2016 AMC 12B 第 21 题

Similar triangles give the recursion $DQ_{i+1}=\dfrac{DQ_i}{1+DQ_i},$ and $DQ_1=\tfrac12$ leads to $DQ_i=\dfrac{1}{i+1}$

The area of $\triangle DQ_iP_i$ is $\tfrac12\cdot\dfrac{1}{i+1}$ $\cdot\dfrac{1}{i+2}$ $=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right);$ the sum telescopes

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 $$\begin{gathered} \text{Area of }\triangle DQ_iP_i=\tfrac12\cdot\dfrac{1}{i+1} \\ \quad{}\cdot\dfrac{1}{i+2} \\ {}=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right). \end{gathered}$$ Summing telescopes to $\tfrac12\cdot\tfrac12=\tfrac14.$

Thus, the correct answer is B.