Problem 3, Iberoamerican Olympiad 2010

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

Problem 3, Iberoamerican Olympiad 2010

G H J

Reveal topic tags

projective geometry

geometry proposed

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

49 posts

#1 h Sep 24, 2010, 6:42 PM • 3 Y

Y by Davi-8191, Adventure10, and 1 other user

The circle $\Gamma$ is inscribed to the scalene triangle $ABC$ . $\Gamma$ is tangent to the sides $BC, CA$ and $AB$ at $D, E$ and $F$ respectively. The line $EF$ intersects the line $BC$ at $G$ . The circle of diameter $GD$ intersects $\Gamma$ in $R$ ( $R\neq D$ ). Let $P$ , $Q$ ( $P\neq R , Q\neq R$ ) be the intersections of $\Gamma$ with $BR$ and $CR$ , respectively. The lines $BQ$ and $CP$ intersects at $X$ . The circumcircle of $CDE$ meets $QR$ at $M$ , and the circumcircle of $BDF$ meet $PR$ at $N$ . Prove that $PM$ , $QN$ and $RX$ are concurrent.

Author: Arnoldo Aguilar, El Salvador

Z K Y

The post below has been deleted. Click to close.

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

515 posts

#2 h Sep 24, 2010, 10:53 PM • 2 Y

Y by Adventure10, Mango247

Lemma: CDEI and BDFI are cyclic (I is the incenter of ABC)
Proof: draw the perpendiculars from I
Then <RNI=<BDI=90 and <RMI=<CDI=90
Then since RN and RM are chords, and I is the center, M and N are the midpoints of PR, QR and MN parallel to QR
Also B,D,C,G are harmonic (wlog AB>AC) and since DR perpendicular to RG because DG is diameter, then DR is bisector of <BDC. Then arc PD=arc QD and ID is perpendicular to PQ, but ID is perpendicular to BC
Therefore MN, PQ, BC are parallel and RX, PM, QN are concurrent (Desargues Theorem)

Z K Y

The post below has been deleted. Click to close.

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

469 posts

#3 h Sep 25, 2010, 7:40 PM • 2 Y

Y by Adventure10, Mango247

Quote:

The circle $\Gamma$ is inscribed to the scalene triangle $ABC$ . $\Gamma$ is tangent to the sides $BC, CA$ and $AB$ at $D, E$ and $F$ respectively. The line $EF$ intersects the line $BC$ at $G$ . The circle of diameter $GD$ intersects $\Gamma$ in $R$ ( $R\neq D$ ). Let $P$ , $Q$ ( $P\neq R , Q\neq R$ ) be the intersections of $\Gamma$ with $BR$ and $CR$ , respectively. The lines $BQ$ and $CP$ intersects at $X$ . The circuncircle of $CDE$ meets $QR$ at $M$ , and the circuncircle of $BDF$ meet $PR$ at $N$ . Prove that $PM$ , $QN$ and $RX$ are concurrents.

Author: Arnoldo Aguilar, El Salvador

Let $K$ be the midpoint of $GD$ . Since $(G,D,B,C)$ is a harmonic division; thus $KD^2=$ $KR^2=$ $\overline {KB}.$ $\overline {KC}$ , which imples $KR$ is a tangent to the circumcircle of $\triangle RBC$ . Hence, $\angle KRB=$ $\angle RCB$ . On the other hand, since $KD$ has already been a tangent from $K$ to $\Gamma$ . As a result, $KR$ is also another tangent from $K$ to $\Gamma$ $\Longrightarrow$ $\angle KRB=$ $\angle RQP$ . Therefore, $\angle RQP=$ $\angle RCB$ , which implies that $RQ$ $\|$ $BC$ . As a consequence, $RX$ will pass through the midpoint of $PQ$ . For that argument, $RX,$ $PM$ and $QN$ will concur at the centroid of $\triangle RQP$ . Our proof is completed then. $\square$

Z K Y

The post below has been deleted. Click to close.

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

157 posts

#4 h Sep 26, 2010, 4:48 PM • 2 Y

Y by Adventure10, Mango247

Z K Y

The post below has been deleted. Click to close.

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

253 posts

#5 h Sep 27, 2010, 3:45 PM • 2 Y

Y by Adventure10, Mango247

If I am not mistaken, second year in a row that Arnoldo gets one of his geometry problems selected as problem 3 in Ibero, so double congrats!!!

