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Question-212822

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Question Number 212822 by ajfour last updated on 24/Oct/24
Commented by ajfour last updated on 24/Oct/24
Consider triangle isosceles about upper vertex.
$${Consider}\:{triangle}\:{isosceles}\:{about} \\ $$$${upper}\:{vertex}. \\ $$
Commented by Ghisom last updated on 26/Oct/24
an approach from the far side: (sorry I changed a⇄b) △_(ppq) ⇒ r=((q(√(2p−q)))/(2(√(2p+q)))) let q=1 ⇒ r=((√(2p−1))/( 2(√(2p+1)))) the vertices of the △ are P, Q, R P= ((((√(2p−1))/( 2(√(2p+1))))),(((√(2p−1))/( 2(√(2p+1))))) ) Q= (((((√(6p+5))+(√(2p−1)))/(2(√(2p+1))))),(0) ) R= ((((p/2)−(1/4)+(((√(6p+5))+(√(2p−1)))/( 4(√(2p+1)))))),((((√((6p+5)(2p−1)))/4)+((√(2p−1))/( 4(√(2p+1)))))) ) C_I = ((((1/4)−(1/(2(2p+1)))+(((√(6p+5))+2(√(2p−1)))/(4(√(2p+1)))))),((((√((6p+5)(2p−1)))/(4(2p+1)))+((√(2p−1))/( 4(√(2p+1)))))) ) ⇒ a=(((√(6p+5))+2(√(2p−1)))/(2(√(2p+1)))) b=((√((6p+5)(2p−1)))/4)+((√(2p−1))/( 4(√(2p+1)))) ∣PC_I ∣=((√p)/( (√(2p+1)))) if we want to match the picture ⇒ a≥b ⇒ (1/2)<p≤1.66853858571 { ((r=((√(2p−1))/( 2(√(2p+1)))))),((a=(((√(6p+5))+2(√(2p−1)))/(2(√(2p+1)))))),((b=((√((6p+5)(2p−1)))/4)+((√(2p−1))/( 4(√(2p+1)))))) :} we can find p for any ratio a:b further we can easily get { ((r=((√(2p−1))/( 2(√(2p+1)))))),((a=2r+(√(1−r^2 )))),((b=((1/2)+((2(√(1−r^2 )))/(1−4r^2 )))r)) :} (2) ⇔ (√(1−r^2 ))=a−2r inserting in (3) & transforming gives r^3 −2(b−1)r^2 −(a+(1/4))r+(b/2)=0 ⇒ r=((2b)/3)−(1/3)+((2(√(−3u)))/3)sin ((arcsin ((3(√3)v)/(2(√(−u^3 )))))/3) where u=−((4b^2 )/3)−a+((8b)/3)−((19)/(12)) v=−((16b^3 )/(27))+((16b^2 )/9)−((2ab)/3)+((2a)/3)−((13b)/9)+((41)/(54)) but a, b are not independent in this approach!
$$\mathrm{an}\:\mathrm{approach}\:\mathrm{from}\:\mathrm{the}\:\mathrm{far}\:\mathrm{side}: \\ $$$$\left(\mathrm{sorry}\:\mathrm{I}\:\mathrm{changed}\:{a}\rightleftarrows{b}\right) \\ $$$$\bigtriangleup_{{ppq}} \:\Rightarrow\:{r}=\frac{{q}\sqrt{\mathrm{2}{p}−{q}}}{\mathrm{2}\sqrt{\mathrm{2}{p}+{q}}} \\ $$$$\mathrm{let}\:{q}=\mathrm{1}\:\Rightarrow\:{r}=\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}} \\ $$$$\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{the}\:\bigtriangleup\:\mathrm{are}\:{P},\:{Q},\:{R} \\ $$$${P}=\begin{pmatrix}{\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\end{pmatrix} \\ $$$${Q}=\begin{pmatrix}{\frac{\sqrt{\mathrm{6}{p}+\mathrm{5}}+\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${R}=\begin{pmatrix}{\frac{{p}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{6}{p}+\mathrm{5}}+\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{4}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{\frac{\sqrt{\left(\mathrm{6}{p}+\mathrm{5}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{4}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\end{pmatrix} \\ $$$${C}_{{I}} =\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{p}+\mathrm{1}\right)}+\frac{\sqrt{\mathrm{6}{p}+\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\mathrm{4}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{\frac{\sqrt{\left(\mathrm{6}{p}+\mathrm{5}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)}}{\mathrm{4}\left(\mathrm{2}{p}+\mathrm{1}\right)}+\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{4}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\end{pmatrix} \\ $$$$\Rightarrow \\ $$$${a}=\frac{\sqrt{\mathrm{6}{p}+\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}} \\ $$$${b}=\frac{\sqrt{\left(\mathrm{6}{p}+\mathrm{5}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{4}\sqrt{\mathrm{2}{p}+\mathrm{1}}} \\ $$$$\mid{PC}_{{I}} \mid=\frac{\sqrt{{p}}}{\:\sqrt{\mathrm{2}{p}+\mathrm{1}}} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{match}\:\mathrm{the}\:\mathrm{picture}\:\Rightarrow \\ $$$${a}\geqslant{b}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}<{p}\leqslant\mathrm{1}.