16th Chinese 2001 Problem A1 ABCD is a cyclic quadrilateral. The center of the circumcircle lies inside ABCD. The shortest side has length √(4 - t 2 ) and the longest side has length t, where √2 < t < 2. The tangents at A and B intersect at A', the tangents at B and C intersect at B', the tangents at C and D intersect at C', and the tangents at D and A intersect at D'. Find the largest and smallest possible values for area A'B'C'D'/area ABCD. Solution
: infinity, 2 are approached arbitrarily closely, but not achieved. Take ABCD to be a rectangle with diameter length 2, and longer side t. Then the two shorter sides have length √(4 - t 2 ) as required and √2 < t < 2. But as t approaches 2, the area of ABCD tends to zero and the area of A'B'C'D' must certainly exceed the area of the circle (which it contains), so the ratio A'B'C'D'/area ABCD tends to infinity. Suppose the circle ABCD has radius R. Then the area ABCD ≤ 2R 2 , and area A'B'C'D' ≥ 4R 2 , so the ratio is at least 2. But this can be approached arbitrarily closely by taking ABCD close to a square side √2. [To prove the statement about ABCD, suppose AB, BC, CD, DA subtend angles w, x, y z at the center, then it is easy to check that area ABCD = R 2 (sin 2w + sin 2x + sin 2y + sin 2z)/2 ≤ 4R 2 /2. To prove the statement about A'B'C'D', we note that area A'B'C'D' = R 2 (tan w + tan x + tan y + tan z). Now if x and y are unequal (and lie between 0 and 90 o , which we is given), then we will show that tan x + tan y is reduced by changing them to their arithmetic mean. It follows that (tan w + tan x + tan y + tan z) is minimised by taking w = x = y = z = 45 o , which gives the value 4. To prove the statement about tan x + tan y, put x = a + b, y = a - b and k = tan a, h = tan b. Then tan x + tan y = (tan a + tan b)/(1 - tan a tan b) + (tan a - tan b)/(1 + tan a tan b) = 2k(1 + h 2 )/(1 - k 2 h 2 ) ≥ 2k as required.] 16th Chinese 2001 © John Scholes jscholes@kalva.demon.co.uk 5 June 2002