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

(→‎Solution 2 (factoring a polynomial))
(redirect)
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{{duplicate|[[2023 AMC 10B Problems/Problem 17|2023 AMC 10B #17]] and [[2023 AMC 12B Problems/Problem 13|2023 AMC 12B #13]]}}
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==Problem==
 
==Problem==
  
Line 4: Line 6:
  
 
==Solution 1 (algebraic manipulation)==
 
==Solution 1 (algebraic manipulation)==
 +
<asy>
 +
import geometry;
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pair A = (-3, 4);
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pair B = (-3, 5);
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pair C = (-1, 4);
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pair D = (-1, 5);
 +
 +
 +
pair AA = (0, 0);
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pair BB = (0, 1);
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pair CC = (2, 0);
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pair DD = (2, 1);
 +
 +
 +
 +
 +
draw(D--AA,dashed);
 +
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draw(A--B);
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draw(A--C);
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draw(B--D);
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draw(C--D);
 +
 +
draw(A--AA);
 +
draw(B--BB);
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draw(C--CC);
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draw(D--DD);
 +
 +
// Dotted vertices
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dot(A); dot(B); dot(C); dot(D);
 +
 +
 +
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dot(AA); dot(BB); dot(CC); dot(DD);
 +
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draw(AA--BB);
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draw(AA--CC);
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draw(BB--DD);
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draw(CC--DD);
 +
 +
 +
label("a",midpoint(D--DD),E);
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label("b",midpoint(CC--DD),E);
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label("c",midpoint(AA--CC),S);
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</asy>
 +
 
We can create three equations using the given information.  
 
We can create three equations using the given information.  
 
<cmath>4a+4b+4c = 13</cmath>
 
<cmath>4a+4b+4c = 13</cmath>
Line 9: Line 57:
 
<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> 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 = \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{\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 = \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{\frac{9}{4}}</math>.
 +
 +
Interestingly, we don't use the fact that the volume is <math>\frac{1}{2}</math>. ~andliu766
  
 
~lprado
 
~lprado
 +
~Technodoggo
 +
~minor edits and add-ons by lucaswujc
  
 
==Solution 2 (factoring a polynomial)==
 
==Solution 2 (factoring a polynomial)==
Line 20: Line 72:
  
 
~lprado
 
~lprado
 +
 +
==Solution 3 (find side lengths)==
 +
 +
Let <math>a,b,c</math> be the edge lengths.
 +
<math>4(a+b+c)=13, a+b+c=13/4</math>
 +
<math>2(ab+bc+ac)=11/2, ab+bc+ac=11/4</math>
 +
<math>abc=1/2</math>
 +
 +
Then, you can notice that these look like results of Vieta's formula:
 +
<math>(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc = x^3-13/4x^2+11/4x-1/2</math>
 +
Finding when this <math>= 0</math> will give us the edge lengths.
 +
We can use RRT to find one of the roots:
 +
One is <math>x=1</math>, dividing gives <math>x^2-9/4x+1/2</math>.
 +
The other 2 roots are <math>2,1/4</math>
 +
 +
Then, once we find the 3 edges being <math>a=1,b=2,</math> and <math>c=1/4</math>, we can plug in to the distance formula to get <math>9/4</math>.
 +
 +
 +
-HIA2020
 +
 +
==Solution 4 (Cheese Method)==
 +
 +
Incorporating the solution above, 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) 9/4}}</math>
 +
 +
 +
~kabbybear
 +
  
 
==See Also==
 
==See Also==
 +
{{AMC12 box|year=2023|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2023|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2023|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:42, 15 November 2023

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 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)

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

Interestingly, we don't use the fact that the volume is $\frac{1}{2}$. ~andliu766

~lprado ~Technodoggo ~minor edits and add-ons by lucaswujc

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

Solution 3 (find side lengths)

Let $a,b,c$ be the edge lengths. $4(a+b+c)=13, a+b+c=13/4$ $2(ab+bc+ac)=11/2, ab+bc+ac=11/4$ $abc=1/2$

Then, you can notice that these look like results of Vieta's formula: $(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc = x^3-13/4x^2+11/4x-1/2$ Finding when this $= 0$ will give us the edge lengths. We can use RRT to find one of the roots: One is $x=1$, dividing gives $x^2-9/4x+1/2$. The other 2 roots are $2,1/4$

Then, once we find the 3 edges being $a=1,b=2,$ and $c=1/4$, we can plug in to the distance formula to get $9/4$.


-HIA2020

Solution 4 (Cheese Method)

Incorporating the solution above, 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 $\sqrt{3}$ $\Rightarrow$ $\boxed {\textbf{(D) 9/4}}$


~kabbybear


See Also

2023 AMC 12B (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 12 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

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