Circles, polygons and geometric constructions

A circle is a boundary determined by a positive radius

Imagine a compass opened to 3 cm around a fixed centre O. A point 3 cm from O lies on the circle, a point 2 cm away is inside, and a point 4 cm away is outside. The centre itself is inside, rather than on, the circle. The circle is the boundary; filling it in gives the disk. Which of those three points would belong to the filled region?

Work in a Euclidean plane with one common length unit. For a fixed centre O and a positive radius r, the circle consists of the points P whose distance from O is exactly r. The closed disk consists of the points whose distance from O is at most r, including the circle and its interior. A point at distance less than r is inside the disk but is not on the circle. Keep this distinction when counting common boundary points: two disks can overlap even when their bounding circles do not intersect.

A radius segment joins O to a point on the circle; the same word also names its length r. A chord joins two distinct points on the circle. A diameter is a chord through O and has length 2r, because its two halves are radii. Two distinct endpoints divide the circle into two arcs, each a curved part of its boundary. A chord is straight, while an arc is curved; sharing endpoints does not make their lengths interchangeable. Diameter endpoints give two equal semicircular arcs.

A secant line meets a circle in two distinct points; its portion between them is a chord. A tangent line meets a circle in exactly one point. These definitions concern the entire straight line, rather than a short segment drawn on the screen. We will prove the radius-perpendicular property of a circle's tangent below. A line that misses the circle is neither a secant nor a tangent.

Require r > 0 before making circle claims or drawing a circle. Radius zero collapses the distance condition to the single point O, which is excluded from this lesson's circle definition. A negative radius is not a permitted distance. Reject those specifications before squaring r, since squaring a negative input must not turn it into an accepted positive-radius circle.

Choose a first reading, then return to the proofs

Begin with circle parts, circumference and disk area, then compare the line and two-circle diagrams. You can also count the triangles in the polygon fan. The linked published lessons on powers and roots, quantities and units, and coordinates can help with square roots, square units and ordered pairs.

The proof sections also use triangle angle sums, the Pythagorean relationship and triangle congruence. Congruent triangles have matching sides and angles: SSS means three corresponding sides agree, and SAS means two sides and their included angle agree. If these ideas are new, first read the concrete examples and diagrams, then return to the midpoint, intersection and construction proofs once you are comfortable with them. The pictures help you follow the argument; they do not supply those theorems by measurement.

Circumference measures the boundary; area measures the disk

Use the established Euclidean circle formulas C = 2πr = πD for circumference and A = πr² for disk area, where D = 2r is the diameter. The constant π is the common ratio C/D for circles of positive diameter. Circumference has length units, such as cm; disk area has square units, such as cm². For radius 3 cm, C = 6π cm and A = 9π cm². These are exact values in terms of π, rather than rounded decimal results.

We use these formulas as established results here. The circumference-to-diameter ratio and the disk-area formula require a treatment of curved length and area beyond finite triangle dissections. Cutting a disk into finitely many sectors and rearranging them can suggest the area formula, but does not produce an exact rectangle and is not a rigorous derivation of πr² by itself. The chord, construction and polygon proofs in this lesson do not depend on treating such a suggestive picture as a proof.

Convert supplied lengths before substitution. Diameter 300 cm means radius 150 cm, or radius 3/2 m. The same disk has circumference 300π cm = 3π m and area 22500π cm² = 9π/4 m². One metre is 100 centimetres, so one square metre is 10000 square centimetres. Giving πD² as the disk area would mistake the diameter for the radius and multiply the correct area by 4.

A uniform enlargement by a positive factor k sends r to kr. Its circumference is 2πkr = kC, while its area is π(kr)² = k²A. Doubling the radius doubles circumference and multiplies area by 4. The factor k > 0 preserves a positive circle radius; zero collapses the figure. Keep π and exact fractions until a requested approximation is needed, then use ≈ and round the final result. Neither 3.14 nor 22/7 equals π exactly.

