Triangles and petals

An equilateral triangle rotates around regular polygons and produces an outline like a flower. What are the perimeters of the different flowers?

Age

14 to 16

Challenge level

Secondary

GEOMETRY

Perimeter, Area and Volume

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Triangles and Petals printable sheet

Look at the equilateral triangle rotating around the equilateral triangle. It produces a flower with three petals:

If each equilateral triangle has side length $r$, can you work out the perimeter of the flower's petals?

Now consider a flower made by the triangle rotating about a square with side length $r$ - what is the perimeter of the petals now?

What is the perimeter when the centre of the flower is a regular pentagon, hexagon, heptagon...?

What can you say about the increase in perimeter as the number of sides of the centre shape increases?

Can you explain this increase?

What would be the perimeter of a flower whose centre is a regular $100$-sided polygon with side length $r$?

It may help to work in terms of $\pi$ throughout this problem.

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 $3\times \pi \times \text{ radius } = 3 \times \pi \times \text{ (side of triangle) }= 3\pi r$.

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 $4 \times \frac{150}{360}\times (2\times \pi \times \text{ radius }) =\frac{10}{3} \times \pi \times \text{ (side of the square) }= \frac{10}{3} \pi r$.

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.

Why do this problem?

This problem offers the opportunity to practise calculating arc lengths, working in terms of $\pi$, and calculating interior angles of regular polygons. It also provides an opportunity to generalise from simple examples, and to explain patterns in terms of the underlying structure of the problem.

Possible approach

This printable worksheet may be useful: Triangles and Petals

Start by showing the animation of the triangle rotating around another triangle. Then hide the animation and ask learners to sketch what they saw. Show the animation once more so they can confirm what they saw.

Explain that we are interested in calculating the perimeter of the flower that is traced out. Give them time in pairs to calculate the perimeter, then share answers and methods.

Now show the animation of the triangle rotating around a square. Again, ask learners to sketch what they saw, and show them the animation once more. Challenge them to work out the perimeter of this flower, and to then sketch and work out the perimeter of flowers whose centres are regular pentagons, hexagons and so on. This is a good opportunity to discuss the benefits of working in terms of $\pi$ rather than calculating a decimal answer on the calculator.

After the class have had some time to engage with the task, collect some results together on the board and encourage them to look for patterns. Can they now work out the perimeter of a flower whose centre is a regular $100$-sided polygon?

Key questions

What angles do you know? What angles can you work out?

How can we work out the angle for each arc on a 3, 4, 5... petalled flower?

Possible support

An Unusual Shape offers opportunities to visualise sectors of circles and to consider simple loci.

Possible extension

Construct a proof for the formula for finding the perimeter of a flower with a regular $n$-sided polygon at its centre.

What will the flower look like if we continue this process indefinitely?

Some students may be interested in exploring Reuleaux triangles.

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