11.134. A regular triangular pyramid is cut by a plane perpendicular to the base and dividing two sides of the base in half. Determine the volume of the cut-off pyramid if the side of the base of the original pyramid is $a$, and the dihedral angle at the base is $45^{\circ}$.
Solution.

Let $MO$ be the height of the regular pyramid $MABC$, $\triangle ETF$ be the section of the pyramid mentioned in the problem (Fig. 11.31), and $D$ be the midpoint of

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0844.jpg?height=652&width=534&top_left_y=990&top_left_x=710)

Fig. 11.31 $AB$. Then $\angle MDO$ is the linear angle of the dihedral angle at edge $AB$, and since $\angle MDO = 45^\circ$, we have $MO = DO = \frac{a \sqrt{3}}{6}$. $EF$ is the midline of

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0845.jpg?height=456&width=567&top_left_y=224&top_left_x=80)

Fig. 11.32

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0845.jpg?height=664&width=656&top_left_y=78&top_left_x=607)

Fig. 11.33

$\triangle ABC$, then $S_{\triangle ECF} = \frac{1}{4} S_{\triangle ABC} = \frac{a^2 \sqrt{3}}{16}$, $\triangle TEC = \triangle TFC$ (by two sides and the included angle). Therefore, $TE = TF$, and $K$ is the intersection of $EF$ and $CD$. Then $K$ is the midpoint of $EF$ and $TK$ is the median and altitude of $\triangle ETF$. Since the plane ETF is perpendicular to the plane $ABC$, $TK$ is the height of the pyramid $TEFC$. $CK = \frac{1}{2} CD$, $CO = \frac{2}{3} CD$. Then $\frac{CK}{CO} = \frac{3}{4}$, $\triangle TKC \sim \triangle MOC$, and $\frac{TK}{MO} = \frac{CK}{CO} = \frac{3}{4}$; $TK = \frac{3}{4} MO = \frac{a \sqrt{3}}{8}$. The volume of the pyramid $ETFC$ $V = \frac{1}{3} S_{\triangle ECF} \cdot TK = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{16} \cdot \frac{a \sqrt{3}}{8} = \frac{a^3}{128}$.

Answer: $\frac{a^3}{128}$.