# Five points in a disk

Why is $$\min_{1\le i,j,k\le5}\frac{\mbox{Area}\left(\triangle P_{i}P_{j}P_{k}\right)}{\mbox{Perimeter}\left(\triangle P_{i}P_{j}P_{k}\right)}<\frac{4}{25}$$ for any five points $P_{1}$, $P_{2}$, $P_{3}$, $P_{4}$ and $P_{5}$ in the unit disk? Thanks in advance for any helpful answers!

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Of course you mean $P_n = \exp(2 i \frac{n\pi}{5})$ – Robert Israel Feb 23 '13 at 0:21
If there are not, how to show that this min is always less than $4/25$ for any five points in the unit disk? Thanks. – Joia Feb 23 '13 at 0:49

Since the area is quadratic in length and the perimeter is linear, you want as big a triangle as possible. The natural points are a regular pentagon on the circle, so $P_n=\exp (i\frac {2n \pi}5)$. I find the ratio about $0.15$ for the triangle with two sides of the pentagon and $0.21$ for the triangle with two diagonals. It seems unlikely that you can do better, but I don't have an easy proof.
I get the area/perimeter ratios to be $(\sqrt{5}-1)/8 \approx .1545084971$ and $(3 \sqrt{5}-5)/8 \approx .2135254914$ for those two triangles. – Robert Israel Feb 23 '13 at 0:36
Even an equilateral triangle inscribed in the unit circle has ratio only $1/4$. I'm pretty sure this is the largest such ratio for any triangle in the unit disk. – Robert Israel Feb 23 '13 at 0:45