The point nearest in total to three corners

A centre is three weights described the classical centres of a triangle — centroid, incentre, circumcentre, orthocentre — as weighted averages of its corners, each with its own weights and its own reason to exist. None of them answers what is perhaps the most practical question a triangle can be asked. Three towns stand at the corners; where should a single depot go so that the total length of the three roads to it is as small as possible?

Pierre de Fermat posed the question to Evangelista Torricelli around 1640, and Torricelli answered it. The answer is a point the classical centres miss, characterised by angles rather than by weights: from it, each side of the triangle is seen at exactly 120°. This essay finds it by construction, checks it by minimising the total directly, proves it by a rotation that lays the three roads end to end, and follows the 120° rule out of the triangle into soap films and the shortest networks that join any number of points.

The construction, and a surprise in it #

Build an equilateral triangle outward on each side of the triangle. Join each corner to the outer apex of the equilateral triangle on the opposite side. The three joining lines meet at a single point, the Fermat point $F$, and at $F$ the three segments to the corners make angles of exactly 120° with one another. The figure computes the meeting point from two of the lines, checks that the third passes through it, measures the three angles, and then — independently — minimises the total distance by Weiszfeld’s iteration, which repeatedly moves a guess to the average of the corners weighted by the reciprocals of their distances. The two answers agree to within $10^{-7}$.

The construction holds a second fact that is easy to miss. The three dashed lines are the same length, and that common length is the least total distance itself, $FA + FB + FC$. In the figure’s triangle it is $7.709$. A sum of three distances, minimised over every point of the plane, equals the length of one straight segment that can be drawn with ruler and compass. The proof of that, below, is also the proof that $F$ is the minimum.

The equilateral triangles on the sides are the same ones whose centres form an equilateral triangle in Napoleon’s theorem. The two results share their construction and are close relatives: the lines from corners to opposite apexes, which cross at the Fermat point, are perpendicular to the sides of Napoleon’s triangle, and that is no coincidence — both facts come from rotating the plane by 60° about the corners.

The landscape of total distance #

Before the proof, it is worth seeing why there is exactly one answer.

The distance from a point to a fixed corner is a cone: its level curves are circles, and it rises at the same rate in every direction. A sum of three cones is a convex function, curving upward everywhere, and a convex function has a single lowest point with no false valleys to trap a search. That is why Weiszfeld’s iteration, or any sensible downhill search, finds $F$ from any start. The level curves round off near $F$ and are pointed only at the corners, where one cone’s tip makes the landscape kink.

The centroid is close to $F$ in this triangle and not the same: its total is $7.790$ against $7.709$. The centroid minimises a different quantity, the sum of squared distances, which averaging rewards; the Fermat point minimises the sum of the distances themselves, and the two agree only for an equilateral triangle.

Three pulls in balance #

The 120° condition has a physical reading that explains it at once. At any point, the rate at which the distance to a corner grows, as the point moves, is greatest in the direction away from that corner, and that rate is exactly one. In the language of calculus the slope of each distance is a unit arrow pointing from the corner to the point. The total distance is lowest where the three slopes cancel — where the three unit arrows add to nothing.

Three arrows of equal length add to nothing exactly when they are 120° apart, like the spokes of a symmetric three-pointed star. So the minimum is at the point where the three directions to the corners make 120° angles — the point from which each side is seen at 120°. A mechanical model makes the same argument with real forces: drill holes at the three corners of a table, pass a string through each, tie the three strings together above the table, and hang equal weights beneath. The knot settles where the three equal pulls balance, which is where the potential energy — proportional to the total length of string above the table — is least. Pierre Varignon described exactly this device, and it finds the Fermat point without any computation.

A rotation lays the roads end to end #

The cleanest proof is Torricelli’s own idea, in a form refined later, and it explains why the least total is one straight line.

Take any point $P$. Rotate the triangle $ABP$ through 60° about $B$, in the direction that carries $A$ to $C'$, the outer apex of the equilateral triangle on $AB$; let $P$ go to $P'$. Rotation preserves lengths, so $C'P' = AP$. And $B$, $P$, $P'$ form an equilateral triangle, because $BP = BP'$ and the angle between them is 60°, so $PP' = PB$. The broken line $C'P'PC$ therefore has length

$$C'P' + P'P + PC = PA + PB + PC.$$

A broken line from $C'$ to $C$ is never shorter than the straight segment $C'C$, so every point has total distance at least $C'C$. The total equals $C'C$ exactly when $C'$, $P'$, $P$, $C$ lie on one line in that order, and a short angle chase turns that into the 120° condition: the straight line through $P'$ and $P$ makes 60° with $PP'$ at both ends, so the angles $\angle BPC$ and $\angle APB$ are each 120°. So the minimum is the Fermat point, the minimum value is the length $C'C$, and by the same argument with the rotations about $A$ and $C$, the other two construction lines have the same length. The figure checks both halves: for a point $P$ chosen arbitrarily the broken line measures $7.891$, its total distance; for $F$ it is straight and measures $7.709$.

