10.3 Euclidean Constructions with Compass and Straightedge

10.3 Euclidean Constructions with Compass and Straightedge

Axiomatic Principles of Euclidean Constructions

Classical Euclidean constructions represent the operational manifestation of synthetic geometry, grounded in the axiomatic framework established in Euclid's Elements (c. 300 BCE). A formal Euclidean construction allows only two idealized mechanical instruments:

1. An unmarked straightedge of arbitrary length, which can only perform two operations: draw a unique straight line through two distinct points (Euclid's Postulate 1) and extend a finite straight line continuously in a straight line (Postulate 2).
2. A compass, which can describe a circle with any given center and radius (Postulate 3).

Under strict Euclidean rules, an examinee is barred from using a graduated ruler to measure lengths, a protractor to measure angles, paper folding (origami axioms), or sliding marked edges (neusis constructions).

The Collapsible Compass vs. Rigid Compass Equivalence

Euclid's third postulate originally envisioned a "collapsible compass"—one that retains its radius setting while drawing a circle on paper, but collapses and loses its setting the instant it is lifted. Modern geometry sets use a "rigid compass" that maintains its opening when moved across the page. In Euclid Book I, Proposition 2, Euclid proved that given an arbitrary segment and a separate point, one can construct an identical segment originating at that point using only a collapsible compass. This Compass Equivalence Theorem establishes that a rigid compass confers zero additional geometric power over Euclid's collapsible compass: any construction executable with a fixed divider can be executed with a collapsible compass.

Algebraic Field Extensions and Classical Impossibilities

In the 19th century, Pierre Wantzel (1837) and Ferdinand von Lindemann (1882) translated Euclidean constructions into abstract algebra. A point $(x, y)$ in the Cartesian plane is constructible from a set of starting points if its coordinates lie in a field obtained by a finite sequence of quadratic field extensions over $\mathbb{Q}$. Consequently, the degree of the minimal polynomial of any constructible number over $\mathbb{Q}$ must be a power of two ($[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2^k$). This algebraic truth resolved the three famous problems of antiquity, proving each mathematically impossible with straightedge and compass:

1. Doubling the Cube (Delian Problem): Requires constructing $\sqrt[3]{2}$, whose minimal polynomial $x^3 - 2 = 0$ has degree 3 (not a power of 2).
2. Trisecting an Arbitrary Angle: Trisecting $60^\circ$ requires constructing $\cos(20^\circ)$. The triple-angle identity $\cos(60^\circ) = 4\cos^3(20^\circ) - 3\cos(20^\circ)$ simplifies to $8x^3 - 6x - 1 = 0$, an irreducible cubic polynomial of degree 3 over $\mathbb{Q}$.
3. Squaring the Circle: Requires constructing a segment of length $\sqrt{\pi}$. Lindemann proved $\pi$ is transcendental (not the root of any non-zero polynomial with rational coefficients), precluding constructibility.

---

Foundational Constructions: Copying and Bisecting Segments and Angles

1. Copying a Line Segment

• Given: Segment $\overline{AB}$ and ray with endpoint $P$.
• Procedure: Set compass tip at $A$ and adjust pencil tip to $B$. Without altering the compass opening, place tip at $P$ and strike an arc intersecting the ray at $Q$.
• Justification: Segment $\overline{PQ} \cong \overline{AB}$ by definition of a circle: all radii from center $P$ to points on the arc are congruent to radius $AB$.

2. Copying an Angle

• Given: Angle $\angle A$ and ray with endpoint $P$.
• Procedure: Center compass at $A$ and draw an arc of radius $r$ intersecting the angle rays at points $B$ and $C$. With identical radius $r$, center compass at $P$ and draw an arc intersecting the given ray at point $D$. Set compass opening to span chord distance $BC$. Centering at $D$, strike an arc of radius $BC$ intersecting the first arc at point $E$. Draw ray $PE$.
• Justification: Consider triangles $\triangle ABC$ and $\triangle PDE$. By construction, $AB = PD = r$, $AC = PE = r$, and $BC = DE = d$. By the Side-Side-Side (SSS) Congruence Criterion, $\triangle ABC \cong \triangle PDE$. By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $\angle BAC \cong \angle DPE$.

3. Bisecting a Line Segment (Perpendicular Bisector)

• Given: Segment $\overline{AB}$.
• Procedure: Open compass to a radius $R > \frac{1}{2}AB$. Centering at $A$, draw arcs above and below $\overline{AB}$. With the identical radius $R$, center compass at $B$ and draw intersecting arcs above and below, labeling intersections $P$ and $Q$. Draw line $PQ$.
• Justification: By construction, $PA = PB = R$ and $QA = QB = R$. According to the Perpendicular Bisector Theorem, a point is equidistant from the endpoints of a segment if and only if it lies on the perpendicular bisector of that segment. Because both $P$ and $Q$ are equidistant from $A$ and $B$, line $PQ$ is the unique perpendicular bisector of $\overline{AB}$, bisecting the segment at midpoint $M$ and forming right angles.

