Grand Conclusion Coming Up! - Blog Posts

Sangaku Saturday #13

Last week, we uncovered this configuration which is also a solution to our "three circles in a triangle" problem, just not the one we were hoping for.

This is something that happens in all isosceles triangles. Draw the inscribed circle, with centre B, and the circle with centre C, tangent to the extended base (ON) and the side [SN] at the same point as the first circle is. Then it can be proved that the circle with centre A, whose diameter completes the height [SO] as our problem demands, is tangent to the circle with centre C.

But that's not what I'm going to concentrate on. Despite this plot twist, we are actually very close to getting what we want. What the above configuration means is that, returning to the initial scaled situation with SO = h = 1 and ON = b, we get

Knowing a solution to a degree 3 equation is extremely powerful, as we can factor the polynomial and leave a degree 2 equation, which has simple formulas for solutions. So, to finish off, can you:

1: prove that

2: solve the equation 2x²-(s-b)x-1 = 0, and deduce the general formulas for p, q and r that fit the configuration we are aiming for;

3: test the formulas for an equilateral triangle, in which s = 2b.

This last question is the one the sangaku tablet claims to solve.