# let s be the square of unit area. consider any quadrileteral which has one vertex on each side of s . if a,b,c,and d denote the length of the sides of quad prove that 2

Vikas TU
14149 Points
4 years ago

Sol. a2 = p2 + s2, b2 = (1 – p)2 + q2
c2 = (1 – q)2 + (1 – r)2, a2 = r2 + (1 – s)2
∴ a2 + b2 + c+ d2 = {p2 + (1 + p)2} + {q2 – (1 – q)2} + {r2} + (1 – r)2} + {s2 + (1 – s)2}
Where p, q, r, s all vary in the interval [0, 1].
Now consider the function
y2 = x2 + (1 – x)2, 0 ≤ x ≤ 1,
2y dy/dx = 2x – 2(1 – x) = 0
⇒ x = 1/2 which d2 y/dx2 = 4 i.e. +ive
Hence y is minimum at x = 1/2 and its minimum
Value is 1/4 + 1/4 = 1/2
Clearly value is maximum at the end pts which is 1.
∴ Minimum value of a2 + b2 + c2 + d2 = 1/2 + 1/2 + 1/2 + 1/2 = 2
And maximum value is 1 + 1 + 1 + 1 = 4. Hence proved.
ALTERNATE SOLUTION :
x2 + y2 ≤ (x + y)2 ≤ 1 if x + y = 1
Here x = p, y = 1 – p ∴ x + y = 1
∴ a2 + b2 + c2 + d2 ≤ 1 + 1 + 1 + 1 = 4
Again x2 + y2 = (x + y)2 – 2xy = 1 – 2xy
NOTE THIS STEP : ∴ Minimum of (x2 + y2) = 1 – 2( maximum of xy).
KEY CONCEPT : Now we know that product of two quantities xy is maximum when the quantities are equal provided their sum is constant.
Here x + y = p + 1 – p = 1 = constant.
∴ xy is maximum when x/1 = y/1 = x + y/2 = 1/2
∴ x = 1/2, y = 1/2
Minimum of x2 + y2 = 1 – 2. 1/2. 1/2 = 1 – 1/2 = 1/2
∴ Minimum value of
a2 + b2 + c2 + d2 = 1/2 + 1/2 + 1/2 + 1/2 = 2
∴ 2 ≤ a2 + b2 + c2 + d2 ≤ 4.