Z K Y

The post below has been deleted. Click to close.

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

1347 posts

WakeUp

#6 h Jan 20, 2011, 5:45 PM • 2 Y

Y by Adventure10, Mango247

Solution by Johan Gurandi: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2093045#p2093045

Z K Y

The post below has been deleted. Click to close.

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

761 posts

#7 h Nov 10, 2011, 4:47 AM • 2 Y

Y by Adventure10, Mango247

it's trivial that $C,D,B,G$ form harmonic sequence.hence $R$ is on the Apollonius circle $(B,C,\frac{BD}{DC})$
by $\angle GRD=90$ we get $RD$ bisects $\angle BRC$
hence $PD=DQ$ , $PQ$ parallel to $BC$
then it's trivial that $PM,QN,RX$ are medians of three sides of triangle $QPR$ .

Z K Y

The post below has been deleted. Click to close.

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

Math_Is_Fun_101

159 posts

Math_Is_Fun_101

#8 h Jul 26, 2021, 2:49 AM

Y by

Let $I$ be the center of $\Gamma$ and $Y$ the second intersection of $\overleftrightarrow{GR}$ and $\Gamma$ . Since $\overline{GD}$ is a diameter of $(GRD)$ , we know that $\angle GRD = 90^{\circ}$ . Thus, $\angle YRD = 90^{\circ}$ , so $\overline{YD}$ is a diameter of $\Gamma$ . Now, note that by Ceva+Menelaus, $(GD;BC)=-1$ . Thus,
$-1=(GD;BC)\stackrel{R}= (YD;PQ).$ Since $\overline{YD}$ is a diameter of $\Gamma$ , it must be the case that $Y$ and $D$ are midpoints of the major and minor arcs $\widehat{PQ}$ . Hence, $\overline{PQ}\parallel\overline{BC}$ since both are perpendicular to $\overline{ID}$ . Now, since $R$ and $X$ are the centers of homothety taking $\overline{PQ}$ to $\overline{BC}$ and $\overline{QP}$ to $\overline{BC}$ , respectively, it is well known that $\overline{RX}$ passes through the midpoint of $\overline{PQ}$ . It is easy to see $I\in(CDE), (BDF)$ , since $\overline{IC}$ and $\overline{IB}$ are the diameters of these circles, respectively. Thus, $\overline{IM}\perp\overline{RQ}$ . Since $I$ is the center of $\Gamma$ , $M$ is the midpoint of $\overline{RQ}$ . Analogously, $N$ is the midpoint of $\overline{RP}$ . This implies $\overline{PM}, \overline{QN}, \overline{RX}$ intersect at the centroid of $\triangle PQR$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

1513 posts

#9 h Sep 29, 2021, 3:15 AM

Y by

Its well known that if $I$ is the incenter of $\triangle ABC$ then: $DIEC,DIFB$ are cyclic and $-1=(G, B; D, C)$
A few claims will be shown here to be used on the resolution.
Claim 1: $BC \parallel PQ$
Proof: Note that since $\angle GRD=90$ and $-1=(G, B; D, C)$ we have that $RD$ bisects $\angle PRQ$ thus $ID \perp PQ$ but since $ID \perp BC$ we have the desired result.
Claim 2: $MN \parallel BC \parallel PQ$
Proof: By Claim 1 we have that $BC \parallel PQ$ , now note that $PI=RI=IQ$ and $\angle BNI=\angle IMC=90$ thus $M,N$ are midpoints of $RP, RQ$ respectvily meaning that by mid-base $MN \parallel PQ$ and we are done!.
Claim 3: Let $K$ the midpoint of $BC$ then if $Y, Z$ are points on $BR, CR$ respectivily such that $YZ \parallel BC$ then $AK,BZ,CY$ are concurrent.
Proof: Trivial by Ceva's theorem.
Now use Claim 3 to show that $A,X,K$ are colinear and then note that $A,X$ and the midpoint of $PQ$ are colinear, thus now the concurrency point is the baricenter of $\triangle PRQ$ thus we are done :blush:

:blush:

This post has been edited 1 time. Last edited by MathLuis, Sep 29, 2021, 3:15 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a