线段AB=6,C是AB上动点,线段AC绕A逆旋转120度,线段BC绕B顺旋转120,DE最小值?

悟玄子_506 |浏览993次
收藏|2022/07/15 17:01

满意回答

2022/07/15 17:09

∵ AC绕A逆时针旋转120°得AD,∴ AD=AC,∠CAD=120°。∴∠CAD=∠ACD=(180°-∠CAD)/2=30°∴ CD=sin120°AC/sin30°=AC√3。同理∠BCE=30°,CE=BC√3。∴∠DCE=180°-∠ACD-∠BCE=120°。设P是AB中点,AC=AB/2-PC=3-PC,∴ BC=AB/2+PC=3+PC。∴ DE²=CD²+CE²-2CD*CEcos∠DCE=3AC²+3BC²+3AC*BC=3(3-PC)²+3(3+PC)²+3(3-PC)(3+PC)=3(27+PC²)≥81等号当且仅当PC=0时成立。当C与P重合,即C是AB中点时,DE取最小值√81=9。

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其他回答(1)
  • 设C为坐标系原点;DC =√3AC;CE =√3BC = √3( 6 - AC ) = 6√3 - √3AC;点D坐标 x1 = -DCcos30° = -√3AC *√3/2 = -3AC/2;y1 =DCsin30° =√3AC/2;点E坐标 x2 = CEcos30° = ( 6√3 - √3AC )* √3/2 = 9 - 3AC/2;y2 = CEsin30° = ( 6√3 - √3AC ) /2 = 3√3 - √3AC/2;ED^2 = ( x1 - x2 )^2 + ( y1 - y2 )^2= [√3AC/2 - (9 - 3AC/2 ) ]^2 + [√3AC/2 - (3√3 - √3AC/2 ) ]^2= 81 + 3AC^2 - 18AC + 27= 3[ ( AC^2 - 6AC + 9 ) + 27 ]=3[ ( AC - 3 )^2 + 27 ];故当AC = 3时,ED^2有最小值 3 * 27 = 81;ED最小值为 9 。
    回答于 2022/07/15 17:21
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