2023 AMC 12B Problems/Problem 13

Problem

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 areas of all 6 faces of P is $\frac{11}{2}$ , and the volume of P is $\frac{1}{2}$ . What is the length of the longets interior diagonal connecting two vertices of P?

Solution 1 (algebraic manipulation)

We can create three equationss using the given information. $4a+4b+4c = 13$ $2ab+2ac+2bc=\frac{11}{2}$ $abc=\frac{1}{2}$ We also know that we want $\sqrt{a^2 + b^2 + c^2}$ . We know that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2ab - 2ac - 2bc$ . $a+b+c = \frac{13}{4}$ . So $a^2 + b^2 + c^2 = (\frac{13}{4})^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$ . So our answer is $\sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}$ .

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2023 AMC 12B Problems/Problem 13

Problem

Solution 1 (algebraic manipulation)