In an acute-angled triangle $A B C$, altitudes $A M$ and $C N$ are drawn, and $O$ is the center of the circumscribed circle. It is known that $\angle B=\beta$, and the area of quadrilateral $N O M B$ is $S$. Find $A C$.

In an acute-angled triangle $A B C$, altitudes $A M$ and $C N$ are drawn, and $O$ is the center of the circumscribed circle. It is known that $\angle B=\beta$, and the area of quadrilateral $N O M B$ is $S$. Find $A C$.
The diagonals of quadrilateral NOMB are perpendicular to each other.

## Solution

On the tangent to the circumscribed circle $\Omega$, drawn at point $B$, mark point $P$ such that it and point $C$ lie on opposite sides of line $A B$. Since $\angle A B P=\angle C=\angle B N M$, then $B P \| M N$, and $O B \perp M N$. Triangle $B M N$ is similar to triangle $B A C$ with a coefficient of $\cos \beta$, so $M N=A C \cos \beta$.
![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-01.jpg?height=482&width=1080&top_left_y=992&top_left_x=480)

Moreover, $A C=2 O B \sin \beta$. Therefore, $S=1 / 2 M N \cdot O B=1 / 4 A C^{2} \operatorname{ctg} \beta$, from which $A C=2$

## Answer

$2 \sqrt{S \operatorname{tg} \beta}$.

## [ Auxiliary circle $\quad]$  Problem 52814 topics: [Angles subtending equal arcs and equal chords] ]

On a plane, two squares $A B C D$ and $B K L N$ are placed such that point $K$ lies on the extension of $A B$ beyond point $B$, and $N$ lies on the ray $B C$.

Find the angle between the lines $D L$ and $A N$.

## Hint

Describe circles around the given squares and consider their point of intersection, different from B.

## Solution

Let $B N<B C$. Describe circles around the given squares and denote by $M$ their point of intersection, different from $B$. Then $\angle B M L=\angle B M D=90^{\circ}$. Therefore, points $D, M$ and $L$ lie on the same line. Since $\angle D M A=\angle D C A=45^{\circ}, \angle L M N=135^{\circ}$, then points $M, N$ and $A$ also lie on the same line. Consequently, the desired angle is the angle $D M N$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-02.jpg?height=277&width=429&top_left_y=-1&top_left_x=820)

## Answer

$45^{\circ}$.