Let $T$ be the answer to question $18$. Rectangle $ZOMR$ has $ZO = 2T$ and $ZR = T$. Point $B$ lies on segment $ZO$, $O'$ lies on segment $OM$, and $E$ lies on segment $RM$ such that $BR = BE = EO'$, and $\angle BEO' = 90^o$. Compute $2(ZO + O'M + ER)$.

PS. You had better calculate it in terms of $T$.
1. **Define the coordinates of the vertices of the rectangle:**
   - Let \( Z(0,0) \), \( O(2T,0) \), \( M(2T,T) \), and \( R(0,T) \).

2. **Place point \( B \) on segment \( ZO \):**
   - Let \( B(\lambda, 0) \).

3. **Place point \( E \) on segment \( RM \):**
   - Let \( E(2\lambda, T) \).

4. **Determine the coordinates of point \( O' \) on segment \( OM \):**
   - Let \( O'(2T, y) \).

5. **Use the condition \( BR = BE = EO' \) and \( \angle BEO' = 90^\circ \):**
   - The distance \( BR \) is given by:
     \[
     BR = \sqrt{(\lambda - 0)^2 + (0 - T)^2} = \sqrt{\lambda^2 + T^2}
     \]
   - The distance \( BE \) is given by:
     \[
     BE = \sqrt{(2\lambda - \lambda)^2 + (T - 0)^2} = \sqrt{\lambda^2 + T^2}
     \]
   - The distance \( EO' \) is given by:
     \[
     EO' = \sqrt{(2T - 2\lambda)^2 + (y - T)^2}
     \]
   - Since \( BR = BE = EO' \), we have:
     \[
     \sqrt{\lambda^2 + T^2} = \sqrt{(2T - 2\lambda)^2 + (y - T)^2}
     \]

6. **Use the condition \( \angle BEO' = 90^\circ \):**
   - This implies that \( BE \) and \( EO' \) are perpendicular. Therefore, the slopes of \( BE \) and \( EO' \) must be negative reciprocals:
     \[
     \text{slope of } BE = \frac{T - 0}{2\lambda - \lambda} = \frac{T}{\lambda}
     \]
     \[
     \text{slope of } EO' = \frac{y - T}{2T - 2\lambda}
     \]
     \[
     \frac{T}{\lambda} \cdot \frac{y - T}{2T - 2\lambda} = -1
     \]
     \[
     T(y - T) = -\lambda(2T - 2\lambda)
     \]
     \[
     Ty - T^2 = -2\lambda T + 2\lambda^2
     \]
     \[
     Ty + 2\lambda T = T^2 + 2\lambda^2
     \]
     \[
     y + 2\lambda = T + \frac{2\lambda^2}{T}
     \]
     \[
     y = T + \frac{2\lambda^2}{T} - 2\lambda
     \]

7. **Solve for \( \lambda \) using the equality of distances:**
   - From \( \sqrt{\lambda^2 + T^2} = \sqrt{(2T - 2\lambda)^2 + (y - T)^2} \):
     \[
     \lambda^2 + T^2 = (2T - 2\lambda)^2 + (y - T)^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 4\lambda^2 + (T + \frac{2\lambda^2}{T} - 2\lambda - T)^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 4\lambda^2 + (\frac{2\lambda^2}{T} - 2\lambda)^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 4\lambda^2 + \frac{4\lambda^4}{T^2} - 8\lambda^3 + 4\lambda^2
     \]
     \[
     \lambda^2 + T^2 = 4T^2 - 8T\lambda + 8\lambda^2 + \frac{4\lambda^4}{T^2} - 8\lambda^3
     \]
     - Solving this equation, we find \( \lambda = \frac{T}{2} \).

8. **Determine the coordinates of \( O' \):**
   - Substituting \( \lambda = \frac{T}{2} \) into the equation for \( y \):
     \[
     y = T + \frac{2(\frac{T}{2})^2}{T} - 2(\frac{T}{2})
     \]
     \[
     y = T + \frac{2(\frac{T^2}{4})}{T} - T
     \]
     \[
     y = T + \frac{T}{2} - T
     \]
     \[
     y = \frac{T}{2}
     \]
     - Therefore, \( O' = (2T, \frac{T}{2}) \).

9. **Calculate the lengths \( ZO \), \( O'M \), and \( ER \):**
   - \( ZO = 2T \)
   - \( O'M = \sqrt{(2T - 2T)^2 + (T - \frac{T}{2})^2} = \frac{T}{2} \)
   - \( ER = \sqrt{(2\lambda - 0)^2 + (T - T)^2} = \sqrt{(T)^2} = T \)

10. **Compute \( 2(ZO + O'M + ER) \):**
    \[
    2(ZO + O'M + ER) = 2(2T + \frac{T}{2} + T) = 2(3.5T) = 7T
    \]

The final answer is \( \boxed{7T} \).