\mathrm{66853858571} \\ $$$$ \\ $$$$\begin{cases}{{r}=\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{{a}=\frac{\sqrt{\mathrm{6}{p}+\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{{b}=\frac{\sqrt{\left(\mathrm{6}{p}+\mathrm{5}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{4}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:{p}\:\mathrm{for}\:\mathrm{any}\:\mathrm{ratio}\:{a}:{b} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{further}\:\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{get} \\ $$$$\begin{cases}{{r}=\frac{\sqrt{\mathrm{2}{p}−\mathrm{1}}}{\:\mathrm{2}\sqrt{\mathrm{2}{p}+\mathrm{1}}}}\\{{a}=\mathrm{2}{r}+\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}\\{{b}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }\right){r}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\Leftrightarrow\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }={a}−\mathrm{2}{r} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\left(\mathrm{3}\right)\:\&\:\mathrm{transforming}\:\mathrm{gives} \\ $$$${r}^{\mathrm{3}} −\mathrm{2}\left({b}−\mathrm{1}\right){r}^{\mathrm{2}} −\left({a}+\frac{\mathrm{1}}{\mathrm{4}}\right){r}+\frac{{b}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${r}=\frac{\mathrm{2}{b}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{−\mathrm{3}{u}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{v}}{\mathrm{2}\sqrt{−{u}^{\mathrm{3}} }}}{\mathrm{3}} \\ $$$$\mathrm{where} \\ $$$${u}=−\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{3}}−{a}+\frac{\mathrm{8}{b}}{\mathrm{3}}−\frac{\mathrm{19}}{\mathrm{12}} \\ $$$${v}=−\frac{\mathrm{16}{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{16}{b}^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{2}{ab}}{\mathrm{3}}+\frac{\mathrm{2}{a}}{\mathrm{3}}−\frac{\mathrm{13}{b}}{\mathrm{9}}+\frac{\mathrm{41}}{\mathrm{54}} \\ $$$$\mathrm{but}\:{a},\:{b}\:\mathrm{are}\:\mathrm{not}\:\mathrm{independent}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{approach}! \\ $$
Commented by Frix last updated on 26/Oct/24
Considering the originally posted picture: We can solve for a(b, r) [degree 2] or for b(a, r) [degree 2 for b^2 ] but not for r(a, b) [degree 6].
$$\mathrm{Considering}\:\mathrm{the}\:\mathrm{originally}\:\mathrm{posted}\:\mathrm{picture}: \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{for}\:{a}\left({b},\:{r}\right)\:\left[\mathrm{degree}\:\mathrm{2}\right]\:\mathrm{or}\:\mathrm{for} \\ $$$${b}\left({a},\:{r}\right)\:\left[\mathrm{degree}\:\mathrm{2}\:\mathrm{for}\:{b}^{\mathrm{2}} \right]\:\mathrm{but}\:\mathrm{not}\:\mathrm{for}\:{r}\left({a},\:{b}\right) \\ $$$$\left[\mathrm{degree}\:\mathrm{6}\right]. \\ $$
Commented by Frix last updated on 27/Oct/24
No problem, I will upload my workings again later.
$$\mathrm{No}\:\mathrm{problem},\:\mathrm{I}\:\mathrm{will}\:\mathrm{upload}\:\mathrm{my}\:\mathrm{workings} \\ $$$$\mathrm{again}\:\mathrm{later}. \\ $$

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