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

Revision as of 09:29, 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

Diagram

~luckuso

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 $\boxed{\textbf{(B) }\frac{1}{4}}$ .

@@ Line 10: / Line 10: @@
 == Solutions ==
+=== Diagram ===
+[[Image:2016_AMC_12B_Problem_21.png|thumb|center|800px| ]]
+~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 === Solution 1 ===
 (By Qwertazertl)

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