Bridge-building

Bridge-building — polyline minimum where the middle segment is a "bridge" of fixed length and fixed direction. Use translation to splice the two variable side-pieces of the polyline into a single segment.

Universal form

$\min\big(\, AE + EF + FB \,\big), \quad |EF| = d \text{ (fixed)},\; \vec{EF} \text{ (fixed direction)}$

Since $EF$ has constant length $d$, this is equivalent to

$\min\big(\, AE + FB \,\big).$

Classical model

• Moving point $E$: on a line $\ell_1$
• Moving point $F$: on a line $\ell_2$ ($\ell_1 \parallel \ell_2$)
• Fixed points $A, B$: separated by the two riverbanks
• Key constraint: $EF$ perpendicular to both banks (more generally: $\vec{EF}$ has a given direction and $|EF|$ a given length)

Intuition: a river (with parallel banks $\ell_1, \ell_2$) separates $A$ from $B$. Build a bridge $EF$ perpendicular to the banks; where do you put the bridge so that the total path from $A$ to $B$ is shortest?

Conclusion

Translate $A$ by the vector $\vec{EF}$ (length $d$, fixed direction) to get $A'$. Then

$\min\big(\, AE + FB \,\big) \;=\; |A'B|, \qquad \min\big(\, AE + EF + FB \,\big) \;=\; |A'B| + d,$

with equality when $A', F, B$ are collinear. At equality, $E$ is recovered by translating back: $\vec{FE} = -\vec{EF}$.

The answer does not depend on where the banks are ($\ell_1, \ell_2$'s positions). It depends only on $A, B$ and the bridge's fixed direction and length.

Proof

Step 1 · Translate. Translate $A$ by the vector $\vec{EF}$ (length $d$) to get $A'$. The quadrilateral $AEFA'$ satisfies

$\vec{AA'} = \vec{EF},$

so $AA' \parallel EF$ and $|AA'| = |EF|$ ⇒ it is a parallelogram (parallelogram tests). By parallelogram properties:

$AE \;=\; A'F.$

Step 2 · Triangle inequality. The polyline $A'\to F\to B$ has length

$A'F + FB \;\geq\; |A'B|,$

directly by [[triangle-inequality]], with equality when $A', F, B$ are collinear ($F$ on segment $A'B$).

Step 3 · Chain them together.

$AE + FB \;=\; A'F + FB \;\geq\; |A'B|.$

Add the constant $EF = d$ to get $AE + EF + FB \geq |A'B| + d$. $\qquad\blacksquare$

When to use / mnemonic

• A polyline minimization where the middle segment is fixed in length and direction (not "pick two points on some line and minimize the middle segment") → bridge-building
• The problem mentions "build a bridge perpendicular to the river bank" or a similar narrative trope → bridge-building
• Signature: the variables are the two side-piece endpoints ($E, F$), but $\vec{EF}$ is a fixed vector

Versus "General drinks his horse"

Model	Middle	Tool
General Drinks His Horse	A point (the same moving point $P$ appears in both side-pieces)	Reflection (axial symmetry)
Bridge-building	A fixed-length fixed-direction bridge $EF$	Translation

Both use "rigid transformation + triangle inequality", but the transformation differs: reflection alters "polyline direction"; translation alters "endpoint positions".

Pitfalls

• The bridge direction must be fixed: if the bridge can tilt (direction changes with $E$), this model does not apply — the problem degenerates into a harder optimization needing calculus or variational methods.
• Translation direction: translate $A$ by $\vec{EF}$ (from $E$ to $F$), not the reverse; flipping the direction gives the symmetric solution (usually not between the banks).
• Result is independent of bank distance: often disguised — the problem may give "river width = 5" as the source of the constant $d$, but after the move the bank positions drop out.
• $E, F$ must be on different banks: if $E, F$ both slide on the same line (not on two parallel banks), the problem degenerates into "General drinks his horse" or a plain two-point distance.

Applications

To be added.

Related

• Translation — translation preserves distance and direction
• Parallelogram properties / Parallelogram tests — basis for $AE = A'F$
• [[triangle-inequality]] — closes the polyline
• General Drinking Horse (axial-symmetry shortest path) — the "reflection" sibling in the same family