46th Putnam 1985 Problem A2 ABC is an acute-angled triangle with area 1. A rectangle R has its vertices R i on the sides of the triangle, R 1 and R 2 on BC, R 3 on AC and R 4 on AB. Another rectangle S has it vertices on the sides of the triangle AR 3 R 4 , two on R 3 R 4 and one one each of the other two sides. What is the maximum total area of R and S over all possible choices of triangle and rectangles? Solution
: 2/3. Let k = AR 3 /AC. The triangle AR 4 R 3 is similar to ABC with all dimensions a factor k smaller. So its area is k 2 . Similarly, the two triangles BR 4 R 1 and CR 3 R 2 together have area (1-k) 2 . So the area of the bottom rectangle is 1 - k 2 - (1-k) 2 = 2k(1-k). Defining h in a similar way, the area of the top rectangle is 2h(1-h) times the area of triangle AR 4 R 3 = 2h(1-h)k 2 . For any given choice of k, we maximise the area of the top rectangle by taking h = 1/2. The area of the two rectangles together is then 2k(1-k) + k 2 /2 = 2/3 - 3/2(k - 2/3) 2 . This evidently has maximum value 2/3, obtained by taking k = 2/3. Note that the maximum is independent of the choice of triangle. 46th Putnam 1985 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001