Napoleon's Theorem: Three Centers That Always Form an Equilateral Triangle

What Napoleon's theorem actually claims

Take any triangle ABC in the Euclidean plane. On each of its three sides build an equilateral triangle that points outward — away from the interior of ABC. Let P, Q and R be the centers of those three new triangles (the one on AB, the one on BC, the one on CA). Napoleon's theorem is the claim that PQR is equilateral.

For an equilateral triangle the word “center” needs no qualification: centroid, circumcenter, incenter and orthocenter all coincide at the same point, two-thirds of the way from each vertex along its median. So P is simply the average of the three corners of its triangle, and there is no hidden choice to argue about.

The hypotheses are remarkably light, which is exactly what makes the result feel like sleight of hand. ABC may be scalene, isosceles, right-angled or violently obtuse. It may be enormous or microscopic. It may even be degenerate: if A, B and C all lie on a single straight line the construction still makes sense, and PQR is still equilateral. Two conditions do real work, and both are easy to miss. First, the plane must be Euclidean — the proofs use the law of cosines and the fact that rotation is a rigid motion with a clean algebraic description. Second, all three equilateral triangles must be erected the same way: all outward, or all inward. Mix the directions and the theorem is flatly false. For one randomly chosen triangle whose all-outward construction gives an equilateral triangle of side 2.10953, flipping just one of the three to the inside produces side lengths 2.10953, 1.83505 and 1.66896 — not close to equal, and not close to a rounding error.

The figure PQR is called the outer Napoleon triangle. Its center coincides exactly with the centroid of ABC, so however you deform the original triangle the two figures stay concentric.

Building the figure — and what dragging it does and does not prove

The construction compresses to one line. List the vertices counterclockwise. For the side running from A to B, the apex of the outward equilateral triangle is

apex = A + Rot(−60°)(B − A),

because for a counterclockwise polygon the outside of a directed edge lies to its right. Average that apex with the two endpoints and you have the center. Using one fixed rotation rather than an orientation test matters in practice: it keeps the construction continuous as the triangle is squashed through the flat, degenerate case, where “outward” briefly loses its meaning.

Two measurements can be read straight off the picture and they are the entire engine of the proof in the next section. The center of the equilateral triangle erected on a side of length c lies at distance c⁄√3 from both endpoints of that side — that is the circumradius of an equilateral triangle of side c. And the segment from an endpoint to that center makes a 30° angle with the side, since it bisects the 60° corner of the equilateral triangle.

An honest caution about the animation. Dragging the original triangle through a hundred shapes and watching the three inner lengths stay equal is a demonstration, not a proof. No finite pile of examples establishes a statement quantified over all triangles, and the displayed numbers are rounded, so in principle they could hide a discrepancy in the fifth decimal. What the dragging honestly does is two things. It rules out the suspicion that the diagram was drawn to flatter one lucky configuration, and it exhibits the mechanism: the three inner lengths do not merely happen to agree at one instant, they move together, in lockstep, through every intermediate shape. The two proofs below are what actually close the gap.

A proof you can do with the law of cosines

Write a = |BC|, b = |CA|, c = |AB|, let S be the area of ABC, and let B stand for the angle at vertex B as well as the vertex itself. The point B is shared by two erected triangles: the one on AB, with center P, and the one on BC, with center Q. So the inner side PQ can be computed inside the single triangle PBQ.

From the two measurements above, |BP| = c⁄√3 and |BQ| = a⁄√3, and because both centers lie outside ABC, the angle at B in triangle PBQ is the original angle plus a 30° wedge on each side:

∠PBQ = 30° + B + 30° = B + 60°.

Now the law of cosines in PBQ:

|PQ|² = ^c²⁄₃ + ^a²⁄₃ − 2 · ^c⁄_√3 · ^a⁄_√3 · cos(B + 60°).

Expand cos(B + 60°) = ½ cos B − (√3⁄2) sin B, then substitute two standard facts about ABC itself: ac cos B = (a² + c² − b²)⁄2, which is just the law of cosines rearranged, and ac sin B = 2S, which is twice the area. Everything collapses to

|PQ|² = ^{a² + b² + c²}⁄₆ + ^2S⁄_√3.

Look hard at what survived. The right-hand side is completely symmetric in a, b and c; the letter B has disappeared. Run the identical computation at vertex C for |QR|² and at vertex A for |RP|² and both land on the same expression. Three equal sides — and the proof hands you the exact size of the Napoleon triangle for free, which no purely synthetic argument does.

One caveat is worth stating rather than hiding. The step ∠PBQ = B + 60° is read off the diagram, and when B exceeds 120° the rays BP and BQ have swung past one another, so the true angle at B is 300° − B. The arithmetic is unharmed, because cos(300° − B) = cos(B + 60°), but a complete write-up owes you that extra line. The complex-number proof below never touches a diagram and so never incurs the case split.

