Facebook Twitter Instagram

Problem of the Week

Problem 29 Solution

Congratulations to Anna Poliakoff from Hitchin Girls’ School, Matthew Whitten from The Cavendish School and Markus Kuan from Northstowe Secondary College for successfully solving the problem!

ai) Label the centre of the larger circle O. Since an angle at the centre of a circle is twice the angle at the circumference, \angle{BOC} = 2A.

OB = OC = R, so \triangle{BOC} is isosceles. Therefore, bisecting \angle{BOC}, creating point D on BC produces two right-angled triangles. BD = R\sin{A}, CD = R\sin{A}. Therefore, BD + CD = a = 2R\sin{A}.

aii) Repeat the process for b = 2R\sin{B} and c = 2R\sin{C}. Area of triangle = \frac{1}{2}ab\sin{C} = \frac{1}{2}(2R\sin{A})(2R\sin{B})\sin{C} = 2R^2\sin{A}\sin{B}\sin{C}.

b) Label the centre of the incircle P. Split the triangle into three triangles \triangle{BPC}, \triangle{APC}, \triangle{APB}, with bases a, b, c respectively with height r. The area of the triangle can therefore be represented by an expression in r:

    \[\frac{ar}{2} + \frac{br}{2} + \frac{cr}{2} = 2R^2\sin{A}\sin{B}\sin{C}\]


    \[Rr\sin{A} + Rr\sin{B} + Rr\sin{C} = 2R^2\sin{A}\sin{B}\sin{C}\]


    \[Rr(\sin{A}+\sin{B}+\sin{C}) = 2R^2\sin{A}\sin{B}\sin{C}\]


    \[r = \frac{2R\sin{A}\sin{B}\sin{C}}{\sin{A}+\sin{B}+\sin{C}}\]

ci)

    \[r = \frac{2R\sin{A}\sin{B}\sin{C}}{\sin{A}+\sin{B}+\sin{C}}\]


    \[\frac{r}{R} = \frac{2\sin{A}\sin{B}\sin{C}}{\sin{A}+\sin{B}+\sin{C}}\]


    \[r : R = (2\sin{A}\sin{B}\sin{C}) : (\sin{A}+\sin{B}+\sin{C})\]

cii) In an equilateral triangle, all three angles are 60 degrees.

Substituting the angles into the above ratio, we can get the following:

    \[r : R = (2\sin{60}\sin{60}\sin{60}) : (\sin{60}+\sin{60}+\sin{60})\]


    \[r : R = 2(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) : (\frac{\sqrt{3}}{2})+(\frac{\sqrt{3}}{2})+(\frac{\sqrt{3}}{2})\]


    \[r : R = \frac{3\sqrt{3}}{4} : \frac{3\sqrt{3}}{2}\]


    \[r : R = 1 : 2\]

Site Search