Suppose that |a-b|+|b-c|+|c-a|=20. What is the maximum possible value of |a-b|?
ya that's right thats one of my AoPs questions and I had the same answer.
|a-b| = 20 - |b-c|-|c-a|
If b>c and c>a
|a-b| = 20 - b + c - c + a =20+(a-b)
if a>b then a-b = 20 + a-b
and a-b = 0
if a < b then b-a = 20 -b+a
2b-2a = 20
|a-b| = 10
try others but I think you will find 10
Ah, mathematics! The land of numbers and equations. Let's tackle this problem together, shall we?
If we're given that |a - b| + |b - c| + |c - a| = 20, we can start by thinking about the properties of absolute values. The absolute value of a number is always positive or zero, right? And since the sum of three absolute values is equal to 20, that means all three absolute values must be positive.
Now, to find the maximum possible value of |a - b|, we should first try to set its value as high as possible. Since the absolute value of a difference is positive, we can assume that a > b. This means that a - b is greater than zero.
If a - b is greater than zero, we can get rid of the absolute value signs around b - c and c - a. So, the equation simplifies to (a - b) + (b - c) + (c - a) = 20.
Now, notice that the (c - a) and (a - b) terms cancel each other out. That leaves us with just b - c = 20.
Since we want the maximum value of |a - b|, we want to find the highest value of b - c. If we let b - c = 20, we can see that the maximum value of |a - b| will be 20.
So, the maximum possible value of |a - b| is 20. I hope that puts a smile on your face!
To find the maximum possible value of |a-b|, we need to analyze different cases based on the given equation |a-b| + |b-c| + |c-a| = 20.
Case 1: a > b > c
In this case, let's assume a = b + x, where x is a positive number, and c = b - y, where y is a positive number. Substituting these values into the given equation, we get |x| + |y| + |x-y| = 20.
Now, the value of |a-b| = |b + x - b| = |x|. Since x is positive, the maximum value of |x| is when x = 10. In this case, |a-b| = 10.
Case 2: a > c > b
In this case, let's assume a = c + x, where x is a positive number, and b = c - y, where y is a positive number. Substituting these values into the given equation, we get |x+y| + |y| + |x| = 20.
Now, the value of |a-b| = |c + x - (c-y)| = |x+y|. Since x and y are both positive, the maximum value of |x+y| is when x = 10 and y = 10. In this case, |a-b| = 20.
Therefore, the maximum possible value of |a-b| is 20 when a > c > b.
To find the maximum possible value of |a-b|, we need to understand that the absolute value of a difference between two numbers represents the distance between those numbers on a number line.
Let's break down the given equation: |a-b| + |b-c| + |c-a| = 20.
Since absolute values always yield non-negative values, we can rewrite the equation as follows:
(a-b) + (b-c) + (c-a) = 20.
Now, let's simplify the equation:
a - b + b - c + c - a = 20.
By canceling out the common terms, we are left with:
-a + b - c + c - a = 20.
Simplifying further:
-2a + b - c = 20.
To find the maximum value of |a-b|, we should minimize the values of a and c while maximizing the value of b.
Considering this, let's assign the values of a = 0, b = 20, and c = 40 to satisfy the equation.
Plugging these values into the equation, we get:
|-20| + |20-40| + |40-0| = 20,
20 + 20 + 40 = 80.
Therefore, the maximum possible value of |a-b| is 20.