## [ Areas and volumes (extremum problems $)$.

$[\quad$ Area of the section

The side of the base $A B C$ of a regular pyramid $P A B C$ is $a$, and the lateral edge is $b$. At what distance from the line $B C$ should a section of the pyramid be made, parallel to the edges $B C$ and $P A$, so that its area is the largest possible? 

#
Let the secant plane intersect the edge $A B$ at point $K$. The plane $A B C$ passes through the line $B C$, which is parallel to the secant plane, and has a common point $K$ with the secant plane. Therefore, these planes intersect along a line $l$, which is parallel to $B C$ and passes through point $K$. If the line $l$ intersects the edge $A C$ at point $N$, then $K N \| B C$. Similarly, the secant plane intersects the plane of the face $A B P$ along a line parallel to $A P$, the plane of the face $B P C$ along a line parallel to $B C$, and the plane of the face $A P C$ along a line parallel to $A P$. Let $L, M$, and $N$ be the points of intersection of the secant plane with the edges $B P, P C$, and $A C$ respectively. Then $K L M N$ is a parallelogram, and since the pyramid $P A B C$ is regular, $A P \perp B C$, so $K L M N$ is a rectangle. Let $\frac{B K}{A K}=x(0 \leqslant x \leqslant 1)$. Then

$$
M N=K L=x \cdot A P=b x
$$

$$
L M=K N=B C \cdot \frac{A K}{A B}=(1-x) B C=a(1-x)
$$

therefore

$$
\begin{gathered}
S_{K L M N}=K L \cdot K N=a b \cdot x(1-x)=a b\left(\frac{\mathbf{1}}{\mathbf{4}}-\frac{\mathbf{1}}{\mathbf{4}}+x-x 2\right)= \\
=a b\left(\frac{\mathbf{1}}{\mathbf{4}}-\left(x-\frac{\mathbf{1}}{\mathbf{2}}\right) 2\right) \leqslant \frac{\mathbf{1}}{\mathbf{4}} a b
\end{gathered}
$$

with equality achieved when $x=\frac{1}{2}$. Let $Q$ be the midpoint of $B C$, and the segments $P Q$ and $A Q$ intersect $L M$ and $K N$ at points $E$ and $F$ respectively. Consider the section of the pyramid by the plane $A P Q$. Since

$$
\frac{P^{\prime} E}{E Q}=\frac{A F^{\prime}}{F Q}=\frac{A K}{K B}=\frac{1}{2}
$$

the segment $E F$ is the midline of the triangle $A P Q$. Therefore, the desired distance $d$ is half the height $Q H$ of the triangle $A P Q$. If $O$ is the center of the equilateral triangle $A B C$, then

$$
A O=\frac{2}{3} A Q=\frac{a \sqrt{3}}{3}, O P=\sqrt{A P^{2}-A O^{2}}=\sqrt{b^{2}-\frac{\alpha^{2}}{3}}
$$

and since $A Q \cdot O P=A P \cdot Q H$, then

$$
Q H=\frac{A Q O P}{A P}=\frac{\frac{a \sqrt{3}}{2} \cdot \sqrt{b^{2}-\frac{a^{2}}{3}}}{b}=\frac{1}{2} \cdot \frac{a \sqrt{3 b^{2}-a^{2}}}{b}
$$

Therefore,

$$
d=\frac{1}{2} Q H=\frac{1}{4} \cdot \frac{a \sqrt{8 b^{2}-a^{2}}}{b}
$$

## Answer

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