In square $A B C D$, point $M$ lies on side $B C$, and point $N$ lies on side $A B$. Lines $A M$ and $D N$ intersect at point $O$. Find the area of the square, given that $D N=4, A M=3$, and the cosine of angle $D O A$ is $q$.

#
Denote

$$
\angle A O D=\alpha, \angle A M B=\angle D A M=\beta, \angle A D N=\gamma
$$

and form a trigonometric equation with respect to $\gamma$.

## Solution

Let the side of the square be $a$. Denote,

$$
\angle A O D=\alpha, \angle A M B=\angle D A M=\beta, \angle A D N=\gamma
$$

From the right triangles $A B M$ and $D A N$, we find that

$$
\sin \beta=\frac{A B}{A M}=\frac{a}{3}, \cos \gamma=\frac{A D}{D N}=\frac{a}{4}
$$

From this, it follows that

$$
4 \cos \gamma=3 \sin \beta=3 \sin \left(180^{\circ}-\alpha-\gamma\right)=3 \sin (\alpha+\gamma)=3 \sin \alpha \cos \gamma+3 \cos \alpha \sin \gamma
$$

Dividing both sides of the equation

$$
4 \cos \gamma=3 \sin \alpha \cos \gamma+3 \cos \alpha \sin \gamma
$$

by $\cos \gamma$, we get

$$
3 \sin \alpha+3 \cos \alpha \operatorname{tg} \gamma=4
$$

from which

$$
\operatorname{tg} \gamma=\frac{4-3 \sin \alpha}{3 \cos \alpha}=\frac{4-3 \sqrt{1-q^{2}}}{3 q}
$$

Then

$$
\cos ^{2} \gamma=\frac{1}{1+\operatorname{tg}^{2} \gamma}=\frac{9 q^{2}}{9 q^{2}+16-24 \sqrt{1-q^{2}}+9\left(1-q^{2}\right)}=\frac{9 q^{2}}{25-24 \sqrt{1-q^{2}}}
$$

Therefore,

$$
S_{\mathrm{ABCD}}=a^{2}=(4 \cos \gamma)^{2}=\frac{144 q^{2}}{25-24 \sqrt{1-q^{2}}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-48.jpg?height=515&width=535&top_left_y=1406&top_left_x=756)

## Answer

$\frac{144 q^{2}}{25-24 \sqrt{1-q^{2}}}$

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