# 2019 AMC 12A Problems/Problem 25

## Problem

Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$, $60^\circ$, and $60.001^\circ$. For each positive integer $n$, define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$. Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$, and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$. What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?

$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

## Solution 1

For all nonnegative integers $n$, let $\angle C_nA_nB_n=x_n$, $\angle A_nB_nC_n=y_n$, and $\angle B_nC_nA_n=z_n$.

Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$. By a similar argument, $\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0$. Thus, $x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0$. By a similar argument, $y_1=180^\circ-2y_0$ and $z_1=180^\circ-2z_0$.

Therefore, for any positive integer $n$, we have $x_n=180^\circ-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to $n$ (and the coefficient of $x_0$ is $(-2)^n$). Hence, we let $x_n=pq^n+r+(-2)^nx_0$. We will solve for $p$, $q$, and $r$ by iterating the recurrence to obtain $x_1=180^\circ-2x_0$, $x_2=4x_0-180^\circ$, and $x_3=540-8x_0$. Letting $n=1,2,3$ respectively, we have \begin{align} pq+r&=180 \\ pq^2+r&=-180 \\ pq^3+r&=540 \end{align}

Subtracting $(1)$ from $(3)$, we have $pq(q^2-1)=360$, and subtracting $(1)$ from $(2)$ gives $pq(q-1)=-360$. Dividing these two equations gives $q+1=-1$, so $q=-2$. Substituting back, we get $p=-60$ and $r=60$.

We will now prove that for all positive integers $n$, $x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60$ via induction. Clearly the base case of $n=1$ holds, so it is left to prove that $x_{n+1}=(-2)^{n+1}(x_0-60)+60$ assuming our inductive hypothesis holds for $n$. Using the recurrence relation, we have \begin{align*} x_{n+1}&=180-2x_n \\ &=180-2((-2)^n(x_0-60)+60) \\ &=(-2)^{n+1}(x_0-60)+60 \end{align*}

Our induction is complete, so for all positive integers $n$, $x_n=(-2)^n(x_0-60)+60$. Identical equalities hold for $y_n$ and $z_n$.

The problem asks for the smallest $n$ such that either $x_n$, $y_n$, or $z_n$ is greater than $90^\circ$. WLOG, let $x_0=60^\circ$, $y_0=59.999^\circ$, and $z_0=60.001^\circ$. Thus, $x_n=60^\circ$ for all $n$, $y_n=-(-2)^n(0.001)+60$, and $z_n=(-2)^n(0.001)+60$. Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^\circ$. Therefore, the answer is $\boxed{\textbf{(E) } 15}$.

## Solution 2

We start from Solution 1 until we reach the recurrence relation $x_n = 180 - 2x_{n - 1}.$ Iterate this again, to get $x_{n - 1} = 180 - 2x_{n - 2}.$ Subtract the two, getting $x_{n} = -x_{n - 1} + 2x_{n - 2}.$ This recurrence has characteristic equation $x^2 + x - 2 = 0 = (x + 2)(x - 1) = 0 \iff x = -2, 1.$ Now, write $$x_n = p + q \cdot (-2)^n.$$ We obtain similar recursions for $y, z$ that can be easily solved by getting $x_1, y_1, z_1$ from the original recursive formula and then using those two values to solve for $p$ and $q.$ Then proceed with the last paragraph of Solution 1.

~ dolphin7