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")
 
m (→‎Problem: use $\mathcal{P}$ instead of $P$, as in the original problem's wording)
(36 intermediate revisions by 16 users not shown)
Line 1: Line 1:
−
I love Ashvin tooooo ahsdhbhsflbglhasfv
+
{{duplicate|[[2023 AMC 10B Problems/Problem 17|2023 AMC 10B #17]] and [[2023 AMC 12B Problems/Problem 13|2023 AMC 12B #13]]}}
 +
 
 +
==Problem==
 +
 
 +
A rectangular box <math>\mathcal{P}</math> 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 <math>\mathcal{P}</math> is <math>13</math>, the areas of all <math>6</math> faces of <math>\mathcal{P}</math> is <math>\frac{11}{2}</math>, and the volume of <math>\mathcal{P}</math> is <math>\frac{1}{2}</math>. What is the length of the longest interior diagonal connecting two vertices of <math>\mathcal{P}</math>?
 +
 
 +
<math>\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}</math>
 +
 
 +
==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.
 +
<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> because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that <math>a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)</math>. We know that <math>a+b+c = \frac{13}{4}</math> and <math>2(ab+ac+bc)=\dfrac{11}2</math>, so <math>a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}</math>. So our answer is <math>\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}</math>.
 +
 
 +
Interestingly, we don't use the fact that the volume is <math>\frac{1}{2}</math>.
 +
 
 +
~lprado
 +
 
 +
~minor edits and add-ons by Technodoggo, lucaswujc, andliu766, and BcMath
 +
 
 +
==Solution 2 (Vieta's)==
 +
We use the equations from Solution 1 and manipulate it a little:
 +
<cmath>a+b+c = \frac{13}{4}</cmath>
 +
<cmath>ab+ac+bc=\frac{11}{4}</cmath>
 +
<cmath>abc=\frac{1}{2}</cmath>
 +
Notice how these are the equations for the vieta's formulas for a polynomial with roots of <math>a</math>, <math>b</math>, and <math>c</math>. Let's create that polynomial. It would be <math>x^3 - \frac{13}{4}x^2 + \frac{11}{4}x - \frac{1}{2}</math>. Multiplying each term by 4 to get rid of fractions, we get <math>4x^3 - 13x^2 + 11x - 2</math>. Notice how the coefficients add up to <math>0</math>. Whenever this happens, that means that <math>(x-1)</math> is a factor and that 1 is a root. After using synthetic division to divide <math>4x^3 - 13x^2 + 11x - 2</math> by <math>x-1</math>, we get <math>4x^2 - 9x + 2</math>. Factoring that, you get <math>(x-2)(4x-1)</math>. This means that this polynomial factors to <math>(x-1)(x-2)(4x-1)</math> and that the roots are <math>1</math>, <math>2</math>, and <math>1/4</math>. Since we're looking for <math>\sqrt{a^2 + b^2 + c^2}</math>, this is equal to <math>\sqrt{1^2 + 2^2 + \frac{1}{4}^2} = \sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}</math>
 +
 
 +
~lprado
 +
 
 +
==Solution 3 (Cheese Method)==
 +
 
 +
Incorporating the solution above, we know <math>a+b+c</math> = <math>13/4</math> <math>\Rightarrow</math> <math>a+b+c > 3</math>. The side lengths are larger than <math>1</math> <math>\cdot</math> <math>1</math> <math>\cdot</math> <math>1</math> (a unit cube). The side length of the interior of a unit cube is <math>\sqrt{3}</math>, and we know that the side lengths are larger than <math>1</math> <math>\cdot</math> <math>1</math> <math>\cdot</math> <math>1</math>, so that means the diagonal has to be larger than <math>\sqrt{3}</math>, and the only answer choice larger than <math>\sqrt{3}</math> <math>\Rightarrow</math> <math>\boxed{\textbf{(D)}~\tfrac94}</math>
 +
 
 +
 
 +
~kabbybear
 +
 
 +
Note that the real number <math>\sqrt{3}</math> is around <math>1.73</math>. Option <math>A</math> is also greater than <math>\sqrt{3}</math> meaning there are two options greater than <math>\sqrt{3}</math>. Option <math>A</math> is an integer so educationally guessing we arrive at answer <math>D</math> <math>\Rightarrow</math> <math>\boxed{\textbf{(D)}~\tfrac94}</math>
 +
 
 +
~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)
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=B|num-b=16|num-a=18}}
 +
{{AMC12 box|year=2023|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Revision as of 17:02, 11 September 2024

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 $\mathcal{P}$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $\mathcal{P}$ is $13$, the areas of all $6$ faces of $\mathcal{P}$ is $\frac{11}{2}$, and the volume of $\mathcal{P}$ is $\frac{1}{2}$. What is the length of the longest interior diagonal connecting two vertices of $\mathcal{P}$?

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

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 BcMath

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 $a+b+c$ = $13/4$ $\Rightarrow$ $a+b+c > 3$. 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)

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_13&oldid=226985"