[ Maximum/Minimum Distance ] Properties of Sections $\quad]$

The side of the base $A B C D$ of a regular pyramid $S A B C D$ is $a$, and the lateral edge is $2 a$. Segments with endpoints on the diagonal $B D$ of the base and the lateral edge $S C$, parallel to the plane $S A D$, are considered. 1) One of these segments passes through the point $M$ on the diagonal $B D$, for which $D M: D B=1: 3$. Find its length. 2) Find the minimum length of all the considered segments.
Through point $M$, we draw a plane parallel to the plane $S A D$ (Fig.1). Let the intersecting plane intersect the edges $A B, C D, S C, S B$ of the given pyramid at points $P, Q, R, T$ respectively. Then the segment $M R$ is parallel to the plane $S A D$, its ends lie on the lines $B D$ and $S C$, and the quadrilateral $P Q R T$ is an isosceles trapezoid. Given that

$$
\frac{D M}{D B}=\frac{1}{3},
$$

we have

$$
\frac{D Q}{D C}=\frac{A P}{A B}=\frac{D M}{D B}=\frac{1}{3},
$$

therefore

$$
\frac{C Q}{C D}=\frac{B P'}{A B}=\frac{2}{3}, \frac{V I'}{B C}=\frac{S T'}{S B}=\frac{A P'}{A B}=\frac{1}{3},
$$

which means

$$
Q R=P T=\frac{2}{3} A S=\frac{4}{3} a, R T=\frac{1}{3} B C=\frac{1}{3} a.
$$

Let $R F$ be the height of the isosceles trapezoid $P Q R T$. Then

$$
F Q=\frac{1}{2}(P Q-R T)=\frac{1}{2}\left(a-\frac{1}{3} a\right)=\frac{1}{3} a
$$

and since $M Q=\frac{1}{3} P Q=\frac{1}{3} a$, in this case point $F$ coincides with point $M$. Therefore,

$$
R M=\sqrt{Q R^{2}-Q M^{2}}=\sqrt{\frac{16 a^{2}}{9}-\frac{a^{2}}{9}}=\frac{a \sqrt{15}}{3}.
$$

Now let point $M$ move along the segment $B D$. For each of its positions, we construct the plane $P Q R T$ parallel to the plane $S A D$. Let $\frac{D M}{D B}=x(0 \leqslant x \leqslant 1)$. Using the method described above (Fig.2), we find that

Then

$$
\begin{gathered}
Q F=\frac{1}{2}(P Q-R T)=\frac{1}{2}(a-a x)=\frac{1}{2} a(1-x) \\
R F 2=Q R 2-Q F 2=4 a 2(1-x) 2-\frac{1}{4} a 2(1-x) 2=\frac{15 a^{2}(1-x)^{2}}{4}, \\
M F=M Q-Q F=a x-\frac{1}{2} a(1-x)=\frac{1}{2} a(3 x-1) \\
M R 2=M F 2+R F 2=\frac{1}{4} a 2(3 x-1) 2+\frac{15}{4} a 2(1-x) 2= \\
=\frac{1}{4} a 2((3 x-1) 2+15(1-x) 2)= \\
=\frac{1}{4} a 2(24 x 2-36 x+16)=a 2(6 x 2-9 x+4)= \\
=a 2\left(6\left(x-\frac{3}{4}\right) 2+\frac{5}{8}\right) \geqslant \frac{5}{5} a 2
\end{gathered}
$$

with equality achieved if $x=\frac{3}{4}$. Therefore, the minimum length of the segment $M R$ is $a \sqrt{\frac{5}{8}}$.

## Answer

1) $\left.\frac{a \sqrt{15}}{3} ; 2\right)^{\frac{a \sqrt{10}}{4}}$