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

Latest revision as of 14:29, 25 July 2025

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

We start with $DQ_i = 1/2$ for $i = 1.$ $\triangle BP_iA \sim \triangle DP_iQ_i$ and $\triangle DP_iQ_{i+1} \sim \triangle DBC$ so we have $\frac{DQ_i}{AB} = \frac{DP_i}{P_iB} = \frac{1}{2} \implies \frac{DP_i}{BD} = \frac{DQ_{i+1}}{DC} = \frac{DQ_{i+1}}{1} = \frac{1}{3} \implies DQ_{i+1} = DQ_{2} = \frac{1}{3}.$ Repeating this same process for subsequent $i$ s yields $DQ_3 = \frac{1}{4}, DQ_4 = \frac{1}{5}, DQ_5 = \frac{1}{6} \dots$ We can generalize this by saying $DQ_i = \frac{1}{i + 1}.$ Then $[ADQ_i] = \frac{1}{2(i + 1)}.$ Let $y = [AP_iD]$ and let $x = [DP_iQ_i].$ $\triangle BP_iA$ and $\triangle DP_iQ_i$ are similar with side length ratio $(i + 1):1$ , so $[BP_iA] = (i + 1)^2x.$ Now, we can express the area of $\triangle ADB$ (which is $1/2$ since it's just half the square) as $y + (i + 1)^2x,$ and we can express the area of $ADQ_i$ as $x + y.$ We have a system of equations:

$[ADQ_i] = x + y = \frac{1}{2(i + 1)}$ $[ADB] = (i + 1)^2x + y = \frac{1}{2}.$

Solving, we get $x = \frac{1}{2(i+1)(i+2)}.$ So now the problem becomes $\sum_{i=1}^{\infty} \frac{1}{2(i+1)(i+2)}$ . We can rewrite this as $\frac{1}{2} \sum_{i=1}^{\infty} \left( \frac{1}{i+1} - \frac{1}{i+2} \right) = \frac{1}{2} \left( \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \dots \right)$

All terms cancel except for the $\frac{1}{2}$ on the inside and $\frac{1}{2}$ on the outside, so the answer is $\boxed{\frac{1}{4}}.$

~grogg007

Solution 2

(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 3

(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 4

(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 5 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 10: / Line 10: @@
 == Solutions ==
-=== Solution 1 ===
+===Solution 1===
+[[File:201612B-P21.png|405px]]
+We start with <math>DQ_i = 1/2</math> for <math>i = 1.</math> <math>\triangle BP_iA \sim \triangle DP_iQ_i</math> and <math>\triangle DP_iQ_{i+1} \sim \triangle DBC</math> so we have <cmath>\frac{DQ_i}{AB} = \frac{DP_i}{P_iB} = \frac{1}{2} \implies \frac{DP_i}{BD} = \frac{DQ_{i+1}}{DC} = \frac{DQ_{i+1}}{1} = \frac{1}{3} \implies DQ_{i+1} = DQ_{2} = \frac{1}{3}.</cmath> Repeating this same process for subsequent <math>i</math>s yields <math>DQ_3 = \frac{1}{4}, DQ_4 = \frac{1}{5}, DQ_5 = \frac{1}{6} \dots</math> We can generalize this by saying <math>DQ_i = \frac{1}{i + 1}.</math> Then <math>[ADQ_i] = \frac{1}{2(i + 1)}.</math> Let <math>y = [AP_iD]</math> and let <math>x = [DP_iQ_i].</math> <math>\triangle BP_iA</math> and <math>\triangle DP_iQ_i</math> are similar with side length ratio <math>(i + 1):1</math>, so <math>[BP_iA] = (i + 1)^2x.</math> Now, we can express the area of <math>\triangle ADB</math> (which is <math>1/2</math> since it's just half the square) as <math>y + (i + 1)^2x,</math> and we can express the area of <math>ADQ_i</math> as <math>x + y.</math> We have a system of equations:
+<cmath>[ADQ_i] = x + y = \frac{1}{2(i + 1)}</cmath>
+<cmath>[ADB] = (i + 1)^2x + y = \frac{1}{2}.</cmath>
+Solving, we get <math>x = \frac{1}{2(i+1)(i+2)}.</math> So now the problem becomes <math>\sum_{i=1}^{\infty} \frac{1}{2(i+1)(i+2)}</math>. We can rewrite this as
+<cmath>\frac{1}{2} \sum_{i=1}^{\infty} \left( \frac{1}{i+1} - \frac{1}{i+2} \right) = \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) + \dots \right)</cmath>
+All terms cancel except for the <math>\frac{1}{2}</math> on the inside and <math>\frac{1}{2}</math> on the outside, so the answer is <math>\boxed{\frac{1}{4}}.</math>
+~[[User:grogg007|grogg007]]
+=== Solution 2===
 (By Qwertazertl)
@@ Line 22: / Line 37: @@
 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 3===
 (By mastermind.hk16)
@@ Line 32: / Line 47: @@
 because <math>Q_i \rightarrow D</math> as <math>i \rightarrow \infty</math>.
+=== Solution 4===
-=== Solution 3 ===
 (By user0003)
@@ Line 46: / Line 60: @@
 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===
+=== Solution 5 Diagram and Detailed Steps===
 [[Image:2016_AMC_12B_Problem_21.png|thumb|center|800px| ]]
@@ Line 83: / Line 97: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
-===Solution 5 (simple)===
-[[File:201612B-P21.png|400px|right]]
-We start with <math>DQ_i = 1/2</math> for <math>i = 1.</math> Then because <math>\triangle BP_iA \sim DP_iQ_i</math> we get <cmath>\frac{DP_i}{P_iB} = \frac{1}{2} \implies \frac{DP_i}{BD} = \frac{DQ_{i+1}}{DC} = \frac{1}{3} \implies DQ_{i+1} = DQ_{2} = \frac{1}{3}.</cmath> We can generalize this by saying <math>DQ_i = \frac{1}{i + 1}.</math> Then <math>[ADQ_i] = \frac{1}{2(i + 1)}.</math> Let <math>y = [AP_iD]</math> and let <math>x = [DP_iQ_i].</math> <math>\triangle BP_iA</math> and <math>\triangle DP_iQ_i</math> are similar with side length ratio <math>(i + 1):1</math>, so we can say <math>[BP_iA] = (i + 1)^2x.</math> Now, we can express the area of <math>\triangle ADB</math> (which is <math>1/2</math> since it's just half the square) as <math>y + (i + 1)^2x,</math> and we can express the area of <math>ADQ_i</math> as <math>x + y.</math> We have a system of equations:
-<cmath>[ADQ_i] = x + y = \frac{1}{2(i + 1)}</cmath>
-<cmath>[ADB] = (i + 1)^2x + y = \frac{1}{2}.</cmath>
-Solving, we get <math>x = \frac{1}{2(i+1)(i+2)}.</math> So now the problem becomes <math>\sum_{i=1}^{\infty} \frac{1}{2(i+1)(i+2)}</math>. We can rewrite this as
-<cmath>\frac{1}{2} \sum_{i=1}^{\infty} \left( \frac{1}{i+1} - \frac{1}{i+2} \right) = \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) + \dots \right)</cmath>
-All terms cancel except for the <math>\frac{1}{2}</math> on the inside and <math>\frac{1}{2}</math> on the outside, so the answer is <math>\boxed{\frac{1}{4}}.</math>
-~[[User:grogg007|grogg007]]
 ==Video Solution by CanadaMath (Problem 21-25)==

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