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Difference between revisions of "2023 AMC 12B Problems/Problem 13"

m (→‎Solution 1 (algebraic manipulation))
m (→‎Solution 1 (algebraic manipulation))
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<cmath>2ab+2ac+2bc=\frac{11}{2}</cmath>
 
<cmath>2ab+2ac+2bc=\frac{11}{2}</cmath>
 
<cmath>abc=\frac{1}{2}</cmath>
 
<cmath>abc=\frac{1}{2}</cmath>
−
We also know that we want <math>\sqrt{a^2 + b^2 + c^2}</math>. We know that <math>a^2 + b^2 + c^2 = (a+b+c)^2 - 2ab - 2ac - 2bc</math>. <math>a+b+c = \frac{13}{4}</math>. So <math>a^2 + b^2 + c^2 = (\frac{13}{4})^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}</math>. So our answer is <math>\sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}</math>.
+
We also know that we want <math>\sqrt{a^2 + b^2 + c^2}</math> because that is the length that can be found from using the Pythagorean Theorem. We know that <math>a^2 + b^2 + c^2 = (a+b+c)^2 - 2ab - 2ac - 2bc</math>. <math>a+b+c = \frac{13}{4}</math>. So <math>a^2 + b^2 + c^2 = (\frac{13}{4})^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}</math>. So our answer is <math>\sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}</math>.
  
 
~lprado
 
~lprado

Revision as of 16:26, 15 November 2023

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 polynomials 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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