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# Three Kissing Circles

If three circles with radii $$x,y$$, and $$z$$ make tangents on each other as shown, what will be the area of the figure(magenta colored) formed between them in terms of $$x,y$$, and $$z$$? I am posting my attempt below $$\frac { 1 }{ 2 } \left\{ \left( { x }^{ 2 }+xy+xz+yz \right) \ast \sin { \left( \cos ^{ -1 }{ \left( \frac { { x }^{ 2 }+xy+xz-yz }{ { x }^{ 2 }+xy+xz+yz } \right) } \right) } -\quad { x }^{ 2 }\cos ^{ -1 }{ \left( \frac { { x }^{ 2 }+xy+xz-yz }{ { x }^{ 2 }+xy+xz+yz } \right) -\quad { y }^{ 2 } } \cos ^{ -1 }{ \left( \frac { { y }^{ 2 }+xy+yz-xz }{ { y }^{ 2 }+xy+yz+xz } \right) -\quad { z }^{ 2 } } \cos ^{ -1 }{ \left( \frac { { z }^{ 2 }+xz+yz-xy }{ { z }^{ 2 }+xz+yz+xy } \right) } \right\}$$

Note by Bilal Akmal
2 years, 2 months ago

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"Update on the formula to calculate the magenta area": This one works too :)(angles are again assumed in radian) $$\sqrt { { x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 } } -\frac { { x }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2y } \right) } +\sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2z } \right) } \right] -\frac { { y }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+z }{ 2x+2y } \right) +\sin ^{ -1 }{ \left( \frac { x+z }{ 2y+2z } \right) } } \right] -\frac { { z }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+y }{ 2x+2z } \right) +\sin ^{ -1 }{ \left( \frac { x+y }{ 2y+2z } \right) } } \right]$$

- 2 years, 2 months ago

Yeah, it works out fine. I get the same result, although not exactly with the same expression.

- 2 years, 2 months ago

Area of triangle - sum of the areas of three sectors

- 2 years, 2 months ago

Can you please elaborate on how you produced the formula.

- 2 years, 2 months ago

Form a triangle with the circles' origins as vertices.This should be equal to the magenta area plus some sectors of each circle. Minus the sectors and you get the above equation(angles assumed in radian.)

- 2 years, 2 months ago