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Turkey NMO 2010 P2

Source:

December 15, 2010

geometrycircumcircleratiogeometry proposed

Problem Statement

Let

P

be an interior point of the triangle

ABC

which is not on the median belonging to

BC

and satisfying

\angle CAP = \angle BCP. \: BP \cap CA = \{B'\} \: , \: CP \cap AB = \{C'\}

and

Q

is the second point of intersection of

AP

and the circumcircle of

ABC. \: B'Q

intersects

CC'

at

R

and

B'Q

intersects the line through

P

parallel to

AC

at

S.

Let

T

be the point of intersection of lines

B'C'

and

QB

and

T

be on the other side of

AB

with respect to

C.

Prove that

\angle BAT = \angle BB'Q \: \Longleftrightarrow \: |SQ|=|RB'|

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