collinearity wanted, related to incenter and circumcircle

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

collinearity wanted, related to incenter and circumcircle

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 16th Thailand Mathematical Olympiad 2019 day 2

The post below has been deleted. Click to close.

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

30629 posts

#1 h May 21, 2019, 2:46 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle such that $AB\ne AC$ and $\omega$ be the circumcircle of this triangle.
Let $I$ be the center of the inscribed circle of $ABC$ which touches $BC$ at $D$ .
Let the circle with diameter $AI$ meets $\omega$ again at $K$ .
If the line $AI$ intersects $\omega$ again at $M$ , show that $K, D, M$ are collinear.

This post has been edited 1 time. Last edited by parmenides51, May 21, 2019, 2:48 PM
Reason: name typo

Z K Y

The post below has been deleted. Click to close.

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

40 posts

#2 h May 21, 2019, 2:46 PM • 3 Y

Y by Com10atorics, Adventure10, Mango247

do you have to chat as soon has i get aops up?

Z K Y

The post below has been deleted. Click to close.

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

30629 posts

#3 h May 21, 2019, 2:53 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

DiamondPuppy22 wrote:

do you have to chat as soon has i get aops up?

I do not understand what you just said, probably you meant to answer somewhere else
here we do not chat

Z K Y

The post below has been deleted. Click to close.

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

252 posts

#4 h May 22, 2019, 8:09 AM • 4 Y

Y by ayamgabut, LoloVN, Adventure10, Mango247

Click to reveal hidden text

Z K Y

The post below has been deleted. Click to close.

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

1475 posts

math90

#5 h May 22, 2019, 8:15 AM • 2 Y

Y by Adventure10, Mango247

Very known result.

Z K Y

The post below has been deleted. Click to close.

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

1594 posts

#6 h May 22, 2019, 10:55 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Solution 1 (Spiral Similarity): Let the incircle touches $AC,AB$ at $E,F$ . Then just notice the spiral similarity $\triangle KBF\stackrel{+}{\sim}\triangle KCE$ thus
$\frac{KB}{KC}=\frac{BF}{CE}=\frac{BD}{DC}$ or $KD$ bisects $\angle BKC$ . This immediately implies $K,D,M$ are colinear.

Solution 2 (Inversion): Again, let $E,F$ be the other two intouch points. Perform inversion around the incircle. We deduce the following facts.

$\triangle A'B'C'$ is medial triangle of $\triangle D'E'F'$ .
$I$ is orthocenter of $\triangle A'B'C'$ .
$M'$ is reflection of $I$ across $B'C'$ .
$K'$ is foot from $D'$ to $E'F'$ .

This means points $\{K',D'\}$ and $\{I,M'\}$ are symmetric across $B'C'$ . So $K'D'M'I$ is isosceles trapezoid which obviously cyclic. Inverting back, we find that $K,D,M$ are colinear.

This post has been edited 1 time. Last edited by MarkBcc168, May 22, 2019, 11:14 AM
Reason: Clean up list

Z K Y

The post below has been deleted. Click to close.

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

1179 posts

#7 h Jun 10, 2019, 10:26 PM • 2 Y

Y by Adventure10, Mango247

Invert about $K$ with radius $KI$ . Let $Z^*$ denote the image of $Z$ under this inversion. By the incenter-excenter lemma $KI=KB=KC$ , so $I,B,C$ are fixed by the inversion. Note that $A\mapsto A^*$ , where $A^*=AI\cap BC$ . Hence the circle with diameter $AI$ maps to the circle with diameter $A^*I$ . Furthermore $(ABC)\mapsto BC$ , so $M^*$ is the intersection of $BC$ and the circle with diameter $A^*I$ . But this intersection is clearly $D$ . Therefore $M^*=D$ , which implies $M$ and $D$ are collinear with the center of inversion. Thus, $M,D,K$ are collinear and we are done.

Z K Y

The post below has been deleted. Click to close.

