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More on diagonals of regular polygons

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Edited by Richard Walker, Sunday, 21 Aug 2022, 11:34

A few days ago I wrote about an elegant proof that a regular dodecahedrons has 4 diagonals that all intersect at a point other than the centre of the polygon, see https://learn1.open.ac.uk/mod/oublog/viewpost.php?post=248909

Since then I have found a paper by Poonen and Rubenstein [1] in which they completely solve the problem of concurrent diagonals in regular polygons. They prove several interesting facts, including:

If the number of sides is odd there can never be 3 concurrent diagonals.

The smallest number of sides that allow 3 diagonals to meet at a point other than the centre is eight, for diagonals twelve sides are needed, and for 5 concurrent diagonals eighteen sides.

To get more than 5 diagonals meeting at a point other than the centre we need to go to thirty sides. The regular triacontagon has sets of 6 and 7 concurrent diagonals.

And then suprisingly it stops. No regular polygon, however many sides it has, can have eight or more diagonals intersecting at a point other than the centre.

[1] https://mathproblems123.files.wordpress.com/2011/03/ngon.pdf

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Richard Walker

Beautiful Maths Problem

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Edited by Richard Walker, Wednesday, 17 Aug 2022, 17:33

Problem 1358 at gogeometry [1] asked for a proof that in a regular 12-sided polygon the four diagonal shown all meet at a point. This is quite surprising; it\'s not hard to find threee diagonals the meet at a single point but four is rarer.

After playing arouind for a while I found a proof which was reasonably neat, but I had to use sines and cosines at one point, and I'd hoped for something simpler; and there was nothing very illuminating about my proof in any case. Stan Fulger came up with something much more insightful. Here is his beautiful answer.

It uses two well-known facts about triangles.

The altitudes of a triangle, i.e. the lines drawn from each vertex at 90° to the opposite side, meet at a point.

The angle bisectors of a triangle, i.e. the lines which divide each angle in half, meet at a point.

For example

Now for a "look and see" proof.

In the picture below three diagonals are angle bisectors in the blue triangle, so they meet at a point. Also three diagonals are altitudes of the orange triangle and therefore meet at a point. Two of the diagonals are both a bisector in one triangle and an altitude in the other. Therefore all four diagonals meet at a point.

I drew the figures above using GeoGebra classic.

[1] Geometry problem 1358 https://gogeometry.com/school-college/4/p1358-dodecagon-regular-concurrency-diagonal-infographic-classes.htm

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