1985 AMC 12 Problem 13

Attempt Problem 13 of the 1985 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 1985 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

13.

Pegs are put in a board 11 unit apart both horizontally and vertically. A rubber band is stretched over 44 pegs as shown in the figure, forming a quadrilateral. Its area in square units is
44
4.54.5
55
5.55.5
66
Answer: E
Concepts:coordinate geometrylattice pointshoelace formula
Small hint:
Assign integer coordinates to the four corner pegs
Big hint:
Use the shoelace formula on (1,3),(1,3), (4,1),(4,1), (3,0),(3,0), and (0,1)(0,1)
Solution:
With the lower-left peg as (0,0),(0,0), the quadrilateral’s vertices in order are (1,3),(1,3), (4,1),(4,1), (3,0),(3,0), and (0,1).(0,1). The shoelace formula gives 12∣(1+0+3+0)−(12+3+0+1)∣=6. \begin{aligned} &\frac12\left|(1+0+3+0)\right.\\ &\quad\left.{}-(12+3+0+1)\right|=6. \end{aligned} Therefore the correct answer is E.
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