In my previous post I presented a geometry problem whimsically called Chopsticks and Bowls. Two circles are contained in rectangle. The distance from the centre of the small circle to the opposite corner of the rectangle is equal to the sum of the diameters of the two circles. The problem is to develop an equation in its simplest form that, when solved, provides the radius of the large circle. The radius of the small circle is given as 1. I am pleased to report that Paul Turner, a retired Mathematics Teacher, has found the solution, shown in the image below.
3+8r-r^2-(4r)(r^.5)=0
Solving this equation is a difficult exercise best suited for a computer algebra system, such as Wolfram Alpha.
The labels in Paul’s solution refer to a diagram by Ed Staples, shown below.
