## Iran NMO 2005/2

For discussing Olympiad level Geometry Problems
tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Iran NMO 2005/2

\$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

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

opsssss sorry 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

### Re: Iran NMO 2005/2

Tahmid wrote:opsssss sorry 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. "Questions we can't answer are far better than answers we can't question"