## PROBLEM 7. SCREENS IN THE CORNER

A corner of a rectangular room is partitioned off by two four-foot screens. At what position will the enclosed area be the largest?
Solution. The solution is based on the double application of the following well-known result.

Lemma. Among the set of triangles with a given base $b$ and the angle $\theta$ opposite to it, the isosceles triangle has the largest area. (It is clear that this set is contained within a segment of a circle, and the height of the isosceles triangle is greater than that of other triangles in this set (Fig. 10).)
![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-17.jpg?height=502&width=1148&top_left_y=608&top_left_x=454)

Fig. 10

Let $O$ be the vertex of the room's angle, and the points of contact of the screens with the sides of the angles be $A$ and $B$. Then the area enclosed by the screens is the sum of the areas of triangles $A O B$ and $A B P$, where $P$ is the point of contact of the screens. (It is clear that at the maximum area, point $P$ will be a boundary point of both screens.)

Now, if $\triangle O A B$ is not isosceles ($O A \neq O B$), then the screens can be moved to the positions $A^{\prime} P^{\prime}$ and $B^{\prime} P^{\prime}$, where $O A^{\prime}=O B^{\prime}$, such that the segment $A^{\prime} B^{\prime}$ equals the segment $A B$. Then the triangles $A^{\prime} B^{\prime} P^{\prime}$ and $A B P$
![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-17.jpg?height=538&width=1080&top_left_y=1774&top_left_x=501)

Fig. 11

are equal, and by the lemma, the area of triangle $O A^{\prime} B^{\prime}$ is greater than the area of triangle $O A B$. Thus, in the arrangement that gives the maximum area, the segment $O A$ must be equal to the segment $O B$.

Since the screens have equal lengths, point $P$ must always lie on the perpendicular bisector of segment $A B$, but if $O A=O B$, then the perpendicular bisector of $A B$ is the bisector of the right angle $O$. Thus, at the maximum area, segment $O P$ must be the bisector of angle $O$ (Fig. 11). Consequently, each screen must cut off a triangle of maximum area from the $45^{\circ}$ angle subtended by this segment. (Regardless of the optimal placement.)

Since the length of segment $A P$ is constant, it follows from the lemma that the area of triangle $O A B$ will be maximal in the case of its isosceles nature ($O A=O P$). In this case, $\angle O A P=67.5^{\circ}$ and the solution becomes obvious. As a simple exercise, we suggest constructing segment $O P$ using a compass and a ruler.