Question

A rectangular box $\mathcal{P}$ has distinct edge lengths $a,$ $b,$ and $c.$ The sum of the lengths of all $12$ edges of $\mathcal{P}$ is $13,$ the sum of the areas of all $6$ faces of $\mathcal{P}$ is $\frac{11}{2},$ and the volume of $\mathcal{P}$ is $\frac{1}{2}.$ What is the length of the longest interior diagonal connecting two vertices of $\mathcal{P}?$

$2$

$\dfrac{3}{8}$

$\dfrac{9}{8}$

$\dfrac{9}{4}$

$\dfrac{3}{2}$

Solution:

The $12$ edges give $4(a + b + c) = 13,$ so $a + b + c = \frac{13}{4}.$ The $6$ faces give $2(ab + bc + ca) = \frac{11}{2},$ so $ab + bc + ca = \frac{11}{4}.$ The space diagonal is $\sqrt{a^2 + b^2 + c^2} = \sqrt{(a+b+c)^2 - 2(ab+bc+ca)} = \sqrt{\frac{169}{16} - \frac{11}{2}} = \sqrt{\frac{81}{16}} = \frac{9}{4}.$ Thus, D is the correct answer.

2023 AMC 10B Problem 17

Solution:

Problem 17 in Other Years