Let $O$ and $I$ be the circumcenter and incenter of triangle $ABC$. The perpendicular from $I$ to $OI$ meets $AB$ and the external bisector of angle $C$ at points $X$ and $Y$ respectively. In what ratio does $I$ divide the segment $XY$?
Problem
Source: Sharygin Geometry Olympiad, Final Round 2016, Problem 3 grade 9
Tags: geometry
anantmudgal09
04.08.2016 17:08
We claim that the ratio is $1:2$. Let $\ell$ be the line through $I$ perpendicular to $OI$. Indeed, consider the excentral triangle $I_CI_BI_A$ and let us dilate $A,B,X$ to $A',B',X'$ about $I$ with ratio $2$. Then, we want $IX'=IY$. However, this follows from the butterfly's theorem in $(I_AI_BI_C)$ for the cyclic quadrilateral $A',B',I_B,I_A$ since $I$ bisects the intercepts of the line $\ell$ with the circle $(I_AI_BI_C)$.
Let $DEF$ be the intouch triangle of $ABC$. Let $C'$ be a point on the ray $CI$ such that $CI=2C'I$. We shall prove that line $OI$ is tangent to the circumcircle of triangle $IFC'$. Consider an inversion about point $I$ which fixes the incircle. It sends $C$ to the midpoint $F_1$ of $DE$ and fixes $F$. Point $C'$ is mapped to a point $T$ on the ray $F_1I$ beyond $I$ so that $IT=2IF'$. We need to prove that $IO$ is parallel to $FT$. We shall use complex numbers in triangle $DEF$, letting $I=0$, $D=d,E=e,F=f$ with modulus $1$. Let $O'$ be the orthocenter of triangle $DEF$. We know that points $I,O,O'$ are collinear. Notice that $T=-(e+d)$. Thus, $FT$ is of direction $f-(-(e+d))=d+e+f$ and $IO'$ is also of the same. The claim holds. Now, let $(IFC')$ meet $AB$ again at $X'$. We see that $\angle X'FI=\angle OIX'=90^{\circ}$. Therefore, $X'=X$.
anantmudgal09
05.10.2016 21:55
We claim that the desired ratio is $2:1$. Indeed, let $DEF$ be the contact triangle of $ABC$ and let $A'B'C'$ be the medial triangle of $DEF$; $G$ be the centroid of triangle $DEF$. It is well-known that the points $I,G,O$ are collinear and so an inversion at $I$ fixes the line $IGO$. Point $C$ is mapped to $C'$ and the point $X$ is mapped to a point $X'$ such that $C'X' \parallel IG$ and $\angle C'X'I=90^{\circ}$; and the point $Y$ is mapped to a point $Y'$ such that $\angle IY'F=90^{\circ}$ and $FY' \parallel IG$. It is known from the conditions that $X',I,'Y'$ are collinear. We obtain that $$\frac{IX}{IY}=\frac{IY'}{IX'}=\frac{FG}{GC'}=2:1$$and the result follows.
TelvCohl
14.12.2016 09:08
Let $ \triangle I_aI_bI_c $ be the excentral triangle of $ \triangle ABC $ and $ \overline{DEF} $ be the perspectrix of $ \triangle ABC, $ $ \triangle I_aI_bI_c. $ It's well-known that $ \overline{DEF} $ is perpendicular to $ OI $ $ \Longrightarrow $ $ \overline{DEF} $ $ \parallel $ $ XY, $ so notice $ F(D,X;C,I_c) = -1 $ we get the reflection $ Z $ of $ Y $ in $ X $ lies on $ I_cF, $ hence combining $ (X,Y;I,Z) = F(A,C;I,I_c) = -1 $ we conclude that $ IY $ $ = $ $ 2IX. $
Attachments:
Itticantdomath
25.12.2016 20:56
TelvCohl wrote: Let $ \triangle I_aI_bI_c $ be the excentral triangle of $ \triangle ABC $ and $ \overline{DEF} $ be the perspectrix of $ \triangle ABC, $ $ \triangle I_aI_bI_c. $ It's well-known that $ \overline{DEF} $ is perpendicular to $ OI $ $ \Longrightarrow $ $ \overline{DEF} $ $ \parallel $ $ XY, $ so notice $ F(D,X;C,I_c) = -1 $ we get the reflection $ Z $ of $ Y $ in $ X $ lies on $ I_cF, $ hence combining $ (X,Y;I,Z) = F(A,C;I,I_c) = -1 $ we conclude that $ IY $ $ = $ $ 2IX. $ Can you give me the proof of the well known identity? Sorry if it's too obvious.
buzzychaoz
25.12.2016 22:11
Let $V$ be the circumcenter of $I_aI_bI_c$, since $DI_b \times DI_c = DB\times DC$, so $D$ lies on the radical axis of $(I_aI_bI_c),(ABC)$, similarly $E,F$ lie on it too. Note $I,O,V$ are collinear on the Euler line of $\triangle I_aI_bI_c$, hence $\overline{DEF} \perp \overline{IOV}$.
k.vasilev
25.07.2017 01:52
Another solution to the problem is as follows: First let $M$ be the midpoint of $IY$. We will prove that $\measuredangle XOI= \measuredangle IOM$ and the conclusion will follow. Let $K$ be the midpoint of line segment $AB$. We have that $\measuredangle OKX= \measuredangle OIX=90^{\circ} \Rightarrow O,X,K,I $ are concyclic. Therefore $\measuredangle BKI= \measuredangle XOI$. Now because $\measuredangle ICY= 90^{\circ}$ point $M$ lies on the perpendicular bisector of $CI$. Now let $AI \cap \omega=M_A, BI \cap \omega=M_B$. It is easy to check that $M_{A}M_{B}$ is the perpendicular bisector of $CI \Rightarrow M \in M_{A}M_{B}$. Let N be the midpoint of line segment $M_{A}M_{B}$. We have that$\measuredangle OLM=90^{\circ}=\measuredangle OIM$. Therefore $O,N,M,I$ are concyclic $\Rightarrow \measuredangle INM= \measuredangle IOM$ , but $\triangle IM_{B}M_{A} \sim \triangle IAB$, which means $\measuredangle ILM=\measuredangle IKB$. Hence $\measuredangle XOI=\measuredangle IOM$ as desired.
jayme
05.06.2018 15:40
Dear Mathlinkers, also at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=160902 Sincerely Jean-Louis
math_pi_rate
21.07.2018 18:21
My solution: Let $AI \cap \odot (ABC) = T, BI \cap \odot (ABC) = Z, ZT \cap XY = M, ZT \cap CI = W$. By Fact 5, $TI = TC$ and $ZI = ZC$ $\Rightarrow ZT$ is the perpendicular bisector of $CI$ $\Rightarrow W$ is the midpoint of $CI$. Also, By Fact 5, $T$ is the midpoint of $II_A$, where $I_A$ is the $A$-excenter of $\triangle ABC$ $\Rightarrow M$ is the midpoint of $CI$. Now, Let $XY \cap \odot (ABC) = U,V \Rightarrow$ As $OI \perp UV, I$ is the midpoint of $UV$. Thus, By Butterfly Theorem on the lines $ZB$ and $AT$, we get that $I$ is the midpoint of $XM \Rightarrow XI:IY = 1:2$
dgreenb801
11.07.2020 05:30
See my explanation of this problem on my Youtube channel here. This was a really nice one! https://www.youtube.com/watch?v=UZ55y4fQbmk