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A cyclic polygon with an even number of sides has all angles equal if and only if the alternate sides are equal (that is, sides 1, 3, 5, … are equal, and sides 2, 4, 6, … are equal). [11] A cyclic pentagon with rational sides and area is known as a Robbins pentagon. In all known cases, its diagonals also have rational lengths, though ...
Thus, by the Pythagorean theorem, x and y satisfy the equation + = Since x 2 = (−x) 2 for all x, and since the reflection of any point on the unit circle about the x - or y-axis is also on the unit circle, the above equation holds for all points (x, y) on the unit circle, not only those in the first quadrant.
Choose two points on the circle, and call them V and A. Draw line OV and extended past O so that it intersects the circle at point B which is diametrically opposite the point V. Draw an angle whose vertex is point V and whose sides pass through points A, B. Draw line OA. Angle ∠BOA is a central angle; call it θ.
Kissing circles. Given three mutually tangent circles (black), there are, in general, two possible answers (red) as to what radius a fourth tangent circle can have. In geometry, Descartes' theorem states that for every four kissing, or mutually tangent, circles, the radii of the circles satisfy a certain quadratic equation. By solving this ...
It can only be used to draw a line segment between two points, or to extend an existing line segment. The compass can have an arbitrarily large radius with no markings on it (unlike certain real-world compasses). Circles and circular arcs can be drawn starting from two given points: the centre and a point on the circle. The compass may or may ...
This is a consequence of Jacobi's two-square theorem, which follows almost immediately from the Jacobi triple product. [ 6 ] A much simpler sum appears if the sum of squares function r 2 ( n ) {\displaystyle r_{2}(n)} is defined as the number of ways of writing the number n {\displaystyle n} as the sum of two squares.
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.
Draw three circumcircles (Miquel's circles) to triangles AB´C´, A´BC´, and A´B´C. Miquel's theorem states that these circles intersect in a single point M, called the Miquel point. In addition, the three angles MA´B, MB´C and MC´A (green in the diagram) are all equal, as are the three supplementary angles MA´C, MB´A and MC´B. [2] [3]