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The area a of the circular segment is equal to the area of the circular sector minus the area of the triangular portion (using the double angle formula to get an equation in terms of ): a = R 2 2 ( θ − sin θ ) {\displaystyle a={\tfrac {R^{2}}{2}}\left(\theta -\sin \theta \right)}
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
In plane geometry, the Conway circle theorem states that when the sides meeting at each vertex of a triangle are extended by the length of the opposite side, the six endpoints of the three resulting line segments lie on a circle whose centre is the incentre of the triangle. The circle on which these six points lie is called the Conway circle of ...
The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and because the sequence tends to a circle, the corresponding formula–that the area is half the circumference times the radius–namely, A = 1 / 2 × 2πr × r, holds for a circle.
An excircle or escribed circle [2] of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.
Proposition one states: The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle. Any circle with a circumference c and a radius r is equal in area with a right triangle with the two legs being c and r.
The nine-point circle of a reference triangle is the circumcircle of both the reference triangle's medial triangle (with vertices at the midpoints of the sides of the reference triangle) and its orthic triangle (with vertices at the feet of the reference triangle's altitudes). [6]: p.153
When the area is at least as large as the circumcircle of the points, the solution is any circle of that area surrounding the points. For smaller areas, the optimal curve will be a circular triangle with the three points as its vertices, and with circular arcs of equal radii as its sides, down to the area at which one of the three interior ...