Compass equivalence theorem

The compass equivalence theorem is an important statement in compass and straightedge constructions. In these constructions it is assumed that whenever a compass is lifted from a page, it collapses, so that it may not be directly used to transfer distances. While this might seem a difficult obstacle to surmount, the compass equivalence theorem states that any construction via a "fixed" compass may be attained with a collapsing compass. In other words, it is possible to construct a circle of equal radius, centered at any given point on the plane. This theorem is known as Proposition II of Book I of Euclid's Elements.

Proof

We are given points A, B, and C, and wish to construct a circle centered at A with the same radius as BC (the first green circle).

• Draw a circle centered at A and passing through B and vice versa (the red circles). They will intersect at point D and form equilateral triangle ABD.
• Extend DB past B and find the intersection of DB and the circle BC, labeled E.
• Create a Circle centered at D and passing through E (the blue circle).
• Extend DA past A and find the Intersection of the DA and the circle DE, labeled F.
• Construct a circle centered at A and passing through F (the second green circle)

• Because E is on the circle BC, BE=BC.
• Because ADB is an equilateral triangle, DA=DB.
• Because E and F are on a circle around D, DE=DF.
• Therefore, AF=BE and AF=BC.

Alternate Proof without Straightedge

It is possible to prove compass equivalence without the use of the straightedge. This justifies the use of "fixed compass" moves in proofs of the Mohr-Mascheroni theorem, which states that any construction possible with straightedge and compass can be accomplished with compass alone.

We are given points A, B, and C, and wish to construct a circle centered at A with the same radius as BC, using only a collapsing compass and no straightedge.

• Draw a circle centered at A and passing through B and vice versa (the blue circles). They will intersect at points D and D'.
• Now draw circles through C with centers at D and D' (the red circles). Find their other intersection and label it E.
• Draw a circle (the green circle) with center A passing through E.

• The line DD' is the perpendicular bisector of AB. Thus A is the reflection of B through line DD'.
• By construction, E is the reflection of C through line DD'.
• Since reflection is an isometry, it follows that AE=BC as desired.

External links

• s:Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/32
• Minnesota State University Mathematics Department