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| − | == Solution ==
| + | #redirect[[2023 AMC 12B Problems/Problem 13]] |
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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 | |
| − | 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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