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

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suggesting that <math>[\Delta DQ_iP_i]</math> is equivalent to <math>\frac{1}{2(i+1)(i+2)}</math>. The sum of this from <math>i=1</math> to <math>\infty</math> is a classic telescoping sequence as in Solution 1 and is equal to <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>.
 
suggesting that <math>[\Delta DQ_iP_i]</math> is equivalent to <math>\frac{1}{2(i+1)(i+2)}</math>. The sum of this from <math>i=1</math> to <math>\infty</math> is a classic telescoping sequence as in Solution 1 and is equal to <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>.
  
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=== Solution 4 Showing Detailed Steps===
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=== Solution 4 Detailed Steps===
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[[Image:2016_AMC_12B_Problem_21.png|thumb|center|800px| ]]
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Midpoint <math>Q_1</math> of <math>\overline{CD}</math>:- <math>Q_1 = \left(\frac{1}{2}, 0\right)</math>
 
Midpoint <math>Q_1</math> of <math>\overline{CD}</math>:- <math>Q_1 = \left(\frac{1}{2}, 0\right)</math>
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   </cmath>
 
   </cmath>
  
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Area of <math>\triangle DQ_kP_k</math>  = <math>\frac{1}{2} \times  DQ_k \times  \text{Height}\ (y\ of\ P_{k+2})= \frac{1}{2} \times \frac{1}{k+1} \times \frac{1}{k+2} = \frac{1}{2(k+1)(k+2)}  = \frac{1}{2} ( \frac{1}{k+1} - \frac{1}{k+2} ) </math>
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Area of <math>\triangle DQ_kP_k</math>  = <math>\frac{1}{2} \times  DQ_k \times  \text{Height}\ (y\ of\ P_{k+2})= \frac{1}{2} \times \frac{1}{k+1} \times \frac{1}{k+2}   = \frac{1}{2} ( \frac{1}{k+1} - \frac{1}{k+2} ) </math>
  
 
This recursive process confirms the telescoping series:
 
This recursive process confirms the telescoping series:
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Most terms cancel, and we are left with: <math>\frac{1}{2} \cdot \frac{1}{2} =  \boxed{\textbf{(B) }\frac{1}{4}}</math>.
 
Most terms cancel, and we are left with: <math>\frac{1}{2} \cdot \frac{1}{2} =  \boxed{\textbf{(B) }\frac{1}{4}}</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2016|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:30, 21 July 2024

Problem

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 \[\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_i P_i \, ?\]

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

Solutions

Solution 1

(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 $[DQ_1P_1]=\frac{1}{2}$ ⋅ $\frac{1}{2}$ ⋅ $\frac{1}{3}$.

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 $[DQ_2P_2]=\frac{1}{2}$ ⋅ $\frac{1}{3}$ ⋅ $\frac{1}{4}$. 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 $i \rightarrow \infty$.


Solution 3

(By user0003)

We plot the figure on a coordinate plane with $D=(0,0)$ and $A$ in the positive y-direction from the origin. If $Q_i=(k, 0)$ for some $k \neq 0$, then the line $AQ_i$ can be represented as $y=-\frac{x}{k}+1$. The intersection of this and $BD$, which is the line $y=x$, is

\[P_i = \left(\frac{k}{k+1}, \frac{k}{k+1}\right)\].

As $Q_{i+1}$ is the projection of $P_i$ onto the x-axis, it lies at $\left(\frac{k}{k+1}, 0\right)$. We have thus established that moving from $Q_i$ to $Q_{i+1}$ is equivalent to the transformation $x \rightarrow \frac{x}{x+1}$ on the x-coordinate. The closed form of of the x-coordinate of $Q_i$ can be deduced to be $\frac{1}{1+i}$, which can be determined empirically and proven via induction on the initial case $Q_1 = \left(\frac{1}{2}, 0\right)$. Now

\[[\Delta DQ_iP_i] = \frac{1}{2}(DQ_i)(Q_{i+1}P_i) = \frac{1}{2}(DQ_i)(DQ_{i+1}),\]

suggesting that $[\Delta DQ_iP_i]$ is equivalent to $\frac{1}{2(i+1)(i+2)}$. The sum of this from $i=1$ to $\infty$ is a classic telescoping sequence as in Solution 1 and is equal to $\boxed{\textbf{(B) }\frac{1}{4}}$.

Solution 4 Detailed Steps

2016 AMC 12B Problem 21.png


Midpoint $Q_1$ of $\overline{CD}$:- $Q_1 = \left(\frac{1}{2}, 0\right)$

Equation of $\overline{AQ_1}$: - Slope $m = \frac{0 - 1}{\frac{1}{2} - 0} = -2$ - Equation: $y = -2x + 1$

Line $\overline{BD}$: - Equation: $y = x$

Intersection $P_1$ of $\overline{AQ_1}$ and $\overline{BD}$: 

  - Solve:$-2x + 1 = x \implies 1 = 3x \implies x = \frac{1}{3}$
  - Therefore, $P_1 = \left(\frac{1}{3}, \frac{1}{3}\right)$

Now, using the pattern for subsequent points $P_k$ and $Q_k$:

General $Q_k$ - For $k \geq 1$, $Q_k = \left(\frac{1}{k+1}, 0\right)$

Equation of $\overline{AQ_k}$ 

  - Slope $m = -(k+1)$
  - Equation: $y = -(k+1)x + 1$

Intersection $P_k$ of $\overline{AQ_k}$ and $\overline{BD}$: 

  - Line $\overline{BD}$: $y = x$
  - Solve:
    \[-(k+1)x + 1 = x \implies 1 = (k+2)x \implies x = \frac{1}{k+2}\]
  - Therefore, $P_k = \left(\frac{1}{k+2}, \frac{1}{k+2}\right)$

$Q_{k+1}$ is the foot of the perpendicular from $P_k$ to $\overline{CD}$, so 

  \[Q_{k+1} = \left(\frac{1}{k+2}, 0\right)\]

Area of $\triangle DQ_kP_k$ = $\frac{1}{2} \times DQ_k \times \text{Height}\ (y\ of\ P_{k+2})= \frac{1}{2} \times \frac{1}{k+1} \times \frac{1}{k+2} = \frac{1}{2} ( \frac{1}{k+1} - \frac{1}{k+2} )$

This recursive process confirms the telescoping series:

\[\sum_{k=1}^{\infty} \frac{1}{2(k+1)(k+2)} = \frac{1}{2} \left( \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots \right)\]

Most terms cancel, and we are left with: $\frac{1}{2} \cdot \frac{1}{2} = \boxed{\textbf{(B) }\frac{1}{4}}$.

~luckuso

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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