The volume of a regular quadrilateral pyramid is $V$, and the angle between a lateral edge and the plane of the base is $30^{\circ}$. Consider regular triangular prisms inscribed in the pyramid such that one of the lateral edges lies on the diagonal of the pyramid's base, one of the lateral faces is parallel to the base of the pyramid, and the vertices of this face lie on the lateral faces of the pyramid. Find: a) the volume of the prism whose lateral face plane divides the height of the pyramid in the ratio 2:3, counting from the vertex; b) the maximum value of the volume of the considered prisms.
a) Let's denote by \(a\) the side of the base \(ABCD\) of the given regular pyramid \(PABCD\). Suppose a plane, parallel to the base of the pyramid and passing through a point \(Q\) on the height \(PO\) of the pyramid, divides the height in the ratio \(\frac{PQ}{Q} = \frac{2}{3}\). Then, in the section of the pyramid by this plane, a square is obtained, with the vertices of the opposite lateral face \(LMM1L1\) of the prism lying on the sides \(A1D1\), \(A1B1\), \(B1C1\), \(C1D1\) of the square \(A1B1C1D1\). From the right triangle \(AOP\), we find that

\[
PO = AO \operatorname{tg} \angle OAP = \frac{a \sqrt{2}}{2} \cdot \operatorname{tg} 30^{\circ} = \frac{a \sqrt{2}}{2 \sqrt{3}}
\]

Then, if \(KF\) is the height of the equilateral triangle \(KLM\), we have

\[
KF = OQ = \frac{3}{5} PO = \frac{3}{5} \cdot \frac{a \sqrt{2}}{2 \sqrt{3}} = \frac{a \sqrt{6}}{10}
\]

Let \(b\) be the side of the base of the prism. Then \(KF = \frac{b \sqrt{3}}{2}\). From the equation \(\frac{\frac{b \sqrt{3}}{2}}{2} = \frac{a \sqrt{6}}{10}\), we find that \(b = \frac{a \sqrt{2}}{5}\). Let \(LL1 = MM1 = KK1 = h\). Since the rectangle \(LMM1L1\) is inscribed in the square \(A1B1C1D1\), and its sides are parallel to the diagonals of the square, the perimeter of the rectangle is equal to the sum of the diagonals of the square, i.e., \(2h + 2b = 2 \cdot \frac{2}{5} a \sqrt{2}\). Therefore,

\[
h = \frac{2}{5} a \sqrt{2} - \frac{a \sqrt{2}}{5} = \frac{a \sqrt{2}}{5}
\]

Thus,

\[
V_{KLMK1L1M1} = S_{\Delta KLM} \cdot KK1 = \frac{b^2 \sqrt{3}}{4} \cdot h = \frac{2a^2}{25} \cdot \frac{\sqrt{3}}{4} \cdot \frac{a \sqrt{2}}{5} = \frac{a^3 \sqrt{6}}{250}
\]

Express the found volume in terms of the volume \(V\) of the given pyramid:

\[
V = V_{PABCD} = \frac{1}{3} S_{ABCD} \cdot OP = \frac{1}{3} a^2 \cdot \frac{a \sqrt{2}}{2 \sqrt{3}} = \frac{a^3 \sqrt{6}}{18}
\]

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-10.jpg?height=106&width=618&top_left_y=2031&top_left_x=720)

b) Let now \(\frac{PQ}{PO} = x (0 < x < 1)\). Then

\[
A1B1 = ax, \quad KF = OQ = (1-x) PO = \frac{(1-x) a \sqrt{6}}{6}, \quad b = LM = \frac{(1-x) a \sqrt{2}}{3}.
\]

From the equation \(h + \frac{(1-x) a \sqrt{2}}{3} = ax \sqrt{2}\), we find that

\[
KK1 = LL1 = h = ax \sqrt{2} - \frac{a(1-x) \sqrt{2}}{3} = ax \sqrt{2} = \frac{a(4x-1) \sqrt{2}}{3}
\]

Then

\[
\begin{gathered}
V_{KLMK1L1M1} = V(x) = S_{\Delta KLM} \cdot KK1 = \\
= \frac{b^2 \sqrt{3}}{4} \cdot h = \frac{(1-x)^2 a^2 \sqrt{3}}{18} \cdot \frac{(4x-1) \sqrt{2}}{3} = \frac{a^3 \sqrt{6}}{54}(1-x)^2(4x-1).
\end{gathered}
\]

It remains to find the maximum value of the function \(V(x)\) on the interval \(\left(\frac{1}{4}; 1\right)\).

First method.

By solving the equation \(V'(x) = 0\), we find the critical points of the function \(V(x)\):

\[
\begin{aligned}
& V'(x) = \frac{a^3 \sqrt{6}}{54}((1-x)^2(4x-1))' = \\
& = \frac{a^3 \sqrt{6}}{54}(-2(1-x)(4x-1) + 4(1-x)^2) = \frac{a^3 \sqrt{6}}{27}(x-1)(6x-3) = 0
\end{aligned}
\]

The interval \(\left(\frac{1}{4}; 1\right)\) contains the root \(x = \frac{1}{2}\). As we pass through the point \(\frac{1}{2}\), the derivative \(V'(x)\) changes sign from positive to negative. Therefore, the function reaches its maximum value on the interval \(\left(\frac{1}{4}; 1\right)\) at \(x = \frac{1}{2}\). Thus,

\[
V_{\text{max}} = V\left(\frac{1}{2}\right) = \frac{a^3 \sqrt{6}}{54}\left(1 - \frac{1}{2}\right)^2\left(4 \cdot \frac{1}{2} - 1\right) = \frac{1}{12} V.
\]

## Second method.

Applying the Cauchy inequality for three numbers, we get

\[
\begin{aligned}
& V(x) = \frac{a^3 \sqrt{6}}{54}(1-x)^2(4x-1) = \frac{a^3 \sqrt{6}}{27}(1-x)^2\left(2x - \frac{1}{2}\right) \leqslant
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-11.jpg?height=100&width=763&top_left_y=1053&top_left_x=653)

with equality if \(1-x = 2x - \frac{1}{2}\), i.e., at \(x = \frac{1}{2}\). This value of \(x\) belongs to the interval \(\left(\frac{1}{4}; 1\right)\).

## Answer

a) \(\frac{9}{125} V\); b) \(\frac{1}{12} V\).