Hello everyone,

I have two questions on which I would greatly appreciate help. I have also included the given solution below.

Problem:

Question #1:For part (iii), is it possible to solve it using calculus?

By Implicit Differentiation, the circle has "slope": $\displaystyle \frac{dy}{dx} = -\frac{x}{y}$.

The required line has slope $\displaystyle m = -h$ and the equation: $\displaystyle y = h(3 - x)$

$\displaystyle \Rightarrow$ The line is tangent to the circle when $\displaystyle -\frac{x}{y} = -h \Rightarrow \frac{x}{y} = h $.

Substitution of the equation of the line at the point of tangency gives:

$\displaystyle \frac{x}{h(3 - x)} = h $.

But this does not appear to help too much.

Question #2: This question is based on part (v), but I am trying to find something else that is not asked by the question.Since $\displaystyle h = \frac{6}{7}$, the vertex of the triangle $\displaystyle \text{ie} \left(1, \frac{12}{7}\right)$ is inside the circle.

This means that we are in the case where $\displaystyle h < \frac{\sqrt{3}}{2}$.

The question does not ask for this, but how would I show that

$\displaystyle h = \frac{6}{7} < \frac{\sqrt{3}}{2} $

By basic algebra:

$\displaystyle \frac{\sqrt{3}}{7} < \frac{\sqrt{3}}{2} $ but $\displaystyle \frac{6}{7} > \frac{\sqrt{3}}{7}$.

Thank you very much for your help.

Given solution: