The base of the unit regular pyramid with a square base $ABCD$ is a unit square, and its apex is $E$. On the base edge $AB$ there is a point $P$, and on the side edge $EC$ there is a point $Q$, such that $PQ$ is perpendicular to both $AB$ and $EC$. We also know that $AP: PB = 6: 1$. What are the lengths of the side edges?
I. solution. Let $O$ be the center of the square $ABCD$, $R$ the point that divides the segment $DC$ into a ratio of $1:6$ closer to $C$, and $S$ the orthogonal projection of point $Q$ onto the plane $ABCD$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6ce90944039be3c9828fg-1.jpg?height=588&width=528&top_left_y=281&top_left_x=793)

From the regularity of the pyramid, the orthogonal projection of line $CE$ onto the plane $ABCD$ is the line $CO$, so $S$ lies on the line $CO$. Since $AP:PB=6:1$, the line $PR$ is parallel to $BC$, and thus perpendicular to $AB$. $PQ$ is also perpendicular to $AB$, so $AB$ is perpendicular to the plane $PQR$. Therefore, the line $QS$ perpendicular to $AB$ lies in the plane $PQR$, which means $S$ lies on the line $PR$.

The triangle $CRS$ is a right-angled and isosceles, so $RS=1/7$ and $CS=CR\sqrt{2}=\frac{\sqrt{2}}{7}$. The triangles $QSC$ and $EOC$ are similar because their corresponding sides are parallel. The ratio of similarity is $CS:CO=\frac{\sqrt{2}}{7}:\frac{\sqrt{2}}{2}=2:7$. If $CQ=y$, then $CE=\frac{7y}{2}$.

In the right-angled triangles $QSC$ and $QSP$, by the Pythagorean theorem,

\[
\begin{aligned}
& QS^2=CQ^2-CS^2=y^2-\frac{2}{49} \\
& QP^2=QS^2+SP^2=\left(y^2-\frac{2}{49}\right)+\left(\frac{6}{7}\right)^2=y^2+\frac{34}{49}
\end{aligned}
\]

Finally, in the right-angled triangles $PQC$ and $PBC$, the square of the common hypotenuse $PC$ can be written by the Pythagorean theorem as

\[
PC^2=CQ^2+QP^2=BC^2+BP^2, \quad \text{that is} \quad y^2+\left(y^2+\frac{34}{49}\right)=1+\frac{1}{49}
\]

Solving this equation, we get $2y^2=\frac{16}{49}$, so $y=\frac{\sqrt{8}}{7}$.

Therefore, the length of the lateral edges of the pyramid is $CE=\frac{7y}{2}=\sqrt{2}$.

II. solution. Let's set up a spatial Cartesian coordinate system such that the vertex $B$ of the square $ABCD$ is the origin, and $A$ and $C$ are on the positive $x$ and $y$ axes, respectively. Then $A(1,0,0)$ and $C(0,1,0)$, and due to the regularity of the pyramid, $E(1/2,1/2,e)$ where $|e| \neq 0$ is the distance of vertex $E$ from the plane $ABCD$. Since $AP:PB=6:1$, $P(1/7,0,0)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6ce90944039be3c9828fg-1.jpg?height=537&width=553&top_left_y=2176&top_left_x=775)

Let $CQ:CE=\lambda$. Then, using the formula for the coordinates of a point dividing a segment in a given ratio, we get $Q\left(\frac{\lambda}{2}, 1-\frac{\lambda}{2}, e\lambda\right)$. Since the segment $PQ$ is perpendicular to the $x$-axis, the first coordinates of $P$ and $Q$ are equal, i.e., $1/7=\lambda/2$, so $\lambda=2/7$. Therefore, $Q\left(\frac{1}{7}, \frac{6}{7}, \frac{2e}{7}\right)$. Hence,

\[
\overrightarrow{PQ}=\left(0, \frac{6}{7}, \frac{2e}{7}\right) \quad \text{and} \quad \overrightarrow{CE}=\left(\frac{1}{2}, -\frac{1}{2}, e\right)
\]

According to the condition, these two vectors are perpendicular to each other, so their dot product is 0. Writing this down, we get

\[
0 \cdot \frac{1}{2} + \frac{6}{7} \cdot \left(-\frac{1}{2}\right) + \frac{2e}{7} \cdot e = 0, \quad \text{that is} \quad e^2 = \frac{3}{2}
\]

The length of the segment $CE$ is, by the distance formula between two points,

\[
CE = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + e^2} = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{3}{2}} = \sqrt{2}
\]

Therefore, the lateral edges of the pyramid are $\sqrt{2}$ in length.