Congruence explains the chord's midpoint and perpendicular

Start with radius 5 and a line 3 units from the centre. Its two circle points are U = (3, 4) and V = (3, −4), and F = (3, 0) lies halfway between them. The right triangle has side lengths 3, 4 and 5. Before using the general formula, predict what happens to the chord when the line moves farther from the centre. The diagram also shows why the distance to the line and half the chord are different lengths.

Let AB be a chord of a positive-radius circle with centre O, and let M be the midpoint of AB. First suppose the chord does not pass through O. Then O, A and B form an isosceles triangle because OA = OB = r, and O, M and A are noncollinear. In triangles OMA and OMB, OA = OB, MA = MB and OM is common. SSS makes the triangles congruent. Thus the adjacent angles OMA and OMB are equal; since A, M and B lie on a straight line, their sum is 180° and each is 90°. The line OM is the perpendicular bisector of AB.

The perpendicular-foot version gives the same midpoint conclusion. Suppose O is not on line AB, and let F be the foot of the perpendicular from O to that line. Use F as coordinate origin, with one axis along AB and the other along FO. The Cartesian squared-distance formula, including its aligned cases, gives OA² = OF² + FA² and OB² = OF² + FB² even if a component were zero. Since OA = OB = r, subtraction gives FA² = FB². The lengths FA and FB are nonnegative, so FA = FB. On line AB the only point equidistant from distinct A and B is their midpoint; both distances are then positive. This proves the foot lies inside the chord before using any nondegenerate right triangles there.

Handle a diameter separately. For a diameter AB, its midpoint is O itself. The segment OM then has length zero and does not determine a line or a nondegenerate isosceles triangle. The diameter's perpendicular bisector is the unique line through O perpendicular to AB. It still passes through the centre, but calling the zero segment OM that line would omit this special case.

Let d be the nonnegative perpendicular distance from O to a secant line, and let h be half the resulting chord length. For 0 < d < r, the perpendicular meets the chord at its midpoint, and a nondegenerate right triangle gives h² + d² = r². Therefore h = √(r² − d²) > 0 and AB = 2√(r² − d²). If d = 0, the chord is a diameter, so h = r and the same formula gives 2r; that case uses the diameter definition rather than a collapsed right triangle. The formula for a genuine chord is thus valid for r > 0 and 0 ≤ d < r.

Distance from the centre classifies a line

For a circle with r > 0, take any line and let F be the foot of the perpendicular from O, with distance d = OF ≥ 0. If d > 0, every other point P on the line forms a right triangle OFP. Its distance satisfies OP² = d² + FP², so OP = r requires FP² = r² − d². If d = 0, the line passes through the centre and the two circle points lie a distance r in opposite directions along it. This separate line-distance argument covers the case without claiming a nondegenerate right triangle at F = O.

When 0 ≤ d < r, there are two points at opposite displacements ±√(r² − d²) from F, and the line is a secant. When d = r > 0, only F has distance r from O: the displacement is zero, so the line is a tangent. This one-point boundary is not a chord with two distinct endpoints. When d > r, all points of the line are farther than r from O and there is no intersection. A negative value proposed for d is not a centre-to-line distance and is rejected before squaring it.

In particular, the line through a circle point T perpendicular to OT has distance r from O, so it meets the circle only at T. Conversely, a line through T that is not perpendicular to OT has its perpendicular foot F different from T. Its distance from O is less than OT = r, by the right-triangle relationship when F ≠ O and directly when F = O. It is therefore a secant with a second circle point. Thus the unique tangent at T is perpendicular to the radius OT.

For the explorer's chord frame, O = (0, 0) and the tested vertical line is x = d, with d required to be a nonnegative distance. Its foot is F = (d, 0). For a secant, the two endpoints are U = (d, h) and V = (d, −h), with exact h = √(r² − d²). At d = r the single point is F. A coordinate line x = −3 exists, but entering −3 as the nonnegative distance d fails this scene's specification; a signed line position and its distance from the centre are different quantities.

