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1962 AMC 12 Problem 37

Attempt Problem 37 of the 1962 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1962 AMC 12 solutions, or check the answer key.

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37.

ABCDABCD is a square with side of unit length. Points EE and FF are taken respectively on sides ABAB and ADAD so that AE=AFAE=AF and the quadrilateral CDFECDFE has maximum area. In square units this maximum area is:

12\dfrac12

916\dfrac9{16}

1932\dfrac{19}{32}

58\dfrac58

23\dfrac23

Answer: D
Concepts:areashoelace formulacompleting the squareoptimization
Difficulty rating: 1730
Small Hint:

Let AE=AF=tAE=AF=t and use coordinates or the shoelace formula

Big Hint:

The area becomes 12(1+t−t2)\frac12(1+t-t^2); complete the square

Solution:

Set A=(0,0),A=(0,0), B=(1,0),B=(1,0), C=(1,1),C=(1,1), and D=(0,1).D=(0,1). Then E=(t,0)E=(t,0) and F=(0,t).F=(0,t). The shoelace formula gives [CDFE]=1+t−t22=58−12(t−12)2. \begin{aligned} [CDFE]&=\frac{1+t-t^2}{2}\\ &=\frac58-\frac12 \left(t-\frac12\right)^2. \end{aligned} Its maximum is 58.\frac{5}{8}.

Thus, the correct answer is D.

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