AMC12 2019 A

Answer (E): For any acute triangle $DEF$, the triangle $D'E'F'$ whose vertices are the feet of the respective altitudes is called its orthic triangle. Its angles are $180^\circ-2\angle D$, $180^\circ-2\angle E$, and $180^\circ-2\angle F$. To see this, note that $\angle FF'D$ and $\angle FD'D$ are right angles, so $D'$ and $F'$ lie on the circle with diameter $DF$. Therefore $DFD'F'$ is a cyclic quadrilateral and thus \[ \angle F'D'E = 180^\circ-\angle F'D'F = \angle D. \] Similarly, $\angle E'D'F = 180^\circ-\angle E'D'E = \angle D$. It follows that \[ \angle E'D'F' = 180^\circ-\angle F'D'E-\angle E'D'F = 180^\circ-2\angle D. \] In this problem $\triangle A_nB_nC_n$ is the orthic triangle of $\triangle A_{n-1}B_{n-1}C_{n-1}$. Thus if $\triangle A_iB_iC_i$ is acute for $0\le i\le n-1$, then $\triangle A_nB_nC_n$ has angles $180^\circ-2\angle A_{n-1}$, $180^\circ-2\angle B_{n-1}$, and $180^\circ-2\angle C_{n-1}$. It follows by mathematical induction that if $\triangle A_iB_iC_i$ is acute for $0\le i\le n-1$, the angles of $\triangle A_nB_nC_n$ are \[ \left(60+\frac{2^n}{1000}\right)^\circ,\quad 60^\circ,\quad \text{and}\quad \left(60-\frac{2^n}{1000}\right)^\circ. \] Thus it suffices to find the least positive integer $n$ such that \[ 60+\frac{2^n}{1000}>90, \] which is equivalent to $2^n>30{,}000$. Because $2^{14}=16{,}384$ and $2^{15}=32{,}768$, the requested value is $15$.