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December 27, 2015

Diagonal of a Quadrilateral | RMO 2015 Mumbai Region

This is a problem from the Regional Mathematics Olympiad, RMO 2015 Mumbai Region based on Diagonal of a Quadrilateral. Try to solve it.

Problem: Diagonal of a Quadrilateral

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.


(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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