Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1 (algebraic manipulation))
m (→‎Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
A rectangular box P has distinct edge lengths <math>a</math>, <math>b</math>, and <math>c</math>. The sum of the lengths of all <math>12</math> edges of P is <math>13</math>, the areas of all 6 faces of P is <math>\frac{11}{2}</math>, and the volume of P is <math>\frac{1}{2}</math>. What is the length of the longets interior diagonal connecting two vertices of P?
+
A rectangular box P has distinct edge lengths <math>a</math>, <math>b</math>, and <math>c</math>. The sum of the lengths of all <math>12</math> edges of P is <math>13</math>, the areas of all 6 faces of P is <math>\frac{11}{2}</math>, and the volume of P is <math>\frac{1}{2}</math>. What is the length of the longest interior diagonal connecting two vertices of P?
  
 
==Solution 1 (algebraic manipulation)==
 
==Solution 1 (algebraic manipulation)==

Revision as of 16:24, 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 equationss using the given information. \[4a+4b+4c = 13\] \[2ab+2ac+2bc=\frac{11}{2}\] \[abc=\frac{1}{2}\] We also know that we want $\sqrt{a^2 + b^2 + c^2}$. 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_13&oldid=203241"