GC-04 · Right Triangles, Egyptian 3–4–5 & Kepler Proportions

Huntley’s 3–4–5 bisector circle, Kepler’s geometric-progression right triangle, and Niemeyer’s linkage

In a 1597 letter to Michael Mästlin, Johannes Kepler called the Pythagorean theorem and the division in extreme and mean ratio the ‘two great treasures of geometry.’ This family unites both treasures directly: H. E. Huntley’s angle-bisector circle in the \(3\text{-}4\text{-}5\) Egyptian triangle, Kepler’s unique \(1 : \sqrt{\phi} : \phi\) right triangle over \(\mathbb{Q}(\sqrt{\phi})\), Jo Niemeyer’s three equal leaning rods, and Van Obel cevian concurrence.

Cataloged Constructions (4)

GC-04.01 · H. E. Huntley’s Golden Ratio in the 3-4-5 Right Triangle (1970)

In a \(3\text{-}4\text{-}5\) right triangle ABC (\(|BC|=3, |AC|=4, |AB|=5\)), let the internal bisector of \(\angle ABC\) meet leg AC at O. Circle cOC centered at O through C intersects line BO at P and Q such that \(|PQ|/|BP| = |BQ|/|PQ| = \phi\).

GC-04.02 · Johannes Kepler’s Geometric-Progression Right Triangle (1597)

Divide hypotenuse AB at H in the golden ratio (\(|AH|/|HB| = \phi\)), and erect perpendicular HC to semicircle cOB on diameter AB. Right triangle ABC has side lengths in geometric progression \(1 : \sqrt{\phi} : \phi\) and side quadrances in progression \(1 : \phi : \phi^2\).

GC-04.03 · Jo Niemeyer’s Three Equal Leaning Segments

Place segment A1B1 perpendicular to line L at A1, lean equal segment B2A2 from midpoint B2 of A1B1 onto L, and lean a third equal segment B3A3 from midpoint B3 of B2A2 onto L. Then A2 divides A1A3 in the golden ratio: \(|A1A2|/|A2A3| = \phi\).

GC-04.04 · Bùi Quang Tuân’s Van Obel Division in a Right Isosceles Triangle

In right isosceles triangle ABC with hypotenuse midpoint M and midpoint O of altitude AM, circle cOM cuts OC and OB at D and F. Cevians BD and CF concur on OM at E, dividing OM in the golden ratio: \(|OE|/|EM| = \phi\).