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

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 symmetric 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 $\boxed{\textbf{(E) } 15}$ .

-soyamyboya

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2019 AMC 12A Problems/Problem 25

Problem

Solution

See Also