When an angle is too wide #

The construction always produces a crossing point. It is not always the answer.

If one angle of the triangle is 120° or more, no point inside can see all three sides at 120°: a point near the wide corner already sees the opposite side at nearly that corner’s angle, and moving the point only makes things worse. The three pulls can no longer balance with all three arrows of unit length. What happens instead is that the minimum moves onto the corner, where one of the three distances is nought and its slope is no longer a unit arrow but anything of length up to one. At the corner the other two arrows, from the far corners, add to something of length at most one exactly when the angle between them is at least 120°, and then a third “arrow” of suitable length can balance them. So for a triangle with an angle of 120° or more, the depot goes at that corner, and the total is the sum of the two sides that meet there. The figure’s minimiser lands on the corner to within $10^{-6}$.

This is a sharp switch in a smooth problem. As one angle of a triangle widens through 120°, the Fermat point slides into that corner and stays there. The total distance changes smoothly; the location of the minimum stops moving.

Unequal traffic: the depot moves #

Real roads do not all carry the same traffic. If the depot ships $w_A$ loads to $A$, $w_B$ to $B$ and $w_C$ to $C$, the cost to minimise is $w_A\,PA + w_B\,PB + w_C\,PC$, and the balance argument carries over with arrows of lengths $w_A$, $w_B$, $w_C$ instead of three unit arrows. Three arrows of given lengths add to nothing only if they can form a triangle, and then the angles between them are fixed by the law of cosines; the depot sits where the corners are seen at those angles. Equal weights give the equilateral arrow triangle and the 120° rule. If one weight exceeds the sum of the other two, no arrow triangle exists and the depot goes straight to the heaviest corner, the analogue of the obtuse case.

Alfred Weber made this the foundation of industrial location theory in 1909, with the weights standing for tonnages of raw material and finished product, and the weighted problem is still called the Weber problem. Its solution behaves in a way that surprises people who expect the answer to be a compromise: past a threshold, a single heavy customer captures the depot entirely, however far away the others are. Three circles that touch and are not the largest found the same kind of surprise in another optimisation problem on a triangle, where the obvious answer was wrong, and the Fermat point is a case in which the obvious answer — the centroid, the balance point of the corners — is merely close.

Where it sits among the centres #

The Fermat point appears as the thirteenth entry, $X(13)$, in Clark Kimberling’s encyclopedia of triangle centres, which now lists tens of thousands, most of them defined by some construction and almost none by a property as simple as least total distance. It has a twin, $X(14)$, obtained by building the equilateral triangles inward instead of outward; the inward construction also gives three concurrent lines, but their meeting point minimises nothing.

The Fermat point is tied to other centres through the partnership that every point has a partner across the bisectors described: reflect the three lines from a point to the corners in the angle bisectors at those corners, and the reflected lines meet again, at the point’s isogonal conjugate. The isogonal conjugate of the Fermat point is the isodynamic point — the point whose distances to the corners are inversely proportional to the opposite sides, and whose pedal triangle is equilateral. So the 120° point and the equilateral pedal triangle are partners, and the two kinds of equilateral triangle in this subject — the ones on the sides and the ones inside — are tied together by the same reflection that ties the classical centres in pairs.

Many points: the shortest network #

The question generalises at once. Given several towns, what is the shortest network of roads joining them all, if junctions may be placed anywhere? The answer is called a Steiner tree, and its junctions obey the Fermat rule.

For the corners of a square, joining by the sides costs 3 and crossing the diagonals costs $2\sqrt2 \approx 2.828$. Adding two junctions, each a Fermat point for three of the points around it — two corners and the other junction — brings the total down to $1 + \sqrt3 \approx 2.732$. In a shortest network every junction has exactly three roads meeting at 120°, by the same balance of pulls that located the Fermat point, and no junction has four. That makes the problem a search over which corners attach to which junctions; for $n$ points the number of possible patterns grows faster than exponentially, and finding the shortest network is NP-hard. Each user pays for its own last link shared out the cost of the cheapest network without extra junctions, the spanning tree, which is easy to find; Steiner junctions make the network shorter and the problem much harder.

