Three Kissing Circles

If three circles with radii x,yx,y, and zz make tangents on each other as shown, what will be the area of the figure(magenta colored) formed between them in terms of x,yx,y, and zz? I am posting my attempt below 12{(x2+xy+xz+yz)sin(cos1(x2+xy+xzyzx2+xy+xz+yz))x2cos1(x2+xy+xzyzx2+xy+xz+yz)y2cos1(y2+xy+yzxzy2+xy+yz+xz)z2cos1(z2+xz+yzxyz2+xz+yz+xy)}\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
4 years ago

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Can you please elaborate on how you produced the formula.

Curtis Clement - 4 years ago

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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.)

Bilal Akmal - 4 years ago

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Area of triangle - sum of the areas of three sectors

Vincent Miller Moral - 4 years ago

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

Michael Mendrin - 4 years ago

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"Update on the formula to calculate the magenta area": This one works too :)(angles are again assumed in radian) x2yz+xy2z+xyz2x22[sin1(y+z2x+2y)+sin1(y+z2x+2z)]y22[sin1(x+z2x+2y)+sin1(x+z2y+2z)]z22[sin1(x+y2x+2z)+sin1(x+y2y+2z)]\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 - 4 years ago

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