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Question-214297




Question Number 214297 by ajfour last updated on 04/Dec/24
Commented by ajfour last updated on 04/Dec/24
the tangents are at right angles to  one another and corner maynot be  at semicircle centre.
$${the}\:{tangents}\:{are}\:{at}\:{right}\:{angles}\:{to} \\ $$$${one}\:{another}\:{and}\:{corner}\:{maynot}\:{be} \\ $$$${at}\:{semicircle}\:{centre}. \\ $$
Commented by mr W last updated on 05/Dec/24
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Commented by ajfour last updated on 05/Dec/24
https://youtu.be/ys6xN29xNUM?si=9rk6nXD0GcmHTp9Z
Answered by mr W last updated on 04/Dec/24
Commented by mr W last updated on 05/Dec/24
a=1  R=3  b=r  (b/(tan (β/2)))=x+(√((R−b)^2 −b^2 ))=x+(√(R^2 −2Rb))  ⇒tan (β/2)=(b/(x+(√(R^2 −2Rb))))  similarly  ⇒tan (α/2)=(a/(−x+(√(R^2 −2Ra))))  (β/2)=(π/4)−(α/2)  tan (β/2)=((1−tan (a/2))/(1+tan (a/2)))  (b/(x+(√(R^2 −2Rb))))=((1−(a/(−x+(√(R^2 −2Ra)))))/(1+(a/(−x+(√(R^2 −2Ra))))))  (b/(x+(√(R^2 −2Rb))))=((−x+(√(R^2 −2Ra))−a)/(−x+(√(R^2 −2Ra))+a))  let α=(a/R), β=(b/R), ξ=(x/R)  (β/(ξ+(√(1−2β))))=((−ξ+(√(1−2α))−α)/(−ξ+(√(1−2α))+α))  let λ=(√(1−2β)), p=(√(1−2α))+α, q=(√(1−2α))−α  ((1−λ^2 )/(2(ξ+λ)))=((q−ξ)/(p−ξ))  (p−ξ)λ^2 +2(q−ξ)λ+(2q+1)ξ−[2ξ^2 −(2q+1)ξ+p]=0  λ=((ξ−q+(√((ξ−q)^2 +(p−ξ)[2ξ^2 −(2q+1)ξ+p])))/(p−ξ))  b={1−{((ξ−q+(√((ξ−q)^2 +(p−ξ)[2ξ^2 −(2q+1)ξ+p])))/(p−ξ))}^2 }(R/2)  b_(max) ≈0.6659 at ξ=(x/R)≈−0.1399  =================
$${a}=\mathrm{1} \\ $$$${R}=\mathrm{3} \\ $$$${b}={r} \\ $$$$\frac{{b}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}={x}+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }={x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}} \\ $$$$\frac{\beta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{tan}\:\frac{{a}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{{a}}{\mathrm{2}}} \\ $$$$\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}=\frac{\mathrm{1}−\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}}}{\mathrm{1}+\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}}} \\ $$$$\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}=\frac{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}−{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}+{a}} \\ $$$${let}\:\alpha=\frac{{a}}{{R}},\:\beta=\frac{{b}}{{R}},\:\xi=\frac{{x}}{{R}} \\ $$$$\frac{\beta}{\xi+\sqrt{\mathrm{1}−\mathrm{2}\beta}}=\frac{−\xi+\sqrt{\mathrm{1}−\mathrm{2}\alpha}−\alpha}{−\xi+\sqrt{\mathrm{1}−\mathrm{2}\alpha}+\alpha} \\ $$$${let}\:\lambda=\sqrt{\mathrm{1}−\mathrm{2}\beta},\:{p}=\sqrt{\mathrm{1}−\mathrm{2}\alpha}+\alpha,\:{q}=\sqrt{\mathrm{1}−\mathrm{2}\alpha}−\alpha \\ $$$$\frac{\mathrm{1}−\lambda^{\mathrm{2}} }{\mathrm{2}\left(\xi+\lambda\right)}=\frac{{q}−\xi}{{p}−\xi} \\ $$$$\left({p}−\xi\right)\lambda^{\mathrm{2}} +\mathrm{2}\left({q}−\xi\right)\lambda+\left(\mathrm{2}{q}+\mathrm{1}\right)\xi−\left[\mathrm{2}\xi^{\mathrm{2}} −\left(\mathrm{2}{q}+\mathrm{1}\right)\xi+{p}\right]=\mathrm{0} \\ $$$$\lambda=\frac{\xi−{q}+\sqrt{\left(\xi−{q}\right)^{\mathrm{2}} +\left({p}−\xi\right)\left[\mathrm{2}\xi^{\mathrm{2}} −\left(\mathrm{2}{q}+\mathrm{1}\right)\xi+{p}\right]}}{{p}−\xi} \\ $$$${b}=\left\{\mathrm{1}−\left\{\frac{\xi−{q}+\sqrt{\left(\xi−{q}\right)^{\mathrm{2}} +\left({p}−\xi\right)\left[\mathrm{2}\xi^{\mathrm{2}} −\left(\mathrm{2}{q}+\mathrm{1}\right)\xi+{p}\right]}}{{p}−\xi}\right\}^{\mathrm{2}} \right\}\frac{{R}}{\mathrm{2}} \\ $$$${b}_{{max}} \approx\mathrm{0}.\mathrm{6659}\:{at}\:\xi=\frac{{x}}{{R}}\approx−\mathrm{0}.\mathrm{1399} \\ $$$$================= \\ $$
Commented by mr W last updated on 05/Dec/24
Commented by mr W last updated on 05/Dec/24

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