The Dowel Problem

2026-01-09

Given that we have a circular dowel as such

{{< tikztwo >}}

\begin{document} \begin{tikzpicture}[scale=3,transform shape] \draw[ultra thick] (0,0) circle (2cm); \end{tikzpicture} \end{document}

{{< /tikztwo >}}

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).

{{< tikztwo >}}

\begin{document} \begin{tikzpicture}[scale=3,transform shape] \draw[ultra thick] (0,0) circle (2cm);

\coordinate (A) at (45:2cm); \coordinate (B) at (145:2cm); \node at (A) [above right] {A}; \node at (B) [above left] {B}; \draw[ultra thick, blue] (A) – (B);

\coordinate ( C) at (270:2cm); \coordinate (D) at (180:2cm); \node at ( C) [below] {C}; \node at (D) [left] {D}; \draw[ultra thick, red] ( C) – (D); \end{tikzpicture} \end{document}

{{< /tikztwo >}}

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.

{{< tikztwo >}} \usetikzlibrary{calc,intersections}

\begin{document} \begin{tikzpicture}[scale=3,transform shape] % circle \draw[ultra thick] (0,0) circle (2cm);

% chord AB \coordinate (A) at (45:2cm); \coordinate (B) at (145:2cm); \draw[ultra thick, blue] (A) – (B); \node[above right] at (A) {A}; \node[above left] at (B) {B};

% chord CD \coordinate ( C) at (270:2cm); \coordinate (D) at (180:2cm); \draw[ultra thick, red] ( C) – (D); \node[below] at ( C) {C}; \node[left] at (D) {D};

% midpoints \coordinate (M) at ($(A)!0.5!(B)$); \coordinate (N) at ($( C)!0.5!(D)$); \fill (M) circle (2pt) node[above right] {$M$}; \fill (N) circle (2pt) node[above right] {$N$};

% perpendicular bisector points via rotation (robust in TikZJax) \coordinate (Mup) at ($(M)!1! 90:(B)$); \coordinate (Mdown) at ($(M)!1!-90:(B)$);

\coordinate (Nup) at ($(N)!1! 90:(D)$); \coordinate (Ndown) at ($(N)!1!-90:(D)$);

% bisectors (named paths) \draw[ultra thick, dashed, blue, name path=bis1] (Mdown) – (Mup); \draw[ultra thick, dashed, red, name path=bis2] (Ndown) – (Nup);

% intersection = center \path [name intersections={of=bis1 and bis2, by=O}]; \fill (O) circle (2.2pt) node[below right] {$O$};

% optional radii %\draw[thin] (O) – (A); %\draw[thin] (O) – (D); \end{tikzpicture} \end{document}

{{< /tikztwo >}}

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 \(OA = OB = R\).

{{< tikztwo >}} \usetikzlibrary{calc}

\begin{document} \begin{tikzpicture}[scale=3,transform shape] \draw[ultra thick] (0,0) circle (2cm);

\coordinate (O) at (0,0); \coordinate (P) at (25:2cm);

\fill (O) circle (2pt) node[below left] {$O$}; \fill (P) circle (2pt) node[above right] {$P$};

\draw[ultra thick] (O) – (P); \node[above] at ($(O)!0.55!(P)$) {$R$};

% (optional) show “P lies on circle” \node at (0,-2.35cm) {$OP=R$}; \end{tikzpicture} \end{document}

{{< /tikztwo >}}

  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 \(\overline{AB}\).

{{< tikztwo >}} \usetikzlibrary{calc}

\begin{document} \begin{tikzpicture}[scale=3, transform shape] % circle (the dowel) \coordinate (O) at (0,0); \draw[ultra thick] (O) circle (2cm);

% chord AB on the circle \coordinate (A) at (210:2cm); \coordinate (B) at (-30:2cm); \draw[ultra thick] (A) – (B); \node[below left] at (A) {$A$}; \node[below right] at (B) {$B$};

% midpoint M of the chord \coordinate (M) at ($(A)!0.5!(B)$); \fill (M) circle (2pt) node[below] {$M$};

