# Concyclic incenters of triangles concerning Apollonius circle

Problem:

$$ABCD$$ is a tangential quadrilateral and $$P$$ is a point such that $$\dfrac{PB}{PD}=\dfrac{AB}{AD}=\dfrac{CB}{CD}.$$ Let $$I_1$$, $$I_2$$, $$I_3$$, $$I_4$$ be the incenters of $$\triangle PAB$$, $$\triangle PBC$$, $$\triangle PCD$$, $$\triangle PDA$$, respectively.

Proof that $$I_1$$, $$I_2$$, $$I_3$$ and $$I_4$$ are concyclic.

I have read this problem Prove that $I_1, I_2, I_3, I_4$ are concyclic, but it is not actually similar to my problem.

I suspect the problem has something to do with Apollonius circles (i.e., circle $$PAC$$), but I don't know how to use this information in any way. I cannot understand how Apollonius circles are connected to incenters.

Edit:

My analysis is that the problem clearly implies two cases: either $$AC$$ is the perpendicular bisector of $$BD$$, or $$AB = CB$$ and $$AD = CD$$ (a kite). The first case is trivial, since $$I_1$$, $$I_2$$, $$I_3$$, $$I_4$$ here forms a rectangular. The real difficult part of this problem is the second case.

Based on observations, I have made out the following results, but I cannot prove them. They may provide some clues for the original problem. I would appreciate it if anyone could help with these conjectures too:

Assume the outer and inner bisectors of $$\angle BAD$$ intersect $$BD$$ at point $$M$$ and $$N$$ respectively (so $$M$$ and $$N$$ lie on the Apollonius circle), then:

1. $$M$$, $$I_1$$ and $$I_4$$ are colinear, likewise are $$M$$, $$I_2$$ and $$I_3$$;
2. $$N$$, $$P$$, $$I_1$$ and $$I_2$$ are concyclic, likewise are $$N$$, $$P$$, $$I_3$$ and $$I_4$$.
• Is this some olyimpiad problem, perhaps from IMO shortlist? Jul 15, 2019 at 15:23
• @Aqua Likely it's an olympiad-level problem. My teacher said he came up with the problem by himself, and he only found a proof that's really complicated. Jul 15, 2019 at 15:44

Partial solution:

Well say $$a=AB$$, $$b=BC$$, $$c=CD$$ and $$d=DA$$, then we have $${a\over d} = {b\over c} =:t \implies a= dt \;\;\wedge \;\; b=ct$$

Since $$ABCD$$ is tangential we have $$a+c = b+d\implies d (t-1)= c(t-1)$$

From here we have two options:

• If $$t=1$$ then $$a=d$$ and $$b=c$$ and thus also $$PB =PD$$. So $$A,C,P$$ lies on a perpendicular bisector for $$BD$$ and $$I_1I_2I_3I_4$$ make rectangle and we are done.
• If $$t\ne 1$$ then $$d=c$$ and $$a=b$$...
• Wang, It ca be shown that in quadrilateral ABCP points $I_1$ and $I_2$ are on a circle. In quadrilateral BCDP points $I_2$ and $I_3$ are on another circle and in APCD points $I_3$ and $I_4$ are on a circle. Jul 16, 2019 at 8:28

COMMENT:

Let angles A and B considered as $$A_1$$for left part and $$A_2$$ for right part; similarly $$C_1$$for left and $$C_2$$ for right part of C. For B and D, $$B_1$$ for the top of line BD and $$B_2$$ for the below. Similarly $$D_1$$ for the top and $$D_2$$ for bellow for angle D.We can write:

$$\angle I_1=180-(\frac{A_1}{2}+\frac{B_1}{2})$$

$$\angle I_2=180-(\frac{B_2}{2}+\frac{C_1}{2})$$

$$\angle I_3=180-(\frac{C_2}{2}+\frac{D_2}{2})$$

$$\angle I_4=180-(\frac{D_1}{2}+\frac{A_2}{2})$$

Summing these relations we get:

$$720-(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}+\frac{D}{2})=720-\frac{360}{2}=540$$

Also:

$$\angle I_2+\angle I_4=180-(\frac{B_2}{2}+\frac{C_1}{2})+180-(\frac{D_1}{2}+\frac{A_2}{2})=360-(\frac{B_2}{2}+\frac{C_1}{2}+\frac{D_1}{2}+\frac{A_2}{2})$$

The sides of these angles cross two by two mutually and construct an octagon.

Now when P is on BD, we have:

$$(\frac{B_2}{2}+\frac{C_1}{2}+\frac{D_1}{2}+\frac{A_2}{2})=\frac{360}{4}=90$$

$$\angle I_2+\angle I_4=360-90=270$$

similarly:

$$\angle I_1+\angle I_3=360-90=270$$

Now we must prove two points:

1: for all positions of P we have:

$$(\frac{B_2}{2}+\frac{C_1}{2}+\frac{D_1}{2}+\frac{A_2}{2})=\frac{360}{4}=90$$

2: In an octagon if the sum of two opposite angle is $$270^o$$ the octagon is cyclic.