Draw a circle and a point $P$ inside the circle. Construct an isosceles triangle inscribed in the circle such that one of its legs passes through the point $P$ and the angle at the vertex is $45^{\circ}$.
I. solution: Since the central angle corresponding to the base of the sought isosceles triangle (which is twice the inscribed angle) is $90^{\circ}$, the central angle corresponding to each leg is $\frac{360^{\circ}-90^{\circ}}{2}=135^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d49c7322dfc53e14e43dg-1.jpg?height=496&width=488&top_left_y=267&top_left_x=797)

Therefore, we construct a chord corresponding to a $135^{\circ}$ central angle in any position and draw a concentric circle tangent to this chord. We need to select the sides of the isosceles triangle passing through point $P$ from the tangents of this circle, i.e., the two tangents drawn from $P$ to the concentric circle (using the Thales circle) will be the sought legs. Since any endpoint of these two legs can be considered as the vertex of the sought isosceles triangle, we get a total of 4 solutions. (On the diagram, 2 results are drawn with thick lines, and 2 with thin lines). A solution exists only if the distance $P O$ is $\geqq$ the radius of the concentric circle. (The radius of the concentric circle is $r \cdot \sin \frac{45^{\circ}}{2}$, where $r$ is the radius of the given circle. Therefore, the solvability condition is: $P O \geqq r \sin 22^{\circ} 30'$. In the case of equality, there are only 2 solutions.)

Deseô Zoltán (Bp. X., I. László g. II. o. t.)

II. solution: Construct an isosceles triangle with a $45^{\circ}$ vertex angle in any position within the circle and determine the (generally) 4 intersection points of this triangle's legs with the concentric circle of radius $O P$. If we rotate each of the four intersection points around $O$ to $P$ and with it the entire adopted triangle (by the same angle), we will obtain the 4 solutions to our problem. This can be done without angle copying, simply by taking the segments established on the legs by the individual intersection points with the compass and drawing arcs from $P$ with these radii, which will intersect to give the sought vertices. The number of solutions is 4, 2, or 0 depending on whether the concentric circle of radius $O P$ intersects the two legs at 4 different points, or touches them at 1-1 point, or does not intersect them at all. (This is equivalent to what we established in the first solution.)

Biczó Géza (Bp. II., Rákóczi g. I. o. t.)

III. solution: Since the center of the given circle is $O$, the bisector of the vertex angle of the sought isosceles triangle necessarily lies on it, so the distance $P O$ is seen from the sought vertices at an angle of $\frac{45^{\circ}}{2}=22.5^{\circ}$. If we therefore construct the geometric locus of points from which the segment $O P$ is seen at an angle of $22.5^{\circ}$ (two $O P$-symmetric "viewing arcs" whose central angle is $315^{\circ}$ and whose endpoints are $O$ and $P$), then the two arcs intersect the given circle at the 4 vertices of the isosceles triangles forming the solution. (The radius of the two arcs $=\frac{O P}{2}: \sin 22.5^{\circ}=\frac{O P}{2 \sin 22.5^{\circ}}$. A solution exists only if the radius $\geqq \frac{r}{2}$, or $O P \geqq r \sin 22.5^{\circ}$, which is the same result as obtained in the first solution.) Schmidt Eligius (Bp. I., Fürst Sándor g. II. o. t.)