Rational Trigonometry: Spread in a circle Part 1
Spread measurement in a circle
Following the definition of spread as the ratio of an opposite quadrance to the hypotenuse quadrance in a right-angled triangle, consider (following Euclid’s theorem on intersecting chords) forming two similar right-angled triangles in the semicircles ACB and AC’B. Let the diameter AB equal 1.
Then, comparing corresponding sides (noting OC = OC’ by symmetry)
x/y (in AOC’) = y/(1 – x) (in COB) and therefore y² = x (1 – x).
But since (in COB) h² = y² + (1 – x)² = x (1 – x) + (1 – x) (1 – x) = (1) (1 – x)
the spread at vertex B, s(B) = y² / h² = x (1 – x) / (1 – x) = x (*)
This only measures spread when one of the lines is a diameter of the circle however, so what about a spread between two oblique lines?
Beginning with chords A’B and C’B, if we rotate A’B onto the diameter AB this takes C’B onto chord CB. We can see that chords/arcs A’A and C’C need to be equal.
The perpendicular to C’B through A’, which meets the circle also at C, is parallel to chord AC’, also perpendicular to C’B (a right angle subtended by a diameter). Parallel chords, with midpoints collinear with the circle centre, are bisected by a common radius which bisects the associated inner and outer arcs A’AC’C and AC’ leaving arcs A’A and C’C equal by symmetry, and making A’AC’ and AC’C equal, This allows spread to again be measured by dropping a perpendicular from C to diameter AB at O.
It thus appears a circle provides direct connection with the geometric measurement of spread. However, it is not the circle that determines this construction but the property of perpendicularity. A circle serves simply as a device for constructing these perpendiculars, suggesting that spread could instead be constructed without an auxiliary circle. Is this possible?
Let us re-examine our basic figure by adding further parallel segments:
Here a second perpendicular, this time to chord A’B through point C’, has been drawn. As expected by symmetry this meets the circle, at point D, on the same perpendicular line through C and O.
We observe that new segments, C’G and A’G, are parallel and equal to existing segments AA’ and AC’ respectively. Since AA’ was equal by symmetry to C’C, C’G equals C’C. Since A’C is perpendicular to C’B, triangles C’GE and C’CE are congruent, making segments GE and CE equal, and E (where perpendicular A’C meets C’B) the midpoint of GC. Likewise, point F (on A’B) is midpoint to GD. Finally, O is necessarily the midpoint of chord CD.
Putting this together, triangle OEF will be medial to triangle CDG, ensuring that perpendicular segments from A to lines C’B or A’B followed by parallel segments C’F or A’E followed by parallel segments from F or E meet on the diameter at the same position as the previously dropped perpendicular from C or D on the circle.
This construction measures the spread of oblique lines without the need for an auxiliary circle. We amend our basic figure accordingly:
Here we leave chord CD of the previous construction visible for reference. The only remaining objects necessary to obtain a measurement of spread are:
- a pair of oblique lines through point B (whose spread is to be measured)
- a reference point A (defining the unit segment AB)
- a pair of perpendiculars through A to the given pair of lines through B
- a segment parallel to the alternate line from A perpendicular to the alternate line through B
- a second segment parallel to the original line from A meeting AB at position O.
The spread at B is then the ratio of segment AO to AB.