Two circle boundaries require both radius and centre-distance guards

Compare a radius-5 circle at (0, 0) with a radius-3 circle at (4, 0). They share (4, 3) and (4, −3): each point is 5 units from the first centre and 3 from the second. Predict whether there will be two contacts, one or none if their centres move farther apart while both radii stay fixed. The distances between the centres and the sum or difference of the radii will explain the answer.

Consider circles with centres O and Q, positive radii r and s, and centre separation d = OQ ≥ 0 in the same length unit. Reject a nonpositive radius or a negative proposed distance first. Next handle d = 0 before any division by d. Concentric circles with r ≠ s have no common boundary point, since one point cannot be at two different distances from the same centre. With r = s they coincide: every point of the circle is common, so the result is infinitely many points rather than two or a tangency.

For distinct centres, d > 0. A common point P gives lengths OP = r, QP = s and OQ = d. The distance inequalities require |r − s| ≤ d ≤ r + s. If d > r + s, the circles are separated externally and have no common point. If 0 < d < |r − s|, one circle lies strictly inside the other's disk without a common boundary point. At d = r + s the boundaries touch externally at one point between the centres. At d = |r − s| > 0 they touch internally at one point on the line of centres. The case |r − s| = d = 0 was already the concentric case and must not be called internal tangency.

To prove these conditions are sufficient and count the points, choose O = (0, 0), Q = (d, 0) with d > 0, and write a common point as P = (x, y). The derived Cartesian distance formula gives x² + y² = r² and (x − d)² + y² = s². Subtracting the equations gives x = (r² − s² + d²)/(2d). The remaining equation is y² = r² − x². Algebra factors this value as ((r + s)² − d²)(d² − (r − s)²)/(4d²). The positive-radius and distinct-centre guards precede every square, division and geometric verdict in this calculation.

Because r + s > |r − s| and d > 0, the factored squared height is positive exactly when |r − s| < d < r + s. Then two points (x, √(r² − x²)) and (x, −√(r² − x²)) satisfy both original distance equations. The height is zero at either admissible tangency boundary, giving one point. It is negative outside the interval, so no real point satisfies both equations. The horizontal coordinate x need not lie between 0 and d: intersections can project onto an extension of the centre line. Use the original equations and this guarded count, rather than forcing the foot into the segment OQ.

Equal circles with r = s = d = 3 have x = 3/2 and squared height 9 − 9/4 = 27/4. Their common points are (3/2, ±3√3/2). Each point forms an equilateral triangle with the two centres because all three sides have length 3. The two choices lie on opposite sides of the centre line. This is the same exact intersection principle used to construct an equilateral triangle on a supplied positive segment.

Two equal-radius circles construct a midpoint and a right angle

With endpoints A = (0, 0) and B = (4, 0), try compass radius 3. The two equal circles meet above and below AB; joining their contacts gives the vertical line through x = 2. Now predict what changes for radius 2 and for radius 1. The circles will explain why the same midpoint can exist even when a particular two-point construction fails.

An ideal unmarked straightedge draws the straight line through two distinct known points; it does not read a numerical length from markings. An ideal compass draws a circle with a known centre and a chosen positive radius, including a radius copied from a known segment. The construction consists of such exact operations and selection of their intersections. A pencil drawing approximates those operations. The mathematical claim needs a proof that the intersections exist and have the required property.

Given distinct endpoints A and B with AB = L > 0, choose a common circle radius ρ > L/2 and draw circles centred at A and B. For example, copying L as the radius always satisfies that condition. Their separation L lies strictly between |ρ − ρ| = 0 and ρ + ρ = 2ρ, so the circles have two distinct intersections P and Q on opposite sides of AB. Draw line PQ. Both P and Q have equal distances to A and B by their defining circle equations.

