Carpets in a Quadrilateral II

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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).

Explanation

This problem was suggested by Bui Quang Tuan. Although simple, I do not have a recollection of seeing it anywhere else. (P.S. Years later I came across this problem in the highly recommended Mathematical Olympiad Treasures by T. Andreescu, B. Enescu (Birkhäuser, 2004).)

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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

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