南瓜配对 I

Paired Pumpkins I

专题: General / 综合
难度: L2
来源: QuantQuestion

个人笔记

题目详情

Dracula 有 3 个南瓜，编号 1..3。已知任意两两重量和分别为 19、21、28（千克）。设重量为 $w_1,w_2,w_3$ 。

若可行，求 $w_1^2+w_2^2+w_3^2$ ；若不可行则输出 -1。

英文原题

Dracula has $3$ pumpkins, labeled $1-3$ . He knows the mass of each pair of pumpkins is (in kgs) $19, 21,$ and $28$ . Let $w_1, w_2,$ and $w_3$ be the weights of the three pumpkins. If possible, find $w_1^2 + w_2^2 + w_3^2$ . If this is impossible, enter $-1$ .

解析

由

w_1+w_2=19,\quad w_2+w_3=21,\quad w_1+w_3=28

解得 $w_1=13,w_2=6,w_3=15$ 。

因此

13^2+6^2+15^2=430.

英文解析

Using the notation here, we can set up the system of equations
$w_1 + w_2 = 19, \quad w_2 + w_3 = 21,$
and
$w_1 + w_3 = 28.$
By subtracting the second equation from the first, we get that
$w_1 - w_3 = -2.$
Adding this new equation to the third equation yields
$2w_1 = 26,$
meaning
$w_1 = 13.$
Substituting this back in, we get
$w_2 = 6$
and
$w_3 = 15.$
Therefore,
$w_1^2 + w_2^2 + w_3^2 = 13^2 + 6^2 + 15^2 = 430.$