A horizontally lying square has 4 trees standing at its 4 vertices, the tops of which are at equal distances from point $O$ inside the square. The height of the first tree is $a$, the second is $b$, and the third is $c$. What is the height of the fourth tree?
Let $ABCD$ be the square, in whose vertices the trees $a, b, c$, and $d$ stand. From point 0 inside the square, we draw lines parallel to the sides of the square, which intersect the sides $AB, BC, CD$, and $DA$ at points $F, G, E$, and $H$. According to the problem, if we denote the distances from the tops of the trees to 0 by $r$:

$$
a^{2}=r^{2}-\overline{D E}^{2}-\overline{B G}^{2}
$$

$$
b^{2}=r^{2}-\overline{E C}^{2}-\overline{B G}^{2}
$$

$$
c^{2}=r^{2}-\overline{E C}^{2}-\overline{C G}^{2}
$$

$$
d^{2}=r^{2}-\overline{D E}^{2}-\overline{C G}^{2}
$$

Substituting the values of $\overline{D E}^{2}$ and $\overline{C G}^{2}$ from (1) and (3) into (4):

$$
d^{2}=r^{2}+a^{2}-r^{2}+\overline{B G}^{2}+c^{2}-r^{2}+\overline{E C}^{2}
$$

or, considering (2):

$$
d^{2}=a^{2}+c^{2}-b^{2}
$$

and thus

$$
d=\sqrt{a^{2}+c^{2}-b^{2}}
$$

(Endre Mellinger, Budapest.)

The problem was also solved by: M. Fekete, H. Fodor, P. Hajdú, Gy. Jánosy, E. Kiss, C. Neubauer, S. Pichler, P. Sárközy, Gy. Schuster, Gy. Schwarz, Term. tud. kör, Budapest. VII. ker.