Weighted Fermat Point (rotation + scaling lemma)

Weighted Fermat point — the Fermat-point generalization when each of the three distances carries its own weight. Spiral similarity stitches the three legs into one polyline; the triangle inequality closes it out.

Universal form

$\min_P\,\big(\, w_1 \cdot |PA| + w_2 \cdot |PB| + w_3 \cdot |PC| \,\big), \quad w_1, w_2, w_3 > 0$

Notation: $w_1$ attached to $A$, $w_2$ to $B$, $w_3$ to $C$ (matching the rotation construction below).

Classical model

• Moving point $P$: anywhere in the plane
• Fixed points $A, B, C$: three fixed points with positive weights $w_1, w_2, w_3$ respectively
• Key constraint: the "weight triangle" $(w_1, w_2, w_3)$ must satisfy the triangle inequality $|w_i - w_j| < w_k < w_i + w_j$ (so the rotation angle $\theta$ exists); otherwise the optimal $P$ falls at some vertex (boundary solution)
• Equal-weight special case: $w_1 = w_2 = w_3$ degenerates to Fermat point

When to use

• Minimizing $f(P) = w_1 \cdot PA + w_2 \cdot PB + w_3 \cdot PC$ (three legs, each with its own weight)
• The equal-weight special case $w_1 = w_2 = w_3$ degenerates to Fermat point (the classical Fermat-Torricelli point)
• Physics / engineering setting: three cities with different demands need a shared transfer hub; the "general delivers water" problem upgraded with speed coefficients

The core move

Apply a spiral similarity (rotation + scaling) at the vertex with weight $w_3$ — that is, at $C$. The three legs splice into a single polyline, and the shortest polyline is a straight segment — which gives the minimum.

The rotation angle $\theta$ equals the angle opposite $w_3$ in the "weight triangle" with side lengths $(w_1, w_2, w_3)$.

Construction

Let $A, B, C$ be the three fixed points with corresponding weights $w_1$ (on $A$), $w_2$ (on $B$), $w_3$ (on $C$). First factor $w_1$ out of $f(P)$ and normalize $w_1 = 1$, setting $k = w_2 / w_1 = w_2$ (so that below the polyline length equals $f(P)$ directly, $|PA|$ no longer carries a coefficient, and $|PB|$, $|PC|$ are absorbed cleanly by the spiral similarity).

1. Compute the rotation angle: $\cos\theta = \dfrac{1 + k^2 - w_3^2}{2k}$ (equivalently, $1 + k^2 - 2k\cos\theta = w_3^2$, which is the law of cosines). The choice of $\theta$ is not arbitrary — see the "Why it works" section: to make $|P''P| = w_3 \cdot |PC|$, apply the law of cosines to $\triangle PCP''$ and solve.

   

2. Apply the spiral similarity (centered at $C$, easier to see as two steps):

   • 2a Rotate by $\theta$: $B \to B'$, $P \to P'$. Properties: $|CB'| = |CB|$, $\angle BCB' = \theta$, $|B'P'| = |BP|$ (rotations preserve lengths).
   • 2b Scale by $k$: $B' \to B''$, $P' \to P''$. Properties: $|CB''| = k \cdot |CB|$, $|B''P''| = k \cdot |BP|$ (uniform scaling).

   Together, $B''$ and $P''$ are the spiral-similarity images (with $|CB''| = k \cdot |CB|$, $\angle BCB'' = \theta$).

   

3. The three legs splice exactly into the polyline $A \to P \to P'' \to B''$:

   • $|AP| = w_1 \cdot |PA|$ (with $w_1 = 1$ after normalization, this leg is just $|PA|$).
   • $|PP''| = w_3 \cdot |PC|$ (law of cosines in $\triangle PCP''$: $|PP''| = |CP|\sqrt{1 + k^2 - 2k\cos\theta} = w_3 \cdot |CP|$).
   • $|P''B''| = k \cdot |PB| = w_2 \cdot |PB|$ (scaling factor $k = w_2$).

   So the polyline length $= |AP| + |PP''| + |P''B''| = f(P)$ exactly.

4. Straighten: the shortest polyline equals $|AB''|$, with equality when $A, P, P'', B''$ are collinear.

5. Locate $P$: $P$ is the intersection of line $AB''$ with the relevant arc / line (worked out from the collinearity condition).

Closed-form minimum

Since $\theta$ is determined directly by the weights $(w_1, w_2, w_3)$, the minimum $|AB''|$ can also be computed without finding $P$ — combine the law of cosines on $\triangle ACB''$ with the law of cosines in the weight triangle $2 w_1 w_2 \cos\theta = w_1^2 + w_2^2 - w_3^2$ and simplify, yielding a closed form involving only "weights + $\triangle ABC$":

$$\bigl(\min f\bigr)^2 = \tfrac{1}{2}\!\left[(w_2^2+w_3^2-w_1^2)\,a^2 + (w_1^2+w_3^2-w_2^2)\,b^2 + (w_1^2+w_2^2-w_3^2)\,c^2\right] + 8\,S_{\triangle}\,S_w.$$

Notation: $a = |BC|, b = |CA|, c = |AB|$ (standard opposite-side notation — $a$ opposite vertex $A$, etc.); $S_{\triangle}$ is the area of $\triangle ABC$; $S_w$ is the area of the weight triangle $(w_1, w_2, w_3)$ (Heron: $16\,S_w^2 = 2(w_1^2 w_2^2 + w_2^2 w_3^2 + w_3^2 w_1^2) - (w_1^4 + w_2^4 + w_3^4)$). Each of the three weight coefficients has the form "sum of squares of the other two weights, minus the square of this side's weight" — exactly $2 w_j w_k \cos(\text{the angle at that vertex})$ in the weight triangle.

