Suppose a,b,c,d are real numbers such that ∣a−b∣+∣c−d∣=99;∣a−c∣+∣b−d∣=1 Determine all possible values of ∣a−d∣+∣b−c∣.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
99 If w≥x≥y≥z are four arbitrary real numbers, then ∣w−z∣+∣x−y∣=∣w−y∣+∣x−z∣=w+x−y−z≥w−x+y−z=∣w−x∣+∣y−z∣. Thus, in our case, two of the three numbers ∣a−b∣+∣c−d∣,∣a−c∣+∣b−d∣,∣a−d∣+∣b−c∣ are equal, and the third one is less than or equal to these two. Since we have a 99 and a 1, the third number must be 99.
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