At a Glance

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

No Right Angle Here

Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.

A Sameness Surely

Triangle ABC has a right angle at C. ACRS and CBPQ are squares. ST and PU are perpendicular to AB produced. Show that ST + PU = AB

Triangles and Petals

Stage: 4 Challenge Level:

Herschel from the European School of Varese sent us this solution:

The first flower has 3 petals corresponding to the 3 corners of the triangle. The completed animation shows that each petal is a semicircle, so the perimeter of the flower is $3\times \pi \times \text{ radius } = 3 \times \pi \times \text{ (side of triangle) }$.

The second flower has 4 petals. This time, each petal is a sector of a circle rather than a simple semicircle. The angle of this sector is 360 - (2 triangle corners) - (1 square corner) = $360 - 2 \times 60 - 90 = 150^\circ$.
Therefore, the total perimeter of this petal is $4 \times \frac{150}{360}\times (2\times \pi \times \text{ radius }) =\frac{10}{3} \times \pi \times \text{ (side of the square)}$.

In general, we need to know 3 key bits of data to work out the perimeter of the flower.
They are:
• The number of sides of the central shape; we'll call this $n$.
• The length of each side in the central shape; we'll call this $r$. (Note that this is equal to the radius of the petals).
• The angle at the centre of each petal. This can be derived from $n$:
$\text{Angle } = 360 - 2 \times 60 - \text{( Corner of shape)}$
$\text{Angle } = 360 - 120 - \frac{180(n-2)}{n}$
$\text{Angle } = 240 - 180 - \frac{360}{n}$
$\text{Angle } = 60 + \frac{360}{n}$
Given these data, we can proceed to work out a general formula:

Perimeter= (number of petals) $\times$ (perimeter of a full circle) $\times \frac{\text{angle at centre of petal}}{360}$
$\text{Perimeter }= n \times 2 \times \pi \times r \times \frac{(60+ \frac{360}{n})}{360}$
$\text{Perimeter }= 2 \times \pi \times n \times r \times (\frac{1}{6}+\frac{1}{n})$
$\text{Perimeter }= 2 \times \pi \times n \times r \times \frac{6+n}{6n}$
$\text{Perimeter }= \pi \times r \times \frac{6+n}{3}$

Using this formula, we find the following results:
$n=3$ (Triangle): Perimeter = $\pi \times r \times \frac{9}{3} = 3 \pi r$
$n=4$ (Square): Perimeter = $\pi \times r \times \frac{10}{3}$
$n=5$ (Pentagon): Perimeter = $\pi \times r \times \frac{11}{3}$
$n=6$ (Hexagon): Perimeter = $\pi \times r \times \frac{12}{3} = 4\pi r$
$n=7$ (Heptagon): Perimeter = $\pi \times r \times \frac{13}{3}$
$n=8$ (Octagon):  Perimeter = $\pi \times r \times \frac{14}{3}$
...
$n=100$: Perimeter = $\pi \times r \times \frac{106}{3}$

So a shape with 100 sides will produce a flower with a perimeter of $\pi \times r \times \frac{106}{3}$.
If each edge of the central shape has a length of 1, the perimeter of the flower will be $35.333 \times \pi$, which is 111.00 to two decimal places.

• Well done to Saif from Havering Sixth Form College, Nina, Jure and Kristjan from Elementary school Loka Crnomelj, Slovenia, Chi from Raynes Park, Rajeev from Haberdashers' Aske's Boys' School, Yun Seok Kang, and Cameron, who also sent in correct solutions. Click here to read Nina, Jure and Kristjan's thoughts.