Inscribed angle — the series

An angle standing on a chord. A circle with a fixed chord and a movable apex on the major arc. The angle at the apex is 60 degrees wherever the apex is put, and the angle the same chord subtends at the centre is 120 degrees.

An angle that does not care where it stands

Fix two points on a circle and look at them from anywhere else on the far arc. The angle is the same from every one of those places, and it is exactly half the angle at the centre.

part 1 · geometry

Every chord through the point cuts into pieces whose product is 16.00. A circle with a point inside it and four chords drawn through the point, each labelled with the lengths of its two pieces, beside four rectangles whose sides are those pieces and whose areas are all equal.

One number for every chord through a point

Draw any line through a point and let it cut a circle twice. The two distances from the point to the circle multiply to the same number whichever line is drawn — inside, outside, or grazing as a tangent. The number belongs to the point, and the reason it does not depend on the line is the inscribed angle: two chords through a point cut out two triangles with the same angles.

part 2 · geometry

Each side over the sine of the opposite angle is the diameter. A triangle inscribed in a circle, with the diameter from one corner drawn and joined to a second corner to make a right-angled triangle; the angle opposite the side at the far end of the diameter equals the triangle's own angle opposite that side.

Every side measured by one diameter

In any triangle, divide each side by the sine of the angle opposite it: the three answers are equal. That much is the law of sines, and it is usually left there. The common answer has a name — it is the diameter of the circle through the triangle's corners — and the reason is the inscribed angle once more. It is also why the sine was first a half-chord, why Ptolemy's theorem is the addition formula, and why a circle can hold infinitely many points all at rational distances from each other.

part 3 · geometry

456 pairs at one distance among 144 grid points. A 12-by-12 grid of points with 456 segments joining every pair at distance √25, in 6 directions.

How many pairs can be one apart

Place n points in the plane and count the pairs exactly one unit apart. A square grid with its spacing as the unit gives two pairs per point; the same grid with the unit chosen as a length it realises in many directions — the square root of 5, of 25, of 65, of 325 — gives three, five, eight. Counting points on circles caps the total at a constant times n to the four-thirds. Between that cap and the grids the truth has not moved since 1984, and at every size anyone can draw, the two look alike.

part 4 · geometry

Wings that cut a chord equally

Take a chord of a circle and its midpoint. Draw any two more chords through the midpoint, join their ends crosswise, and the two crossing lines — the butterfly's wings — cut the first chord at equal distances from the middle. The proof is the inscribed angle and the power of a point working together; move the point off the middle and what survives is a law about reciprocals; replace the circle by any conic and the theorem does not notice.

part 5 · geometry

Two ways to cut a hexagon, one total of radii. A cyclic hexagon triangulated two ways with the incircles drawn; the inradii sum to 1.1622 both times.

A total hung in a temple

Put a polygon's corners on a circle, cut it into triangles, and add up the radii of the circles inscribed in the triangles. Cut it a different way and the total is the same — for all fourteen ways of cutting a hexagon, to every digit. The fact was painted on a wooden tablet in a Japanese temple around 1800, and the reason for it is a theorem about one triangle and the distances from its circumcentre to its sides.

part 6 · geometry

Two triangulations, the same sum of cos 3x, different sums of cos 2x. Cyclic heptagon, triangulations 0 and 29: Σcos3x 1.52896873 / 1.52896873; Σcos2x -0.789261 / -0.467086.

The sums a triangulation cannot change

The Japanese theorem says the inradii of any triangulation of a cyclic polygon add to the same total. Write the inradius as a sum of cosines of the angles, and the theorem becomes one case of a rule that fits in a line: a sum of f over every angle of every triangle ignores the triangulation exactly when f(x) + f(π − x) is constant. cos 3x qualifies, cos 2x does not, and the proof is the inscribed-angle theorem applied to a single flipped diagonal.

part 7 · geometry