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# An irregular tiling

A triangle divided by three line segments. What fraction is blue?

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## Solution

The blue fraction is 2/7.

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## 6 replies on “An irregular tiling”

Dr. Wolfgang Ludwickisays:

Solution with ROUTH’s Theorem

Marshall Bucksays:

If an interior triangle has vertices given by barycentric coordinates
(a,b,c), (d,e,f), (g,h,i), then the fraction of the whole triangle is the 3×3 determinant formed from those 9 values.
In this case, the barycentric coordinates are (4/7,2/7,/1/7) and its cyclic rotations, so the 3×3 matrix is a circulant matrix formed by rotating those three. The determinant of a circulant matrix whose first row is (a,b,c) is just a^3+b^3+c^3 = 3abc. Substituting (4/7,2/7,1/7) gives the fraction 1/7.
Similarly, one shows that each of the smaller triangles has fraction 1/21, for a total area of 2/7 for all 4 triangles combined.

Marshall Bucksays:

There was a typo: the determinant of the circulant matrix is
a^3+b^3+c^3 – 3abc

Dr. Wolfgang Ludwickisays:

Solution without ROUTH’s Theorem

Peter H.says:

One more way

Peter H.says:

Even easier: