USAMO 1991 Problem 5
- Day 2
- Answer
is a point on the side of the triangle . Take the other common tangent (apart from ) to the incircles of ABX and ACX which intersects the segments AB and AC. Let it meet AX at Y. Show that the locus of , as varies, is the arc of a circle.
Dataset text NuminaMath olympiads. Quoted with source.
Show the answer
AY = \frac{AB + AC - BC}{2}
Solutions
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