# Collinearity in bicentric pentagon

Can you provide a proof for the following claim:

Claim. The circumcenter, the incenter, and the excenter of the pentagon formed by diagonals in a bicentric pentagon are collinear.

GeoGebra applet that demonstrates this claim can be found here.

My idea is to show that $$|OJ|=|OI|+|IJ|$$ . By Fuss formula we know that $$r(R-|OI|)=(R+|OI|)\sqrt{(R-r+|OI|)(R-r-|OI|)}+(R+|OI|)\sqrt{2R(R-r-|OI|)}$$

where $$R$$ and $$r$$ are circumradius and inradius of bicentric pentagon, respectively. But how to express lengths $$|OJ|$$ and $$|IJ|$$ in terms of $$R$$ and $$r$$ ?

• Could you give a reference of Fuss formula in this case ? (I know Fuss formula for quadrilaterals, not for pentagons) Commented Nov 7, 2023 at 0:55
• @JeanMarie I added a link. Commented Nov 7, 2023 at 17:51
• @brainjam Thank you ! Commented Nov 7, 2023 at 18:13

The short answer is that the three circles are coaxal, and therefore their centers are collinear. The circles are coaxal because of the following lemma (screen shot from Johnson, Modern Geometry aka Advanced Euclidean Geometry, pg 92):

Proof later, but you might take a look at another question and answers.

All of this is related to Poncelet's Porism, one general version of which is (excerpted from Hraskó, Poncelet's theorem, well worth reading in full):

Poncelet's General Theorem: Let $$e$$ be a circle of a non-intersecting pencil and let $$a_1,a_2,\ldots,a_n$$ be (not necessarily different) oriented circles in the interior of $$e$$ that belong to the same pencil. Starting at an arbitrary point $$A_0$$ of the circle $$e$$, the points $$A_1,A_2,\ldots,A_n$$ are constructed on the same circle, such that the lines $$A_0A_1, A_1A_2, \ldots, A_{n-1}A_n$$ touch the circles $$a_1,a_2,\ldots, a_n$$, respectively, in the appropriate direction. It may happen that at the end of the construction, we get back to the starting point, that is, $$A_n=A_0$$. The theorem states that in that case, we will always get back to the starting point in the $$n$$-th step, whichever point of $$e$$ we start from. We do not even need to take care to draw the tangents to the circles in a fixed order.

Here's a page from Poncelet's Traité des propriétés projectives des figures, 1827 that illustrates the lemma.

But we don't need the general theorem, other than to observe that a Poncelet Porism will involve (or generate) circles in a coaxal system.

By Poncelet's porism, you can move the pentagon in the OP continuously with its vertices on the circumcircle and sides touching the incircle. Along for the ride will go triangles like $$A_1A_2A_3$$ with sides $$A_2A_1,A_2A_3$$ tangent to the incircle and line $$d_2$$ which must be tangent to the diagonals incircle (it's shape will vary). By the lemma, the diagonals incircle must be coaxal with the other two circles.

The proof of the lemma is excerpted here (Johnson, pg 93):

Consider the effect of a continuous movement of $$A_1,A_2$$ to $$B_1,B_2$$, keeping the points on the circumcircle and maintaining tangency with the incircle. Let $$A_1A_2$$ be tangential at $$P_1$$ and let $$B_1B_2$$ be tangential at $$P_2$$.

Let line $$P_1P_2$$ cut $$A_1B_1$$ at $$Q_1$$ and cut $$A_2B_2$$ at $$Q_2$$. Then $$A_1B_1$$ and $$A_2B_2$$ are equally inclined to line $$Q_1Q_2$$ and so there is a circle $$\mathscr{C}$$ tangential at $$Q_1$$ and $$Q_2$$ to the lines $$A_1B_1$$ and $$A_2B_2$$.

Let the line $$Q_1Q_2$$ make successive angles $$p,q,q,p$$ with the four tangents at $$Q_1,P_1,P_2,Q_2$$. Then, applying the sine rule, the ratio of the powers of each of $$A_1,A_2,B_1,B_2$$ with respect to the two circles is $$\frac {\sin q}{\sin p}\text{, a constant}.$$ The circumcircle, through these four points, is therefore coaxal with $$\mathscr{C}$$ and the incircle.

Similarly, we can determine a further coaxal circle, $$\mathscr{C'}$$, for the movement of $$A_2,A_3$$ to $$B_2,B_3$$. The coaxal circles $$\mathscr{C}$$ and $$\mathscr{C'}$$ have a common tangent, $$A_2B_2$$ and so are either equal or on opposite side of the radical axis of the system but the latter possibility is impossible by continuity.

We can now apply the same idea to the movement of $$A_1,A_3$$ to $$B_1,B_3$$ with respect to the circumcircle and $$\mathscr{C}$$. Then $$A_1A_3$$ and $$B_1B_3$$ are both tangential to a further coaxal circle $$\mathscr{D}$$.

We can now consider successive moves of each $$A_i$$ to $$A_{i+1}$$, considering the indices modulo $$5$$.Then each edge $$A_iA_{i+2}$$ is tangential to $$\mathscr{D}$$ and so the incentre of the pentagon is the coaxal circle $$\mathscr{D}$$.