5. In a rectangular parallelepiped $\mathrm{ABCDA}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1} \mathrm{AB}=1$ cm, $A D=2, \mathrm{AA}_{1}=1$. Find the smallest area of triangle $\mathrm{PA}_{1} \mathrm{C}$, where vertex $\mathrm{P}$ lies on the line $\mathrm{AB}_{1}$.
# Solution

The area of triangle $\mathrm{PA}_{1} \mathrm{C}$ is $S=\frac{1}{2} A_{1} C \times P H$, where $P H$ is the distance from point $\mathrm{P}$, taken on line $\mathrm{AB}_{1}$, to line $\mathrm{A}_{1} \mathrm{C}$. The area will be the smallest when the length of segment $P H$ is the smallest, i.e., when $P H$ is the common perpendicular to lines $\mathrm{A}_{1} \mathrm{C}$ and $\mathrm{AB}_{1}$.

Construct the common perpendicular to lines $\mathrm{A}_{1} \mathrm{C}$ and $\mathrm{AB}_{1}$. Through line $\mathrm{A}_{1} \mathrm{C}$, construct a plane parallel to line $\mathrm{AB}_{1}$. For this, through point $\mathrm{A}_{1}$, draw a line parallel to $\mathrm{AB}_{1}$. Point $\mathrm{B}_{2}$ is the intersection of this line with line $\mathrm{BB}_{1}$. The desired plane is $\left(\mathrm{A}_{1} \mathrm{~B}_{2} \mathrm{C}\right)$.

Consider the plane $\left(\mathrm{QB}_{1} \mathrm{~F}\right)$: line $\mathrm{QB}_{1}$ is perpendicular to line $\mathrm{A}_{1} \mathrm{~B}_{2}$, which in turn is parallel to line $\mathrm{AB}_{1}$, and line $\mathrm{B}_{1} \mathrm{~F}$ is perpendicular to line $\mathrm{AB}_{1}$. Therefore, plane $\left(\mathrm{QB}_{1} \mathrm{~F}\right)$ is perpendicular to line $\mathrm{AB}_{1}$.

The projection of line $\mathrm{AB}_{1}$ onto plane $\left(\mathrm{QB}_{1} \mathrm{~F}\right)$ is point $\mathrm{B}_{1}$, and the projection of line $\mathrm{A}_{1} \mathrm{C}$ onto this same plane is line QF. Thus, the desired distance between the skew lines $\mathrm{A}_{1} \mathrm{C}$ and $\mathrm{AB}_{1}$ is the height of triangle $\mathrm{QB}_{1} \mathrm{~F}$, dropped from vertex $\mathrm{B}_{1}$ to side $\mathrm{QF}$. On the diagram, this is segment $\mathrm{B}_{1} \mathrm{M}$.

From triangle $\mathrm{A}_{1} \mathrm{~B}_{2} \mathrm{~B}_{1}$, we find $\mathrm{QB}_{1}=\frac{\sqrt{2}}{2}$. From triangle $\mathrm{B}_{1} \mathrm{FQ}$, we find $\mathrm{B}_{1} \mathrm{M}=\frac{\sqrt{3}}{3}$.

Thus, the desired area is

![](https://cdn.mathpix.com/cropped/2024_05_06_d90ef84ca6dbd2e3b7dbg-3.jpg?height=888&width=985&top_left_y=1315&top_left_x=227)

$$
S=\frac{1}{2} A_{1} C \times P H=\frac{\sqrt{6}}{2} \times \frac{\sqrt{3}}{3}=\frac{\sqrt{2}}{2}
$$

## Grading Criteria

7 points for obtaining the correct answer based on correct reasoning.

1 point for concluding that the area will be the smallest when the height of the triangle is the distance between the skew lines $\mathrm{A}_{1} \mathrm{C}$ and $\mathrm{AB}_{1}$.

3 points for correctly identifying (constructing) the segment whose length is the distance between the lines $\mathrm{A}_{1} \mathrm{C}$ and $\mathrm{AB}_{1}$.