2023 AMC 12B Problems/Problem 13

The following problem is from both the 2023 AMC 10B #17 and 2023 AMC 12B #13, so both problems redirect to this page.

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 longest interior diagonal connecting two vertices of $P$?

$\textbf{(A)}~2\qquad\textbf{(B)}~\frac{3}{8}\qquad\textbf{(C)}~\frac{9}{8}\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~\frac{3}{2}$

Solution 1 (algebraic manipulation)

[asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5);   pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1);     draw(D--AA,dashed);  draw(A--B); draw(A--C); draw(B--D); draw(C--D);  draw(A--AA); draw(B--BB); draw(C--CC); draw(D--DD);  // Dotted vertices dot(A); dot(B); dot(C); dot(D);    dot(AA); dot(BB); dot(CC); dot(DD);  draw(AA--BB); draw(AA--CC); draw(BB--DD); draw(CC--DD);   label("a",midpoint(D--DD),E); label("b",midpoint(CC--DD),E); label("c",midpoint(AA--CC),S); [/asy]

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 $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.

Interestingly, we don't use the fact that the volume is $\frac{1}{2}$.

~lprado

~minor edits and add-ons by Technodoggo, lucaswujc, andliu766, and Sotowa

Solution 2 (vieta's)

We use the equations from Solution 1 and manipulate it a little: \[a+b+c = \frac{13}{4}\] \[ab+ac+bc=\frac{11}{4}\] \[abc=\frac{1}{2}\] Notice how these are the equations for the vieta's formulas for a polynomial with roots of $a$, $b$, and $c$. Let's create that polynomial. It would be $x^3 - \frac{13}{4}x^2 + \frac{11}{4}x - \frac{1}{2}$. Multiplying each term by 4 to get rid of fractions, we get $4x^3 - 13x^2 + 11x - 2$. Notice how the coefficients add up to $0$. Whenever this happens, that means that $(x-1)$ is a factor and that 1 is a root. After using synthetic division to divide $4x^3 - 13x^2 + 11x - 2$ by $x-1$, we get $4x^2 - 9x + 2$. Factoring that, you get $(x-2)(4x-1)$. This means that this polynomial factors to $(x-1)(x-2)(4x-1)$ and that the roots are $1$, $2$, and $1/4$. Since we're looking for $\sqrt{a^2 + b^2 + c^2}$, this is equal to $\sqrt{1^2 + 2^2 + \frac{1}{4}^2} = \sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$

~lprado

Solution 3 (Cheese Method)

Incorporating the solution above, we know $a+b+c$ = $14/4$ $\Rightarrow$ $a+b+c > 3$. The side lengths are larger than $1$ $\cdot$ $1$ $\cdot$ $1$ (a unit cube). The side length of the interior of a unit cube is $\sqrt{3}$, and we know that the side lengths are larger than $1$ $\cdot$ $1$ $\cdot$ $1$, so that means the diagonal has to be larger than $\sqrt{3}$, and the only answer choice larger than $\sqrt{3}$ $\Rightarrow$ $\boxed{\textbf{(D)}~\tfrac94}$


~kabbybear

Note that the real number $\sqrt{3}$ is around $1.73$. Option $A$ is also greater than $\sqrt{3}$ meaning there are two options greater than $\sqrt{3}$. Option $A$ is an integer so educationally guessing we arrive at answer $D$ $\Rightarrow$ $\boxed{\textbf{(D)}~\tfrac94}$

~atictacksh


Video Solution 1 by OmegaLearn

https://youtu.be/bXbOPnIAKPo

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=lVkvcCmY9uM

Video Solution 3 by MegaMath

https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s

~megahertz13

Video Solution

https://youtu.be/4jjWyikA7mg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2023 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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