Hard Geometry

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hqthao

71 posts

hqthao

#1 h Oct 16, 2010, 10:00 AM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle and $(O)$ its circumcircle. $P \in AC,Q \in AB$ . Let $M$ be the midpoint of $BP$ , $N$ be the midpoint of $CQ$ , $J$ be the midpoint of $PQ$ . the circumcircle pass through $M N J$ meets $PQ$ at $R$ . Prove: $OR \perp PQ$

thanks

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vntbqpqh234

286 posts

vntbqpqh234

#2 h Oct 16, 2010, 11:29 AM • 2 Y

Y by Adventure10, Mango247

I think it is not true. IMO 2009 near it but it is not true

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jayme

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jayme

#3 h Oct 16, 2010, 11:49 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
I made the figure and after toying with my programm I think that this conjecture is true...
Sincerely
Jean-Louis

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hqthao

71 posts

hqthao

#4 h Oct 16, 2010, 12:30 PM • 2 Y

Y by Adventure10, Mango247

@vnbqh234: why do you think my problem doesn't true, huh

If you just only draw on your paper (and maybe wrong) and think that wrong, you should prove your statement before saying something. If you draw on computer, please give me the figure.

It is TRUE

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hqthao

71 posts

hqthao

#5 h Oct 16, 2010, 6:15 PM • 2 Y

Y by Adventure10, Mango247

Noone solve my problem

I think I can use complex number but it's so complicate. who has any ideas for this problem

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rogueknight

35 posts

rogueknight

#6 h Oct 17, 2010, 9:00 AM • 2 Y

Y by Adventure10, Mango247

This is a nice geometry and and I think Its true. I am learning inversion but I cannot solve this by using inversion. who can solve this problem by inversion, please give me

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jayme

9775 posts

jayme

#7 h Oct 17, 2010, 1:08 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
you can see
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288839
Sincerely
Jean-Louis

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jayme

9775 posts

jayme

#8 h Oct 17, 2010, 1:28 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
the challenge is to research the most easy synthetic proof...
Any ideas?
Sincerely
Jean-Louis

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hqthao

71 posts

hqthao

#9 h Oct 17, 2010, 3:42 PM • 2 Y

Y by Adventure10, Mango247

oh, I see. IMO 2009 is the special case of my problem. So, when the perpendicular of $0$ to $PQ$ same with $J$ , it's will be IMO 2009 problem 2. so, any idea now, I still thinking and cannot find how to apply that problem for this

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armpist

527 posts

armpist

#10 h Oct 17, 2010, 5:14 PM • 2 Y

Y by Adventure10, Mango247

http://www.artofproblemsolving.com/Forum/posting.php?mode=edit&f=46&&t=372184&p=2054142

This post has been edited 1 time. Last edited by armpist, Oct 17, 2010, 11:09 PM

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skytin

418 posts

skytin

#11 h Oct 17, 2010, 6:04 PM • 2 Y

Y by Adventure10, Mango247

We need to prove this two interesting Lemmas :
(1) Trangle ABC , points X Y Z are on BC AC AB and (XYC) intersect (XZB) at points X and P , than YZAP is cyclic.
(2) Circles a and b intersects at points P and Q line through point P intersect a at A and b at B , and line through Q intersect a at A' and b at B' . Then AA' || BB'.
Let X is midpoint of AB and Y is midpoint of AC , let circles (RQX) and (RPY) intersects at R and O'. Use (1) for circles (RQX) (RPY) and (AXY) and trangle APQ , so O' is on (AXY). JN || AC , so use (2) for (NRM) and for (YRP) , so YN intersect (YPR) at point K and K is on (NRM). Like the same XM intersect (RQX) at point L and L is on (NRM). Let PB intersect (YPR) at point S , use (2) for (KMZ) and (YPR) (were Z is midpoint of BC) ZM || YP , so KZMS is cyclic , so angle ZSM = ZKM = NJM = A = ZXb , so ZSXB is cyclic and for the end use (1) fot circles (ZSXB) (YPR) and (RQX) and trangle PQB , so O' is on (ZSXB) , so O' is on (ZXB) like the same O' is on (YZC) so O' = O . done

