
2. Let $A B C D$ be a rectangle such that $A B=a \geqslant b=B C$. Construct points $P, Q$ on the line $B D$ such that $A P=P Q=Q C$. Determine the number of solutions in terms of $a, b$.

(Jaroslav Svrček)


Solution. Let $A^{\prime}, C^{\prime}$ be perpendicular projections of $A, C$ onto $B D$, respectively. Note that $A A^{\prime}=C C^{\prime}$ and that $A^{\prime}$ and $C^{\prime}$ are symmetric about the center $S$ of the rectangle.

Fix $d>0$. We investigate whether there exist points $P, Q$ on $B D$ such that $A P=P Q=Q C=d$.

Point $P$ has to lie on the line $B D$ and on the circle $k_{A}$ with center $A$ and radius $d$. Similarly, point $Q$ has to lie on the line $B D$ and on the circle $k_{C}$ with center $C$ and radius $d$.

If $dA A^{\prime}$ then $k_{A}$ intersects line $B D$ at two points. Denote by $A_{1}$ the one on the ray $A^{\prime} D$ and by $A_{2}$ the one on the ray $A^{\prime} B$. Similarly, let $B_{1}, B_{2}$ be the intersections of $k_{B}$ with rays $C^{\prime} D, C^{\prime} B$, respectively (Fig. 1). Since triangles $A A^{\prime} A_{1}$, $A A^{\prime} A_{2}, C C^{\prime} C_{1}, C C^{\prime} C_{2}$ are all congruent (HL), we have $A_{1} B_{1}=A_{2} B_{2}=A^{\prime} C^{\prime}$ and the pairs of points $\left(A_{1}, B_{2}\right)$ and $\left(A_{2}, B_{1}\right)$ are symmetric about $S$ (whereas pairs $\left(A_{1}, B_{1}\right),\left(A_{2}, B_{2}\right)$ are not). It remains to investigate which of the pairs $(P, Q) \in$ $\left\{\left(A_{1}, B_{1}\right),\left(A_{2}, B_{2}\right),\left(A_{1}, B_{2}\right),\left(A_{2}, B_{1}\right)\right\}$ are a valid solution.

![](https://cdn.mathpix.com/cropped/2024_04_17_447c393b5df10bbb25bcg-17.jpg?height=299&width=633&top_left_y=995&top_left_x=386)

Fig. 1

(a) Since $A_{1} B_{1}=A_{2} B_{2}=A^{\prime} C^{\prime}$, pairs $\left(A_{1}, B_{1}\right),\left(A_{2}, B_{2}\right)$ are a valid solution if and only if $d=A^{\prime} C^{\prime}$. We obtain those two solutions whenever $A^{\prime} C^{\prime}>A A^{\prime}$. By the same computation as above, the inequality $A^{\prime} C^{\prime}>A A^{\prime}$ holds if and only if $a / b>\frac{1}{2}(1+\sqrt{5})$. To construct the solutions, we simply construct circles $k_{A}, k_{B}$ with radii $A^{\prime} C^{\prime}$ and take the appropriate intersections with line $B D$.

(b) Since points $A_{1}, B_{2}$ are symmetric about $S$, a pair $\left(A_{1}, B_{2}\right)$ is a valid solution if and only if $A A_{1}=A_{1} B_{2}=2 \cdot A_{1} S$. Likewise, a pair $\left(A_{2}, B_{1}\right)$ is a valid solution if and only if $A A_{2}=2 \cdot A_{2} S$. Hence we get an extra solution for every point $X$ on $B D$ such that $A X=2 X S$. Regardless of $a$ and $b$, there are always two such points and they can be constructed for example as the intersections of line $B D$ with the Apollonius circle corresponding to points $A, S$ and ratio 2. Note that this includes the solution obtained earlier in the case $d=A A^{\prime}$.

To summarize, there are two solutions when $1 \leqslant a / b \leqslant \frac{1}{2}(1+\sqrt{5})$ and four solutions when $\frac{1}{2}(1+\sqrt{5})<a / b$.
