Equal area on a sphere, without a rectangle

In the plane, two polygons of the same area can always be cut into each other with finitely many straight cuts. On a sphere the same statement is true, with great-circle arcs in place of straight lines and rotations of the sphere in place of rigid motions of the plane. The essay on hinged dissections recorded that fact in a sentence and remarked that the plane argument survives the change of geometry “with care”.

The care is most of the argument. The plane proof has three moves — turn a triangle into a rectangle, turn a rectangle into one of a chosen width, and stack rectangles of the same width — and on a sphere the second and third cannot be made, because the objects they use do not exist.

What replaces the rectangle is a quadrilateral with two right angles and two equal legs, and the whole spherical theorem can be run through it.

The plane proof, and what it leans on #

The plane argument is worth setting out, because each step leans on a fact about flat space that is easy to use without noticing.

A triangle becomes a rectangle. Cut across at half the height, parallel to the base, and turn the two top pieces down beside the bottom one. The result has four right angles.

A rectangle becomes one of any chosen width. Slide it along a parallel strip, cutting and re-gluing, which keeps the area and changes the proportions — a move that rests on similar triangles of different sizes and on the fact that a rectangle’s area is its width times its height.

Rectangles of the same width stack. Two of them placed end to end make a third, which is what lets every triangle in a polygon’s triangulation be combined into one final rectangle and then compared with another polygon’s.

Every one of those moves uses parallel lines, and two of them use rectangles or similar figures of different sizes. A sphere has no parallel great circles — any two meet twice — and, as the next figure shows, no rectangles.

A sphere has no rectangles #

Try to build one. Start at a corner, run two arcs out at right angles, turn through a right angle at the end of each, and see where the last two sides meet.

The fourth angle is always obtuse, and by exactly the amount the shape encloses. That is Girard’s theorem — a spherical polygon’s area is its angle sum minus what a flat polygon with the same number of corners would have — applied to a quadrilateral. A quadrilateral with four right angles would have area nought, so no rectangle of any positive size exists on a sphere.

The table makes the point by degrees. At a side of 10° the fourth angle is 91.73°, barely distinguishable from a right angle; at 30° it is 104.48°; at 50° it is 125.93°. A small patch of sphere is nearly flat, and a nearly rectangular quadrilateral exists there — but “nearly” is exactly what a proof cannot use.

The same fact kills the second plane move as well. On a sphere a triangle’s area is fixed by its angles, so two triangles with the same angles have the same area: there is no scaling, and similar triangles of different sizes do not exist. The strip argument had nothing to slide.

A triangle cut along its midline #

The replacement comes from a construction that works in any geometry where midpoints and perpendiculars make sense, and it is old — the quadrilateral it produces carries the name of Saccheri, who studied such figures in 1733 while trying to prove the parallel postulate.

Take a triangle $ABC$, and let $M$ and $N$ be the midpoints of $AB$ and $AC$. Draw the great-circle arc through $M$ and $N$ — the midline. From each of the three corners drop a perpendicular to it, meeting it at $D$ (from $A$), $E$ (from $B$) and $F$ (from $C$).

The two triangles at $A$ reappear at $B$ and $C$. Triangle $ADM$ has a right angle at $D$; triangle $BEM$ has a right angle at $E$; the two have equal hypotenuses $AM$ and $BM$, because $M$ is the midpoint, and the angles at $M$ are vertical angles, so they are equal. Right triangles agreeing in a hypotenuse and an acute angle are congruent, on a sphere as in the plane. The same argument makes $ADN$ congruent to $CFN$.

So cut the triangle along the midline and along the perpendicular from $A$, carry the piece $ADM$ over to $BEM$ and the piece $ADN$ over to $CFN$, and what remains is the quadrilateral $BCFE$. It has right angles at $E$ and $F$, and its legs $BE$ and $CF$ are both equal to $AD$. In the hero figure the legs are 25.2° each. The figure computes every one of those claims on the sphere itself: the midpoints as equal arc lengths, the right angles from the tangent directions at the feet, the congruences from matching sides, and the legs from their lengths.

And the area is the triangle’s, because the same pieces make both. The figure checks it twice, by routes that share nothing: once from the angles, by Girard’s excess, and once from the corners, by the solid-angle formula of Van Oosterom and Strackee, which never mentions an angle at all.

What the quadrilateral remembers #

The quadrilateral forgets almost everything about the triangle, and the one thing it keeps is the thing that matters.

