A, B, C each have integral coordinates and lie on a circle radius R. Show that at least one of the distances AB, BC, CA exceeds R1/3.
Solution
Let the sides have lengths a, b, c as usual. The question suggests that we use some relationship of the form abc = constant x R. That would give the power of 1/3. If you happen to know the relation abc = 4ΔR, where Δ is the area of the triangle, then the rest is easy. If not, you will have a rough time.
To prove the relation, let O be the centre of the circumcircle. Project AO to meet the circle again at K. Let AH be the altitude. Then angle ABC = angle AKC, so triangles ABH and AKC are similar. Hence AB/AH = AK/AC or c/AH = 2R/b. Hence abc = 2R·a·AH = 4ΔR.
So far we have not used the fact that the triangle has integral coordinates. Take the vectors along two sides. The area is half the magnitude of the cross product of these vectors. But the vectors have integral components, so the cross product has also. Hence its magnitude is at least 1 and the area Δ ≥ 1/2, so abc ≥ 2R > R, hence at least one of a, b, c > R1/3.
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001