Equilateral Triangles on Sides of a Quadrilateral
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A Mathematical Droodle

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Discussion

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Copyright © 1996-2018 Alexander Bogomolny

The applet may suggest the following statement [Yaglom, p. 39]:

On the sides of an arbitrary quadrilateral ABCD equilateral triangles ABP, BCQ, CDR, DAS are constructed, so that the orientation of the first and the third is different from that of the second and the fourth. Then the quadrilateral PQRS is a parallelogram.

(In [Yaglom] quadrilateral ABCD is unnecessarily required to be convex, whereas both the theorem and the proof remain valid even for self-intersecting quadrilaterals.)

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Rotations will play an important role in the following. So let's introduce a convenience notation. For a point M and an angle α, let (M, α) denote a rotation around M through angle α. The sum of any two rotations is either a rotation or a translation. If, in the latter case, the sum has a fixed point, then it's the identical transformation, i.e. the transformation that leaves all the points unmoved.

Consider the four rotations: TP = (P, 60°), TQ = (Q, -60°), TR = (R, 60°), and TS = (S, -60°). The sum (or composition) of the four rotations is the rotation through angle 0°, i.e. a translation. In addition, successive execution of the four rotations maps A back to itself. Therefore, the sum of the four is the identical transformation.

For the same reason, the sum of the first two is a translation, as is the sum of the remaining two rotations. The two translations add up to the identity, or "zero" translation:

(TP + TQ) + (TR + TS) = 0,

which simply means that they are equal as vectors but point in opposite directions.

How can we visualize the two translations? Under a translation, all points move by the same vector. What happens, for example, to the point P under the sum TP + TQ? Obviously P is the fixed point of TP: TP(P) = P. Let TQ(P) = P'. Under the sum of the two rotations, P maps onto P'. The sum is, therefore, equivalent to a translation by vector PP'. Note that ΔPQP' is equilateral.

Similarly, if R' is the image of R under the sum TR + TS, then ΔRSR' is equilateral. The sides PP' and RR' of the two triangles are equal, parallel, but point in the opposite direction. The same holds for the other pairs of corresponding sides, e.g., PQ and RS. Which is just what is needed: PQ and RS are equal and parallel and point in opposite directions. The quadrilateral PQRS is indeed a parallelogram.

(This theorem has a natural generalization.)

References

  1. I. M. Yaglom, Geometric Transformations I, MAA, 1962

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  • Euler Line Cuts Off Equilateral Triangle
  • Four Incircles in Equilateral Triangle
  • Problem in Equilateral Triangle
  • Problem in Equilateral Triangle II
  • Sum of Squares in Equilateral Triangle
  • Triangle Classification
  • Isoperimetric Property of Equilateral Triangles
  • Maximum Area Property of Equilateral Triangles
  • Angle Trisectors on Circumcircle
  • Equilateral Triangles On Sides of a Parallelogram
  • Pompeiu's Theorem
  • Circle of Apollonius in Equilateral Triangle

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