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 \(\frac { 1 }{ 2 } \left\{ \left( { x }^{ 2 }+xy+xz+yz \right) \ast \sin { \left( \cos ^{ -1 }{ \left( \frac { { x }^{ 2 }+xy+xz-yz }{ { x }^{ 2 }+xy+xz+yz } \right) } \right) } -\quad { x }^{ 2 }\cos ^{ -1 }{ \left( \frac { { x }^{ 2 }+xy+xz-yz }{ { x }^{ 2 }+xy+xz+yz } \right) -\quad { y }^{ 2 } } \cos ^{ -1 }{ \left( \frac { { y }^{ 2 }+xy+yz-xz }{ { y }^{ 2 }+xy+yz+xz } \right) -\quad { z }^{ 2 } } \cos ^{ -1 }{ \left( \frac { { z }^{ 2 }+xz+yz-xy }{ { z }^{ 2 }+xz+yz+xy } \right) } \right\} \)

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TopNewest"Update on the formula to calculate the magenta area": This one works too :)(angles are again assumed in radian) \(\sqrt { { x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 } } -\frac { { x }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2y } \right) } +\sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2z } \right) } \right] -\frac { { y }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+z }{ 2x+2y } \right) +\sin ^{ -1 }{ \left( \frac { x+z }{ 2y+2z } \right) } } \right] -\frac { { z }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+y }{ 2x+2z } \right) +\sin ^{ -1 }{ \left( \frac { x+y }{ 2y+2z } \right) } } \right] \) – Bilal Akmal · 1 year, 10 months ago

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Yeah, it works out fine. I get the same result, although not exactly with the same expression. – Michael Mendrin · 1 year, 10 months ago

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Area of triangle - sum of the areas of three sectors – Vincent Miller Moral · 1 year, 10 months ago

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Can you please elaborate on how you produced the formula. – Curtis Clement · 1 year, 10 months ago

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– Bilal Akmal · 1 year, 10 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.)Log in to reply