Difference between revisions of "PaperMath’s circles"

Revision as of 19:48, 27 March 2024

PaperMath’s circles

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 $r=\frac 12$ , so $\lim_{n \to \infty} \frac {1-\cos(\frac{90(n-2}n)}{2\cos(\frac{90(n-2)}n)} = \infty$ . This makes sense since we have apeirogon such that each side has length $1$ . We can visualize this as a circle, except each point has length $1$ , which may not make any sense, but infinity doesn’t make any sense, so bear with me. This incircle of this polygon would be the radius of the apeirogon, or infinite. But since we can also add some calculus thoughts in, and determine that this polygon’s circumference over the radius is $2\pi$ , or at least I hope so. I am not a master of theoretical math, but a circle with points that have length doesn’t make much sense. Plus, we probably aren’t even in 2D math anymore, since for each line segment, we put two circles on each ends, but when we have a point, where are the ends?? Do we stack them?? Infinity is weird. All you need to know is this function is divergent, and thats all.

Also, what happens when $n=4$ ? Short answer just solve it yourself. It’s really simple 45-45-90 triangle. Have fun mathing! Formulas can’t save you on $n=4$ .

Notes

PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.

@@ Line 59: / Line 59: @@
 ==Fun stuff==
 Let <math>r=\frac 12</math>, so <math>\lim_{n \to \infty} \frac {1-\cos(\frac{90(n-2}n)}{2\cos(\frac{90(n-2)}n)} = \infty</math>. This makes sense since we have apeirogon such that each side has length <math>1</math>. We can visualize this as a circle, except each point has length <math>1</math>, which may not make any sense, but infinity doesn’t make any sense, so bear with me. This incircle of this polygon would be the radius of the apeirogon, or infinite. But since we can also add some calculus thoughts in, and determine that this polygon’s circumference over the radius is <math>2\pi</math>, or at least I hope so. I am not a master of theoretical math, but a circle with points that have length doesn’t make much sense. Plus, we probably aren’t even in 2D math anymore, since for each line segment, we put two circles on each ends, but when we have a point, where are the ends?? Do we stack them?? Infinity is weird. All you need to know is this function is divergent, and thats all.
+Also, what happens when <math>n=4</math>? Short answer just solve it yourself. It’s really simple 45-45-90 triangle. Have fun mathing! Formulas can’t save you on <math>n=4</math>.
 ==Notes==

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