2024 AIME I Problem 15

Question

Let

\mathcal{B}

be the set of rectangular boxes with surface area

54

and volume

23.

Let

r

be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of

\mathcal{B}.

The value of

r^2

can be written as

\frac{p}{q},

where

p

and

q

are relatively prime positive integers. Find

p + q.

Answer

Answer: 721

Concepts:Vieta’s Formulas rectangular prism optimization

Difficulty rating: 3370

Small hint:

The constraints say

ab + bc + ca = 27

and

abc = 23,

and the sphere must contain the space diagonal, so maximize

a^2 + b^2 + c^2

Big hint:

a^2 + b^2 + c^2 = (a + b + c)^2 - 54,

and the extreme of

a + b + c

occurs when two dimensions are equal — solve the resulting cubic in that common value

Solution:

For a box with dimensions

a, b, c,

the conditions are

2(ab + bc + ca) = 54

and

abc = 23,

so

ab + bc + ca = 27.

The smallest sphere containing a box has the box’s space diagonal as a diameter, so

\begin{aligned} &r^2 = \max_{\mathcal{B}} \frac{a^2 + b^2 + c^2}{4} \\ &= \max_{\mathcal{B}} \frac{(a + b + c)^2 - 54}{4}. \end{aligned}

With

ab + bc + ca

and

abc

fixed,

s = a + b + c

ranges over an interval, and at an endpoint the cubic

t^3 - st^2 + 27t - 23

has a double root, meaning two dimensions coincide. Setting

b = c:

2ab + b^2 = 27

and

ab^2 = 23,

so eliminating

a

gives

\frac{b(27 - b^2)}{2} = 23,

i.e.

b^3 - 27b + 46 = 0,

which factors as

(b - 2)(b^2 + 2b - 23) = 0.

The roots are

b = 2

and

b = 2\sqrt{6} - 1.

For

b = 2,

a = \frac{23}{4}

and

s = \frac{23}{4} + 4 = \frac{39}{4} = 9.75;

for

b = 2\sqrt{6} - 1,

s \approx 9.31

is smaller. So the maximum of

a^2 + b^2 + c^2

is

\left(\frac{39}{4}\right)^2 - 54 = \frac{657}{16},

giving

r^2 = \frac{657}{64}

and

p + q = 657 + 64 = 721.