Given four weights in geometric progression and an equal arm balance, show how to find the heaviest weight using the balance only twice.
1. **Define the weights in geometric progression:**
   Let the weights be \( x, xr, xr^2, xr^3 \) where \( r \) is the common ratio of the geometric progression.

2. **First weighing:**
   Weigh \( x \) and \( xr \) against \( xr^2 \) and \( xr^3 \).
   - If \( x + xr \) is heavier, then the heaviest weight is among \( x \) and \( xr \).
   - If \( xr^2 + xr^3 \) is heavier, then the heaviest weight is among \( xr^2 \) and \( xr^3 \).

3. **Second weighing:**
   - If \( x + xr \) was heavier, weigh \( x \) against \( xr \).
     - If \( x \) is heavier, then \( x \) is the heaviest.
     - If \( xr \) is heavier, then \( xr \) is the heaviest.
   - If \( xr^2 + xr^3 \) was heavier, weigh \( xr^2 \) against \( xr^3 \).
     - If \( xr^2 \) is heavier, then \( xr^2 \) is the heaviest.
     - If \( xr^3 \) is heavier, then \( xr^3 \) is the heaviest.

4. **Proof of the first weighing:**
   Consider the case where we put \( xr^3 \) and \( x \) on one side and \( xr \) and \( xr^2 \) on the other side. We need to show that the side containing \( xr^3 \) will always be heavier.
   \[
   xr^3 + x \geq xr + xr^2
   \]
   Simplifying the inequality:
   \[
   xr^3 + x - xr - xr^2 = xr^2(r-1) - x(r-1) = x(r-1)(r^2-1) = x(r-1)(r-1)(r+1) = x(r-1)^2(r+1) \geq 0
   \]
   Since \( x \) and \( r \) are positive, \( (r-1)^2 \geq 0 \) and \( (r+1) > 0 \), the inequality holds true.

The final answer is \( \boxed{ xr^3 } \) or \( xr^2 \) or \( xr \) or \( x \) depending on the second weighing.