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? $\phantom{}$

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

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

2019 AMC 12A (Problems • Answer Key • Resources)
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2019 AMC 12A Problems/Problem 25

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