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(1) Let a_1, a_2, a_3 be real numbers such that
|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_1| = 1.
What is the largest possible value of |a_1 - a_2|?


(2) Let a_1, a_2, a_3, \dots, a_{10} be real numbers such that
|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_4| + 4 |a_4 - a_5 | + \dots + 9 |a_9 - a_{10}| + 10 |a_{10} - a_1| = 1.
What is the largest possible value of |a_1 - a_6|?

 Mar 9, 2024

Best Answer 

 #1
avatar+394 
+4

1)

To simplify this equation, we want to eliminate some of the terms. Set 

\(x=a_1-a_2\)

\(y=a_2-a_3\)

Therefore:

\(a_3-a_1=-(a_1-a_3)=-(x+y)=-x-y\).

Now our equation is:

\(|x|+2|y|+3|-x-y|=1\)

\(|x|+2|y|+3|x+y|=1\).

One way to approach this is by graphing. 

We split this into 3 sections:

One for \(x \ge 0, x < 0\)

\(y\ge0, y<0\)

For \(|x+y|\) we have

\(x>-y, x < -y\).

Here is a graph, with \(x=0, y=0, x=-y\) labeled. (The green one is \(|x|+2|y|+3|x+y|=1\)).

We see the largest possible value of \(|x|\) in this graph is \(\frac{1}{3}\).

Therefore, the maximum value is \(|a_1-a_2|=\frac{1}{3}\).

 Mar 9, 2024
edited by hairyberry  Mar 9, 2024
 #1
avatar+394 
+4
Best Answer

1)

To simplify this equation, we want to eliminate some of the terms. Set 

\(x=a_1-a_2\)

\(y=a_2-a_3\)

Therefore:

\(a_3-a_1=-(a_1-a_3)=-(x+y)=-x-y\).

Now our equation is:

\(|x|+2|y|+3|-x-y|=1\)

\(|x|+2|y|+3|x+y|=1\).

One way to approach this is by graphing. 

We split this into 3 sections:

One for \(x \ge 0, x < 0\)

\(y\ge0, y<0\)

For \(|x+y|\) we have

\(x>-y, x < -y\).

Here is a graph, with \(x=0, y=0, x=-y\) labeled. (The green one is \(|x|+2|y|+3|x+y|=1\)).

We see the largest possible value of \(|x|\) in this graph is \(\frac{1}{3}\).

Therefore, the maximum value is \(|a_1-a_2|=\frac{1}{3}\).

hairyberry Mar 9, 2024
edited by hairyberry  Mar 9, 2024

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