# Diagonal of a Quadrilateral (RMO 2015 Mumbai Region Solution)

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.