I’m Up and Down, and Round and Round

Draw AB = 5 cm. At A draw a ray making 70° with AB; at B draw a ray making 60° with BA, on the same side. Their intersection is C.

Construct perpendicular bisectors of AB and AC, meeting at O. Draw the circle with centre O and radius OA.

The third angle is 50°, so the triangle is acute.

Draw AB = 5 cm. At A construct 100° and mark AC = 4 cm along the ray; join BC.

Intersect perpendicular bisectors of two sides to get O, then draw the circle with radius OA.

Draw AB = 6 cm. Arcs of radius 7 cm from A and B intersect at C. Join AC and BC.

Construct the perpendicular bisectors; their intersection is the centre.

The height is √(7² − 3²) = 2√10. Using radius R = abc/(4Δ), Δ = 6√10 gives R = 49/(4√10).

For any centre O, AB ≤ OA + OB = 2r.

Equality is reached at the midpoint of AB, making AB a diameter.

Both segments from the centre to the chord endpoints are radii of the same circle.

In the same circle, both triangles have two sides equal to the radius, and their bases are equal by assumption.

Let OM be perpendicular to AB. Right triangles OMA and OMB share leg OM and have equal hypotenuses OA = OB.

RHS congruence gives AM = BM.

In an isosceles triangle with AB = AC, the altitude from A to BC is also the perpendicular bisector of BC.

The centre is equidistant from B and C, so it lies on that perpendicular bisector.

For length 6, the half-chord is 3 and distance from centre is √(25 − 9) = 4 cm.

For length 8, the half-chord is 4 and distance is √(25 − 16) = 3 cm.

The midpoints lie on opposite sides along the same perpendicular.

For chord length L in a circle of radius r, the perpendicular distance satisfies d² = r² − (L/2)².

Equal lengths give equal nonnegative distances.

Treat H as the foot on GF, as shown in Fig. 5.15; the text’s “CH perpendicular to GH” identifies the same line.

The perpendiculars bisect AB and GF. Right triangles CEA and CHF have equal hypotenuses and equal perpendicular legs CE = CH.

RHS gives AE = FH; doubling gives AB = GF.

AE² = r² − CE² and FH² = r² − CH².

CE = CH implies AE = FH, and full chords are twice these lengths.

Half-chord length = √(7² − 6²) = √13.

The central perpendicular bisects the chord, producing a right triangle with hypotenuse r, one leg d and the other leg L/2.

Pythagoras gives (L/2)² = r² − d².

The chord lengths depend on √(r² − d²), not linearly on d.

Take r = 5, distances of AB and CD equal to 4 and 2. Then AB = 6 and CD = 2√21, which is not 12.

OA = OB = 12, so the two base angles of OAB are (180 − 60)/2 = 60°.

All three angles are equal, making the triangle equilateral.

(i) Points on the same side of AB and on the circle lie in the same segment, so the angles cannot differ.

(ii) Interpreting “same side” as same side of chord AB: this is generally true, but when AB is a diameter, angles from either semicircle are 90°, giving a counterexample.

(iii) Equal angles imply concyclicity if X and Y lie on the same side of AB. Without that condition it is not guaranteed.

For a counterexample, take A(−1, 0), B(1, 0), X(0, √3), Y(0, −√3). Both angles are 60°, but the circle through A, B, X does not pass through Y.

The four vertices are on a circle, so opposite angles add to 180°.

x = 180° − 100°.

Half-chord = √(13² − 5²) = √144 = 12.

At a point on the complementary arc, the inscribed angle is half the angle the specified arc subtends at the centre.

Radius = 26/2 = 13; half-chord = 12.

d = √(169 − 144).

Half-chord = √(225 − 81) = 12.

The centre has equal distances to both chord endpoints.

The locus of points equidistant from the two endpoints is their perpendicular bisector.

The relevant semicircular arc has central angle 180°.

The angle at C, distinct from the endpoints, is half of this.

Opposite angles sum to 180°: C = 180 − 75; D = 180 − 110.

(2x + 10) + (3x − 20) = 180 gives 5x = 190.

At x = 38, P = 86° and R = 94°.

Half-chord = 8; perpendicular distance = 6.

r² = 8² + 6² = 100.

For a cyclic quadrilateral use Brahmagupta’s formula K = √[(s − a)(s − b)(s − c)(s − d)].

s = (5 + 5 + 12 + 12)/2 = 17.

K = √(12×12×5×5) = 60.

Construct perpendicular bisectors of any two nonparallel sides; their intersection is the circumcentre.

Compare this point with the quadrilateral’s interior. This construction finds the centre without needing the full circumcircle.

A centre on a side is a boundary case; do not force it into an inside/outside answer.

Let AB and CD meet at P, with total length L. Write PA = u, PB = L − u, PC = v, PD = L − v.

The intersecting-chords relation follows from similar triangles: PA·PB = PC·PD.

u(L − u) = v(L − v) simplifies to (u − v)(L − u − v) = 0.

Thus either u = v or u = L − v; the unordered pairs of piece lengths are the same.

Draw AB = 6 cm and its midpoint M. Construct a perpendicular at M and mark O with OM = 3 cm.

Draw the circle centred at O with radius OA = √(3² + 3²) = 3√2 cm.

