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

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Copyright © 1996-2017 Alexander Bogomolny

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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Copyright © 1996-2017 Alexander Bogomolny

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