Spiral similarity

technique

The composition of "rotation by θ\theta about a fixed point" with "central dilation by kk from the same point"; a similarity transformation that simultaneously absorbs one rotation angle and one scaling factor — the core tool for "weighted-distance + angle" minimization.

Spiral similarity — the composite of "rotation about a fixed point" and "central dilation about the same point". A Swiss-army tool for problems whose target is a weighted sum of distances.

Definition

A spiral similarity in the plane is determined by three parameters:

  • a centre MM (a fixed point)
  • a rotation angle θ\theta
  • a dilation ratio k>0k > 0

It sends any point PP to PP' with

MP=kMP,PMP=θ.MP' = k \cdot MP, \qquad \angle PMP' = \theta.

Equivalently, compose Rotation properties (rotate by θ\theta around MM) with Homothety (central similarity) (dilate from MM by factor kk). The two factors commute when they share their centre, so "rotate then dilate" equals "dilate then rotate".

Spiral similarity: M is the centre, and P \mapsto P' satisfies MP' = k\,MP and \angle PMP' = \theta

Key properties

  • Conformal: a spiral similarity is a similarity transformation — it preserves all angles and scales every distance by the same factor kk.
  • Unique determination: given two points P,PP, P' and two images Q,QQ, Q' (with PQ0PQ \ne 0), there is a unique spiral similarity sending PPP \mapsto P' and QQQ \mapsto Q'. The centre is fixed by "perpendicular bisectors of the two segment pairs + matched rotation arcs".
  • SAS similarity comes for free: if MP1Q1\triangle MP_1Q_1 and MP2Q2\triangle MP_2Q_2 satisfy MP2/MP1=MQ2/MQ1=kMP_2/MP_1 = MQ_2/MQ_1 = k and P1MP2=Q1MQ2=θ\angle P_1MP_2 = \angle Q_1MQ_2 = \theta, then they are SAS-similar (see SSS and SAS similarity tests), and the spiral similarity above sends one to the other.

When to use

When the problem displays both of the following:

  1. Weighted target with an angle: minimise w1d1+w2d2w_1 \cdot d_1 + w_2 \cdot d_2 or similar, with weights wiw_i NOT in the "lookup table" {1,2,3}\{1, \sqrt 2, \sqrt 3\};
  2. Two coupled moving points: two moving points slide on different lines / circles, but are locked together by some ratio V1F:V2E=1:kV_1 F : V_2 E = 1 : k.

In such a problem, rotating-and-scaling about a shared vertex glues two weighted segments into a single polyline, after which [[triangle-inequality]] closes it out.

Classical uses

Use 1 · Weighted Fermat point with arbitrary weights

Target wADA+wBDB+wCDCw_A \cdot DA + w_B \cdot DB + w_C \cdot DC minimised, with all three weights arbitrary (not in the 1/2/31/\sqrt 2/\sqrt 3 table). Spin around one of the weight endpoints (say BB) by angle θ\theta and scale by k=wC/wAk = w_C / w_A, sending the whole BPC\triangle BPC to BP2C2\triangle BP_2 C_2. Then

  • dilation ratio k=wC/wAk = w_C / w_A;
  • rotation angle θ\theta comes from the law of cosines: cosθ=1+k2(wB/wA)22k\cos\theta = \dfrac{1 + k^{2} - (w_B / w_A)^{2}}{2k}.

Minimum target =wAAC2= w_A \cdot |A C_2|.

See Weighted Fermat Point (rotation + scaling lemma) (special weights) for the natural generalisation.

Use 2 · Minimum of a segment between two moving points

Target wDF+EFw \cdot DF + EF, where E,FE, F are two moving points (sliding on adjacent sides of a triangle) coupled by V1F:V2E=1:kV_1 F : V_2 E = 1 : k.

Find a fixed pivot MM such that the spiral similarity sends EFE \mapsto F (rotation angle V1MV2\angle V_1 M V_2, dilation ratio 1/w1/w); then FM=EF/wFM = EF / w, and the target becomes w(DF+FM)wDMw(DF + FM) \ge w \cdot DM (equality when the three points are collinear).

An equilateral triangle with w=3w = \sqrt 3, k=2k = 2 makes V1V2M\triangle V_1 V_2 M degenerate to a 30609030{-}60{-}90 right triangle — a common configuration when locating MM by hand.

Pitfalls

  • Wrong centre / wrong weighted endpoint: the centre of the spiral similarity must be the point shared by the two moving points, or the common endpoint of the two weighted segments. Pick the wrong one and you do not get the SAS similarity you need.
  • Signed rotation angle: θ\theta is signed. The "clockwise vs counter-clockwise" choice in the problem decides which side of the figure the image P2/C2P_2 / C_2 lands on; the wrong choice still produces a similar figure but moves the equality configuration.
  • Direction of kk: k=wC/wAk = w_C / w_A or wA/wCw_A / w_C depending on which coefficient in the target you want to eliminate. Write down the weighted sum first, decide which coefficient to absorb, then pick kk.
  • k1k \ne 1 does not reduce to pure rotation: pure rotation (k=1k = 1) gives SAS congruence; spiral similarity with k1k \ne 1 only gives SAS similarity — you get "ratios of corresponding sides 1:k1 : k", not equal corresponding sides.

Applications

To be added.

Related