IA idea · Pure maths, number & proof
The product of a regular polygon's diagonals
Research question
For a regular n-gon inscribed in a unit circle, why is the product of the distances from one vertex to all the others exactly n?
Adapt it: change the place, the data or the comparison until the question is yours.
Why it makes a good exploration
Measure the chords of a regular polygon and multiply them — you always get n. The proof uses roots of unity and a factorisation of zⁿ − 1, one of the most elegant uses of complex numbers in AA HL.
The mathematics you'll need
- Complex numbers in polar and Euler form
- Roots of unity and the factorisation of zⁿ − 1
- Modulus of products
- Limits and derivatives (the polynomial 1 + z + … + zⁿ⁻¹ at z = 1)
- Trigonometric product identities
Course labels show where a technique sits; using maths from outside your course is fine if you explain it clearly and say it is new to you.
Where the data comes from
Verify numerically and by measurement on drawn polygons in GeoGebra.
- GeoGebra — Free geometry and graphing software — Voronoi diagrams, loci and regression built in.
Cite every source in a footnote where you use it and in your bibliography. Check the licence of any dataset you download.
A possible outline
- Measure for small n and conjecture.
- Represent vertices as roots of unity.
- Factorise and evaluate at z = 1.
- Derive a trigonometric identity as a corollary.
- Reflect on the power of changing representation.
Pitfalls that cost marks
- Numerical verification presented as proof.
- Skipping the factorisation step.
- Losing track of which roots are included.
Showing personal engagement
- Discover the result yourself with GeoGebra before reading about it.
- Find the corresponding result for sums of squared distances.
- Explain the proof to a classmate and note what confused them.
See Criterion C: personal engagement for what examiners look for.
Taking it further
Prove the related identity for ∏ sin(kπ/n) and explore what happens for points not on a vertex.