*
Exercise 2.2.* The Rusty Compass Theorem**
***Printout*

*It is easier to square the circle than to get round a
mathematician.
—*

**The error may be more difficult to find
in this problem than with the previous exercise, since the proof is valid if additional restrictions are made. One of
the hardest tasks to do is to find errors in proofs of statements that we know
are correct; it is too easy to base proofs on diagrams and preconceptions. This
proof is an example of a very common type of error that even good mathematicians
make. Also, this illustrates how dynamic
geometry software may be used to show someone a proof is incorrect when it is
difficult to explain the error in logic. (The incorrect proof below was used in
a college geometry textbook as the proof for the theorem.)**

* The
Rusty Compass Theorem (or
Compass Equivalence Theorem). Given a circle
centered at a point B with radius r. Let A be any point distinct from B. A
straightedge and collapsing compass **(Euclidean straightedge and compass) can be used to construct a circle centered
at A that is congruent to the given circle centered at B with radius r.*

(b) Identify and write a definition for each term used in the statement of the theorem and proof.

(c) Identify and state any assumptions made in the proof.

(d) Identify and state any theorems used in the proof.

(e) The Compass Equivalence Theorem or Rusty Compass Theorem is valid in Euclidean geometry, though the following proof is incorrect. Is it valid in hyperbolic geometry? Explain.

* **Proof*. Consider circle *C*(*B*, *r*) and point *A* distinct from *B*.

Our objective is to construct a circle *C*(*A,
r*) using only a straightedge and a collapsing compass. Construct circles *C*(*A,
AB*) and *C*(*B*, *BA*) and let *C* and *D* be the points of intersection of *C*(*A, AB*) and *C*(*B*,
*BA*). Let *E* be one of the points of intersection of *C*(*B, r*) and *C*(*A,
AB*).

Next we construct circle *C*(*C,
CE*) and let *P* be a point of
intersection of circle *C*(*C, CE*) with *C*(*B, BA*). We claim that *AP* = *r*,
so that *C*(*A*,* AP*) is the desired
circle implied by the statement of the theorem.

To show that *AP* = *r*, it will suffice
to show that *AP* = *BE* = *r*,
since segment *BE* and *r* are radii of the same circle. We will
show that by SAS. Since radii of a circle are congruent,
and Hence, it only remains for us to show that

Since ,
And since and ,
we have by SSS. Thus, Hence,

Thus, which implies that *AP* = *BE* = *r*. Therefore, *C*(*A, AP*) is congruent to *C(B, r*).//

© Copyright 2005, 2006 - Timothy Peil |