Let's draw the orthic triangle of triangle $ABC$. What are the angles of the original triangle if the angles of the orthic triangle are in the ratio $2: 6: 7$?
Let the angles of the pedal triangle $A_{1} B_{1} C_{1}$ be $2 x, 6 x, 7 x$. Then

$$
2 x+6 x+7 x=180^{\circ},
$$

from which $x=12^{\circ}$, so

$$
\begin{gathered}
\angle B_{1} A_{1} C_{1}=\angle A_{1}=24^{\circ}, \\
\angle A_{1} B_{1} C_{1}=\angle B_{1}=72^{\circ} \text { and } \angle A_{1} C_{1} B_{1}=\angle C_{1}=84^{\circ} .
\end{gathered}
$$

The quadrilateral $A B A_{1} B_{1}$ is a cyclic quadrilateral because both $A_{1}$ and $B_{1}$ lie on the semicircle drawn over the diameter $A B$, thus

$$
\angle A=180^{\circ}-\angle B A_{1} B_{1} .
$$

Similarly, from the cyclic quadrilateral $A C A_{1} C_{1}$,

$$
\angle A=180^{\circ}-\angle C A_{1} C_{1},
$$

so

$$
2 \angle A=360^{\circ}-\left(\angle B A_{1} B_{1}+\angle C A_{1} C_{1}\right)=360^{\circ}-\left(180^{\circ}+\angle A_{1}\right),
$$

which means

$$
\angle A=90^{\circ}-\frac{\angle A_{1}}{2}=90^{\circ}-12^{\circ}=78^{\circ} .
$$

Similarly,

$$
\angle B=90^{\circ}-\frac{\angle B_{1}}{2}=90^{\circ}-36^{\circ}=54^{\circ}
$$

and

$$
\angle C=90^{\circ}-\frac{\angle C_{1}}{2}=90^{\circ}-42^{\circ}=48^{\circ} .
$$

(Tibor Pálos, Budapest.)