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# help algebra

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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

#1
+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
+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 $$|a_1-a_2|=\frac{1}{3}$$.

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