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

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Subtracting <math>(1)</math> from <math>(3)</math>, we have <math>pq(q^2-1)=360</math>, and subtracting <math>(1)</math> from <math>(2)</math> gives <math>pq(q-1)=-360</math>. Dividing these two equations gives <math>q+1=-1</math>, so <math>q=-2</math>. Substituting back, we get <math>p=-60</math> and <math>r=60</math>.
 
Subtracting <math>(1)</math> from <math>(3)</math>, we have <math>pq(q^2-1)=360</math>, and subtracting <math>(1)</math> from <math>(2)</math> gives <math>pq(q-1)=-360</math>. Dividing these two equations gives <math>q+1=-1</math>, so <math>q=-2</math>. Substituting back, we get <math>p=-60</math> and <math>r=60</math>.
  
−
We will now prove that for all positive integers <math>n</math>, <math>x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60</math> via induction. Clearly the base case of <math>n=1</math> holds, so it is left to prove that <math>x_{n+1}=(-2)^{n+1}(x_0-60)+60</math> assuming our inductive hypothesis holds for <math>n</math>. Using the recurrence relation, we have <math></math>\begin{align*}
+
We will now prove that for all positive integers <math>n</math>, <math>x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60</math> via induction. Clearly the base case of <math>n=1</math> holds, so it is left to prove that <math>x_{n+1}=(-2)^{n+1}(x_0-60)+60</math> assuming our inductive hypothesis holds for <math>n</math>. Using the recurrence relation, we have <cmath>\begin{align*}
 
x_{n+1}&=180-2x_n \\
 
x_{n+1}&=180-2x_n \\
 
&=180-2((-2)^n(x_0-60)+60) \\
 
&=180-2((-2)^n(x_0-60)+60) \\
 
&=(-2)^{n+1}(x_0-60)+60
 
&=(-2)^{n+1}(x_0-60)+60
−
\end{align*}
+
\end{align*}</cmath>
  
 
Our induction is complete, so for all positive integers <math>n</math>, <math>x_n=(-2)^n(x_0-60)+60</math>. Identical equalities hold for <math>y_n</math> and <math>z_n</math>
 
Our induction is complete, so for all positive integers <math>n</math>, <math>x_n=(-2)^n(x_0-60)+60</math>. Identical equalities hold for <math>y_n</math> and <math>z_n</math>

Revision as of 21:57, 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 $x_n=180^\circ-2x_{n-1}$ (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 $\boxed{\bold{(E) 15}}$.

-soyamyboya

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions

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