% perpendicular bisector line through M \coordinate (Mup) at ($(M)!2! 90:(B)$); \coordinate (Mdown) at ($(M)!1!-90:(B)$); \draw[ultra thick, dashed] (Mdown) – (Mup); \node[right] at ($(M)!0.95!90:(B)$) {$\ell_{AB}$};

% choose a point X on the perpendicular bisector (inside the circle) \coordinate (X) at ($(M)!1.0!90:(B)$); \fill (X) circle (2pt) node[above left] {$X$};

% show that X is equidistant from A and B \draw[thin] (X) – (A); \draw[thin] (X) – (B); \node[left] at ($(X)!0.55!(A)$) {$XA$}; \node[right] at ($(X)!0.55!(B)$) {$XB$}; \node[above] at ($(X)+(0,0.5)$) {$XA=XB$};

% (optional) show the center just for context \fill (O) circle (2pt) node[below left] {$O$}; \end{tikzpicture} \end{document}

{{< /tikztwo >}}

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

    {{< tikztwo >}} \usetikzlibrary{calc}

    \begin{document} \begin{tikzpicture}[scale=3,transform shape] \coordinate (O) at (0,0); \draw[ultra thick] (O) circle (2cm);

    \coordinate (A) at (35:2cm); \coordinate (B) at (155:2cm); \draw[ultra thick] (A) – (B); \node[above right] at (A) {$A$}; \node[above left] at (B) {$B$};

    \coordinate (M) at ($(A)!0.5!(B)$); \fill (M) circle (2pt) node[above] {$M$};

    % perpendicular bisector through M \coordinate (Mup) at ($(M)!1! 90:(B)$); \coordinate (Mdown) at ($(M)!1!-90:(B)$); \draw[ultra thick, dashed] (Mdown) – (Mup); \node[above] at ($(M)!0.55!90:(B)$) {$\ell_{AB}$};

    % radii \draw[ultra thick] (O) – (A); \draw[ultra thick] (O) – (B); \fill (O) circle (2pt) node[below right] {$O$}; \node at ($(O)!0.55!(A)$) {$R$}; \node at ($(O)!0.55!(B)$) {$R$}; \end{tikzpicture} \end{document}

    {{< /tikztwo >}}

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

    {{< tikztwo >}} \usetikzlibrary{calc,intersections}

    \begin{document} \begin{tikzpicture}[scale=3,transform shape] % Tight bounding box (prevents floating/huge whitespace) \path[use as bounding box] (-2.6cm,-2.6cm) rectangle (2.6cm,2.6cm);

    % circle \draw[ultra thick] (0,0) circle (2cm);

    % chord AB \coordinate (A) at (45:2cm); \coordinate (B) at (145:2cm); \draw[ultra thick, blue] (A) – (B); \node[above right] at (A) {$A$}; \node[above left] at (B) {$B$};

    % chord CD (clearly non-parallel) \coordinate ( C) at (250:2cm); \coordinate (D) at (330:2cm); \draw[ultra thick, red] ( C) – (D); \node[below left] at ( C) {$C$}; \node[below right] at (D) {$D$};

    % midpoints \coordinate (M) at ($(A)!0.5!(B)$); \coordinate (N) at ($( C)!0.5!(D)$); \fill (M) circle (2pt) node[above right] {$M$}; \fill (N) circle (2pt) node[above] {$N$};

    % Make bisectors long enough to cross, but not ridiculous \def\k{6}

    % perpendicular bisector endpoints (via rotation, TikZJax-safe) \coordinate (Mup) at ($(M)!\k! 90:(B)$); \coordinate (Mdown) at ($(M)!\k!-90:(B)$);

    \coordinate (Nup) at ($(N)!\k! 90:(D)$); \coordinate (Ndown) at ($(N)!\k!-90:(D)$);

    % helper paths for intersection (overlay avoids affecting bounding box) \path[overlay, name path=bis1] (Mdown) – (Mup); \path[overlay, name path=bis2] (Ndown) – (Nup);

    % intersection = center \path[overlay, name intersections={of=bis1 and bis2, by={O}}]; \fill (O) circle (2.2pt) node[above left] {$O$};