In a parallelogram opposite angles are equal. In a cyclic quadrilateral they add to 180°.

Equal angles with sum 180° are each 90°. Conversely every rectangle has supplementary opposite angles.

A rectangle’s diagonals are equal and bisect one another.

Their intersection is therefore equally distant from all four vertices and is the unique circumcentre.

For fixed chord length L, each midpoint is at distance d = √(r² − L²/4) from O.

Rotating the chord through all orientations traces every point at this distance.

Triangles OAB and OAC have OA common, OB = OC (radii), and AB = AC (given).

SSS congruence gives ∠BAO = ∠OAC.

Let the offset of the 24 cm chord be d. The 10 cm chord is shorter, so its offset is d + 7.

r² = d² + 12² = (d + 7)² + 5².

144 = 14d + 49 + 25, so d = 5; r² = 25 + 144 = 169.

The six central angles are 360°/6 = 60°, making each centre-side triangle equilateral.

Each side length is r. Its offset is √(r² − (r/2)²) = (√3/2)r.

MN is a diameter, so ∠MON = ∠MPN = 90°. These are not the two angles asked for.

The requested ∠MOP and ∠MNP are opposite angles of cyclic MNOP, so their sum is 180°; neither is fixed individually.

The exterior angle at D equals 180° − ∠CDA.

Since ∠CDA + ∠ABC = 180°, this is ∠ABC.

For chord endpoints A, B, the triangle inequality gives AB ≤ AO + OB = 2r.

Equality occurs only when A, O, B are collinear, i.e. the chord is a diameter.

A line through A has perpendicular distance d from O with d ≤ OA.

Chord length 2√(r² − d²) decreases as d increases.

The maximum possible d is OA, attained by the line through A perpendicular to OA.

The two triangles formed by joining O to the top vertex have two equal radii each.

Their top angles equal the base angles a and b. The whole triangle therefore has top angle a + b.

Its angle sum is 2(a + b) = 180°.

Reflection across diameter AB exchanges C with C′ and D with D′, since AB bisects both perpendicular chords.

It therefore exchanges the midpoint M of CD with the midpoint M′ of C′D′.

A point and its distinct reflection are joined by a segment perpendicular to the reflecting line.

Join the centre to the four vertices. A radius creates isosceles triangles; their equal base angles show that an inscribed angle is half the central measure of its intercepted arc.

If the centre lies outside an angle, subtract the adjacent isosceles-triangle angles instead of adding them. Thus the arc-angle rule covers that position too.

The arcs intercepted by a pair of opposite vertices together form the full circle. Half of their total 360° is 180°.

Examples include water ripples, a cut stem and the outline of the full moon. These are approximations to an ideal mathematical circle.

Fold the paper circle so its edges coincide; the crease is a diameter. Repeat along a different direction.

A regular n-gon has n rotation symmetries modulo a full turn and n reflection axes.

Square: 0°, 90°, 180°, 270°; pentagon: multiples of 72°; hexagon: multiples of 60°.

The diameter is the longest chord: 2×5 = 10.

Distinct endpoints can be arbitrarily close, so positive chord lengths approach 0 without a smallest positive value.

Every point equidistant from A and B lies on their perpendicular bisector.

Conversely, if P is on that line through midpoint M, right triangles PMA and PMB have matching legs, so PA = PB.

For AB = L, centres lie anywhere on its perpendicular bisector. At offset t from the midpoint the radius is √((L/2)² + t²).

There are infinitely many circles. The minimum is L/2; there is no maximum, and radii below L/2 are impossible.

Moving farther from the midpoint increases radius. Larger-radius arcs appear flatter over a fixed chord.

With A and B merely on the boundary, there are infinitely many squares.

With A and B both vertices, either AB is a side (two choices of side of the line) or a diagonal (one choice).

For three distinct collinear points, the perpendicular bisectors of adjacent segments are distinct parallel lines, so no point is equally distant from all three.

A line meets a circle in at most two distinct points; no circle passes through three collinear points.

Rotating or reflecting an inscribed triangle about its circle’s centre gives congruent triangles sharing the same circle.

Folding to superpose a chord’s endpoints constructs its perpendicular bisector, meeting it at 90°.

An equal-length chord can be made by rotating the first chord around O; its offset remains unchanged.

Sample measurements in a radius-5 circle: chord 6 gives offset 4; chord 8 gives offset 3; chord 10 gives offset 0.

Use a protractor on the rays in Fig. 5.19, measuring along the indicated arc rather than automatically choosing the smaller angle.

From the drawing, AKB is about 100° and CLD about 205°; these are approximate drawing measurements, not supplied exact angles.

Choose three points outside a fixed arc, draw rays to its endpoints and measure.

Each measured angle is half the arc’s central angle, subject to drawing error.

80° + 100° = 180° and 110° + 70° = 180°.

Both opposite pairs satisfy the cyclic condition.

If C lies inside the circle through A, B, D, extend DC beyond C to meet the circle at E.

The cyclic quadrilateral ABED gives ∠BAD + ∠BED = 180°. The hypothesis gives ∠BAD + ∠BCD = 180°, so ∠BED = ∠BCD.

But ∠BCD is an exterior angle of triangle BCE, strictly greater than ∠BEC = ∠BED. This contradiction excludes the inside case.