, 2013

Since $ABC$ is a quarter of a circular pizza with centre $A$ and radius 20 cm, then$AC=AB=20$ cm. We are also told that $\angle CAB = 90^\circ$ (one-quarter of $360^\circ$). Since $\angle CAB = 90^\circ$ and $A$, $B$ and $C$ are all on the circumference of the circle, then $CB$ is a diameter of the pan. (This is a property of circles: if $X$, $Y$ and $Z$ are three points on a circle with $\angle ZXY = 90^\circ$, then $YZ$ must be a diameter of the circle.) Since $\triangle CAB$ is right-angled and isosceles, then $CB = \sqrt{2}AC = 20\sqrt{2}$ cm. Therefore, the radius of the circular plate is $\frac{1}{2}CB$ or $10\sqrt{2}$ cm. Thus, the area of the circular pan is $\text{(10 2 cm ) 2 = 200 cm 2}$. The area of the slice of pizza is one-quarter of the area of a circle with radius 20 cm, or $\text{1 4 (20 cm ) 2 = 100 cm 2}$. Finally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\text{100 cm 2 200 cm 2 = 1 2}$. Suppose that the length of $AF$ is $x$ m. Since the length of $AB$ is 8 m, then the length of $FB$ is $(8-x)$ m. Since $\triangle MAF$ is right-angled and has an angle of $60^\circ$, then it is $30^\circ$-$60^\circ$-$90^\circ$ triangle. Therefore, $MF = \sqrt{3}AF$, since $MF$ is opposite the $60^\circ$ angle and $AF$ is opposite the $30^\circ$ angle. Thus, $MF = \sqrt{3}x$ m. Since $MP = 2$ m, then $PF = MF - MP = (\sqrt{3}x-2)$ m. We can now look at $\triangle BFP$ which is right-angled at $F$. We have $$\text{= PF FB = ( 3 x-2) m (8-x) m = 3 x-2 8-x}$$ Therefore, $(8-x)\tan \theta = \sqrt{3}x-2$ or $8\tan\theta + 2 = \sqrt{3}x+(\tan \theta)x$. This gives $8\tan\theta + 2 = x(\sqrt{3} + \tan \theta)$ or $x = \dfrac{8\tan\theta+2}{\tan\theta+\sqrt{3}}$. Finally, $MF = \sqrt{3}x = \dfrac{8\sqrt{3}\tan\theta+2\sqrt{3}}{\tan\theta+\sqrt{3}}$ m.