In Euclidean geometry, the Mohr–Mascheroni theorem states that any geometric construction that can be performed by a compass and straightedge can be performed by a compass alone.
This theorem refers to geometric constructions which only involve points and circles, since it is not possible to draw straight lines without a straightedge. However, a line is considered to be determined if two distinct points on that line are given or constructed, even if the line itself is not drawn.
Although the use of a straightedge can make certain constructions significantly easier, the theorem shows that these constructions are possible even without the use of it. This means the only use of a straightedge is for the aesthetics of drawing straight lines, and is functionally unnecessary for the purposes of construction.
History The result was originally published by Georg Mohr in 1672,Georg Mohr, Euclides Danicus (Amsterdam: Jacob van Velsen, 1672). but his proof languished in obscurity until 1928.Hjelmslev, J. (1928) "Om et af den danske matematiker Georg Mohr udgivet skrift Euclides Danicus, udkommet i Amsterdam i 1672" [Of a memoir Euclides Danicus published by the Danish mathematician Georg Mohr in 1672 in Amsterdam], Matematisk Tidsskrift B, pages 1–7.Schogt, J. H. (1938) "Om Georg Mohr's Euclides Danicus," Matematisk Tidsskrift A, pages 34–36. The theorem was independently discovered by Lorenzo Mascheroni in 1797 and it was known as Mascheroni's Theorem until Mohr's work was rediscovered.Lorenzo Mascheroni, La Geometria del Compasso (Pavia: Pietro Galeazzi, 1797). 1901 edition.
Several proofs of the result are known. Mascheroni's proof of 1797 was generally based on the idea of using reflection in a line as the major tool. Mohr's solution was different. In 1890, August Adler published a proof using the inversion transformation.
An algebraic approach uses the isomorphism between the Euclidean plane and the real coordinate space . In this way, a stronger version of the theorem was proven in 1990.Arnon Avron, "On strict strong constructibility with a compass alone", Journal of Geometry (1990) 38: 12. It also shows the dependence of the theorem on Archimedes' axiom (which cannot be formulated in a first-order language).
Constructive proof Outline To prove the Mohr–Mascheroni theorem, it suffices to show that each of the basic constructions of compass and straightedge is possible using a compass alone, as these are the foundations of all other constructions. All constructions can be written as a series of steps involving these five basic constructions:
Creating the line through two existing points Creating the circle through one point with centre another point Creating the point which is the intersection of two existing, non-parallel lines Creating the one or two points in the intersection of a line and a circle (if they intersect) Creating the one or two points in the intersection of two circles (if they intersect).
Constructions (2) and (5) can be done with a compass alone. For construction (1), a line is considered to be given by any two points. It is understood that the line itself cannot be drawn without a straightedge, so the proof of the theorem lies in showing that constructions (3) and (4) are possible using only a compass. Once this is done, it follows that every compass-straightedge construction can be done under the restrictions of the theorem.
Notation The following notation will be used throughout this article. A circle whose center is located at point and that passes through point will be denoted by . A circle with center and radius specified by a number, , or a line segment will be denoted by or , respectively.
Some preliminary constructions To prove the above constructions (3) and (4), a few necessary intermediary constructions are also explained below since they are used and referenced frequently. These are also compass-only constructions.
Compass equivalence theorem (circle translation)
The modern compass with its fixable (non-collapsing) aperture can be used to transfer distances directly, while a collapsing compass cannot. The compass equivalence theorem states that, while a "modern compass" appears to be a more powerful instrument, it can be simulated with a collapsing compass alone. This justifies the use of "fixed compass" moves (constructing a circle of a given radius at a different location) for the proof of this theorem.
Given points , , and , construct a circle centered at with the radius , using only a collapsing compass. Draw a circle centered at and passing through and vice versa (the blue circles). They will intersect at points and . Draw circles through with centers at and (the red circles). Label their other intersection . Draw a circle (the green circle) with center passing through . This is the required circle.
Reflecting a point across a line
Given a line determined by two points and , and an arbitrary point , construct the image of upon reflection across this line: Construct two circles: one centered at and one centered at , both passing through . The other point of intersection of the two circles, , is the reflection of across the line . If (that is, there is a unique point of intersection of the two circles), then is its own reflection and lies on the line .
Extending the length of a line segment
Given a line determined by two points and , construct the point on the line such that is the midpoint of line segment . Construct point as the intersection of circles and . ( is an equilateral triangle.) Construct point as the intersection of circles and . ( is an equilateral triangle.) Finally, construct point as the intersection of circles and . ( is an equilateral triangle, and the three angles at show that are collinear.)
