利用者:ShuBraque/sandbox/アポロニウスの問題

交点理論

現代的な代数幾何学（特に交点理論）の技法を使用してアポロニウスの問題を解くことができる。このアプローチでは、アポロニウスの問題は複素射影平面（英語版）内の円に関する命題に再解釈される。複素数が絡む解であっても許容され、縮退した場合は重複度を込めて数えられる。これによって、アポロニウスの問題に対していつも8つの解を得ることができる^[1]。

$X$ 、 $Y$ 、 $Z$ で示される二次方程式はただ一つの円錐と軌跡の消失点を定める。逆に言えば、複素射影平面におけるあらゆる円錐は方程式によって表すことが可能であり、（方程式のスケールを変えてもその方程式の軌跡の消失点は変わらないため）この方程式は全体でひとつのスケール因子まで一意となる。そのため、すべての円錐の集合は5次の複素射影平面 $P 5$ によってパラメータ表示が可能であり、それを表すのが以下の式である。

\{[X:Y:Z]\in \mathbf {P} ^{2}\colon AX^{2}+BXY+CY^{2}+DXZ+EYZ+FZ^{2}=0\}\leftrightarrow [A:B:C:D:E:F]\in \mathbf {P} ^{5}.

複素射影平面上の円は2点 $O + = [1 : i : 0]$ と $O - = [1 : - i : 0]$ （ただし $i$ は $-1$ の平方根を示す）を通る円錐として定義される。この2点 $O +$ と $O -$ は Circular points と呼ばれる。すべての円の射影多様体はcircular pointsを通る円錐に対応する点から成る $P 5$ の部分多様体である。円錐の一般式にcircular pointsを代入することで次の式を得る。

A+Bi-C=0,

A-Bi-C=0.

式の変更を行うことで次の条件に等しいことが導かれる。

A=C

and

B=0

.

よって、すべての円のvarietyは $P 5$ の3次元線型部分空間である。スケールの変更と平方完成を行うと、circular pointsを通る任意の円錐は次の形式の方程式で表される。

(X-aZ)^{2}+(Y-bZ)^{2}=r^{2}Z^{2},

これはアフィン平面の円の通常の等式の斉次化である。そのため、上記の意味での円を議論するのは点灯機な意味での円を議論するのにほぼ等しい。唯一の違いは上記の意味での円は、２直線の和集合である円の重複を許容する点である。重複しない円は Smooth circles と呼ばれ、重複する円は Singular circles と呼ばれる。Singular circleには２種類あり、ひとつは $Z = 0$

There are two types of singular circles.  One is the union of the line at infinity  $Z = 0$  with another line in the projective plane (possibly the line at infinity again), and the other is union of two lines in the projective plane, one through each of the two circular points.  These are the limits of smooth circles as the radius  $r$  tends to  $+\infty$  and  $0$ , respectively.  In the latter case, no point on either of the two lines has real coordinates except for the origin  $[0 : 0 : 1]$ .

Let $D$ be a fixed smooth circle. If $C$ is any other circle, then, by the definition of a circle, $C$ and $D$ intersect at the circular points $O +$ and $O -$ . Because $C$ and $D$ are conics, Bézout's theorem implies $C$ and $D$ intersect in four points total, when those points are counted with the proper intersection multiplicity. That is, there are four points of intersection $O +$ , $O -$ , $P$ , and $Q$ , but some of these points might collide. Appolonius' problem is concerned with the situation where $P = Q$ , meaning that the intersection multiplicity at that point is $2$ ; if $P$ is also equal to a circular point, this should be interpreted as the intersection multiplicity being $3$ .

Let $Z D$ be the variety of circles tangent to $D$ . This variety is a quadric cone in the $P 3$ of all circles. To see this, consider the incidence correspondence

\Phi =\{(r,C)\in D\times \mathbf {P} ^{3}\colon C\ {\text{is tangent to}}\ D\ {\text{at}}\ r\}.

