Carpets in a Quadrilateral II
E, F, G, H are the midpoints of sides AB, BC, CD, DA of a convex quadrilateral ABCD. P is the intersection of AF, DE, Q the intersection of AF, BG, R the intersection of BG, CH, and S the intersection of CH, DE. Then Area(ΔAEP) + Area(ΔBFQ) + Area(ΔCGR) + Area(ΔDHS) = Area(PQRS).  |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander BogomolnyThis problem was suggested by Bui Quang Tuan. Although simple, I do not have a recollection of seeing it anywhere else.
The solution is a direct consequence of the Carpet's Theorem. We split the quadrilateral in two ways S1 ∪ S2 = T1 ∪ T2, where S1 is the quadrilateral AFCH and S2 is the union of the remaining two triangles. Similarly, T1 is the quadrilateral BGDE and T2 its complement in the quadrilateral. Note that the area of each of the four regions is half that of the quadrilateral. For example, Area(ΔADE) = Area(ΔBDE) and Area(ΔBCG) = Area(ΔBDG). Adding the two gives Area(ΔADE) + Area(ΔBCG) = Area(ΔBDE) + Area(ΔBDG) = Area(BGDE). By the Carpet's Theorem, Area(S1 ∩ T1) = Area(S2 ∩ T2). But S1 ∩ T1 is the quadrilateral PQRS, whereas S2 ∩ T2 is the union of triangles AEP, BFQ, CGR, and DHS, as required. Carpets Theorem
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
| 40619558 |