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

2125 posts

#8 h Jul 9, 2019, 7:02 PM • 1 Y

Y by Adventure10

Let $\Delta DEF$ be the intouch triangle WRT $\Delta ABC$ with $I$ as incenter and $M,M'$ as midpoints of arc $BC$ not containing $A$ and arc $BAC$ . Let $AI \cap BC$ $=$ $L$ . Let $M'I$ $\cap$ $\odot (ABC)$ $=$ $T$ . Let $\odot (AFE)$ $\cap$ $\odot (ABC)$ $=$ $K$

Obviously $\angle AKI=90^{\circ}$ , so if $KI \cap \odot (ABC)$ $=$ $P$ , then, $P$ is $A-$ antipode. The line through $I$ perpendicular to $AI$ is ofcourse tangent to $\odot (AFE)$ and $\odot (IBC)$ . Also, $T$ is the $A-$ mixtilinear incircle touch point. By Radical Axes Theorem, $AK,$ $I-\text {tangent}$ and $BC$ are concurrent, say at $G$ . Let $N$ be other tangent from $G$ to $\odot (BIC)$ . So, from La Hire's Theorem $\implies$ $N$ $\in$ $M'I$ . Since, $MT \perp IN$ $\implies$ $IT=NT$ $\implies$ $G \in TM$ . Also, $GKIT$ is cyclic.
$\angle MKI=\angle MAO=\angle AMO=\angle ITD=\angle DKI$

Z K Y

The post below has been deleted. Click to close.

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

2125 posts

#9 h Jul 18, 2019, 9:52 AM • 3 Y

Y by WolfusA, Adventure10, Mango247

Much Shorter: Perform Inversion about $\odot (BIC)$ and note, $K,D$ get swapped $\implies$ $M \in KD$ $\qquad \blacksquare$

Detailed Proof

This post has been edited 1 time. Last edited by AlastorMoody, Jul 18, 2019, 9:55 AM

Z K Y

The post below has been deleted. Click to close.

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

1080 posts

#11 h May 4, 2022, 8:16 PM • 1 Y

Y by centslordm

Invert about the incircle. Notice $\triangle A'B'C'$ is the medial triangle of $\triangle DEF,$ $K'$ is the foot from $D$ to $\overline{EF},$ and $I$ is the orthocenter of $\triangle A'B'C'.$ Hence, $\overline{K'I}$ is the reflection of $\overline{DM'}$ in $\overline{B'C'}$ so $IM'DK'$ is cyclic. $\square$

Z K Y

The post below has been deleted. Click to close.

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

295 posts

#12 h Apr 24, 2023, 11:39 AM • 1 Y

Y by Math.1234

parmenides51 wrote:

Let $ABC$ be a triangle such that $AB\ne AC$ and $\omega$ be the circumcircle of this triangle.
Let $I$ be the center of the inscribed circle of $ABC$ which touches $BC$ at $D$ .
Let the circle with diameter $AI$ meets $\omega$ again at $K$ .
If the line $AI$ intersects $\omega$ again at $M$ , show that $K, D, M$ are collinear.

Without inversion and spiral simirality.We find with angle chansing $\angle$ $MID$ $=$ $2$ $×$ $\beta$ $+$ $\alpha$ $-90$ $(1)$ . $K^{'}$ is intersection of $MD$ and circumcircle of $ABC$ .We must show that $K^{'}=K$ .Then $\angle$ $MK^{'}B$ $=$ $\angle$ $CMK$ $=$ $\angle$ $MBC$ then from triangle similarity $MB^2=MD$ $\times$ $MK^{'}$ besides we know that $MB=MI=MC$ (well known) then $MI^2=MD$ $\times$ $MK^{'}$ $(2)$
We get from $(1)$ and $(2)$ $\angle$ $MID$ $=$ $\angle$ $IK^{'}D$ $=$ $2×\beta+\alpha-90$ $(3)$
On the other hand $\angle$ $MIA$ $=$ $\angle$ $MBC$ $+$ $\angle$ $CBA$ $=$ $\alpha$ $+$ $2$ $\beta$ $(4)$
After we get from $(3)$ and $(4)$ $\angle$ $IK^{'}A$ $=$ $90$ then $K^{'}=K$

This post has been edited 4 times. Last edited by Mathlover_1, Apr 24, 2023, 11:50 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a