Let $ABCD$ be a circumscribed quadrilateral. Prove that the common point of the diagonals, the incenter of triangle $ABC$ and the centre of excircle of triangle $CDA$ touching the side $AC$ are collinear.
Problem
Source: Sharygin 2018 Grade 9 Day 2 Problem 9.6
Tags: incenter, excenter, diagonals, geometry, circumscribed quadrilateral
01.08.2018 21:40
My solution: Let $I$ and $I_1$ be the incenter of $\triangle ABC$ and the $D$-excenter of $\triangle CDA$. Let $O$ be the incenter of quadrilateral $ABCD$. WLOG assume that $AB \leq CD$. Finally let $AC \cap BD = P,AD \cap BC = Q,AI_1 \cap CI =T$. Then, $T$ is the incenter of $\triangle ACQ$. As $O$ is the incenter of $\triangle QCD$, we get that $T$ lies on $QO$. Now, Notice that $I$ lies on $BO$ and $I_1$ lies on $DO$. Also, by our previous claim, we get that $\triangle BOD$ and $\triangle CTA$ are perspective from a point. Thus, these triangles should also be perspective from a line. Hence, $P,I,I_1$ are collinear. Remark: The official solution used Monge's Theorem on the 3 circles given to show that $P$ is the exsimilicenter of the incircle of $\triangle ABC$ and the $D$-excircle of $\triangle CDA$.
01.08.2018 22:41
This fact was already known (at least to me) before the competition, anyways here is the solution with Monge's: Let $(J_D)$, $(I)$, $(I_B)$ and $(I_D)$ be the D-excircle of $ACD$, incircle of $ABCD$, incircle of $ABC$ and incircle of $ADC$ respectively. From (the positive variant of) Monge's on $(I, I_B, J_D)$, we see that the exsimilicenter of $I_B$ and $J_D$ lies on $BD$. Now use the fact that $(I_B)$ and $(I_D)$ are tangent to $AC$ at the same point( see ISL 2017/G7 or Brazil 2017 ) and use (the negative variant of) Monge's on $(I_B, I_D, J_D)$ to see that the exsimilcenter of $J_D$ and $I_B$ also lies on $AC$, thus it is $P$, the intersection of the diagonals. Note that all the discussion above yields the nice fact that $IP$ bisects $I_BI_D$.
02.08.2018 13:09
26 persons solved it during the contest.
24.07.2022 21:56
Let $L = AC \cap BD$, $X = BI \cap AC$, $Y = DI \cap AC$, $I_B$ be incenter of $\triangle ABC$, $I_D$ be $D$-excenter of $\triangle ADC$. Look at lines $BI,DI$. By Prism Lemma, we obtain our problem is equivalent to (projecting through $L$) $$ (B,X ; I,I_B) = (D,Y ; I,I_D) $$ But it is easy to show that $$ A(B,X ; I,I_B) = A(D,Y ; I,I_D) $$by writing everything in terms of Sines (everything just cancels out each other). $\blacksquare$