IMO 1991 SL 22

Origin: USA

Problem

Real constants a, b, c are such that there is exactly one square all of whose vertices lie on the cubic curve y = x3 + ax2 + bx + c. Prove that the square has sides of length 4\sqrt 72.

Solution

Suppose w.l.o.g. that the center of the square is at the origin O(0, 0). We denote the curve y = f(x) = x3 + ax2 + bx + c by \gamma and the vertices of the square by A, B, C, D in this order. At first, the symmetry with respect to the point O maps \gamma into the curve \gamma (y = f(−x) = x3 −ax2 + bx −c). Obviously \gamma also passes through A, B, C, D, and thus has four different intersection points with \gamma. Then 2ax2 + 2c has at least four distinct solution, which implies a = c = 0. Particularly, \gamma passes through O and intersects all quadrants, and hence b < 0. Further, the curve \gamma′, obtained by rotation of \gamma around O for 90◦, has an equation −x = f(y) and also contains the points A, B, C, D and O. The intersection points (x, y) of \gamma \cap\gamma′ are determined by −x = f(f(x)), and hence they are roots of a polynomial p(x) = f(f(x)) + x of 9-th degree.

But the number of times that one cubic actually crosses the other in each quadrant is in the general case even (draw the picture!), and since ABCD is the only square lying on \gamma \cap\gamma′, the intersection points A, B, C, D must be double. It follows that p(x) = x[(x −r)(x + r)(x −s)(x + s)]2, (1) where r, s are the x-coordinates of A and B. On the other hand, p(x) is defined by (x3+bx)3+b(x3+bx)+x, and therefore equating of coefficients with (1) yields 3b = −2(r2 + s2), 3b2 = (r2 + s2)2 + 2r2s2, b(b2 + 1) = −2r2s2(r2 + s2), b2 + 1 = r4s4. Straightforward solving this system of equations gives b = − \sqrt 8 and r2 + s2 = \sqrt 18. The line segment from O to (r, s) is half a diagonal of the square, and thus a side of the square has length a =

2(r2 + s2) = 4\sqrt 72.