General Drinking Horse (axial-symmetry shortest path)

Use axial symmetry to turn "sum of distances from a moving point on a line to two fixed points" into a single straight-segment problem between two fixed points.

Axial-symmetry shortest path (General Drinks His Horse) — the standard model for a moving point on a line with an "equal-coefficient two-leg distance sum". Reflect $A$ across $\ell$ to get $A'$ , which straightens the polyline into the segment $|A'B|$ .

Universal form

$\min_{P \in \ell}\,\big(\, PA + PB \,\big)$

Classical model

Moving point $P$ : on a line $\ell$
Fixed points $A, B$ : two fixed points on the same side of $\ell$
Key constraint: both legs carry weight $1$ (use Hu-Bu-Gui (Different-Speed Minimum, sin-Angle Method) for non-unit coefficients); in the opposite-side case, just connect $AB$ and intersect with $\ell$ — no reflection needed (see "Variants / generalizations" below)

Moving point P (orange, sliding along ℓ) + two fixed points A, B on the same side of ℓ; minimize PA + PB

When to use

A moving point $P$ lies on a line (river / road / mirror), and you want to minimize $PA + PB$ (with $A, B$ two fixed points)
More generally: a sum of two distances tied to a single moving point, or any "polyline" wrapped around a set of fixed points
The problem often mentions "General drinks his horse at the river" (将军饮马), "light path", "mirror reflection", or "fetch water from a river then deliver to B"

The core move

Move one of the two points to the opposite side — reflect $A$ across $\ell$ . The polyline then straightens out into $|A'B|$ .

A and B on the same side of ℓ, plus A' the reflection of A across ℓ: the intersection of the segment A'B with ℓ is the optimal P

Construction

Given a line $\ell$ in the plane with two fixed points $A, B$ on the same side and a moving point $P \in \ell$ :

Reflect: take $A'$ , the reflection of $A$ across $\ell$ (so $\ell$ is the perpendicular bisector of $AA'$ ).
Swap distances: for any $P \in \ell$ , $PA = PA'$ , so $PA + PB = PA' + PB$ .
Straighten: by [[triangle-inequality]], $PA' + PB \geq |A'B|$ , with equality when $A', P, B$ are collinear.
Locate $P$ : $P$ is the intersection of segment $A'B$ with $\ell$ . The minimum equals $|A'B|$ .

Why it works

Axial symmetry is a distance-preserving isometry (a reflection): every point on $\ell$ has the same distance to $A$ as to $A'$ . So the unmanageable distance "from the moving point $P$ to $A$ " can be replaced, with no loss, by "from $P$ to $A'$ ". After the swap, both legs start at $P$ , so the triangle inequality (polyline $\geq$ straight segment) applies directly.

Worked examples

[[0001-shortest-path]] — the canonical "General drinks his horse" problem
Light leaves $A$ , reflects off a mirror $\ell$ , and reaches $B$ — find the shortest light path (Fermat's principle = General Drinking Horse)

Variants / generalizations

When A and B are on opposite sides of ℓ, simply connect AB and take its intersection with ℓ — no reflection needed

Opposite-side case: if $A, B$ lie on opposite sides of $\ell$ , just connect $AB$ and intersect with $\ell$ — no reflection needed.

Two lines, two reflections: unfold the polyline into a straight segment

Two lines + one moving point each: a moving point $P$ on $\ell_1$ and another $Q$ on $\ell_2$ ; minimize $AP + PQ + QB$ — reflect $A$ across $\ell_1$ , $B$ across $\ell_2$ , then connect with a straight line.
One line + several moving points: chain reflections to "unfold" the path into a single straight segment.
Weighted variant: when $w_1 \cdot PA + w_2 \cdot PB$ has unequal coefficients, reflection no longer applies — switch to Hu-Bu-Gui (Different-Speed Minimum, sin-Angle Method) / Weighted Fermat Point (rotation + scaling lemma).