This construction can be repeated as often as necessary to find a point so that the length of line segment is times the length of line segment for any positive integer .
Inversion in a circle
Given a circle , for some radius (in black) and a point , construct the point that is the inverse of about the circle. Naturally there is no inversion for a point . Draw a circle (in red). Assume that the red circle intersects the black circle at and If the circles do not intersect in two points, see below for an alternative construction. If the circles intersect in only one point, , it is possible to invert simply by doubling the length of (quadrupling the length of ). Reflect the circle center across the line : Construct two new circles and (in light blue). The light blue circles intersect at and at another point . Point is the desired inverse of in the black circle.
This point lies on line and satisfies .
In the event that the above construction fails (that is, the red circle and the black circle do not intersect in two points), find a point on the line so that the length of line segment is a positive integral multiple, say , of the length of and is greater than . Find the inverse of in circle as above (the red and black circles must now intersect in two points). The point is now obtained by extending so that = .
The existence of such an integer relies on Archimedes' axiom. As a result, this construction may require an unbounded number of iterations depending on the ratio of to .
Determining the center of a circle through three points
Given three non-collinear points , and , construct the center of the circle they determine. Construct point , the inverse of in the circle . Reflect in the line to the point . is the inverse of in the circle .
Intersection of two non-parallel lines
The third basic construction concerns the intersection of two non-parallel lines.
Given non-parallel lines and determined by points , , , , construct their point of intersection, . Select circle of arbitrary radius whose center does not lie on either line. Invert points and in circle to points and respectively. The line is inverted to the circle passing through , and . Find the center of this circle. Invert points and in circle to points and respectively. The line is inverted to the circle passing through , and . Find the center of this circle. Let be the intersection of circles and . is the inverse of in the circle .
Intersection of a line and a circle The fourth basic construction concerns the intersection of a line and a circle. The construction below breaks into two cases depending upon whether the center of the circle is or is not collinear with the line.
Circle center is not collinear with the line Assume that center of the circle does not lie on the line.
Given a circle (in black) and a line , construct the points of intersection, and , between them (if they exist). Construct the point , which is the reflection of point across line . (See above.) Under the assumption of this case, . If in fact then this construction will fail, and we have verification of collinearity. Construct a circle (in red). (See above, compass equivalence.) The intersections of circle and the new red circle are points and . If the two circles are (externally) tangential then . If the two circles do not intersect then neither does the circle with the line. Points and are the intersection points of circle and the line . If then the line is tangential to the circle .
An alternate construction, using circle inversion can also be given.
Invert points and in circle to points and respectively. Under the assumption of this case, points , , and are not collinear. Find the center of the circle passing through points , , and . Construct circle , which represents the inversion of the line into circle . and are the intersection points of circles and .Pedoe carries out one more inversion at this point, but the points and are on the circle of inversion and so are invariant under this last unneeded inversion. If the two circles are (internally) tangential then , and the line is also tangential.
Circle center is collinear with the line
Given the circle whose center lies on the line , construct the points and , the intersection points of the circle and the line. Choose an arbitrary point on the circle. Construct point as the reflection of across line . Construct point as the intersection of circles and . ( is the fourth vertex of parallelogram .) Construct point as the intersection of circles and . ( is the fourth vertex of parallelogram .) Construct point as an intersection of circles and . ( lies on .) Points and are the intersections of circles and .
Conclusion
Since all five basic constructions have been shown to be achievable with only a compass, this proves the Mohr–Mascheroni theorem. Any compass-straightedge construction may be achieved with the compass alone by describing their constructive steps in terms of the five basic constructions.
Validity of the theorem
Dono Kijne points out that the Mohr–Mascheroni theorem fundamentally relies on Archimedes' axiom. As a result, any proof of Mohr–Mascheroni theorem must inherently involve an unbounded number of steps. This raises some questions about what constitutes a valid geometric construction.
Most geometric constructions can be thought of as "straight-line programs", a list of elementary instructions with a fixed number of steps. Under this model, the Mohr–Mascheroni theorem would not qualify as a valid result because it has no a priori bound on the number of iterations required.
To address this, Erwin Engeler suggested that geometric constructions be defined as "programs with loops", a list of instructions that allow conditionals and control flow. This saves the Mohr–Mascheroni theorem, but introduces new issues: