Question-213323

Question Number 213323 by Spillover last updated on 02/Nov/24

Commented by Frix last updated on 03/Nov/24

No nice numbers Let r=1 ⇒ ρ_(q.c.) ≈4.87164313433 [ _(polynome of 8^(th) degree )^(Solution of a) ] ρ_(s.c.) ≈3.49740513991 R≈1.66629777247

$$\mathrm{No}\:\mathrm{nice}\:\mathrm{numbers} \\ $$$$\mathrm{Let}\:{r}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\rho_{{q}.{c}.} \approx\mathrm{4}.\mathrm{87164313433}\:\left[\:_{\mathrm{polynome}\:\mathrm{of}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{degree}\:} ^{\mathrm{Solution}\:\mathrm{of}\:\mathrm{a}} \right] \\ $$$$\rho_{{s}.{c}.} \approx\mathrm{3}.\mathrm{49740513991} \\ $$$${R}\approx\mathrm{1}.\mathrm{66629777247} \\ $$

Answered by mr W last updated on 03/Nov/24

Commented by mr W last updated on 03/Nov/24

a=radius of quarter circle b=radius of semicircle (a−r)^2 −r^2 =(b+r)^2 −(b−r)^2 ⇒ a^2 −2ar=4br ...(i) (√((a−R)^2 −R^2 ))+(√((b−R)^2 −R^2 ))=b ⇒ (√(a^2 −2aR))+(√(b^2 −2bR))=b ...(ii) (√((a+r)^2 −r^2 ))=b+(√((b−r)^2 −r^2 )) ⇒ (√(a^2 +2ar))=b+(√(b^2 −2br)) ...(iii) let α=(a/r), β=(b/r), λ=(R/r) α^2 −2α=4β ...(i) (√(α^2 −2αλ))+(√(β^2 −2βλ))=β ...(ii) (√(α^2 +2α))=β+(√(β^2 −2β)) ...(iii) ⇒λ≈1.666297772

$${a}={radius}\:{of}\:{quarter}\:{circle} \\ $$$${b}={radius}\:{of}\:{semicircle} \\ $$$$\left({a}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\left({b}+{r}\right)^{\mathrm{2}} −\left({b}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} −\mathrm{2}{ar}=\mathrm{4}{br}\:\:\:\:…\left({i}\right) \\ $$$$\sqrt{\left({a}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }+\sqrt{\left({b}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }={b} \\ $$$$\Rightarrow\:\sqrt{{a}^{\mathrm{2}} −\mathrm{2}{aR}}+\sqrt{{b}^{\mathrm{2}} −\mathrm{2}{bR}}={b}\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\sqrt{\left({a}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }={b}+\sqrt{\left({b}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{ar}}={b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{2}{br}}\:\:\:\:\:…\left({iii}\right) \\ $$$${let}\:\alpha=\frac{{a}}{{r}},\:\beta=\frac{{b}}{{r}},\:\lambda=\frac{{R}}{{r}} \\ $$$$\alpha^{\mathrm{2}} −\mathrm{2}\alpha=\mathrm{4}\beta\:\:\:\:…\left({i}\right) \\ $$$$\:\sqrt{\alpha^{\mathrm{2}} −\mathrm{2}\alpha\lambda}+\sqrt{\beta^{\mathrm{2}} −\mathrm{2}\beta\lambda}=\beta\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\sqrt{\alpha^{\mathrm{2}} +\mathrm{2}\alpha}=\beta+\sqrt{\beta^{\mathrm{2}} −\mathrm{2}\beta}\:\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\lambda\approx\mathrm{1}.\mathrm{666297772} \\ $$

Commented by Spillover last updated on 03/Nov/24

$${perfect} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply