Difference between revisions of "PaperMath’s circles"
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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. | ||
==Notes== | ==Notes== | ||
| − | + | PaperMath’s circles was discovered by the aops user PaperMath, as the name implies. | |
| + | |||
==See also== | ==See also== | ||
*[[PaperMath’s sum]] | *[[PaperMath’s sum]] | ||
[[Category:Geometry]] | [[Category:Geometry]] | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Revision as of 17:22, 27 March 2024
Contents
PaperMath’s circles
This theorem states that for a 
tangent externally tangent circles with equal radii in the shape of a -gon, the radius of the circle that is externally tangent to all the other circles can be written as 
and the radius of the circle that is internally tangent to all the other circles can be written as 
Where 
is the radius of one of the congruent circles and where is the number of tangent circles. The formula for the radius of the externally tangent circle is true for all values of 
, since there would obviously be no circle that could be drawn internally tangent to the other circles at 
Proof
We can let be the radius of one of the congruent circles, and let 
be the radius of the externally tangent circle, which means the side length of the -gon is 
. We can draw an apothem of the -gon, which bisects the side length, forming a right triangle. The length of the base is half of , or , and the hypotenuse is 
. The angle adjacent to the base is half of an angle of a regular -gon. We know the angle of a regular -gon to be 
, so half of that would be 
. Let 
for simplicity. We now have 
, or 
. Multiply both sides by and we get 
, and then a bit of manipulation later you get that 
, or when you plug in , you get . Add to find the radius of the internally tangent circle to get , and we are done.
Notes
PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.
