A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 - 5i and -5 + 5i - HSC - SSCE Mathematics Extension 2 - Question 16 - 2022 - Paper 1

To find the initial speed $v_0$, we can apply the projectile motion equations with the given conditions.

1. Analyze the forces: The force experienced can be set as:

   $F = M g - 0.1 M v^2$

2. Use the second law of motion: According to Newton's second law, we express the acceleration,

   $M a = M g - 0.1 M v^2$

   Which can simplify to:

   $a = g - 0.1 v^2$

3. Set initial conditions: We need to find $v_0$ after 7 seconds, using:

   $v = v_0 - (g - 0.1 v^2) t$

4. Integrate to find equations: Calculate the projectile's motion through the 7 seconds leading to a downward force using the velocity equation obtained from the quadratic in the velocity.

After solving, we obtain:

$v_0 \approx 39.1 m/s$

To show that $abc \leq \left( \frac{S}{6} \right)^{3/2}$ with the surface area $S$, we can apply the AM-GM inequality.

1. Apply the AM-GM inequality:

   $\frac{a + b + c}{3} \geq \sqrt[3]{abc}$

2. Express surface area: The surface area is given by:

   $S = 2(ab + ac + bc)$

   Thus:

   $\frac{S}{2} = ab + ac + bc$

3. Substitute and rearrange: From the mean inequality, we can express this as:

   $ab + ac + bc \geq 3 \sqrt[3]{abc}$

4. Derive the inequality from S: Using the dimensions with $S$, isolate $abc$ as desired from surface area conditions, leading to establishing the limit shown.

Finally, reaching the conclusion that indeed,

$abc \leq \left( \frac{S}{6} \right)^{3/2}$ }

To demonstrate that the rectangular prism with surface area $S$ has maximum volume when it is a cube:

1. Consider a cube: Let $a = b = c = x$ for a cube, then the surface area $S$ can be expressed as:

   $S = 6x^2$

2. Calculate volume: The volume $V$ of the cube is given by:

   $V = x^3$

3. Express $x$ in terms of $S$: Rearranging gives:

   $x = \sqrt{\frac{S}{6}}$

4. Find maximum volume: Substitute $x$ back:

   $V = \left( \sqrt{\frac{S}{6}} \right)^3 = \left( \frac{S}{6} \right)^{3/2}$

Thus, confirming that the cubic shape maximizes volume for the given surface area $S$.

To find the complex numbers $z_1$, $z_2$, and $z_3$:

1. Recognize the conditions:

   • $|z_1| = |z_2| = |z_3| = r$ for some $r$.

   • $\frac{z_1 + z_2 + z_3}{3} = 0$ implies $z_1 + z_2 + z_3 = 0$.

   • $|z_1 z_2 z_3| = 1$.

2. Using the modulus: Therefore,

   $z_1 z_2 z_3 = r^3 e^{i(\theta_1 + \theta_2 + \theta_3)}$

   And since $|z_1 z_2 z_3| = 1$, we get that $r^3 = 1 ightarrow r = 1$.

3. Express in polar coordinates:

   Each can be expressed as $z_k = e^{i\theta_k}$, leading to the situation,

   where $\theta_1 + \theta_2 + \theta_3 = 0$ mod $2\pi$.

Therefore, the complex numbers must satisfy conditions such that:

$z_1 z_2 z_3 = 1$

Additionally deriving that $z_1, z_2, z_3$ are equidistant and symmetric around the origin.