As a land surveyor I'm commonly tasked with calculating "lot width". Lot width is defined as the shortest distance between two side lot lines measured along a line segment that passes through a point on the principal structure. When the side lot lines are not parallel, I've only been able to approximate this measurement through trial and error. What is the mathematical solution?

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    $\begingroup$ (+$\bigstar$) I like this question really much: it is a geometry (with the archaic meaning related with land measurement) problem that does not have a solution through straightedge and compass, but still has an approximated very simple solution that can be implemented in real life. $\endgroup$ – Jack D'Aurizio Feb 3 '17 at 13:49

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Interesting question. We may assume that $r$ and $s$ are two rays with common origin $O$, $P$ is a fixed point inside the $rOs$ angle and $P_r, P_s$ are the projections of $P$ on $r$ and $s$. We are looking for $R\in r$ and $S\in s$ such that $P\in RS$ and the length of $RS$ is minimal.

This is a simple minimization problem in one variable if we assume that $\widehat{RPP_r}=\theta$. Let $\widehat{SOR}=\alpha$.

We have: $$ RS = \frac{PP_r}{\cos\theta}+\frac{PP_s}{\cos(\alpha-\theta)} $$ whose minimum is achieved at the angle $\theta$ such that $$ PP_s\,\frac{\sin(\alpha-\theta)}{\cos^2(\alpha-\theta)} = PP_r\frac{\sin\theta}{\cos^2\theta}.$$ The exact solution is so dependent on the root of a cubic equation and it is, in general, not constructible through straightedge and compass only. For practical purposes, however, we may consider that:

The shortest $RS$ segment is approximately perpendicular to the line $q$,
given by the reflection of $OP$ with respect to the angle bisector of $rOS$.



I should suppose a measurement method should be constructible more as a convention nominally rather than be strictly found by calculation. Suggesting here is a simple construction to find distance connecting feet of perpendiculars to make an inner (red) line... lot-width as a minor diagonal of a cyclic quadrilateral can be found approximately from a formula:

$$ EF= (OF\cdot ED + OE\cdot DF)/OD $$


  • $\begingroup$ Not really a logic: that is not an accurate approximation of the shortest segment joining two points on the given lines and going through $D$. Try to compute lengths in Geogebra to figure that. The shortest segment, in general, is not perpendicular to the angle bisector of the given angle. $\endgroup$ – Jack D'Aurizio Feb 3 '17 at 17:46
  • $\begingroup$ Another point is that the width could even be curved computable by variational approach, especially with the transversality constraints. $\endgroup$ – Narasimham Feb 3 '17 at 18:58

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