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Difference between revisions of "PaperMath’s circles"
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==PaperMath’s circles==
 
==PaperMath’s circles==
−
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>
+
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==
 
==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.
+
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==
 
==Notes==
−
Papermath’s sum was discovered by the aops user Papermath, as the name implies.
+
PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.
 +
 
 
==See also==
 
==See also==
 +
*[[PaperMath’s sum]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Latest revision as of 23:04, 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 $r=1$, 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!

Notes

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

See also

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