We have three (different) points. Construct a rectangle such that one of its vertices is one of the given points, one of its side lines passes through each of the remaining two points, and its diagonal makes a $30^{\circ}$ angle with the longer side. - How many such rectangles are possible?
I. Let's select a point $A$ from the given points, which will be the front vertex of the sought rectangle $A B C D=T$, and the other two points will be $P_{2}$ and $P_{3}$. The latter will have two different side lines of $T$ passing through them, denoted by $e_{2}$ and $e_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f0e6310b31440a4f1901g-1.jpg?height=737&width=897&top_left_y=301&top_left_x=603)

a) First, we seek a solution where two adjacent side lines of $T$ pass through $P_{2}$ and $P_{3}$, and neither $e_{2}$ nor $e_{3}$ passes through $A$. In this case, they intersect at the vertex $C$ of $T$ opposite to $A$ (1st figure), and $\angle P_{2} C P_{3} = 90^{\circ}$, and the value of $\angle A C P_{2}$ is one of $30^{\circ}, 150^{\circ}, 60^{\circ}$, and $120^{\circ}$. Accordingly, $C$ can be determined from the Thales circle $k$ with diameter $P_{2} P_{3}$ using the viewing circles $k_{1}, k_{1}^{\prime}$, and $k_{2}, k_{2}^{\prime}$ with angles $30^{\circ}$ and $60^{\circ}$ from the segment $A P_{2}$. $C$ uniquely determines $T$, as it gives the vertices $B$ and $D$ as the projections of $A$ onto $C P_{2}$ and $C P_{3}$, respectively.

b) Let's still assume that $e_{2} \perp e_{3}$, but one of them passes through $A$, let this be $e_{2}$. Then the line $A P_{2} = e_{2}$ is already a side line of $T$, and the vertex $B$ on this line is determined by the line $e_{3}$, which is perpendicular to it and passes through $P_{3}$. $C$ is determined from $e_{3}$ by the diagonal line $d$ passing through $A$ and making an angle of $30^{\circ}$ or $60^{\circ}$ with $e_{2}$, and thus $D$ is also determined (2nd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_f0e6310b31440a4f1901g-1.jpg?height=377&width=792&top_left_y=1514&top_left_x=664)

c) Finally, we seek a solution where parallel sides of $T$ pass through $P_{2}$ and $P_{3}$, and one of them passes through $A$, let this again be $e_{2}$. Then $A P_{2} = e_{2}$ is again a side line (let it be $A B$), $C D$ is parallel to it and passes through $P_{3}$, $D$ is the projection of $A$ onto $e_{3}$, $C$ is determined by the line $d$ mentioned above, and thus $B$ is also determined (3rd figure). - In all the above cases, we have provided the construction for every possible mutual position of $A, e_{2}$, and $e_{3}$. In each case, it is obvious that the rectangle, if it forms, meets the given conditions.

![](https://cdn.mathpix.com/cropped/2024_05_02_f0e6310b31440a4f1901g-1.jpg?height=323&width=735&top_left_y=2240&top_left_x=706)

II. In the solution of type a), $P_{2}$ is a common point of the mentioned 5 circles, so each of $k_{1}, k_{1}^{\prime}, k_{2}, k_{2}^{\prime}$ generally intersects $k$ again, and $C$ and $T$ are formed. If $k$ touches any of the other circles - it can only touch one of them, obviously - $P_{2}$ itself can also serve as $C$, in which case the line $C D$ is $P_{2} P_{3}$, and the line $C B$ is the common tangent, because in this case the line $C A = P_{2} A$ makes an angle of $30^{\circ}$ or $60^{\circ}$ with $C D = P_{2} P_{3}$. If $k$ passes through $A$, the other 4 circles have no common point with $k$ other than $A$ and $P_{2}$, but $P_{2}$ can only serve as $C$ if $\angle P_{3} P_{2} A = 30^{\circ}$ or $60^{\circ}$. In this case, however, one of the 4 circles is identical to $k$, and $C$ can be any point of $k$ different from $A$ (4th figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_f0e6310b31440a4f1901g-2.jpg?height=621&width=591&top_left_y=408&top_left_x=694)

4th figure

- Here we mention that if $k$ passes through $A$, there is also a special, previously unmentioned type of solution where the side lines $A B$ and $A D$ pass through $P_{2}$ and $P_{3}$; in this case, any point of the line $d$ determined in type b) different from $A$ can serve as $C$.

The solutions of types b) and c) always exist, but the solution of type c) degenerates into a line segment if $P_{3}$ is on $A P_{2}$.

III. The number of possible rectangles in type a), for a fixed $A$, is at most 4 from each of $k_{1}, k_{1}^{\prime}, k_{2}, k_{2}^{\prime}$, and in the cases of types b) and c), taking $A P_{2}$ as a side line, 4, because 4 lines can serve as $d$, and the same number if we take $A P_{3}$ as $A B$. Thus, (ignoring special cases) the number of suitable rectangles, taking each of the three points in turn as $A$, is $3(4 + 2 \cdot 4 + 2 \cdot 4) = 60$.

Juhász Ágnes (Budapest, Berzsenyi D. g. II. o. t.) Moson Péter (Budapest, Fazekas M. gyak. g. II. o. t.)