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<\/figure><\/div>\n\n\n\nLet’s connect several lines as the following, with line $\\overline{GI}$ and $\\overline{JK}$ perpendicular to each other and intersecting at point $L$, and point $H$ as the center of the full circle:<\/p>\n\n\n\n
![\"\"](\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2019-11-23-at-4.22.20-PM.png\")
<\/figure><\/div>\n\n\n\nTherefore, both $\\triangle{GLH}$ and $\\triangle{CJH}$ are right triangles. \nAccording to Pythagoras theorem, we have:\n$$\\overline{GL}^2+\\overline{HL}^2=\\overline{GH}^2,\\ \\ \\\n\\overline{CJ}^2+\\overline{HJ}^2=\\overline{CH}^2$$\nBased on the result of the previous post in this series, the radius of the semi-circle is $\\dfrac{1}{4}$. Additionally, since point $C$ is the center of the quarter circle and point $G$ is the center of the semi-circle, both of them are tangent with the full circle centered at point $H$, by assuming the radius of the full circle is $r$ and $\\overline{GL} = t$, we have:\n$$\\overline{GH}=\\dfrac{1}{4}+r,\\ \\ \\ \\overline{HL}=\\dfrac{1}{4}-r$$\n$$\\overline{CH}=1+r, \\ \\ \\ \\overline{CJ}=1-t. \\ \\ \\ \\overline{HJ}=1-r$$\nTherefore, we have the following equations:\n$$t^2+(\\dfrac{1}{4})^2=(\\dfrac{1}{4}+r)^2$$\n$$(1-t)^2+(1-r)^2=(1+r)^2$$\nSimplify the above equations, we have: $$t^2=r, \\ \\ \\ (1-t)^2=4r$$\nWe have two solutions $$t=\\dfrac{1}{3}, \\ \\ \\ r=\\dfrac{1}{9}$$\n$$t=\u22121, \\ \\ \\ r=1$$\nBy ignoring invalid answer $r=1$, we have the answer $r=\\dfrac{1}{9}$.\n<\/p>\n\n\n\n
To be continued…<\/em><\/p>\n","protected":false},"excerpt":{"rendered":" Continue the topic in the previous post of this series, we add another circle inscribed in the area bounded by one side of the unit squares, the simi-circle and the quarter-circle, as the following. What is the radius of the … Continue reading