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Question-222636

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Question Number 222636 by Mingma last updated on 02/Jul/25
Answered by mr W last updated on 04/Jul/25
d=(√(a^2 −(a−x)^2 ))+x+(√(b^2 −(b−x)^2 )) =(√(2ax−x^2 ))+x+(√(2bx−x^2 )) d=(√((a+b)^2 −(b−a)^2 ))=2(√(ab)) (√(2ax−x^2 ))+x+(√(2bx−x^2 ))=2(√(ab)) (√(2ax−x^2 ))+(√(2bx−x^2 ))=2(√(ab))−x 2(√((2ax−x^2 )(2bx−x^2 )))=3x^2 −2((√a)+(√b))^2 x+4ab 4x^4 −8(a+b)x^3 +16abx^2 =9x^4 +4((√a)+(√b))^4 x^2 +16a^2 b^2 −12((√a)+(√b))^2 x^3 +24abx^2 −16ab((√a)+(√b))^2 x 5x^4 −4(a+b+6(√(ab)))x^3 +4(a^2 +b^2 +8ab+4(a+b)(√(ab)))x^2 −16ab(a+b+2(√(ab)))x+16a^2 b^2 =0 ....
$${d}=\sqrt{{a}^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }+{x}+\sqrt{{b}^{\mathrm{2}} −\left({b}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\:=\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}} } \\ $$$${d}=\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{ab}} \\ $$$$\:\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{{ab}} \\ $$$$\:\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{{ab}}−{x} \\ $$$$\:\mathrm{2}\sqrt{\left(\mathrm{2}{ax}−{x}^{\mathrm{2}} \right)\left(\mathrm{2}{bx}−{x}^{\mathrm{2}} \right)}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} {x}+\mathrm{4}{ab} \\ $$$$\:\mathrm{4}{x}^{\mathrm{4}} −\mathrm{8}\left({a}+{b}\right){x}^{\mathrm{3}} +\mathrm{16}{abx}^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{4}} +\mathrm{4}\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{4}} {x}^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{12}\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{24}{abx}^{\mathrm{2}} −\mathrm{16}{ab}\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} {x} \\ $$$$\mathrm{5}{x}^{\mathrm{4}} −\mathrm{4}\left({a}+{b}+\mathrm{6}\sqrt{{ab}}\right){x}^{\mathrm{3}} +\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{8}{ab}+\mathrm{4}\left({a}+{b}\right)\sqrt{{ab}}\right){x}^{\mathrm{2}} −\mathrm{16}{ab}\left({a}+{b}+\mathrm{2}\sqrt{{ab}}\right){x}+\mathrm{16}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$…. \\ $$
Commented by Tawa11 last updated on 04/Jul/25
Sir, please check, question 222646. The part (c)
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{check},\:\mathrm{question}\:\mathrm{222646}. \\ $$$$\mathrm{The}\:\mathrm{part}\:\left(\mathrm{c}\right) \\ $$
Answered by fantastic last updated on 03/Jul/25
Commented by fantastic last updated on 03/Jul/25
AF=a−x AG=a FG=(√(a^2 −(a−x)^2 ))=(√(a^2 −a^2 −x^2 +2ax))=(√(2ax−x^2 )) BD=b−x BE=b DE=(√(b^2 −(b−x)^2 ))=(√(2bx−x^2 )) ∴FD=(√(2ax−x^2 ))+x+(√(2bx−x^2 )) AB=a+b BC=b−a ∴AC=(√((b+a)^2 −(b−a)^2 ))=(√(4ab))=2(√(ab)) FD=AC ∴(√(2ax−x^2 ))+x+(√(2bx−x^2 ))=2(√(ab)) ⇒(√(2×((√2)/4)×x−x^2 ))+x+(√(2×(1/2)x−x^2 ))=2(√(((√2)/4)×(1/2))) (√((x/( (√2)))−x^2 ))+x+(√(x−x^2 ))=(1/( (√(√2)))) This equation will take alot effort Im approximating this x≈0.167 235 19 So x≈0.167 235 19
$${AF}={a}−{x} \\ $$$${AG}={a} \\ $$$${FG}=\sqrt{{a}^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} −{a}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{2}{ax}}=\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} } \\ $$$${BD}={b}−{x} \\ $$$${BE}={b} \\ $$$${DE}=\sqrt{{b}^{\mathrm{2}} −\left({b}−{x}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}} } \\ $$$$\therefore{FD}=\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}} } \\ $$$${AB}={a}+{b} \\ $$$${BC}={b}−{a} \\ $$$$\therefore{AC}=\sqrt{\left({b}+{a}\right)^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}{ab}}=\mathrm{2}\sqrt{{ab}} \\ $$$${FD}={AC} \\ $$$$\therefore\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{{ab}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}×{x}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{x}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\sqrt{\frac{{x}}{\:\sqrt{\mathrm{2}}}−{x}^{\mathrm{2}} }+{x}+\sqrt{{x}−{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}} \\ $$$${This}\:{equation}\:{will}\:{take}\:{alot}\:{effort} \\ $$$${Im}\:{approximating}\:{this} \\ $$$${x}\approx\mathrm{0}.\mathrm{167}\:\mathrm{235}\:\mathrm{19} \\ $$$${So}\:{x}\approx\mathrm{0}.\mathrm{167}\:\mathrm{235}\:\mathrm{19} \\ $$
