Let $ABCD$ be a cyclic quadrilateral with circumcentre $O_1$. The diagonals $AC$ and $BD$ meet at point $P$. Suppose the four incentres of triangles $PAB, PBC, PCD, PDA$ lie on a circle with centre $O_2$. Prove that $P, O_1, O_2$ are collinear. Proposed by Shantanu Nene
Problem
Source: The 1st India-Iran Friendly Competition Problem 2
Tags: geometry, Incentres, incenter
math_comb01
13.06.2024 00:42
Nice Problem! (will add in details later, late at night currently) Call the incenters of $PAB,PBC,PCD,PDA$; $I_1,I_2,I_3,I_4$ respectively. Claim 1: $I_1I_2I_3I_4$ is cyclic iff $ABCD$ is tangential quadrilateral.
Unfortunately the only proof I can see right now is a trig+length bash, compute $\tan(PI_1I_2)$ and $\tan(PI_3I_4)$ and then menelaus bash gives $I_1I_2,I_3I_4,AC$ concur, now applying the following lemma and radax finishes:
Lemma: Let $\triangle ABC$ be a triangle with circumcircle $\omega$. Let $P$ be a variable point on the arc $BC$ of $\omega$ not containing $A$. Call the incenters of $ABP$ and $ACP$ to be $I_1$ and $I_2$, respectively. Then the circumcircle of $\triangle I_1PI_2$ passes through incircle touchpoint.
the idea is to drop perpendiculars from the incenters to $AC$ and then notice that by some trig the ratio of radii comes same which is enough to imply that the lines mentioned in proof 1 concur due to exsimiliscenters
Now notice $P,I_1,I_3$ and $P,I_2,I_4$ are collinear and $I_1I_3 \perp I_2I_4$, notice that due to isogonal conjugates in quadrilateral lemma, $O_2,P$ are isogonal conjugates in $I_1I_2I_3I_4$, we finish by observing that the pedal quadrilateral of $P$ is homothetic to $ABCD$. Remark: An alternate (artificial) finish after Claim 1 is to fit the diagram in incircle configuration.
Usernumbersomething
20.06.2024 08:55
Page 72 here for converse of claim 1: https://forumgeom.fau.edu/FG2011volume11/FG201108.pdf The iff statement is apparently well-known though I don’t know whether it’s citable.
sami1618
31.07.2024 17:45
Let $M_{AB}$, $M_{BC}$, $M_{CD}$, and $M_{DA}$ be the respective arc midpoints of $(ABCD)$. Let $I_a$, $I_b$, $I_c$, and $I_d$ be the incenters of triangles $ABD$, $BCA$, $CDB$, and $DAC$. Claim: $I_a I_b I_c I_d $ is a rectangle
We have that $ABI_aI_b$ is cyclic. So $\angle I_bI_aI_1=\angle I_1 M_{DA}M_{BC}$ thus $I_aI_b\parallel M_{BC} M_{DA}$. Similar statements on other sides combined with the fact that $M_{AB}M_{CD}\perp M_{BC}M_{DA}$ finish the claim.
Claim: $I_aI_bI_cI_d$ and $I_1I_2I_3I_4$ share a circumcircle
Let $M_{AB}M_{CD}$ intersect $M_{BC}M_{DA}$ at $O_2'$. As $I_2$ and $I_4$ are reflections about $M_{AB}M_{CD}$ we have that $O_2'I_2=O_2'I_4$. Similarly, $O_2'I_1=O_2'I_3$ so $O_2'=O_2$. As $M_{AB}M_{CD}$ is perpendicular to $I_aI_b$ and $\angle I_a M_{AB}M_{CD}=\angle I_bM_{AB}M_{CD}$ we have that $M_{AB}M_{CD}$ bisects $I_aI_b$. Cyclic statements imply that $O_2$ is also the center of $I_1I_2I_3I_4$. Thus $(I_aI_bI_cI_d)$ and $(I_1I_2I_3I_4)$ are concentric so $(I_1I_2I_3I_4)$ either coincides with $(I_aI_bI_cI_d)$ or lies completely outside or inside of it. As $\angle I_aI_1I_b+\angle I_bI_2I_c=270^{\circ}$ we must have that the circumcircles coincide.
Claim: $I_1I_2I_3I_4$ and $M_{AB}M_{BC}M_{CD}M_{DA}$ are homothetic
As their diagonals are parallel, it suffices to show that they have parallel sides. Notice $$\angle I_2I_1I_b=\angle I_2I_cI_b=\angle I_2 M_{DC}M_{AB}=\angle I_bM_{BC}M_{AB}$$
Let $H$ be the center of the homothety sending $I_1I_2I_3I_4$ to $M_{AB}M_{BC}M_{CD}M_{DA}$. Their circumcenters are also homothetic so $H$, $O_2$, and $O_1$ are collinear. Also the intersection points of their diagonals are homothetic so $H$, $P$, and $O_2$, are collinear. Thus, $P$, $O_1$, and $O_2$ are collinear, as desired.
Attachments:
Pyramix
23.09.2024 15:49
Define $I_a, I_b, I_c, I_d$ to be incenters of $PAB, PBC, PCD, PDA$, respectively. The internal angle-bisectors of $\angle APB (=\angle CPD)$ and those of $\angle BPC (=\angle DPA)$ are mutually perpendicular. Choose $I_aI_c$ and $I_bI_d$ as $x-$ and $y-$axes, respectively, with $P$ as origin. So, $AC$ and $BD$ are symmetric about $x-$ and $y-$ axes. Let $A'$ be point on line $I_bI_d$ such that $A',I_a,B$ are collinear. Note that \[\angle AI_aB=90^{\circ}+\angle API_a\Longrightarrow \angle A'I_aA = 90^{\circ}-\angle API_a=\angle A'PA.\]Hence, $A,P,I_a,A'$ are concyclic. We now use coordinates. It can be shown that the coordinates of $A$ are a linear combination of those of $I_a,I_d$. The problem statement follows from this. $\blacksquare$