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

Given that T is the total time taken to drain the cooler, we reassess the equation to be:

$y = y_0\left(1 - \frac{t}{7}\right).$

Since half the cooler drains in 10 seconds, we can set:

$y = \frac{y_0}{2} \text{ at } t = 10.$

Using the derived equation:

$\frac{y_0}{2} = y_0\left(1 - \frac{t}{T}\right),$

we simplify to:

$\frac{1}{2} = 1 - \frac{10}{T}.$

Thus:

$\frac{10}{T} = \frac{1}{2},$

leading to T = 20 seconds. Therefore, it takes 20 seconds to empty the cooler.

We need to show that the angles formed by the points P₀, P₁, and P₂ are equal. By using vector addition, we observe that:

heta_0 consists of the rotation from OP_0 to OP_1, while 	heta_1 is from OP_1 to OP_2. By the symmetry of the unit circle and the periodic nature of the angles involved, it follows that:

$θ_0 = θ_1 = θ_2 = β.$

The angles can be shown to be equal due to the alternate segment theorem. Since OP₀P₁P₂ are points on a circle, we note that:

∠P₀OP₁ is equal to

the angle subtended by the same arc at point P₂, hence:

∠P₀OP₁ = ∠P₀P₁P₂.

Thus, OP₀, P₁, and P₂ form a cyclic quadrilateral.

Given z₀ + z₁ + z₂ + z₃ = 0, if we plot these points regularly on the Argand plane, we can establish that:

The cyclic nature of quadrilaterals holds as P₀, P₁, P₂, and P₃ maintain equal angles about point O, confirming they lie on a single circumcircle, therefore showing that:

P₀P₁P₂P₃ is a cyclic quadrilateral.

Given that the weights of the complex numbers must sum to zero and that they are equally spaced, we can set

Let each arc subtended by angles be given as:

$nβ$. By symmetry and the total circle measurements:

Since the sum pertains to angles around a circle, it leads to the conclusion that for m distinct roots, we arrive at:

$β = \frac{2π}{5}.$