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.

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PaperMath’s circles

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