The centre a compass finds in six circles

The centre a straightedge cannot find took a circle whose centre had been lost and proved that a straightedge alone cannot recover it. To a straightedge a circle has no centre: the maps that keep a circle and all straight lines are the motions of the hyperbolic plane, and among those the centre is a point like any other. The straightedge buys nothing proved the opposite fact about the other instrument: whatever a compass and straightedge can construct together, the compass can construct alone.

Put the two essays together and they promise something definite. The compass must be able to find a lost centre; the only question is how. This is the puzzle traditionally called Napoleon’s problem — in its compass-only form, find the centre of a given circle with the compass alone — and this essay answers it in the sharpest way available: the fewest circles that can do it, found by searching every construction, together with the reason the winning construction works.

The rules, and why two points are needed #

The starting position is a drawn circle $\Gamma$ and two marked points on it, $A$ and $B$. A move draws a circle centred at a known point and passing through another known point — a compass set by two points on the page, not the fixed opening of the compass that will not open and not one that carries a length from elsewhere. Every point where the new circle crosses something already drawn becomes known. The construction succeeds when the centre $O$ of $\Gamma$ is known.

Two given points are the least that can work. With $\Gamma$ and a single point on it nothing can be drawn at all: a compass needs a centre and a second point to fix its opening, and one point gives a circle of radius nought. The classical statement of the problem lets the solver “choose a point on the circle” and “open the compass to any radius”, which amounts to the same thing as two given points; marking both makes the problem exact, so that the fewest moves is a well-defined number. And it is a number that depends on where the two points are, which is the subject of the next essay. Here the arc between them is 37.3°, a size with no special relation to anything.

Where the problem comes from #

Mascheroni’s Geometria del compasso of 1797, the book in which the compass-alone theorem first reached a wide audience, was dedicated to Napoleon Bonaparte, whom Mascheroni had met and who took an interest in it. The name “Napoleon’s problem” has attached itself to two of the book’s exercises: dividing a circle with known centre into four equal arcs, which the straightedge buys nothing described, and finding the centre of a circle whose centre is unknown, which is the one solved here. Whether Napoleon ever posed either is doubtful; the story that he challenged the mathematicians of the Institut with them is told often and documented poorly.

The attribution aside, the problem earned its place. It is the natural test of the compass-alone theorem, because the centre is precisely the point the straightedge cannot reach, so a compass construction of it cannot be a straightedge construction translated move by move. Georg Mohr’s Euclides Danicus of 1672 had already proved the theorem and was forgotten until 1928; Mascheroni’s six-circle solution is the one that has come down in textbooks. The search here confirms that it is not merely a solution but a cheapest one, for points in general position.

Inversion as the compass’s straightedge #

The construction is one instance of a general method. Whenever a straightedge would draw the line through two points, a compass can instead work with that line’s image under an inversion, which is a circle through the inversion’s centre; whenever a straightedge construction would intersect two lines, the compass intersects the two image circles and inverts the result back. Because inverting a point needs only three circles, every line-and-circle construction can be simulated with circles alone, at a cost of a bounded number of circles per original move. That is one standard proof of the Mohr–Mascheroni theorem, and it is the same idea that one circle and a straightedge turned round, when a single drawn circle let a straightedge simulate every compass.

The general simulation is expensive: translated move by move, the five-move straightedge construction above would cost far more than six circles. What the search shows is that for this problem the translation is wasteful by a wide margin, and that a construction designed for the compass from the start — inverting the given circle itself rather than translating someone else’s lines — gets within one move of the full instrument set. That is typical of the compass-alone constructions that have been optimised by hand: the general theorem guarantees existence, and the good constructions exploit inversion directly.

The construction, step by step #

The six circles of the opening figure are, in order:

1. centred at $A$ through $B$, crossing $\Gamma$ again at a point $D$ — and, by symmetry, at $B$;
2. centred at $B$ through $A$;
3. centred at a point where the first two cross, through $A$;
4. and 5. centred at two points the third circle creates, each through $A$;
5. centred at a further crossing, through $A$, which passes through the centre.

