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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
1 year, 7 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) $$\sqrt { { x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 } } -\frac { { x }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2y } \right) } +\sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2z } \right) } \right] -\frac { { y }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+z }{ 2x+2y } \right) +\sin ^{ -1 }{ \left( \frac { x+z }{ 2y+2z } \right) } } \right] -\frac { { z }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+y }{ 2x+2z } \right) +\sin ^{ -1 }{ \left( \frac { x+y }{ 2y+2z } \right) } } \right]$$ · 1 year, 7 months ago

Yeah, it works out fine. I get the same result, although not exactly with the same expression. · 1 year, 7 months ago

Area of triangle - sum of the areas of three sectors · 1 year, 7 months ago

Can you please elaborate on how you produced the formula. · 1 year, 7 months ago