Question Number 207533 by mr W last updated on 18/May/24
Commented by mr W last updated on 18/May/24
$${find}\:{the}\:{radius}\:{of}\:{circumcircle} \\ $$
Answered by mr W last updated on 20/May/24
![OA=p=R−a OB=q=R−b OC=r=R−c X=q^2 +r^2 −u^2 =(R−b)^2 +(R−c)^2 −u^2 =2R^2 −2(b+c)R+b^2 +c^2 −u^2 Y=r^2 +p^2 −v^2 =(R−c)^2 +(R−a)^2 −v^2 =2R^2 −2(c+a)R+c^2 +a^2 −v^2 Z=p^2 +q^2 −w^2 =(R−a)^2 +(R−b)^2 −w^2 =2R^2 −2(a+b)R+a^2 +b^2 −w^2 volume of tetrahedron O−ABC should be zero. V=((√(4p^2 q^2 r^2 +XYZ−p^2 X^2 −q^2 Y^2 −r^2 Z^2 ))/(12))=0 4p^2 q^2 r^2 +XYZ−p^2 X^2 −q^2 Y^2 −r^2 Z^2 =0 4(R−a)^2 (R−b)^2 (R−c)^2 +XYZ−(R−a)^2 X^2 −(R−b)^2 Y^2 −(R−c)^2 Z^2 =0 after expansion we get a quadratic equation for R: {−(u^4 +v^4 +w^4 )+2(u^2 v^2 +v^2 w^2 +w^2 u^2 )+4[a(b+c)−bc−a^2 ]u^2 +4[b(c+a)−ca−b^2 ]v^2 +4[c(a+b)−ab−c^2 ]w^2 }R^2 +2{au^4 +bv^4 +cw^4 +2(a^3 u^2 +b^3 v^2 +c^3 w^2 )−(a+b)(u^2 v^2 +c^2 w^2 )−(b+c)(v^2 w^2 +a^2 u^2 )−(c+a)(w^2 u^2 +b^2 v^2 )+[bc(b+c)−a(b^2 +c^2 )]u^2 +[ca(c+a)−b(c^2 +a^2 )]v^2 +[ab(a+b)−c(a^2 +b^2 )]w^2 }R −{a^2 u^2 (a^2 +u^2 )+b^2 v^2 (b^2 +v^2 )+c^2 w^2 (c^2 +w^2 )−[a^2 (b^2 +c^2 )−b^2 c^2 ]u^2 −[b^2 (c^2 +a^2 )−c^2 a^2 ]v^2 −[c^2 (a^2 +b^2 )−a^2 b^2 ]w^2 −(a^2 +b^2 )u^2 v^2 −(b^2 +c^2 )v^2 w^2 −(c^2 +a^2 )w^2 u^2 +u^2 v^2 w^2 }=0 with A=−(u^4 +v^4 +w^4 )+2(u^2 v^2 +v^2 w^2 +w^2 u^2 )+4[a(b+c)−bc−a^2 ]u^2 +4[b(c+a)−ca−b^2 ]v^2 +4[c(a+b)−ab−c^2 ]w^2 B=au^4 +bv^4 +cw^4 +2(a^3 u^2 +b^3 v^2 +c^3 w^2 )−(a+b)(u^2 v^2 +c^2 w^2 )−(b+c)(v^2 w^2 +a^2 u^2 )−(c+a)(w^2 u^2 +b^2 v^2 )+[bc(b+c)−a(b^2 +c^2 )]u^2 +[ca(c+a)−b(c^2 +a^2 )]v^2 +[ab(a+b)−c(a^2 +b^2 )]w^2 C=a^2 u^2 (a^2 +u^2 )+b^2 v^2 (b^2 +v^2 )+c^2 w^2 (c^2 +w^2 )−[a^2 (b^2 +c^2 )−b^2 c^2 ]u^2 −[b^2 (c^2 +a^2 )−c^2 a^2 ]v^2 −[c^2 (a^2 +b^2 )−a^2 b^2 ]w^2 −(a^2 +b^2 )u^2 v^2 −(b^2 +c^2 )v^2 w^2 −(c^2 +a^2 )w^2 u^2 +u^2 v^2 w^2 ⇒AR^2 +2BR−C=0 ⇒R=((−B±(√(B^2 +AC)))/A) example: u=7, v=6, w=8 a=3, b=2, c=1 A=−(7^4 +6^4 +8^4 )+2(7^2 6^2 +6^2 8^2 +8^2 7^2 )+4[3(2+1)−2×1−3^2 ]7^2 +4[2(1+3)−1×3−2^2 ]6^2 +4[1(3+2)−3×2−1^2 ]8^2 =5855 B=3×7^4 +2×6^4 +1×8^4 +2(3^3 7^2 +2^3 6^2 +1^3 8^2 )−(3+2)(7^2 6^2 +1^2 8^2 )−(2+1)(6^2 8^2 +3^2 7^2 )−(1+3)(8^2 7^2 +2^2 6^2 )+[2×1(2+1)−3(2^2 +1^2 )]7^2 +[1×3(1+3)−2(1^2 +3^2 )]6^2 +[3×2(3+2)−1(3^2 +2^2 )]8^2 =−12895 C=3^2 7^2 (3^2 +7^2 )+2^2 6^2 (2^2 +6^2 )+1^2 8^2 (1^2 +8^2 )−[3^2 (2^2 +1^2 )−2^2 1^2 ]7^2 −[2^2 (1^2 +3^2 )−1^2 3^2 ]6^2 −[1^2 (3^2 +2^2 )−3^2 2^2 ]8^2 −(3^2 +2^2 )7^2 6^2 −(2^2 +1^2 )6^2 8^2 −(1^2 +3^2 )8^2 7^2 +7^2 6^2 8^2 =80929 R=((12895±(√(12895^2 +5855×80929)))/(5855)) =((2579)/(1171))±((3024(√(70)))/(5855)) ≈6.523586 / −2.118804](https://www.tinkutara.com/question/Q207534.png)
