Problem 908

Let the sum of the two diagonals be $2 s$, and the given side be $a$. To the distance $R A=s$, we draw a $45^{\circ}$ angle at $R$. The other side of this angle intersects the circle drawn from $A$ with radius $a$ at points $B$ and $B_{1}$. Then, we draw circles from $B$ and $B_{1}$ with radius $a$. These circles intersect $R A$ outside point $A$ at points $C$ and $C_{1}$. Points $A, B$, and $C$, as well as $A, B_{1}$, and $C_{1}$, are the vertices of the sought rhombus. Indeed, let the feet of the perpendiculars from $B$ and $B_{1}$ to $A C$ be $O$ and $O_{1}$, then

$$A R=A O+O R=A O+O B=s$$

and

$$A R=A O_{1}+O_{1} R=A O_{1}+O_{1} B_{1}=s$$

(Jenő Silbermann, Nagyvárad.)