In the equilateral triangle $ABC$, on the side $AB$, we draw the perpendicular $AD$ at $A$ such that $AD=AB$. Calculate the angles, sides, and area of the obtuse triangle $ADC$ if $AB=a$.

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In the obtuse isosceles triangle $A D C$, $\angle C A D = 60^{\circ} + 90^{\circ} = 150^{\circ}$, and each of the base angles is $15^{\circ}$. Given that $A C = A D = a$, to calculate $C D$, draw a line through $D$ parallel to $A B$, which intersects the perpendicular from $C$ to $A B$ at $E$.

In the right triangle $D C E$, $C E = a + \frac{a}{2} \sqrt{3}$, $D E = \frac{a}{2}$, and thus

$$
\overline{C D}^{2} = \left(\frac{a}{2}\right)^{2} + \left(a + \frac{a}{2} \sqrt{3}\right)^{2}
$$

from which

$$
C D = a \sqrt{2 + \sqrt{3}}
$$

The area of triangle $C D E$ is

$$
\frac{A D \times D E}{2} = \frac{a \times a}{2 \times 2} = \frac{a^{2}}{4}
$$

(Kornél Miklóssy, Arad.)

The problem was also solved by: V. Erdős, M. Fekete, B. Friedländer, P. Füstös, P. Heimlich, Gy. Jánosy, J. Kiss, J. Martini, Gy. Meleghy, L. Neumann, E. Peikert, S. Pichler, E. Rajz, E. Sárközy, Gy. Schuster, J. Steiger, E. Tandlich, B. Tóth, E. Vetter, V. Wáhl.