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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.

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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:

enter image description here

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 $I_4$.
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  • $\begingroup$ Is this some olyimpiad problem, perhaps from IMO shortlist? $\endgroup$
    – nonuser
    Jul 15, 2019 at 15:23
  • $\begingroup$ @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. $\endgroup$ Jul 15, 2019 at 15:44

2 Answers 2

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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 $a=b$...
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  • $\begingroup$ 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. $\endgroup$
    – sirous
    Jul 16, 2019 at 8:28
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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.

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