35. Suppose $a, b, c, d$ are real numbers such that
$$
|a-b|+|c-d|=99 ; \quad|a-c|+|b-d|=1 .
$$

Determine all possible values of $|a-d|+|b-c|$.
Solution: 99 If $w \geq x \geq y \geq z$ are four arbitrary real numbers, then $|w-z|+|x-y|=$ $|w-y|+|x-z|=w+x-y-z \geq w-x+y-z=|w-x|+|y-z|$. 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 .