ABC is a triangle with circumradius R and inradius r. If p is the inradius of the orthic triangle, show that p/R ≤ 1 - (1/3) (1 + r/R)^{2}). [The *orthic* triangle has as vertices the feet of the altitudes of ABC.]

**Solution**

∠HA'B = ∠HC'C = 90^{o}, so HA'BC' is cyclic. Hence ∠HA'C' = ∠HBC' = 90^{o} - ∠A. Similarly, ∠HA'B' = 90^{o} - ∠A, so H is the incenter of the orthic triangle.

HA' = A'B tan HBA' = A'B cot C = c cos B cot C. Hence p = HA' sin HA'C' = HA' cos A = c cos A cos B cos C/sin C. If O is the circumcenter, then ∠AOB = 2 ∠C, so c/2R = sin C. Hence p = 2R cos A cos B cos C.

We have 4 cos A cos B cos C = 2 (cos(A+B) + cos(A-B)) cos C = cos(A+B+C) + cos(A+B-C) + cos(A-B+C) + cos(A-B-C) = cos 180^{o} + cos(180^{o}-2C) + cos(180^{o}-2B) + cos(2A-180^{o}) = -1 - cos 2C - cos 2B - cos 2A = 2 - 2(cos^{2}A + cos^{2}B + cos^{2}C). Hence p/R = 1 - (cos^{2}A + cos^{2}B + cos^{2}C).

Now by Cauchy we have (1^{2}+1^{2}+1^{2})(cos^{2}A + cos^{2}B + cos^{2}C) ≥ (cos A + cos B + cos C)^{2}, so p/R ≤ 1 - (cos A + cos B + cos C)^{2}/3.

Finally, we need the result that cos A + cos B + cos C = 1 + r/R. We have cos A = (b^{2}+c^{2}-a^{2})/2bc, cos B = (a^{2}+c^{2}-b^{2})/2ac, cos C = (a^{2}+b^{2}-c^{2})/2ab, so cos A + cos B + cos C = (ab^{2} + a^{2}b + b^{2}c + bc^{2} + c^{2}a + ca^{2} - a^{3} - b^{3} - c^{3})/2abc = (8(s-a)(s-b)(s-c) + 2abc)/2abc = 4(area ABC)^{2}/(abcs) + 1 (using Heron). But we also have area ABC = ½ab sin C = abc/4R, and area ABC = rs. So 4(area ABC)^{2}/(abcs) = r/R, giving cos A + cos B + cos C = 1 + r/R, as required.

© John Scholes

jscholes@kalva.demon.co.uk

8 Nov 2003

Last corrected/updated 8 Nov 03