Ratio of division of line perpendicular to $OI$

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

anantmudgal09

1979 posts

anantmudgal09

#1 h Aug 4, 2016, 1:22 PM • 3 Y

Y by artsolver, Adventure10, Mango247

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 $XY$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

anantmudgal09

1979 posts

anantmudgal09

#2 h Aug 4, 2016, 2:08 PM • 1 Y

Y by Adventure10

Solution 1.

Solution 2.

This post has been edited 2 times. Last edited by anantmudgal09, Aug 4, 2016, 2:33 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

anantmudgal09

1979 posts

anantmudgal09

#3 h Oct 5, 2016, 6:55 PM • 1 Y

Y by Adventure10

Solution 3.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

TelvCohl

2312 posts

TelvCohl

#4 h Dec 14, 2016, 6:08 AM • 6 Y

Y by anantmudgal09, Itticantdomath, enhanced, Gaussian_cyber, Adventure10, Om245

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:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Itticantdomath

25 posts

Itticantdomath

#5 h Dec 25, 2016, 5:56 PM • 2 Y

Y by Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

buzzychaoz

178 posts

buzzychaoz

#6 h Dec 25, 2016, 7:11 PM • 2 Y

Y by Adventure10, Mango247

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 $\overline{DEF} \perp \overline{IOV}$ .

This post has been edited 1 time. Last edited by buzzychaoz, Dec 25, 2016, 7:13 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

k.vasilev

66 posts

k.vasilev

#7 h Jul 24, 2017, 10:52 PM • 2 Y

Y by Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

jayme

9775 posts

jayme

#8 h Jun 5, 2018, 12:40 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=160902

Sincerely
Jean-Louis

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

math_pi_rate

1218 posts

math_pi_rate

#9 h Jul 21, 2018, 3:21 PM • 2 Y

Y by Adventure10, Mango247

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 $XM \Rightarrow XI:IY = 1:2$

This post has been edited 1 time. Last edited by math_pi_rate, Jul 21, 2018, 3:22 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

dgreenb801

1896 posts

dgreenb801

#10 h Jul 11, 2020, 2:30 AM • 1 Y

Y by lilavati_2005

See my explanation of this problem on my Youtube channel here. This was a really nice one!

https://www.youtube.com/watch?v=UZ55y4fQbmk

Z K Y