[ Similar triangles (etc.)
$\left[\begin{array}{l}\text { Isosceles, inscribed, and circumscribed trapezoids }\end{array}\right]$
$\left[\begin{array}{l}\text { Area of a trapezoid }\end{array}\right]$
$[$ Application of trigonometric formulas (geometry).]

Vertex $C$ of rectangle $A B C D$ lies on side $K M$ of isosceles trapezoid $A B K M$ ( $B K \| A M$ ), $P$ is the point of intersection of segments $A M$ and $C D$.

Find the angles of the trapezoid and the ratio of the areas of the rectangle and the trapezoid, if $A B=2 B C, A P=3 B K$.
Through point $C$, draw a line parallel to $A M$, until it intersects $A B$ at point $T$, and consider the isosceles trapezoid $T B K C$.

## Solution

Let $T$ be a point on side $A B$ such that $T C \| A M$, and let $O$ be the point of intersection of the diagonals of the isosceles trapezoid $T B K C$. Denote $T K = B C = 4 x$, $\angle B A M = \alpha$. Then $A B = 2 B C = 8 x$, $S_{A B C D} = A B \cdot B C = 32 x^{2}$.

From the similarity of triangles $B O K$ and $C O T$, it follows that $B O : O C = B K : T C = B K : A P = 32 x^{2}$. Therefore, $B O = x$, $T O = 3 x$, $B T^{2} = T O^{2} - B O^{2} = 8 x^{2}$. Consequently,
$\operatorname{tg} \alpha = \operatorname{tg} \angle B T C = B C / B T = \sqrt{2}$, $T C = B C / \sin \alpha = 4 x / \sin \alpha$.

Let $B F$ be the height of trapezoid $A B K M$. Then $B F = A B \sin \alpha = 8 x \sin \alpha$, $A F = 8 x \cos \alpha$, and the midline of this trapezoid is $B K + A F = \frac{T C}{3} + A F = \frac{4 x}{3 \sin \alpha} + 8 x \cos \alpha = \frac{4 x (1 + 3 \sin 2 \alpha)}{3 \sin \alpha}$, $S_{A B K M} = \frac{4 x (1 + 3 \sin 2 \alpha)}{3 \sin \alpha} \cdot 8 x \sin \alpha = \frac{32 x^{2} (1 + 3 \sin 2 \alpha)}{3} = \frac{32 x^{2} (1 + 2 \sqrt{2})}{3}$. Therefore, $\frac{S_{A B C D}}{S_{A B K M}} = \frac{3}{1 + 2 \sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-42.jpg?height=435&width=531&top_left_y=477&top_left_x=772)

## Answer

$\angle B A M = \operatorname{arctg} \sqrt{2} ; \frac{3}{1 + 2 \sqrt{2}}$.