12.3. The acute isosceles triangle \(ABC\), with \(m(\angle B) = m(\angle C) = \alpha\), is the base of the prism \(ABC A_1 B_1 C_1\). The lateral edge \(A_1 A\) is perpendicular to the edge \(AC\), and \(m(\angle A_1 A B) = \beta < 90^\circ\). Determine the lateral area of the prism, if \(A_1 A = BC = a\).
## Solution:

Given that $A_{1} A C C_{1}$ is a rectangle, $A C=A B=\frac{a}{2 \cos \alpha}$, we obtain $\mathcal{A}_{A_{1} A C C_{1}}=\frac{a^{2}}{2 \cos \alpha}$. $\mathcal{A}_{A_{1} A B B_{1}}=\frac{a^{2}}{2 \cos \alpha} \sin \beta$.

Consider the line $d \| A B, C \in d$. Let $K \in(A B C),\left(C_{1} K\right) \perp(A B C)$. Let $T \in d,(K T) \perp d$.

Then $m\left(\angle C_{1} C T\right)=\beta$ and $C T=a \cos \beta$.

$(K C) \perp(A C) \Rightarrow m(\angle K C T)=2 \alpha-90^{\circ}$

and $m(\angle K C B)=90^{\circ}-\alpha$.

Then $K C=\frac{C T}{\sin (2 \alpha)}=\frac{a \cos \beta}{\sin (2 \alpha)}$ and

$C_{1} K^{2}=C_{1} C^{2}-K C^{2}=a^{2}-\frac{a^{2} \cos ^{2} \beta}{\sin ^{2}(2 \alpha)}$.

Let $M \in B C,(K M) \perp(B C)$. Then

![](https://cdn.mathpix.com/cropped/2024_06_06_5d61f345a4435a18791ag-2.jpg?height=848&width=899&top_left_y=172&top_left_x=987)

$K M=K C \sin \left(90^{\circ}-\alpha\right)=\frac{a \cos \beta}{\sin (2 \alpha)} \cos \alpha$.

$$
\begin{aligned}
& C_{1} M^{2}=C_{1} K^{2}+K M^{2} \\
&=a^{2}\left(1-\frac{\cos ^{2} \beta}{\sin ^{2}(2 \alpha)}\right. \\
&+\left.\frac{\cos ^{2} \beta}{\sin ^{2}(2 \alpha)} \cos ^{2} \alpha\right)= \\
& \quad=a^{2}\left(1-\frac{\cos ^{2} \beta}{4 \cos ^{2} \alpha}\right)=a^{2} \frac{4 \cos ^{2} \alpha-\cos ^{2} \beta}{4 \cos ^{2} \alpha}
\end{aligned}
$$

Then $\mathcal{A}_{B C C_{1} B_{1}}=B C \cdot C_{1} M=\frac{a^{2} \sqrt{4 \cos ^{2} \alpha-\cos ^{2} \beta}}{2 \cos \alpha}$.

We obtain

$$
\mathcal{A}_{\text {lat. }}=\frac{a^{2}}{2 \cos \alpha}\left(1+\sin \beta+\sqrt{4 \cos ^{2} \alpha-\cos ^{2} \beta}\right)
$$