The vertices of an equilateral triangle, with perimeter P and area A , lie on a circle with radius r . Find an expression for $\frac{P}{A}$ in the form $\frac{k}{r}$, where $k \in {\mathbb{Z}^ + }$.

let the length of one side of the triangle be x

consider the triangle consisting of a side of the triangle and two radii

EITHER

${x^2} = {r^2} + {r^2} - 2{r^2}\cos 120^\circ$     M1

$= 3{r^2}$

OR

$x = 2r\cos 30^\circ$     M1

THEN

$x = r\sqrt 3$     A1

so perimeter $= 3\sqrt 3 r$     A1

now consider the area of the triangle

area $= 3 \times \frac{1}{2}{r^2}\sin 120^\circ$     M1

$= 3 \times \frac{{\sqrt 3 }}{4}{r^2}$     A1

$\frac{P}{A} = \frac{{3\sqrt 3 r}}{{\frac{{3\sqrt 3 }}{4}{r^2}}}$

$= \frac{4}{r}$     A1

Note: Accept alternative methods

[6 marks]

It was pleasing to see some very slick solutions to this question. There were various reasons for the less successful attempts: not drawing a diagram; drawing a diagram, but putting one vertex of the triangle at the centre of the circle; drawing the circle inside the triangle; the side of the triangle being denoted by r the symbol used in the question for the radius of the circle.