Question Number 35871 by MJS last updated on 25/May/18
![new idea (and solution) to questions 35178 & 35195 triangle: ABC; a=BC, b=CA, c=AB; α=∠CAB, β=∠ABC, γ=∠BCA d=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) put it as this: A= ((0),(0) ), B= ((c),(0) ), C= ((((−a^2 +b^2 +c^2 )/(2c))),((d/(2c))) ) circumcircle: center=M_1 = (((c/2)),((((a^2 +b^2 −c^2 )c)/(2d))) ) radius=R=((abc)/d) (calculated by intersection of circles with centers A, B, C or of symmetry−axes of AB and AC) 2 circles touching b, c and circumcircle, one from inside, the other from outside: center=M_2 lies on y=kx with k=tan (α/2) M_2 = ((x),((xtan (α/2))) ) radius=r_1 =R−∣M_1 M_2 ∣=xtan (α/2) (inside) r_2 =∣M_1 M_2 ∣−R=xtan (α/2) (outside) (obviously any circle with center M_2 (x) and touching the x−axis has radius xtan (α/2) and also obviously the touching point of 2 circles is located on the line connecting their centers) 1. ∣M_1 M_2 ∣=R−xtan (α/2) M_1 M_2 =(R−xtan (α/2))^2 2. ∣M_1 M_2 ∣=R+xtan (α/2) M_1 M_2 =(R+xtan (α/2))^2 tan (α/2)=((sin (α/2))/(cos (α/2)))=((√((1−cos α)/2))/( (√((1+cos α)/2))))=(√((1−cos α)/(1+cos α)))= [cos α=((−a^2 +b^2 +c^2 )/(2bc))] =(√((a^2 −b^2 +2bc−c^2 )/(−a^2 +b^2 +2bc+c^2 )))=(√(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))) M_1 M_2 =(m_1 −m_2 )^2 +(n_1 −n_2 )^2 = =((c/2)−x)^2 +((((a^2 +b^2 −c^2 )c)/(2d))−xtan (α/2))^2 = [after some transformation work] =((4bc)/((a+b+c)(−a+b+c)))x^2 −((2bc(b+c))/((a+b+c)(−a+b+c)))x+((a^2 b^2 c^2 )/d^2 ) [((a^2 b^2 c^2 )/d^2 )=R^2 ] (R±xtan (α/2))^2 =x^2 tan^2 (α/2)±2Rxtan (α/2)+R^2 = =(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))x^2 ±((2abc)/((a+b+c)(−a+b+c)))x+R^2 so we have ((4bc)/((a+b+c)(−a+b+c)))x^2 −((2bc(b+c))/((a+b+c)(−a+b+c)))x= =(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))x^2 ±((2abc)/((a+b+c)(−a+b+c)))x which leads to x_3 =0 (as I explained before, the point A can be seen as a circle with radius 0 still meeting the requirements) x_1 =((2bc)/(a+b+c)) ⇒ r_1 =2bc(√(((a+b−c)(a−b+c))/((a+b+c)^3 (−a+b+c)))) x_2 =((2bc)/(−a+b+c)) ⇒ r_2 =2bc(√(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)^3 ))) for the circles corresponding with the points B and C just interchange {a, b, c} with {b, c, a} and {c, a, b}](https://www.tinkutara.com/question/Q35871.png)
$$\mathrm{new}\:\mathrm{idea}\:\left(\mathrm{and}\:\mathrm{solution}\right)\:\mathrm{to}\:\mathrm{questions}\:\mathrm{35178}\:\&\:\:\mathrm{35195} \\ $$$$ \\ $$$$\mathrm{triangle}: \\ $$$${ABC};\:{a}={BC},\:{b}={CA},\:{c}={AB}; \\ $$$$\alpha=\angle{CAB},\:\beta=\angle{ABC},\:\gamma=\angle{BCA} \\ $$$${d}=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$ \\ $$$$\mathrm{put}\:\mathrm{it}\:\mathrm{as}\:\mathrm{this}: \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix},\:{B}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix},\:{C}=\begin{pmatrix}{\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{c}}}\\{\frac{{d}}{\mathrm{2}{c}}}\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{circumcircle}: \\ $$$$\mathrm{center}={M}_{\mathrm{1}} =\begin{pmatrix}{\frac{{c}}{\mathrm{2}}}\\{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){c}}{\mathrm{2}{d}}}\end{pmatrix} \\ $$$$\mathrm{radius}={R}=\frac{{abc}}{{d}} \\ $$$$\left(\mathrm{calculated}\:\mathrm{by}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{circles}\:\mathrm{with}\right. \\ $$$$\mathrm{centers}\:{A},\:{B},\:{C}\:\mathrm{or}\:\mathrm{of}\:\mathrm{symmetry}−\mathrm{axes}\:\mathrm{of} \\ $$$$\left.{AB}\:\mathrm{and}\:{AC}\right) \\ $$$$ \\ $$$$\mathrm{2}\:\mathrm{circles}\:\mathrm{touching}\:{b},\:{c}\:\mathrm{and}\:\mathrm{circumcircle}, \\ $$$$\mathrm{one}\:\mathrm{from}\:\mathrm{inside},\:\mathrm{the}\:\mathrm{other}\:\mathrm{from}\:\mathrm{outside}: \\ $$$$\mathrm{center}={M}_{\mathrm{2}} \:\mathrm{lies}\:\mathrm{on}\:{y}={kx}\:\mathrm{with}\:{k}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$${M}_{\mathrm{2}} =\begin{pmatrix}{{x}}\\{{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{radius}={r}_{\mathrm{1}} ={R}−\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid={x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\left(\mathrm{inside}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{r}_{\mathrm{2}} =\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid−{R}={x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\left(\mathrm{outside}\right) \\ $$$$ \\ $$$$\left(\mathrm{obviously}\:\mathrm{any}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:{M}_{\mathrm{2}} \left({x}\right)\right. \\ $$$$\mathrm{and}\:\mathrm{touching}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{has}\:\mathrm{radius}\:{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{touching}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{2}\:\mathrm{circles}\:\mathrm{is}\:\mathrm{located}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{connecting} \\ $$$$\left.