2023 AMC 12B Problem 13

Question

A rectangular box

P

has distinct edge lengths

a,

b,

and

c.

The sum of the lengths of all

12

edges of

P

is

13,

the sum of the areas of all

6

faces of

P

is

\tfrac{11}{2},

and the volume of

P

is

\tfrac{1}{2}.

What is the length of the longest interior diagonal connecting two vertices of

P?

2

\dfrac{3}{8}

\dfrac{9}{8}

\dfrac{9}{4}

\dfrac{3}{2}

The diagonal is

\sqrt{a^2+b^2+c^2};

you are given

a+b+c

and

ab+bc+ca

Use

a^2+b^2+c^2

=(a+b+c)^2-2(ab+bc+ca)

From the edges,

4(a+b+c)=13,

so

a+b+c=\tfrac{13}{4}.

From the faces,

2(ab+bc+ca)=\tfrac{11}{2},

so

ab+bc+ca=\tfrac{11}{4}.

Then

\begin{aligned} a^2+b^2+c^2 &=\left(\tfrac{13}{4}\right)^2-2\cdot\tfrac{11}{4} \\ &=\tfrac{169}{16}-\tfrac{88}{16} \\ &=\tfrac{81}{16}, \end{aligned}

so the diagonal is

\sqrt{\tfrac{81}{16}}=\tfrac{9}{4}.

Thus, the correct answer is D.

Problem 13 in other years