Given an isosceles right triangle $A B S$ with the base $A B$. On the circle with center at point $S$ and passing through points $A$ and $B$, there is a point $C$ such that triangle $A B C$ is isosceles.

Determine how many points $C$ satisfy the given conditions, and construct all such points.

(K. Pazourek)

Hint. What is an isosceles triangle?
Side $AB$ of the isosceles triangle $ABC$ can be either its base or one of its legs. Depending on this, we divide the solution into two parts.

a) Side $AB$ is the base of the isosceles triangle $ABC$. In this case, $C$ is the main vertex of the triangle $ABC$ and lies on its axis of symmetry. The axis of symmetry of the triangle $ABC$ is the perpendicular bisector of segment $AB$. This line intersects the given circle at two points, which are two possible solutions to the problem, denoted as $C_{1}$ and $C_{2}$.
b) Side $AB$ is a leg of the isosceles triangle $ABC$. In this case, the main vertex of the triangle $ABC$ can be either point $A$ or point $B$. If the main vertex were point $A$, then side $AC$ would be a leg and point $C$ would be equidistant from point $A$ as point $B$. Therefore, point $C$ would lie on the circle centered at point $A$ passing through point $B$. This circle intersects the given circle at one additional point, which we denote as $C_{3}$.

Similarly, if the main vertex were point $B$, then the remaining vertex of the triangle would lie on the circle centered at point $B$ passing through point $A$. The corresponding intersection with the given circle is denoted as $C_{4}$.

On the given circle, there are four points $C_{1}, C_{2}, C_{3}, C_{4}$ that satisfy the conditions of the problem. The construction of all points follows from the previous description: points $C_{3}$ and $C_{4}$ are the intersections of the given circle with two auxiliary circles; the perpendicular bisector of segment $AB$, and thus points $C_{1}$ and $C_{2}$, is determined by the common points of these two auxiliary circles.

![](https://cdn.mathpix.com/cropped/2024_04_17_848c3145926e37c65b12g-5.jpg?height=757&width=920&top_left_y=855&top_left_x=568)

Note. The right-angled triangle $ABS$ implies that points $A, B, C_{3}, C_{4}$ form the vertices of a square. From this, alternative constructions of the relevant points can be derived.