From a point outside a circle, tangents and a secant are drawn, and the points of tangency and the points of intersection of the secant with the circle are the vertices of a certain trapezoid. Find the ratio of the bases of the trapezoid, given that the angle between the tangents is $60^{\circ}$.
Let $S$ be a given point, $S B D$ a secant, $A$ and $C$ the points of tangency, and $A B \| C D$. Then triangle $S B C$ is similar to triangle $S C D$, and triangle $S B A$ is similar to triangle $S A D$.

## Solution

Let $S$ be a given point, $S B D$ a secant, $S A$ and $S C$ tangents, and $A B \| D C$. Denote $A B = x, C D = y$. From the similarity of triangles $S B C$ and $S C D$ (by two angles), we have $\frac{B C}{D C} = \frac{S B}{C S}$, and from the similarity of triangles $S B A$ and $S A D$, we have $\frac{A B}{A D} = \frac{S B}{A S}$.

Since $C S = A S$, we have $\frac{A B}{A D} = \frac{B C}{D C}$, or $A B \cdot D C = x y = B C^2$. From this, we find that $B C = \sqrt{x y}$. Additionally,

$$
\angle B C D = \angle A D C = 60^{\circ}.
$$

If $P$ is the projection of point $B$ onto the base $C D$ of trapezoid $A B C D$, then $P C = \frac{y - x}{2}$. In triangle $B P C$, we have:

$$
P C = \frac{1}{2} B C \text{, or } \frac{y - x}{2} = \frac{\sqrt{x y}}{2} \text{, or } \frac{x}{y} + \sqrt{\frac{x}{y}} - 1 = 0 \text{.}
$$

From this equation, we find that

$$
\frac{x}{y} = \frac{3 - \sqrt{5}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-21.jpg?height=449&width=783&top_left_y=2182&top_left_x=640)

Answer

$\frac{3 - \sqrt{5}}{2}$.

Using a compass and a straightedge, draw a line through a given point that cuts off a triangle of a given perimeter from a given angle.

## Hint

The desired line is a tangent to a certain circle inscribed in this angle.

## Solution

Let $A$ be the vertex of the given angle, and $2 p$ the given perimeter. On the sides of the angle, mark points $B$ and $C$ such that $A B = A C = p$. Inscribe a circle in the angle, touching its sides at points $B$ and $C$.

Let $M$ be a point inside the given angle. If point $M$ is inside the curvilinear triangle $A B C$ bounded by segments $A B, A C$ and the smaller arc $B C$ of the circle, then draw tangents to the circle through point $M$.

Let one of the drawn tangents intersect the sides $A B$ and $A C$ of the given angle at points $P$ and $Q$ respectively and touch the circle at point $K$. Then

$$
\begin{gathered}
A P + P Q + Q A = A P + (P K + K Q) + Q A = (A P + P K) + (K Q + Q A) = \\
= (A P + P B) + (C Q + Q A) = A B + A C = 2 p
\end{gathered}
$$

Similarly for the second tangent. Thus, in this case, the problem has two solutions (if point $M$ lies on the smaller arc $B C$ and is different from points $B$ and $C$, then these solutions coincide).

If point $M$ lies inside the angle but outside the specified curvilinear triangle, then the problem has no solutions. Now let point $M$ be located outside the angle. If one of the tangents to the constructed circle cuts off a triangle from the given angle for which this circle is an exscribed circle, then the cut-off triangle is the desired one (the proof is similar to the one given above). In this case, the problem has a unique solution. In other cases, there are no solutions.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-22.jpg?height=414&width=688&top_left_y=1371&top_left_x=685)

Problem