Unit V: Axial Reflection & Chapter III — The Theory of Revolution

Julius Petersen — Methods and Theories (Authorized Translation by Sophus Haagensen, 1879 · pp. 60–68)

Chapter II (cont.) & Chapter III — Axial Reflection & Rotation Around a Point

Revolution Around an Axis (Chapter II.B cont., pp. 60–64, Problems 336–348)

[p. 60] Revolution around an axis (reflection across a straight line of symmetry) is a special case of Replacing, yet so fruitful that it warrants independent treatment. By revolving one half of a figure around an axis of symmetry, symmetric intersections become direct intersections of known curves, and broken equal-angle paths straighten into single segments:

  • Problem 336 (p. 61): To a given line to draw a perpendicular, of which two given curves shall cut off equal parts from the line. Revolve one curve around the given line as axis; its intersections with the other curve determine the sought perpendiculars immediately.

  • Problem 338 — Hero’s Reflection Problem (p. 61, Hero of Alexandria): In a given line to find a point X, from which lines drawn to two given points A and B (situated on the same side of the line) shall form equal angles with the given line. - Construction & Remark: Revolve A around the given line to \(A_1\); the straight line \(B X A_1\) determines \(X\). As Petersen notes (p. 61), an elastic body or a ray of light striking a plane mirror reflects at an angle equal to the angle of incidence; since \(AX + XB = A_1 X + XB = A_1 B\) while any other point \(X'\) yields a broken path \(A_1 X' + X' B > A_1 B\), the distance travelled by light is a minimum.

Petersen Problem 338 — Hero's Mirror Reflection & Shortest Optical Path [ M1 M2 ]: y = 0 [ A A1 ]: x - 2 = 0 [ A1 B ]: -16*x + 12*y + 104 = 0 [ A X ]: 6*x + 9*y/2 - 39 = 0 M1 = (0, 0) M2 = (16, 0) A = (2, 6) B = (14, 10) A1 = (2, -6) F = (2, 0) X = (13/2, 0) A B A1 F X

Figure 5.1a — Petersen’s Problem 338 (p. 61): Axial Reflection & Hero’s Shortest Optical Path. Revolving A across mirror axis M1M2 into A1 straightens the broken path into line A1B, determining X \(= (13/2, 0)\) where segments AX and XB make equal angles with the mirror and achieve minimum path length \(20\).

  • Problem 339 — Polygonal Billiards (pp. 61–62): Two balls M and N lie on an n-sided polygonal billiard table; roll M against side AB so that after striking all n sides successively it hits N. Unfold \(M \mapsto M_1 \mapsto M_2 \mapsto \dots \mapsto M_n\) by reflecting successively across each side in order, reducing the \(n\)-cushion path to a single straight line!

  • Problem 340 — Minimum-Perimeter Inscribed Polygon (p. 62): In a given n-gon to inscribe another n-gon of minimum perimeter. Consecutive sides must form equal angles with each host side; unfolding the entire polygon through \(n\) axial reflections straightens the perimeter into one segment, proving why odd-\(n\) polygons admit a unique determinate solution (for an acute triangle, its orthic triangle) whereas even-\(n\) polygons are either impossible or indeterminate (porismatic)!

Chapter III — The Theory of Revolution (pp. 65–68, §§ 1–5 & Problems 349–362)

[p. 65] 1. From a given point \(O\), let straight lines be drawn to all points \(a\) of a given curve \(k\); revolve every such ray around \(O\) through a given constant angle \(v\), increasing (or diminishing) its length simultaneously in a given constant ratio \(f\), so that the extremity \(A\) satisfies

\[\angle aOA = v \qquad\text{and}\qquad \frac{OA}{Oa} = f = \text{const.}\]

The locus of \(A\) is a new curve \(K\) directly similar to \(k\) (obtained by composing pure rotation through \(v\) around \(O\) with homothetic multiplication by \(f\) around \(O\)). Petersen names (p. 65):

For every pair of similarly situated points \(a \in k\) and \(A \in K\), the triangle \(\triangle aOA\) preserves a constant shape; thus \(K\) is generated by one vertex \(A\) of a triangle of invariant form rotating and scaling around its fixed vertex \(O\) while the third vertex \(a\) traces \(k\)!

[p. 66] 3. When \((O, v, f)\) are given, any system of straight lines and circles can be revolved by straight edge and compasses: a straight line is revolved by revolving its foot from \(O\) (turning the line by angle \(v\)), and a circle is revolved by revolving its centre and radius.

4. Fundamental Triangle-Placement Theorem (p. 66):

To place a triangle, similar to a given triangle, with one vertex at a given point O and the two other vertices lying on two given curves \(k_1\) and \(k_2\).

Make \(O\) the pivot point, take the angle of the triangle at \(O\) for \(v\), and take the ratio of the two sides enclosing angle \(O\) for \(f\). Revolve \(k_1\) by \((O, v, f)\) into \(K_1\); each intersection of \(K_1\) with \(k_2\) furnishes the second vertex, and revolving back gives the vertex on \(k_1\)!

Petersen Problem 349 — Equilateral Triangle Across Three Parallel Lines ( P R ): x**2 + y**2 - 28/3 = 0 ( R P ): (x - sqrt(3)/3)**2 + (y - 3)**2 - 28/3 = 0 [ P A1 ]: y = 0 [ B1 B2 ]: y - 1 = 0 [ C1 C2 ]: y - 3 = 0 [ B1p B2p ]: -3*sqrt(3)*x/2 + 3*y/2 - 3 = 0 [ P Q ]: -x + 5*sqrt(3)*y/3 = 0 [ P R ]: -3*x + sqrt(3)*y/3 = 0 P = (0, 0) A1 = (4, 0) B1 = (0, 1) B2 = (3, 1) C1 = (0, 3) C2 = (4, 3) R = (sqrt(3)/3, 3) Q = (5*sqrt(3)/3, 1) P R Q

Figure 5.1 — Petersen’s Problem 349 (p. 67): Equilateral Triangle Across Three Parallel Lines. Choosing P on PA1 (\(L_1: y = 0\)) as pivot point with \(v = 60^\circ, f = 1\), revolving middle parallel B1B2 (\(L_2: y = 1\)) by \(+60^\circ\) yields locus line B1pB2p, which meets upper parallel C1C2 (\(L_3: y = 3\)) at R \(= (\sqrt{3}/3, 3)\). Revolving back via equilateral arcs PR and RP places Q \(= (5\sqrt{3}/3, 1)\) on \(L_2\), forming equilateral PQR of exact quadrance \(28/3\).

Canonical Rotation & Spiral-Similarity Problems (pp. 67–68)

  • Problem 349 (p. 67): To place an equilateral triangle with its vertices in three parallel lines. Solution: Take any point P in one of the parallels as one vertex; this point will be the pivot point; \(v = 60^\circ, f = 1\).

  • Problem 350 (p. 67): To place an equilateral triangle with its vertices in three concentric circles. Take any point on one circle as pivot point with \(v = 60^\circ, f = 1\) (revolving the second circle by \(60^\circ\) around that point until it meets the third circle!).

  • Problem 352 (p. 67): In a given triangle to inscribe another triangle similar to a given one and having one vertex at a given point P on one side. Pivot at \(P\) with \((v, f)\) taken from the prototype triangle!

  • § 5 — Existence & Construction of the Pivot Point (p. 68): Any two directly similar figures in the plane admit a unique pivot point \(O\) (unless congruent and parallel-translated): if corresponding lines \(AB\) and \(ab\) meet at \(P\), then \(O\) is the second intersection of circles \(PAa\) and \(PBb\) (Miquel’s pivot theorem).