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Question-70452

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Question Number 70452 by ajfour last updated on 04/Oct/19
Commented by ajfour last updated on 04/Oct/19
ABCD is a square. Find minimum side length s of equilateral △DEF in terms of quarter circle radii a and b. Also find range of (a/b) so that such an equilateral triangle be certainly possible.
$${ABCD}\:{is}\:{a}\:{square}.\:{Find}\:{minimum} \\ $$$${side}\:{length}\:\boldsymbol{{s}}\:{of}\:{equilateral}\:\bigtriangleup{DEF} \\ $$$${in}\:{terms}\:{of}\:{quarter}\:{circle}\:{radii} \\ $$$${a}\:{and}\:{b}.\:{Also}\:{find}\:{range}\:{of}\: \\ $$$$\:\frac{{a}}{{b}}\:{so}\:{that}\:{such}\:{an}\:{equilateral} \\ $$$${triangle}\:{be}\:{certainly}\:{possible}. \\ $$
Commented by MJS last updated on 05/Oct/19
I tried with coordinate method A= ((0),(1) ) B= ((0),(0) ) C= ((1),(0) ) D= ((1),(1) ) E= (((acos α)),((1+asin α)) ) F= ((((1−a)cos β)),(((1−a)sin β)) ) ∣DE∣=∣EF∣=∣FD∣ easily leads to a=((1−(sin β +cos β))/(1−(cos α +sin β +cos β))) then I put α=2arctan t β=2arctan u which leads to t^4 −((8u(u−1)^2 (3u−1))/((3u^2 −1)(5u^2 −4u+1)))t^3 −((2(7u^4 −4u^3 −2u^2 +4u−1))/((3u^2 −1)(5u^2 −4u+1)))t^2 −((8u(u−1)^2 (u+1))/((3u^2 −1)(5u^2 −4u+1)))t−(((u^2 +1)(u^2 −4u+1))/((3u^2 −1)(5u^2 −4u+1)))=0 and we can solve this! t_1 =(((3+2(√3))u^2 −2(1+(√3))u+1)/(3u^2 −1)) t_2 =(((3−3(√2))u^2 −2(1−(√3))u+1)/(3u^2 −1)) t_(3, 4) =−((u^2 +2u−1)/(5u^2 −4u+1))±((2u(u−1))/(5u^2 −4u+1))i I′m running out of time, please try to continue
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{with}\:\mathrm{coordinate}\:\mathrm{method} \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${E}=\begin{pmatrix}{{a}\mathrm{cos}\:\alpha}\\{\mathrm{1}+{a}\mathrm{sin}\:\alpha}\end{pmatrix}\:\:{F}=\begin{pmatrix}{\left(\mathrm{1}−{a}\right)\mathrm{cos}\:\beta}\\{\left(\mathrm{1}−{a}\right)\mathrm{sin}\:\beta}\end{pmatrix} \\ $$$$\mid{DE}\mid=\mid{EF}\mid=\mid{FD}\mid \\ $$$$\mathrm{easily}\:\mathrm{leads}\:\mathrm{to} \\ $$$${a}=\frac{\mathrm{1}−\left(\mathrm{sin}\:\beta\:+\mathrm{cos}\:\beta\right)}{\mathrm{1}−\left(\mathrm{cos}\:\alpha\:+\mathrm{sin}\:\beta\:+\mathrm{cos}\:\beta\right)} \\ $$$$\mathrm{then}\:\mathrm{I}\:\mathrm{put} \\ $$$$\alpha=\mathrm{2arctan}\:{t} \\ $$$$\beta=\mathrm{2arctan}\:{u} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$${t}^{\mathrm{4}} −\frac{\mathrm{8}{u}\left({u}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{3}{u}−\mathrm{1}\right)}{\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{5}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}\right)}{t}^{\mathrm{3}} −\frac{\mathrm{2}\left(\mathrm{7}{u}^{\mathrm{4}} −\mathrm{4}{u}^{\mathrm{3}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{4}{u}−\mathrm{1}\right)}{\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{5}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}\right)}{t}^{\mathrm{2}} −\frac{\mathrm{8}{u}\left({u}−\mathrm{1}\right)^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{5}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}\right)}{t}−\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}\right)}{\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{5}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}\right)}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}! \\ $$$${t}_{\mathrm{1}} =\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\right){u}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){u}+\mathrm{1}}{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$${t}_{\mathrm{2}} =\frac{\left(\mathrm{3}−\mathrm{3}\sqrt{\mathrm{2}}\right){u}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){u}+\mathrm{1}}{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$${t}_{\mathrm{3},\:\mathrm{4}} =−\frac{{u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{1}}{\mathrm{5}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}}\pm\frac{\mathrm{2}{u}\left({u}−\mathrm{1}\right)}{\mathrm{5}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}}\mathrm{i} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{running}\:\mathrm{out}\:\mathrm{of}\:\mathrm{time},\:\mathrm{please}\:\mathrm{try}\:\mathrm{to}\:\mathrm{continue} \\ $$
Commented by mr W last updated on 05/Oct/19
Commented by mr W last updated on 05/Oct/19
a geometrical consideration: the green circle can intersect with the smaller quarter−circle at − no point ⇒ no solution − one point ⇒ one solution − two points ⇒ two solutions
$${a}\:{geometrical}\:{consideration}: \\ $$$${the}\:{green}\:{circle}\:{can}\:{intersect}\:{with}\:{the} \\ $$$${smaller}\:{quarter}−{circle}\:{at} \\ $$$$−\:{no}\:{point}\:\Rightarrow\:{no}\:{solution} \\ $$$$−\:{one}\:{point}\:\Rightarrow\:{one}\:{solution} \\ $$$$−\:{two}\:{points}\:\Rightarrow\:{two}\:{solutions} \\ $$
Answered by mr W last updated on 10/Oct/19
Answered by ajfour last updated on 04/Oct/19

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