2016 AMC 12B Problem 21

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

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.

Problem 21 in other years