For the proof, let M denote the mathematical midpoint of AB, whose existence on a positive segment is independent of whether we have yet constructed it with tools. For either intersection P, triangles AMP and BMP have AM = BM, AP = BP and common MP. They are nondegenerate because the two-circle intersections lie off AB. SSS makes their adjacent angles at M equal, hence each is 90°. The same argument places Q on the unique perpendicular line through M. Therefore PQ is the perpendicular bisector of AB. Its intersection with AB constructs M, and its crossing with AB constructs a right angle.

The strict radius condition belongs to this two-intersection method. If ρ = L/2 > 0, the circles are externally tangent at M: there is only one intersection, so no line through distinct P and Q is determined. If 0 < ρ < L/2, the circles do not intersect. A nonpositive radius is rejected before squaring. The midpoint still exists when the chosen radius fails; it is this attempted construction that lacks the required two points.

For A = (0, 0) and B = (4, 0), valid radii satisfy ρ > 2. The two intersections are (2, ±√(ρ² − 4)). With ρ = 3 they are (2, ±√5), and their joining line is x = 2. With ρ = 5/2 they are (2, ±3/2). In either case the constructed midpoint is (2, 0). Exact root coordinates certify the geometry even if the screen plots their positions using rounded numerical values.

Equal distances justify equilateral triangles and angle bisection

To construct an equilateral triangle on distinct A and B, draw circles centred at A and B with common radius AB. The positive separation AB is less than twice that radius, so there are two intersections off AB. Choose either as C and join AC and BC. The first circle gives AC = AB and the second gives BC = AB, hence all three sides are equal. Its three equal angles sum to 180°, so each is 60°. The construction produces two mirror choices for C, each with the same certificate.

For an internal angle bisector, start with two rays from O forming a nonzero, nonstraight interior angle θ, where 0° < θ < 180°. Draw a positive-radius circle centred at O; let it cut the rays at A and B, so OA = OB = t > 0. The rays are distinct and not opposite, so 0 < AB < 2t. Draw equal circles centred at A and B with radius t. They meet at O and a second point P on the opposite side of AB from O. Draw the ray OP.

The companion diagram uses an exact 60° angle with O = (0, 0), A = (√3, 1) and B = (√3, −1). Both marks are 2 units from O, since 3 + 1 = 4. Their radius-2 circles meet at O and P = (2√3, 0), with midpoint M = (√3, 0) on ray OP. The two equilateral triangles OAB and PAB put P across AB from O. The diagram shows the internal ray; the following congruence argument explains why its two parts are 30°. A supplied 80° angle is handled by the same general proof, giving 40° halves.

The perpendicular-bisector argument places O and P on the line through the midpoint M of AB perpendicular to AB. Because P lies across AB from O, ray OP passes through M, which lies inside angle AOB. The ray is therefore the internal one, rather than the opposite extension. Triangles OAP and OBP have OA = OB, AP = BP and common OP; SSS gives ∠AOP = ∠POB. Since they partition θ, each is θ/2. For a supplied 80° angle, each half is exactly 40°; no protractor reading is needed to justify the equality.

The hypotheses prevent hidden degeneracy. At θ = 0° the rays coincide, A = B and the two equal circles coincide rather than defining a second intersection by this method. At θ = 180°, A and B are opposite circle points and their two radius-t circles touch only at O, so P is missing. A straight angle does have perpendicular rays that split it into two right angles, obtainable by a different construction, but this nonstraight-angle procedure cannot be used unchanged. A zero compass radius also fails before any circle is drawn.

A convex polygon's angle sum follows from a fan of triangles

Take a convex pentagon and draw two diagonals from one vertex to the two non-neighbouring vertices. The fan has three triangles, so its interior-angle total is 3 × 180° = 540°. In the diagram, each shaded triangle fills one part of the polygon. Predict how many triangles a hexagon would give before generalising to n sides. The vertex positions are approximate drawing coordinates; the triangle count and angle totals are exact.

