User:Jacobolus/Descartes

A draft fork of part of Descartes' theorem

From triangles

Given three points in the plane in general position (a scalene triangle), it is possible to construct a triple of mutually tangent circles centered at the respective points in four distinct ways; in each case, by Descartes' theorem, these three circles define a pair of additional circles which are tangent to all three, sometimes called the Soddy circles of the triangle.^[1]

Letting $A,B,C$ be the three points, $a,b,c$ be the lengths of the opposite sides, and ${\textstyle s={\tfrac {1}{2}}(a+b+c)}$ be the semiperimeter, the radii of the respective circles in each triple are: $(s-a,s-b,s-c),$ $(-s,s-c,s-b),$ $(s-c,-s,s-a),$ $(s-b,s-a,-s),$ where a negative radius indicates that the circle is tangent to the other two in its interior.^[1]

For the first triple the three circles are externally tangent to each-other. Each of the three circles is orthogonal to two of the sides of the triangle (and does not intersect the third), and all three are orthogonal to the incircle of the triangle. The three points of tangency between the circles are also the intouch points of the triangle, where the incircle is tangent to each respective side. The two Soddy centers are at triple intersections of the three hyperbolae with two triangle vertices for foci and passing through the third vertex.^[2]^[3]

One of the Soddy circles is in the interior of the incircle and the other is in the exterior. The inner Soddy center is always on the branch of the hyperbolae consisting of points $P$ satisfying the equations $|AP|-|BP|=|AC|-|BC|,$ $|BP|-|CP|=|BA|-|CA|,$ and $|AP|-|CP|=|AB|-|CB|,$ making it an equal detour point: the shortest path between any two triangle vertices passing through the point is longer than the direct path between those vertices by an amount that does not depend on which two vertices are chosen.^[4]

Descartes' theorem can be used to relate the inner Soddy circle's curvature to the triangle's area $\Delta$ , semiperimeter $s$ , circumradius $R$ , and inradius $r.$ In terms of these variables, the inner curvature is ${\textstyle (4R+r+2s)/\Delta .}$ ^[5]

The outer Soddy circle has curvature ${\textstyle (4R+r-2s)/\Delta .}$ Depending on the position of the three points, this can be positive (externally tangent to the three circles), zero (degenerate to a straight line), or negative (internally tangent).^[5] When the curvature is positive, its center is on the same branch of the three hyperbolae as the inner Soddy center, and is also an equal detour point; otherwise the equal detour point is unique.^[4] When the outer Soddy circle has negative curvature, the three hyperbolae have their second intersection on the other branch where $|AP|-|BP|=|BC|-|AC|,$ $|BP|-|CP|=|CA|-|BA|,$ and $|AP|-|CP|=|CB|-|AB|,$ making the outer Soddy center the isoperimetric point of the triangle: the three triangles formed by this center and two vertices of the starting triangle all have the same perimeter.^[4] Triangles where the outer circle degenerates to a straight line (zero curvature) have been called "Soddyian triangles".^[5]

The line passing through both Soddy centers is called the Soddy line. It also passes through the incenter of the triangle, which is the homothetic center of the two Soddy circles,^[1] as well as several other triangle centers:^[6]

The three lines connecting the intouch points of the triangle to the opposite vertices concur at the Gergonne point, which lies on the Soddy line.
The reflection of the orthocenter of the triangle about the circumcenter is the de Longchamps point, which lies on the Soddy line.^[6]
Four mutually tangent circles define six points of tangency, which can be grouped in three pairs of tangent points, each pair coming from two disjoint pairs of circles. The three lines through these three pairs of tangent points are concurrent, and the points of concurrency defined in this way from the inner and outer circles define two more triangle centers called the Eppstein points that also lie on the Soddy line.^[6]^[7]

Instead of forming three externally tangent circles centered on the vertices of a triangle, one of the circles can enclose the other two, in three different ways. The pairs of tangent circles to these three triples of circles behave in analogous ways to the pair of inner and outer circles. In particular, the lines through their pairs of centers, analogous to the Soddy line, all meet in the de Longchamps point. Each pair of tangent circles has one of the excenters of the triangle as a center of similitude.^[1]

Proof idea

One of the many proofs of Descartes' theorem uses this connection to triangle geometry, and Heron's formula, that a triangle with sides $a$ , $b$ , and $c$ , and semiperimeter $s$ , has area ${\textstyle {\sqrt {s(s-a)(s-b)(s-c)}}}$ . When a triangle is formed by the centers of three externally tangent circles, its sides are the sums of two of the circle radii, and its semiperimeter is the sum of all three. The centers of four externally tangent circles of radii $r_{i}$ ( $i\in \{1,2,3,4\}$ ) form four triangles, whose areas are given by this formula as

${\sqrt {r_{i}r_{j}r_{k}(r_{i}+r_{j}+r_{k})}}$

for each of the four ways of choosing three indexes from the four. One of the centers must within the triangle formed by the other three centers. That triangle is subdivided into three parts by the other three triangles, and its area equals the sum of the areas of the parts. Thus, for instance, if circle 4 lies within the triangle of the other three circle centers, this leads to the equality

\displaystyle {\begin{aligned}{\sqrt {r_{1}r_{2}r_{3}(r_{1}+r_{2}+r_{3})}}&{}={\sqrt {r_{1}r_{2}r_{4}(r_{1}+r_{2}+r_{4})}}\\&{}+{\sqrt {r_{1}r_{3}r_{4}(r_{1}+r_{3}+r_{4})}}\\&{}+{\sqrt {r_{2}r_{3}r_{4}(r_{1}+r_{3}+r_{4})}}.\\\end{aligned}}

The square roots can be eliminated by careful squaring and algebraic manipulation, giving the formula in Descartes' theorem. The case in which one of the circles is internally tangent to the other three (possibly with four centers in a convex quadrilateral, covered in two ways by pairs of triangles) uses similar reasoning. This method does not apply directly to the cases of the theorem in which one of the circles degenerates to a line, but those can be handled as a limiting case of circles.^[8]

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