2009 JBMO Shortlist G5

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

2009 JBMO Shortlist G5

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2009 JBMO Shortlist G5

The post below has been deleted. Click to close.

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

30629 posts

#1 h Oct 8, 2017, 1:10 PM • 2 Y

Y by Adventure10, Mango247

Let ${A, B, C}$ and ${O}$ be four points in plane, such that $\angle ABC>{{90}^{{}^\circ }}$ and ${OA=OB=OC}$ .Define the point ${D\in AB}$ and the line ${l}$ such that ${D\in l, AC\perp DC}$ and ${l\perp AO}$ . Line ${l}$ cuts ${AC}$ at ${E}$ and circumcircle of ${ABC}$ at ${F}$ . Prove that the circumcircles of triangles ${BEF}$ and ${CFD}$ are tangent at ${F}$ .

Z K Y

The post below has been deleted. Click to close.

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

252 posts

#3 h Oct 8, 2017, 3:53 PM • 2 Y

Y by Adventure10, Mango247

So,It is easy to see $O$ circumcenter of $(ABC) \implies \angle OAC=B-90^\circ$ .
If $l \cap (ABC)=F$ at arc $AC$ containing $B$ and $l \cap AO=K,I\cap AC=E$ .
We need to show $\angle EBF+\angle FDC=\angle EFC \implies\angle EBF+\angle FDC+\angle FEC+\angle FCE=180^\circ$ .
Given $\angle AKD=\angle ACD=90^\circ$ .
So,quadrilateral $ADCK$ cyclic.
So, $\angle OAC=\angle KDC=B-90^\circ$ .
In $\triangle ADK$ $\angle ADK=C$ .
So,quadrilateral $BDCE$ cyclic.Then, $\angle EBC=B-90^\circ$ .In triangle $EDC$ $\angle CED=180^\circ-B$ .So,
$\angle EBF+\angle FDC+\angle FEC+\angle FCE=B-90^\circ+\angle CBF+B-90^\circ+180^\circ+C+\angle BCF=B+C+A=180^\circ$ .
Proved.

This post has been edited 2 times. Last edited by Davrbek, Oct 8, 2017, 3:57 PM
Reason: Sorry

Z K Y

The post below has been deleted. Click to close.

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

1772 posts

#4 h Oct 8, 2017, 4:52 PM • 1 Y

Y by Adventure10

Any solution with inversion?

Z K Y

The post below has been deleted. Click to close.

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

68 posts

#5 h Oct 9, 2017, 9:32 AM • 2 Y

Y by claserken, Adventure10

claserken wrote:

Any solution with inversion?

Here it is

Z K Y

The post below has been deleted. Click to close.

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

MathematicalPhysicist

179 posts

MathematicalPhysicist

#6 h Oct 15, 2018, 5:49 PM • 2 Y

Y by Adventure10, Mango247

Nothing...

This post has been edited 2 times. Last edited by MathematicalPhysicist, Oct 15, 2018, 6:07 PM
Reason: iq low

Z K Y

The post below has been deleted. Click to close.

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

1218 posts

#7 h Oct 16, 2018, 7:04 AM • 2 Y

Y by Adventure10, Mango247

This is just a restated problem. Anyway, here's my solution: Note that $O$ is the circumcenter of $\triangle ABC$ . Let $A'$ be the antipode of $A$ in $\odot (ABC)$ . Then $\angle A'CA=90^{\circ} \Rightarrow D,C,A'$ are collinear. Also, as $DE \perp AA'$ and $AE \perp A'D$ , we get that $E$ is the orthocenter of $\triangle ADA'$ . But $\angle A'BA=90^{\circ}$ , giving $A',E,B$ are collinear. Thus $\angle EFC=\angle CDF+\angle DCF=\angle CDF+\angle A'BF=\angle CDF+\angle EBF$ This easily gives that $\odot (CDF)$ and $\odot (BEF)$ are tangent to each other at $F$ .

REMARK: If one states the problem in terms of $\triangle ADA'$ , then the problem becomes quite well known.

This post has been edited 1 time. Last edited by math_pi_rate, Oct 16, 2018, 7:05 AM

Z K Y

The post below has been deleted. Click to close.

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

1219 posts

#8 h Oct 20, 2023, 8:00 AM

Y by

Fun fact, this is Balkan MO 2012/1

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a