Difference between revisions of "PaperMath’s circles"

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==Annular Steiner chains==
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#REDIRECT [[Annular Steiner chains]]
 
 
See [https://en.m.wikipedia.org/wiki/Steiner_chain# Steiner chain (Annular case)]
 
 
 
 
 
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.
 
 
 
Here is a diagram of what <math>n=5</math> would look like.
 
 
 
<asy>
 
size(10cm);  //Asymptote by PaperMath
 
real s = 0.218;
 
pair A, B, C, D, E;
 
A = dir(90 + 0*72)*s/cos(36);
 
B = dir(90 + 1*72)*s/cos(36);
 
C = dir(90 + 2*72)*s/cos(36);
 
D = dir(90 + 3*72)*s/cos(36);
 
E = dir(90 + 4*72)*s/cos(36);
 
draw(A--B--C--D--E--cycle);
 
real r = 1;  // Radius of the congruent circles is 1 unit
 
draw(circle(A, r));
 
draw(circle(B, r));
 
draw(circle(C, r));
 
draw(circle(D, r));
 
draw(circle(E, r));
 
pair P_center = (A + B + C + D + E) / 5;
 
real R_central = 1/cos(pi/180*54) - 1; 
 
draw(circle(P_center, R_central));
 
</asy>
 
 
 
Here is a diagram of what <math>n=8</math> would look like.
 
 
 
<asy>
 
size(10cm); // Asymptote by PaperMath
 
real s = 2.28;
 
pair A, B, C, D, E, F, G, H;
 
A = dir(90 + 0*45)*s/cos(22.5);
 
B = dir(90 + 1*45)*s/cos(22.5);
 
C = dir(90 + 2*45)*s/cos(22.5);
 
D = dir(90 + 3*45)*s/cos(22.5);
 
E = dir(90 + 4*45)*s/cos(22.5);
 
F = dir(90 + 5*45)*s/cos(22.5);
 
G = dir(90 + 6*45)*s/cos(22.5);
 
H = dir(90 + 7*45)*s/cos(22.5);
 
draw(A--B--C--D--E--F--G--H--cycle);
 
real r = 1; // Radius of the congruent circles is 1 unit
 
draw(circle(A, r));
 
draw(circle(B, r));
 
draw(circle(C, r));
 
draw(circle(D, r));
 
draw(circle(E, r));
 
draw(circle(F, r));
 
draw(circle(G, r));
 
draw(circle(H, r));
 
pair P_center = (A + B + C + D + E + F + G + H) / 8;
 
real R_central = 1/cos(pi/180*67.5) - 1; // Updated radius of the central circle
 
draw(circle(P_center, R_central));
 
</asy>
 
 
 
==Proof==
 
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.
 
 
 
==Fun stuff==
 
Let <math>r=1</math>, then <math>\lim_{n \to \infty} \frac {1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)} = \infty</math>. How? Just plug in infinity to find out!
 
 
 
==Notes==
 
PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.
 
 
 
==See also==
 
*[[PaperMath’s sum]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 

Latest revision as of 22:40, 1 September 2024