Given a triangle and a set of three angles, the celebrated geometrical theorem of Jacobi produces a new triangle in perspective with the first. If this second triangle is related to the first by another set of three angles then these two triangles are said to be reciprocal Jacobi triangles. It is shown that the locus of the perspector is then the McCay cubic. 1. Jacobi Triangles. With ABC being any triangle, construct the points P, Q, R so that ∠RAB = ∠QAC = α, ∠P BC = ∠RBA = β and ∠QCA = ∠P CB

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... = γ. These points form a Jacobi triangle for ABC and Jacobi's theorem states that the lines AP, BQ and CR are concurrent (at the point K), see Figure 1. To quote [5], this result 'was seemingly discovered by Carl Friedrich Andreas Jacobi (not to be confused with Carl Gustav Jacob Jacobi), and published in 1825 in Latin'. Many proofs of this result are available, e.g. [4] and [1, pp. 55-56], but one is given here because some results from it will be needed later. 1.1. A Proof of Jacobi's Theorem. With reference to Figure 1, let the lines AP and BC meet at P. The sine rule applied to triangles BP P and CP P gives BP P C = sin γ sin β sin ∠BP P sin ∠CP P (1) and applied to triangles ABP and ACP , sin ∠BP A sin ∠CP A = c b sin(B + β) sin(C + γ) = sin ∠BP P sin ∠CP P. (2)