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Set: there exist some lambda in [0,1]

Source: ISI(BS) 2007 #9

March 31, 2012
algebra unsolvedalgebra

Problem Statement

Let X⊂R2X \subset \mathbb{R}^2 be a set satisfying the following properties: (i) if (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) are any two distinct elements in XX, then  either,   x1>x2 and y1>y2 or,   x1<x2 and y1<y2\text{ either, }\ \ x_1>x_2 \text{ and } y_1>y_2\\ \text{ or, } \ \ x_1<x_2 \text{ and } y_1<y_2 (ii) there are two elements (a1,b1)(a_1,b_1) and (a2,b2)(a_2,b_2) in XX such that for any (x,y)∈X(x,y) \in X, a1≤x≤a2 and b1≤y≤b2a_1\le x \le a_2 \text{ and } b_1\le y \le b_2 (iii) if (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) are two elements of XX, then for all λ∈[0,1]\lambda \in [0,1], (λx1+(1−λ)x2,λy1+(1−λ)y2)∈X\left(\lambda x_1+(1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2\right) \in X
Show that if (x,y)∈X(x,y) \in X, then for some λ∈[0,1]\lambda \in [0,1], x=λa1+(1−λ)a2,y=λb1+(1−λ)b2x=\lambda a_1 +(1-\lambda)a_2, y=\lambda b_1 +(1-\lambda)b_2
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