D is a point inside the equilateral triangle ABC. E is a point inside DBC. Show that area DBC/(perimeter DBC)2 > area EBC/(perimeter EBC)2.
Solution
Let us find an expression for t = (area DBC)/(perimeter DBC)2. Let angle B = 2x, angle C = 2y and angle D = 2z and let the inradius be r. Then area DBC = r/2 x perimeter DBC. Also BC = r cot x + r cot y, and similarly for the other sides, so perimeter = 2r(cot x + cot y + cot z). Hence t = 1/(4 (cot x + cot y + cot z) ). Putting z = 90o - x - y, we get 1/4t = cot x + cot y + (cot x + cot y)/(cot x cot y - 1). Now EBC has both x and y smaller than DBC, and cot x is a decreasing function of x (certainly for x in the range 0 to 30o). So writing u = cot x, v = cot y, it is sufficient to show that u + v + (u + v)/(uv - 1) is an increasing function of u. [It is symmetric, so it follows that it is also an increasing function of v.] We have x, y < 30o, and hence u, v > √3, so we need the result at least for u, v > √3.
We have u + v + (u + v)/(uv - 1) = u + v + 1/v (uv - 1)/(uv - 1) + (v + 1/v)/(uv - 1). In considering the dependence on u, we can ignore the terms that do not depend on u, so we have u + (1 + 1/v2)/(u - 1/v). Put k = u - 1/v, then we have to consider k + h2/k, where h2 = 1 + 1/v2. But this is an increasing function for k > h (see below). Thus u + v + (u + v)/(uv - 1) is an increasing function of u for u - 1/v > (1 + 1/v2)1/2, or for u > 1/v + (1 + 1/v2)1/2. This bound is highest for the smallest v in other words for v = √3, when it is √3. So for v > √3, u + v + (u + v)/(uv - 1) is an increasing function of u for u > √3.
[To see that k + h2/k is increasing for k > h, take k' > k > h. Then k' + h2/k' - k - h2/k = (k' - k) - h2(1/k - 1/k') = (k' - k)(1 - h2/kk') > 0.]
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002