Iran NMO 2005/2

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tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Iran NMO 2005/2

Unread post by tanmoy » Sun Jun 28, 2015 7:29 pm

$BC$ is a diameter of a circle and the points $X,Y$ are on the circle such that $XY \perp BC$.The points $P,M$ are on $XY,CY$(or their stretches),respectively,such that $CY \parallel PB$ and $CX \parallel PM$.Let $K$ be the meet point of the lines $XC,BP$.Prove that $PB \perp MK$.
(An easy problem.Just a little angle chasing)
"Questions we can't answer are far better than answers we can't question"

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: Iran NMO 2005/2

Unread post by Tahmid » Mon Jun 29, 2015 6:13 pm

are you sure about the statement?
tanmoy wrote: $CX \parallel PM$.
this line is confusing. :?

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: Iran NMO 2005/2

Unread post by Tahmid » Mon Jun 29, 2015 6:52 pm

opsssss sorry :oops: my mistake

here is the solution:
$PM=YM$ [because $PM||XC$]
so,$KC=YM$

$\angle KBC=\angle BCY=\angle KCB$
$\therefore KB=KC$
$KB=KC=MY$

$\therefore MYBK$ is a parallelgram.
$\therefore \angle MKB=\angle MYB =90$
so, $MK$ is perpendicular to $PB$ .

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: Iran NMO 2005/2

Unread post by tanmoy » Tue Jun 30, 2015 12:07 pm

Tahmid wrote:opsssss sorry :oops: my mistake

here is the solution:
$PM=YM$ [because $PM||XC$]
so,$KC=YM$

$\angle KBC=\angle BCY=\angle KCB$
$\therefore KB=KC$
$KB=KC=MY$

$\therefore MYBK$ is a parallelgram.
$\therefore \angle MKB=\angle MYB =90$
so, $MK$ is perpendicular to $PB$ .
Same as my solution. :mrgreen:
"Questions we can't answer are far better than answers we can't question"

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