1962 AMC 12 Problem 37

Question

ABCD

is a square with side of unit length. Points

E

and

F

are taken respectively on sides

AB

and

AD

so that

AE=AF

and the quadrilateral

CDFE

has maximum area. In square units this maximum area is:

\dfrac12

\dfrac9{16}

\dfrac{19}{32}

\dfrac58

\dfrac23

Let

AE=AF=t

and use coordinates or the shoelace formula

The area becomes

\frac12(1+t-t^2)

; complete the square

Set

A=(0,0),

B=(1,0),

C=(1,1),

and

D=(0,1).

Then

E=(t,0)

and

F=(0,t).

The shoelace formula gives

\begin{aligned} [CDFE]&=\frac{1+t-t^2}{2}\\ &=\frac58-\frac12 \left(t-\frac12\right)^2. \end{aligned}

Its maximum is

\frac{5}{8}.

Thus, the correct answer is D.

Problem 37 in other years