## December 2003

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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?

- no two boxes have the same dimensions;
- they all have the same volume;
- the length of the piece of string tied around each box (four times the sum of the box's dimensions) are the same.

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*Solution*

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, c^{2} - (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.

*Corollary*:

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

X^{3} - 3LX^{2} + 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: a^{3} = b(b^{2} + (d-c)b -
cd) ……… (2)

As per (1) d-c=3(a-b). Replacing in (2) we have a^{3} - b^{3} = 3b^{2} (a-b) - bcd

(3) ………… (a-b) (a^{2} + ab - 2^{2}) = - 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 a^{2} + ab - 2b^{2} ≤0. As a>b>0,
we have
a^{2} > b^{2} and ab ≥ b^{2}

therefore a^{2} + ab > 2b^{2} and a^{2} + ab - 2b^{2} > 0, a contradiction.

• case 2: a < b. (3) becomes a^{2} + ab - 2b^{2} >0. As b>a>0, we
have
a^{2} < b^{2} and ab ≤ b^{2}. Therefore a^{2} + ab ≤ 2b^{2} and
a^{2} + ab - 2b^{2}
< 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 a^{2} b = c^{2} b + 2c^{2} (a-c), i.e. b(a-c)(a+c) = 2 c^{2} (a-c).

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

c = ¼ ( b +
sqrt(b^{2} + 8ab))

or

c = ¼ ( b -
sqrt(b^{2} + 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(b^{2} + 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.