All Three Concurrency
Geometry
Level
5
Triangle \(ABC\) is drawn with \(\angle B=90^{\circ}\) and \(AC=1\).
An angle bisector is drawn from \(A\) hitting \(BC\) at \(D\). An altitude is drawn from \(B\) hitting \(AC\) at \(E\). Finally, a median is drawn from \(C\) hitting \(AB\) at \(F\).
Given that \(AD\), \(BE\), and \(CF\) are concurrent, then the area of triangle \(DEF\) can be represented by \[\dfrac{1}{\sqrt{a+b\sqrt{c}}}\] where \(a,b,c\) are integers with \(c\) square-free. Find \(a+b+c\).
by
Daniel Liu