Daily Math Puzzle: 2026-04-17

Five-eighths (5/8) — Let the side length of the square temple plot be 'L'. Its total area 'A' is L*L = L^2. Imagine the square with corners at (0,L), (L,L), (L,0), and (0,0) in a coordinate plane. The four sacred points are placed as follows: 1. Starting from the top-left corner (0,L) and moving clockwise: The first point is on the top side (y=L), one-quarter of the way towards (L,L). This point P1 is at (L/4, L). 2. Starting from the top-right corner (L,L) and moving clockwise: The second point is on the right side (x=L), one-quarter of the way towards (L,0). This point P2 is at (L, 3L/4). 3. Starting from the bottom-right corner (L,0) and moving clockwise: The third point is on the bottom side (y=0), one-quarter of the way towards (0,0). This point P3 is at (3L/4, 0). 4. Starting from the bottom-left corner (0,0) and moving clockwise: The fourth point is on the left side (x=0), one-quarter of the way towards (0,L). This point P4 is at (0, L/4). When these four points (P1, P2, P3, P4) are connected, they form an inner square. The easiest way to find the area of this inner square is to subtract the areas of the four right-angled triangles at the corners of the main square from the total area 'A'. Consider one of these corner triangles, for example, the one in the top-left corner, formed by the original corner (0,L), point P1(L/4, L), and point P4(0, L/4). * The base of this triangle (along the top edge) is the distance from (0,L) to (L/4, L), which is L/4. * The height of this triangle (along the left edge) is the distance from (0,L) to (0, L/4), which is L - L/4 = 3L/4. The area of this single corner triangle is (1/2) * base * height = (1/2) * (L/4) * (3L/4) = 3L^2 / 32. Due to the symmetry of the placement, all four corner triangles are identical in size. So, the total area of the four corner triangles is 4 * (3L^2 / 32) = 12L^2 / 32 = 3L^2 / 8. Finally, the area of the inner sacred quadrilateral is the total area of the temple plot minus the area of the four corner triangles: Inner Area = A - (3/8)A = L^2 - (3L^2 / 8) = (8L^2 / 8) - (3L^2 / 8) = 5L^2 / 8. Therefore, the area of the inner sacred quadrilateral is 5/8 of the total area 'A'.