# A hard problem from regular $n$-gon

I found the following theorem in a book of mine without a proof. Could someone show me a proof of it?

Given a regular $n$-gon, with $n$ odd and vertices $v_1,\ldots,v_n$, and $C$ its circumcircle. At each $v_i$ draw a circle that is internally tangent to $C$ at $v_i$, and suppose all these tangent circles are congruent. Let $P$ be any point on the minor arc from $v_1$ to $v_n$ and let $t_i$ be the length of the tangent from $P$ to the circle tangent to $C$ at $v_i$. Then $\sum_{i=1}^n(-1)^it_i=0$.

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Sorry but drawing is too difficult for me as I have a cerebral palsy. – GeometryFan May 11 '12 at 13:27
"the length of the tangent from $P$ to the circle tangent to $C$ at $v_i$ But there are two possible tangents, no? – leonbloy May 11 '12 at 13:59
@leonbloy: But they both have the same length. – TonyK May 11 '12 at 14:05
The description seems pretty straightforward to me. But I have no idea how to prove the thing. – TonyK May 11 '12 at 14:06
The theorem was stated of the book "Which Way did the Bicycle Go" and it was said that this is a generalization of the Casey's theorem. Furthermore, the book says the proof can be found on the book "Ptolemy's Legacy" but I don't have that book. – Jaakko Seppälä May 11 '12 at 21:54

A diagram :

An attempt using complex arithmetic:

Let $R=$radius of C (circumcircle, center at the origin of the complex plane), $r=$radius of inner circles, $c_k$ their centers ($k=1\cdots n$). Further, let

$$\begin{array}{cl} q &=&e^{i 2 \pi /n}\\ c_k &=& s \, q^k , \hskip{1cm} s = R -r\\ P &=& R \, q^u , \hskip{1cm} u \in (0,1)\\ \end{array}$$

For each circle $k$, let $l_k = c_k -P$ (distance from the center to the point P), let $t_k$ the length of the tangent from P to the circle. Because the tangent is normal to the radius, we have:

$$|l_k|^2 = t_k ^2 + r^2$$

$$t_k^2 = |s q^k - R q^u|^2 -r ^2=s^2+R^2 - s \,R (q^{k-u} + q^{u-k}) -r^2= s \, R \, \left[ 2 - (q^{k-u} + q^{u-k})\right]$$

Using $(z^y-z^{-y})^2 = z^{2y} + z^{-2y} -2$, we get

$$t_k^2 = - s \, R \, \left[ q^{(k-u)/2} - q^{(u-k)/2}\right]^2$$

or, setting $p=q^{1/2}$ and multiplying by $q^{nk}$:

$$(p^{n k } t_k)^2 = s \, R \left[ -i p^{n k } \left( p^{k-u} - p^{u-k} \right)\right]^2= s \, R \left[ p^{n k} 2 \, Im\left( p^{k-u} \right)\right]^2$$

But $Im\left( p^{k-u} \right) \ge 0$ for $k=1\cdots n$ and $u \in (0,1)$; and $p^{nk}=(-1)^k$. Hence inside the squares on each side we have real numbers with the same sign (for each $k$), and so we can take safely the square root and sum over $k$. On the right side we have a geometric sum:

$$\sum_{k=1}^n p^{n k} p^{k-u} = p^{-u} \sum_{k=1}^n p^{(n+1) k} = p^{-u} p^{n+1} \frac{1-p^{(n+1)n}}{1-p^{n+1}}$$

But $$p^{(n+1)n}= q^{\frac{(n+1)n}{2}}=1$$ for $n$ odd integer, so finally

$$\sum_{k=1}^n (-1)^k t_k = 0$$

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