You may also like

problem icon

Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

problem icon

Ladder and Cube

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

problem icon

At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

Round and Round

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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

Circle inside semicircle

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 Circle inside quarter circle
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 $$\frac{\text{area of inner circle}}{\text{area of quadrant}} = \frac{4}{3 + \sqrt{2}} = 68.6\%$$

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

Circle inside sector of angle 2 alpha


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 $$\frac{\text{area of inner circle}}{\text{area of sector}} = \frac{\pi\sin^{2}\alpha}{\alpha(1 + \sin\alpha)^2}$$ 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 $$\frac{3\pi/4}{(\pi/3)(1+\sqrt{3}/2)^2} = \frac{9}{2+\sqrt{3}} = 9(2 - \sqrt{3})^2 = 64.6 \%$$

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