Problem: Let ABCD be a convex quadrilateral with AB = a, BC = b, CD = c and DA = d. Suppose a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da and the area of ABCD is 60 square units. If the length of one of the diagonals is 30 unit, determine the length of the other diagonal.

Discussion:

(Solution suggested in class by Megha Chakraborty)

It is given that a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da
Multiplying 2 to both sides we have

2a^2 + 2b^2 + 2c^2 + 2d^2 = 2ab + 2bc + 2cd + 2da
\Rightarrow a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2cd + d^2 + d^2 - 2ad + a^2 = 0
\Rightarrow (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 = 0

But sum of squares can be 0 if and only if each square is individually 0. This implies a = b = c = d. Hence the quadrilateral is a rhombus (in a special case, a square).

The area of a rhombus is \frac{1}{2}\times{d_1}\times {d_2} where d_1 and d_2 are the diagonals. It is given that one of the diagonals is 30 and area is 60. Hence we have
\frac{1}{2}\times 30\times {d_2} =60
\Rightarrow {d_2} = 4

Hence the length of the other diagonal is 4.

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