Hu-Bu-Gui (Different-Speed Minimum, sin-Angle Method)

Hu-Bu-Gui — the standard model for a "weighted sum of distances" with a moving point on a line. Through a fixed point $B$, draw an auxiliary line tilted by $\arcsin k$ to swap $k \cdot PB$ for a perpendicular segment, then close out via "polyline $\geq$ point-to-line distance".

Universal form

$\min_{P \in \ell}\,\big(\, PA + k \cdot PB \,\big), \quad 0 < k < 1$

Classical model

• Moving point $P$: on a line $\ell$
• Fixed point $A$: anywhere on one side of $\ell$
• Fixed point $B$: on $\ell$ (or with a fixed projection direction onto $\ell$)
• Key constraint: $k \in (0, 1)$ (if $k > 1$, rewrite as $\dfrac{1}{k}\cdot PA + PB$ and draw the auxiliary line through the $PA$ leg instead)

When to use

• Minimizing $PA + k \cdot PB$, where $k \in (0, 1)$ and $P$ is a moving point on some line $\ell$
• The problem often mentions "two legs at different speeds", "sand vs. paved road", or "light through two media" — in essence, minimizing total time as $\sum (\text{distance} / \text{speed})$
• The coefficient $k$ is the given "speed ratio" or "discount ratio" — the story-form version is called 胡不归 (Hu-Bu-Gui)

The core move

Set $\sin\alpha = k$ and draw an auxiliary line tilted by $\alpha$ through the fixed point — this replaces $k \cdot PB$ with a perpendicular segment, and the whole sum collapses into "distance from $A$ to that auxiliary line".

Construction

Let $\ell$ be the line on which the moving point $P$ lies, $A$ a fixed point on one side of $\ell$, and $B$ a fixed point on $\ell$ (or with a fixed projection direction onto $\ell$). We want $\min_{P \in \ell} (PA + k \cdot PB)$ with $k \in (0, 1)$:

1. Fix the tilt angle: take $\alpha = \arcsin k$ (i.e. $\sin\alpha = k$).

2. Draw the auxiliary line: through $B$, draw a line $m$ making angle $\alpha$ with $\ell$, on the same side as $A$.

   

3. Trade for a perpendicular segment: for any $P \in \ell$, drop a perpendicular from $P$ to $m$ with foot $H$. Then $PH = PB \sin\alpha = k \cdot PB$.

   

4. Fold into a polyline: $PA + k \cdot PB = PA + PH \geq d(A, m)$ (the distance from $A$ to the line $m$, i.e. the perpendicular segment from $A$ to $m$).

5. Equality: holds when $A, P, H$ are collinear and $AH \perp m$ — i.e. $P$ is the intersection of $\ell$ with the perpendicular from $A$ to $m$. The minimum equals $d(A, m)$.

Why it works

The leg with coefficient $k$ cannot be handled by the usual reflection trick (General Drinking Horse (axial-symmetry shortest path) only works for $k = 1$). But multiplying "distance from $P$ to $B$" by $\sin\alpha$ gives exactly "distance from $P$ to a line $m$ that passes through $B$ at tilt $\alpha$" — this spreads the weight $k$ into the geometry. After the swap, what remains is the everyday fact that a polyline is at least as long as a straight segment (distance from a point to a line).

Worked examples

• In $\triangle ABC$ with $\angle B = 30°$, $D$ moves on $BC$; minimize $AD + \tfrac{1}{2} BD$ (take $\sin\alpha = \tfrac{1}{2}$, so $\alpha = 30°$ — this happens to coincide with $\angle B$, and the auxiliary line is the line containing $AB$).
• Sand-and-road problem: from $A$, with sand speed $v_1$ and road speed $v_2$ ($v_1 < v_2$), the road is the line $\ell$ and the target $B$ lies on the road. Minimizing total time is equivalent to minimizing $PA + (v_1 / v_2) \cdot PB$.

Variants / generalizations

• The case $k > 1$: $\sin\alpha = k$ has no solution. Switch to $\dfrac{1}{k} \cdot PA + PB$ and run the same playbook (the coefficient $\dfrac{1}{k} < 1$ now applies to the $PA$ leg).

• Three weighted legs ($w_1 \cdot PA + w_2 \cdot PB + w_3 \cdot PC$): a single auxiliary line is not enough — switch to Weighted Fermat Point (rotation + scaling lemma) (rotation + scaling).

• Physical meaning: $\sin\alpha = v_1/v_2$ is exactly Snell's law of refraction; Hu-Bu-Gui is, in essence, Fermat's principle at a two-medium boundary turned into an $\arg\min$ construction.