IMO 1995 · Problem 3

Determine all integers \(n>3\) for which there exist \(n\) points \(A_1,\ldots,A_n\) in the plane, no three collinear, and real numbers \(r_1,\ldots,r_n\) such that for \(1\le i<j<k\le n\), the area of \(\triangle A_iA_jA_k\) is \(r_i+r_j+r_k\).

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{n=4}. \]

Solution

For \(n=4\), take the vertices of a unit square and set \[ r_1=r_2=r_3=r_4=\frac16. \] Every triangle formed by three vertices has area \(\frac12\), so this satisfies the requirements.

We prove that five points cannot satisfy the condition. This also excludes every \(n>5\), since any five points of such a configuration would inherit the condition.

Write \([XYZ]\) for the area of \(\triangle XYZ\), and denote the real number assigned to a point by the corresponding lowercase letter. Thus \[ [XYZ]=x+y+z. \] All these triangle areas are strictly positive.

We first record two consequences.

  • If \(X,Y,Z,W\) form a convex quadrilateral in that cyclic order, its two diagonal decompositions give \[ [XYZ]+[XZW]=[XYW]+[YZW], \] and hence \[ x+z=y+w. \tag{1} \]

  • If \(P\) lies strictly inside \(\triangle XYZ\), then \[ [XYZ]=[PXY]+[PYZ]+[PZX]. \] Substitution gives \[ x+y+z+3p=0. \tag{2} \] In particular, \(p<0\).

The convex hull of five points with no three collinear has three, four, or five vertices. We treat each possibility.

Case 1: The convex hull is a triangle \(ABC\).

The remaining points \(P,Q\) lie strictly inside it. Put \[ T=a+b+c=[ABC]>0. \] By (2), \[ p=q=-\frac{T}{3}. \] Consequently, \[ [APQ]+[BPQ]+[CPQ] =a+b+c+3p+3q =-T<0, \] contradicting the positivity of the three areas.

Case 2: The convex hull is a quadrilateral \(ABCD\).

Let \(P\) be the remaining point and let \(O=AC\cap BD\). Since no three points are collinear, \(P\) lies on neither diagonal. The diagonals partition the quadrilateral into four triangles, so, after cyclically relabelling the vertices, we may assume that \(P\) lies strictly inside \(\triangle ABO\). It therefore lies strictly inside both \(\triangle ABC\) and \(\triangle ABD\).

Applying (2) to those two triangles gives \[ a+b+c+3p=0,\qquad a+b+d+3p=0, \] so \(c=d\). Equation (1) for \(ABCD\) then gives \(a=b\).

Write \[ a=b=x,\qquad c=d=y. \] We have \[ 2x+y+3p=0,\qquad p<0. \tag{3} \] On the other hand, calculating the quadrilateral’s area using diagonal \(AC\) gives \[ [ABCD]=[ABC]+[ACD]=3(x+y). \] Calculating it using the four triangles with vertex \(P\) gives \[ \begin{aligned} [ABCD] &=[ABP]+[BCP]+[CDP]+[DAP]\\ &=(2x+p)+(x+y+p)+(2y+p)+(x+y+p)\\ &=4(x+y+p). \end{aligned} \] Therefore \[ x+y+4p=0. \] Subtracting this equation from (3) yields \(x=p\). But then \[ [ABP]=2x+p=3p<0, \] again a contradiction.

Case 3: The convex hull is a pentagon \(ABCDE\).

The vertices are listed in cyclic order. Applying (1) to the quadrilaterals \(ABCD\), \(ABCE\), \(ABDE\), and \(ACDE\), respectively, gives \[ \begin{aligned} a+c&=b+d,\\ a+c&=b+e,\\ a+d&=b+e,\\ a+d&=c+e. \end{aligned} \] The first two equations give \(d=e\). The third then gives \(a=b\), and the fourth gives \(a=c\). Substitution into the first gives \(c=d\). Thus \[ a=b=c=d=e. \] In particular, \[ [ABC]=[ABD]=[ABE]. \] These triangles have the same base \(AB\), and \(C,D,E\) all lie on the same side of the line \(AB\), because \(AB\) is an edge of the convex hull. Their equal areas therefore imply that \(C,D,E\) have equal perpendicular distances from \(AB\). They lie on one line parallel to \(AB\), contradicting the assumption that no three points are collinear.

All possible configurations of five points have been excluded. Together with the square construction, this proves that the only answer is \(n=4\).

The solution was checked against John Scholes' IMO solution archive (Kalva).