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GEOMETRICAL THEOREM

By Rev. Hamnet Holditch

If a chord of a closed curve, of constant length ${\displaystyle c+c'}$, be divided into two parts of lengths ${\displaystyle c}$, ${\displaystyle c'}$ respectively, the difference between the areas of the closed curve, and of the locus of the dividing point, will be ${\displaystyle \pi cc'}$.

  Solution. Let ${\displaystyle AB}$ be the chord in any position; ${\displaystyle P}$ the dividing point, so that ${\displaystyle AP=c}$, ${\displaystyle BP=c'}$; let ${\displaystyle Q}$ be the point in which the chord intersects its consecutive position; let ${\displaystyle [A]}$ be the area of the given curve, ${\displaystyle [P]}$, ${\displaystyle [Q]}$, those of the loci of ${\displaystyle P}$, ${\displaystyle Q}$, respectively; ${\displaystyle AQ=r}$, ${\displaystyle BQ=c+c'-r}$.

  Then    ${\displaystyle [A]-[Q]={\frac {1}{2}}\int _{0}^{2\pi }r^{2}\,d\theta }$(1),

but also    ${\displaystyle [A]-[Q]={\frac {1}{2}}\int _{0}^{2\pi }(c+c'-r)^{2}\,d\theta }$;

therefore    ${\displaystyle {\frac {1}{2}}\int _{0}^{2\pi }r^{2}\,d\theta ={\frac {1}{2}}\int _{0}^{2\pi }(c+c'-r)^{2}\,d\theta }$,

or    ${\displaystyle (c+c')\int _{0}^{2\pi }r\,d\theta ={\frac {1}{2}}\int _{0}^{2\pi }(c+c')^{2}\,d\theta }$;

therefore    ${\displaystyle \int _{0}^{2\pi }r\,d\theta =\pi (c+c')}$(2),

  Also    ${\displaystyle [P]-[Q]={\frac {1}{2}}\int _{0}^{2\pi }(c-r)^{2}\,d\theta }$,

therefore, by (1),

   ${\displaystyle {\begin{aligned}\lbrack A]-[P]&={\frac {1}{2}}\int _{0}^{2\pi }(2cr-c^{2})\,d\theta \\&=c\int _{0}^{2\pi }r\,d\theta -\pi c^{2}\\&=\pi c(c+c')-\pi c^{2}{\text{, by (2),}}\\&=\pi cc'\end{aligned}}}$.

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