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PrintIndija mo 2011
India 2011 geometry
Problem
Let be a convex hexagon in which the diagonals , , are concurrent at . Suppose the area of triangle is the geometric mean of those of and ; and the area of triangle is the geometric mean of those of and . Prove that the area of triangle is the geometric mean of those of and .

Solution
Let , , , , , , , , , , and . We are given that , and we have to prove that . Since , we have
Multiplying these three equalities, we get . Hence
This gives , as desired.
Multiplying these three equalities, we get . Hence
This gives , as desired.
Techniques
TrigonometryAngle chasingTriangle trigonometry