# Napoleon on Hinges

Napoleon's theorem claims that the centers A', B', C', of the equilateral triangles A''BC, AB''C, ABC'', erected on the sides (either all inwardly or all outwardly) of a given triangle ABC form an equilateral triangle. The applet below serves to illustrate a very simple proof of this result.

Use the scroll bar at the bottom of the applet to rotate some triangles 120°. Which triangles are get rotated depends on which of the three boxes Swivel at "B, C", "A, C", or "A, B", is checked. Rotations are actually around A', B', or C'.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

The proof below is a slight modification of a recently published proof by Alex Anderson (the proof has appeared a decade earlier in a Russian popular math and physics journal Kvant.) Below we assume that none of the angles in ΔABC exceeds 120°. The proof will have to be modified otherwise.

Assume the box "B, C" is checked so that we rotate triangles A'BC' (around C') and A'B'C (around B'). B and C are rotated onto A (because triangles ABC' and AB'C are isosceles with the apex angles at C' and B', respectively, equal to 120°.) Now, let's check the angles:

 ∠A'BC' = ∠ABC + ∠ABC' + ∠CBA' = ∠ABC + 30° + 30° = ∠ABC + 60°.

Similarly, ∠A'CB' = ∠ACB + 60° and ∠B'AC' = ∠BAC + 60°.

At the end of rotations when triangles A'BC' and A'B'C meet at A, the sum of angles at A is exactly 360°:

 ∠A'BC' + ∠ A'CB' + ∠ B'AC' = ∠ABC + ∠ACB + ∠BAC + 3×60° = 360°.

This means that, after the rotation, the sides A'B and A'C coalesce into a segment which we denote AA*. Observe that by SSS, ΔA*B'C' = ΔA'B'C'. Let &alpha, β &gamma denote the respective angles of ΔA'B'C'. Then ∠A*C'B' = 120° - γ while, on the other hand, ∠A*C'B' = ∠A'C'B' = γ. It follows that γ = 60°. Similarly, α = β = 60°. Napoleon's triangle is indeed equiangular, hence, also equilateral.

As we just saw, Napoleon's triangle A'B'C' is the union of the triangles congruent to triangles AB'C', A'BC', and A'B'C. This dissection can be obtained by reflecting A in B'C', B in A'C', and C in A'B', telling us that all three coincide. The common point, say X, is the Fermat point of ΔABC. Indeed,

XB' = AB' = CB' = B'B'',

Implying that X is on the circumcircle of ΔAB''C. Similarly, it is on the circumcircles of triangles ABC'' and A''BC and hence is their common point.

### References

1. A. Anderson, A New Proof for Napoleon's Theorem, Mathematical Reflections 3 (2007)
2. A. Savin, Geometric Surprises, Kvant 2 (1996) (in Russian)