2
$\begingroup$

This question is closely related to my previous question.

Can you provide a proof for the following claim:

In any regular convex pentagon $ABCDE$ construct an arbitrary tangent to the incircle of pentagon . Let $d_1,d_2,d_3,d_4,d_5$ be a signed distances from vertices $A,B,C,D,E$ to tangent line respectively, such that distances to a tangent from points on opposite sides are opposite in sign, while those from points on the same side have the same sign. Denote the side length of pentagon by $a$ and the area of pentagon by $K$ ,then $a(d_1+d_2+d_3+d_4+d_5)=2K$

enter image description here

GeoGebra applet that demonstrates this claim can be found here.

$\endgroup$

1 Answer 1

1
$\begingroup$

$\newcommand{real}{\operatorname{Re}}$Set up a coordinate system where the origin is the pentagon centre; without loss of generality take the pentagon's inradius as $1$ and the tangent line as $x=-1$. Then $a=2\tan\frac\pi5$, the pentagon circumradius $R=\sec\frac\pi5$ and $K=5\tan\frac\pi5$.

The argument of $A$ may be any angle $\theta$, but then $d_1=1+\real(Re^{i\theta})=1+R\real(e^{i\theta})$. Thus $$d_1+d_2+d_3+d_4+d_5=5+R\real\left(\sum_{k=0}^4e^{i(\theta+2k\pi/5)}\right)$$ The terms in the sum of exponentials are the roots of $z^5=e^{5i\theta}$, so by Viète's relations that sum is the negative of the $z^4$ coefficient, which is zero. Therefore $d_1+d_2+d_3+d_4+d_5=5$ and the equation to prove reduces to $5a=2K$, which is easily seen to be true.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .