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Difference between revisions of "2016 AMC 12B Problems/Problem 21"

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<cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n(n+1)}</cmath>  
 
<cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n(n+1)}</cmath>  
 
or  
 
or  
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<cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n} - \frac{1}{n+1}</cmath>
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<cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right)</cmath>
−
 
 
−
This is known as a telescoping series because we can see that every term after the first <math>\frac{1}{n}</math> is going to cancel out. Thus, the the summation is equal to <math>\frac{1}{2}</math> and after multiplying by the half out in front, we find that the answer is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>.
 
  
 +
This is known as a telescoping series because we can see that every term after the first <math>\frac{1}{n}</math> is going to cancel out. Thus, the summation is equal to <math>\frac{1}{2}</math> and after multiplying by the half out in front, we find that the answer is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 10:53, 3 February 2019

Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}$. For $i=1,2,\dots,$ let $P_i$ be the intersection of $\overline{AQ_i}$ and $\overline{BD}$, and let $Q_{i+1}$ be the foot of the perpendicular from $P_i$ to $\overline{CD}$. What is \[\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_i P_i \, ?\]

$\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{1}{3} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ 1$

Solution

(By Qwertazertl)

We are tasked with finding the sum of the areas of every $\triangle DQ_i^{}P_i^{}$ where $i$ is a positive integer. We can start by finding the area of the first triangle, $\triangle DQ_1^{}P_1^{}$. This is equal to $\frac{1}{2}$ ⋅ $DQ_1^{}$ ⋅ $P_1^{}Q_2^{}$. Notice that since triangle $\triangle DQ_1^{}P_1^{}$ is similar to triangle $\triangle ABP_1^{}$ in a 1 : 2 ratio, $P_1^{}Q_2^{}$ must equal $\frac{1}{3}$ (since we are dealing with a unit square whose side lengths are 1). $DQ_1^{}$ is of course equal to $\frac{1}{2}$ as it is the mid-point of CD. Thus, the area of the first triangle is $\frac{1}{2}$ ⋅ $\frac{1}{2}$ ⋅ $\frac{1}{3}$.


The second triangle has a base $DQ_2^{}$ equal to that of $P_1^{}Q_2^{}$ (see that $\triangle DQ_2^{}P_1^{}$ ~ $\triangle DCB$) and using the same similar triangle logic as with the first triangle, we find the area to be $\frac{1}{2}$ ⋅ $\frac{1}{3}$ ⋅ $\frac{1}{4}$. If we continue and test the next few triangles, we will find that the sum of all $\triangle DQ_i^{}P_i^{}$ is equal to \[\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n(n+1)}\] or \[\frac{1}{2} \sum\limits_{n=2}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right)\]

This is known as a telescoping series because we can see that every term after the first $\frac{1}{n}$ is going to cancel out. Thus, the summation is equal to $\frac{1}{2}$ and after multiplying by the half out in front, we find that the answer is $\boxed{\textbf{(B) }\frac{1}{4}}$.

Solution 2

(By mastermind.hk16)

Note that $AD \|\ P_iQ_{i+1}\ \forall i \in \mathbb{N}$. So $\triangle ADQ_i \sim \triangle P_{i}Q_{i+1}Q_{i} \ \forall i \in \mathbb{N}$

Hence $\frac{Q_iQ_{i+1}}{DQ_{i}}=\frac{P_{i}Q_{i+1}}{AD} \ \ \Longrightarrow DQ_i \cdot P_iQ_{i+1}=Q_iQ_{i+1}$

We compute $\frac{1}{2} \sum_{i=1}^{\infty}DQ_i \cdot P_iQ_{i+1}= \frac{1}{2} \sum_{i=1}^{\infty}Q_iQ_{i+1}=\frac{1}{2} \cdot DQ_1 =\frac{1}{4}$ because $Q_i \rightarrow D$ as $i \rightarrow \infty$.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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