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.