$${OA}={p}={R}−{a} \\ $$$${OB}={q}={R}−{b} \\ $$$${OC}={r}={R}−{c} \\ $$$${X}={q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{u}^{\mathrm{2}} =\left({R}−{b}\right)^{\mathrm{2}} +\left({R}−{c}\right)^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}\left({b}+{c}\right){R}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{u}^{\mathrm{2}} \\ $$$${Y}={r}^{\mathrm{2}} +{p}^{\mathrm{2}} −{v}^{\mathrm{2}} =\left({R}−{c}\right)^{\mathrm{2}} +\left({R}−{a}\right)^{\mathrm{2}} −{v}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}\left({c}+{a}\right){R}+{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{v}^{\mathrm{2}} \\ $$$${Z}={p}^{\mathrm{2}} +{q}^{\mathrm{2}} −{w}^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} −{w}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){R}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{w}^{\mathrm{2}} \\ $$$${volume}\:{of}\:{tetrahedron}\:{O}−{ABC} \\ $$$${should}\:{be}\:{zero}. \\ $$$${V}=\frac{\sqrt{\mathrm{4}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} +{XYZ}−{p}^{\mathrm{2}} {X}^{\mathrm{2}} −{q}^{\mathrm{2}} {Y}^{\mathrm{2}} −{r}^{\mathrm{2}} {Z}^{\mathrm{2}} }}{\mathrm{12}}=\mathrm{0} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} +{XYZ}−{p}^{\mathrm{2}} {X}^{\mathrm{2}} −{q}^{\mathrm{2}} {Y}^{\mathrm{2}} −{r}^{\mathrm{2}} {Z}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}\left({R}−{a}\right)^{\mathrm{2}} \left({R}−{b}\right)^{\mathrm{2}} \left({R}−{c}\right)^{\mathrm{2}} +{XYZ}−\left({R}−{a}\right)^{\mathrm{2}} {X}^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} {Y}^{\mathrm{2}} −\left({R}−{c}\right)^{\mathrm{2}} {Z}^{\mathrm{2}} =\mathrm{0} \\ $$$${after}\:{expansion}\:{we}\:{get}\:{a}\:{quadratic} \\ $$$${equation}\:{for}\:{R}: \\ $$$$\left\{−\left({u}^{\mathrm{4}} +{v}^{\mathrm{4}} +{w}^{\mathrm{4}} \right)+\mathrm{2}\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{v}^{\mathrm{2}} {w}^{\mathrm{2}} +{w}^{\mathrm{2}} {u}^{\mathrm{2}} \right)+\mathrm{4}\left[{a}\left({b}+{c}\right)−{bc}−{a}^{\mathrm{2}} \right]{u}^{\mathrm{2}} +\mathrm{4}\left[{b}\left({c}+{a}\right)−{ca}−{b}^{\mathrm{2}} \right]{v}^{\mathrm{2}} +\mathrm{4}\left[{c}\left({a}+{b}\right)−{ab}−{c}^{\mathrm{2}} \right]{w}^{\mathrm{2}} \right\}{R}^{\mathrm{2}} \\ $$$$+\mathrm{2}\left\{{au}^{\mathrm{4}} +{bv}^{\mathrm{4}} +{cw}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{3}} {u}^{\mathrm{2}} +{b}^{\mathrm{3}} {v}^{\mathrm{2}} +{c}^{\mathrm{3}} {w}^{\mathrm{2}} \right)−\left({a}+{b}\right)\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{c}^{\mathrm{2}} {w}^{\mathrm{2}} \right)−\left({b}+{c}\right)\left({v}^{\mathrm{2}} {w}^{\mathrm{2}} +{a}^{\mathrm{2}} {u}^{\mathrm{2}} \right)−\left({c}+{a}\right)\left({w}^{\mathrm{2}} {u}^{\mathrm{2}} +{b}^{\mathrm{2}} {v}^{\mathrm{2}} \right)+\left[{bc}\left({b}+{c}\right)−{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right]{u}^{\mathrm{2}} +\left[{ca}\left({c}+{a}\right)−{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\right]{v}^{\mathrm{2}} +\left[{ab}\left({a}+{b}\right)−{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right]{w}^{\mathrm{2}} \right\}{R} \\ $$$$−\left\{{a}^{\mathrm{2}} {u}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{u}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {v}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {w}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{w}^{\mathrm{2}} \right)−\left[{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right]{u}^{\mathrm{2}} −\left[{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right]{v}^{\mathrm{2}} −\left[{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right]{w}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){u}^{\mathrm{2}} {v}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){v}^{\mathrm{2}} {w}^{\mathrm{2}} −\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right){w}^{\mathrm{2}} {u}^{\mathrm{2}} +{u}^{\mathrm{2}} {v}^{\mathrm{2}} {w}^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$${with} \\ $$$${A}=−\left({u}^{\mathrm{4}} +{v}^{\mathrm{4}} +{w}^{\mathrm{4}} \right)+\mathrm{2}\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{v}^{\mathrm{2}} {w}^{\mathrm{2}} +{w}^{\mathrm{2}} {u}^{\mathrm{2}} \right)+\mathrm{4}\left[{a}\left({b}+{c}\right)−{bc}−{a}^{\mathrm{2}} \right]{u}^{\mathrm{2}} +\mathrm{4}\left[{b}\left({c}+{a}\right)−{ca}−{b}^{\mathrm{2}} \right]{v}^{\mathrm{2}} +\mathrm{4}\left[{c}\left({a}+{b}\right)−{ab}−{c}^{\mathrm{2}} \right]{w}^{\mathrm{2}} \\ $$$${B}={au}^{\mathrm{4}} +{bv}^{\mathrm{4}} +{cw}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{3}} {u}^{\mathrm{2}} +{b}^{\mathrm{3}} {v}^{\mathrm{2}} +{c}^{\mathrm{3}} {w}^{\mathrm{2}} \right)−\left({a}+{b}\right)\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{c}^{\mathrm{2}} {w}^{\mathrm{2}} \right)−\left({b}+{c}\right)\left({v}^{\mathrm{2}} {w}^{\mathrm{2}} +{a}^{\mathrm{2}} {u}^{\mathrm{2}} \right)−\left({c}+{a}\right)\left({w}^{\mathrm{2}} {u}^{\mathrm{2}} +{b}^{\mathrm{2}} {v}^{\mathrm{2}} \right)+\left[{bc}\left({b}+{c}\right)−{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right]{u}^{\mathrm{2}} +\left[{ca}\left({c}+{a}\right)−{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\right]{v}^{\mathrm{2}} +\left[{ab}\left({a}+{b}\right)−{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right]{w}^{\mathrm{2}} \\ $$$${C}={a}^{\mathrm{2}} {u}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{u}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {v}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {w}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{w}^{\mathrm{2}} \right)−\left[{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right]{u}^{\mathrm{2}} −\left[{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right]{v}^{\mathrm{2}} −\left[{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right]{w}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){u}^{\mathrm{2}} {v}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){v}^{\mathrm{2}} {w}^{\mathrm{2}} −\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right){w}^{\mathrm{2}} {u}^{\mathrm{2}} +{u}^{\mathrm{2}} {v}^{\mathrm{2}} {w}^{\mathrm{2}} \\ $$$$\Rightarrow{AR}^{\mathrm{2}} +\mathrm{2}{BR}−{C}=\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{−{B}\pm\sqrt{{B}^{\mathrm{2}} +{AC}}}{{A}} \\ $$$$ \\ $$$${example}: \\ $$$${u}=\mathrm{7},\:{v}=\mathrm{6},\:{w}=\mathrm{8} \\ $$$${a}=\mathrm{3},\:{b}=\mathrm{2},\:{c}=\mathrm{1} \\ $$$${A}=−\left(\mathrm{7}^{\mathrm{4}} +\mathrm{6}^{\mathrm{4}} +\mathrm{8}^{\mathrm{4}} \right)+\mathrm{2}\left(\mathrm{7}^{\mathrm{2}} \mathrm{6}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \mathrm{8}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}} \right)+\mathrm{4}\left[\mathrm{3}\left(\mathrm{2}+\mathrm{1}\right)−\mathrm{2}×\mathrm{1}−\mathrm{3}^{\mathrm{2}} \right]\mathrm{7}^{\mathrm{2}} +\mathrm{4}\left[\mathrm{2}\left(\mathrm{1}+\mathrm{3}\right)−\mathrm{1}×\mathrm{3}−\mathrm{2}^{\mathrm{2}} \right]\mathrm{6}^{\mathrm{2}} +\mathrm{4}\left[\mathrm{1}\left(\mathrm{3}+\mathrm{2}\right)−\mathrm{3}×\mathrm{2}−\mathrm{1}^{\mathrm{2}} \right]\mathrm{8}^{\mathrm{2}} =\mathrm{5855} \\ $$$${B}=\mathrm{3}×\mathrm{7}^{\mathrm{4}} +\mathrm{2}×\mathrm{6}^{\mathrm{4}} +\mathrm{1}×\mathrm{8}^{\mathrm{4}} +\mathrm{2}\left(\mathrm{3}^{\mathrm{3}} \mathrm{7}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} \mathrm{6}^{\mathrm{2}} +\mathrm{1}^{\mathrm{3}} \mathrm{8}^{\mathrm{2}} \right)−\left(\mathrm{3}+\mathrm{2}\right)\left(\mathrm{7}^{\mathrm{2}} \mathrm{6}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \mathrm{8}^{\mathrm{2}} \right)−\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{6}^{\mathrm{2}} \mathrm{8}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}} \right)−\left(\mathrm{1}+\mathrm{3}\right)\left(\mathrm{8}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \mathrm{6}^{\mathrm{2}} \right)+\left[\mathrm{2}×\mathrm{1}\left(\mathrm{2}+\mathrm{1}\right)−\mathrm{3}\left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \right)\right]\mathrm{7}^{\mathrm{2}} +\left[\mathrm{1}×\mathrm{3}\left(\mathrm{1}+\mathrm{3}\right)−\mathrm{2}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\right]\mathrm{6}^{\mathrm{2}} +\left[\mathrm{3}×\mathrm{2}\left(\mathrm{3}+\mathrm{2}\right)−\mathrm{1}\left(\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)\right]\mathrm{8}^{\mathrm{2}} =−\mathrm{12895} \\ $$$${C}=\mathrm{3}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}} \left(\mathrm{3}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \right)+\mathrm{2}^{\mathrm{2}} \mathrm{6}^{\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \right)+\mathrm{1}^{\mathrm{2}} \mathrm{8}^{\mathrm{2}} \left(\mathrm{1}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} \right)−\left[\mathrm{3}^{\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \right)−\mathrm{2}^{\mathrm{2}} \mathrm{1}^{\mathrm{2}} \right]\mathrm{7}^{\mathrm{2}} −\left[\mathrm{2}^{\mathrm{2}} \left(\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)−\mathrm{1}^{\mathrm{2}} \mathrm{3}^{\mathrm{2}} \right]\mathrm{6}^{\mathrm{2}} −\left[\mathrm{1}^{\mathrm{2}} \left(\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)−\mathrm{3}^{\mathrm{2}} \mathrm{2}^{\mathrm{2}} \right]\mathrm{8}^{\mathrm{2}} −\left(\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)\mathrm{7}^{\mathrm{2}} \mathrm{6}^{\mathrm{2}} −\left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \right)\mathrm{6}^{\mathrm{2}} \mathrm{8}^{\mathrm{2}} −\left(\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\mathrm{8}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \mathrm{6}^{\mathrm{2}} \mathrm{8}^{\mathrm{2}} =\mathrm{80929} \\ $$$${R}=\frac{\mathrm{12895}\pm\sqrt{\mathrm{12895}^{\mathrm{2}} +\mathrm{5855}×\mathrm{80929}}}{\mathrm{5855}} \\ $$$$\:\:\:\:=\frac{\mathrm{2579}}{\mathrm{1171}}\pm\frac{\mathrm{3024}\sqrt{\mathrm{70}}}{\mathrm{5855}} \\ $$$$\:\:\:\:\approx\mathrm{6}.\mathrm{523586}\:/\:−\mathrm{2}.\mathrm{118804} \\ $$
Commented by mr W last updated on 19/May/24
Answered by ajfour last updated on 18/May/24
$${say}\:{A}\:{is}\:{origin}.\:{AO}\:\:{y}\:{axis}. \\ $$$${red}\:{circle}\:{x}^{\mathrm{2}} +\left({y}−{R}+{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${x}_{{B}} ^{\mathrm{2}} +{y}_{{B}} ^{\mathrm{2}} ={w}^{\mathrm{2}} \\ $$$${x}_{{C}} ^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} ={v}^{\mathrm{2}} \\ $$$${x}_{{B}} ^{\mathrm{2}} +\left({y}_{{B}} −{R}+{a}\right)^{\mathrm{2}} =\left({R}−{b}\right)^{\mathrm{2}} \\ $$$${x}_{{C}} ^{\mathrm{2}} +\left({y}_{{C}} −{R}+{a}\right)^{\mathrm{2}} =\left({R}−{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{w}^{\mathrm{2}} −\mathrm{2}\left({R}−{a}\right){y}_{{B}} =\left({R}−{b}\right)^{\mathrm{2}} \\ $$$$\&\:{v}^{\mathrm{2}} −\mathrm{2}\left({R}−{a}\right){y}_{{c}} =\left({R}−{c}\right)^{\mathrm{2}} \\ $$$${x}_{{B}} =\sqrt{{w}^{\mathrm{2}} −{y}_{{B}} ^{\mathrm{2}} } \\ $$$${x}_{{C}} =−\sqrt{{v}^{\mathrm{2}} −{y}_{{C}} ^{\mathrm{2}} } \\ $$$$\left({x}_{{B}} −{x}_{{C}} \right)^{\mathrm{2}} +\left({y}_{{B}} −{y}_{{C}} \right)^{\mathrm{2}} ={u}^{\mathrm{2}} \\ $$$$\bigstar \\ $$
Commented by mr W last updated on 18/May/24
$${welcome}\:{back}\:{sir}! \\ $$
Commented by ajfour last updated on 18/May/24
$${thanks},\:{but}\:{could}\:{u}\:{follow}\:{my}\:{solution} \\ $$$${sir}.. \\ $$
Commented by mr W last updated on 19/May/24
$${your}\:{path}\:{is}\:{right},\:{thanks}\:{sir}! \\ $$
Commented by ajfour last updated on 18/May/24
$${can}\:{this}\:{do}? \\ $$$$\Sigma\mathrm{cos}^{−\mathrm{1}} \frac{\left({R}−{b}\right)^{\mathrm{2}} +\left({R}−{c}\right)^{\mathrm{2}} −{u}^{\mathrm{2}} }{\mathrm{2}\left({R}−{b}\right)\left({R}−{c}\right)}=\mathrm{2}\pi \\ $$
Commented by mr W last updated on 19/May/24
$${yes},\:{this}\:{works}\:{either}.\:{but}\:{this} \\ $$$${equation}\:{is}\:{not}\:{polynomial}. \\ $$$${the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{the} \\ $$$${vertices}\:{of}\:{a}\:{triangle}\:{fulfills} \\ $$$${a}\:{polynomial}\:{equation}. \\ $$
Commented by mr W last updated on 18/May/24
Commented by mr W last updated on 19/May/24
$${the}\:{final}\:{equation}\:{can}\:{be}\:{simplied}\:{to} \\ $$$${a}\:{quadratic}\:{equation},\:{so}\:{we}\:{can}\:{get} \\ $$$${the}\:{exact}\:{solution}.\:{see}\:{above}. \\ $$