My Math Forum Straight wires and average

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 March 27th, 2014, 01:38 AM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 429 Thanks: 18 Straight wires and average Question) If you are given two straight pieces of wire, is it possible to cut one of them into two pieces so that the length of one of the three pieces is the average of the lengths of the other two? Explain My attempt: The question doesn't say whether the wires are equal or unequal, so I have to check both cases Case 1: both straight wires are of equal length = a |-----(x)------|----(a - x)-----| wire 1 |------------(a)-----------------| wire 2 Situation 1: a = [x + (a - x)]/2 then 2a = a which is impossible Situation 2: x = [a + (a - x)]/2 then x = 2a/3 which is acceptable Situation 3: a - x = (a + x)/2 then x = a/3 This is acceptable So it is possible to cut one of two equal length wires in such a way that one is the average of the other two. It can be done by cutting at 1/3 or 2/3 of the length. Case 2: both straight wires are of unequal length I tried each situation as above and there was no contradiction. So if the wires are unequal, there's no problem in the task of cutting one in such a way that one piece is the average of the other. Am I right? Is there a better way to do this problem? Is there an underlying mathematical idea behind the problem? Thanks
March 27th, 2014, 07:07 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Straight wires and average

Hello, shunya!

Quote:
 If you are given two straight pieces of wire, is it possible to cut one of them into two pieces so that the length of one of the three pieces is the average of the lengths of the other two?

$\text{W\!e are given two wire of lengths }a\text{ and }b.
\text{Cut the }b\text{-wire into two parts: }\,x\text{ and }b-x.$

Code:
      : - - a - - :
*-----------*

: - - - - b - - - - - :
*-------+-------------*
: - x - : - - b-x - - :
$\text{Suppose }x\text{ is the average of the other two pieces.}
\text{W\!e have: }\:x \:=\:\frac{a\,+\,(b-x)}{2} \;\;\;\Rightarrow\;\;\;2x \:=\:a\,+\,b\,-\,x \;\;\;\Rightarrow\;\;\;3x \:=\:a\,+\,b$

$\text{Therefore: }\:x \;=\;\frac{a\,+\,b}{3}\;.\;.\;.\text{ Yes, it is possible!}$

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