Forgot password?
 Create new account
Search
View: 78|Reply: 9

[几何] 圆I是内切圆,圆P是旁切圆,求作经过C点并与前两圆都相切的圆

[Copy link]

120

Threads

448

Posts

3195

Credits

Credits
3195

Show all posts

TSC999 Post time 2023-9-17 20:41 |Read mode
I 是△ABC 的内心,P 是与 BC 边相切的旁切圆圆心。问如何作出圆D,它与圆I、圆P 都相切并且通过 C 点。
如何作圆D.png

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-17 21:09

PCC

本帖最后由 hbghlyj 于 2023-9-19 09:55 编辑 Problem of Apollonius - PCC
cut-the-knot PCC: Apollonius' Problem with Two Circles and a Point
MSE Apollonius's Problem: why PCC can be reduced to PPC

第一种作法见3#
第二种作法见4#,5#,6#
第三种作法见7#,最简单

Comments

链接网打不开。不知何故?  Post time 2023-9-18 11:04

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-18 16:15
本帖最后由 hbghlyj 于 2023-9-19 09:51 编辑 Wikipedia摘抄:
Shrinking one given circle to a point
In the first approach, the given circles are shrunk or swelled (appropriately to their tangency) until one given circle is shrunk to a point P.[37] In that case, Apollonius' problem degenerates to the CCP limiting case, which is the problem of finding a solution circle tangent to the two remaining given circles that passes through the point P. Inversion in a circle centered on P transforms the two given circles into new circles, and the solution circle into a line. Therefore, the transformed solution is a line that is tangent to the two transformed given circles. There are four such solution lines, which may be constructed from the external and internal homothetic centers of the two circles. Re-inversion in P and undoing the resizing transforms such a solution line into the desired solution circle of the original Apollonius problem. All eight general solutions can be obtained by shrinking and swelling the circles according to the differing internal and external tangencies of each solution; however, different given circles may be shrunk to a point for different solutions.

根據這段描述,得出第1种作法:以$C$为中心、任意长为半徑作圆,圆$I$、$P$关于它的反演为圆$I'$、$P'$,作出圆$I'$、$P'$的4條公切线$L_1,L_2,L_3,L_4$,其中$L_2$就是直線$AC$、$L_4$就是直線$BC$
Untitled.png
再反演回去,得到$L_1',L_2',L_3',L_4'$就是4个切圆,其中$L_1'$就是圆$D$,剩下的分別是直线$BC$、$AC$和另一个圆$L_3'$,它与圆$I$外切、与圆$P$内切。
Untitled.png

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-18 16:33
近代欧氏几何学 [美]约翰逊 著 2012年版
近代欧氏几何学 [(美)约翰逊 著] 2012年版_Page_092.jpg 近代欧氏几何学 [(美)约翰逊 著] 2012年版_Page_093.jpg

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-18 17:33
TSC999 发表于 2023-9-17 20:41
I 是△ABC 的内心,P 是与 BC 边相切的旁切圆圆心。问如何作出圆D,它与圆I、圆P 都相切并且通过 C 点。

如图标记$E,F,G,H$切点,$C'$为圆$D$与$AC$的交点,
则$AGH$共线,
由$AC\cdot AC'=AG\cdot AH=AE\cdot AF$可作出$C'$
Untitled.png

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-18 17:46
本帖最后由 hbghlyj 于 2023-9-19 09:52 编辑 得到$C'$后转化为§169问题(作一个圆通过两个已知点,并且与一个已知圆相切)
作一个圆通过两个点C,C'并且与圆I相切:
过$C,C'$作圆与圆$I$相交于两点,
连接这两个交点与$CC'$交于$O$,
过$O$作圆$I$切线,切点为$Q$,
过$C$、$C'$、$Q$作出圆$D$
Untitled.png

$type tid11444.ggb (20.64 KB, Downloads: 0)

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-18 19:09
本帖最后由 hbghlyj 于 2023-9-19 11:22 编辑
《近代欧氏几何学》§49:以三个圆的根心为中心可作一圆,与三个圆都正交.
§171:关于这个公共正交圆的反演保持三个已知圆在原地不动,而公切圆互换.
又见Pairs of solutions by inversion
Untitled.png
由此得出第3种作法:从CE中点Y引CI的垂线,从CF中点Z引CP的垂线,交于X.
以X为中心、CX为半径作圆,则AC关于它的反演就是所求的圆D,BC的反演是另一个公切圆.
Untitled.png

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-19 09:59
本帖最后由 hbghlyj 于 2023-9-19 11:45 编辑 从3#、7#都看出有4个解。
见Wikipedia Table 1的第7行:一般的CCP有4个解。
一般的Apollonius问题可能有0,1,2,3,4,5,6,8个解
不可能7个解的证明
Pedoe 1970
The circle $C_0$ which is orthogonal to each of the $C_i$ is uniquely defined, unless the $C_i$ belong to a coaxal system (pencil) of circles. In the latter case the only tangent circles are two point-circles, in the case when the coaxal system is of the intersecting type. The circle $C_0$ plays a very special rôle with regard to the $C_i$ since inversion (transformation by reciprocal radii) in $C_0$ maps each $C_i$ onto itself, and maps a tangent circle $C$ onto a tangent circle $C^{\prime}$. When there are 8 tangent circles (which may be called the general case) these can be split into 4 pairs. We shall call the circles in a pair conjugate circles. (For all this, proved algebraically, see Pedoe [3]). If we wish to specialize the $C_i$ so that there are only 7 tangent circles, the specialization must aim at making a pair of conjugate circles identical, since if two tangent circles which are not conjugate become identical the conjugates also become identical, and the number of tangent circles would reduce to 6, at most.

