Sangaku Geometry Problem!

Label the diagram as follows, with $O$ at the origin:

Then $AB = 72 + 32 = 104$ , and $AG = 72 - 32 = 40$ , and by the Pythagorean Theorem on $\triangle AGB$ , $GB = 96$ .

Therefore, $A$ has coordinates $(72, 72)$ and $B$ has coordinates $(168, 32)$ .

Let the radius of the large circle be $r = DI = DJ = DK$ and let $y = OD$ . Then $DA = r - 72$ and $DB = r - 32$ , and by the distance formula on both $DA$ and $DB$ , $(168 - 0)^2 + (32 - y)^2 = (r - 32)^2$ and $(72 - 0)^2 + (72 - y)^2 = (r - 72)^2$ , which solves to $y = 63$ and $r = 225$ .

Also, $AC = c + 72$ , and $AE = 72$ , and by the Pythagorean Theorem on $\triangle ACE$ , $CE = \sqrt{(c + 72)^2 - 72^2}$ .

Finally, $DI = IC + CE + EO + OD$ , and substituting the above values gives $225 = c + \sqrt{(c + 72)^2 - 72^2} + 72 + 63$ , which solves to $c = \boxed{25}$ .