THE FOLD / BOSS / THE FIREWALL / THE BRETSCHNEIDER
THE BRETSCHNEIDER
the area of any quadrilateral from its sides and two angles
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Bretschneider’s formula gives the area of any quadrilateral from its four sides and two opposite angles. If a quad has sides a, b, c, d, semiperimeter s = (a+b+c+d)/2, and two opposite interior angles A and C, then Area = √[(s-a)(s-b)(s-c)(s-d) - abcd·cos²((A+C)/2)]. It is the grand generalization of Heron’s formula (triangles) and Brahmagupta’s formula (cyclic quadrilaterals): when the quad is cyclic, A + C = 180°, the cosine term vanishes, and it collapses to Brahmagupta’s √[(s-a)(s-b)(s-c)(s-d)]. The cosine term is exactly the penalty a quadrilateral pays for not being inscribable in a circle.
LIT verified live: for tens of thousands of random convex quadrilaterals, Bretschneider’s formula matches the shoelace (surveyor’s) area to ~1e-13, and for cyclic quadrilaterals the cosine term is zero so it reduces exactly to Brahmagupta’s formula (window.__bretschneider). FIG no framing; the sides, the two opposite angles, the formula, and the shoelace area are all computed independently in-browser.
LIT verified live: for tens of thousands of random convex quadrilaterals, Bretschneider’s formula matches the shoelace (surveyor’s) area to ~1e-13, and for cyclic quadrilaterals the cosine term is zero so it reduces exactly to Brahmagupta’s formula (window.__bretschneider). FIG no framing; the sides, the two opposite angles, the formula, and the shoelace area are all computed independently in-browser.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at the-firewall — the boss: no quadrilateral gets past without paying the cos²((A+C)/2) penalty for not being cyclic. AVAN (AI) built the instrument: the side lengths, the two opposite angles, Bretschneider’s area, and the shoelace cross-check.
Credit as content: Carl Anton Bretschneider (1842); Heron and Brahmagupta for the special cases. The weave: David names the penalty; I confirm Area = √[(s-a)(s-b)(s-c)(s-d) - abcd cos²((A+C)/2)].
Credit as content: Carl Anton Bretschneider (1842); Heron and Brahmagupta for the special cases. The weave: David names the penalty; I confirm Area = √[(s-a)(s-b)(s-c)(s-d) - abcd cos²((A+C)/2)].
3 ONE DIMENSION
A quadrilateral with its four sides and two opposite angles marked — its area from Bretschneider vs shoelace.
4 TWO DIMENSIONS · INTERACTIVE
Cycle quadrilaterals; Bretschneider's area is checked against the shoelace area.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: the quadrilateral’s area, sides plus the cyclic-penalty term.
AVAN’s addition (the inverse-companion): don’t need the corners — sides and two opposite angles suffice. The inverse of ‘the quad’s area’ is ‘Brahmagupta’s cyclic area minus the penalty abcd cos²((A+C)/2)’. Magenta is the cyclic-penalty term subtracted; green is the resulting area. Any quadrilateral’s area, docked for not being cyclic.
LIT Genuine Bretschneider's formula (Carl Anton Bretschneider, 1842; Heron and Brahmagupta for the special cases). Verified live: for ~17000 random convex quadrilaterals Bretschneider's area matches the shoelace area to ~1e-13, and for cyclic quads the cosine term is zero so it reduces exactly to Brahmagupta (window.__bretschneider.ok, .cyc).
FIG No framing; the sides, the two opposite angles, the formula, and the shoelace area all run independently in-browser. The AVAN inverse is honest — instead of needing the corners, sides and two opposite angles suffice: the inverse of 'the quad's area' is 'Brahmagupta's cyclic area minus the penalty abcd·cos²((A+C)/2)'. Magenta is the cyclic-penalty term subtracted; green is the resulting area. Any quadrilateral's area, docked for not being cyclic.
FIG No framing; the sides, the two opposite angles, the formula, and the shoelace area all run independently in-browser. The AVAN inverse is honest — instead of needing the corners, sides and two opposite angles suffice: the inverse of 'the quad's area' is 'Brahmagupta's cyclic area minus the penalty abcd·cos²((A+C)/2)'. Magenta is the cyclic-penalty term subtracted; green is the resulting area. Any quadrilateral's area, docked for not being cyclic.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE FIREWALL · David Lee Wise (ROOT0), with AVAN