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Difference between revisions of "2019 AMC 12A Problems/Problem 25"

(Created page with "==Problem== Let <math>\triangle A_0B_0C_0</math> be a triangle whose angle measures are exactly <math>59.999^\circ</math>, <math>60^\circ</math>, and <math>60.001^\circ</math...")
 
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==Solution==
 
==Solution==
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For all nonnegative integers <math>n</math>, let <math>\angle C_nA_nB_n=x_n</math>, <math>\angle A_nB_nC_n=y_n</math>, and <math>\angle B_nC_nA_n=z_n</math>.
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Note that quadrilateral <math>A_0B_0A_1B_1</math> is cyclic since <math>\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ</math>; thus, <math>\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0</math>. By a similar argument, <math>\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0</math>. Thus, <math>x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0</math>. By a symmetric argument, <math>y_1=180^\circ-2y_0</math> and <math>z_1=180^\circ-2z_0</math>.
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Therefore, for any positive integer <math>n</math>, we have
  
 
==See Also==
 
==See Also==

Revision as of 21:11, 9 February 2019

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

See Also

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

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