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[几何] 塞瓦三角形反演得到一个共点

本帖最后由 hbghlyj 于 2020-8-4 23:01 编辑

K是△ABC的欧拉线上一点,其塞瓦三角形为△DEF,D,E,F关于外接圆的反演点分别为M,N,L,证明: (1)AM,BN,CL共点K'; (2)K'的轨迹是一锥线.(by zjm)
(1)的证明:(by踏雪无痕)
200521000849253bb8d700100f(1).jpg
2020-8-4 17:56

设圆OCD再交AC于P,圆OBD再交AB于Q
$\because\frac{BM}{CM}=\frac{BD}{CD}=\frac{S_{\triangle AKB}}{S_{\triangle AKC}}=\frac{S_{\triangle AOB}+3 kS_{\triangle AGB}}{S_{\triangle AOC}+3 kS_{\triangle AGC}}=\frac{R^2 \sin 2 C+4 kR^2 \sin A \sin B\sin C}{R^2 \sin 2 B+4 {kR}^{2} \sin {A} \sin {B} \sin {C}}=\frac{ \sin 2 C+4 k \sin A \sin B\sin C}{ \sin 2 B+4 k \sin {A} \sin {B} \sin {C}},$
$\therefore\frac{2 DO'}{BC}=\frac{\sin 2 {C}-\sin 2 B}{\sin 2 {B}+\sin 2 {C}+8 {k} \sin {A} \sin {B} \sin C}=\frac{\cos {A} \sin ({B}-{C})}{\sin A\cos({B}-{C})+4 {k} \sin {A} \sin {B} \sin C}$
$\therefore2 {AQ}={c}+2 {O'D} \cdot \frac{\sin \angle ABO}{\sin \angle OBO'}={c}+{a} \frac{\cos {A} \sin ({B}-{C})}{\sin A \cos({B}-{C})+4 {k} \sin {A} \sin {B} \sin C} \cdot \frac{\cos C}{\cos A}={c}+{a} \frac{\cos {C} \sin ({B}-{C})}{\sin A \cos({B}-{C})+4 {k} \sin {A} \sin {B} \sin C}$
$\therefore\frac{\sin \angle {ABM}}{\sin \angle {ACM}}=\frac{\sin (\angle {ABO}+\angle{OBM})}{\sin (\angle {ACO}+\angle {OCM})}=\frac{\sin (\angle {ABO}+\angle {ODB})}{\sin (\angle {ACO}+\angle {ODC})}=\frac{\sin \angle {AOQ}}{\sin \angle {AOP}}=\frac{AQ}{AP}=\frac{c+a \frac{\cos C \sin (B-C)}{\sin A \cos (B-C)+4 k \sin A \sin B \sin C}}{b+a \frac{\cos B \sin (C-B)}{\sin A \cos (C-B)+4 k \sin A \sin B \sin C}}=\frac{\sin C+\frac{\cos C \sin (B-C)}{\cos (B-C)+4 k \sin B \sin C}}{\sin B-\frac{\cos B \sin (B-C)}{\cos (B-C)+4 k \sin B \sin C}}=\frac{\sin {C} \cos ({B}-{C})+4 {k} \sin {B} \sin ^{2} {C}+\cos {C} \sin ({B}-{C})}{\sin {B} \cos ({B}-{C})+4 {k} \sin ^{2} {B} \sin {C}-\cos {B} \sin ({B}-{C})}=\frac{\sin {B}}{\sin C} \cdot \frac{1+4 {k} \sin ^{2} C}{1+4 {k} \sin ^{2} B}$
$\because\frac{\sin \angle{BAM}}{\sin \angle {CAM}}= \frac{BM}{CM}\cdot \frac{\sin \angle {ABM}}{\sin \angle {ACM}}$
$\therefore\prod\frac{\sin \angle{BAM}}{\sin \angle {CAM}}=\prod \frac{BD}{CD} \cdot \prod \frac{\sin {B}}{\sin {C}} \cdot \prod \frac{1+4 k\sin ^2C}{1+4 k \sin ^2B}=1$,由塞瓦定理,AM,BN,CL共点K'

200521000849253bb8d700100f.jpg
2020-8-4 22:46

第二问怎么做呢?

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本帖最后由 hbghlyj 于 2020-8-6 02:51 编辑

题外话:
反过来考虑,使得AM,BN,CL共点的K的轨迹是欧拉线与外接圆的并集.
200521000849253bb8d700100f(1).jpg
2020-8-6 02:41

A'B'C'为O的垂足三角形,PA,PB,PC关于B'C',C'A',A'B'的反射像共点Q,当且仅当P∈耶拉贝克双曲线∪无穷远线,当且仅当Q∈欧拉线∪外接圆
设T为P的等角共轭,我们就得到一个欧拉线到自身的双射Q↔T
这个映射的不动点是X5000和X5001,即欧拉线与外接圆的交点
把O换成别的点,P的集合是什么?比如O=H
见"风信子"429
设欧拉线上一点T满足HT:TO=t,则Q满足HQ:QO=t'=k/(k+t),其中k = 4 cosA cosB cosC. 则t'=t当且仅当T=X5000或X5001.
见"风信子"431

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