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# Help!!!

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a.  Suppose that

|a - b| + |b - c| + |c - a| = 20.

What is the maximum possible value of |a-b|?

b.  Suppose that

|a - b| + |b - c| + |c - d| +....+ |m-n| + |n-o| +....+ |x - y| + |y - z| + |z - a| = 20

What is the maximum possible value of |a-n|?

p.s: the two lines mean absolute value.

Guest Aug 19, 2018
#1
+2

Part a)

It's simplest if you regard a, b, c as being on a line (say the x axis).

Then if c is between a and b, that is either $$a<=c<=b\text{ or } a>=c>=b$$, we have

$$|a - c| + |c - b| = |a - b|$$.

You may have to draw a picture to see this.

Otherwise, c is not between a and b and either |a - c| > |a - b| or |c - b| > |a - b| (or both!), so
$$|a - c| + |c - b| > |a - b|.$$

Again, a picture may help.

It follows from the above two cases that wherever c is in relation to a and b we must have:

$$|a - c| + |c - b| \ge |a - b|,$$

with equality if and only if c is between a and b.

Adding $$|a - b|$$ to both sides of this inequality, and remembering that $$|c-b|=|b-c|$$ etc, we see that

$$|a-b|+ |b - c| +|c-a| \ge2 |a - b|.$$

Therefore, if

$$|a-b|+ |b - c| +|c-a|=20$$

we must have

$$|a-b|\le10$$

with equality if and only if c is between a and b.

If no one else has done it I might look at part b tomorrow.

Guest Aug 19, 2018
#2
+93299
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Thanks for answering. I want to look at you answer more thoroughly when I have the time.

It looks interesting :)

Melody  Aug 20, 2018
#3
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Thanks for the encouragement: I had more or less written this one off as a dud!

The answer looks more complicated than it really is: it's a consequence of the fact that the circumference of a triangle must have at least twice the length of any one side – only in this case the triangle is degenerate (all on one line).

Guest Aug 20, 2018
#4
+88830
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|a - b| + |b - c| + |c - a| = 20

Let us suppose that  a > c > b      with  a, b and  c  >  0

So...we actually have

( a - b) + (c - b) + (a - c)  = 20    simplify

2a - 2b   = 20

a - b   = 10   and  since  a > b

Then

l a - b  l  = 10

CPhill  Aug 20, 2018
#6
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Would part be be 10 as well?

Guest Aug 20, 2018
#9
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@CPhill: You say "let us suppose that  $$a>c>b$$", but what if this isn't true? You show that that if you do suppose it you can achieve the limit, but you haven't proved that you can't do even better by supposing something else. This is the tricky bit!

Guest Aug 20, 2018
edited by Guest  Aug 20, 2018
#5
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Would part b be 10 as well?

Guest Aug 20, 2018
#7
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Short answer: yes. Long answer to come.

Guest Aug 20, 2018
#8
+1

Part b:

The answer to both parts really depends on the idea that the circumference of a polygon is at least twice the linear distance between any two vertices. So for the second part you just have to show that

$$|a - b| + |b - c| + |c - d| +\cdots+ |m-n|\ge |a-n|$$

and

$$|n-o| +\cdots+ |x - y| + |y - z| + |z - a| \ge |a-n|.$$

Then

$$|a - b| + |b - c| + |c - d| +\cdots+ |x - y| + |y - z| + |z - a|\ge 2|a-n|.$$

The details are essentially the same as in the previous case.

Guest Aug 20, 2018
edited by Guest  Aug 20, 2018