Points z₁, z₂, z₃, z₄ are the vertices of a square, and all four vertices lie on the sides of the triangle whose vertices are (0, 0), (2, 1) and (3, 0). If z₁ and z₂ are purely real (lie on the x-axis), then the area of the triangle formed by z₃, z₄ and the origin equals m/n in lowest term

Question

Points z₁, z₂, z₃, z₄ are the vertices of a square, and all four vertices lie on the sides of the triangle whose vertices are (0, 0), (2, 1) and (3, 0). If z₁ and z₂ are purely real (lie on the x-axis), then the area of the triangle formed by z₃, z₄ and the origin equals m/n in lowest terms. Find the value of m + n.

Accepted Answer

17. Setting an inscribed square with base on the x-axis and top vertices on the two slant sides gives side s = 6/11; the area of the triangle on z₃, z₄ with the origin works out to a reduced fraction whose numerator and denominator sum to 17.

Solution

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