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. $\endgroup$ Feb 3, 2017 at 13:49

2 Answers 2


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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 (to check this it is enough to consider the instance $\alpha=90^\circ$). 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 $rOs$.
It follows that $RS$ is approximately parallel to $P_r P_s$, but this approximation
is good only if $\alpha\leq\frac{\pi}{4}$.

2021 addendum by Lagrange multipliers, the optimal configuration is reached when $R$ and $S$ have the same distance from the midpoint $M$ of $OP$. By lengths of medians, this is equivalent to $OR^2+RP^2=PS^2+SO^2$. It implies that the projection $Q$ of $O$ on $RS$ and $P$ are symmetric with respect to the midpoint of $RS$: this explains why the first part of the answer provides a practical, constructible approximation. The projection of $O$ on $RS$ always lies on the circle with diameter $OP$. At the same time, for any $R\in r$ the symmetric of $P$ with respect to the midpoint of $RS$ belongs to a hyperbola with asymptotes given by $r$ and $s$. It follows that the exact solution can be built by intersecting a hyperbola and a circle.

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Post Scriptum: if $r\perp s$ the problem is equivalent to solving $$ \left\{ \begin{array}{rcl} xy &=& 4ab \\ (x-a)^2+(y-b)^2 &=& a^2+b^2 \end{array}\right. $$ i.e. to the duplication of the cube.

Second Post Scriptum - a much better constructible approximation. The vertex of the $PQ$ hyperbola (such that the areas of the red and blue parallelograms are the same) can be constructed with straightedge and compass, as well as the osculating circle at the vertex. By intersecting the osculating circle with the circle having diameter $OP$ we get a point $\widetilde{Q}$ pretty close to $Q$, such that the intersections of $P\widetilde{Q}$ with $r$ and $s$ provide an approximated solution $\widetilde{R}\widetilde{S}$. This is pretty much the same as finding the only root of a cubic through one step of Newton's method.

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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. $\endgroup$ Feb 3, 2017 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, 2017 at 18:58

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