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lym

175 posts

lym

#12 h Oct 18, 2010, 6:09 AM • 4 Y

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

As a macro explanation，you just need consider Butterfly Theorem .
But this problem，also have easy way to prove：

Let $D,E$ be midpoints of $AP,AQ$ ， $PP_1\perp AB$ ， $QQ_1\perp AC$ ， $S,T,U,V$ be midpoints of $BQ,CP,AB,AC$ ，
$O'$ be the circumcircle of $\odot MNJ$ ， $JDP_1F,JEQ_1G$ be parallelogram， $O''F\perp AQ$ ， $O''G\perp AP$ .

Obviously， $P_1F=\frac{AQ}{2}$ ， $Q_1G=\frac{AP}{2}$ ，the 9-point center $Z$ is the circumcircle of $\odot AFG$ ，so， $Z$ is midpoint of $AO''$ .
Hence， $O''J\perp PQ$ . On another hand，we can easy found $MJP_1S$ is a isosceles trapezoid，and $SU=\frac{AQ}{2}=P_1F$ ，
So， $O'$ is on perpendicular bisector of $FU$ ，the same way， $O'$ is also on perpendicular bisector of $GV$ .
So， $O'$ is midpoint of $OO''$ ，therefore，we get $OR\perp PQ$ . Q.E.D

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jayme

9775 posts

jayme

#13 h Oct 21, 2010, 7:41 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
an article IN ENGLISH concerning this problem has been put on my website
http://perso.orange.fr/jl.ayme vol. 10 , A generalization of the problem 2 of the IMO 2009; an unexpected proof with the midcircle"
Sincerely
Jean-Louis

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armpist

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armpist

#14 h Dec 12, 2010, 5:52 AM • 2 Y

Y by Adventure10, Mango247

Dec 11, 2010

M.T.

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SCP

1502 posts

SCP

#15 h Dec 12, 2010, 9:00 AM • 1 Y

Y by Adventure10

lym wrote:

As a macro explanation，you just need consider Butterfly Theorem .
But this problem，also have easy way to prove：

Let $D,E$ be midpoints of $AP,AQ$ ， $PP_1\perp AB$ ， $QQ_1\perp AC$ ， $S,T,U,V$ be midpoints of $BQ,CP,AB,AC$ ，
$O'$ be the circumcircle of $\odot MNJ$ ， $JDP_1F,JEQ_1G$ be parallelogram， $O''F\perp AQ$ ， $O''G\perp AP$ .

Obviously， $P_1F=\frac{AQ}{2}$ ， $Q_1G=\frac{AP}{2}$ ，the 9-point center $Z$ is the circumcircle of $\odot AFG$ ，so， $Z$ is midpoint of $AO''$ .
Hence， $O''J\perp PQ$ . On another hand，we can easy found $MJP_1S$ is a isosceles trapezoid，and $SU=\frac{AQ}{2}=P_1F$ ，
So， $O'$ is on perpendicular bisector of $FU$ ，the same way， $O'$ is also on perpendicular bisector of $GV$ .
So， $O'$ is midpoint of $OO''$ ，therefore，we get $OR\perp PQ$ . Q.E.D

HOw about that butterfly theorem, I don't know.

