Difference between revisions of "PaperMath’s circles"
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==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. |
==Fun stuff== | ==Fun stuff== |
Revision as of 22:04, 27 March 2024
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.
Here is a diagram of what would look like.
Here is a diagram of what would look like.
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.
Fun stuff
Let , then . How? Just plug in infinity to find out!
Notes
PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.