\mathrm{their}\:\mathrm{centers}\right) \\ $$$$ \\ $$$$\mathrm{1}.\:\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid={R}−{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:{M}_{\mathrm{1}} {M}_{\mathrm{2}} =\left({R}−{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}.\:\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid={R}+{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:{M}_{\mathrm{1}} {M}_{\mathrm{2}} =\left({R}+{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}=\frac{\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{\mathrm{2}}}}{\:\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\alpha}{\mathrm{2}}}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{cos}\:\alpha}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{cos}\:\alpha=\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{bc}}\right] \\ $$$$=\sqrt{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{bc}−{c}^{\mathrm{2}} }{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{bc}+{c}^{\mathrm{2}} }}=\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}} \\ $$$$ \\ $$$${M}_{\mathrm{1}} {M}_{\mathrm{2}} =\left({m}_{\mathrm{1}} −{m}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({n}_{\mathrm{1}} −{n}_{\mathrm{2}} \right)^{\mathrm{2}} = \\ $$$$=\left(\frac{{c}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} +\left(\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){c}}{\mathrm{2}{d}}−{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} = \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{after}\:\mathrm{some}\:\mathrm{transformation}\:\mathrm{work}\right] \\ $$$$=\frac{\mathrm{4}{bc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} −\frac{\mathrm{2}{bc}\left({b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}+\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{d}^{\mathrm{2}} }={R}^{\mathrm{2}} \right] \\ $$$$ \\ $$$$\left({R}\pm{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}\pm\mathrm{2}{Rx}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}+{R}^{\mathrm{2}} = \\ $$$$=\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} \pm\frac{\mathrm{2}{abc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}+{R}^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{4}{bc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} −\frac{\mathrm{2}{bc}\left({b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}= \\ $$$$\:\:\:\:\:=\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} \pm\frac{\mathrm{2}{abc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}_{\mathrm{3}} =\mathrm{0}\:\left(\mathrm{as}\:\mathrm{I}\:\mathrm{explained}\:\mathrm{before},\:\mathrm{the}\:\mathrm{point}\:{A}\:\mathrm{can}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{be}\:\mathrm{seen}\:\mathrm{as}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{0}\:\mathrm{still} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{meeting}\:\mathrm{the}\:\mathrm{requirements}\right) \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}{bc}}{{a}+{b}+{c}}\:\Rightarrow\:{r}_{\mathrm{1}} =\mathrm{2}{bc}\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)^{\mathrm{3}} \left(−{a}+{b}+{c}\right)}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}{bc}}{−{a}+{b}+{c}}\:\Rightarrow\:{r}_{\mathrm{2}} =\mathrm{2}{bc}\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)^{\mathrm{3}} }} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{circles}\:\mathrm{corresponding}\:\mathrm{with}\:\mathrm{the}\:\mathrm{points} \\ $$$${B}\:\mathrm{and}\:{C}\:\mathrm{just}\:\mathrm{interchange}\:\left\{{a},\:{b},\:{c}\right\}\:\mathrm{with} \\ $$$$\left\{{b},\:{c},\:{a}\right\}\:\mathrm{and}\:\left\{{c},\:{a},\:{b}\right\} \\ $$