Sanity check (Fermat point, $w_1 = w_2 = w_3 = 1$): all three weight coefficients are $1$, $S_w = \sqrt{3}/4$, and the formula gives

$$(\min f)^2 = \tfrac{a^2 + b^2 + c^2}{2} + 2\sqrt{3}\,S_{\triangle},$$

i.e. the classical closed form for the Fermat-Torricelli minimum.

Applicability: the weight triangle $(w_1, w_2, w_3)$ must satisfy the triangle inequality (so $\theta$ exists), and the optimal $P$ must lie inside $\triangle ABC$; otherwise the minimum is attained at a vertex (boundary solution), and you must compare the three vertex values of $f$ directly.

Why it works

Rotations preserve angles and scalings preserve ratios; together, a spiral similarity sends $\triangle BCP$ to $\triangle B''CP''$ so that $|B''P''| = k \cdot |BP|$, and it converts the leg "$P$ to $C$" into the leg "$P$ to $P''$" — whose length is exactly $w_3 \cdot |PC|$ by our choice of $\theta$ (via the law of cosines). The three legs with different weights are thereby translated into three consecutive segments of one polyline ($|AP|$ stays in place because $w_1 = 1$); the rest is just "polyline $\geq$ straight segment".

Recap of the law of cosines: in any $\triangle XYZ$ with side lengths $a = |YZ|, b = |XZ|, c = |XY|$, the angle $\angle Z$ opposite side $c$ satisfies

$$c^2 = a^2 + b^2 - 2ab\cos\angle Z.$$

Where the Step 1 formula comes from: plug the three sides of the weight triangle $(w_1, w_2, w_3)$ — after normalization $(1, k, w_3)$ with $k = w_2$ — into the law of cosines. The angle opposite $w_3$ is $\theta$, so

$$w_3^2 = 1 + k^2 - 2k\cos\theta \;\Longrightarrow\; \cos\theta = \frac{1 + k^2 - w_3^2}{2k}.$$

Step 3's $|P''P| = w_3 \cdot |CP|$ is the same idea: in $\triangle PCP''$, the sides $|CP|$ and $|CP''| = k \cdot |CP|$ emanate from $C$ with included angle $\angle PCP'' = \theta$. Law of cosines again:

$$|P''P|^2 = |CP|^2 + (k|CP|)^2 - 2 \cdot |CP| \cdot k|CP| \cdot \cos\theta = |CP|^2 (1 + k^2 - 2k\cos\theta) = |CP|^2 \cdot w_3^2.$$

So $|P''P| = w_3 \cdot |CP|$ — and that equality is exactly what $\theta$ was chosen to enforce.

Geometric intuition: "In the weight triangle $(w_1, w_2, w_3)$, the angle opposite $w_3$ is the rotation angle." $(1, 1, 1) \Rightarrow \theta = 60°$ (the classical Fermat point) $(1, 1, \sqrt{2}) \Rightarrow \theta = 90°$ (the isoceles right-angled weight triangle) $(3, 4, 5) \Rightarrow \theta = 90°$ (the right angle is opposite the hypotenuse)

Worked examples

• The classical Fermat point: $w_1 = w_2 = w_3 = 1$, rotate by $60°$ to build an equilateral triangle — the optimal $P$ sees all three sides at $120°$.
• Weights $(1, 1, \sqrt{2})$: minimize $|PA| + |PB| + \sqrt{2} \cdot |PC|$. After factoring out $w_1 = 1$, the spiral similarity sends $B$ to $B''$ with $\theta = 90°$, $k = 1$; the problem reduces to $|AB''|$.
• Weights $(3, 4, 5)$: rotate by $90°$ — the weight triangle is the $3{-}4{-}5$ right triangle.

Variants / generalizations

• Special-angle weight triangles: when $(w_1, w_2, w_3)$ corresponds to special angles such as $30{-}60{-}90$ or $45{-}45{-}90$, the rotation angle is immediate and the polyline construction is easy.
• Degenerate case: if the weight-triangle inequality $|w_1 - w_2| < w_3 < w_1 + w_2$ fails, the minimum is attained at one of the vertices (a "boundary solution"), and the spiral-similarity trick no longer applies.
• Dimension-reduction: for the single-leg weighted variant, see Hu-Bu-Gui (Different-Speed Minimum, sin-Angle Method); for the two-leg equal-weight variant, see General Drinking Horse (axial-symmetry shortest path).
• Naming note: this lemma is known in Chinese olympiad circles as the "weighted Fermat-point rotation method"; in the English literature it appears as the weighted rotation lemma or under names like Tellier's construction / spiral similarity. It is not a named theorem but the standard lemma for the weighted Fermat-Torricelli problem — re-derive it on the fly with the law of cosines rather than citing it.