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Classical

Napoleon and Van Aubel through Complex Numbers

Two regular side constructions encoded by fixed rotations

Complex multiplication encodes the external equilateral triangles of Napoleon’s theorem and the external squares of Van Aubel’s theorem, reducing both conclusions to rotation identities.

Methods Complex numbers · Direct similarity

Napoleon’s theorem

Theorem

Construct outward equilateral triangles on the sides BC,CA,ABBC,CA,AB of a triangle ABCABC, and let their centers be P,Q,RP,Q,R, respectively. Then PQRPQR is equilateral.

Triangle ABC with external equilateral triangles and their centers P, Q, R joined to form an equilateral triangle.
Figure 1. The centers P, Q, R of the three external equilateral triangles form the Napoleon triangle.

Proof

Identify the plane with the complex plane, orient ABCABC counterclockwise, and write a,b,ca,b,c for the coordinates of its vertices. Multiplication by −i-i gives the outward normal to each directed side. The center of an equilateral triangle on a side of length ℓ\ell lies at distance ℓ/(23)\ell/(2\sqrt3) from the midpoint of that side. Hence the coordinates p,q,rp,q,r of P,Q,RP,Q,R are

p=b+c2−i23(c−b),q=c+a2−i23(a−c),r=a+b2−i23(b−a).\begin{aligned} p&=\frac{b+c}{2}-\frac{i}{2\sqrt3}(c-b),\\ q&=\frac{c+a}{2}-\frac{i}{2\sqrt3}(a-c),\\ r&=\frac{a+b}{2}-\frac{i}{2\sqrt3}(b-a). \end{aligned}

Put

ε=eπi/3.\varepsilon=e^{\pi i/3}.

Direct subtraction gives

q−r=ε(p−r).q-r=\varepsilon(p-r).

Thus the vector RPRP is carried to RQRQ by a rotation through 60∘60^\circ. The two vectors have the same length, and therefore PQRPQR is equilateral.

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Van Aubel’s theorem

Theorem

On the sides AB,BC,CD,DAAB,BC,CD,DA of a convex quadrilateral ABCDABCD, construct squares externally, and let P,Q,R,SP,Q,R,S be their centers in the same order. Then the segments PRPR and QSQS are equal in length and perpendicular.

Quadrilateral ABCD with external squares and their centers P, Q, R, S; the segments PR and QS are equal in length and meet at a right angle.
Figure 2. The segments joining the centers of opposite external squares are equal and perpendicular.

Proof

Write a,b,c,da,b,c,d for the complex coordinates of the counterclockwise vertices A,B,C,DA,B,C,D. The center of the external square on an oriented side is obtained by adding half of its clockwise quarter-turn to the side midpoint. Consequently,

p=a+b2−i2(b−a),q=b+c2−i2(c−b),r=c+d2−i2(d−c),s=d+a2−i2(a−d).\begin{aligned} p&=\frac{a+b}{2}-\frac i2(b-a),\\ q&=\frac{b+c}{2}-\frac i2(c-b),\\ r&=\frac{c+d}{2}-\frac i2(d-c),\\ s&=\frac{d+a}{2}-\frac i2(a-d). \end{aligned}

Subtracting the formulas above directly gives

r−p=12((1+i)(c−a)+(1−i)(d−b)),s−q=12((1−i)(a−c)+(1+i)(d−b))=i(r−p).\begin{aligned} r-p&=\frac12\bigl((1+i)(c-a)+(1-i)(d-b)\bigr),\\ s-q&=\frac12\bigl((1-i)(a-c)+(1+i)(d-b)\bigr)\\ &=i(r-p). \end{aligned}

Multiplication by ii is a quarter-turn and preserves length. Hence QSQS is perpendicular to PRPR and has the same length.

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Both proofs encode each regular side construction as the midpoint of a side plus a fixed complex multiple of its side vector. After the endpoint terms are collected, the first calculation leaves a 60∘60^\circ rotation between two sides of PQRPQR, while the second leaves a 90∘90^\circ rotation between PRPR and QSQS.