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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.