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The four-vertex theorem was first proved for convex curves (i.e. curves with strictly positive curvature) in 1909 by Syamadas Mukhopadhyaya. [8] His proof utilizes the fact that a point on the curve is an extremum of the curvature function if and only if the osculating circle at that point has fourth-order contact with the curve; in general the osculating circle has only third-order contact ...
As a base case, the result is trivial when n = 3 and a, b and c are the only vertices in G. Thus, we may assume that n ≥ 4. By Euler's formula for planar graphs, G has 3n − 6 edges; equivalently, if one defines the deficiency of a vertex v in G to be 6 − deg(v), the sum of the deficiencies is 12.
Circular triangles give the solution to an isoperimetric problem in which one seeks a curve of minimum length that encloses three given points and has a prescribed area. . When the area is at least as large as the circumcircle of the points, the solution is any circle of that area surrounding the poi
Solution of triangles (Latin: solutio triangulorum) is the main trigonometric problem of finding the characteristics of a triangle (angles and lengths of sides), when some of these are known. The triangle can be located on a plane or on a sphere .
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. [3]
Case 3: two sides and an opposite angle given (SSA). The sine rule gives C and then we have Case 7. There are either one or two solutions. Case 4: two angles and an included side given (ASA). The four-part cotangent formulae for sets (cBaC) and (BaCb) give c and b, then A follows from the sine rule. Case 5: two angles and an opposite side given ...
At any point during this rotation, two of the corners of the Reuleaux triangle touch two adjacent sides of the square, while the third corner of the triangle traces out a curve near the opposite vertex of the square. The shape traced out by the rotating Reuleaux triangle covers approximately 98.8% of the area of the square. [29]
for triangle sides a, b, c and area Δ. This can be seen in the figure at the top of this article, with interior point P partitioning triangle ABC into three triangles PBC, PCA, PAB with respective areas ′, ′, ′.