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Label the three sides of the given triangle as a, b, and c, and label the three bitangents that are not angle bisectors as x, y, and z, where x is the bitangent to the two circles that do not touch side a, y is the bitangent to the two circles that do not touch side b, and z is the bitangent to the two circles that do not touch side c. Then the ...
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.
Circle packing in a square is a packing problem in recreational mathematics, where the aim is to pack n unit circles into the smallest possible square. Equivalently, the problem is to arrange n points in a unit square aiming to get the greatest minimal separation, d n , between points. [ 1 ]
In trigonometry, Mollweide's formula is a pair of relationships between sides and angles in a triangle. [1] [2] A variant in more geometrical style was first published by Isaac Newton in 1707 and then by Friedrich Wilhelm von Oppel in 1746. Thomas Simpson published the now-standard expression in 1748.
The two circles tangent to these three circles are separated by the incircle, one interior to it and one exterior. The Soddy centers lie at the common intersections of three hyperbolas, each having two triangle vertices as foci and passing through the third vertex. [1] [2] [3]
Given a unit sphere, a "spherical triangle" on the surface of the sphere is defined by the great circles connecting three points u, v, and w on the sphere (shown at right). If the lengths of these three sides are a (from u to v ), b (from u to w ), and c (from v to w ), and the angle of the corner opposite c is C , then the (first) spherical ...
In general, the same inversion transforms the given line L and given circle C into two new circles, c 1 and c 2. Thus, the problem becomes that of finding a solution line tangent to the two inverted circles, which was solved above. There are four such lines, and re-inversion transforms them into the four solution circles of the Apollonius problem.
Then, the image of the -excircle under is a circle internally tangent to sides , and the circumcircle of , that is, the -mixtilinear incircle. Therefore, the A {\displaystyle A} -mixtilinear incircle exists and is unique, and a similar argument can prove the same for the mixtilinear incircles corresponding to B {\displaystyle B} and C ...