Problem — Locus of the endpoints of congruent parallel segments from a line

Text

A line $m$ is given. From each point of $m$ a segment of fixed length $\ell$ is drawn, so that all these segments are parallel to one another, congruent, and with the same orientation. Determine the locus described by the free endpoints of such segments and prove it.

(original figure: line $m$ and the congruent parallel segments, with the free endpoints highlighted)

Solution

For each point $X$ of $m$ let $X'$ denote the free endpoint of the segment $XX'$; by hypothesis all the segments $XX'$ have the same length $\ell$, the same direction and the same orientation. Fix a point $X_0$ of $m$ with its endpoint $X_0'$. We claim that the required locus is the line $m'$ through $X_0'$ parallel to $m$.

Every endpoint $X'$ lies on $m'$. Let $X$ be another point of $m$. The segments $XX'$ and $X_0X_0'$ are congruent, parallel and with the same orientation: hence the quadrilateral $X X' X_0' X_0$ has the pair of sides $XX'$ and $X_0X_0'$ parallel and congruent, and is therefore a parallelogram. It follows that $X'X_0' \parallel X X_0$; but $XX_0$ lies on $m$, hence $X'X_0' \parallel m$: the point $X'$ lies on the line through $X_0'$ parallel to $m$, that is $m'$.

Every point of $m'$ is an endpoint. Let $Y$ be any point of $m'$. Construct the point $X$ as the fourth vertex of the parallelogram $X_0\,X_0'\,Y\,X$, i.e. such that $\overrightarrow{X_0X} = \overrightarrow{X_0'Y}$. Then $X_0X \parallel X_0'Y$ and $X_0X \cong X_0'Y$; since $X_0'Y$ lies on $m'$, the line $X_0X$ is parallel to $m'$ and passes through $X_0 \in m$, hence it coincides with $m$: therefore $X \in m$. Moreover, by construction, the sides $XY$ and $X_0X_0'$ of the parallelogram are parallel, congruent and with the same orientation, that is $XY$ is the fixed segment drawn from $X$. Hence $Y = X'$ is the endpoint of the segment issuing from $X \in m$, and it belongs to the locus.

The required locus is therefore the line $m'$ parallel to $m$ (shifted by $\ell$ in the direction and orientation of the segments). $\;\blacksquare$