Question

Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}.$ For $i=1,2,\ldots,$ 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

$\sum_{i=1}^{\infty}\text{Area of }\triangle DQ_iP_i?$

$\dfrac16$

$\dfrac14$

$\dfrac13$

$\dfrac12$

$1$

Solution:

Place $D=(0,0),$ $C=(1,0),$ $B=(1,1),$ $A=(0,1),$ and let $q_i=DQ_i.$ Intersecting line $AQ_i$ with $\overline{BD}$ (the line $y=x$ ) gives $P_i$ with both coordinates $\dfrac{q_i}{1+q_i},$ so $q_{i+1}=\dfrac{q_i}{1+q_i}.$ From $q_1=\tfrac12$ this yields $q_i=\dfrac{1}{i+1}.$ The base of $\triangle DQ_iP_i$ is $DQ_i=\dfrac{1}{i+1}$ and its height is the $y$ -coordinate of $P_i,$ which is $q_{i+1}=\dfrac{1}{i+2}.$ Then $\text{Area of }\triangle DQ_iP_i=\tfrac12\cdot\dfrac{1}{i+1}\cdot \dfrac{1}{i+2}=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right).$ Summing telescopes to $\tfrac12\cdot\tfrac12=\tfrac14.$

Thus, the correct answer is B.

2016 AMC 12B Problem 21

Solution:

Problem 21 in Other Years