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

Latest revision as of 23:28, 9 November 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 Diagram and Detailed Steps

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: Therefore,

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 :  Solve:
  - Therefore,

$Q_{k+1}$ 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

Video Solution by CanadaMath (Problem 21-25)

https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s

~THEMATHCANADIAN

@@ Line 1: / Line 1: @@
+== Problem ==
 Let <math>ABCD</math> be a unit square. Let <math>Q_1</math> be the midpoint of <math>\overline{CD}</math>. For <math>i=1,2,\dots,</math> let <math>P_i</math> be the intersection of <math>\overline{AQ_i}</math> and <math>\overline{BD}</math>, and let <math>Q_{i+1}</math> be the foot of the perpendicular from <math>P_i</math> to <math>\overline{CD}</math>. What is
 <cmath>\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_i P_i \, ?</cmath>
@@ Line 8: / Line 9: @@
 \textbf{(E)}\ 1</math>
-==Solution==
+== Solutions ==
+=== Solution 1 ===
+(By Qwertazertl)
+We are tasked with finding the sum of the areas of every <math>\triangle DQ_i^{}P_i^{}</math> where <math>i</math> is a positive integer. We can start by finding the area of the first triangle, <math>\triangle DQ_1^{}P_1^{}</math>. This is equal to <math>\frac{1}{2}</math> &sdot; <math>DQ_1^{}</math> &sdot; <math>P_1^{}Q_2^{}</math>. Notice that since triangle <math>\triangle DQ_1^{}P_1^{}</math> is similar to triangle <math>\triangle ABP_1^{}</math> in a 1 : 2 ratio, <math>P_1^{}Q_2^{}</math> must equal <math>\frac{1}{3}</math> (since we are dealing with a unit square whose side lengths are 1). <math>DQ_1^{}</math> is of course equal to <math>\frac{1}{2}</math> as it is the mid-point of CD. Thus, the area of the first triangle is <math>[DQ_1P_1]=\frac{1}{2}</math> &sdot; <math>\frac{1}{2}</math> &sdot; <math>\frac{1}{3}</math>.
+The second triangle has a base <math>DQ_2^{}</math> equal to that of <math>P_1^{}Q_2^{}</math> (see that <math>\triangle DQ_2^{}P_1^{}</math> ~ <math>\triangle DCB</math>) and using the same similar triangle logic as with the first triangle, we find the area to be <math>[DQ_2P_2]=\frac{1}{2}</math> &sdot; <math>\frac{1}{3}</math> &sdot; <math>\frac{1}{4}</math>. If we continue and test the next few triangles, we will find that the sum of all <math>\triangle DQ_i^{}P_i^{}</math> is equal to
+<cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n(n+1)}</cmath>
+or
+<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 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 ===
+(By mastermind.hk16)
+Note that <math>AD \|\ P_iQ_{i+1}\  \forall i \in \mathbb{N}</math>. So <math>\triangle ADQ_i \sim \triangle P_{i}Q_{i+1}Q_{i} \ \forall i \in \mathbb{N}</math>
+Hence <math>\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}</math>
+We compute <math>\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}</math>
+because <math>Q_i \rightarrow D</math> as <math>i \rightarrow \infty</math>.
+=== Solution 3 ===
+(By user0003)
+We plot the figure on a coordinate plane with <math>D=(0,0)</math> and <math>A</math> in the positive y-direction from the origin. If <math>Q_i=(k, 0)</math> for some <math>k \neq 0</math>, then the line <math>AQ_i</math> can be represented as <math>y=-\frac{x}{k}+1</math>. The intersection of this and <math>BD</math>, which is the line <math>y=x</math>, is
+<cmath>P_i = \left(\frac{k}{k+1}, \frac{k}{k+1}\right)</cmath>.
+As <math>Q_{i+1}</math> is the projection of <math>P_i</math> onto the x-axis, it lies at <math>\left(\frac{k}{k+1}, 0\right)</math>. We have thus established that moving from <math>Q_i</math> to <math>Q_{i+1}</math> is equivalent to the transformation <math>x \rightarrow \frac{x}{x+1}</math> on the x-coordinate. The closed form of of the x-coordinate of <math>Q_i</math> can be deduced to be <math>\frac{1}{1+i}</math>, which can be determined empirically and proven via induction on the initial case <math>Q_1 = \left(\frac{1}{2}, 0\right)</math>. Now
+<cmath>[\Delta DQ_iP_i] = \frac{1}{2}(DQ_i)(Q_{i+1}P_i) = \frac{1}{2}(DQ_i)(DQ_{i+1}),</cmath>
+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>.
+=== Solution 4 Diagram and Detailed Steps===
+[[Image:2016_AMC_12B_Problem_21.png|thumb|center|800px| ]]
+Midpoint <math>Q_1</math> of <math>\overline{CD}</math>: <math>Q_1 = \left(\frac{1}{2}, 0\right)</math>
+Equation of <math>\overline{AQ_1}</math>:  Slope <math>m = \frac{0 - 1}{\frac{1}{2} - 0} = -2</math>  Equation: <math>y = -2x + 1</math>
+Line <math>\overline{BD}</math>:   - Equation: <math>y = x</math>
+Intersection <math>P_1</math> of <math>\overline{AQ_1}</math> and <math>\overline{BD}</math>:
+   - Solve:<math>-2x + 1 = x \implies 1 = 3x \implies x = \frac{1}{3}</math> Therefore, <math>P_1 = \left(\frac{1}{3}, \frac{1}{3}\right)</math>
+Now, using the pattern for subsequent points <math>P_k</math> and <math>Q_k</math>:
+General <math>Q_k</math> - For <math>k \geq 1</math>, <math>Q_k = \left(\frac{1}{k+1}, 0\right)</math>
+Equation of <math>\overline{AQ_k}</math>  Slope <math>m = -(k+1)</math>  Equation: <math>y = -(k+1)x + 1</math>
+Intersection <math>P_k</math> of <math>\overline{AQ_k}</math> and <math>\overline{BD}</math>:
+   - Line <math>\overline{BD}</math>: <math>y = x</math> Solve:<math>    -(k+1)x + 1 = x \implies 1 = (k+2)x \implies x = \frac{1}{k+2} </math>
+   - Therefore, <math>P_k = \left(\frac{1}{k+2}, \frac{1}{k+2}\right)</math>
+<math>Q_{k+1}</math> is the foot of the perpendicular from <math>P_k</math> to <math>\overline{CD}</math>, so <math> Q_{k+1} = \left(\frac{1}{k+2}, 0\right)</math>
+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:
+<cmath>
+\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)
+</cmath>
+Most terms cancel, and we are left with: <math>\frac{1}{2} \cdot \frac{1}{2} =  \boxed{\textbf{(B) }\frac{1}{4}}</math>.
+~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+==Video Solution by CanadaMath (Problem 21-25)==
+https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s
+~THEMATHCANADIAN
 ==See Also==
 {{AMC12 box|year=2016|ab=B|num-b=20|num-a=22}}
 {{MAA Notice}}

2016 AMC 12B (Problems • Answer Key • Resources)
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