Question

已知线段AB=m,在直线AB上取一点P,恰好使AP/PB =n,点Q为PB的中点，求线段AQ的长。

Answer 1

AP=M*n/(n+1)
BP=M*1/(n+1)
AQ=M*n/(n+1)+(1/2)M*1/(n+1)
=M*(2n+1)/(2n+2)

参考思路

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过程

追答

设BP为x ，ap=nx ab=nx+x=m，x=bp=m*1/(n+1)
同理就能求ap=m*n/(n+1)

Q为PB的中点 所以pq=1/2*m*1/(n+1)

aq=ap+pq=m*n/(n+1)+1/2*m*1/(n+1)=m*(2n+1)/(2n+2)