 1/8. But each So we can assume that the point (1,1) is on the same side of 1/8(x_1) the line connecting (x_1,y_1) and 1/8(y_2) (x_2.y_2) is 1/8 on the other side of the line and the above expression is the negative of the area of a the triangle formed by three consecutive mentioned above. Now the farther apart the points of are the pentagon. Since we have been using transformations that preserve greater the ratio area of areas the triangle so if we assume maximum separation we get x_1 and y_2 = zero and substituting this into the above we have 1/8(-x_2-y_1 + (x_1)(y_1) -1 is less than zeroor 1/8(y_1-1)(x_2-1) is less than zero but since x_1+y_1 is greater than one and x_1 is zero we have chosen y_1 is greater than one similarly we have x_2 is greater than one and the three points product is positive. This follows Speyer's so far proof, I thought I could get the 1/8 as the difference between the midpoint polygon and 3/4 the original polygon but in some cases it is less.

So we have transformed to the following the area of the midpoint polygon is less than 3/4 of the original polygon furthermore the difference between 3/4 of the original polygon and the area of the midpoint polygon is 1/8-plus or minus the area of the triangle formed by (0,1), x_1,y_1),(1,1),(x_2.y_2) depending on which side of the line connecting (0,0) x_1,y_1) and (1,0) x_2.y_2) is (1,1).I want to have maximal find the difference between 3/4 of the area 1/8 and the area of the midpoint polygon in terms of the areas of geometric figures in the original polygon.

Tracing this back to the original polygon the area of either will be less than 1/8 and we have the midpoint polygon is 3/4 of the original polygon -1/4 of the area of triangle ABC plus another triangle which has a point N I need to construct. Here is the construction take the midpoint pentagon of CA, M connect it to B then extend the line BM to twice its length to get point N then if N is on the same side of DE as A subtract 1/4 of the area of END otherwise add it from the above the result will always be less than 3/4 of the area of the original pentagon and we are donepolygon or less from the above. This should hold for any three consecutive vertices of any polygon.

7 I have redone the proof.

I think I can prove this. I start by getting three consecutive points of the pentagon equal to (0, 1),(0, 0) (1, 0) and the remainder with coordinates (x_1,y_1) and (x_2,y_2) both with sum of coordinates greater than one and both with both coordinates positive using area preserving transformations as mentioned in a comment to another post. I start by choosing as three consecutive points the three consecutive points which form the greatest area, if there is I tie, I pick one.

I expand one dimension and contract an orthoganal direction by the same amount. Doing this I first get the angle to be right then after this I use the same type of transformation contracting along one ray of the now right angle and expanding along the other ray. Doing this I can any ratio I want. Then I translate the middle point to the origin and rotate till I have the three desired points in the desired position and the other conditions on the remaining points come from the convexity of the polygon which is preserved by the transformations. I think I also have to note that the following proof will not be changed by an expansion of both coordinates by the same amount because while I can get an equilateral right trianangle by area preserving transformations The area of the original triangle may not be 1/2.

Then I use the fact that the area in the plane of the triangle formed by any two points with coordinates (x_1,y_1) and (x_2,y_2) and the origin is the one half of the determinant of the matrix containing the two points. I think you have to be careful the points are going in clockwise order or else you will get a negative number. The points being in clockwise order means that y_1/x_1 > y_2/x_2. I apply this to the pentagon and get the area the 1/4 of determinant of (x_1,y_1) and (x_2,y_2) plus 1/4(x_1) + 1/4(y_1). I sweep clockwise from the origin and count the areas of the five coordinates formed by the midpoints this overestimates by 1/8 which I will have to subtract. I look at the determinants. For the second and fourth of these triangles it becomes clearer if I subtract one row from another. In any case the result of all this is the area of the midpoint polygon is 1/4 of the determinant of (x_1,y_1) and (x_2,y_2) plus 1/4(x_2) + 1/4(y_1) +1/8(x_1)+1/8(y_2). 1/8(x_1)+1/8(y_2)-1/8. Now the fact noted above that y_1/x_1 > y_2/x_2 means either x_1 is less than x_2 or y_2 is less than y_1 in either case we can add the x_1 term to the x_2 term getting 3/8 or the y_2 term to the y_1 term. This will raise one of 1/4(x_2) + 1/4(y_1) to 3/8. Now if the other term is also less then its corresponding term 1/8 we are done .

If not because we use the inequality generated by the line connecting its two neighborswill subtract 1/8. Because it is convex it must be separated from the orgin by this line and that gives us an equality that implies the following: area of the midpoint pentagon is less than 3/8 of determinant But each of (x_1,y_1) 1/8(x_1) and (x_2,y_2) plus 3/8(x_11/8(y_2) + 3/8(y_1)-1/8. This means the overall ratio is less than 3/4 and we are done. I think this proof gives more: since the 1/8 was originally a triangle connecting the two midpoints adjacent to a vertex. I think we have the following: the area of the midpoint pentagon is less than or equal to 3/4 of the area of the orignal pentagon minus the area of any of the triangles a triangle formed by a vertex and the two midpoints on the edges that meet at that vertex. Which doesn't work with some three consecutive points of the previous examplespentagon. In David Speyer's proof he gets the inequality to hold if the following triangle is positive (x_1,y_1),(1,1),(x_2,y_2).

If I could try and see what this is in Since we have been using transformations that preserve the original configuration I could get a bound in any configuration in terms ratio of a geometrically derived configuration. Clearly (x_1,y_1) areas and (x_2,y_2) are not going to be problems but I need since we have chosen the three points we have transformed to define (1,1) in terms that will remain constant throughout the various transformations. If I connect 0,1), (0,1) 0,0) and (1,0) and take the midpoint I get (1/2,1/2) call this x if I extend the line connecting (0,0) to x to y where the distance xy is equal to the distance from (0,0) to x I get the point 1,1.I think that the property of being a midpoint should survive all have maximal area 1/8 of the transformations I used above so area of either will be less than 1/8 and we should have the following:

For any pentagon the area of the midpoint pentagon is less than or equal to 3/4 of the area of the original pentagon minus the area of any triangle NDE formed the following way:Let A,B,C be three consecutive vertices connect A to C take the midpoint M extend BM to N so that MN equals BM then form the triangle NDEand we are done.