In triangle \(\triangle ABC\) with orthocenter \(H\), the internal angle bisector of \(\angle BAC\) intersects \(\overline{BC}\) at \(Y\). Given that \(AH=4\), \(AY=6\), and the distance from \(Y\) to \(\overline{AC}\) is \(\sqrt{15}\), compute \(BC\).
Solution
Answer: \(4\sqrt{35}\)
Let \(E\) be the foot of the altitude from \(B\) to \(AC\), and let \(R\) be the circumradius. Using the altitude relationships and the extended law of sines gives
\[4=AH=\frac{AE}{\sin\angle AHE}=\frac{AB\cos A}{\sin C}=2R\cos A.\]
The distance from \(Y\) to \(AC\) is \(\sqrt{15}\), while \(AY=6\). Thus \(\cos\frac A2=\frac{\sqrt{21}}6\), and
\[\cos A=2\cos^2\frac A2-1=2\left(\frac{\sqrt{21}}6\right)^2-1=\frac16.\]
Substituting gives \(R=12\) and \(\sin A=\frac{\sqrt{35}}6\). By the extended law of sines, \(BC=2R\sin A=4\sqrt{35}\).