1970 AMC 12 Problem 20

Attempt Problem 20 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

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20.

Lines HKHK and BCBC lie in a plane. MM is the midpoint of line segment BC,BC, and BHBH and CKCK are perpendicular to HK.HK. Then we:
always have MH=MKMH=MK
always have MH>BKMH\gt BK
sometimes have MH=MKMH=MK but not always
always have MH>MBMH\gt MB
always have BH<BCBH\lt BC
Answer: A
Concepts:coordinate geometrymidpointperpendicular bisector
Small hint:
Put HKHK on the xx-axis
Big hint:
Use H=(h,0),H=(h,0), B=(h,u),B=(h,u), K=(k,0),K=(k,0), and C=(k,v)C=(k,v); compare squared distances from their midpoint
Solution:
Choose coordinates H=(h,0),B=(h,u),K=(k,0),C=(k,v). \begin{gathered} H=(h,0),\quad B=(h,u),\\ K=(k,0),\quad C=(k,v). \end{gathered} Then M=(h+k2,u+v2). M=\left(\frac{h+k}{2},\frac{u+v}{2}\right). The horizontal displacements from MM to HH and KK are opposites, while the vertical displacements are equal. Therefore MH2=MK2,MH^2=MK^2, so MH=MKMH=MK always.
Therefore, the correct answer is A.
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