»  Solutions to puzzles in my National Review Online Diary

  December 2003


In my December diary I posed the following brain-teaser.

Three children get three different Christmas presents, packed in three boxes. All of these boxes have dimensions which are an integer number of centimeters, due to production constraints at the boxing company. All dimensions are less than a meter.

To their amazement, the children note that
Can you find the possible dimensions of the boxes? Are there any other coincidences to be found among the three boxes? For example, in the surface area of the paper wrapping them? Can any box be a cube? Have a square side?



François Charton, who sent me the puzzle, adds the following theorems and proofs. The second theorem explains why you cannot have two parcels with the same surface, the third why you will never have a cube as a solution.

Theorem 1:  No two parcels can have a common dimension.

Proof:  Suppose we have two parcels of dimensions (a,b,c) and (a,d,e), as their volume and sum of the their dimensions are equal, we have  abc=ade and a+b+c=a+d+e. As a is not equal to zero, this yields bc=de and b+c=d+e, which means b=d+e-c and, replacing this value in de-bc=0, c2 - (d+e)c +de=0.

Now this means (c-e)(c-d)=0 and so c is either equal to d or e, and b is equal to e or d: the two parcels are the same, a contradiction.


No two parcels can have one side with the same surface, or perimeter. For (a,b,c) and (d,e,f), ab=de yields (given abc=def) c=f, and a+b=d+e yields (given a+b+c=d+e+f) c=f.

Theorem 2:  No two parcels can have the same surface.

Proof:  Let (a,b,c) be the dimension of a parcel, its surface is 2ab+2bc+2ac. Suppose two parcels (a,b,c) and (d,e,f) have the same surface, let V=abc=def and L=a+b+c=d+e+f (by hypothesis), and S=ab+bc+ac=de+ef+df (as they have the same surface). Then let P(X) be the polynomial

           X3 - 3LX2 + 3SX - V

P(X), a third degree polynomial, has three roots. And can be factored in two ways as P(X)=(X-a)(X-b)(X-c)=(X-d)(X-e)(X-f), which means the parcels have to be the same.

Theorem 3:  No parcel can be a cube

Proof:  (a little bit more involved). Let (a,a,a) be a cubic parcel and (b-c,b,b+d) (c,d≥0) another one. 

The equality of the sums yields 3a=3b+d-c ……… (1)

And that of the volumes:  a3 = b(b2 + (d-c)b - cd) ……… (2)

As per (1) d-c=3(a-b). Replacing in (2) we have a3 - b3 = 3b2 (a-b) - bcd

(3) ………… (a-b) (a2 + ab - 22) = - bcd ≤0 (per hypothesis)

we know that a cannot be equal to b (theorem 1), we now have two cases:

•  case 1 : a > b.  Dividing (3) by (a-b) we get a2 + ab - 2b2 ≤0. As a>b>0, we have a2 > b2 and ab ≥ b2

therefore a2 + ab > 2b2 and a2 + ab - 2b2 > 0, a contradiction.

•  case 2:  a < b.  (3) becomes a2 + ab - 2b2 >0. As b>a>0, we have a2 < b2 and ab ≤ b2. Therefore a2 + ab ≤ 2b2 and a2 + ab - 2b2 < 0, a contradiction, which completes the proof.

(Visually, this means that a cube is the smallest possible volume for a given sum of length in a parcel, and the minimum is strict.)

Theorem 4:  At least one parcel has no square face.

Proof:  Suppose the three parcels have dimensions (a a b) (c c d) (e e f). We have 2a+b = 2c+d, which yields d = 2 (a-c)+b.

then a2 b = c2 b + 2c2 (a-c), i.e. b(a-c)(a+c) = 2 c2 (a-c).

dividing by a-c (per theorem 1 they cannot be equal): 2 c2 - bc - ba = 0. Solving this in c yields

                c = ¼ ( b + sqrt(b2 + 8ab))


                c = ¼ ( b - sqrt(b2 + 8ab))

and the same argument holds for the value of e, as c and e cannot be equal, we have to have either c or e equal to

                ¼ ( b - sqrt(b2 + 8ab)),

which is a negative number, a contradiction.

Note that all the three first theorems do not make use of the fact that we have three parcels (they work with only two), and that none of the four use the fact that the dimensions are integers.