2014 AMC 10A Problem 18

Attempt Problem 18 of the 2014 AMC 10A 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 2014 AMC 10A solutions, or check the answer key.

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

18.

A square in the coordinate plane has vertices whose yy-coordinates are 0,0, 1,1, 4,4, and 5.5. What is the area of the square?
1616
1717
2525
2626
2727
Answer: B
Concepts:coordinate geometrysquare (geometry)vector
Small hint:
Choose adjacent vertices whose y-coordinates differ by 11
Big hint:
A side vector rotated 90∘90^\circ creates a y-change of 44
Solution:
Opposite vertices of a square have the same average yy-coordinate. Thus the opposite pairs must have yy-coordinates (0,5)(0,5) and (1,4)(1,4), the two pairs with equal sum. In particular, a vertex with yy-coordinate 00 is adjacent to vertices with yy-coordinates 11 and 44.
Let one vertex be A=(0,0)A=(0,0), and let its adjacent vertex with yy-coordinate 11 be B=(x,1)B=(x,1) with x>0x>0.
Rotating the side vector (x,1)(x,1) by 90∘90^\circ gives the next side vector (−1,x)(-1,x), so another vertex has y-coordinate xx. The remaining y-coordinates are 44 and 55, hence x=4x=4.
The side length squared is AB2=x2+12=42+1=17AB^2=x^2+1^2=4^2+1=17, which is the area of the square.
Thus, B is the correct answer.
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