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Let S be a square of unit area. Consider a quadrilateral which has one vertex on each side of the square S. If a,b,c,d denote the lengths of sides of quadrilateral then prove that 2<= a 2 +b 2 +c 2 +d 2 <= 4

Let S be a square of unit area. Consider a quadrilateral which has one vertex on each side of the square S. If a,b,c,d denote the lengths of sides of quadrilateral then prove that
2<= a2 +b2 +c2 +d<= 4

Grade:11

1 Answers

Arun
25763 Points
3 years ago
Let the center of the unit square be at the origin. 
Let P (p, 1/2), Q (1/2, q), R (r, -1/2) and S (-1/2, s) be the 
vertices of the quadrilateral on the sides of the square. 

=> a^2 + b^2 + c^2 + d^2 
= PQ^2 + QR^2 + RS^2 + SP^2 
= (p - 1/2)^2 + (q - 1/2)^2 + (r - 1/2)^2 + (q + 1/2)^2 
+ (r + 1/2)^2 + (s + 1/2)^2 + (p + 1/2)^2 + (s - 1/2)^2 
= 2 (p^2 + q^2 + r^2 + s^2 + 1). 

a^2 + b^2 + c^2 + d^2 = 2 is minimum when p = q = r = s = 0, 
i.e., when P, Q, R and S are the mid-points of the sides of the square. 

As each of p, q, r and s lies between - 1/2 and 1/2, 
maximum values of each of p^2, q^2, r^2 and s^2 = (1/2)^2 = 1/4 
=> maximum value of a^2 + b^2 + c^2 + d^2 = 2 (1/4 + 1/4 + 1/4 + 1/4 + 1) = 4. 

=> 2 ≤ a² + b² + c² + d² ≤ 4.

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