The one-line proof with complex numbers

Identify the plane with ℂ and let ω = e^2πi⁄3 be a primitive cube root of unity, so that ω³ = 1 and 1 + ω + ω² = 0. The tool is a purely algebraic test for equilateral: three complex numbers z₁, z₂, z₃ are the vertices of an equilateral triangle, traversed in one particular orientation, exactly when

z₁ + ω z₂ + ω² z₃ = 0.

Rotation by −60° is multiplication by e^−iπ⁄3, which equals −ω. So with the vertices written as complex numbers a, b, c, the apex of the outward equilateral triangle on the side a → b is a − ω(b − a), and its center is the average of the three corners:

P = ¹⁄₃ [a + b + (a − ω(b − a))] = ¹⁄₃ [(2 + ω)a + (1 − ω)b].

The same formula, cycled, gives Q = ⅓[(2 + ω)b + (1 − ω)c] and R = ⅓[(2 + ω)c + (1 − ω)a]. Feed them into the test and collect terms:

3(P + ωQ + ω²R) = [(2+ω) + ω²(1−ω)] a + [(1−ω) + ω(2+ω)] b + [ω(1−ω) + ω²(2+ω)] c.

Every bracket vanishes. The first is 2 + ω + ω² − ω³ = 2 + ω + ω² − 1 = 1 + ω + ω² = 0. The second is 1 − ω + 2ω + ω² = 1 + ω + ω² = 0. The third is ω − ω² + 2ω² + ω³ = 1 + ω + ω² = 0. Hence P + ωQ + ω²R = 0 identically in a, b and c.

That word identically is the whole payoff. The computation never divides by anything, never assumes the three points are distinct, and never consults a picture. It therefore proves the theorem for every triple of complex numbers — collinear triples included, and even the totally collapsed case a = b = c. It also throws in a bonus: since (2 + ω) + (1 − ω) = 3, adding the three centers gives P + Q + R = a + b + c, so the Napoleon triangle and the original have exactly the same centroid.

A third proof, entirely synthetic, is worth carrying around. A 120° rotation about P carries A to B; one about Q carries B to C; one about R carries C to A. Compose all three, all turning the same way, and A comes back to A. The total turning is 360°, so the composition is a translation, and a translation with a fixed point is the identity. A composition of three 120° rotations is the identity only when the three centers form an equilateral triangle — which is the theorem, with no coordinates at all.

The inner triangle, an area identity, and Weitzenböck's inequality

Erect the three equilateral triangles inward instead and every argument above runs unchanged with −60° replaced by +60°. The three centers again form an equilateral triangle — the inner Napoleon triangle — with the same center as before, the centroid of ABC, but the opposite orientation. Its side obeys the mirror formula:

L_inner² = ^a²+b²+c²⁄₆ − ^2S⁄_√3    against    L_outer² = ^a²+b²+c²⁄₆ + ^2S⁄_√3.

Turn each into an area with the equilateral formula (√3⁄4)L² and subtract. The symmetric part cancels and the √3 factors cancel too, leaving one of the prettiest identities in triangle geometry:

area(outer Napoleon) − area(inner Napoleon) = area(ABC).

For a 3-4-5 right triangle that reads 6.60844 − 0.60844 = 6, which is the triangle's area on the nose. Two edge cases show the identity flexing. When ABC is equilateral of side s, the inner triangle collapses to a single point and the outer one is congruent to ABC — in fact it is ABC rotated 60° about its own centroid — so the difference is area(ABC) − 0. When ABC is degenerate, with area 0, the inner and outer triangles have the same side length (both √[(a²+b²+c²)⁄6]) but opposite turn, and the difference is 0.

The inner formula also hides a famous inequality. L_inner is an honest distance between two honest points, so L_inner² cannot be negative. That forces

a² + b² + c² ≥ 4√3 · S,

with equality precisely when the inner triangle shrinks to a point, that is, when ABC is equilateral. This is Weitzenböck's inequality, published by Roland Weitzenböck in Mathematische Zeitschrift in 1919 — and here it falls out as a by-product of Napoleon's construction rather than as a separate theorem. The slack is a measure of how far from equilateral you are: the 3-4-5 triangle gives 50 ≥ 41.569 (slack 8.431), a 2-3-4 triangle gives 29 ≥ 20.125 (slack 8.875), and a long thin isosceles triangle with sides 1, 1, 1.9 gives 5.61 ≥ 2.055 (slack 3.555).

Worked numbers, end to end

Take the 3-4-5 right triangle, with a = 3, b = 4, c = 5 and area S = 6. Then a² + b² + c² = 50, so the symmetric term is 50⁄6 = 8.33333, and the area term is 2 · 6⁄√3 = 6.92820.

L_outer² = 8.33333 + 6.92820 = 15.26154  ⇒  L_outer = 3.90660
L_inner² = 8.33333 − 6.92820 = 1.40513  ⇒  L_inner = 1.18538

Their areas are (√3⁄4)(15.26154) = 6.60844 and (√3⁄4)(1.40513) = 0.60844, differing by exactly 6. Notice how large the outer triangle is: with a side of 3.9066 it is nearly as long as the longest side of the triangle that generated it, and its area exceeds the original's. The Napoleon triangle is not a small detail tucked inside the figure; it is a comparably sized object sitting across it.

