1969 IMO Problems/Problem 5
Contents
[hide]Problem
Given points in the plane such that no three are collinear, prove that there are at least 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) = convex quadrilaterals.
The above solution was posted and copyrighted by Philip_Leszczynski. The original thread for this problem can be found here: [1]
Remarks (added by pf02, August 2024)
The solution given above is incorrect. This is pointed out very clearly and explicitly by DHu on https://aops.com/community/p364185. I will not repeat the argument here. DHu also gives the idea for a correct proof.
On the same page (https://aops.com/community/p364185), manuel gives an idea for a proof of a stronger result. His proof is somewhere between vague, incorrect and incomplete.
Below, I will give two solutions. The first is just an expansion of DHu's idea. The second one is carrying out manuel's idea.
Solution based on DHu's idea
We start by proving that given points, there is at least one convex quadrilateral formed by of them. Denote the points by . Without loss of generality, we can chose to label the points so that are on one side of the line . Further, we chose to label the point so that are on one side of . It follows that are inside , with segments and extended beyond and respectively.
Note that from the choice of it follows that . Also, we could have chosen by considering the convex hull of the set of 5 points. It is either a triangle, or a convex quadrilateral, or a convex pentagon. We choose to be any of the vertices, and to be the adjacent vertices. (We don't need these notes in the proof, I just wanted to make things easier to visualize.)
The picture below shows this choice of labels.
The line must intersect at least one of and . Without loss of generality, we can assume it intersects . In general, it will also intersect . The argument we are going to give is for the case when intersects both and . The argument needed in the limiting case when is identical, and is left as an exercise to the reader.
There are three cases:
Case 1, when intersects , but it does not intersect . This is shown on the first 3 pictures (one picture would have been enough, but I made all of them, for emphasis). In this case the quadrilateral formed by the points is convex. (Note that this also applies to the case when .)
Case 2, when intersects both and . This is shown in the 4th picture. In this case the quadrilateral formed by the points is convex.
Case 3, when does not intersect either of or . This is shown in the 5th picture. In this case the quadrilateral formed by the points is convex. (Note that this also applies to the case when .)
We thus proved that in any group of 5 points, there is a subgroup of 4 points which forms a convex quadrilateral. Note that in general, given a group of 5 points, more than one convex quadrilateral can be found by taking subgroups of 4 points.
Important remark: In fact, we proved more: if are inside the angle formed by and (with segments and extended beyond and respectively), then we can choose the convex quadrilateral so that are both part of it.
Now the statement of the problem is easy to prove. Assume points are given. Choose of them, and label them , so that all the other points are inside . There are such points. Now choose all the groups of points from these points. There are ways of doing this. From what we proved before, we know that for each choice of 2 points, call them , there is a group of points from among which forms a convex quadrilateral, such that are of the points. A different choice will yield a different quadrilateral because .
This concludes the proof.
Solution 2, based on manuel's idea
As we did in the previous solution, we first prove that given a group of points, there is a subgroup of of them which forms a convex quadrilateral. Unlike in the previous solution, we don't care whether specific points are part of it or not. Note that generally there might be several convex quadrilaterals formed from the group of points.
Now, given points, create a list of all the groups of points chosen from the given points. This can be done in ways, so the list has elements. For each group of points, choose a convex quadrilateral formed by a subgroup of the points. Thus, create a list of convex quadrilaterals. This list has elements.
However, note that some convex quadrilaterals in are not unique, they might appear several times in . A given quadrilateral could appear up to times, as chosen from a group of points which contains the points of the quadrilateral, and one point from the remaining points. We want to form the list , which is a subset of , such that from each group of duplicates, we keep only one. Let be the number of elements in .
Since each quadrilateral in appears at most times, it follows that , in other words, there are at least convex quadrilaterals formed from points in the given points.
This is a stronger result than the one in the statement of the problem.
To prove the statement of the problem, we will prove the assertion I just made above, by showing that . This amounts to proving that
for . Carry out the computations, and we get that we need to prove that for . The polynomial has roots from which it follows that
is true, and it is an equality for and a strict inequality for .
The original thread for this problem can be found here: https://aops.com/community/p364185
See Also
1969 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |