Difference between revisions of "PaperMath’s circles"

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Papermath’s sum was discovered by the aops user Papermath, as the name implies.
 
Papermath’s sum was discovered by the aops user Papermath, as the name implies.
 
==See also==
 
==See also==
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*[[Papermath’s sum]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 16:21, 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 $n \leq 4$

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 sum was discovered by the aops user Papermath, as the name implies.

See also