Ptolemy's Riddle IV

Let the center of the circle be $O$ , its radius $r=62.5$ , $\angle AOB = \theta$ and $AB=BE=x$ . Since $\triangle ABO$ and $\triangle ABE$ are similar, $\angle ABE = \angle AOB = \theta$ . Because $ABCD$ is a cyclic quadrilateral the diagonal angles $\angle ABD=\angle ACD=\theta$ . Therefore $\triangle AOB$ , $\triangle ABE$ , and $\triangle CDE$ are similar isosceles triangles.

Then $\dfrac {AE}{AB} = \dfrac {AB}{AO}$ $\implies AE = \dfrac {x^2}r$ . And $\dfrac {ED}{EC} = \dfrac {ED}{2r-AE} = \dfrac xr$ $\implies ED = 2x - \dfrac {x^3}{r^2}$ . Since $BD=BE+ED = 3x - \dfrac {x^3}{r^2} = 117$ $\implies 4x^3 -46875x + 1828125 = 0$ . Solving the equation, we get $x=AB = 75$ . $CD = \dfrac rx \times DE = \dfrac {62.5}{75} (117-75) = 35$ . $BC =\sqrt{125^2-75^2} = 100$ , and $DA =\sqrt{125^2-35^2} = 120$ .

Therefore, the perimeter of $ABCD$ is $AB+BC+CD+DA = 75+100+35+120 = \boxed{330}$ .

Let,

(1) AB=BE=x

(2) EC=CD=y

(3) AE=z

Write using given problem, segment theorem, and solve the following Diophantine equations using WolframAlpha

z+y=125, x+(z*y/x)=117 , get

x=75, y=35 and z=90.

The perimeter is

75+ 35+ √{(125)^2-(75)^2} + √{(125)^2-(35)^2

= 75 + 35 + 100 + 120 = 330

Answer=330