Problem 10533
(American Mathematical Monthly, Vol.103, June-July 1996) Proposed by A. Flores (USA).
On a parallelogram P construct exterior squares on the sides. The centers of these squares form a square QE. On the same parallelogram construct the interior squares on the sides. The centers of these squares form another square QI.
(a) Show that Area(QE) − Area(QI) = 2Area(P ).
(b) Is there a generalization when P is replaced by an arbitrary convex quadrilateral?
Solution proposed by Roberto Tauraso, Scuola Normale Superiore, piazza dei Cavalieri, 56100 Pisa, Italy.
We will face directly the general case of a convex quadrilateral Q.
First note that if ABC is a counter-clockwise oriented triangle in C then Area(ABC) = 12Re{i(B − A)(C − B)} (⋆).
Let Q = ABCD be a convex counter-clockwise oriented quadrilateral in C. If z1= B − A, z2= C − B, z3= D − C, z4= A − D,
then the vertices of the exterior quadrilateral QE = P1P2P3P4 and the interior quadrilateral QI = P1′P2′P3′P4′ are:
P1= 1−i2 z1
P2= z1+1−i2 z2
P3= z1+ z2+1−i2 z3
P4= z1+ z2+ z3+1−i2 z4
P1′= 1+i2 z1
P2′= z1+1+i2 z1
P3′= z1+ z2+1+i2 z1
P4′= z1+ z2+ z3+1+i2 z1
By (⋆), the area of the counter-clockwise oriented quadrilateral QE is
Area(QE) = 1
2Re{i[(P2− P1)(P3− P2) + (P4− P3)(P1− P4)]}
= 1
8Re{i[(1 + i)z1+ (1 − i)z2][(1 + i)z2+ (1 − i)z3] + +i[(1 + i)z3+ (1 − i)z4][(1 + i)z4+ (1 − i)z1]} =
= 1
4Re{i[z1z2+ z2z3+ z3z4+ z4z1]} +1
4(|z2|2+ |z4|2− 2Re{z1z3})
= Area(Q) +1
8(|z1− z3|2+ |z2− z4|2).
In a similar way, it is possible to check that the area of the clockwise oriented quadrilateral QI is Area(QI) = −Area(Q) +1
8(|z1− z3|2+ |z2− z4|2).
Therefore, the following equality holds
Area(QE) − Area(QI) = 2Area(Q).
which generalizes the particular case (a) when z1= −z3 and z2= −z4.
