Euler's Eyes
2026-10-11
Eccentric angles
When the vertex of the angle is inside or outside the circle, but not at the center or on it, the angle is eccentric — and its measure is an average of two arcs.
Who discovered
The formulas for eccentric angles are not stated in The Elements, but come directly from the proposition III.20 of Euclid (inscribed angle) combined with I.32 (external angle of a triangle). One of the most famous problems linked to them has an author: in 1471, the German astronomer Regiomontanus asked from what distance a statue on top of a pedestal is seen from the largest angle possible.
The story
Regiomontanus’ answer uses an exterior eccentric angle. From any point on the ground, the statue is seen at an angle that is half the difference of two arches; the greatest angle occurs when the circumference that passes through the foot and the top of the statue tanges the ground. It was one of the first maximum problems in history, solved without calculation.
The same reasoning appears in rugby: after a try, the kicker chooses, along a line, the point from which he sees the posts from the greatest possible angle.
In everyday life
- Museums. The best distance to see a painting hanging above the eyes is from the Regiomontanus problem.
- Rugby. The choice of the conversion point follows the same principle.
- Cinema. In the first rows, the screen is seen from a large but very inclined angle; the eccentric angles explain why the middle of the room is best.
- Photography. To frame a tall object, the photographer looks for the point where it occupies the greatest angle of the field of view.
Sources: Regiomontanus (MacTutor) · Elements, proposition III.20 · Elements, proposition I.32.
Theory and constructions
TeoremaInner and outer eccentric angles
- Interior: if two chords intersect at an interior point, distinct from the center, the angle between them measures the semisoma of the arcs that it and its opposite vertex see: $x = \dfrac{a + b}{2}$.
- Exterior: if two secant rays (or tangents, or one of each) depart from an exterior point, the angle between them measures the semi-difference of the intercepted arcs: $x = \dfrac{a - b}{2}$.
Justification. In the interior case, $x$ is the exterior angle of a triangle whose two non-adjacent interior angles are inscribed, with measurements $\dfrac{a}{2}$ and $\dfrac{b}{2}$. Abroad, $\dfrac{a}{2}$ is the exterior angle of a triangle with interior angles $x$ and $\dfrac{b}{2}$.
