The base of the right prism $A B C A_{1} B_{1} C_{1}$ is an isosceles triangle $A B C$, where $A B=B C=5$, $\angle A B C=2 \arcsin 3 / 5$. A plane perpendicular to the line $A_{1} C$ intersects the edges $A C$ and $A_{1} C_{1}$ at points $D$ and $E$ respectively, such that $A D=1 / 3 A C, E C_{1}=1 / 3 A_{1} C_{1}$. Find the area of the section of the prism by this plane.
Let $M$ be the midpoint of the base $AC$ of the isosceles triangle $ABC$ (Fig. 1). From the right triangle $AMB$ we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-13.jpg?height=771&width=955&top_left_y=2130&top_left_x=1066)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-14.jpg?height=1441&width=946&top_left_y=63&top_left_x=1065)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-14.jpg?height=704&width=972&top_left_y=1577&top_left_x=1072)

Fig. 3

$AM = AB \sin \angle ABM = 5 \cdot \begin{aligned} & 3 \\ & 5\end{aligned} = 3, \quad BM = \sqrt{AB^2 - AM^2} = \sqrt{25 - 9} = 4$.

Then

$AC = 2AM = 6, \quad AD = \frac{1}{3}AC = 2, \quad C_1E = \frac{1}{3}A_1C_1 = 2$.

Since the line $A_1C$ is perpendicular to the cutting plane (Fig. 2), then $A_1C \perp DE$. Let $O$ be the intersection point of segments $A_1C$ and $DE$. From the equality of right triangles $A_1OE$ and $COD$ (by hypotenuse and acute angle), it follows that $O$ is the midpoint of $A_1C$. Therefore, the perpendicular dropped from point $O$ to $AC$ passes through point $M$, and since the prism is straight, this perpendicular lies in the plane of the face $AA_1C_1C$.

Let $K$ be the intersection point of the cutting plane with the line $AB$. The line $A_1C$ is perpendicular to the cutting plane containing the line $DK$, so $A_1C \perp DK$. By the theorem of three perpendiculars, $DK \perp AC$, since $AC$ is the orthogonal projection of the inclined line $A_1C$ onto the plane $ABC$. Triangles

$ADK$ and $AMB$ are similar with a coefficient $k = \frac{AD}{AM} = \frac{2}{3}$, so

$AK = k \cdot AB = \frac{2}{3} AB < AB$.

Thus, point $K$ lies on the segment $AB$. Similarly, point $P$ of the intersection of the plane with the line $B_1C_1$ lies on the edge $B_1C_1$. In this case, $EP \parallel DK$ and $EP = DK$.

Let the lines $ED$ and $AA_1$ intersect at point $L$, and the lines $LK$ and $BB_1$ intersect at point $N$. Then the section of the prism in question is the pentagon $DKNPE$. Its orthogonal projection onto the plane $ABC$ is the pentagon $DKBP_2E_2$ (Fig. 1, 2), where $P_2$ and $E_2$ are the orthogonal projections of points $P$ and $E$ onto the plane $ABC$. Then

$S_{\triangle CE_2P_2} = S_{\triangle ADK} = k^2 \cdot S_{\triangle AMB} = k^2 \cdot \frac{1}{2} \cdot AM \cdot BM = \left(\frac{2}{3}\right)^2 \cdot \frac{1}{2} \cdot 3 \cdot 4 = \frac{8}{3}$,

$S_{DKBP_2E_2} = S_{\triangle ABC} - 2S_{\triangle ADK} = \frac{1}{2} \cdot AC \cdot BM - 2S_{\triangle ADK} = \frac{1}{2} \cdot 6 \cdot 4 - 2 \cdot \frac{8}{3} = \frac{20}{3}$.

Let $h$ be the height (lateral edge) of the prism. Consider the face $AA_1C_1C$ (Fig. 3). Denote $\angle A_1CC_1 = \angle EE_2D = \alpha$. From the right triangles $CA_1C_1$ and $DEE_2$ we find that

$\tan \alpha = \frac{A_1C_1}{CC_1} = \frac{6}{h}$ and $\tan \alpha = \frac{EE_2}{DE_2} = \frac{h}{2}$,

from which $h = 2\sqrt{3}$ and $\tan \alpha = \sqrt{3}$. Therefore, $\alpha = 60^\circ$.

Since $CD \perp KD$ and $ED \perp KD$, the angle $EDC$ is the linear angle of the dihedral angle between the cutting plane and the plane $ABC$, and since $\angle EDC = \alpha = 60^\circ$, then, by the theorem of the area of the orthogonal projection of a polygon,

$S_{DKNPE} = \frac{S_{DKBP_2E_2}}{\cos \alpha} = \frac{\frac{20}{3}}{\frac{1}{2}} = \frac{40}{3}$.