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

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== Problem ==
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#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
 
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 ?
 
 
 
== 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 <math>a,b,</math> and <math>c</math> be the sides of the box, we get
 
 
 
<cmath>\begin{align*}
 
  4(a+b+c) &= 13\\
 
2(ab+bc+ca) &= \dfrac{11}{2}\\
 
abc &= \dfrac{1}{2}
 
\end{align*}</cmath>
 
 
 
 
 
The diagonal of the box is
 
 
 
<cmath>\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*}
 
</cmath>
 
 
 
~Technodoggo
 

Latest revision as of 20:43, 15 November 2023