Question-227937

Question Number 227937 by mr W last updated on 03/Mar/26

Commented by fantastic2 last updated on 04/Mar/26

$${what}\:{a}\:{coincidenepce}!\:{I}\:{was}\:{about}\:{to}\:{give}\:{that} \\ $$$${question}.\:{taken}\:{from}\:{phantom}\:{of}\:{maths} \\ $$

Answered by TonyCWX last updated on 04/Mar/26

Let a = Longer side, b = Shorter side O = Center of original circle O′ = Center of folded segment circle P = Tangent Point r = ((a+b)/2) O′P = r = ((a+b)/2) O′B = (√((((a+b)/2))^2 +b^2 )) = (√((a^2 +2ab+5b^2 )/4)) = ((√(a^2 +2ab+5b^2 ))/2) O′O = O′B = ((√(a^2 +2ab+5b^2 ))/2) ((X/2))^2 +(((√(a^2 +2ab+5b^2 ))/4))^2 = (((a+b)/2))^2 (X^2 /4) = ((3a^2 +6ab−b^2 )/(16)) X^2 = ((3a^2 +6ab−b^2 )/4) X>0, determinant (((X = ((√(3a^2 +6ab−b^2 ))/2), a>b))) From the question, a = 3 & b = 1 ⇒ X = ((√(3(3^2 )+6(3)(1)−1^2 ))/2) = (√(11))

$$\mathrm{Let}\:{a}\:=\:\mathrm{Longer}\:\mathrm{side},\:{b}\:=\:\mathrm{Shorter}\:\mathrm{side} \\ $$$${O}\:=\:\mathrm{Center}\:\mathrm{of}\:\mathrm{original}\:\mathrm{circle} \\ $$$${O}'\:=\:\mathrm{Center}\:\mathrm{of}\:\mathrm{folded}\:\mathrm{segment}\:\mathrm{circle} \\ $$$${P}\:=\:\mathrm{Tangent}\:\mathrm{Point} \\ $$$$ \\ $$$${r}\:=\:\frac{{a}+{b}}{\mathrm{2}} \\ $$$${O}'{P}\:=\:{r}\:=\:\frac{{a}+{b}}{\mathrm{2}} \\ $$$${O}'{B}\:=\:\sqrt{\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:\sqrt{\frac{{a}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{5}{b}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{5}{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${O}'{O}\:=\:{O}'{B}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{5}{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$ \\ $$$$\left(\frac{{X}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{5}{b}^{\mathrm{2}} }}{\mathrm{4}}\right)^{\mathrm{2}} \:=\:\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\frac{{X}^{\mathrm{2}} }{\mathrm{4}}\:=\:\frac{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{6}{ab}−{b}^{\mathrm{2}} }{\mathrm{16}} \\ $$$${X}^{\mathrm{2}} \:=\:\frac{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{6}{ab}−{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${X}>\mathrm{0},\begin{array}{|c|}{{X}\:=\:\frac{\sqrt{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{6}{ab}−{b}^{\mathrm{2}} }}{\mathrm{2}},\:{a}>{b}}\\\hline\end{array} \\ $$$$ \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{question},\:{a}\:=\:\mathrm{3}\:\&\:{b}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:{X}\:=\:\frac{\sqrt{\mathrm{3}\left(\mathrm{3}^{\mathrm{2}} \right)+\mathrm{6}\left(\mathrm{3}\right)\left(\mathrm{1}\right)−\mathrm{1}^{\mathrm{2}} }}{\mathrm{2}}\:=\:\sqrt{\mathrm{11}} \\ $$

Commented by TonyCWX last updated on 04/Mar/26

Commented by mr W last updated on 04/Mar/26

$${great}! \\ $$

Answered by mr W last updated on 04/Mar/26

Commented by mr W last updated on 04/Mar/26

R=((3+1)/2)=2 AB=(√(1^2 +2^2 ))=(√5) ((X/2))^2 +(((AB)/2))^2 =R^2 X=2(√(2^2 −(((√5)/2))^2 ))=(√(11))

$${R}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{2}}=\mathrm{2} \\ $$$${AB}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$$\left(\frac{{X}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{AB}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${X}=\mathrm{2}\sqrt{\mathrm{2}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{11}} \\ $$

Commented by TonyCWX last updated on 04/Mar/26

$${Great}\:{work}\:{sir}! \\ $$

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