Pentagon Star


In a regular pentagon the diagonals are joined to form a star.

What fraction of the pentagon does the star occupy?


Consider the following diagam.

As the interior angle of the pentagon, $a = 108^o$, $2p = 180 - 108 = 72$, therefore $p = 36$.

But $a = 2p + q$, $108 = 72 + q \implies q = 36$; that is, the two diagonals in a regular pentagon trisect the interior angle.

In the diagram below, $M$ is the midpoint of $DE$ and $O$ is the centre.

It should be clear that we only need consider the fraction of triangle $ODM$ that is shaded in order to determine the fraction of the whole pentagon that is occupied by the star.

In addition, as $\Delta ODM$ and $\Delta PDM$ share the same base we need only consider the ratio of their respective heights, $PM$ and $OM$.

As $OD$ bisects $\angle ADB$, $\angle ODM = 18 + 36 = 54$, so $OM = DM \tan(54)$, and $PM = DM \tan(36)$.

Therefore $\Delta PDM$ occupies $\dfrac{\tan(36)}{\tan(54)} \approx 53$% of triangle $ODM$, and as this represents the fraction that is not shaded, we deduce that the star occupies approximately 47% of pentagon.

Prove without the use of a calculator that $\dfrac{\tan(36)}{\tan(54)} \gt \dfrac{1}{2}$.

Problem ID: 271 (16 Feb 2006)     Difficulty: 3 Star

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