2023 AMC 12B Problems/Problem 13

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?

Solution 1 (algebraic manipulation)

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 know that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2ab - 2ac - 2bc$ . $a+b+c = \frac{13}{4}$ . So $a^2 + b^2 + c^2 = (\frac{13}{4})^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$ . So our answer is $\sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}$ .

~lprado

Solution 2 (factoring a polynomial)

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 $\sqrt{1^2 + 2^2 + \frac{1}{4}^2} = \sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}$

~lprado

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2023 AMC 12B Problems/Problem 13

Problem

Solution 1 (algebraic manipulation)

Solution 2 (factoring a polynomial)