Round and round

Well done Kang Hong Joo from the Chinese High School, Singapore; Lucinda Hearth from Stamford High School; Jessica Zhang; Matthew Hodgetts from King Edward VI Camp Hill School, Birmingham; Tom Davie and Michael Grey from Madras College, St. Andrews.

Case 1: a semicircle

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Let the radius of the inner circle be $r$; then its area is $\pi{r^2}$. The area of the semicircle is $\pi(2r)^2/2$, which is $2\pi{r^2}$. The percentage of the semicircle covered by the inner circle is 50\%.

Case 2: a quadrant

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The area of the inner circle is $\pi{r^2}$; the radius of the quadrant is $r(1 + \sqrt{2})$, and the area of the quadrant is $$\frac{1}{4}\pi r^2(1+\sqrt{2}{)}^2=\frac{1}{4}\pi r^2(3+2\sqrt{2})$$ Therefore $$\frac{\text{area of innercircle}}{\text{area of quadrant}}=\frac{4}{3+\sqrt{2}}=68.6\mathrm{\%}$$

Case 3: a sector of angle $2\alpha$

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The area of the inner circle is $\pi{r^2}$. Taking the radius of the sector to be 1 unit, then $$\sin \alpha =\frac{r}{1-r}$$ Hence $$r=\frac{\sin \alpha }{1+\sin \alpha }$$ The area of the sector is ${1\over2} 1^2 (2a) = a$ or, equivalently, ${\pi{\alpha}\over180}$ if $\alpha$ is measured in degrees.

Therefore $$\frac{\text{area of inner circle}}{\text{area ofsector}}=\frac{\pi {\sin }^2\alpha }{\alpha (1+\sin \alpha {)}^2}$$ When $\alpha = {\pi\over6} = 30^{\circ}$, then $\sin\alpha = {1\over2}$, and this ratio is $2/3$.

When $\alpha = {\pi\over3} = 60^{\circ}$, then $\sin\alpha = {\sqrt{3}\over2}$, and this ratio is $$\frac{3\pi /4}{(\pi /3)(1+\sqrt{3}/2{)}^2}=\frac{9}{2+\sqrt{3}}=9(2-\sqrt{3}{)}^2=64.6\mathrm{\%}$$

a	30 o	45 o	60 o	90 o
Ratio (nearest 1%)	67%	69%	65%	50%