Five of the six circles pass through $A$, and that is no accident. The construction is organised around $A$ the way an inversion is organised around its centre, and the next section shows that it is an inversion. The point $O$ appears as the second crossing of two of the circles through $A$ — the circles meet at $A$ itself and at exactly one other point, which is the centre.

Why it works: the circle becomes a line #

The classical version of the construction, which Lorenzo Mascheroni gave in 1797, makes the reason transparent, and the figure below draws it with every point labelled.

Inversion in a circle $K$ of centre $A$ and radius $r$ sends each point $P$ to the point $P^*$ on the ray from $A$ through $P$ with $AP \cdot AP^* = r^2$. The map that trades circles for lines studied what it does: circles through $A$ become straight lines, and points on $K$ stay put. The given circle $\Gamma$ passes through $A$, so it becomes a straight line, and since $\Gamma$ meets $K$ at $D$ and $E$, which stay put, the line is $DE$.

Now the key fact. Inversion carries the centre of a circle through $A$ to the mirror image of $A$ in the image line — the centre and $A$ are mirror-symmetric with respect to the circle, so their images are mirror-symmetric with respect to the line. The mirror image of $A$ in the line $DE$ is easy to construct with a compass: it is the second crossing of the circles centred $D$ and $E$ through $A$, which is the point $F$. So $O$ is the inverse of $F$ in $K$. And the inverse of a point can be constructed with the compass alone — the trick at the heart of Mohr and Mascheroni’s theorem: draw the circle centred $F$ through $A$, let it meet $K$ at $G$ and $H$, and the circles centred $G$ and $H$ through $A$ cross again at the inverse of $F$. The figure checks the inversion directly: $AO \cdot AF$ equals $r^2$ to within $10^{-9}$.

Six circles: $K$, the two through $A$ centred at $D$ and $E$, the one centred at $F$, and the two centred at $G$ and $H$. The construction the search found for the opening figure is a rearrangement of the same idea, with $B$ playing the part that the arbitrary radius plays in Mascheroni’s.

No five circles will do #

That six circles suffice is a construction; that none fewer suffice is a search. Starting from $\Gamma$, $A$ and $B$, the search tries every circle that can be drawn, then every circle that can be drawn after that, and so on, treating two constructions that draw the same set of circles in different orders as one.

The numbers grow quickly: the search examines three configurations at depth one, eleven at depth two, a hundred and twenty-one at three, about five thousand at four and nearly six hundred thousand at five. Every one of them fails to produce the centre. Six circles succeed, so six is the fewest. The search is the same kind of argument the fewest moves to draw it used for the midpoint and the square, and it is a proof in exactly the same sense: a finite set of possibilities, each checked.

The growth is the reason the search stops where it does. Each new circle can cross every circle already drawn in up to two points, so after $k$ circles the number of known points grows roughly like $k^2$, and the number of circles that can be drawn next — one for every ordered pair of known points — grows like the square of that. The counts bear it out: from depth four to depth five the number of configurations multiplies by about 117. One more depth at that rate would be some seventy million configurations, and the depth after that several billion, which is why six moves can be settled by search here and seven or eight is roughly the edge of what any search of this kind reaches. The straightedge, oddly, makes the tree smaller at depth five — about 272,000 configurations against 584,000 — because with lines available a construction exists at depth five, and the search stops the moment it meets one, so most of that depth is never examined.

One detail of the bookkeeping matters for the proof’s validity. Two points are treated as the same if they agree to seven decimal places, so a construction that produced a point near the centre without producing the centre itself would be counted as a failure only if it were more than $10^{-7}$ away. Constructions are exact, and a point of a construction either is the centre or is at a distance from it determined by the algebra of the configuration; nothing in the search came within that tolerance without being the centre.

The same six moves at every arc #

The construction does not depend on the particular arc.

When the search was run separately at 37.3°, 71.9°, 103.7° and 141.1°, it returned the same sequence of moves, written as “the circle centred at the third known point through the first” and so on. The circles grow and shrink with the arc, and the picture changes shape, but the pattern is fixed. That is what the inversion argument predicts: it uses nothing about the arc except that the circle centred $A$ through $B$ meets $\Gamma$, which it always does. For a generic arc, six is the price, and one construction pays it everywhere. Arcs of special sizes can be cheaper, because they give the compass coincidences to exploit; those are the subject of the next essay.

