Given two non-intersecting circles. Construct a right-angled triangle, whose hypotenuse is the common external tangent of the two circles, the legs touch one of the given circles each, and one of the common internal tangents of the two circles passes through the right-angle vertex.
We ignore those trivial solutions where one of the legs is itself a common internal tangent (i.e., this leg touches both circles). The problem clearly does not refer to these cases, because in such a case, it would be superfluous to separately require that "one of the common internal tangents passes through the right angle vertex."

Our diagram shows the solved problem and the labeling.

![](https://cdn.mathpix.com/cropped/2024_05_02_2eca3b782308e1cbf905g-1.jpg?height=700&width=897&top_left_y=360&top_left_x=603)

Clearly, it is sufficient to determine the right-angle vertex $C$ of the right triangle on the common internal tangent and the perpendicular lines through it, each of which touches one of the circles. Each common external tangent provides a solution with each of these pairs of perpendicular lines.

Consider a solution $C_{1}$ (on the common internal tangent $t_{1}$) where $O_{1}$ and $O_{2}$ fall into one of the four right-angle regions defined by the two tangents. In this case, the angle subtended by one circle from $C_{1}$ is $2 \alpha$, the angle subtended by the other circle is $2 \beta$, and the segment $O_{1} O_{2}$ is seen at an angle of $\alpha + \beta$. Since $2 \alpha + 2 \beta = 90^{\circ}$, it follows that $\alpha + \beta = 45^{\circ}$.

Consider another solution $C_{1}^{\prime}$ (also on the tangent $t_{1}$) where the points $O_{1}$ and $O_{2}$ do not satisfy the previous conditions. In this case, the angle subtended by one circle from $C_{1}^{\prime}$ is $2 \gamma$, the angle subtended by the other circle is $2 \delta$, and the segment $O_{1} O_{2}$ is seen at an angle of $\gamma + \delta + \varepsilon$. We have $2 \gamma + \varepsilon = 180^{\circ}$ and $2 \delta + \varepsilon = 90^{\circ}$, so $2 \gamma + 2 \delta + 2 \varepsilon = 270^{\circ}$, which means $\gamma + \delta + \varepsilon = 135^{\circ}$.

Therefore, the geometric locus of the points $C$ is the entire circle from which the segment $O_{1} O_{2}$ is seen at an angle of $45^{\circ}$ or $135^{\circ}$. Circle $\mathrm{E}$ intersects 4 points $\left(C_{1}, C_{1}^{\prime}, C_{2}, C_{2}^{\prime}\right)$ from the common internal tangents $t_{1}$ and $t_{2}$. These 4 points determine 4 pairs of perpendicular tangents, each of which, together with one of the common external tangents, provides a solution. Thus, a total of 8 (differing in size, shape, and position) right triangles can be constructed, some of which may be degenerate, where a common external tangent is parallel to one of the legs. (If we also count the reflections with respect to the line $O_{1} O_{2}$, the problem has a total of 16 solutions.)

Péter Szántó (Bp. I., Petőfi g. II. o. t.)

A complete solution is not provided beyond this.