For a curve that is the vanishing locus of a single equation $f = 0$ , the condition that the curve meets $D$ at $r$ with multiplicity $m$ means that the Taylor series expansion of $f | D$ vanishes to order $m$ at $r$ ; it is therefore $m$ linear conditions on the coefficients of $f$ . This shows that, for each $r$ , the fiber of $Φ$ over $r$ is a $P 1$ cut out by two linear equations in the space of circles. Consequently, $Φ$ is irreducible of dimension $2$ . Since it is possible to exhibit a circle that is tangent to $D$ at only a single point, a generic element of $Z D$ must be tangent at only a single point. Therefore, the projection $Φ \to P 2$ sending $(r, C)$ to $C$ is a birational morphism. It follows that the image of $Φ$ , which is $Z D$ , is also irreducible and two dimensional.

To determine the shape of $Z D$ , fix two distinct circles $C 0$ and $C \infty$ , not necessarily tangent to $D$ . These two circles determine a pencil, meaning a line $L$ in the $P 3$ of circles. If the equations of $C 0$ and $C \infty$ are $f$ and $g$ , respectively, then the points on $L$ correspond to the circles whose equations are $Sf + Tg$ , where $[S : T]$ is a point of $P 1$ . The points where $L$ meets $Z D$ are precisely the circles in the pencil that are tangent to $D$ .

There are two possibilities for the number of points of intersections. One is that either $f$ or $g$ , say $f$ , is the equation for $D$ . In this case, $L$ is a line through $D$ . If $C \infty$ is tangent to $D$ , then so is every circle in the pencil, and therefore $L$ is contained in $Z D$ . The other possibility is that neither $f$ nor $g$ is the equation for $D$ . In this case, the function $(f / g)| D$ is a quotient of quadratics, neither of which vanishes identically. Therefore, it vanishes at two points and has poles at two points. These are the points in $C 0 \cap D$ and $C \infty \cap D$ , respectively, counted with multiplicity and with the circular points deducted. The rational function determines a morphism $D \to P 1$ of degree two. The fiber over $[S : T] \in P 1$ is the set of points $P$ for which $f (P) T = g (P) S$ . These are precisely the points at which the circle whose equation is $Tf - Sg$ meets $D$ . The branch points of this morphism are the circles tangent to $D$ . By the Riemann–Hurwitz formula, there are precisely two branch points, and therefore $L$ meets $Z D$ in two points. Together, these two possibilities for the intersection of $L$ and $Z D$ demonstrate that $Z D$ is a quadric cone. All such cones in $P 3$ are the same up to a change of coordinates, so this completely determines the shape of $Z D$ .

To conclude the argument, let $D 1$ , $D 2$ , and $D 3$ be three circles. If the intersection $Z D 1 \cap Z D 2 \cap Z D 3$ is finite, then it has degree $23 = 8$ , and therefore there are eight solutions to the problem of Apollonius, counted with multiplicity. To prove that the intersection is generically finite, consider the incidence correspondence

\Psi =\{(D_{1},D_{2},D_{3},C)\in (\mathbf {P} ^{3})^{4}\colon C\ {\text{is tangent to all}}\ D_{i}\}.

There is a morphism which projects $Ψ$ onto its final factor of $P 3$ . The fiber over $C$ is $Z C 3$ . This has dimension $6$ , so $Ψ$ has dimension $9$ . Because $(P 3) 3$ also has dimension $9$ , the generic fiber of the projection from $Ψ$ to the first three factors cannot have positive dimension. This proves that generically, there are eight solutions counted with multiplicity. Since it is possible to exhibit a configuration where the eight solutions are distinct, the generic configuration must have all eight solutions distinct.

^ Eisenbud, David and Harris, Joe, 3264 and All That: A Second Course in Algebraic Geometry. Cambridge University Press, 2016. ISBN 978-1107602724. pp. 66–68.

[eisenbud_harris_2016-1] Eisenbud, David and Harris, Joe, 3264 and All That: A Second Course in Algebraic Geometry. Cambridge University Press, 2016. ISBN 978-1107602724. pp. 66–68.

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