With a straightedge, one move fewer #

The comparison with the full instrument set is short.

The textbook method uses the fact that the perpendicular bisector of any chord passes through the centre: bisect two chords and intersect the bisectors. That costs six moves — two circles and a line for each bisector. The search finds five, by bisecting only one chord: after the two circles and the perpendicular bisector of $AB$, which passes through the centre without marking it, the search finds two lines, each joining one of the marked points to a point the bisector created, that cross exactly at the centre. No construction of four moves exists, which the search also checks.

So the full instrument set needs five moves, the compass alone six, and the straightedge alone cannot do it at all. The price of a construction measured the cost of giving up an instrument in Lemoine’s units, counting every movement of the hand; measured in whole moves, the price of giving up the straightedge for this problem is exactly one.

What the search does and does not establish #

The lower bound of six is established for the arc of 37.3° and for the other arcs on which the search was run, and the construction’s uniformity suggests it holds for every arc with no special relation to the circle. It is not proved here for every arc; the next essay shows that special arcs are cheaper, and an argument that every generic arc costs six would have to say precisely which arcs are special — something the search finds case by case and does not explain in general.

The rules also matter. The search allows only circles centred at a known point through another known point. A compass that can carry a length — draw a circle centred at one point with the radius given by two others — is a stronger instrument, and with it some constructions shorten; the classical “collapsing compass” of Euclid’s own rules is weaker. The answer six belongs to the rules stated, which are the natural ones for a compass that is set by two points and lifted.

Two shortcuts make the search feasible, and both are safe. The first is identifying a construction with the set of circles it has drawn: the points known after drawing a set of circles do not depend on the order in which they were drawn, so exploring the same set twice can never find anything new, and the search records every set it has seen together with how many moves it still had available there. The second is discarding crossings far outside the picture, more than twenty radii away. That one is a convention rather than a theorem: such points exist and are constructible, and the bound of six is strictly a bound for constructions that stay within twenty radii. No construction known by hand leaves the picture by anything like that much, and raising the cut to forty radii at depth five changes nothing the search reports, but a fully unconditional statement would need the cut removed altogether. What the search cannot do is certify its own floating-point arithmetic: two different constructions whose points coincided to seven decimal places would be merged. That is a theoretical gap rather than a practical one — the coordinates involved are algebraic numbers of small degree, which cannot agree to seven places without being equal unless their defining polynomials have very large coefficients — and closing it would mean redoing the search in exact arithmetic, which is possible and slower.

Still open: a lower bound without search #

Like every result of the search for the fewest moves, the bound here comes from trying everything, and it does not extend beyond the reach of the computer — which, for this problem, is about seven or eight moves before the number of configurations exhausts memory. Several natural questions lie just beyond it: the fewest compass circles to find the centre when the two given points are close together, where the cheap constructions fail and the search runs out of room; the fewest to find the centre of a circle given only three points on it, without the circle drawn; the fewest to inscribe a regular pentagon in a circle whose centre is unknown.

What is missing, as before, is a quantity that grows by a bounded amount with each move and is far from the centre at the start — a potential that would prove a lower bound without enumerating constructions. Algebraic degree is the natural candidate, and it fails here, because the centre of a circle through two given points is reached by square roots of degree two at most; every point in the constructions above lies in a field of small degree over the starting data. A lower bound on the number of moves has to see something finer than degree, and no such measure is known.

A compass that remembers where the centre was #

The straightedge cannot find a lost centre because it cannot tell a circle from any other conic in its own geometry: everything it does is unchanged by maps that move the centre. The compass can, because a compass draws circles, and circles carry their centres with them. The six-circle construction makes that concrete through inversion: a circle through the compass point is a line in disguise, its centre is a mirror image in disguise, and mirror images and inverses are exactly what a compass constructs.

Six is the cost. Mohr and Mascheroni’s theorem promised that the compass could do anything the pair of instruments can; the search says what that promise is worth for one of the oldest small problems in construction — one circle more than the ruler and compass together, and nothing for the straightedge alone.