Here a polygon has an integer number n ≥ 3 of distinct vertices joined consecutively by positive straight sides and closed by the final side. It is simple: boundary sides meet only at their shared endpoints when consecutive, with no crossings. We restrict the angle proof to convex polygons without straight corners, so each interior angle is less than 180° and the whole segment joining any two points of the region stays in the region. A crossing star or a collapsed list of collinear points does not meet these hypotheses.

Label the vertices in boundary order V₀, V₁, …, Vₙ₋₁. From V₀ draw diagonals to V₂ through Vₙ₋₂. Convexity puts these diagonals inside the polygon and gives a fan of n − 2 nondegenerate triangles with disjoint interiors, filling the region. For n = 3 the fan is just the original triangle and no diagonal is needed. At each polygon vertex, the fan's triangle angles partition the interior angle. Adding all triangle angle sums therefore counts each polygon interior angle exactly once, giving total (n − 2) × 180°.

To count exterior angles, travel once around the boundary in one consistent direction. At each vertex, use the turn from the incoming directed side to the outgoing directed side, taking the same orientation throughout. For this convex polygon each turn has magnitude 180° minus the adjacent interior angle. Their total is n × 180° − (n − 2) × 180° = 360°. Reversing the direction reverses the signs of the directed turns; their magnitudes still total 360°. Counting both exterior choices at each vertex or changing orientation halfway would be a different calculation.

A regular polygon in this lesson is simple and convex with every side equal and every interior angle equal. Thus each interior angle is ((n − 2) × 180°)/n = 180° − 360°/n, and each exterior turn has magnitude 360°/n. Division by n occurs only after checking that n is an integer at least 3. A proposed n = 2 does not specify a polygon here, and n = 5/2 cannot count its vertices. Equal side lengths alone do not imply regularity: a rhombus can have unequal adjacent angles.

Selected regular polygons have exact construction certificates

The two-circle construction already gives a regular triangle on any supplied positive side. To construct a square in a given positive-radius circle, draw a diameter AC through its known centre O. Construct the perpendicular bisector of AC, which passes through O, and let its intersections with the given circle be B and D. Read A, B, C and D consecutively around the circle and join them. The four central angles are right angles, and every radius has length r.

Each of triangles AOB, BOC, COD and DOA has two sides r with an included 90° angle, so SAS makes them congruent. Each is isosceles with base angles 45°, since its remaining two angles sum to 90°. Adjacent base angles at each polygon vertex give an interior angle 45° + 45° = 90°, while congruence gives four equal polygon sides. Thus the polygon is a square. The Pythagorean theorem gives each side √(r² + r²) = r√2. This proof uses right triangles and congruence, without assuming an angle-at-the-circumference theorem.

For a regular hexagon in a positive-radius circle, choose an initial circle point A and keep the compass opening equal to the radius r. A circle centred at A with radius r cuts the original circle in two points; choose one as the next vertex B. At each new vertex, select the next intersection in the same direction around the original circle, rather than returning to the preceding vertex. The centre and every consecutive pair of selected vertices form an equilateral triangle, since its two radii and joining chord all have length r. Therefore each consecutive central angle is 60°.

Six such positive turns fill 360° and return to A, with six distinct vertices before closure. Joining consecutive vertices gives a simple convex hexagon with six equal sides r. At every vertex, the two neighbouring equilateral triangles contribute base angles 60° + 60° = 120°, so all interior angles are equal as well. Its perimeter is 6r. The return after six steps follows from the exact 60° angle certificate, rather than from almost meeting a starting pencil mark.

An exact formula for a regular polygon's angles does not automatically supply a straightedge-and-compass construction of it. We have justified the triangle, square and hexagon constructions above. Not every regular polygon can be constructed exactly with only those ideal tools; this lesson does not claim or prove a general constructibility criterion. Dividing a full turn into n equal parts with a protractor can produce a useful approximate drawing, but a decimal angle reading is not an exact compass-and-straightedge certificate.

