Force up–down mirror symmetry on the yantra while keeping all 29 triple points exact, and two of its bases are pulled into one line, carrying one point on the axis and a mirror pair: 1 + 2.
+Put ½ at the centre instead of the bindu. The critical line becomes the strut through 0 and ∞, the functional equation a sign flip through the centre. Swap ζ for a function with the same symmetry and no Euler product (Davenport–Heilbronn) and the zeros step off the strut: symmetry alone does not pin them, the arithmetic is the extra incidence.
+ζ, ξ and Davenport–Heilbronn painted in the s-plane, in w = 1/(s − ½), and on the Riemann sphere with ½ in front; Hardy's Z(t) along the strut, with the off-line zero at t ≈ 85.7 marked.
+ open explorer → + + +The earlier picture: ζ on a ball of diameter ½ resting at ½, turned by quaternion quarter-turns; 3b1b's grid as the ball's own latitude–longitude, bent by η.
+ open explorer → + +four upward triangles · five downward · the bindu
Triangle coordinates follow a numerical solution of the yantra's concurrency conditions (after Kulaichev and Huet), drawn as 27 lines with 29 triple points.