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

(Solution)
(Solution)
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label("c",midpoint(AA--CC),S);
 
label("c",midpoint(AA--CC),S);
 
</asy>
 
</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}\\
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abc &= \dfrac{1}{2}
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\end{align*}</cmath>
 +
  
 
The diagonal of the box is
 
The diagonal of the box is
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   \sqrt{a^2+b^2+c^2}&=\sqrt{(a+b+c)^2-2(ab+bc+ca)}\\
 
   \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{13}{4})^2-\dfrac{11}{2}}\\
 +
&=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\
 +
&=\sqrt{\dfrac{81}{16}}\\
 
&=\dfrac{9}{4}
 
&=\dfrac{9}{4}
 
\end{align*}
 
\end{align*}

Revision as of 15:32, 15 November 2023

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