Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "PaperMath’s circles"

(→‎Notes)
(→‎PaperMath’s circles)
Line 1: Line 1:
 
==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. 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>
 +
<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 the pentagon
 +
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>
 +
 
==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.

Revision as of 19:18, 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 $n \leq 4$ [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 the pentagon 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]

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=PaperMath’s_circles&oldid=217500"