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| == Problem ==
| | #redirect[[2023 AMC 12B Problems/Problem 13]] |
| A rectangular box 𝒫 has distinct edge lengths 𝑎, 𝑏, and 𝑐. The sum of the lengths of
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| all 12 edges of 𝒫 is 13, the sum of the areas of all 6 faces of 𝒫 is <math>\dfrac{11}{2}</math>, and the volume of 𝒫 is <math>\dfrac{1}{2}</math>. What is the length of the longest interior diagonal connecting two vertices of 𝒫 ?
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| == Solution 1==
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| <asy>
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| 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);
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| 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);
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| 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);
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| draw(A--AA);
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| draw(B--BB);
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| draw(C--CC);
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| draw(D--DD);
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| // 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);
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| 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>
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| Let <math>a,b,</math> and <math>c</math> be the sides of the box, we get
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| <cmath>\begin{align*}
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| 4(a+b+c) &= 13\\
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| 2(ab+bc+ca) &= \dfrac{11}{2}\\
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| abc &= \dfrac{1}{2}
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| \end{align*}</cmath>
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| The diagonal of the box is
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| <cmath>\begin{align*}
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| \sqrt{a^2+b^2+c^2}&=\sqrt{(a+b+c)^2-2(ab+bc+ca)}\\
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| &=\sqrt{(\dfrac{13}{4})^2-\dfrac{11}{2}}\\
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| &=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\
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| &=\sqrt{\dfrac{81}{16}}\\
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| &=\dfrac{9}{4}
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| \end{align*}
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| </cmath>
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| ~Technodoggo
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| ==Note==
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| Interestingly, we don't use the fact that the volume is <math>\frac{1}{2}</math> ~andliu766
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| ==Solution 2 (find side lengths)==
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| Let <math>a,b,c</math> be the edge lengths.
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| <math>4(a+b+c)=13, a+b+c=13/4</math>
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| <math>2(ab+bc+ac)=11/2, ab+bc+ac=11/4</math>
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| <math>abc=1/2</math>
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| Then, you can notice that these look like results of Vieta's formula:
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| <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>
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| Finding when this <math>= 0</math> will give us the edge lengths.
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| We can use RRT to find one of the roots:
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| One is <math>x=1</math>, dividing gives <math>x^2-9/4x+1/2</math>.
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| The other 2 roots are <math>2,1/4</math>
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| 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>.
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| -HIA2020
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| ==Solution 3 (Cheese method)==
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| Incorporating the solution above, the side lengths are larger than <math>1x1x1</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>1x1x1</math>, so that means the diagonal has to be larger than <math>\sqrt{3}</math>, and there is only one answer choice larger than <math>\sqrt{3}</math> <math>\Rightarrow</math> <math>\boxed {\textbf{(D) 9/4}}</math> ~ kabbybear
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