Let $ABCD$ be a cyclic quadrilateral, and $M$, $N$ be the midpoints of arcs $AB$ and $CD$ respectively. Prove that $MN$ bisects the segment between the incenters of triangles $ABC$ and $ADC$.
Problem
Source: Sharygin 2018
Tags: geometry
04.04.2018 22:29
Use the following fact:The incentres of triangles $ABC$, $BCD$, $ACD$ and $ABD$ form a rectangle. The rest should be easy.
04.04.2018 22:35
Let $X,Y$ be the midpoints of arc $BC,AD$ and let $I_1,I_2$ be the incenters of $\Delta ABC,\Delta ADC$ and $T=MN\cap XY$ ,now pascal theorem on $(ANMCYX),\implies T,I_1,I_2$ are collinear now simple use of sine law gives $TI_1=TI_2$ $\blacksquare$
04.04.2018 22:43
Define $I_a$ to be incenter of $\triangle BCD$, similarly define $I_b, I_c, I_d$. $I_a, I_b, C, D$ are concyclic, quick angle chase gives us that $I_aI_bI_cI_d$ is a rectangle, one of the perpendicular bisector of which is $MN$, result follows.
05.04.2018 08:14
Can also be done with complex numbers.
05.04.2018 08:21
Drunken_Master wrote: Can also be done with complex numbers. You have to be very careful there since dealing with midpoints of arcs of cyclic quadrilaterals is very difficult, lucky for us, because of the rectangle $I_aI_bI_cI_d$ form, that doesn't cause a problem in this case but the proof isn't essentially correct since we swapped the midpoints of arcs. This blog post by evan chen clarifies all the doubts Vrangr wrote: Define $I_a$ to be incenter of $\triangle BCD$, similarly define $I_b, I_c, I_d$. $I_a, I_b, C, D$ are concyclic, quick angle chase gives us that $I_aI_bI_cI_d$ is a rectangle, one of the perpendicular bisector of which is $MN$, result follows. Expanding upon my previous sketch, First remember the incentre-excentre lemma. Let $M_{AB}$ refer to the midpoint of the arc $AB$ not containing $C, D$, similarly define $M_{AD}$, $M_{CD}$ and $M_{BC}$, note that in the question's notation, $M \equiv M_{AB}$ and $N \equiv M_{CD}$. Let $I_A$ refer to the incenter of $\triangle BCD$, similarly define $I_B, I_C$ and $I_D$.
05.04.2018 09:46
05.04.2018 14:20
This is a really beautiful problem in the fact that it shows the importance of "completing the diagram".
20.07.2018 14:48
A different solution:
20.05.2021 10:37
We can also use Complex numbers easily , by letting - $a= x^2 $ $b= y^2$ $c= z^2 $ & $d = w^2$ $I_1 = - (xy + yz +zx) $ $I_2 = -(xz +zw + xw)$ $m= - xy$ $n = -wz$ And then using Complex Intersection Formula by letting $ Z= MN \cap I_1 I_2$ to find $z$ , and prove it equal to $ \frac{I_1 + I_2}{2} $...
28.08.2022 02:20
This is so nice. Call $I_A,I_B,I_C,I_D$ the incenters of $\triangle DAC, \triangle DBC, \triangle ADB, \triangle ACB$ respectivily, its known that $I_AI_BI_CI_D$ is a rectangle, now let $M_{AB}, M_{CD}$ the midpoints of arcs $AB,CD$ respectivily then by I-E Lemma applied 4 times we get that $M_{AB}$ lies in the perpendicular bisector of $I_DI_C$ and $M_{CD}$ lies in the perpendicular bisector of $I_AI_B$ and since $I_AI_BI_CI_D$ is a rectangle we get that $I_DI_C$ and $I_AI_B$ share perpendicular bisectors hence $M_{AB}M_{CD}$ bisects $I_AI_B$ thus we are done