1969 IMO Problems/Problem 5

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Problem

Given $n > 4$ points in the plane such that no three are collinear, prove that there are at least $(n-3)/2$ convex quadrilaterals whose vertices are four of the given points.

Solution

Orient the points so that one of them corresponds to the origin (A), another lies on the y-axis (B), and all the others are in either quandrant I or IV. (In other words, you are creating a boundary line.) Select one of the points (C) which minimizes the angle BAC. You cannot have two of them because then they would be collinear. Also no points lie inside the area of ABC because if such a point D did exist, then the angle DAC would be less than BAC - contradiction. So there are none of the other n-3 points inside the area of ABC.

Now we will show that for any two of the remaining n-3 points, you can create a convex quadrilateral containing two of them and two of A,B,C.

Now select two arbitrary points from the remaining n-3. Call them E and F, and draw a line though them; call this line l. If l does not intersect ABC, then ABCEF is convex and you choose the quadrilateral ABEF. The other case is that it separates ABC such that one of A,B,C is one one side of l and the other two are on the other side. Without loss of generality, A is on one side of l while B and C are on the other side. Consider the quadrilateral BCEF (or CBEF, depending on orientation.) We know that BC are on the same side of EF. Also, since no points are in the third quadrant, we know that EF are on the same side of BC. Now we can choose either BCEF or CBEF to be our convex quadrilateral.

So we now have that for the given ABC, all pairs of the other n points give a distinct convex quadrilateral. So we have n-3 points to choose from, and we must choose two of them. So this gives us at least C(n-3,2) = $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals.

The above solution was posted and copyrighted by Philip_Leszczynski. The original thread for this problem can be found here: [1]