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}


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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}$
$new idea (and solution) to questions 35178 & 35195$ $triangle :$ $A B C; a = B C, b = C A, c = A B;$ $α = ∠ C A B, β = ∠ A B C, γ = ∠ B C A$ $d = \sqrt{(a + b + c) (a + b - c) (a - b + c) (- a + b + c)}$ $put it as this :$ $A = (\begin{matrix} 0 \\ 0 \end{matrix}), B = (\begin{matrix} c \\ 0 \end{matrix}), C = (\begin{matrix} \frac{- a^{2} + b^{2} + c^{2}}{2 c} \\ \frac{d}{2 c} \end{matrix})$ $circumcircle :$ $center = M_{1} = (\begin{matrix} \frac{c}{2} \\ \frac{(a^{2} + b^{2} - c^{2}) c}{2 d} \end{matrix})$ $radius = R = \frac{a b c}{d}$ $(calculated by intersection of circles with$ $centers A, B, C or of symmetry - axes of$ $A B and A C)$ $2 circles touching b, c and circumcircle,$ $one from inside, the other from outside :$ $center = M_{2} lies on y = k x with k = \tan \frac{α}{2}$ $M_{2} = (\begin{matrix} x \\ x \tan \frac{α}{2} \end{matrix})$ $radius = r_{1} = R - ∣ M_{1} M_{2} ∣= x \tan \frac{α}{2} (inside)$ $r_{2} =∣ M_{1} M_{2} ∣ - R = x \tan \frac{α}{2} (outside)$ $(obviously any circle with center M_{2} (x)$ $and touching the x - axis has radius x \tan \frac{α}{2}$ $and also obviously the touching point of$ $2 circles is located on the line connecting$ $their centers)$ $1 . ∣ M_{1} M_{2} ∣= R - x \tan \frac{α}{2}$ $M_{1} M_{2} = {(R - x \tan \frac{α}{2})}^{2}$ $2 . ∣ M_{1} M_{2} ∣= R + x \tan \frac{α}{2}$ $M_{1} M_{2} = {(R + x \tan \frac{α}{2})}^{2}$ $\tan \frac{α}{2} = \frac{\sin \frac{α}{2}}{\cos \frac{α}{2}} = \frac{\sqrt{\frac{1 - \cos α}{2}}}{\sqrt{\frac{1 + \cos α}{2}}} = \sqrt{\frac{1 - \cos α}{1 + \cos α}} =$ $[\cos α = \frac{- a^{2} + b^{2} + c^{2}}{2 b c}]$ $= \sqrt{\frac{a^{2} - b^{2} + 2 b c - c^{2}}{- a^{2} + b^{2} + 2 b c + c^{2}}} = \sqrt{\frac{(a + b - c) (a - b + c)}{(a + b + c) (- a + b + c)}}$ $M_{1} M_{2} = {(m_{1} - m_{2})}^{2} + {(n_{1} - n_{2})}^{2} =$ $= {(\frac{c}{2} - x)}^{2} + {(\frac{(a^{2} + b^{2} - c^{2}) c}{2 d} - x \tan \frac{α}{2})}^{2} =$ $[after some transformation work]$ $= \frac{4 b c}{(a + b + c) (- a + b + c)} x^{2} - \frac{2 b c (b + c)}{(a + b + c) (- a + b + c)} x + \frac{a^{2} b^{2} c^{2}}{d^{2}}$ $[\frac{a^{2} b^{2} c^{2}}{d^{2}} = R^{2}]$ ${(R \pm x \tan \frac{α}{2})}^{2} = x^{2} \tan^{2} \frac{α}{2} \pm 2 R x \tan \frac{α}{2} + R^{2} =$ $= \frac{(a + b - c) (a - b + c)}{(a + b + c) (- a + b + c)} x^{2} \pm \frac{2 a b c}{(a + b + c) (- a + b + c)} x + R^{2}$ $so we have$ $\frac{4 b c}{(a + b + c) (- a + b + c)} x^{2} - \frac{2 b c (b + c)}{(a + b + c) (- a + b + c)} x =$ $= \frac{(a + b - c) (a - b + c)}{(a + b + c) (- a + b + c)} x^{2} \pm \frac{2 a b c}{(a + b + c) (- a + b + c)} x$ $which leads to$ $x_{3} = 0 (as I explained before, the point A can$ $be seen as a circle with radius 0 still$ $meeting the requirements)$ $x_{1} = \frac{2 b c}{a + b + c} \Rightarrow r_{1} = 2 b c \sqrt{\frac{(a + b - c) (a - b + c)}{{(a + b + c)}^{3} (- a + b + c)}}$ $x_{2} = \frac{2 b c}{- a + b + c} \Rightarrow r_{2} = 2 b c \sqrt{\frac{(a + b - c) (a - b + c)}{(a + b + c) {(- a + b + c)}^{3}}}$ $for the circles corresponding with the points$ $B and C just interchange {a, b, c} with$ ${b, c, a} and {c, a, b}$
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