What are vectors, by the way?
This is the last problem of filling-blanks section for a time-pressing exam and has defeated many, many Chinese students without truly understanding of knowledge. But in fact, it's a level 1 problem for you folks if you have taken the courses of brilliant.org and truly understand the essence of vectors. So let's give it a shot why not?
A square A B C D has the side length 1 , and for every possible combinations of λ i ( i = 1 , 2 , 3 , 4 , 5 , 6 ) where λ i = 1 or λ i = − 1 , the expression ∣ λ 1 A B + λ 2 B C + λ 3 C D + λ 4 D A + λ 5 A C + λ 6 B D ∣ has the maximum value M and the minimum value N .
Submit ⌊ 1 0 0 0 ( M − N ) ⌋ .
Note: ∣ n ∣ notes the length of the vector on the Euclidean plane. i.e. The Euclidean norm.
The answer is 4472.
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2 solutions
The problem statement requires the reduction of a vector to a real number and that requires a normed vector space . Wiikipedia also listed a number of applicable norms .
Using a few samples of the applicable norms available in Wolfram Mathematica , the following answers to the problem result: ⎝ ⎜ ⎜ ⎛ 1 2 3 ∞ 6 0 0 0 . 4 4 7 2 . 1 4 4 1 6 0 . 1 7 4 0 0 0 . ⎠ ⎟ ⎟ ⎞
Since this solution was posted, the author added a note clarifying what norm is being used.
ab = { 0 , 1 } ; bc = { 1 , 0 } ; cd = { 0 , − 1 } ; da = { − 1 , 0 } ; ac = { 1 , 1 } ; bd = { 1 , − 1 } ;
v = Flatten [ Table [ ab i + bc j + cd k + da l + ac m + bd n , { i , { − 1 , 1 } } , { j , { − 1 , 1 } } , { k , { − 1 , 1 } } , { l , { − 1 , 1 } } , { m , { − 1 , 1 } } , { n , { − 1 , 1 } } ] , 5 ]
⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ − 2 0 0 2 − 4 − 2 − 2 0 − 2 0 0 2 − 4 − 2 − 2 0 0 2 2 4 − 2 0 0 2 0 2 2 4 − 2 0 0 2 − 2 0 0 2 − 4 − 2 − 2 0 − 2 0 0 2 − 4 − 2 − 2 0 0 2 2 4 − 2 0 0 2 0 2 2 4 − 2 0 0 2 0 − 2 2 0 0 − 2 2 0 − 2 − 4 0 − 2 − 2 − 4 0 − 2 0 − 2 2 0 0 − 2 2 0 − 2 − 4 0 − 2 − 2 − 4 0 − 2 2 0 4 2 2 0 4 2 0 − 2 2 0 0 − 2 2 0 2 0 4 2 2 0 4 2 0 − 2 2 0 0 − 2 2 0 ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
Table [ { p , 1 0 0 0 N [ max ( Table [ ∥ t ∥ p , { t , v } ] ) ] } , { p , { 1 , 2 , 3 , ∞ } } ] which generates the answer table above.
Very sorry about that. I should have noted the norm is the Euclidean norm (because in senior high courses in China we use it as the definition of the norm). Knowing that, can you think of a better solution to make this problem a level 1 one?
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What was your approach??
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Hint: Treat vectors as an act of the movement of the points.
Your note is quite sufficient. Yes, it is thye usual norm. At Brilliant, often enough authors use concepts not realizing that multiple definitions exist or that areas of the world use different conventions. Having worked in massively multi-lingual, multi-cultural environments (AT&T Bell Laboratories in the 1980s and Fermi National Acceleration Laboratoryin the 1990s and 2000s), I encountered such confusions many times. With your note, I will delete my report. Your note is the correct response to my report. A comment on a report is not sufficient as the reporter can not see that comment until the problem is answered or failed.
In Wolfram Mathematica, Norm[vectorValue] is Norm[vectorValue,2] and is the Euclidean norm.
WLOG A is the origin, B = ( 0 , 1 ) , C = ( 1 , 1 ) , D = ( 1 , 0 ) . Then the expression becomes the norm of the vector ( λ 1 − λ 3 + λ 5 + λ 6 , λ 2 − λ 4 + λ 5 − λ 6 ) . It's not hard to make this zero: e.g. λ 1 = − 1 , λ 2 = λ 3 = λ 4 = λ 5 = λ 6 = 1 . So N = 0 .
It's easy to see immediately that the entries of the vector are both even, and have absolute value ≤ 4 .
But the entries cannot both have absolute value 4 ; if the absolute value of the x -coordinate is 4 , then λ 5 = λ 6 and so the y -coordinate is λ 2 − λ 4 , which has absolute value at most 2 .
The largest norm subject to the given conditions is 4 2 + 2 2 , which is attainable by setting λ 1 = λ 2 = λ 5 = λ 6 = 1 and λ 3 = λ 4 = − 1 . So M = 4 2 + 2 2 = 2 5 , and the answer is ⌊ 2 0 0 0 5 ⌋ = 4 4 7 2 .