AMC10 2023 B

AMC10 2023 B · Q17

AMC10 2023 B · Q17. It mainly tests Linear equations, 3D geometry (volume).

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 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 $\mathcal{P}$?

一个长方体$\mathcal{P}$有不同的边长$a$、$b$和$c$。$\mathcal{P}$所有$12$条边的长度和为$13$，所有$6$个面的面积和为$\frac{11}{2}$，体积为$\frac{1}{2}$。求$\mathcal{P}$连接两个顶点的 longest interior diagonal 的长度？

(A) 2 2

(B) \frac{3}{8} \frac{3}{8}

(D) \frac{9}{4} \frac{9}{4}

(E) \frac{3}{2} \frac{3}{2}

Answer

Correct choice: (D)

正确答案：（D）

Solution

We can create three equations using the given information. \[4a+4b+4c = 13\] \[2ab+2ac+2bc=\frac{11}{2}\] \[abc=\frac{1}{2}\] We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$. We know that $a+b+c = \frac{13}{4}$ and $2(ab+ac+bc)=\dfrac{11}2$, so $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$. So our answer is $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.

我们用给定的信息建立三个方程。 \[4a+4b+4c = 13\] \[2ab+2ac+2bc=\frac{11}{2}\] \[abc=\frac{1}{2}\] 我们知道要找$\sqrt{a^2 + b^2 + c^2}$，因为那是使用勾股定理可得的长度。我们巧妙地注意到$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$。我们知道$a+b+c = \frac{13}{4}$且$2(ab+ac+bc)=\frac{11}{2}$，所以$a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$。因此答案为$\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$。

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