线段AC上有一点B。以AB和BC为边,分别构造等边三角形XAB和YBC。证明,以下七个点共圆:点B,XC和YA的交点,以及线段AC、XC、YA、XB、YB的中点。
如图,S为AB的中点,T为BC的中点。首先注意到,△ABY和△XBC全等(SAS),也就是说△XBC是△ABY绕B点旋转60°得到的,因此AY和XC的夹角∠AMX=60°。注意到中位线LQ∥AX,中位线KQ∥CY,我们立即得到∠KQL也等于60°,于是K、L、M、Q四点共圆。
另外,中位线NL∥BC,中位线KP∥AB,因此线段NL、KP和AC全都平行。又中位线PT∥YC,YC∥XB,而XB与中位线LT平行,因此PT∥LT,即P、L、T三点共线。于是我们得知∠LPK=∠LTB=60°。类似地,∠NKP也等于60°,因此四边形KNLP是一个等腰梯形,这四个点共圆。而∠LPK也是弦KL所对的圆周角,因此这个圆和前面的那个圆是同一个圆,L、N、K、Q、P、M都在这个圆上。又∠NBP=60°=∠NKP,它们都是NP所对的圆周角,因此B也在这个圆上。
来源:http://www.cut-the-knot.org/Curriculum/Geometry/TwoEquilateralBumps.shtml
a sofa
先占个座位
翻译得不错 似乎那个网站很棒啊 看原文更有感觉 呵呵
看见SAS总是让我想起初中时无穷无尽的几何证明题
不错,感觉上这个很有意思。
我记得……狗狗给我说过个类似的……
=.=好像叫做九点共圆吧……
九点共圆~~诡异的东东~~MS和三角形滴几条线有关~
还有个“五点共圆”:
http://www.cyol.com/zqb/content/2001-04/06/content_200107.htm
《中国青年报》有个“五点共圆”问题:
http://www.cyol.com/zqb/content/2001-04/06/content_200107.htm
中国青年报 China Youth Daily 2001年4月6日 星期五
为什么网址贴不上来:
h t t p : / / w w w . c y o l . c o m / z q b / c o n t e n t / 2 0 0 1 – 0 4 / 0 6 / c o n t e n t _ 2 0 0 1 0 7 . h t m
看了那么多字母就晕了……
虽然能看懂……
NL∥AC∥KP , KN∥AX∥LQ∥PB , LP∥XB
NK=0.5(AX-BY)=0.5(XB-YC)=LP => NKPL等腰梯形 => NKPL共圆
NL=0.5BC , KQ=0.5YC , BP=0.5BY => NKQL等腰梯形,NBPL等腰梯形 => NKQL共圆,NBPL共圆
△YAB≌△CXB => ∠YAB=∠CXB => ∠LMK=∠XBA=∠KPL => KPML共圆
故NKQBPML共圆
NL∥AC∥KP , KN∥AX∥LQ∥PB , LP∥XB
NK=0.5(AX-BY)=0.5(XB-YC)=LP => NKPL等腰梯形 => NKPL共圆
NL=0.5BC , KQ=0.5YC , BP=0.5BY => NKQL等腰梯形,NBPL等腰梯形 => NKQL共圆,NBPL共圆
△YAB≌△CXB => ∠YAB=∠CXB => ∠LMK=∠XBA=∠KPL => KPML共圆
故NKQBPML共圆
话说其实不能算美妙,等边三角形和中点群太imba了..
NL//AC//KP , KN//AX//LQ//PB , LP//XB
NK=0.5(AX-BY)=0.5(XB-YC)=LP => NKPL等腰梯形 => NKPL共圆
NL=0.5BC , KQ=0.5YC , BP=0.5BY => NKQL等腰梯形,NBPL等腰梯形 => NKQL共圆,NBPL共圆
△YAB全等△CXB => ∠YAB=∠CXB => ∠LMK=∠XBA=∠KPL => KPML共圆
故NKQBPML共圆
这数学题是什么时候的,记得初中也有类似的吧。
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