Fermat point

Fermat point — the point inside a triangle that minimizes the sum of distances to the three vertices.

Universal form

$\min_P\,\big(\, PA + PB + PC \,\big)$

Classical model

• Moving point $P$: anywhere in the plane (interior and boundary of $\triangle ABC$ included)
• Fixed points $A, B, C$: the three vertices of $\triangle ABC$
• Key constraint: all three weights are $1$ (use Weighted Fermat Point (rotation + scaling lemma) if weighted); when some interior angle is $\geq 120°$, the optimal point degenerates to that vertex

Conclusion

Condition	Optimal $P$	Angles at $P$
All interior angles $< 120°$	Fermat point (strictly inside the triangle)	$\angle APB = \angle BPC = \angle CPA = 120°$
Some angle $\geq 120°$	The obtuse vertex	—

The 60° rotation construction (the standard recipe)

On any side of $\triangle ABC$ (say $BC$), build an equilateral $\triangle BCA'$ outside $\triangle ABC$. Then:

$\min(PA + PB + PC) = |AA'|$

Proof sketch: rotate $\triangle BPC$ about $B$ clockwise by $60°$: $P \to P'$, $C \to A'$. Then:

• $\triangle BPP'$ is equilateral ($BP = BP'$ with included angle $60°$) ⇒ $PP' = PB$
• Rotation preserves length ⇒ $P'A' = PC$
• Hence $PA + PB + PC = PA + PP' + P'A'$

The right-hand side is the length of the polyline from $A$ to $A'$. By [[triangle-inequality]]:

$PA + PP' + P'A' \geq |AA'|$

with equality when $A, P, P', A'$ are collinear — at which point $P$ is the Fermat point.

Applications

• [[0002-fermat-point]] — the standard isosceles-acute-triangle problem