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Difference between revisions of "PaperMath’s circles"

(→‎PaperMath’s circles)
(→‎PaperMath’s circles)
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<asy>
 
<asy>
 
size(10cm);  //Asymptote by PaperMath  
 
size(10cm);  //Asymptote by PaperMath  
−
 
−
 
 
real s = 0.218;
 
real s = 0.218;
 
pair A, B, C, D, E;
 
pair A, B, C, D, E;
−
 
−
 
 
A = dir(90 + 0*72)*s/cos(36);
 
A = dir(90 + 0*72)*s/cos(36);
 
B = dir(90 + 1*72)*s/cos(36);
 
B = dir(90 + 1*72)*s/cos(36);
Line 17: Line 13:
 
D = dir(90 + 3*72)*s/cos(36);
 
D = dir(90 + 3*72)*s/cos(36);
 
E = dir(90 + 4*72)*s/cos(36);
 
E = dir(90 + 4*72)*s/cos(36);
−
 
−
// Draw the pentagon
 
 
draw(A--B--C--D--E--cycle);
 
draw(A--B--C--D--E--cycle);
−
 
 
real r = 1;  // Radius of the congruent circles is 1 unit
 
real r = 1;  // Radius of the congruent circles is 1 unit
 
draw(circle(A, r));
 
draw(circle(A, r));
Line 27: Line 20:
 
draw(circle(D, r));
 
draw(circle(D, r));
 
draw(circle(E, r));
 
draw(circle(E, r));
−
 
−
 
 
pair P_center = (A + B + C + D + E) / 5;
 
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.
  
−
real R_central = 1/cos(pi/180*54) - 1;
+
<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));
 
draw(circle(P_center, R_central));
 
</asy>
 
</asy>

Revision as of 19:29, 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. The formula for the radius of the externally tangent circle is true for all values of $n>4$, since there would obviously be no circle that could be drawn internally tangent to the other circles at $n \leq 4$

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.

Notes

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

See also

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