mathematics – Sangaku Triangle Puzzle – why I am getting the wrong answer?

Last night I was looking through Can You Solve My Problems by Alex Bellos and came across this teaser (problem 36):

Prove that the radius of the large circle (R_l) on the left is twice that of the smallest circle (R_s). (I sketched this up in Geogebra.) We also have a square at the bottom right and a right triangle enclosing everything.

However, I keep getting the incorrect answer that R_l = 6/(2+sqrt(2)) * R_s. Can someone help me see where the error in my thinking is?

MY THOUGHT PROCESS

First, with Pythagoras’s help and a bit of slightly ugly algebra, it is fairly straightforward to show that R_m (the radius of the mid-size circle) = 3/2 * R_s. (This follows from setting up this tringle and solving: )
That is correct – it says so in the back of the book.

Then, the trick becomes expressing R_l in terms of R_m. First, I noticed that the big triangle and the two smaller triangles bounded by the large one and the square are all similar. (Each smaller one has one of the same angles as the big one plus a 90 degree angle, meaning the third angle must also be the same). So this means that, with the following labels: , R_m/b = R_l/a. It’s fairly obvious that a=4R_m, so this becomes R_l = R_m*(4R_m)/b. To solve for b first we see this:, then, b = sqrt(2)*(1+sqrt(2))*R_m (the sqrt(2) in the image and in the formula for b both the result of Pythagoras again). Then, the formula for R_l reduces to R_l = R_m * 4 / (2+sqrt(2)). Substituting what we already know about R_s, this gives my final, though incorrect, answer: R_l = 6 / (2+sqrt(2)) * R_s.

My answer is not far from correct (6+(2+sqrt(2)) ~= 1.757), but it’s not right. Where did I go wrong?

P.S. (I’m new here, so I apologize if this is the wrong type of question!)