Problem — The shortest chord through an interior point and the geometric mean

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Let $A$ be a point inside a circle. Through $A$ draw the chord perpendicular to the radius through $A$: call this chord $EF$, of which $A$ is the midpoint, and call $\overline{AF}$ its half-chord.

• (a) Prove that the half-chord $\overline{AF}$ is the geometric mean of the two segments into which $A$ divides any other chord through $A$.
• (b) A chord through $A$ is divided by $A$ into two segments of lengths $4$ and $9$. Compute $\overline{AF}$.

Solution

(a) Let $PQ$ be any chord through $A$, with $\overline{AP}=p$ and $\overline{AQ}=q$. Consider the triangles $AFP$ and $AQE$, where $E$ and $F$ are the endpoints of the chord perpendicular to the radius ($\overline{AE}=\overline{AF}$).

• $\widehat{FAP}\cong\widehat{EAQ}$ (vertical angles at $A$);
• $\widehat{AFP}\cong\widehat{AQE}$ (inscribed angles subtending the same arc $EP$).

By the first criterion the triangles $AFP$ and $AQE$ are similar, so the homologous sides are proportional: $\frac{\overline{AF}}{\overline{AQ}}=\frac{\overline{AP}}{\overline{AE}}.$ Since $\overline{AE}=\overline{AF}$, substituting gives $\frac{\overline{AF}}{q}=\frac{p}{\overline{AF}}\ \Longrightarrow\ \overline{AF}^{\,2}=p\cdot q,$ that is, $\overline{AF}$ is the geometric mean of $p$ and $q$. This product $p\cdot q$ is the same for every chord through $A$ (power of the point), so $EF$ is indeed the shortest chord through $A$ and is bisected by $A$. $\blacksquare$

(b) With $p=4$ and $q=9$: $\overline{AF}=\sqrt{p\cdot q}=\sqrt{4\cdot 9}=\sqrt{36}=6.$ $\boxed{\overline{AF}=6.}$