### 52nd Putnam 1991

**Problem A1**

The rectangle with vertices (0, 0), (0, 3), (2, 0) and (2, 3) is rotated clockwise through a right angle about the point (2, 0), then about (5, 0), then about (7, 0), and finally about (10, 0). The net effect is to translate it a distance 10 along the x-axis. The point initially at (1, 1) traces out a curve. Find the area under this curve (in other words, the area of the region bounded by the curve, the x-axis and the lines parallel to the y-axis through (1, 0) and (11, 0) ).

**Solution**

Answer: 7π/2 + 6.

Trivial - in the great tradition of Putnam problems designed to stop the dunces scoring nil.

Draw a diagram. The required area is comprised of two quarter-disks radius √2, two quarter disks radius √5, four right-angle triangles sides 1, 1, √2, and four right-angle triangles sides 1, 2, √5. Hence area = π + 5/2 π + 2 + 4.

52nd Putnam 1991

© John Scholes

jscholes@kalva.demon.co.uk

21 Sep 1999