Subdividing each side of a quadrilateral give sixth

[Guest]

Good after noon

If you subdivide each side of a quadrilateral in 3 equal parts, you get 9 sub-quadrilaterals. Prove that the area of the initial quadrilateral is 9 times the area of the central sub-quadrilateral.

[Guest]

Probably easiest to prove this using vectors.

Let a, b, c ,d be the vectors from some (arbitrary) origin to A, B, C, D respectively.
E is the vector a+(b-a)/3 = (2a+b)/3
J is the vector d+(c-d)/3 = (2d+c)/3
M is the vector a+(d-a)/3 = (2a+d)/3
H is the vector b+(c-b)/3 = (2b+c)/3
The point 1/3 down EJ is (2a+b)/3+((2d+c)/3-(2a+b)/3)/3 = (4a+2b+c+2d)/9
The point 1/3 down MH is (2a+d)/3+((2b+c)/3-(2a+d)/3)/3 = (4a+2b+c+2d)/9
So the point 1/3 down EJ = the point 1/3 down MH and therefore must be the intersection point of EJ and MH, i.e. the point N.
We can get similar results for O, P, and Q.

Let the vectors A to B, B to C, C to D, D to A, be given by [tex]w, x, y, z[/tex] respectively.
twice the area of ABCD is given by the magnitude of the vector [tex]w\times x + y\times z[/tex] (the direction of the vector will be perpendicular to the plane containing the quadrilateral).
Also twice the area is given by the magnitude of [tex]x\times y+z\times w[/tex].

We can apply the same formula to calculate the area of NOPQ.
The vector N to O is 1/3 of the vector M to H (by our previous work on finding intersection points).
M to H [tex]= z/3+w+x/3[/tex]
So N to O [tex]= (z+3w+x)/9[/tex]
We get similar results for O to P, P to Q, Q to N.
So twice the area of NOPQ is given by the magnitude of
[tex](z+3w+x)/9 \times (w+3x+y)/9 + (x+3y+z)/9 \times (y+3z+w)/9[/tex]
[tex]=((z+3w+x) \times (w+3x+y) + (x+3y+z) \times (y+3z+w))/81[/tex]
[tex]=(2z \times w+7w \times x+2x \times y+7y \times z)/81[/tex] (Remember that [tex]u\times v = -v\times u[/tex] and [tex]u\times u=0[/tex].)
[tex]=(7(w \times x+y \times z) + 2(x \times y+z \times w))/81[/tex]
[tex]=(w \times x+y \times z)/9[/tex] (Remember that [tex]w \times x+y \times z = x \times y+z \times w[/tex].)

So twice the area of NOPQ is 1/9 of twice the area of ABCD, or equivalently ABCD is 9 times the area of NOPQ.

Hope this helped,

R. Baber.

[Guest]

Good morning

Thnk you Baber.

I will work your solution.