11None....no geometry-lovers?
It is not of this year...
ABc is a triangle with AB=AC...D is the mid-point of BC, P is any other point on AD. PE is perpendicular to AC.
\frac{AP}{PD}=\frac{BP}{PE}=\lambda \\ \ \frac{BD}{AD}=m\\ \ z=m^2(1+\lambda)\\ \text {Prove} \\ \boxed{z^2-(\lambda^3-\lambda^2-2)z+1=0}
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