Not 1803 Sangaku Problem

Let $y$ be the radius of the yellow circle and $b$ be the side of the square, and label the diagram (not drawn to scale) as follows:

Then $OB = 20$ as the radius of the big circle, $AB = b$ as the side of the square, and $OA = VD + DA - VO = 2y + b - 20$ . By the Pythagorean Theorem on $\triangle OAB$ , $(2y + b - 20)^2 + b^2 = 400$ , which rearranges to $b = \sqrt{-y^2 + 20y + 100} - y + 10$ .

Also, $PQ = PR + RQ = y + 5$ , $OQ = OU - QU = 20 - 5 = 15$ , and $PO = VO - VP = 20 - y$ . By the law of cosines on $\triangle QPO$ , $\cos \angle QPO = \frac{(20 - y)^2 + (y + 5)^2 - 15^2}{(20 - y)(y + 5)}$ . Letting $k = PT$ and $h = QT$ , on $\triangle QPT$ , $\cos \angle QPO = \frac{k}{y + 5}$ and $k^2 + h^2 = (y + 5)^2$ . These three equations simplify to $k = \frac{y^2 - 15y + 100}{20 - y}$ and $h = \frac{20\sqrt{-y^2 + 15y}}{20 - y}$ .

Also, $QC = 5$ as the radius of the green circle, $QS = QT - ST = h - b$ , and $SC = TD = PD - PT = y - k$ . By the Pythagorean Theorem on $\triangle QSC$ , $(y - k)^2 + (h - b)^2 = 25$ , which after substituting the above values, $(y - \frac{y^2 - 15y + 100}{20 - y})^2 + (\frac{20\sqrt{-y^2 + 15y}}{20 - y} - (\sqrt{-y^2 + 20y + 100} - y + 10))^2 = 25$ .

If $p = \sqrt{-y^2 + 15y}$ and $q = \sqrt{-y^2 + 20y + 100}$ , the above equation becomes $(y - \frac{y^2 - 15y + 100}{20 - y})^2 + (\frac{20p}{20 - y} - (q - y + 10))^2 = 25$ . Solving this for $q$ and simplifying gives $q = \frac{-20p \pm 2p(y-10)}{y - 20} + y - 10$ , which means either $q = 2p + y - 10$ or $q = \frac{-2py}{y - 20} + y - 10$ . If $q = 2p + y - 10$ , then $q^2 = 4p^2 + (4y - 40)p + y^2 - 20y + 100$ , or $-y^2 + 20y + 100 = 4(-y^2 + 15y) + (4y - 40)p + y^2 - 20y + 100$ . Solving this for $p$ and simplifying gives $p = \frac{1}{2}y$ , which means $p^2 = \frac{1}{4}y^2$ , or $-y^2 + 15y = \frac{1}{4}y^2$ , which simplifies to $y(y - 12) = 0$ , so that $y = 12$ for $y > 0$ . If $q = \frac{-2py}{y - 20} + y - 10$ , the same steps above will lead to $y \approx 14.857$ , which would make the green circle overlap the square. Therefore, $y = 12$ .

If $y = 12$ , then $b = \sqrt{-y^2 + 20y + 100} - y + 10 = \sqrt{-12^2 + 20 \cdot 12 + 100} - 12 + 10 = 12$ , and the area of the square is $b^2 = 12^2 = \boxed{144}$ .

The only way i found to solve to problem is to assume that A (the center of the green circle) is on the perpendiculare bisector of the diameter of the biggest circle.
Then we use the pythagorean theorem to calculate AI, since we know that AJ=15 and JI=20 and we find that AI=25:

• Next, AJ is parallel do EC, then the triangles AJI and EIC are similar
• We denote FG as R and EC as c
• we write : $\frac{AJ}{EC}$ = $\frac{IJ}{IC}$ <=> $\frac{15}{c}$ = $\frac{20}{40-2R}$
• This gives us this equation : R= $\frac{60-2c}{3}$
• The product of chords in a circle gives us this second equation : c²=(2R+c)(40-2R-c)
• We use substitution de solve the system and find c=12 -The area of the square is c²=12²=144

I solved it by drawing it on graph paper on iPad using ipencil with all the conditions of the problem.