The proof belongs to the specification, not the picture's pixels

A geometric construction has three questions: which operations are permitted, whether their required distinct intersections exist, and why the resulting figure has the desired property. The circle definitions supply equal distances; SSS and SAS transfer angle and length information; positivity and intersection inequalities establish existence. A sketch helps organise these reasons but does not replace them. Looking perpendicular or measuring two nearly equal sides cannot establish an exact right angle or equality.

Exact fractions and symbolic roots can certify a plotted construction. With AB = 4 and ρ = 3, an intersection has coordinates (2, √5). Its squared distances to A and B are both 4 + 5 = 9, exactly the required radius square. The screen must approximate √5 to place a point, but the proof uses the exact expression. Similarly, the circle-pair count comes from exact signs and equalities, rather than a numerical tolerance that turns near tangency into tangency.

Keep the distinction between exact mathematical decimals and rounded measurements. With exact radii 3 and 2 and exact centre separation 5, the circles are externally tangent because d = r + s. Replacing the exact separation by 5.001 gives d > r + s and no common point. If 5 is a rounded measured separation instead, it does not by itself prove exact tangency. A physical drawing or installation also needs a stated measurement model and precision.

Use equal length scales on the two axes when a Cartesian diagram represents distances. Stretching only one display axis can make a circle look elliptical or a right angle look incorrect even though the exact coordinate specification is unchanged. Polygon vertices plotted from approximate angle coordinates illustrate the selected regular polygon; their exact angle values come from the proved formulas, and the plot does not claim a construction for every possible n.

Predict a change, then compare the exact result

Keep radius 5 in the line scene and compare distances 3, 0, 5 and 6. Predict the number of common points and chord length before checking the result. A diameter is a separate case, tangency has only one point, and a missing line gives no real height. Try a rejected input and explain which positive-radius or unsigned-distance condition it fails.

In the circle-pair scene, compare the intersecting, tangent, separated and equal-centre choices. Use the exact radius bounds to explain each count. In the polygon scene, compare a pentagon with a hexagon and an invalid side count; the angle formulas require a valid integer count. In the bisector scene, compare radii 3, 2 and 1 for AB = 4 and ask which choice supplies the two distinct contacts needed for their joining line. Keep the exact table values separate from the approximate drawing positions.

Worked examples

Try it yourself

1. A circle has radius 7/2 cm. State its diameter, exact circumference and closed-disk area. Is its centre on the circle or in the disk? Does a chord with distinct endpoints have the same definition as the arc joining those endpoints? What are the circumference and disk-area factors when the radius is tripled?

   Show the answer and reasoning

   The diameter is 7 cm, circumference 7π cm and disk area 49π/4 cm². The centre lies in the disk's interior but not on the circle. A chord is a straight segment and an arc is a curved boundary portion. Tripling the radius multiplies circumference by 3 and disk area by 9.

   The radius is positive. The diameter is 2 × 7/2 = 7, C = 2π × 7/2 = 7π, and A = π × (7/2)² = 49π/4. The centre's distance from itself is 0, less than 7/2, so it belongs to the interior rather than the boundary. Two distinct endpoints define one straight chord and two possible boundary arcs; a chosen arc must be identified. Positive scaling k = 3 sends 2πr to 3(2πr) and πr² to 9(πr²), with length and square units preserved.

2. For a circle with radius r = 3/2 and a line at perpendicular distance d = 1 from its centre, find the half-chord and full chord exactly. Classify the line when d is instead 0, 3/2 or 2. Does proposed distance d = −1 give another permitted chord case?

   Show the answer and reasoning

   At d = 1 the half-chord is √5/2 and the chord length is √5 units. At d = 0 the line gives a diameter chord of length 3; at d = 3/2 it is tangent with one point and no genuine chord; at d = 2 it misses the circle. The negative distance is rejected.

   With r > 0 and 0 < 1 < 3/2, h² = 9/4 − 1 = 5/4 and h = √5/2 > 0. Doubling h gives √5 for the full chord. At d = 0, the endpoints are opposite radius points and length 2r = 3 follows directly, without a nondegenerate right triangle at the centre. The equality d = r leaves zero displacement and one intersection, while d > r leaves none. A negative input is not a perpendicular distance; it is rejected before squaring, even though a separately specified signed coordinate line can lie on either side of the centre.

