On each side of a given (arbitrary) triangle describe an equilateral triangle exterior to the given one, and join the centers of the three thus obtained equilateral triangles. Show that the resulting triangle is also equilateral.
It's indeed quite surprising that the shape of the resulting triangle does not depend on the shape of the original one. However it appears to depend on the shape of the constructed triangles: it's equilateral whenever the latter are equilateral. Herein lies an opportunity for a generalization:
On sides of an arbitrary triangle, exterior to it, construct (directly) similar triangles subject to two conditions:
The apex angles of the three triangles are all different.
The triangle of apices has the same orientation as the three triangles.
Connect centroids of the three triangles. Thus obtained triangle is similar to the constructed three.
Actually it's not even necessary to connect the centers. Any three corresponding (in the sense of similarity) points, when connected, define a triangle similar to the constructed ones [Wells, pp. 178-181]. Perhaps less surprisingly by now, the triangles can be constructed on the same side as the original triangle.
The original problem is traditionally ascribed to Napoleon Bonaparte who was known to be an amateur mathematician. Two of its proofs were kindly sent to me by Dr. Brodie. The problem relates to Fermat's question: find a point with the minimal total distance to the vertices of a given triangle, and to a slew of construction problems. For example, given three centroids of equilateral triangles constructed on the sides of the given triangle in its exterior - reconstruct the original triangle. Further on, one may consider n points and attempt to reconstruct a polygon. Centers of the equilateral triangles together with two base vertices form isosceles triangles. We actually may start with arbitrary isosceles triangles constructed on the sides of a given polygon. Given apex angles of the triangles, reconstruct the original polygon. We already discussed a construction problem where all such angles equaled 180° so that the triangles degenerated into straight lines.
There is a third proof based on the complex number arithmetic which, nonetheless, very much retains the geometric background. Another application of complex numbers leads to the shortest proof I am aware of. A fifth proof is derived from a simple yet general property of circumscribed circles.
- H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, NY, 1961
- G. Polya, Mathematical Discovery, John Wiley and Sons, 1981.
- D. Wells, You Are a Mathematician, John Wiley & Sons, 1997
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