+2115 (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|?
+394 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 \(|a_1-a_2|=\frac{1}{3}\).
+394 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 \(|a_1-a_2|=\frac{1}{3}\).
