[ Maximum and minimum problems ] [ Linear dependence of vectors $]$

The side of the base $A B C D$ of a regular quadrilateral pyramid $A B C D P$ is equal to $a$, and the lateral edges are equal to $2a$. Segments with endpoints on the edges $A D$ and $P C$, parallel to the plane $P A B$, are considered. a) One of these segments passes through a point $M$ on the edge $A D$ such that $A M: A D=3: 4$. Find its length. b) Find the minimum length of the considered segments.
a) Through point $M$, we draw a plane parallel to the plane $A P B$ (Fig.1). Let this plane intersect the edges $P D, P C$, and $B C$ at points $K, L$, and $N$ respectively. According to the theorem of the intersection of two parallel planes by a third, $K M\|A P, K L\| C D$, and $L N \| P B$. The segment $M L$ lies in a plane parallel to the plane $A P B$, so $M L$ is parallel to the plane $A P B$, and its ends lie on the lines $A D$ and $P C$. Therefore, $M L$ is the desired segment. From the similarity of triangles $D M K$ and $D A P, P K L$ and $P D C, C N L$ and $C B P$, it follows that

$$
\begin{gathered}
K M=A P \cdot \frac{D M}{A D}=2 a \cdot \frac{\frac{1}{4}}{4}=\frac{\frac{a}{2}}{2} \\
K L=D C \cdot \frac{P^{\prime} K}{P D}=D C \cdot \frac{A M}{A D}=a \cdot \frac{3}{4}=\frac{3 a}{4}
\end{gathered}
$$

$$
L N=B P \cdot \frac{C N}{C B}=B P \cdot \frac{D M}{A D}=2 a \cdot \frac{1}{4}=\frac{a}{2}
$$

Since the line $M N$ is parallel to the plane $C P D$, and the intersecting plane passes through $M N$ and intersects the plane $CPD$ along the line $K L$, then $M N \| K L$. In the isosceles trapezoid $M K L N$ (Fig.2), the bases $M N$ $=a, K L=\frac{3 a}{4}$ and the lateral sides $M K=N L=\frac{a}{2}$ are known. We will find its diagonal $M L$. Let $F$ be the base of the perpendicular dropped from point $L$ to $M N$. Then

$$
\begin{aligned}
& N F=\frac{1}{2}(M N-K L)=\frac{\frac{a}{8}}{8}, M F=\frac{1}{2}(M N+K L)=\frac{7 a}{8} \\
& L F 2=L N 2-N F 2=\left(\frac{\mathbf{a}}{(\mathbf{2}}\right) 2-\left(\frac{\left.\frac{\mathbf{a}}{\overline{8}}\right)}{}\right)=\frac{\mathbf{1 5 a ^ { 2 }}}{64}, \\
& M L 2=M F 2+L F 2=\frac{49 a^{2}}{64}+\frac{15 a^{2}}{64}=a 2
\end{aligned}
$$

Therefore, $M L=a$. b) Now let $X Y$ be an arbitrary segment with ends on the edges $A D$ and $C P$ (Fig.3). Project it onto the plane $A P B$ parallel to the line $A D$. In this case, point $X$ on the edge $A D$ will move to point $A$, and point $Y$ on the edge $C P$ - to some point $Z$ on the edge $B P$. Since $X Y$ is parallel to the plane $A P B, A Z=X Y$.

The segment $A Z$ will be the smallest when $Z$ is the foot of the height of the isosceles triangle $A P B$, dropped from vertex $A$ (Fig.4). In this case, $B P \cdot A Z=A B \cdot P E$, where $E$ is the midpoint of $A B$. From this, we find that

$$
A Z=\frac{A B \cdot P^{P} E P^{\prime}}{B P^{2}}=\frac{\Delta \sqrt{4 a^{2}-\frac{a^{2}}{1}}}{2 a}=\frac{a \sqrt{15}}{4} .
$$

## Answer

a) $a ;$ b) $\frac{\underline{a \sqrt{15}}}{\mathbf{4}}$.