2023 AMC 10B Problem 17

Attempt Problem 17 of the 2023 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

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}?
22
38\dfrac{3}{8}
98\dfrac{9}{8}
94\dfrac{9}{4}
32\dfrac{3}{2}
Answer: D
Concepts:rectangular prismalgebraic manipulationPythagorean Theorem
Small hint:
The edge sum gives a+b+ca + b + c and the face-area sum gives ab+bc+caab + bc + ca
Big hint:
The diagonal is a2+b2+c2;\sqrt{a^2+b^2+c^2}; expand (a+b+c)2(a+b+c)^2 to recover its radicand from the two known symmetric sums
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)2−2(ab+bc+ca)=16916−112=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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