The Dowel Problem

2026-02-23
geometry

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:

Definition (Circle)

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 \(OA = OB = R\).

Definition (Perpendicular Bisector)

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 \(\overline{AB}\).

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

Lemma (Perpendicular Bisector Passes Through Center)

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 \(O\). \(\square\)

Lemma (Intersection of Perpendicular Bisectors is the Center)

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 \(CD\). \(\square\)

Lemma (Non-parallel Chords Give Unique Intersection)

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 (Center of the Circle)

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

Definition (Tangent Line)

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.

Lemma (Radius Perpendicular to Tangent)

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 \(t_A\). \(\square\)

Lemma (Two Tangents Determine the Center)

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 \(O\). \(\square\)

Lemma (Connection to Chords)

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 \(\overline{AB}\). \(\square\)