Bounded area for any triangle formed by polygonsLet P 1 , P 2 , P 3 be closed polygons on the plane. Suppose that for any points A ∈ P 1 (meaning A can be inside or on the boudary of P 1 ), B ∈ P 2 , C ∈ P 3 , we have [ A B C ] ≤ 1. Is it possible that two of P 1 , P 2 , P 3 have area ≥ 4? What about all three having area ≥ 4?Here, [X] denotes the area of the polygon X.

Question

Bounded area for any triangle formed by polygonsLet       P    1    ,      P    2    ,      P    3   be closed polygons on the plane. Suppose that for any points   A  ∈      P    1   (meaning A can be inside or on the boudary of       P    1  ),   B  ∈      P    2    ,  C  ∈      P    3  , we have   [  A  B  C  ]  ≤  1. Is it possible that two of       P    1    ,      P    2    ,      P    3   have area   ≥  4? What about all three having area   ≥  4?Here, [X] denotes the area of the polygon X.

Neil Short · Accepted Answer

Step 1The following is an argument that two polygons can have area of 4. We will work with circles and at the end replace them by many-sided polygons.Let C1 and C2 be circles centered at the origin both of radius r and C3, a circle also at the origin of radius s for   s  ≤  r. Pick A in C3, B in C1 and C in C2 so the triangle has largest area with these conditions imposed. By symmetry we can pick A to have coordinates (0,-s), B to have coordinates (r cos x, r sin x) and C to have coordinates (- r cos x, r sin x) for some angle x between 0 and pi/2.Let   q  =  s      /    r. Then straightforward calculations show that the area of triangle ABC is       r    2    (  q  +  sin  ⁡  x  )  cos  ⁡  x. In order to have maximum area its derivative with respect to x is 0. This gives the equation:      cos    2    ⁡  x  =  q  sin  ⁡  x  +  cos  ⁡  x  sin  ⁡  x  .Step 2We want the area of triangle ABC to be 1, so   1  =      r    2    (  q  +  sin  ⁡  x  )  cos  ⁡  x  .We now ask if there is a solution to these two equations where   π      r    2    =  4. Eliminating r and q from these now three equations, we get the equation:  π      /    4  =  (      cos    2    ⁡  x      /    sin  ⁡  x  –  cos  ⁡  x  +  sin  ⁡  x  )  cos  ⁡  x  .The expression on the right is infinity at   x  =  0 and 0 at 0 at x   =  π      /    2, so by continuity it does have a solution.Excel tells me that when x is   &amp;gt;  0.45 radians roughly then q is less than 1 and when x is   &amp;gt;  0.5 radians roughly the expression on the right above is approximately   π      /    4. So there is a solution for x approximately .5 radians.Now replace the circles by polygons with many sides approximating them to get the final conclusion.I suspect there is no solution for all three polygons but don’t know how to proceed.

Let P_1, P_2, P_3 be closed polygons on the plane. Suppose that for any points A in P_1 (meaning A can be inside or on the boudary of P_1), B in P_2, C in P_3, we have [ABC] leq 1. Is it possible that two of P_1, P_2, P_3 have area geq 4? What about all three having area geq 4?

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