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<\/figure><\/div>\n\n\n\nThe problem is harder than the previous one, but it is still solvable by following the same steps.<\/p>\n\n\n\n
![\"\"](\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2019-11-23-at-4.49.21-PM.png\")
<\/figure><\/div>\n\n\n\nObviously, the radius of the semi-circle\u00a0is $\\dfrac{1}{2}$. Assume\u00a0radius of the quarter\u00a0circle is $q$. According to Pythagoras theorem, $$(\\overline{BF}+\\overline{EF})^2=\\overline{BK}^2+\\overline{EK}^2$$ We have $$(q+\\dfrac{1}{2})^2=(\\dfrac{1}{2})^2+1^2$$ The above equation can be solved as $$q=\\dfrac{\\pm\\sqrt{5}-1}{2}$$ Ignoring the negative $q$ value, we have $$q=\\dfrac{\\sqrt{5}-1}{2}$$ Let’s\u00a0denote the radius of the full circle as\u00a0$r$, and the length of line $\\overline{IL}$ and $\\overline{MK}$ as $t$.\u00a0Since $\\triangle{BIL}$ and $\\triangle{EIM}$ are right\u00a0triangles, we have $$(q+r)^2 = t^2 + (1-r)^2$$ $$(\\dfrac{1}{2}+r)^2=(1-t)^2+(\\dfrac{1}{2}-r)^2$$ The above equations can be solved as $$r=\\dfrac{3-2\\sqrt{2}}{3-\\sqrt{5}}, \\ \\ \\ t=\\dfrac{2+\\sqrt{2}-\\sqrt{10}}{3-\\sqrt{5}}$$ Simplifying the above values, we have the radius of the full circle as $$ r = \\dfrac{9-6\\sqrt{2}+3\\sqrt{5}-2\\sqrt{10}}{4} \\approx 0.2245918095$$ <\/p>\n\n\n\n
To be continued…<\/em><\/p>\n","protected":false},"excerpt":{"rendered":" Continue the topic of the previous post in this series, if we construct a semi-circle first, then a quarter circle, and finally a full circle, as the following, what the radius of the full circle? The problem is harder than the … Continue reading