Problem 960

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,

!

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

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

!

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) $\frac{1}{12} V$.