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"

(Created page with "I love Ashvin tooooo ahsdhbhsflbglhasfv")
 
Line 1: Line 1:
−
I love Ashvin tooooo ahsdhbhsflbglhasfv
+
==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?
 +
 
 +
==Solution 1 (algebraic manipulation)==
 +
We can create three equationss using the given information.
 +
<cmath>4a+4b+4c = 13</cmath>
 +
<cmath>2ab+2ac+2bc=\frac{11}{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>.

Revision as of 14:21, 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 longets 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 $\sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_13&oldid=203086"