Problem:
A quadrilateral has vertices P(a,b),Q(b,a),R(−a,−b), and S(−b,−a), where a and b are integers with a>b>0. The area of PQRS is 16. What is a+b ?
Answer Choices:
A. 4
B. 5
C. 6
D. 12
E. 13 Solution:
The slopes of PQ​ and RS are -1 , and the slopes of QR​ and PS are 1 , so the figure is a rectangle. The side lengths are PQ=(a−b)2​ and PS=(a+b)2​, so the area is 2(a−b)(a+b)=2(a2−b2)=16. Therefore a2−b2=8. The only perfect squares whose difference is 8 are 9 and 1 , so a=3, b=1, and a+b=4​.
