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 $P$ ?

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

Video Solution by MegaMath

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

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

https://youtu.be/4jjWyikA7mg

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

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

2023 AMC 12B (Problems • Answer Key • Resources)
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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