smc

As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $△ABP,△BCP,△CDP,△DAP,$ respectively. By SAS, we conclude that $△G_1{G}_{2}P\sim △M_1{M}_{2}P,△G_2{G}_{3}P\sim △M_2{M}_{3}P,△G_3{G}_{4}P\sim △M_3{M}_{4}P,$ and $△G_4{G}_{1}P\sim △M_4{M}_{1}P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac{2}{3}.$ Note that quadrilateral $M_1{M}_2{M}_3{M}_4$ is a square of side-length $15\sqrt{2}.$ It follows that: 1. Since $\overline{G_1{G}_2}\parallel \overline{M_1{M}_2},\overline{G_2{G}_3}\parallel \overline{M_2{M}_3},\overline{G_3{G}_4}\parallel \overline{M_3{M}_4},$ and $\overline{G_4{G}_1}\parallel \overline{M_4{M}_1}$ by the Converse of the Corresponding Angles Postulate, we have $\angle G_1{G}_2{G}_3=\angle G_2{G}_3{G}_4=\angle G_3{G}_4{G}_1=\angle G_4{G}_1{G}_2=90^{\circ }.$ 2. Since $G_1{G}_2=\frac{2}{3}{M}_1{M}_2,G_2{G}_3=\frac{2}{3}{M}_2{M}_3,G_3{G}_4=\frac{2}{3}{M}_3{M}_4,$ and $G_4{G}_1=\frac{2}{3}{M}_4{M}_1$ by the ratio of similitude, we have $G_1{G}_2=G_2{G}_3=G_3{G}_4=G_4{G}_1=10\sqrt{2}.$ Together, quadrilateral $G_1{G}_2{G}_3{G}_4$ is a square of side-length $10\sqrt{2},$ so its area is ${\left(10\sqrt{2}\right)}^2=\boxed{200}.$ Remark This solution shows that, if point $P$ is within square $ABCD,$ then the shape and the area of quadrilateral $G_1{G}_2{G}_3{G}_4$ are independent of the location of $P.$ Let the brackets denote areas. More generally, $G_1{G}_2{G}_3{G}_4$ is always a square of area $$[G_1{G}_2{G}_3{G}_4]={\left(\frac{2}{3}\right)}^2[M_1{M}_2{M}_3{M}_4]=\frac{4}{9}[M_1{M}_2{M}_3{M}_4]=\frac{2}{9}[ABCD].$$ On the other hand, the location of $G_1{G}_2{G}_3{G}_4$ is dependent on the location of $P.$