Maths Olympiad Prep

Track / Stage 5 / 377 of 400 #977 of 1964

Problem 977

AIME late
Algebra Difficulty 5.9 Prove it

Three. (25 points) Given five numbers x1,x2,x3,x4,x5x_{1}, x_{2}, x_{3}, x_{4}, x_{5} (x1⩽x2⩽x3⩽x4⩽x5)\left(x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}\right), find the sum of every two numbers, resulting in 10 numbers, denoted as a1,a2,⋯ ,a10a_{1}, a_{2}, \cdots, a_{10}. Can the original five numbers x1,x2,x3,x4,x5x_{1}, x_{2}, x_{3}, x_{4}, x_{5} be determined if the 10 numbers a1,a2,⋯ ,a10a_{1}, a_{2}, \cdots, a_{10} are known (but it is not known which two xi(i=1,2,⋯ ,5)x_{i}(i=1,2, \cdots, 5) each aia_{i} is the sum of)? Please explain your reasoning.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Three, let S=14(a1+a2+⋯+a10)S=\frac{1}{4}\left(a_{1}+a_{2}+\cdots+a_{10}\right).
From the problem, SS is equal to the sum of the original 5 numbers x1,x2,x3,x4,x5x_{1}, x_{2}, x_{3}, x_{4}, x_{5}.
Given x1⩽x2⩽x3⩽x4⩽x5x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}, we have
x1+x2⩽x1+x3⩽⋯⩽x3+x5⩽x4+x5.  x_{1}+x_{2} \leqslant x_{1}+x_{3} \leqslant \cdots \leqslant x_{3}+x_{5} \leqslant x_{4}+x_{5} \text {. }

If we arrange a1,a2,⋯ ,a10a_{1}, a_{2}, \cdots, a_{10} in ascending order, let's assume a1⩽a2⩽⋯⩽a9⩽a10a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{9} \leqslant a_{10}, then
a1=x1+x2,a2=x1+x3,⋯⋯a9=x3+x5,a10=x4+x5. Hence x3=(x1+x2+x3+x4+x5)−(x1+x2+x4+x5)=S−(a1+a10). \begin{array}{l} a_{1}=x_{1}+x_{2}, a_{2}=x_{1}+x_{3}, \\ \cdots \cdots \\ a_{9}=x_{3}+x_{5}, a_{10}=x_{4}+x_{5} . \\ \text { Hence } x_{3}=\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)- \\ \quad\left(x_{1}+x_{2}+x_{4}+x_{5}\right) \\ =S-\left(a_{1}+a_{10}\right) . \end{array}

Furthermore, x1=a2−x3=a2−S+a1+a10x_{1}=a_{2}-x_{3}=a_{2}-S+a_{1}+a_{10},
x2=a1−x1=a1−a2+S−a1−a10=S−a2−a10,x5=a9−x3=a9−S+a1+a10,x4=a10−x5=a10−a9+S−a1−a10=S−a1−a9. \begin{array}{l} x_{2}=a_{1}-x_{1}=a_{1}-a_{2}+S-a_{1}-a_{10}=S-a_{2}-a_{10}, \\ x_{5}=a_{9}-x_{3}=a_{9}-S+a_{1}+a_{10}, \\ x_{4}=a_{10}-x_{5}=a_{10}-a_{9}+S-a_{1}-a_{10}=S-a_{1}-a_{9} . \end{array}

Therefore, if we know a1,a2,⋯ ,a10a_{1}, a_{2}, \cdots, a_{10}, we can definitely find the original 5 numbers x1,x2,x3,x4,x5x_{1}, x_{2}, x_{3}, x_{4}, x_{5}.
(Li Yaowen, Shuzhuang No. 40 Middle School, Shandong Province, 277200)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.