Relative Area Of A Quadrilateral
is a quadrilateral with known side lengths . The black points divide each side of the quadrilateral into eight equal parts.
Which area is bigger, the sum of the red quadrilaterals or the sum of the blue ones?
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1 solution
First, let's just do a 2x2 cut of the quadrilateral ABCD. Let's look at it in affine space (or vector space).
The division points on the sides are then (A+B)/2, (B+C)/2, ... The point inside where the two cutting lines intersect is (A+B+C+D)/4.
The area of a parallelogram is the exterior product (or wedge product, ∧) of its two sides (as vectors), but they have to be so that the angle from the first vector to the second one is between 0° and 180° counterclockwise; if it's clockwise (or counterclockwise reflex angle), the result is negative area (x∧y = -y∧x). So the area of a quadrilateral is half the exterior product of its diagonals (e.g. the area of ABCD is 1/2 (C-A) ∧ (D-B)).
The two blue quadrilaterals (the one at A and the one at C) have areas 1/2 ((A+B+C+D)/4 - A) ∧ ((A+D)/2 - (A+B)/2) and 1/2 (C - (A+B+C+D)/4) ∧ ((C+D)/2 - (B+C)/2). They add up to 1/4 (C-A) ∧ (D-B). Same for the two red ones.
Now to the original task: ABCD is actually divided into 4×4 quadrilaterals, each of which is cut the way I'd cut ABCD before, into 4: 2 red and 2 blue, which, as just demonstrated, add up to the same area. So the total blue area and the total red area are the same.