ABC is an acute-angled triangle area 1. Show that the triangle whose vertices are the feet of the perpendiculars from the centroid to AB, BC, CA has area between 4/27 and 1/4.
Solution
Let the centroid be G and the feet of the perpendiculars to BC, CA, AB be D, E, F respectively. Since G is 1/3 of the way along the median from the midpoint, area BGC = 1/3 area ABC = 1/3. But area BGC = a.GD/2, hence GD = 2/(3a). Similarly, GE = 2/(3b) and GF = 2/(3c). Now area GEF = (GE.GF.sin EFG)/2 = (2 sin A)/(9bc). But area ABC = (bc sin A)/2, so bc = 2/sin A. So area GEF = (sin2A)/9. Hence Area DEF = (sin2A + sin2B + sin2C)/9. So we need to show that 4/3 ≤ (sin2A + sin2B + sin2C) ≤ 9/4.
Take A ≤ B ≤ C. We have C ≤ π/2, so A + B ≥ π/2, so B ≥ π/4. Now sin2x has decreasing derivative between π/4 and π/2, so sin2B + sin2C ≤ 2 sin2((B+C)/2) = 2 cos2A/2 = cos A + 1. Hence (sin2A + sin2B + sin2C) ≤ sin2A + cos A + 1 = 2 - cos2A + cos A = 9/4 - (cos A - 1/2)2 ≤ 9/4. Note that we can only have equality at the last step if cos A = 1/2 or A = π/3, which implies the triangle is equilateral. It is easy to check that in this case (sin2A + sin2B + sin2C) = 9/4.
In the other direction we have in fact that sin2A + sin2B + sin2C >= 2, which is significantly stronger than what we need. This is slightly awkward to prove. Take C = π/2 - 2y, B = π/4 + y + x, A = π/4 + y - x. So we are assuming that x, y >= 0, and x + 3y ≤ π/4. Now sin2A + sin2B + sin2C = (1/√2 cos(x - y) - 1/√2 sin(x - y) )2 + (1/√2 cos(x + y) + 1/√2 sin(x + y) )2 + cos22y = 1 - cos(x - y)sin(x - y) + cos(x + y)sin(x + y) + cos22y = 1 + cos22y - 1/2 sin2(x - y) + 1/2 sin2(x + y) = 1 + cos22y + cos 2x sin 2y = 2 + sin 2y (cos 2x - sin 2y). But sin 2y ≥ 0 and sin 2y < sin 6y ≤ sin(π/2 - 2x) = cos 2x. Hence sin 2y (cos 2x - sin 2y) ≥ 0 (with equality iff y = 0, so that the triangle is right-angled).
Comment. It is curious that the question made the inequality weaker than necessary. It should have been "between 2/9 and 1/4".
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002