ok... let's say, the pieces are "a", "b", and "c". "a" being the shortest and "c" being the longest we know the longest is 38 more than the shortest, so c = a + 38 and the third piece, b, is half the longest, or c/2 we know the three pieces come from the 102in ribbon, thus a + b + c = 102 $\bf a+b+c=102\quad \begin{cases} c=a+38\\ b=\frac{c}{2}\\ \quad =\frac{a+38}{2} \end{cases}\implies a+(a+38)+\left( \frac{a+38}{2} \right)=102$ solve for "a". c = a + 38, and b = c/2