Difference between revisions of "1999 JBMO Problems/Problem 4"
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Revision as of 18:00, 2 December 2018
Its easy to see that , , are collinear (since angle = = ).
Applying Sine rule in triangle , we get:
Since and are cyclic quadrilaterals, anlge = anlge and
So,
So Thus, (the circumcirlcles are congruent).
From right traingles and , we have:
So
Since is the midpoint of , is perpendicular to and hence is parallel to .
So area of traiangle = area of traingle and hence is independent of position of on .
By