tangent circle

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

tangent circle

G H J

Reveal topic tags

geometric transformation

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: JBMO 2010 Shortlist

The post below has been deleted. Click to close.

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

131 posts

#1 h Oct 18, 2012, 7:25 AM • 2 Y

Y by Adventure10 and 1 other user

Consider a triangle $ABC$ with $\angle ACB=90^{\circ}$ . Let $F$ be the foot of the altitude from $C$ . Circle $\omega$ touches the line segment $FB$ at point $P$ , the altitude $CF$ at point $Q$ and the circumcircle of $ABC$ at point $R$ . Prove that point $A,Q,R$ are collinear and $AP=AC$ .

Z K Y

The post below has been deleted. Click to close.

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

4379 posts

#2 h Oct 18, 2012, 8:45 AM • 3 Y

Y by MinatoF, Adventure10, Mango247

Hint: use inversion (of pole $A$ and power $AC^2$ ).

Best regards,
sunken rock

Z K Y

The post below has been deleted. Click to close.

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

76 posts

duanby

#3 h Oct 19, 2012, 2:08 PM • 4 Y

Y by MinatoF, Adventure10, Mango247, and 1 other user

Also we can use Casey theorem let C'be the intersection of the attitude from C with the circumcircle
Use Casey theorem for A C w B and A C' w B add the two we have 2AP*BC=4Sabc
SoAP=AC

Z K Y

The post below has been deleted. Click to close.

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

207 posts

#4 h Oct 20, 2012, 1:24 PM • 3 Y

Y by MinatoF, Adventure10, Mango247

Let $D$ is the circumcenter of $w$ , $I$ is the intersection of $w$ and $BR$ , $d$ is the tangent of the circumcircle of triangle $ABC$ at $R$ . We have to prove following results:
- $QI$ is parallel to $BC$
- $QDPF$ is a square $\Rightarrow \widehat{QPF}=\widehat{QIP}=\widehat{IPB}=45^o \Rightarrow \widehat{QPI}=90^o \Rightarrow \widehat{QRI}=90^o \Rightarrow$ $A,Q,R$ is collinear
- $AC^2=AQ.AR=AP^2$
Hence proved!

Z K Y

The post below has been deleted. Click to close.

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

59 posts

bozzio

#5 h Apr 22, 2013, 5:48 PM • 2 Y

Y by Adventure10, Mango247

Sunken rock how do you prove that $\omega$ is fixed?

Z K Y

The post below has been deleted. Click to close.

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

4379 posts

#6 h Apr 22, 2013, 6:13 PM • 1 Y

Y by Adventure10

@bozzio:

Take $C'$ reflection of $C$ in $AB$ ; well known, $RQ$ is the bisector of $\angle CRC'$ , hence $A-Q-R$ collinear; since $Q$ is the image of $R$ , we have also $AQ\cdot AR=AC^2$ . But $AP$ is tangent to this circle, hence $AC^2=AQ\cdot AR= AP^2$ .
The before said inversion sends the circle $\odot (ABC)$ to $CF$ and the circle $\omega$ to 'another' circle, tangent to $CF$ , after the rule that the common point $R$ is sent to $Q$ . AAs it is not possible to have to circle passing through $R,Q$ and being tangent to $CF$ , the answer is straight forward.

Best regards,
sunken rock

Z K Y

The post below has been deleted. Click to close.

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

252 posts

MMEEvN

#7 h Apr 22, 2013, 6:30 PM • 3 Y

Y by Mathlover_1, Adventure10, Mango247

Consider $\Delta RAB$ It is well known that $RP$ bisects $\angle ARB$ .Hence $\angle PRA =45$ .But $\angle PRQ=\angle QPF=45 \Rightarrow Q,A,R$ are collinear . $AC^2=AF.AB=AQ.AR=AP^2$

Z K Y

The post below has been deleted. Click to close.

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

59 posts

bozzio

#8 h Apr 22, 2013, 7:00 PM • 1 Y

Y by Adventure10

Ok now I understand, you don't prove that A,Q,R are collinear with inversion ( or I am wrong?).

Z K Y

The post below has been deleted. Click to close.

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

129 posts

#9 h Jun 6, 2013, 12:38 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of $AB$ and $N$ be the center of $w$ . Since $M$ is the center of the circumcircle of triangle $ABC, M,N,R$ are collinear. We have $QN||AM, \angle AMR=\angle QNR$ so triangles $AMR,QNR$ are isosceles which equal sides are the radii of the circumcircle of $ABC$ and $w$ . They have an equal common angle, so their rest angles are also equal. Now, $\angle MRA=\angle NRQ$ so $A,Q,R$ are collinear.

Z K Y

The post below has been deleted. Click to close.

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

129 posts

#10 h Jun 6, 2013, 12:42 PM • 2 Y

Y by Adventure10, Mango247

Part two:
$AFQ$ is similar to $ARB$ so $AQ:AB=AF:AR$ or $AQ*AR=AF*AB$ . From power of point of $A$ to $w$ we have $AP^2=AQ*AR$ . From the similarity of $ABC$ and $ACF$ we get $AF*AB=AC^2$ . Now $AP^2=AQ*AR=AF*AB=AC^2, Q.E.D$

Z K Y

The post below has been deleted. Click to close.

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

132 posts

#11 h Jun 9, 2013, 9:52 AM • 1 Y

Y by Adventure10

it can easily be done by inversion at the center $A$ and radius $AC^{2}$
you see that the circle would remain fix and so the point $P$ and that results:
$AP=AC$ and $A,Q,R$ are collinear.

Z K Y

The post below has been deleted. Click to close.

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

111 posts

minmin

#12 h Jun 12, 2013, 12:57 AM • 1 Y

Y by Adventure10

Nguyenhuyhoang wrote:

Let $D$ is the circumcenter of $w$ , $I$ is the intersection of $w$ and $BR$ , $d$ is the tangent of the circumcircle of triangle $ABC$ at $R$ . We have to prove following results:
- $QI$ is parallel to $BC$

Can you explain how to prove $QI \parallel BC$ ?

Z K Y

The post below has been deleted. Click to close.

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

618 posts

#13 h Dec 3, 2018, 11:42 AM • 2 Y

Y by Adventure10, Mango247

This is also Serbia 2016 first task.

Z K Y

The post below has been deleted. Click to close.

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

9775 posts

jayme

#14 h Dec 3, 2018, 11:48 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
I think also to a converse of the Reim's theorem...

Sincerely
Jean-Louis

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a