3. Classify the common boundary points for the circle specifications (r, s, d) = (5, 5, 6), (5, 3, 4), (3, 3, 3), (3, 2, 5), (5, 2, 3), (3, 2, 6), (5, 2, 2), (5, 2, 0), (3, 3, 0) and (0, 2, 3). Which case makes each common point an equilateral-triangle vertex with the two centres?

   Show the answer and reasoning

   The first three have two common points. The next two have one point each: external tangency for (3, 2, 5) and internal tangency for (5, 2, 3). The next three have none: external separation, strict containment and concentric unequal circles, respectively. The (3, 3, 0) circles coincide with infinitely many common points. The final zero-radius specification is rejected. The (3, 3, 3) case gives an equilateral triangle at either intersection.

   For distinct positive-radius centres, compare d with |r − s| and r + s: the first three satisfy 0 < 6 < 10, 2 < 4 < 8 and 0 < 3 < 6. The tangencies have 5 = 3 + 2 and 3 = |5 − 2| > 0. Next, 6 > 3 + 2 gives separation and 0 < 2 < |5 − 2| gives containment without boundary contact. At d = 0, unequal radii cannot hold at one common point, while equal radii define exactly the same circle; handle these before dividing by d. A radius 0 fails the definition before squaring. With r = s = d = 3, the two radii to an intersection and the centre segment are all length 3, so either point supplies an equilateral triangle.

4. Fix A = (0, 0) and B = (4, 0). For common compass radii 3, 2, 1, 0, −3 and 5/2, say whether the two-circle procedure constructs the perpendicular bisector. Give exact intersection coordinates for every successful choice, and explain why a single midpoint contact does not determine the required line.

   Show the answer and reasoning

   Only radii 3 and 5/2 succeed. They give (2, ±√5) and (2, ±3/2), respectively, and both determine x = 2. Radius 2 gives only (2, 0), radius 1 gives no points, and radii 0 and −3 are invalid. One point alone lies on many lines, so the contact point does not determine the bisector by this procedure.

   The segment has positive length 4. Two equal-radius circles with separation 4 have two intersections exactly when ρ > 2. For successful choices, subtraction gives x = 2 and y² = ρ² − 4, giving 5 for ρ = 3 and 9/4 for ρ = 5/2. Both points are equidistant from A and B, and the SSS midpoint proof gives their joining perpendicular line. At ρ = 2 the squared height is zero and the two written signs refer to the same point, so they cannot be used as distinct straightedge points. At positive ρ = 1 the circles are externally disjoint. Nonpositive radii are rejected before computing squares or intersections.

5. Two rays from O form an exact 80° interior angle. A positive-radius circle centred at O marks A and B at equal distances t along the rays. Equal radius-t circles centred at A and B meet at O and a second point P across AB. Explain why ray OP is the internal angle bisector and state each half-angle. Why does this same procedure fail for coincident rays and for opposite rays?

   Show the answer and reasoning

   Ray OP passes through the midpoint of AB inside the angle, and SSS in triangles OAP and OBP makes the two parts equal. Each part is 40°. Coincident rays give coincident auxiliary circles; opposite rays give just one contact point O, so neither case provides the distinct second intersection required by this procedure.

   Because 0° < 80° < 180°, the marked points satisfy 0 < AB < 2t. The equal circles therefore have two points, with P distinct from O. Both lie on the perpendicular bisector of AB; choosing P across AB makes ray OP pass through its midpoint, hence inside the given angle. The equal sides OA = OB, AP = BP and common OP give SSS congruence, so ∠AOP = ∠POB and their sum 80° gives 40° each. At 0° the marks A and B coincide. At 180° their separation is 2t and the circles are externally tangent at O. A straight angle can instead be bisected using a constructed perpendicular, but the two-intersection angle method has failed its stated hypotheses.