4. Bisecting an Angle

• Given: Angle with vertex $V$.
• Procedure: Center compass at $V$, draw an arc of radius $r$ intersecting the angle rays at $A$ and $B$. Centering at $A$ and $B$ with equal radius $R > \frac{1}{2}AB$, draw intersecting arcs in the angle interior meeting at point $P$. Draw ray $VP$.
• Justification: In triangles $\triangle VAP$ and $\triangle VBP$, $VA = VB = r$ (radii of circle $V$), $AP = BP = R$ (radii of congruent circles $A$ and $B$), and $\overline{VP} \cong \overline{VP}$ (reflexive property). By SSS congruence, $\triangle VAP \cong \triangle VBP$, yielding $\angle AVP \cong \angle BVP$ by CPCTC.

---

Perpendicular and Parallel Line Constructions

Perpendicular Through a Point ON the Line

• Algorithm: Given line $L$ and point $P \in L$. Center compass at $P$, strike arcs of radius $r$ on both sides of $P$ intersecting $L$ at $A$ and $B$. $P$ is now the midpoint of $\overline{AB}$. Set compass radius $R > r$. Draw intersecting arcs centered at $A$ and $B$ meeting at $Q$. Draw line $PQ$.
• Justification: Triangle $\triangle QAB$ is isosceles with $QA = QB = R$. Point $P$ is the midpoint of base $\overline{AB}$. In an isosceles triangle, the median to the base is also the altitude, guaranteeing $PQ \perp L$.

Perpendicular Through a Point NOT on the Line

• Algorithm: Given line $L$ and external point $P \notin L$. Center compass at $P$, draw an arc intersecting line $L$ at two distinct points $A$ and $B$. Centering at $A$ and $B$ with equal radius $R > \frac{1}{2}AB$, draw intersecting arcs on the opposite side of $L$ meeting at point $Q$. Draw line $PQ$.
• Justification: Quadrilateral $PAQB$ satisfies $PA = PB$ and $QA = QB$. By definition, a quadrilateral with two distinct pairs of adjacent congruent sides is a kite (or rhombus if $R = PA$). A fundamental property of kites is that the main diagonal connecting the vertices between equal sides ($PQ$) is the perpendicular bisector of the cross diagonal ($AB$). Because $AB$ lies on line $L$, $PQ \perp L$.

Parallel Line Through an External Point

• Algorithm: Given line $L$ and external point $P$. Draw an arbitrary transversal line through $P$ intersecting line $L$ at point $Q$. At vertex $P$, construct an angle congruent to corresponding angle $\angle PQR$ on the same side of the transversal line.
• Justification: The Corresponding Angles Converse Postulate states that if two lines are cut by a transversal such that corresponding angles are congruent, the lines are parallel.

---

Constructing Inscribed Regular Polygons: Triangle, Square, and Hexagon

Regular Hexagon Inscribed in a Circle

• Algorithm: Given circle $\mathcal{C}(O, r)$. Maintain the compass opening set to the circle's radius $r$. Select any starting point $A$ on the circumference. Centering at $A$, strike an arc on the circumference to locate $B$. Continuing sequentially around the circle with radius $r$, mark points $C, D, E, F$. Connect adjacent vertices $A-B-C-D-E-F-A$.
• Justification: Each chord has length equal to radius $r$. Triangle $\triangle OAB$ has three sides of length $r$, making it an equilateral triangle with central angle $60^\circ$. Six consecutive $60^\circ$ central angles complete $360^\circ$, producing an inscribed regular hexagon with side length $s = r$.

Equilateral Triangle Inscribed in a Circle

• Algorithm: Execute the regular hexagon construction to establish six equidistant vertices on the circle ($60^\circ$ apart). Connect alternating vertices (e.g., $A, C, E$) with straight segments, bypassing vertices $B, D, F$.
• Justification: Connecting alternating vertices subtends a central angle of $60^\circ + 60^\circ = 120^\circ$. The chords have congruent lengths $s = 2r\sin(60^\circ) = r\sqrt{3}$. Three congruent chords spanning $360^\circ$ form an inscribed equilateral triangle.

Square Inscribed in a Circle

• Algorithm: Draw a line through center $O$ to form a diameter $\overline{AB}$. Construct the perpendicular bisector of diameter $\overline{AB}$. The perpendicular bisector passes through $O$ and intersects the circle at points $C$ and $D$, forming a second diameter $\overline{CD} \perp \overline{AB}$. Connect adjacent endpoints $A-C-B-D-A$.
• Justification: Diameters $\overline{AB}$ and $\overline{CD}$ are congruent, perpendicular, and bisect each other at center $O$. A quadrilateral whose diagonals are congruent perpendicular bisectors of each other is identically a square. The side length is $s = r\sqrt{2}$, and each central angle is $90^\circ$.

---

Euclidean Construction Algorithms and Justifications Reference Table

---

Deductive Justification Exemplar: Synthetic Proof of the Angle Bisector Construction

To demonstrate the deductive rigor required by secondary mathematics standards, we present a formal two-column synthetic proof establishing the validity of the angle bisector construction:

• Given: Angle $\angle AVB$ with vertex $V$.
• Construction:
  1. Circle centered at $V$ with radius $r$ intersects ray $VA$ at $X$ and ray $VB$ at $Y$.
  2. Circles centered at $X$ and $Y$ with radius $R > \frac{1}{2}XY$ intersect in the interior of $\angle AVB$ at point $P$.
  3. Ray $VP$ is drawn.
• Prove: Ray $VP$ bisects $\angle AVB$ (i.e., $\angle XVP \cong \angle YVP$).