The two cuboid boxes have integer dimensions that are all distinct from one another. But they both have the same diagonal length (in red) and the same surface area.
What is the least possible surface area of such boxes?
Note: Figures are not drawn to scale.
Inspired by Sum & Square Sum
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From the conditions in the problem, we can set up equations as:
a 2 + b 2 + c 2 = d 2 + e 2 + f 2 for the same diagonal constraint.
2 ( a b + b c + c a ) = 2 ( d e + e f + f d ) for the same area constraint.
Adding both equations together, we will get:
a 2 + b 2 + c 2 + 2 ( a b + b c + c d ) = d 2 + e 2 + f 2 + 2 ( d e + e f + f d )
( a + b + c ) 2 = ( d + e + f ) 2
Because all variables are natural numbers, we can conclude a + b + c = d + e + f .
With a + b + c = d + e + f , we are looking for the number partition into six different numbers. Taking a , b , c , d , e , f ∈ S for some space set of natural numbers, if S = { 1 , 2 , 3 , 4 , 5 , 6 } , whose elements are as least as possible, no equal summation is possible because 1 + 2 + 3 + 4 + 5 + 6 = 2 1 , which can't be halved.
Hence, 7 must be involved in S . Then if S = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } , we will have 1 + 2 + 3 + 4 + 5 + 6 + 7 = 2 8 . Thus, the summation will be less than 2 2 8 = 1 4 because only six numbers can be selected, and the excluded one must be even to follow the constraint.
To yield the least summation, we will attempt to examine the trials by excluding the most possible even integer. First, by taking 6 out, we will reach 1 1 = 7 + 3 + 1 = 5 + 4 + 2 as the only possible partition. However, 7 2 + 3 2 + 1 2 = 5 2 + 4 2 + 2 2 . It's not our desired solution.
Then by taking 4 out, we will reach 1 2 = 7 + 3 + 2 = 6 + 5 + 1 as the only possible partition, and 7 2 + 3 2 + 2 2 = 6 2 + 5 2 + 1 2 = 6 2 , which works perfectly.
As a result, the least possible total surface area equals 2 ( 7 × 3 + 3 × 2 + 2 × 7 ) = 2 ( 6 × 5 + 5 × 1 + 1 × 6 ) = 8 2 .