IMO 1968 SL 8

Origin: ROM

Problem

Given an oriented line $\Delta$ and a fixed point $A$ on it, consider all trapezoids $ABCD$ one of whose bases $AB$ lies on $\Delta$, in the positive direction. Let $E, F$ be the midpoints of $AB$ and $CD$ respectively. Find the loci of vertices $B, C, D$ of trapezoids that satisfy the following:

(i) $|AB| \leq a$ ($a$ fixed);

(ii) $|EF| = l$ ($l$ fixed);

(iii) the sum of squares of the nonparallel sides of the trapezoid is constant.

Remark. The constants are chosen so that such trapezoids exist.

Solution

Let $G$ be the point such that $BCDG$ is a parallelogram and let $H$ be the midpoint of $AG$. Obviously $HEFD$ is also a parallelogram, and thus $DH = EF = l$. If $AD^2 + BC^2 = m^2$ is fixed, then from the Stewart theorem we have

$$DH^2 = \frac{2DA^2 + 2DG^2 - AG^2}{4} = \frac{2m^2 - AG^2}{4},$$

which is fixed.

Thus $G$ and $H$ are fixed points, and from here the locus of $D$ is a circle with center $H$ and radius $l$. The locus of $B$ is the segment $(GI]$, where $I \in \Delta$ is a point in the positive direction such that $AI = a$. Finally, the locus of $C$ is a region of the plane consisting of a rectangle sandwiched between two semicircles of radius $l$ centered at points $H$ and $H'$, where $H'$ is a point such that $\overrightarrow{IH'} = \overrightarrow{GH}$.