Now the degenerate case, which is the one most statements of the theorem quietly skip. Put three points on a line at spacings 2 and 3, so that a = 2, b = 3, c = 5 and S = 0. Then a² + b² + c² = 38, and

L_outer = L_inner = √(38⁄6) = 2.51661.

Three collinear points, no triangle to speak of, and the construction still returns a genuine equilateral triangle of side 2.51661 — two of them, in fact, the same size, sharing a center, rotated 60° from each other. This is precisely the case the complex-number proof covers without comment and the diagram-based proof has to be talked into.

A useful reality check while implementing any of this: compute all three inner side lengths independently rather than computing one and assuming. In double precision they agree to about 1 part in 10¹⁵, which is floating-point noise. If your implementation returns three visibly different numbers, you have almost certainly erected one of the three equilateral triangles on the wrong side.

Where the name came from, and how far the result stretches

The attribution is almost certainly wrong. The earliest recorded appearance of the statement is an 1825 question in The Ladies' Diary, the long-running English almanac with a serious mathematical section, credited there to W. Rutherford — who posed it rather than claimed it. Napoleon Bonaparte had died in 1821. The earliest known source that attaches his name to the result is the 17th edition of Aurelio Faifofer's Elementi di geometria, published in 1911, ninety years after his death. In Geometry Revisited (1967), H. S. M. Coxeter and S. L. Greitzer rated the odds of Napoleon knowing enough geometry for the feat as roughly the odds of his knowing enough English to compose the palindrome ABLE WAS I ERE I SAW ELBA.

What is genuinely true is that Napoleon cultivated mathematics and mathematicians. Lorenzo Mascheroni dedicated La geometria del compasso (1797) — the book proving that every straightedge-and-compass construction can be done with compasses alone — to him, and he was close to Gaspard Monge and Pierre-Simon Laplace. Patronage is not authorship.

Generalizations. Napoleon's theorem is the smallest case of the Petr–Douglas–Neumann theorem, found by Karel Petr in 1908 and rediscovered independently by Jesse Douglas in 1940 and Bernhard Neumann in 1941. Starting from an arbitrary n-gon, a sequence of n − 2 apex-constructions — erecting isosceles triangles with apex angles 2πk⁄n on the sides and keeping the apexes — ends on a regular n-gon, and the order of the angles does not matter. The quadrilateral cousin is Van Aubel's theorem (H. H. van Aubel, 1878): build squares outward on the four sides of any quadrilateral, and the two segments joining the centers of opposite squares are equal in length and perpendicular to each other.

Close relatives inside the same figure. The three circumcircles of the outward equilateral triangles — the Torricelli circles — always pass through one common point, the first isogonic center. The three segments joining each vertex of ABC to the far apex of the equilateral triangle on the opposite side are concurrent at that same point and are all exactly equal in length. If every angle of ABC is under 120°, that point is the Fermat point, the unique point minimizing the total distance to the three vertices, and the common segment length equals that minimum. If one angle reaches or exceeds 120°, the minimizing point is that obtuse vertex instead and the equality breaks: in a sample triangle with a 132.3° angle the three concurrent segments all measure 4.36220, while the true minimum total distance is 4.38616, attained at the obtuse vertex and equal to the sum of the two sides meeting there. Join each vertex instead to the center of the opposite erected triangle and you get three more concurrent lines, meeting at the first Napoleon point, catalogued as X(17) in Clark Kimberling's Encyclopedia of Triangle Centers; the inward construction gives X(18).

What breaks. This is a theorem of the Euclidean plane and the proofs lean on that fact hard: the complex-number argument needs rotation to be multiplication by a unit complex number, and the trigonometric one needs the Euclidean law of cosines and a 180° angle sum. In hyperbolic or spherical geometry the construction can still be carried out, but the three centers do not in general form an equilateral triangle. Analogues have been worked out in both settings, and they are interesting, but each requires restating the construction; none reproduces the unconditional Euclidean identity. Nor is the theorem affine: a shear preserves collinearity and ratios but not angles, so it does not preserve equilateral triangles, and the result does not survive one.

Five very different starting triangles, all evaluated from L² = (a²+b²+c²)/6 ± 2S/√3. In every row the outer Napoleon area minus the inner Napoleon area comes out exactly equal to S — for the 3-4-5 triangle, 6.60844 − 0.60844 = 6.

Original triangle (a, b, c)	Area S	Outer Napoleon side	Inner Napoleon side
Equilateral (1, 1, 1)	0.43301	1.00000 — congruent to the original, turned 60°	0 — collapses to a single point
Right (3, 4, 5)	6	3.90660	1.18538
Scalene (2, 3, 4)	2.90474	2.86137	1.21624
Thin isosceles (1, 1, 1.9)	0.29664	1.13028	0.76972
Degenerate (2, 3, 5), collinear	0	2.51661	2.51661 — same size, opposite turn