Lean Napoleon's Triangles: What is this about?
A Mathematical Droodle

Explanation

The applet attempts to suggest the following statement [Garfunkel, Stahl]:

Proof

The proof is computational but not tedious.

Similarly, ΔAMB' is right and (AB')² = b²/3.

We are now in a position to apply the Law of Cosines in ΔAB'C' to determine the length of B'C'. Note that B'AC' = A + 60°.

where S is the area of ΔABC. (We again used Law of Cosines and also a formula for the area of a triangle.) The expression is symmetric in a, b, c so that (A'B')² and (A'C')² are bound to be equal to the same quantity, proving that indeed ΔA'B'C' is equilateral.

Observe that the above derivation only works when triangles KLC', MNB', and PQA' are drawn outwardly on the sides of ΔABC. If they are drawn inwardly, then B'AC' may equal either A - 60° or 60° - A. In both cases we would obtain

with the same conclusion. Thus we are led to two equilateral triangles related to a base ΔABC. From the above relations it follows that the difference in areas of the two triangles is exactly that of ΔABC.

References

1. J. Garfunkel, S. Stahl, The Triangle Reinvestigated, Am Math Monthly, Vol. 72, No. 1. (Jan., 1965), pp. 12-20

Napoleon's Theorem

• A proof by tessellation
• A proof with complex numbers
• A second proof with complex numbers
• Two proofs
• Inscribed angles
• Douglas' Generalization
• Napoleon's Theorem by Transformation
• A Generalization
• Napoleon's Propeller
• Fermat point
• Kiepert's theorem
• Lean Napoleon's Triangles