In dividing a equilateral triangle into n equal parts with equal areas and equal shapes...

People of Brilliant, how are you all? I came across an interesting puzzle that I want some help solving.

Suppose we want to divide an equilateral triangle into $n$ equal pieces, with the same area and the same shape. Is it possible to divide such a triangle into $5$ equal pieces? Is it possible to do it for an arbitrary value of $n$ ?

My investigations showed me that 2 and 3 are the most basic divisions available, and then onward I could find ways to divide a triangle into $n^2$ pieces, and then either into $2n^2, 3n^2$ and $6n^2$ . Any hints as to how I can either do it, or prove there is no way?

#Geometry

Easy Math Editor

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Comments

If the pieces needn't be contiguous, it turns out it can be done as shown here !!

Anirudh Sreekumar - 3 years, 6 months ago

While this is not exactly a solution I wanted, there is a reference to a paper in another answer which seems to state it is impossible. Thanks for the find!

Alexandre Miquilino - 3 years, 6 months ago

Hi,may I ask that where do you live,which country?

Frost Constantin - 3 years, 5 months ago

4 join all the midpoints of the sides forming 4 equilateral triangles each of equal area

U K Swamy - 3 years, 4 months ago

4 falls in the $n^2$ category, but thanks for the input.

Alexandre Miquilino - 3 years, 4 months ago

oh sorry

U K Swamy - 3 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$