原题:
等边三角形RQP中有一点离三顶点的距离分别为a, b, c, 边长为 x试证 x=u+v+w
证明:
所 以A,D,E,F 共线
AF=u+v+w
以AF为一边作等边三角形AFG,只要证明BG=b
我们来证明三角形BFG 和CFA 全等:
CF=BF=a, AF=FG,
故BG=AC=b
How do you know that x is unique? |
The original question only showed two pictures, and said to prove x=u+v+w. I think I should have interpreted the problem like following: There is a equilateral triangle with the length of u+v+w in whch there is point such that the distances from the point to the three vertices are a,b,c respectively. As in the book, there three proofs, no one proved the uniqueness. I think QFT's proof showed the uniqueness. |
not written a proof. It was hard for me to draw a nice picture, so I gave up. This problem is very cute. |
it for you! I am glad you like the problem. |
feel it is too easy to be bothered or you are lazy to write or to draw. |
I never knew how to prove geometry problems, so I don't even read them . Now I need to draw pictures about the cube kissing problem. |
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