Mohr–Mascheroni theorem

To prove the theorem, each of the basic constructions of compass and straightedge need to be proven to be possible by using a compass alone, as these are the foundations of, or elementary steps for, all other constructions. These are:

1. Creating the line through two existing points
2. Creating the circle through one point with centre another point
3. Creating the point which is the intersection of two existing, non-parallel lines
4. Creating the one or two points in the intersection of a line and a circle (if they intersect)
5. Creating the one or two points in the intersection of two circles (if they intersect).

#1 - A line through two points

It is understood that a straight line cannot be drawn without a straightedge. A line is considered to be given by any two points, as any such pair define a unique line. In keeping with the intent of the theorem which we aim to prove, the actual line need not be drawn but for aesthetic reasons.

#2 - A circle through one point with defined center

This can be done with a compass alone. A straightedge is not required for this.

#5 - Intersection of two circles

This construction can also be done directly with a compass.

#3, #4 - The other constructions

Thus, to prove the theorem, only compass-only constructions for #3 and #4 need to be given.

The following notation will be used throughout this article. A circle whose center is located at point U and that passes through point V will be denoted by U(V). A circle with center U and radius specified by a number, r, or a line segment AB will be denoted by U(r) or U(AB), respectively.[9]

In general constructions there are often several variations that will produce the same result. The choices made in such a variant can be made without loss of generality. However, when a construction is being used to prove that something can be done, it is not necessary to describe all these various choices and, for the sake of clarity of exposition, only one variant will be given below. However, many constructions come in different forms depending on whether or not they use circle inversion and these alternatives will be given if possible.

It is also important to note that some of the constructions proving the Mohr-Mascheroni theorem require the arbitrary placement of points in space, such as finding the center of a circle when not already provided (see construction below). In some construction paradigms - such as in the geometric definition of the constructible number - this may be prohibited. In such a paradigm, however, for example, circles without their centers will not be provided by hypothesis, thus there is no issue.

To prove the above constructions #3 and #4, which are included below, a few necessary intermediary constructions are also explained below since they are used and referenced frequently. These are also compass-only constructions. All constructions below rely on #1,#2,#5, and any other construction that is listed prior to it.

The ability to translate, or copy, a circle to a new center is vital in these proofs and fundamental to establishing the veracity of the theorem. The creation of a new circle with the same radius as the first, but centered at a different point, is the key feature distinguishing the collapsing compass from the modern, rigid compass. With the rigid compass this is a triviality, but with the collapsing compass it is a question of construction possibility. The equivalence of a collapsing compass and a rigid compass was proved by Euclid (Book I Proposition 2 of The Elements) using straightedge and collapsing compass when he, essentially, constructs a copy of a circle with a different center. This equivalence can also be established with (collapsing) compass alone, a proof of which can be found in the main article.

Point symmetry

• Given a line segment AB and a point C not on the line determined by that segment, construct the image of C upon reflection across this line.

1. Construct two circles: one centered at A and one centered at B, both passing through C.
2. D, the other point of intersection of the two circles, is the reflection of C across the line AB.
   • If C = D (that is, there is a unique point of intersection of the two circles), then C is its own reflection and lies on the line AB (contrary to the assumption), and the two circles are internally tangential.

A compass-only construction of doubling the length of segment AB

• Given a line segment AB find a point C on the line AB such that B is the midpoint of line segment AC.[10]

1. Construct point D as the intersection of circles A(B) and B(A). (∆ABD is an equilateral triangle.)
2. Construct point E ≠ A as the intersection of circles D(B) and B(D). (∆DBE is an equilateral triangle.)
3. Finally, construct point C ≠ D as the intersection of circles B(E) and E(B). (∆EBC is an equilateral triangle, and the three angles at B show that A, B and C are collinear.)

This construction can be repeated as often as necessary to find a point Q so that the length of line segment AQ = n⋅ length of line segment AB for any positive integer n.

Point inversion in a circle

• Given a circle B(r), for some radius r (in black) and a point D (≠ B) construct the point I that is the inverse of D in the circle.[11] Naturally there is no inversion for a point ${\textstyle D=B}$.

