A geometry problem by Syed Hamza Khalid

Consider the point $P$ inside the square. If it's moved by some distance in either the $x$ or the $y$ direction, then the sum of the areas of opposite quadrilaterals remains constant. Hence, the two pairs of opposite quadrilaterals have the same sum, regardless of where the point $P$ is. Therefore, the missing area can be computed $16+32-20=28$ .

This is a problem relayed here by Youtube channel "Mind your Decisions"

Denote the square by ABCD, A at lower left, and clockwise. Let the side be S and P the point of intersection inside the square. Let h be a line from P perpendicular to AB, and H a line from P perpendicular to AD. Consider the quadeilateral whose area is 16. It can be computed from the expression: hH + (1/2)(H)(S/2 - h) +(1/2)(h)(S/2 - H) = 16. Simplifying HS + Sh = 64. Next, consider the quadrilateral whose area is 20. We have: h(S - H) - (1/2)(h)(S/2 - H) + (1/2(S - H)(S/2 -h) = 20. Simplifying, S^2 + hS - SH = 80. The quadrilateral with area 32 is given by (S -h)(S - H) - (1/2)(S - H)(S/2 - h) - (1/2)(S - h)(S/2 - H), which simplifies to 2S^2 - SH - Sh = 128. Substituting the first equation into the third,(notice we didn't need the second equation), we get 2S^2 -64 = 128, or s^2 = 96. Then the blue area is 96 -16 - 20 - 32 = 28. Ed Gray