Four holes must lie on one circle. Or a round insert must touch four straight edges. Both problems involve a circle and a quadrilateral, but their conditions differ: the first concerns angles, the second side lengths.

An inscribed angle is half the degree measure of the arc not containing its vertex. A convex quadrilateral has a common circle through its vertices exactly when opposite angles sum to 180°; it has an interior circle tangent to all sides exactly when opposite side-length sums are equal. Prerequisites: triangle angles and circles and tangents.

One chord bounds two arcs

An angle with vertex at centre O is a central angle. Radii OA and OB bound two angular regions and two arcs. A full circle measures 360°; a minor arc might measure 120°, leaving a major arc of 240°. This is the arc’s degree measure, not its length.

An inscribed angle has vertex P on the circle and sides through two other points A and B. Choose arc AB without P. If it measures θ:

∠APB = (θ)/(2).

A 120° arc gives 60°; a 240° arc gives 120°. The second case uses the corresponding central angle exceeding 180°, rather than automatically choosing the smaller angle between the radii. Euclid III.20.

120° arc → 60° inscribed angle

ABOP120°60°

240° arc → 120° inscribed angle

ABOP240°120°

Four vertices on a circle

ABCD100°80°

Circle inside, sides tangent

ABCDIabcd
Each panel uses a uniform scale without a physical length unit. In the first two, P is not on the highlighted orange arc: the same AB has different chosen arcs, 120° and 240°. Third: ∠A=100°, ∠C=80°. Fourth: I is the centre; the circle touches the interior of all four sides; a,b,c,d are consecutive side lengths.

The factor of two comes from isosceles triangles: join O to the inscribed-angle vertex, obtaining equal radii and equal base angles. An exterior angle of such a triangle is the sum of two equal angles. Depending on the centre’s position these pieces are added or subtracted; the half-angle relationship remains.

Inscribed angles intercepting the same arc are equal. But angles on opposite sides of one chord can be 60° and 120°: they intercept different arcs. An angle intercepting a 180° semicircle is right. Euclid III.21.

Four vertices on a circle

We now use convex quadrilaterals, without self-intersections or interior angles of 180° or more. ABCD is cyclic if all four vertices lie on one circle in order.

Opposite angles A and C intercept arcs that together complete the circle. Their sum is therefore half of 360°:

∠A + ∠C = 180°.

Likewise, ∠B+∠D=180°. This is necessary. The converse also holds: if a given convex quadrilateral has supplementary opposite angles, one circle passes through its vertices. The condition is therefore sufficient too. Euclid III.22, AMSI’s converse criterion.

Opposite angles of 100° and 80° pass the test. Two equal angles alone, without specifying their positions, are insufficient.

A circle touching four sides

A quadrilateral is tangential, or circumscribed about a circle, when the circle is inside and every side is tangent to it. The contact points lie on the side segments, not their extensions.

Two tangent segments from one external point are equal: radii to their contact points are perpendicular, and the resulting right triangles share a hypotenuse from the point to the centre and have equal radius legs. They are congruent by the hypotenuse-leg criterion, so their tangent segments have equal lengths.

Each side consists of two such segments from neighbouring vertices. Each opposite-side sum includes all four tangent lengths once. If a,b,c,d are consecutive sides:

a + c = b + d.

This is necessary. For a given convex quadrilateral it is also sufficient, but the converse is a separate theorem, not a consequence of merely regrouping terms. AMSI, Pitot’s theorem in the Mathematical Gazette.

Keep the two tests distinct: 180° concerns vertices on a circle; equal length sums concern sides tangent to a circle. Even when both properties hold, the incircle and circumcircle are different objects.