6th Chinese 1991 Problem A1 For which convex quadrilaterals ABCD can we find a point in the plane P such that the four triangles PAB, PBC, PCD, PDA all have equal area? What is the maximum possible number of such points for a given quadrilateral? Solution
: 3. Suppose P is inside ABCD. We use the fact that area PBA = area PBC. Let the diagonal AC meet the line PB at X. Then area PBA = 1/2 PB.AX sin PXA, area PBC = 1/2 PB.CX sin PXC. Hence AX = CX, so X is the midpoint of AC. Similarly, considering PDA and PDC, we conclude that the midpoint of AC must lie on the line PD. So either (1) P = X, or (2) P and X lie on BD. In case (1), considering PAB and PAD, we see that the midpoint of BD must lie on AC, or put another way, area ABC = area ADC, and this is also a sufficient condition. In case (2), considering PDA and PBA, we see that P must be the midpoint of BD. So both cases are essentially the same: for a point to exist the midpoint of one diagonal must lie on the other and the unique point P is then the midpoint of the other. Now suppose P is outside ABCD. It must lie in one of the four areas bounded by a side and the extension of two adjacent sides (otherwise one of the P triangles would lie inside another). So suppose P lies outside the side AB. Then since area PAB = area PAD, BD must be parallel to PA. Similarly, PB must be parallel to AC. So if the diagonals AC, BD intersect at O, then PAOB is a parallelogram and area PAB = area ABO. But we also have that area PAB + area ABCD = area PAD area PCD + area PBC and hence area ABO = area PAB = 1/2 area ABCD. Thus for P to exist one of the triangles ABO, BCO, CDO and DAO must have half the area of the quadrilateral and the corresponding position of P is then uniquely fixed. If there are two such triangles, then one point of the quadrilateral must be the midpoint of another two. So we might have, for example, C the midpoint of BD. In this case, there are two points P (the intersections of the line through A parallel to BD with the lines parallel to AC through B and D). In this case, there is also a third point P inside ABCD, namely the midpoint of AC. In general, however, there is at most one internal point and at most one external point. We show, in fact, that the case just dealt with is the only case where we can have both an internal and an external point. For if there is an internal point, then one diagonal bisects the area of the quadrilateral. If also one of the four triangles has area half the quadrilateral, then two triangles must have zero area and hence one vertex (say C) must line on the segment joining two others (say B and D). Then since AC bisects the area, C must be the midpoint of BD (which is the case already dealt with). 6th Chinese 1991 © John Scholes jscholes@kalva.demon.co.uk 1 May 2002