    % draw bisectors, but visually only inside the dowel \begin{scope} \clip (0,0) circle (2cm); \draw[ultra thick, dashed, blue] (Mdown) – (Mup); \draw[ultra thick, dashed, red] (Ndown) – (Nup); \end{scope}

    % labels for the bisectors (inside-ish) \node[blue] at ($(M)+(-0.4,0.35)$) {$\ell_{AB}$}; \node[red] at ($(N)+(-0.5,0.15)$) {$\ell_{CD}$};

    % radii to emphasise “center” \draw[thin] (O) – (A); \draw[thin] (O) – (D); \node at ($(O)!0.55!(A)$) {$R$}; \node at ($(O)!0.55!(D)$) {$R$}; \end{tikzpicture} \end{document}

    {{< /tikztwo >}}

    1. (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.

    {{< tikztwo >}} \usetikzlibrary{calc}

    \begin{document} \begin{tikzpicture}[scale=3,transform shape] % tight bounding box \path[use as bounding box] (-2.7cm,-2.6cm) rectangle (2.7cm,2.6cm);

    % circle + center \coordinate (O) at (0,0); \draw[ultra thick] (O) circle (2cm); \fill (O) circle (2pt) node[below left] {$O$};

    % point of tangency \coordinate (A) at (40:2cm); \fill (A) circle (2pt) node[above right] {$A$};

    % tangent at A: direction is (O-A) rotated by 90 degrees \def\k{3.2} \coordinate (Aup) at ($(A)!\k! 90:(O)$); \coordinate (Adown) at ($(A)!\k!-90:(O)$); \draw[ultra thick, dashed] (Aup) – (Adown); \node[above] at ($(A)!0.55!(Aup)$) {$t_A$};

    % (optional) small highlight that the line touches at A \draw[thin] (O) – (A); \node at ($(O)!0.52!(A)$) {$R$}; \end{tikzpicture} \end{document}

    {{< /tikztwo >}}

  • 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 \(t_A\). \(\square\)

    {{< tikztwo >}} \usetikzlibrary{calc}

    \begin{document} \begin{tikzpicture}[scale=3,transform shape] \path[use as bounding box] (-2.7cm,-2.6cm) rectangle (2.7cm,2.6cm);

    \coordinate (O) at (0,0); \draw[ultra thick] (O) circle (2cm); \fill (O) circle (2pt) node[below left] {$O$};

    \coordinate (A) at (40:2cm); \fill (A) circle (2pt) node[above right] {$A$};

    % radius OA \draw[ultra thick] (O) – (A); \node at ($(O)!0.55!(A)$) {$R$};

    % tangent at A \def\k{3.2} \coordinate (Aup) at ($(A)!\k! 90:(O)$); \coordinate (Adown) at ($(A)!\k!-90:(O)$); \draw[ultra thick, dashed] (Aup) – (Adown); \node[above] at ($(A)!0.55!(Aup)$) {$t_A$};

    % right-angle marker at A between OA and t_A \coordinate (P) at ($(A)!0.2cm!(O)$); % along radius toward O \coordinate (Q) at ($(A)!0.2cm!(Aup)$); % along tangent \coordinate (S) at ($(P)+(Q)-(A)$); \draw[thin] (P) – (S) – (Q);

    \node[right] at ($(A)+(0.35,0.05)$) {$OA\perp t_A$}; \end{tikzpicture} \end{document}

    {{< /tikztwo >}}

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

    {{< tikztwo >}} \usetikzlibrary{calc}

    \begin{document} \begin{tikzpicture}[scale=3,transform shape] \path[use as bounding box] (-2.8cm,-2.7cm) rectangle (2.8cm,2.7cm);

    \coordinate (O) at (0,0); \draw[ultra thick] (O) circle (2cm); \fill (O) circle (2pt) node[below left] {$O$};

    % two tangency points \coordinate (A) at (25:2cm); \coordinate ( C) at (155:2cm); \fill (A) circle (2pt) node[above right] {$A$}; \fill ( C) circle (2pt) node[above left] {$C$};