1. Draw a circle D(B) (in red).
2. Assume that the red circle intersects the black circle at E and E'
   • if the circles do not intersect in two points see below for an alternative construction.
   • if the circles intersect in only one point, ${\displaystyle E=E'}$, it is possible to invert ${\displaystyle D}$ simply by doubling the length of ${\displaystyle EB}$ (quadrupling the length of ${\displaystyle DB}$).
3. Reflect the circle center ${\displaystyle B}$ across the line ${\displaystyle EE'}$:
   1. Construct two new circles E(B) and E' (B) (in light blue).
   2. The light blue circles intersect at B and at another point I ≠ B.
4. Point I is the desired inverse of D in the black circle.

Point I is such that the radius r of B(r) is to IB as DB is to the radius; or IB / r = r / DB.

In the event that the above construction fails (that is, the red circle and the black circle do not intersect in two points),[12] find a point Q on the line BD so that the length of line segment BQ is a positive integral multiple, say n, of the length of BD and is greater than r / 2 (this is possible by Archimede's axiom). Find Q' the inverse of Q in circle B(r) as above (the red and black circles must now intersect in two points). The point I is now obtained by extending BQ' so that BI = n ⋅ BQ' .

Compass-only construction of the center of a circle through three points (A,B,C)

• Given three non-collinear points A, B and C, find the center O of the circle they determine.[13]

1. Construct point D, the inverse of C in the circle A(B).
2. Reflect A in the line BD to the point X.
3. O is the inverse of X in the circle A(B).

Compass-only construction of the intersection of two lines (not all construction steps shown)

• Given non-parallel lines AB and CD, find their point of intersection, X.[13]

1. Select circle O(r) of arbitrary radius whose center O does not lie on either line.
2. Invert points A and B in circle O(r) to points A' and B' respectively.
3. The line AB is inverted to the circle passing through O, A' and B'. Find the center E of this circle.
4. Invert points C and D in circle O(r) to points C' and D' respectively.
5. The line CD is inverted to the circle passing through O, C' and D'. Find the center F of this circle.
6. Let Y ≠ O be the intersection of circles E(O) and F(O).
7. X is the inverse of Y in the circle O(r).

The compass-only construction of the intersection points of a line and a circle breaks into two cases depending upon whether the center of the circle is or is not collinear with the line.

Assume that center of the circle does not lie on the line.

Line-circle intersection (non-collinear case)

• Given a circle C(r) (in black) and a line AB. We wish to construct the points of intersection, P and Q, between them (if they exist).[14][15]

1. Construct the point D, which is the reflection of point C across line AB. (See above.)
   • Under the assumption of this case, C ≠ D.
2. Construct a circle D(r) (in red). (See above, compass equivalence.)
3. The intersections of circle C(r) and the new red circle D(r) are points P and Q.
   • If the two circles are (externally) tangential then ${\displaystyle P=Q}$.
4. Points P and Q are the intersection points of circle C(r) and the line AB.
   • If ${\displaystyle P=Q}$ then the line is tangential to the circle ${\displaystyle C(r)}$.

An alternate construction, using circle inversion can also be given.[13]

• Given a circle C(r) and a line AB. We wish to construct the points of intersection, P and Q, between them (if they exist).

1. Invert points A and B in circle C(r) to points A' and B' respectively.
   • Under the assumption of this case, points A', B', and C are not collinear.
2. Find the center E of the circle passing through points C, A', and B'.
3. Construct circle E(C), which represents the inversion of the line AB into circle C(r).
4. P and Q are the intersection points of circles C(r) and E(C).[16]
   • If the two circles are (internally) tangential then ${\displaystyle P=Q}$, and the line is also tangential.

Compass only construction of intersection of a circle and a line (circle center on line)

• Given the circle C(D) whose center C lies on the line AB, find the points P and Q, the intersection points of the circle and the line.[17]

1. Construct point D' ≠ D as the other intersection of circles A(D) and C(D).
2. Construct point F as the intersection of circles C(DD' ) and D(C). (F is the fourth vertex of parallelogram CD'DF.)
3. Construct point F' as the intersection of circles C(DD' ) and D' (C). (F' is the fourth vertex of parallelogram CDD'F'.)
4. Construct point M as an intersection of circles F(D' ) and F' (D). (M lies on AB.)
5. Points P and Q are the intersections of circles F(CM) and C(D).

Thus it has been shown that all of the basic construction one can perform with a straightedge and compass can be done with a compass alone, provided that it is understood that a line cannot be literally drawn but merely defined by two points.