Its summit angles — the angles at $B$ and $C$ — are equal to each other, and together they make up the triangle’s whole angle sum. In the hero figure the triangle’s angles add to 218.0°, and the quadrilateral’s summit angles are 109.0° each. The angle at $B$ in the quadrilateral is the triangle’s angle at $B$ plus the angle the moved piece brings with it, which is part of the triangle’s angle at $A$; the angle at $C$ picks up the rest of $A$.

That is the point of the construction. A quadrilateral with two right angles at its base and equal legs is determined, up to moving it about the sphere, by its summit and its summit angles. The summit angles encode the area, since by Girard’s theorem the area is the angle sum less a half turn. So two triangles of equal area produce quadrilaterals with equal summit angles, and the only thing left that can differ between them is the length of the summit.

An apex that slides along a circle #

One observation turns the construction into a tool: two triangles that produce the same quadrilateral can be cut into each other, by cutting each into that quadrilateral and running one of the cuttings backwards. The question is how to find triangles that share a quadrilateral, and the plane has a familiar answer. Keep a triangle’s base, slide its apex along a line parallel to the base, and the area does not change, because neither the base nor the height does. The sphere has no parallel lines and no heights of that kind, and the move survives anyway, in a form the midline supplies directly.

Keep $B$, $C$ and the midline. A half-turn of the sphere about any point $P$ of the midline — a rotation through 180° about the axis through $P$ — carries $B$ to a new point $A'$, and $P$ is then the midpoint of $A'B$. Because $B$ and $C$ are the same distance from the midline, the midpoint of $A'C$ lands on the midline as well. So the triangle $A'BC$ has the same midline as $ABC$, the same perpendiculars from $B$ and $C$, and the same quadrilateral $BCFE$. It has the same area, and no area had to be computed to know it.

The apexes do not run along a great circle. They run along the circle of points at the legs’ distance beyond the midline, which is a small circle, and it has a property that can be checked without drawing it: the point opposite $B$ is on it, and so is the point opposite $C$, both round the back of the sphere. That is Lexell’s theorem of 1781 — the apexes of all the spherical triangles with a given base and a given area lie on one small circle through the points opposite the base’s two ends — arrived at here as a by-product of cutting, with no trigonometry in it. In the plane the corresponding curve is the straight line parallel to the base, and the two opposite points have gone off to infinity along it.

What the slide buys is control over a side. The side $A'B$ is twice the arc from $B$ to $P$, and as $P$ moves along the midline away from $E$ that arc grows from the leg upward. In the figure the leg is 25.2°, so the side from the apex to $B$ can be set to any length from 50.4° up to the size at which the triangle stops fitting comfortably inside a hemisphere, and 52.0° and 84.0° are two such choices. The figure checks, for each apex, that both midpoints lie on the midline, that the apex is on the circle, that the side has the length asked for, and that the area agrees with the original’s by Girard’s excess and again from the corners.

One length for every triangle, and one triangle for every polygon #

The rest of the proof is two uses of the slide, and together they are the spherical counterpart of choosing a rectangle’s width and stacking rectangles.

Every triangle can be given a side of one common length. Fix a length $\ell$, longer than twice every leg that will occur. Slide each triangle’s apex until one of its sides is $\ell$ long, and then run the midline construction again with that side as the summit, using the midpoints of the other two sides. Two triangles of equal area treated this way produce quadrilaterals with the same summit, $\ell$, and the same summit angles, half their equal angle sums, and quadrilaterals that agree in both are congruent. So any two triangles of equal area can be cut into each other.

Two triangles sharing a side become one triangle. Place two triangles $ABC$ and $ACD$ along a common side $AC$, on opposite sides of it. Keep the base $AC$ of the second and slide its apex $D$ along its own circle until it reaches the great circle through $B$ and $A$, just beyond $A$ — which, for triangles small enough, it always does, because that great circle leaves $A$ heading across the second triangle’s midline. The second triangle keeps its area, and now the two together form a single triangle $BCD'$ with $A$ lying on its side $BD'$. Repeating the merge, a polygon cut into any number of small triangles becomes one triangle of the same total area.

Then the two polygons meet. Each is now a single triangle, the two triangles have equal areas because the polygons did, and by the first step they can be cut into each other. Every cut in the chain is along a great-circle arc and every move is a rotation of the sphere, so two polygons of equal area are cut into each other with finitely many pieces. The rectangles were never essential. They were the plane’s version of a quadrilateral with a fixed base and fixed base angles, and the sphere has its own — and the parallel strip was the plane’s version of a circle of apexes.

The same argument runs in the hyperbolic plane, with the excess replaced by a deficit: there a triangle’s area is a half turn less its angle sum, the Saccheri quadrilateral’s summit angles are acute rather than obtuse, the apexes slide along a curve at constant distance from the midline, and every congruence above holds as stated. The construction was made for a setting in which the parallel postulate is not assumed, and both of the geometries that result — or, strictly, the hyperbolic plane and a sphere restricted to small enough regions — are within its reach.

Where the argument needs its hypotheses #

Triangles must be small enough. On a sphere two points have a unique shortest arc between them only if they are not opposite, and a point has a unique nearest point on a great circle only if it is not the circle’s pole. The construction is sound for triangles inside a hemisphere, which is why the figures place their corners well within one, and a large polygon is first cut into small triangles.

The pieces are moved by rotations of the sphere, including reflections if they are allowed. The congruences above are rigid motions of the sphere, and the theorem is stated with them. Restricting the motions, as the plane’s translation-only version did, changes the question on the sphere too.

The theorem says a dissection exists and not how many pieces it takes. The construction above is explicit, and running it on two given polygons produces a definite dissection, but nobody would call its piece count small — the situation the plane’s piece counts already described.

And the drawing is a projection. The sphere is drawn as a disc seen from far away, which preserves nothing about angles, so the right angles at $D$, $E$ and $F$ do not look like right angles and the quadrilateral’s congruent pieces do not look congruent. The figures compute every claim on the sphere and only then draw it; the drawing is evidence of the construction’s shape, not of its measurements.

Saccheri’s quadrilateral, used for something else #

Girolamo Saccheri published Euclid Freed of Every Flaw in 1733. His aim was to prove the parallel postulate by contradiction, and his method was to take exactly this quadrilateral — two equal legs perpendicular to a base — and ask whether its summit angles are right, obtuse or acute. He showed the obtuse case contradicts the other postulates, which is correct in the geometry he was working in, and believed he had refuted the acute case too, which he had not: he had developed a large part of hyperbolic geometry without accepting it.

Johann Heinrich Lambert studied the quadrilateral with three right angles in 1766, and noticed that in the acute case the area of a triangle would be proportional to its angle deficit, much as a spherical triangle’s is to its excess — an observation that points straight at the non-Euclidean geometries a lifetime before they were accepted.

The quadrilaterals were built to prove that only one geometry is possible, and they turned into the tool that shows the dissection theorem holds in all three. Once the question was no longer whether the other geometries exist but what is true in them, the figures designed as instruments of refutation became instruments of construction.

What a sphere drawn flat cannot show #

Every figure here is a sphere seen from a distance and flattened onto a page. Lengths near the rim are foreshortened and angles are distorted everywhere except at the centre of the view, so the congruent pieces are drawn at visibly different sizes and the right angles are drawn at visibly different openings. The figures’ statements are about the sphere, where every such claim was checked, and not about the drawing.

Nor do the figures show the whole theorem. They show one triangle becoming one quadrilateral, twice, a quadrilateral that cannot close, and an apex sliding along its circle. The step that merges two triangles into one is described and not drawn, and a reader has it on the strength of the slide rather than of a picture of it.

And the curvature itself is invisible. The sphere looks like a disc, and the one quantity that makes the spherical case different from the plane — that the angles of a closed figure add to more than a flat figure’s — has to be read off numbers in the margin, because a projection to a flat page cannot display it.

Still open: curved space in three dimensions #

In three dimensions the flat case is settled: volume and the Dehn invariant decide everything. In three-dimensional spherical and hyperbolic space the Dehn invariant still exists and is still an obstruction, and whether it is a complete one — whether volume and the Dehn invariant together decide which polyhedra can be cut into which — is not known. The problem is tied to difficult questions about the structure of the groups of motions of those spaces, and it is one of the places where an elementary question about cutting things up leads directly to research mathematics. Whether the hinged version of the spherical theorem holds is open too.

A theorem whose proof fails and whose conclusion survives #

The habit is about what to do when a proof stops working in a new setting.

The plane proof of the dissection theorem fails on a sphere at almost every step, and the theorem is still true there. That combination is common and easy to misread in both directions: a proof that fails is not evidence that its conclusion fails, and a conclusion that survives is not evidence that the proof secretly does. When a proof breaks in a new setting, ask which of its objects were essential and which were the old setting’s convenient version of something more general. Here the rectangle was a convenience, the quadrilateral with a fixed base and fixed base angles was the essential object, and a rearrangement proof of Pythagoras, which does depend on squares, is an example of a flat-space argument whose conclusion does not survive at all.