Problem — Locus of the vertex of a triangle with fixed base and altitude

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A fixed segment $UV$ is given. Consider all triangles $UVW$ having $UV$ as base and altitude (relative to that base) of fixed measure $h$. Determine the locus described by the vertex $W$ and prove it.

(original figure: segment $UV$ and the two lines parallel to it at distance $h$)

Solution

The altitude of the triangle $UVW$ relative to the base $UV$ is the distance of the vertex $W$ from the line containing $UV$; call this line $r$. The condition “altitude $= h$” is therefore equivalent to: $W$ lies at distance $h$ from the line $r$.

The locus of the points at distance $h$ from a line $r$ consists of the two lines parallel to $r$ at distance $h$, one on each side of $r$; call them $a$ and $b$.

Every admissible vertex $W$ lies on $a$ or $b$. If $UVW$ is one of the triangles considered, its altitude equals $h$, i.e. $W$ lies at distance $h$ from $r$: hence $W \in a \cup b$.

Every point of $a\cup b$ is an admissible vertex. If $W \in a\cup b$, then $W$ lies at distance $h$ from $r$ and, since $a$ and $b$ are parallel to and distinct from $r$, the point $W$ does not lie on $r$: the three points $U$, $V$, $W$ are not collinear and form a triangle $UVW$ with base $UV$ and altitude $h$.

The locus of the vertex $W$ therefore consists of the two lines parallel to the line $UV$, placed at distance $h$ from it, one in each half-plane. $\;\blacksquare$