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2023 AMC 10B Problem 17

Problem 17 of 25IntermediateAlgebraGeometry

A rectangular box P\mathcal{P} has distinct edge lengths a,a, b,b, and c.c. The sum of the lengths of all 1212 edges of P\mathcal{P} is 13,13, the sum of the areas of all 66 faces of P\mathcal{P} is 112,\frac{11}{2}, and the volume of P\mathcal{P} is 12.\frac{1}{2}. What is the length of the longest interior diagonal connecting two vertices of P?\mathcal{P}?

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Solution

The 1212 edges give 4(a+b+c)=13,4(a + b + c) = 13, so a+b+c=134.a + b + c = \frac{13}{4}. The 66 faces give 2(ab+bc+ca)=112,2(ab + bc + ca) = \frac{11}{2}, so ab+bc+ca=114.ab + bc + ca = \frac{11}{4}. The space diagonal is a2+b2+c2=(a+b+c)22(ab+bc+ca)=16916112=8116=94. \begin{gathered} \sqrt{a^2 + b^2 + c^2} \\ = \small \sqrt{(a+b+c)^2 - 2(ab+bc+ca)} \\ = \sqrt{\frac{169}{16} - \frac{11}{2}} \\ = \sqrt{\frac{81}{16}} = \frac{9}{4}. \end{gathered} Thus, D is the correct answer.

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Concepts: rectangular prism · algebraic manipulation · Pythagorean Theorem

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