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| − | ==PaperMath’s circles==
| + | #REDIRECT [[Annular Steiner chains]] |
| − | This theorem states that for a <math>n</math> tangent externally tangent circles with equal radii in the shape of a <math>n</math>-gon, the radius of the circle that is externally tangent to all the other circles can be written as <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}</math> and the radius of the circle that is internally tangent to all the other circles can be written as <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}+2r</math> Where <math>r</math> is the radius of one of the congruent circles and where <math>n</math> is the number of tangent circles. The formula for the radius of the externally tangent circle is true for all values of <math>n>4</math>, since there would obviously be no circle that could be drawn internally tangent to the other circles at <math>n \leq 4</math>
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| − | <asy>
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| − | size(10cm); //Asymptote by PaperMath
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| − | real s = 0.218;
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| − | pair A, B, C, D, E;
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| − | A = dir(90 + 0*72)*s/cos(36);
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| − | B = dir(90 + 1*72)*s/cos(36);
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| − | C = dir(90 + 2*72)*s/cos(36);
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| − | D = dir(90 + 3*72)*s/cos(36);
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| − | E = dir(90 + 4*72)*s/cos(36);
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| − | // Draw the pentagon
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| − | draw(A--B--C--D--E--cycle);
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| − | real r = 1; // Radius of the congruent circles is 1 unit
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| − | draw(circle(A, r));
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| − | draw(circle(B, r));
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| − | draw(circle(C, r));
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| − | draw(circle(D, r));
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| − | draw(circle(E, r));
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| − | pair P_center = (A + B + C + D + E) / 5;
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| − | real R_central = 1/cos(pi/180*54) - 1;
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| − | draw(circle(P_center, R_central));
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| − | </asy>
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| − | ==Proof==
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| − | We can let <math>r</math> be the radius of one of the congruent circles, and let <math>x</math> be the radius of the externally tangent circle, which means the side length of the <math>n</math>-gon is <math>2r</math>. We can draw an apothem of the <math>n</math>-gon, which bisects the side length, forming a right triangle. The length of the base is half of <math>2r</math>, or <math>r</math>, and the hypotenuse is <math>x+r</math>. The angle adjacent to the base is half of an angle of a regular <math>n</math>-gon. We know the angle of a regular <math>n</math>-gon to be <math>\frac {180(n-2)}n</math>, so half of that would be <math>\frac {90(n-2)}n</math>. Let <math>a=\frac {90(n-2)}n</math> for simplicity. We now have <math>\cos a=\frac {adj}{hyp}</math>, or <math>\cos a = \frac {r}{x+r}</math>. Multiply both sides by <math>x+r</math> and we get <math>\cos a~x+\cos a~r=r</math>, and then a bit of manipulation later you get that <math>x=\frac {r(1-\cos a}{cos a}</math>, or when you plug in <math>a=\frac {90(n-2)}n</math>, you get <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}</math>. Add <math>2r</math> to find the radius of the internally tangent circle to get <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}+2r</math>, and we are done.
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| − | ==Notes==
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| − | PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.
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| − | ==See also==
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| − | *[[PaperMath’s sum]]
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| − | [[Category:Geometry]]
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| − | [[Category:Theorems]]
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