6. For regular convex polygons with n = 5, 8 and 12, find the interior-angle sum, each interior angle and each exterior turn. Explain the n − 2 triangle count and the total 360° exterior turn. Are proposed side counts 2 and 5/2 permitted, or can their angle quotients be used as polygon answers?

   Show the answer and reasoning

   For n = 5 the values are 540°, 108° and 72°; for n = 8 they are 1080°, 135° and 45°; for n = 12 they are 1800°, 150° and 30°. A convex fan has n − 2 triangles, and one consistent exterior turn per vertex totals 360°. Neither 2 nor 5/2 is a permitted side count, so neither receives a polygon angle verdict.

   From one vertex, the n − 3 internal diagonals partition the convex polygon into n − 2 nondegenerate triangles. Their angle sums partition all polygon interior angles, giving (n − 2)180°. Equal regular angles divide that sum by valid integer n. Each exterior turn is the supplement, 360°/n; summing one consistently oriented turn at every vertex gives n180° − (n − 2)180° = 360°. Substituting 5, 8 and 12 gives the stated values. A polygon here must have an integer count at least 3, with distinct vertices and a simple convex boundary. An algebraic quotient for a rejected input does not establish such a figure.

7. A circle has exact radius 2 cm. Compare the square formed by two perpendicular diameters with the hexagon formed by stepping that radius around the circle in a consistent direction. Give each polygon's side length, perimeter and interior angle, and justify the constructions. Does a nearly closed compass drawing or a plotted regular polygon prove every regular n is straightedge-and-compass constructible?

   Show the answer and reasoning

   The square has side 2√2 cm, perimeter 8√2 cm and interior angle 90°. The regular hexagon has side 2 cm, perimeter 12 cm and interior angle 120°. Congruent isosceles right triangles certify the square, and six equilateral central triangles certify the hexagon. Approximate closure or a plot does not prove exactness or general constructibility.

   The square's four central triangles have two radii of length 2 and an included 90° angle. SAS gives equal chords, the Pythagorean theorem gives side √(4 + 4) = 2√2, and two 45° base angles give each 90° interior angle. For the hexagon, each consecutive chord is a copied radius, so the centre and its endpoints form an equilateral triangle of side 2 with central angle 60°. Six consistent central steps exactly total 360°, and two 60° base angles give each 120° polygon angle. Summing equal side lengths gives the stated perimeters. Screen coordinates and pencil marks approximate these ideal operations; the exact angle and distance certificates justify them. The existence of angle formulas or an approximate regular-polygon plot supplies no blanket compass-and-straightedge construction claim.

What to carry forward

• A circle is the positive-radius boundary OP = r; its closed disk satisfies OP ≤ r. Radii and diameters name segments or their lengths, while a straight chord and a curved arc have different meanings.
• Use the established formulas C = 2πr = πD and A = πr² with consistent length and square units. Keep π exact; a positive enlargement k multiplies circumference by k and disk area by k².
• SSS proves the perpendicular-midpoint property for a chord not through the centre. A diameter is a separate case; a genuine chord at distance 0 ≤ d < r has length 2√(r² − d²).
• Check r > 0 and d ≥ 0 before classifying a line. Distance d < r gives two points, d = r gives one tangent point, and d > r gives none; a circle's tangent is perpendicular to the radius at contact.
• For two positive-radius circles, handle concentric centres before division. Distinct centres give two points for |r − s| < d < r + s, one at either admissible boundary, and none outside; equal concentric circles coincide.
• A two-circle perpendicular-bisector construction needs a positive segment and common radius greater than half its length. Congruence also certifies an equilateral triangle and the internal bisector of a nonzero, nonstraight angle.
• A simple convex integer-n polygon has interior-angle sum (n − 2)180° and consistently counted exterior turns totalling 360°. Regular angles divide those totals by valid n; exact formulas and plotted figures do not imply universal straightedge-and-compass constructibility.