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Question Number 35871 by MJS last updated on 25/May/18

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}