    % tangents \def\k{3.3} \coordinate (Aup) at ($(A)!\k! 90:(O)$); \coordinate (Adown) at ($(A)!\k!-90:(O)$); \draw[ultra thick, dashed, blue] (Aup) – (Adown); \node[blue] at ($(A)!0.6!(Aup)$) {$t_A$};

    \coordinate (Cup) at ($( C)!\k! 90:(O)$); \coordinate (Cdown) at ($( C)!\k!-90:(O)$); \draw[ultra thick, dashed, red] (Cup) – (Cdown); \node[red] at ($( C)!0.6!(Cup)$) {$t_C$};

    % lines through A,C perpendicular to tangents (these are the radii) \draw[ultra thick] (O) – (A); \draw[ultra thick] (O) – ( C); \node at ($(O)!0.55!(A)$) {$R$}; \node at ($(O)!0.55!( C)$) {$R$};

    % emphasize the “perpendicular through A/C” idea \node[below right] at ($(A)!0.35!(O)$) {$\perp$}; \node[below left] at ($( C)!0.35!(O)$) {$\perp$};

    \node at (0,-2.35cm) {perpendiculars meet at $O$}; \end{tikzpicture} \end{document}

    {{< /tikztwo >}}

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

    {{< tikztwo >}} \usetikzlibrary{calc,intersections} \usepackage{amsmath}

    \begin{document} \begin{tikzpicture}[scale=3,transform shape] \path[use as bounding box] (-3.1cm,-2.8cm) rectangle (3.1cm,3.2cm);

    \coordinate (O) at (0,0); \draw[ultra thick] (O) circle (2cm); \fill (O) circle (2pt) node[below left] {$O$};

    % choose A,B symmetric about the y-axis for a clean picture \coordinate (A) at (40:2cm); \coordinate (B) at (140:2cm); \fill (A) circle (2pt) node[above right] {$A$}; \fill (B) circle (2pt) node[above left] {$B$};

    % chord AB \draw[ultra thick] (A) – (B); \node[above left] at ($(A)!0.5!(B)$) {$\tiny \overline{AB}$};

    % midpoint M of the chord \coordinate (M) at ($(A)!0.5!(B)$); \fill (M) circle (2pt) node[below right] {$M$};

    % tangents at A and B \def\k{3.8} \coordinate (Aup) at ($(A)!\k! 90:(O)$); \coordinate (Adown) at ($(A)!\k!-90:(O)$); \coordinate (Bup) at ($(B)!\k! 90:(O)$); \coordinate (Bdown) at ($(B)!\k!-90:(O)$);

    % named paths for intersection T \path[overlay, name path=tA] (Aup) – (Adown); \path[overlay, name path=tB] (Bup) – (Bdown); \path[overlay, name intersections={of=tA and tB, by=T}];

    % draw tangents \draw[ultra thick, dashed, blue] (Aup) – (Adown); \draw[ultra thick, dashed, red] (Bup) – (Bdown); \node[blue] at ($(A)!0.65!(Aup)$) {$t_A$}; \node[red] at ($(B)!0.65!(Bup)$) {$t_B$};

    % intersection of tangents \fill (T) circle (2pt) node[above] {$T$};

    % segments TA and TB (to suggest equal tangents) \draw[thin] (T) – (A); \draw[thin] (T) – (B); \node[right] at ($(T)!0.55!(A)$) {$TA$}; \node[left] at ($(T)!0.55!(B)$) {$TB$};

    % line OT (claimed to be perpendicular bisector) \draw[ultra thick, dashed] (O) – (T); \node[right] at ($(O)!0.55!(T)$) {$\small OT$};

    % right-angle marker at M between chord AB and OT \coordinate (P) at ($(M)!0.14!(A)$); % along chord \coordinate (Q) at ($(M)!0.14!(O)$); % along OT (towards O) \coordinate (S) at ($(P)+(Q)-(M)$); \draw (P) – (S) – (Q);

    \node at (0,-2.35cm) {$OT\ \text{bisects and is }\perp\ \overline{AB}$}; \end{tikzpicture} \end{document}

    {{< /tikztwo >}}