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vntbqpqh234

286 posts

vntbqpqh234

#16 h Dec 12, 2010, 12:35 PM • 2 Y

Y by Adventure10, Mango247

let R' lie in PQ such that $OR'\perp PQ$
CR' meet (O) at K, BR' meet (O) at H
Apply Pascal theorem we have:KQ meet HP at X then X lie on (O)
CR' meet (O) at Y. YB meet PQ at G,YC meet PQ at E
Apply butterfly theorem
we have $R'G=R'P$ , $R'Q=R'E$
hence $MR' \parallel YB$ , $NR' \parallel YC$
Then $\angle MR'N=\angle BYC=\angle BAC=\angle MJN$
Hence $R\equiv R'$
WE have QED.
sorry,

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litongyang

387 posts

litongyang

#17 h Dec 27, 2010, 10:11 AM • 2 Y

Y by Adventure10, Mango247

vntbqpqh234 wrote:

let R' lie in PQ such that $OR'\perp PQ$
CR' meet (O) at K, BR' meet (O) at H
Apply Pascal theorem we have:KQ meet HP at X then X lie on (O)
CR' meet (O) at Y. YB meet PQ at G,YC meet PQ at E
Apply butterfly theorem
we have $R'G=R'P$ , $R'Q=R'E$
hence $MR' \parallel YB$ , $NR' \parallel YC$
Then $\angle MR'N=\angle BYC=\angle BAC=\angle MJN$
Hence $R\equiv R'$
WE have QED.

Very very nice solution! But there is a little typo, the black part should be XR'.

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ndk09

162 posts

ndk09

#18 h Jan 2, 2011, 12:23 PM • 1 Y

Y by Adventure10

I think don't need to use Pascan's theorem . Let $K$ is the food of the perpendicular line from $B$ to $PQ$ .
Let $d$ is the line pass throught $B$ and paralley $MK$ , $d$ meet $(O)$ at $T$ and meet $PQ$ at $X$ , we have $K$ is the midpoint of $XP$ ,
Let $TC$ meet $PQ$ at $Y$ , apply "Butterfly Theorem" we have $K$ is the midpoint of $YQ$ so we have $TC\parallel KN$ so $\measuredangle CAB =\measuredangle CTB = \measuredangle MKN=\measuredangle MJN$
$Q.E.D$

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vntbqpqh234

286 posts

vntbqpqh234

#19 h Jan 3, 2011, 12:31 PM • 1 Y

Y by Adventure10

I think your prove not complete because if apply general butterfly theorem then not have all conditions.We don't have K is the midpoint of QY and XP.

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ndk09

162 posts

ndk09

#20 h Jan 3, 2011, 1:33 PM • 2 Y

Y by Adventure10, Mango247

vntbqpqh234 wrote:

I think your prove not complete because if apply general butterfly theorem then not have all conditions.We don't have K is the midpoint of QY and XP.

I think you have a mistake, because we only need $K$ is the midpoint of $XP \Longrightarrow K$ is the mid point of $YQ$ , you should read "Butterfly Theorem" again.
First, we have $K$ is the midpoint of $XP$ because $XB\parallel MK$ and $M$ is midpoint of $BP$

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skytin

418 posts

skytin

#21 h Feb 15, 2011, 2:06 PM • 2 Y

Y by livetolove212, Adventure10

You can get another solution from this picture :

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skytin

418 posts

skytin

#22 h Feb 15, 2011, 4:25 PM • 3 Y

Y by livetolove212, Adventure10, Mango247

You can get new solution of Gauss line if you use result of this problem twice

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yetti

2643 posts

yetti

#23 h Feb 15, 2011, 4:54 PM • 2 Y

Y by Adventure10, Mango247

hqthao wrote:

Let $ABC$ be a triangle and $(O)$ its circumcircle. $P \in AC,Q \in AB$ . Let $M$ be the midpoint of $BP$ , $N$ be the midpoint of $CQ$ , $J$ be the midpoint of $PQ$ . the circumcircle pass through $M N J$ meets $PQ$ at $R$ . Prove: $OR \perp PQ$

This problem was posted before at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=291269.

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lym

175 posts

lym

#24 h Feb 15, 2011, 8:12 PM • 2 Y

Y by Adventure10, Mango247

Nice solution ！

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