The same rule governs soap films. The least wall for equal rooms described Plateau’s laws: soap films meet three at a time along curves at 120°. A film is a surface minimising its area, and where three sheets meet, their surface tensions pull equally, exactly like the three strings of Varignon’s table. A soap film stretched between two glass plates with pegs at the corners of a square forms the drawn network, junctions and all — a computer for Steiner trees made of detergent.

How much the junctions save #

For the square, the Steiner network of length $1 + \sqrt3$ is 8.9% shorter than the best network without extra junctions — the spanning tree of three sides, length 3 — and 3.4% shorter than the two diagonals, which already use their crossing as a free junction but meet there at 90° rather than 120°. How much a Steiner network can save over the cheapest spanning tree, for any configuration of points in the plane, is measured by the Steiner ratio: the least possible value of (Steiner length) ÷ (spanning-tree length). Three points at the corners of an equilateral triangle achieve $\sqrt3/2 \approx 0.866$: the spanning tree uses two sides, length 2, and the Fermat point connects them with total $\sqrt3$. Edgar Gilbert and Henry Pollak conjectured in 1968 that no configuration does better.

The 120° rule also fixes the shape of a shortest network more tightly than it first appears. In a network where every given point is an end of exactly one road — a full Steiner tree — $n$ points need exactly $n - 2$ junctions and $2n - 3$ roads, as the square’s two junctions and five roads illustrate. The search is then over which pairs of points share a junction, and the number of such patterns for $n$ points is $1 \cdot 3 \cdot 5 \cdots (2n - 5)$: three for four points, fifteen for five, over two million for ten. For the square only two of the three patterns can be drawn with 120° junctions — the third would pair opposite corners and cross itself — and the two that can are mirror images of equal length. Each pattern that can be realised is solved by repeated Fermat-point constructions, which is the sense in which the Fermat point is the whole local theory of shortest networks.

What the figures check and what they assume #

Each figure computes the Fermat point twice — by the construction and by direct minimisation — and the agreement is a check on both, not a proof of the theorem, which is the rotation argument. The level curves are drawn from a sampled grid, and their exact shapes near the corners, where the landscape has a cone-shaped kink, are approximated by the grid’s resolution; the location of the minimum does not depend on that approximation, since it is computed separately.

The Steiner network for the square is drawn from its known shape and checked for its length and its 120° angles; the figure does not search the alternatives, and the claim that $1 + \sqrt3$ is the least possible length rests on the general theory of Steiner trees, under which the two junctions and their 120° angles are forced and the remaining choice is between two mirror-image layouts of equal length. For larger sets of points nothing here computes a Steiner tree, and the difficulty of doing so is the subject of the next section.

Still open: the Steiner ratio #

Gilbert and Pollak’s conjecture — that in the plane a Steiner network is never shorter than $\sqrt3/2$ times the shortest spanning tree, so that the equilateral triangle is the worst case — was announced as proved by Ding-Zhu Du and Frank Hwang in 1990, and the proof was widely accepted for two decades. In 2012 Alexander Ivanov and Alexei Tuzhilin pointed out a gap in it, and the conjecture is again regarded as open; the best fully established bounds are a little below $\sqrt3/2$.

Exact Steiner trees are another frontier. For points in the plane they can be found for a few thousand points by programs that exploit the geometry, notably the GeoSteiner code of David Warme, Pawel Winter and Martin Zachariasen, but the problem is NP-hard, and in three dimensions, where the junctions still meet at 120° but the possible patterns are far richer, even modest sets of points defeat exact methods. The Fermat point is the one case in which the shortest network is completely understood by a construction with ruler and compass.

A centre defined by angles #

The classical centres of a triangle are averages. The Fermat point is a balance: the place where three equal pulls cancel, which is a statement about directions rather than distances, and it is pinned down by the angles it sees the sides at rather than by the lengths of anything. That is why the construction uses equilateral triangles — they are what 60° rotations make — and why the least total turns out to be a single straight line, the image of a broken path under rotation.

It is also why the same point keeps reappearing outside geometry. Wherever three things pull equally and the configuration settles to a minimum — three strings over a table, three soap films along a line, three roads at a junction of the shortest possible network — the angles are 120°, and the Fermat point is where the triangle’s corners put them.