The diagram represents a vertical cylindrical water cooler of constant cross-sectional area A - HSC - SSCE Mathematics Extension 2 - Question 7 - 2002 - Paper 1

From the earlier derived equation, we can express dy/dt in terms of y:
$\frac{dy}{\sqrt{y}} = -\frac{k}{A} dt$
Integrating both sides gives:
$2\sqrt{y} = -\frac{k}{A}t + C$
Letting C = 2\sqrt{y_0}$, and solving for y yields:
$\sqrt{y} = \frac{-kt}{2A} + \sqrt{y_0}$
Squaring both sides gives:
$y = (\sqrt{y_0} - \frac{kt}{2A})^2$
Given that T is the time taken for the water to drain to zero, we find that
y = 0 at t = T, leading to the relation
y = y_0(1 - t/T) for 0 ≤ t ≤ T.

If it takes 10 seconds to drain to half, we have:
$y(10) = \frac{y_0}{2}$
Using the earlier derived equation, we can substitute t = 10 to derive T as follows:
$\frac{y_0}{2} = y_0\left(1 - \frac{10}{T}\right)$
Solving gives:
$1 - \frac{10}{T} = \frac{1}{2} \Rightarrow \frac{10}{T} = \frac{1}{2} \Rightarrow T = 20 \text{ seconds}$
Hence, to empty the full cooler takes 20 seconds.

Using vector addition, we note that the complex numbers correspond to positions in the Argand diagram. The angles between the vectors OP₀, OP₁, and OP₂ are determined by the angles formed at those points, which are equal due to the uniform increment of β in the definitions of z_n. This uniformity implies that θ₀, θ₁, and θ₂ correspond to the same angle β.

Since the angles at P₀, P₁, and P₂ are defined by the external angles, it follows that the angles ∠P₀P₁P₂ and ∠P₄P₃P₂ are equal, because corresponding angles formed by the intersecting lines maintain equality. As a result, since opposite angles sum up to 180°, OP₀P₁P₂ forms a cyclic quadrilateral.

The cyclic nature of quadrilaterals is indicated by two of its angles summing to 180°. Since we have already shown that ∠P₀P₁P₂ = ∠P₄P₃P₂, and ∠OP₀P₃ = ∠OP₁P₂, this satisfies the cyclic quadrilateral condition for P₀P₁P₂P₃. This also confirms that points O, P₀, P₁, P₂, and P₃ reside on the same circumcircle.

Given that z₀, z₁, z₂, z₃, and z₄ are symmetrically spaced around the unit circle since they represent the circumferential points formed by angle β, their average must equate to zero. Since these points are evenly distributed, the angle β can be determined through the relation:
$\beta = \frac{2\pi}{n}$ where n is the number of points, here n=5. Thus,
$\beta = \frac{2\pi}{5}.$