Answered by Ghisom last updated on 03/Jul/25
C_1 : (x+p)^2 +(y−a)^2 =a^2 C_2 : (x−p)^2 +(y−b)^2 =b^2 C_1 ∩C_2 must have exactly 1 solution ⇒ p=(√(ab)) relevant half circles: C_1 : x=−(√(ab))+(√(2ay−y^2 )) C_2 : x=(√(ab))−(√(2by−y^2 )) wr need to find y so that ((√(ab))−(√(2by−y^2 )))−(−(√(ab))+(√(2ay−y^2 )))=y 2(√(ab))=(√(2ay−y^2 ))+(√(2by−y^2 )) this leads to y^4 −((4(a+b+6(√(ab))))/5)y^3 +((4(a^2 +8ab+b^2 +4(a+b)(√(ab))))/5)y^2 −((16ab(a+b+2(√(ab))))/5)y+((16a^2 b^2 )/5)=0 with a=((√2)/4)∧b=(1/2) an exact solution only is possible if we get a “nice” exact value for t from: t^3 +((8×2^(3/4) −60×2^(1/2) +40×2^(1/4) +15)/(300))+((284×2^(3/4) −1174×2^(1/2) +164×2^(1/4) +909)/(13500))=0 which I don′t think we can... ⇒ y≈0.167235190955 [= side length of square]
$${C}_{\mathrm{1}} :\:\left({x}+{p}\right)^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} :\:\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${C}_{\mathrm{1}} \cap{C}_{\mathrm{2}} \:\mathrm{must}\:\mathrm{have}\:\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution} \\ $$$$\Rightarrow\:{p}=\sqrt{{ab}} \\ $$$$\mathrm{relevant}\:\mathrm{half}\:\mathrm{circles}: \\ $$$${C}_{\mathrm{1}} :\:{x}=−\sqrt{{ab}}+\sqrt{\mathrm{2}{ay}−{y}^{\mathrm{2}} } \\ $$$${C}_{\mathrm{2}} :\:{x}=\sqrt{{ab}}−\sqrt{\mathrm{2}{by}−{y}^{\mathrm{2}} } \\ $$$$\mathrm{wr}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:{y}\:\mathrm{so}\:\mathrm{that} \\ $$$$\left(\sqrt{{ab}}−\sqrt{\mathrm{2}{by}−{y}^{\mathrm{2}} }\right)−\left(−\sqrt{{ab}}+\sqrt{\mathrm{2}{ay}−{y}^{\mathrm{2}} }\right)={y} \\ $$$$\mathrm{2}\sqrt{{ab}}=\sqrt{\mathrm{2}{ay}−{y}^{\mathrm{2}} }+\sqrt{\mathrm{2}{by}−{y}^{\mathrm{2}} } \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${y}^{\mathrm{4}} −\frac{\mathrm{4}\left({a}+{b}+\mathrm{6}\sqrt{{ab}}\right)}{\mathrm{5}}{y}^{\mathrm{3}} +\frac{\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{8}{ab}+{b}^{\mathrm{2}} +\mathrm{4}\left({a}+{b}\right)\sqrt{{ab}}\right)}{\mathrm{5}}{y}^{\mathrm{2}} −\frac{\mathrm{16}{ab}\left({a}+{b}+\mathrm{2}\sqrt{{ab}}\right)}{\mathrm{5}}{y}+\frac{\mathrm{16}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{0} \\ $$$$\mathrm{with}\:{a}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\wedge{b}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{only} \\ $$$$\mathrm{is}\:\mathrm{possible}\:\mathrm{if}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:“\mathrm{nice}''\:\mathrm{exact}\:\mathrm{value} \\ $$$$\mathrm{for}\:{t}\:\mathrm{from}: \\ $$$${t}^{\mathrm{3}} +\frac{\mathrm{8}×\mathrm{2}^{\mathrm{3}/\mathrm{4}} −\mathrm{60}×\mathrm{2}^{\mathrm{1}/\mathrm{2}} +\mathrm{40}×\mathrm{2}^{\mathrm{1}/\mathrm{4}} +\mathrm{15}}{\mathrm{300}}+\frac{\mathrm{284}×\mathrm{2}^{\mathrm{3}/\mathrm{4}} −\mathrm{1174}×\mathrm{2}^{\mathrm{1}/\mathrm{2}} +\mathrm{164}×\mathrm{2}^{\mathrm{1}/\mathrm{4}} +\mathrm{909}}{\mathrm{13500}}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}… \\ $$$$ \\ $$$$\Rightarrow\:{y}\approx\mathrm{0}.\mathrm{167235190955}\:\left[=\:\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{square}\right] \\ $$
Commented by Tawa11 last updated on 14/Jul/25
Sir, Please do transform of a^(1/5) = b^(1/5) + c^(1/5)
$$\mathrm{Sir}, \\ $$$$\mathrm{Please}\:\mathrm{do}\:\mathrm{transform}\:\mathrm{of}\:\:\:\:\mathrm{a}^{\mathrm{1}/\mathrm{5}} \:\:=\:\:\mathrm{b}^{\mathrm{1}/\mathrm{5}} \:\:+\:\:\mathrm{c}^{\mathrm{1}/\mathrm{5}} \\ $$

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