Difference between revisions of "PaperMath’s circles"

Revision as of 22:32, 1 September 2024

Annular Steiner chains

This theorem states that for a $n$ tangent externally tangent circles with equal radii in the shape of a $n$ -gon, the radius of the circle that is externally tangent to all the other circles can be written as $\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}$ and the radius of the circle that is internally tangent to all the other circles can be written as $\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}+2r$ Where $r$ is the radius of one of the congruent circles and where $n$ is the number of tangent circles.

Here is a diagram of what $n=5$ 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 $n=8$ 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 $r$ be the radius of one of the congruent circles, and let $x$ be the radius of the externally tangent circle, which means the side length of the $n$ -gon is $2r$ . We can draw an apothem of the $n$ -gon, which bisects the side length, forming a right triangle. The length of the base is half of $2r$ , or $r$ , and the hypotenuse is $x+r$ . The angle adjacent to the base is half of an angle of a regular $n$ -gon. We know the angle of a regular $n$ -gon to be $\frac {180(n-2)}n$ , so half of that would be $\frac {90(n-2)}n$ . Let $a=\frac {90(n-2)}n$ for simplicity. We now have $\cos a=\frac {adj}{hyp}$ , or $\cos a = \frac {r}{x+r}$ . Multiply both sides by $x+r$ and we get $\cos a~x+\cos a~r=r$ , and then a bit of manipulation later you get that $x=\frac {r(1-\cos a)}{\cos a}$ , or when you plug in $a=\frac {90(n-2)}n$ , you get $\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}$ . Add $2r$ to find the radius of the internally tangent circle to get $\frac {r(1-\cos)(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}+2r$ , and we are done.

Fun stuff

Let $r=1$ , then $\lim_{n \to \infty} \frac {1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)} = \infty$ . How? Just plug in infinity to find out!

@@ Line 64: / Line 64: @@
 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!
-==See also==
-*[[PaperMath’s sum]]
 [[Category:Geometry]]
 [[Category:Theorems]]

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Difference between revisions of "PaperMath’s circles"

Revision as of 22:32, 1 September 2024

Annular Steiner chains

Proof

Fun stuff