Given: The area of triangle \( \triangle ABC = 32 \, \text{square units} \), and side \( BC \) lies on the vertical line \( x = 4 \), with length \( 8 \) units.
To find the shortest distance between point \( A \) and the origin \( (0, 0) \), we start by using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, base \( BC = 8 \) units and the area is given as 32, so we plug in:
\[ 32 = \frac{1}{2} \times 8 \times \text{height} \]
Solving for height:
\[ \text{height} = \frac{32 \times 2}{8} = 8 \, \text{units} \]
Since line \( BC \) lies on the vertical line \( x = 4 \), any point directly above or below it forming the height must also lie on a vertical line.
Thus, point \( A \) must lie on a line parallel to \( BC \), also at \( x = 4 \), so A’s x-coordinate is 4.
The shortest distance from point \( A \) to the origin \( (0, 0) \) is the horizontal distance, since both points lie on vertical lines:
\[ \text{Distance} = |4 - 0| = 4 \, \text{units} \]
Therefore, the shortest possible distance between A and the origin is:
\[ \boxed{4 \text{ units}} \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: