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[几何] $HH_aH_bH_c$为平行四边形

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hbghlyj Posted 2020-4-19 15:59 |Read mode
等面四面体ABCD,P是空间中一点,作P到各面∆ABC,DBC,DCA,DAB上的投影,再转移到∆ABC中去,得到$H,H_a,H_b,H_c$,求所有点P的集合,使四边形$HH_aH_bH_c$为平行四边形
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这道题我暂时没筭出来。但是容易得到下面的小结论:
P与A的对踬点A’重合时,$H_a$与垂心重合,$H,H_c,H_b$为弧中点.
P在AA’上运动时,$H_a$在欧拉线上运动,$H,H_c,H_b$在角平分线上运动.
P与D重合时,A与$H_a$重合,B与$H_b$重合,C与$H_c$重合,H与$X_{20}$(De Longchamps点)重合;
P与A重合时,A与H重合,B与$H_c$重合,C与$H_b$重合,$H_a$与$X_{20}$重合;
P与B重合时,A与$H_c$重合,B与H重合,C与$H_a$重合,$H_b$与$X_{20}$重合;
P与C重合时,A与$H_b$重合,B与$H_a$重合,C与H重合,$H_c$与$X_{20}$重合.
P在一双对棱AB,CD中点连线上运动时,$H_a,H_b$重合,$H,H_c$重合,这两点关于O对称,且在AB中垂线上.
P与四面体ABCD的重心(即外心)重合时,$O,H,H_a,H_b,H_c$重合.

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 Author| hbghlyj Posted 2022-3-3 08:26
Last edited by hbghlyj 2025-3-19 18:20The de Longchamps point ($L$) is the the orthocenter ($H$) reflected through the circumcenter ($C$).

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