A rhombus has three consecutive vertices on the sides $AB$, $BC$, and $CD$ of a given square $ABCD$. What is the area of the locus of all points that can be the fourth vertex of such a rhombus?
Let the vertices on the sides $AB$, $BC$, $CD$ of the rhombus be denoted by $P$, $Q$, $R$ respectively, and let $E$, $F$ be the midpoints of the sides $BC$, $AD$ of the square. The center $K$ of the rhombus - as the midpoint of the segment $PR$ - falls on the segment $EF$, and the fourth vertex $S$ in question is the reflection of $Q$ over $K$. Consequently, $S$ will be a point of the rectangle $BC'B'C$, where $B'C'$ is the reflection of the side $BC$ over $F$; but $S$ is not a point on the segment $BC$.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf111f232494678e4312g-1.jpg?height=791&width=1334&top_left_y=347&top_left_x=382)

To describe the set of points traced by $S$ in a coordinate system where the origin is $E$, and $F$ is the point $(1; 0)$, and thus the coordinates of $B$ are $(0, 1/2)$. Let $S(x, y)$ be a point in the rectangle $BC'B'C$, i.e., the inequalities

$$
0 < x \leq 2, \quad -1/2 \leq y \leq 1/2
$$

hold. We need to determine what additional conditions must be placed on $x$ and $y$ for $S$ to be the fourth vertex of a suitable rhombus.

Since $K$ bisects the segment $QS$ and lies on the line $EF$, the coordinates of $Q$ and $K$ must necessarily be

$$
K(x/2, 0), \quad Q(0, -y)
$$

By assumption (1), $Q$ will be a point on the segment $BC$. The vertices $P$ and $R$ of the rhombus are the points where the line perpendicular to $QS$ at $K$ intersects the lines $AB$ and $CD$, respectively, with coordinates

$$
P\left(\frac{x}{2} - \frac{y}{x}, \frac{1}{2}\right), \quad R\left(\frac{x}{2} + \frac{y}{x}, -\frac{1}{2}\right)
$$

If $S$ is the fourth vertex of a suitable rhombus, its other vertices can only be the points $P$, $Q$, $R$ obtained above. Therefore, $S(x, y)$ belongs to the sought region if and only if, in addition to (1), the conditions

$$
0 \leq \frac{x}{2} - \frac{y}{x} \leq 1, \quad 0 \leq \frac{x}{2} + \frac{y}{x} \leq 1
$$

are also satisfied. Thus, our task is to determine the area of the region defined by the points $(x, y)$ that satisfy both (1) and (2).

Multiplying the inequalities (2) by the positive $x$ from (1) and rearranging, we get

$$
-x^2 \leq 2y \leq x^2, \quad -1 + (1-x)^2 \leq 2y \leq 1 - (1-x)^2
$$

The region defined by the inequalities (3) is shown in the second figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf111f232494678e4312g-2.jpg?height=805&width=1293&top_left_y=140&top_left_x=400)

The curved lines are congruent parabolas. Since this is part of the rectangle $BC'B'C$, only the points within this rectangle satisfy both (1) and (2). Therefore, we need to determine the area of this region.

From (3), it is clear that the part of the region that falls within the rectangle $AC'GF$ is congruent to the part missing from the rectangle $EFDC$. Indeed, draw a segment of length $1/2$ parallel to $AF$ and at a distance $t$ from $AF$ in the rectangle $AC'GF$. This segment divides the region into two parts: the lower part

$$
\frac{1}{2}\left(1 - [1 - (1+t)]^2\right) = \frac{1}{2}\left(1 - t^2\right)
$$

belongs to the region, while the upper part, of length $t^2 / 2$, does not. Similarly, cutting the rectangle $EFDC$, the upper part of length $t^2 / 2$ belongs to the region, and the lower part of length $(1 - t^2) / 2$ does not. Therefore, the parts of the region in the two rectangles have the same area as the rectangle $EFDC$.

Similarly, it can be shown that the parts of the region in the rectangles $FGB'D$ and $BAFE$ together have the same area as the rectangle $BAFE$. The area of the sought region is thus equal to the area of the square $ABCD$.