We therefore specialize the $C_i$ so that a conjugate pair $C$ and $C^{\prime}$ become the same circle $D$, say. This means that inversion in $C_0$ maps the tangent circle $D$ onto itself. If this is the case, $D$ must be orthogonal to $C_0$. We therefore find ourselves with three circles $C_i$, a circle $C_0$ orthogonal to the $C_i$, and a circle $D$ which touches the $C_i$ and is also orthogonal to $C_0$. We show that this means that two of the $C_i$ touch each other.

Invert with respect to a centre of inversion on $C_0$. We obtain three circles $C_i^{\prime}$, with diameters which lie along the line $C_0^{\prime}$.

These three circles are touched by a circle $D^{\prime}$ whose diameter also lies along $C_0^{\prime}$. If two circles with diameters along the same line touch at a point not on this line, they have the same centre, and must therefore coincide. If the circles are distinct contact can only take place at an endpoint of a diameter. Since $D^{\prime}$ has only 2 points of intersection with the line $C_0^{\prime}$, and has to touch each of $C_1^{\prime}, C_2^{\prime}$ and $C_3^{\prime}$ at a point on $C_0^{\prime}$, the three points of contact cannot be distinct. Hence at least two of the circles $C_i^{\prime}$ intersect $C_0^{\prime}$ at the same point. That is, at least two of the circles $C_i^{\prime}$ touch each other. But if at least two of the circles $C_i$ touch each other, the number of circles tangent to the three $C_i$ is readily seen to be 6, at most.

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-19 10:01
本帖最后由 hbghlyj 于 2023-9-19 12:58 编辑 \[圓D的半徑=\frac{a^2 (2 b-c)+(b-c)^2 (2 b+c)}{16 S}\]
推導
7#圖中$YZ=\frac{EF}2=\frac a2$
直角三角形$XYZ$中,$YZ=\frac a2$,$\angle XZY=\frac C2$,得出$X$到$YZ$的距離$$XH=XY\cos\frac C2=YZ\sin\frac C2\cos\frac C2=\frac{YZ\sin C}2=\frac{a\sin C}4$$
$YH=XY\sin\frac C2=YZ\sin^2\frac C2=\frac{a\sin^2\frac C2}2=\frac{a(1-\cos C)}4$
$HC=YC-YH=\frac{(a+b-c)-a(1-\cos C)}4=\frac{b-c}4+\frac{a\cos C}4$
$XC^2=XH^2+HC^2=(\frac{a\sin C}4)^2+(\frac{b-c}4+\frac{a\cos C}4)^2=(\frac a4)^2+(\frac{b-c}4)^2+\frac{(b-c)a\cos C}8$
\[圓D的半徑=\frac{XC^2}{2XH}\]
代入$2XH=\frac{a\sin C}2=\frac{ab\sin C}{2b}=\frac{S}b$、$\cos C=\frac{a^2+b^2-c^2}{2ab}$得
\[圓D的半徑={(\frac a4)^2+(\frac{b-c}4)^2+\frac{(b-c)a\frac{a^2+b^2-c^2}{2ab}}8\over\frac{S}b}=\frac{a^2 (2 b-c)+(b-c)^2 (2 b+c)}{16 S}\]

同理,記7#圖中較大紅色圓為圓$D_1$,则
\[圓D_1的半徑=\frac{a^2 (2 b-c)+(b-c)^2 (2 b+c)}{16 S |b-c|/a}\]
推導
如圖標記$E_1,F_1,Y_1,Z_1,H_1$,则
\[圓D_1的半徑=\frac{XC^2}{2XH_1}\]前面已算出$XC^2$,只需算出$XH_1$
直角三角形$XY_1Z_1$中,$Y_1Z_1=\frac{E_1F_1}2=\frac{|b-c|}2$,$\angle XZ_1Y_1=\frac C2$,得出$X$到$Y_1Z_1$的距離$$XH_1=XY_1\cos\frac C2=Y_1Z_1\sin\frac C2\cos\frac C2=\frac{Y_1Z_1\sin C}2=\frac{|b-c|\sin C}4=\frac{|b-c|S}{2ab}$$
代入前面的公式
\[圓D_1的半徑={(\frac a4)^2+(\frac{b-c}4)^2+\frac{(b-c)a\frac{a^2+b^2-c^2}{2ab}}8\over\frac{S|b-c|}{ab}}=\frac{a^2 (2 b-c)+(b-c)^2 (2 b+c)}{16 S|b-c|/a}\]
Untitled.png

手机版|悠闲数学娱乐论坛(第3版)

2025-3-6 21:17 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list