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Let p be an interior point of the disk, and let n be a multiple of 4 that is greater than or equal to 8. Form n sectors of the disk with equal angles by choosing an arbitrary line through p, rotating the line n / 2 − 1 times by an angle of 2 π / n radians, and slicing the disk on each of the resulting n / 2 lines ...
Suppose that the area C enclosed by the circle is greater than the area T = cr/2 of the triangle. Let E denote the excess amount. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those gaps, G 4, is greater than E, split each arc in
The lemma establishes an important property for solving the problem. By employing an inductive proof, one can arrive at a formula for f(n) in terms of f(n − 1).. Proof. In the figure the dark lines are connecting points 1 through 4 dividing the circle into 8 total regions (i.e., f(4) = 8).
If the smaller circle depends on the larger one (Case I), the larger circle's motion forces the smaller to traverse the larger’s circumference. If the larger circle depends on the smaller one (Case II), then the smaller circle's motion forces the larger circle to traverse the smaller circle’s circumference. This is the simplest solution.
Gauss's circle problem asks how many points there are inside this circle of the form (,) where and are both integers. Since the equation of this circle is given in Cartesian coordinates by x 2 + y 2 = r 2 {\displaystyle x^{2}+y^{2}=r^{2}} , the question is equivalently asking how many pairs of integers m and n there are such that
This can be seen from the area formula πr 2 and the circumference formula 2πr. The area of a regular polygon is half its perimeter times the apothem (where the apothem is the distance from the center to the nearest point on any side).
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The area within a circle is equal to the radius multiplied by half the circumference, or A = r x C /2 = r x r x π.. Liu Hui argued: "Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the ...