Given a convex angle and a point $P$ outside it. Draw a line through $P$ that cuts off a triangle of a given perimeter from the angle domain. What is the condition for constructibility?

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Let $A$ be the vertex of the convex angle, $e, f$ the rays. Suppose we have already drawn the desired line through $P$. Let this line intersect $e$ and $f$ at points $B$ and $C$, respectively. Consider the circle $k$ inscribed to the side $BC$ of triangle $ABC$, touching $e$ and $f$ at points $E$ and $F$, respectively. It is known that if $2s$ denotes the perimeter of triangle $ABC$, then $AE = AF = s$.

![](https://cdn.mathpix.com/cropped/2024_05_02_31e6c52144d65c500843g-1.jpg?height=664&width=962&top_left_y=356&top_left_x=560)

Based on this, the construction is as follows: the points $E$ and $F$ can be constructed from the given perimeter. Then we construct the circle that touches the sides of the angle at points $E$ and $F$. The line $EF$ divides the plane into two half-planes, one of which has $A$ as an interior point. Drawing a tangent from $P$ to the arc of circle $k$ that lies in this half-plane gives us the line of side $BC$. The perimeter of this triangle is the desired one, since $k$ is the circle inscribed to the side $BC$ of triangle $ABC$.

![](https://cdn.mathpix.com/cropped/2024_05_02_31e6c52144d65c500843g-1.jpg?height=631&width=1071&top_left_y=1300&top_left_x=516)

Since $k$ is always constructible, the condition for solvability is that a tangent can be drawn from $P$ to the arc $\sum$. If we draw the tangent at every point of $\sum$, the resulting set of lines covers every point of the plane outside the angle sector, but not any point of the vertex angle or its boundary. Therefore, the condition for solvability is that $P$ is not in the vertex angle sector or on its boundary.