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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 $\geq 4$?

Here, $[X]$ denotes the area of the polygon $X$.

[Source: Hungarian competition problem]

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The 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.

We 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 pi r^2 = 4. Eliminating r and q from these now three equations, we get the equation:

pi/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 = pi/2, so by continuity it does have a solution.

Excel tells me that when x is > .45 radians roughly then q is less than 1 and when x is > .5 radians roughly the expression on the right above is approximately pi/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.

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