Given that we have a circular dowel as such

a)

Find the exact center point using only a ruler and a marker

Solution

Draw 2 arbitrary chords on the dowel: $AB$ and $CD$ (make sure they are not parallel).

Then find the midpoints of both these chords (call them $M$ and $N$), and draw the perpendicular bisector of each chord (i.e. the line through the midpoint that is perpendicular to the chord).

  • Using a ruler: measure the length of the chord, mark the halfway point along the chord, and label it the midpoint.

Naturally, the two perpendicular bisectors intersect at a single point $O$, and that point is the center of the circle. That is where you should drill the pilot hole.

b)

Give a proof for the correctness of your method.

Solution

We start with some definitions so that we are all on the same page:

Definitions
  1. A circle with center $O$ and radius $R$ is the set \[ \{P : OP = R\}. \] In particular, if $A$ and $B$ lie on the circle, then $OA = OB = R$.
  1. The perpendicular bisector of the segment $\overline{AB}$ can be described as the set (locus) \[ \ell_{AB} := \{X : XA = XB\}. \] Geometrically, this is exactly the line through the midpoint $M$ of $\overline{AB}$ that is perpendicular to $\overline{AB}$.
Lemmas

Next, we prove a couple of auxillary lemmas that will make the conclusion obvious:

1. (the perpendicular bisector passes through the center)
Claim
Let $A,B$ be arbitrary points on the circle with center $O$. Then $O$ lies on the perpendicular bisector of $\overline{AB}$.
Proof
Since $A$ and $B$ are on the circle, $OA = OB$ (both equal the radius). By the locus definition of the perpendicular bisector,

\[ OA = OB \quad \Longrightarrow \quad O \in \ell_{AB}. \] So the perpendicular bisector of $\overline{AB}$ passes through $O$. \(\square\)

2. (the intersection of 2 perpendicular bisectors is the unique center of the circle)
Claim
If $AB$ and $CD$ are two chords, then the center $O$ lies on both perpendicular bisectors $\ell_{AB}$ and $\ell_{CD}$.
Proof.
Apply Lemma 1 to chord $AB$ and again to chord $CD$. \(\square\)

3. (non-parallel chords give a unique intersection of bisectors)
Claim
If $AB$ and $CD$ are two chords with $AB \nparallel CD$, then the perpendicular bisectors $\ell_{AB}$ and $\ell_{CD}$ intersect in exactly one point.
Proof
  1. Since $\ell_{AB}\perp AB$ and $\ell_{CD}\perp CD$, if $\ell_{AB}\parallel \ell_{CD}$ then $AB$ and $CD$ would both be perpendicular to the same direction, hence $AB\parallel CD$. This contradicts $AB \nparallel CD$. Therefore $\ell_{AB}\nparallel \ell_{CD}$.
  2. In Euclidean geometry, two non-parallel lines intersect.
  3. They intersect in at most one point: if two lines shared two distinct points, they would be the same line (a line is uniquely determined by two points).

Hence $\ell_{AB}$ and $\ell_{CD}$ intersect in exactly one point. \(\square\)

Proof

In our construction, we draw two chords $AB$ and $CD$ with $AB \nparallel CD$, then construct their perpendicular bisectors $\ell_{AB}$ and $\ell_{CD}$, and finally take their intersection point \[ P := \ell_{AB}\cap \ell_{CD}. \]

By Lemma 2, the true center $O$ lies on both perpendicular bisectors: \[ O\in \ell_{AB} \quad\text{and}\quad O\in \ell_{CD}. \] Therefore $O\in \ell_{AB}\cap \ell_{CD}$.

By Lemma 3, since $AB \nparallel CD$, the lines $\ell_{AB}$ and $\ell_{CD}$ intersect in exactly one point. Thus the intersection $\ell_{AB}\cap \ell_{CD}$ contains a unique point. Since $O$ lies in this intersection, it must be that unique point; hence $P=O$.

Therefore the intersection point of the two perpendicular bisectors is exactly the center of the circle. \(\square\)

Alternative approach via tangents
Definitions
  1. A tangent line to a circle at a point $A$ on the circle is a line $t_A$ that meets the circle at $A$ and locally "just touches" it there.
Lemmas
1. (radius is perpendicular to tangent at the point of tangency)
Claim
If $t_A$ is tangent to the circle at $A$ and $O$ is the center, then $OA \perp t_A$.
Proof
Among all points $X$ on the tangent line $t_A$, the point $A$ is the unique point on the circle, hence the unique point on $t_A$ at distance exactly $R$ from $O$.

If $OA$ were not perpendicular to $t_A$, then moving a tiny amount along the line $t_A$ from $A$ would initially decrease the distance to $O$, producing points $X$ on $t_A$ with $OX < OA = R$, contradicting that $A$ is the closest point of $t_A$ to $O$. Therefore $OA$ must be perpendicular to $t_A$. \(\square\)

2. (two tangents determine the center)
Claim
Let $A$ and $C$ be two distinct points on the circle such that the tangents $t_A$ and $t_C$ are not parallel. Then the lines through $A$ and $C$ perpendicular to $t_A$ and $t_C$ intersect at the unique center $O$.
Proof
By Lemma 1, the center $O$ lies on the perpendicular to $t_A$ through $A$, and also lies on the perpendicular to $t_C$ through $C$.

If $t_A \nparallel t_C$, then these two perpendicular lines are also not parallel, hence intersect in exactly one point. That intersection point must be $O$. \(\square\)

3. (connection to chords, if you want to start from a chord)
Claim
Given a chord $\overline{AB}$, let $t_A$ and $t_B$ be the tangents at $A$ and $B$, and let $T := t_A \cap t_B$ (assuming they are not parallel). Then the line $OT$ is the perpendicular bisector of $\overline{AB}$.
Proof
By Lemma 1, $OA \perp t_A$ and $OB \perp t_B$. Also, tangents from the same external point have equal lengths, so $TA = TB$.

Thus the right triangles $\triangle OAT$ and $\triangle OBT$ have

  • $OA = OB$ (radii),
  • $TA = TB$ (tangents from $T$),
  • right angles at $A$ and $B$.

So they are congruent, giving that $T$ is symmetric with respect to swapping $A$ and $B$ and hence $OT$ is the symmetry axis of the kite $OATB$. Therefore $OT$ meets $\overline{AB}$ at its midpoint and is perpendicular to it; i.e. $OT$ is the perpendicular bisector of $\overline{AB}$. \(\square\)