This question is closely related to my previous question.

Can you prove the claim given below? The following claim is a conjectured generalization of Harcourt's theorem.

Claim. Let $A_1,A_2 \ldots A_n$ be the vertices of $n$-sided bicentric polygon with semiperimeter $s$ and area $K$. Let $d_i$ be the length of the line segment whose endpoints are polygon's vertices adjacent to the vertex $A_i$ . Let line $t$ be the tangent to the polygon's incircle at any point on that circle. Denote the signed perpendicular distance of the vertex $A_i$ from the tangent line as $n_i$ , with a distance being negative if and only if the vertex is on the opposite side of the tangent line from the incenter. Then, $$K=\frac{s \cdot \displaystyle\sum_{i=1}^n n_i \cdot d_i}{\displaystyle\sum_{i=1}^n d_i}$$

Picture for the case $n=5$:

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GeoGebra applets that demonstrate this claim in the case of $n=5$ and $n=6$: Area of bicentric pentagon , Area of bicentric hexagon. The proofs for the case $n=3$ and the case $n=4$ can be found here and here ,respectively.

  • $\begingroup$ $K/s$ is just the inradius, right? $\endgroup$ Feb 6 '21 at 9:28
  • $\begingroup$ @FedorPetrov Of course. $\endgroup$ Feb 6 '21 at 9:40
  • $\begingroup$ Then the formulation $r=(\sum n_id_i)/(\sum d_i)$ looks more natural for me. $\endgroup$ Feb 6 '21 at 9:48
  • $\begingroup$ @FedorPetrov I have formulated the claim in this way because Harcourt's theorem is about area of a triangle. $\endgroup$ Feb 6 '21 at 10:07
  • $\begingroup$ This is just a small suggestion which applies to all the versions of Harcourt's theorem. A back of the envelope calculation shows that the rhs of the equation at the end of your claim is, as a function of a point $(x,y)$, a quadratic form whose contours are circles with centre at the incentre. This reduces the proof to showing that the formula holds for one of the points of contact. It also allows extensions to results about other circles (i.e., other than the incircle). $\endgroup$ Feb 9 '21 at 12:32

So we need to show $r\cdot \sum d_i=\sum n_id_i$, where $r$ denotes the inradius. Rewrite this as $0=\sum (n_i-r)d_i$, denote $n_i-r=m_i$. Then $m_i=:f(A_i)$ is the signed distance from $A_i$ to the line passing through $I$ and parallel to $t$. I claim $$\sum d_iA_i=(\sum d_i)I,\quad\quad\quad\quad(\heartsuit)$$ this yields $0=(\sum d_i)f(I)=f(\sum d_iA_i)=\sum d_if(A_i)=\sum d_im_i$.

Note that $d_i=2R\sin \angle A_{i-1}A_iA_{i+1}$ where $R$ is the circumradius. Denote by $B_i$ the tangency point of $A_iA_{i+1}$ and the incircle. Then $IB_{i-1}A_iB_i$ is a cyclic quadrilateral with diameter $IA_i$, thus $B_{i-1}B_i=IA_i\cdot \sin \angle A_{i-1}A_iA_{i+1}$, and the vector $(A_i-I)\cdot \sin \angle A_{i-1}A_iA_{i+1}$ is obtained from $B_i-B_{i-1}$ by a rotation on angle $\pi/2$ (clockwise on your picture). Since the sum of vectors $B_i-B_{i-1}$ is zero, the sum of rotated vectors is also 0, as needed.


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