[

The Law of Cosines

[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part

A tangent $A M$ (where $M$ is the point of tangency) is drawn from point $A$ to a circle with radius $R$. A secant passing through point $A$ intersects the circle at points $K$ and $L$, with $L$ being the midpoint of segment $A K$, and the angle $A M K$ is $60^{\circ}$. Find the area of triangle $A M K$.

#
Apply the tangent-secant theorem and the cosine rule.

## Solution

Let $O$ be the center of the circle. Then $\angle M O K=120^{\circ}$ and $M K=R \sqrt{3}$. Let $A M=x$.

By the tangent-secant theorem,

$$
A M^{2}=A K \cdot A L=2 A L^{2}
$$

Therefore, $A L=\frac{x}{\sqrt{2}}$ and $A K=x \sqrt{2}$.

By the cosine rule in triangle $A M K$, we find that

$$
x^{2}+3 R^{2}-x R \sqrt{3}=2 x^{2}, \text{ or } x^{2}+x R \sqrt{3}-3 R^{2}=0
$$

Thus, $x=\frac{R(\sqrt{15}-\sqrt{3})}{2}$ (the only positive root of the obtained equation),

$$
S_{\triangle \mathrm{AMK}}=\frac{1}{2} A M \cdot M K \sin 60^{\circ}=\frac{3 R^{2}(\sqrt{15}-\sqrt{3})}{8}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-46.jpg?height=420&width=569&top_left_y=-1&top_left_x=739)

## Answer

$\frac{3 R^{2}(\sqrt{15}-\sqrt{3})}{8}$.