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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 longest diagonal of prism will be the space diagonal, which can be found by summing the squares of all the sides and square rooting them.
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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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| − | Note the expansion of <cmath>\begin{align*}
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| − | & (a+b+c)^2 = a \cdot(a+b+c) + b \cdot(a+b+c) + c\cdot(a+b+c)\\
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| − | &=\ a^2 + ab + ac+ ab + b^2 + bc+ac+bc+c^2\\
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| − | &=\ a^2+b^2+c^2 + 2ab + 2bc + 2ac)
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| − | \end{align*}
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| − | </cmath>
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| − | ~Technodoggo ~minor edits and add-ons by lucaswujc
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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>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
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