Difference between revisions of "2023 AMC 10B Problems/Problem 17"

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== Problem ==
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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 ==
 
== Solution ==
 
 
  
 
<asy>
 
<asy>

Revision as of 15:37, 15 November 2023

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 sum of the areas of all 6 faces of 𝒫 is $\dfrac{11}{2}$, and the volume of 𝒫 is $\dfrac{1}{2}$. What is the length of the longest interior diagonal connecting two vertices of 𝒫 ?

Solution

[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] Let $a,b,$ and $c$ be the sides of the box, we get

\begin{align*}    4(a+b+c) &= 13\\ 2(ab+bc+ca) &= \dfrac{11}{2}\\ abc &= \dfrac{1}{2} \end{align*}


The diagonal of the box is

\begin{align*}    \sqrt{a^2+b^2+c^2}&=\sqrt{(a+b+c)^2-2(ab+bc+ca)}\\ &=\sqrt{(\dfrac{13}{4})^2-\dfrac{11}{2}}\\ &=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\ &=\sqrt{\dfrac{81}{16}}\\ &=\dfrac{9}{4} \end{align*}

~Technodoggo